-- I hope you guys are a bit friendlier than the folks at rec.math... 24 people.. 12 males, 12 females. If they reproduce and their children reproduce and their children reproduce, after 100 yrs, what would be the population of their group? They start reproducing when they are 18 and its possible to do so until they are 45 but I really only want them to have about 5 kids at a time..so they would stop repro'ing at about 25 I suppose..I said 45 as its more down to earth. There is enough food and very little desease. The folks live to about 90yrs old. The original 24 are 18yrs old. have to avoid cross breeding. this isn't Tassie :) This isn't a test. I'm just curious about the outcome..No hidden agendas or whatever. hope you can help gra --- ==== --------------------------------------------------------------------- I came up with this while an undergrad in 1974. I showed it to a couple of profs who suggested I look it up in Riordan's book. I didn't find anything like it there. Has anyone seen the equivalent of it? The identity is (apologies for the notation) n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} Here's how I came up with it. I was churning through Fermat's Last Theorem and examining the differences of x^n. The n! on the left is the nth derivative of x^n. The business on the right comes from constructing the nth difference: (x+n)^n-(x+n-1)^n [(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n] {[(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n]}-{[(x+n-1)^n-(x+n-2)^n]-[(x+n-2)^ n-(x+n-3)^n]} : : : Proving this (if indeed it holds) amounts to proving that the coefficient for [(x+n-k)^n] is {[(-1)^k] times [n choose k]}. Heuristically that seems obvious but I have never come up with a rigorous proof. Can anyone offer one, or a counterexample? ==== > The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} The right sides have x in them, while the left sides don't... do you want to extract the coefficient of x^n here? J ==== No, but one may set x=0 The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} The right sides have x in them, while the left sides don't... do you > want to extract the coefficient of x^n here? J > ==== > I came up with this while an undergrad in 1974. I showed it to a couple > of profs who suggested I look it up in Riordan's book. I didn't find > anything like it there. Has anyone seen the equivalent of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by > [(k!)(n-k)!]} Here's how I came up with it. I was churning through Fermat's Last > Theorem and examining the differences of x^n. The n! on the left is the > nth derivative of x^n. The business on the right comes from constructing > the nth difference: (x+n)^n-(x+n-1)^n > [(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n] > {[(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n]}-{[(x+n-1)^n-(x+n-2)^n]-[(x+n-2)^ n-(x+n-3)^n]} > : > : > : Proving this (if indeed it holds) amounts to proving that the coefficient > for [(x+n-k)^n] is {[(-1)^k] times [n choose k]}. Heuristically that > seems obvious but I have never come up with a rigorous proof. Can anyone > offer one, or a counterexample? Just for grins, I entered this into xmaxima, which I have just started playing around with, and I couldn't get it to perform any simplification. However, I tested it for a few values of n, including n = 125, and the identity held everywhere. It returned almost immediately for n = 25, but when I jumped to n = 125, it had to think for the better part of a minute on my 533 MHz AlphaPC. My guess is that you're absolutely right, but I have never seen that identity. What a great find! Have you tried putting this into any other more powerful symbolic math packages like Maple or Mathematica? Perhaps one of them would recognize the identity and then show you the way to a proof. ==== Robert, My n_C_k is your [n choose k]. Consider the binomial expansion: (1-x)^n = Sum(k=0 to n) (-x)^k n_C_k = Sum(k=0 to n) (-1)^k x^k n_C_k (1) For n>0 and x=1 this gives 0 = Sum(k=0 to n) (-1)^k n_C_k (2) Take the first derivative of (1): -n (1-x)^(n-1) = Sum(k=0 to n) (-1)^k k x^(k-1) n_C_k For n>1 and x=1 this gives 0 = Sum(k=0 to n) (-1)^k k n_C_k (3) Take the second derivative of (1) n (n-1) (1-x)^(n-2) = Sum(k=0 to n) (-1)^k k (k-1) x^(k-2) n_C_k Note that actually the k=0 and k=1 terms on the right are both zero, but I am formally keeping them. For a good proof you will have to justify doing this, so that when you use k (k-1) = k^2 - k and separate the sum into the sum of two sums, you can then use (3) to kill the sum with the k. For n>2 and x=1, and using (3) 0 = Sum(k=0 to n) (-1)^k k^2 n_C_k The nth derivative will give (-1)^n n! = Sum(k=0 to n) (-1)^k k (k-1)... (k-n+1) x^(k-n) n_C_k So when x=1 and using all the previous identities we have (-1)^n n! = Sum(k=0 to n) (-1)^k k^n n_C_k Now your sum your sum = Sum(k=0 to n) (-1)^k (x+n-k)^n n_C_k can be rewritten using the identity n_C_n-k = n_C_k, and letting k -> n-k, as your sum = (-1)^n Sum(k=0 to n) (-1)^k (x+k)^n n_C_k (4) Use the binomial expansion as (x+k)^n = Sum(j=0 to n) k^j x^(n-j) n_C_j and put that into (4) and rearrange to get your sum = (-1)^n Sum(j=0 to n) x^(n-j) n_C_j Sum(k=0 to n) (-1)^k k^j n_C_k In that sum on k my identities show that the sum is zero for j = 0, 1, 2, ..., (n-1) and is (-1)^n n! when j=n, so your sum = (-1)^n x^0 n_C_n (-1)^n n! = n! as you surmised. Bob > I came up with this while an undergrad in 1974. I showed it to a couple of > profs who suggested I look it up in Riordan's book. I didn't find anything > like it there. Has anyone seen the equivalent of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} Alternatively: > 1=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] divided by [(k!)(n-k)!]} Here's how I came up with it. I was churning through Fermat's Last Theorem > and examining the differences of x^n. > The n! on the left is the nth derivative of x^n. > The business on the right comes from constructing the nth difference: > (x+n)^n-(x+n-1)^n > [(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n] {[(x+n)^n-(x+n-1)^n]-[(x+n-1)^n-(x+n-2)^n]}-{[(x+n-1)^n-(x+n-2)^n]-[(x+n-2)^n - > (x+n-3)^n]} > : > : > : Proving this (if indeed it holds) amounts to proving that the coefficient for > [(x+n-k)^n] is {[(-1)^k] times [n choose k]}. Heuristically that seems > obvious but I have never come up with a rigorous proof. Can anyone offer one, or a counterexample? > ==== 2, ..., (n-1) and is (-1)^n n! when j=n, so I know there is a great combinatorial argument for this, but it escapes me at the moment. I had just come around to basically this same proof when I noticed yours. Nicely done. Basically, sum for i = 0 to n of (-1)^(n-i) * C(n,i) * i^j is 0 when j is less than n, and nonzero for j >= n -- where I'm using C(n,i) to mean n choose i or (n!/(i!(n-i)!)). Now that I think about it, I think that it's something like j * (j-1) * (j-2) * .... (j-n+1), which is 0 when j is less than n, and greater than zero when j is greater than or equal to n. Now the trick is to remember the combinatorial argument. So is this a Bob/Robert discussion or what? ==== I came up with this while an undergrad in 1974. I showed it to a > couple of profs who suggested I look it up in Riordan's book. I > didn't find anything like it there. Has anyone seen the equivalent > of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} You can simplify things by replacing (x+n) by x -- since the result should not depend on x this should not change the answer. In this form it is Sum[k = 0 to n]((-1)^k * C(n,k) * (x-k)^n), which you can find as equation (19) of: http://mathworld.wolfram.com/BinomialSums.html (Note that they ascribe the method of solution to exactly the sort of process you described.) Geoff. ---------------------------------------------------------------------------- - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ---------------------------------------------------------------------------- - ==== Messrs Kimble, Delaney, & Bailey, I pore over Mr. Delaney's equations, but I will do so. Mr. Bailey is indeed a wonder to come up with a web site with the particular equation; I do not think I will play poker with him anytime soon. Ruiz 1996! Flattering. I was unaware of mathworld.wolfram.com, xmaxima, and Maple. Looking forward to exploring those. I am getting back into math following a 25 year career in programming (mainly APL). Robert Pullman > I came up with this while an undergrad in 1974. I showed it to a > couple of profs who suggested I look it up in Riordan's book. I > didn't find anything like it there. Has anyone seen the equivalent > of it? The identity is (apologies for the notation) > n!=Sum over k=0 to n of {[(-1)^k] times [(x+n-k)^n] times [n choose k]} You can simplify things by replacing (x+n) by x -- since the result > should not depend on x this should not change the answer. In this form > it is Sum[k = 0 to n]((-1)^k * C(n,k) * (x-k)^n), which you can find as > equation (19) of: http://mathworld.wolfram.com/BinomialSums.html (Note that they ascribe the method of solution to exactly the sort of > process you described.) Geoff. -------------------------------------------------------------------------- --- > Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- > ftww@maths.usyd.edu.au | Gameplayer by vocation. > -------------------------------------------------------------------------- --- > ==== I need an algoritm to solve the following problem by programming in Visual Basic. We play a competition with this basic assumptions: - In that competition will play N teams (e.g. 14). - There are S sessions ( e.g. 6). - A session counts R rounds. After each round the teams get another opponent. - A team will meet as many as other teams as there are rounds. this are mostly 6 teams. - When the number of teams for a session is odd, during each round there is one team that don't play and thus don't meet another team. - Which teams will meet each other depends on the place of the teams in the play-scheme. For each number of teams that will participate there is a scheme. - It is possible that one or more teams don't participate in a session. The number of absent teams is rarely 4 or more. - We count the frequency of the meetings between the teams. - Before the start of a session we want the present teams place in the scheme so that the frequenty of the meetings between the teams will optimize. For further explanation an example: After 2 sessions the matrix of the meetings between the teams A to L of a competition with 12 teams may be: A B C D E F G H I J K L A - 2 0 2 1 2 2 0 1 0 1 1 B 2 1 0 1 2 0 1 0 2 1 1 1 C 0 0 0 1 1 1 0 0 1 1 0 1 (This team has played only one session) D 2 1 1 1 1 1 0 0 1 2 1 1 E 1 2 1 1 0 2 1 0 0 1 1 2 F 2 0 1 1 2 1 1 0 1 2 1 0 G 2 1 0 0 1 1 1 0 1 2 2 1 H 0 0 0 0 0 0 0 - 0 0 0 0 (This team didn't play the first 2 sessions) I 1 2 1 1 0 1 1 0 0 1 2 2 J 0 1 1 2 1 2 2 0 1 0 1 1 K 1 1 0 1 1 1 2 0 2 1 1 1 L 1 1 1 1 2 0 1 0 2 1 1 1 When a team cannot play a round, because the number of teams for a session is odd, we see in the matrix a meeting with the team itself. See for example the 1 in the element (B, B). Suppose that all 12 teams will be present during session 3. The play-scheme for 12 teams is: Team number meet the teams 1 2, 4, 5, 7, 11, 12 2 1, 3, 5, 7, 9, 11 3 2, 4, 6, 8, 10, 12 4 1, 3, 5, 7, 9,10 5 1, 2, 4, 6, 8, 12 6 3, 5, 7, 9, 10, 11 7 1, 2, 4, 6, 8, 12 8 3, 5, 7, 9, 10, 11 9 2, 4, 6, 8, 10, 12 10 3, 4, 6, 8, 9, 11 11 1, 2, 6, 8, 10, 12 12 1, 3, 5, 7, 9, 11 Which teamnumbers must we attribute to the teams A to L to optimize the meetings between the teams? An attribution is for example: A gets number 3/ B to 5/ C to 1/ D to 12/ .... etc. Has anyone a solution for this optimize-problem? May be based on experience with this kind of competitions. I am very interested. Jan van der Meulen ==== > I need an algoritm to solve the following problem by programming in Visual > Basic. > Similar school timetabling problems are now solved using genetic algorithms/simulated annealing http://www-2.cs.cmu.edu/Groups/AI/html/faqs/ai/genetic/part3/faq-doc-1.html gtoomey Free ASX end of day data www.float.com.au/data ==== I am trying to find out whether the function below has ben recognoized, described, its properies listed, etc. The function has one parameter (nu) and two independent variables -- z and tau. I denote it by k_nu(z,tau). Its integral representation is k_nu(z,tau) = int_{0}^{tau} t^{nu-1} exp(-t-z^2/(4t)} dt { integral between t=0 and t=tau of t to the power of (nu-1) times the exponent of [-t - z^2/(4t)] }. This is, of course, completely analogous to the modified Bessel function, K_nu (z), whose integral representation is the above integral with tau -> infinity. Thus, if tau It has been many years since I solved such problems so I'm pretty > rusty. A vessel in the shape of an inverted cone with its axis vertical is > full of water. It is being emptied through a hole at the vertex at a > rate proportional to the square root of the dept at any instant. > Initially the water is 10 meters deep and falls to 9 meters in the > first minute. Find out how long it takes for the vessel to empty. Could you please show all the steps to the solution. > you this problem? ==== Could anyone help, any functions (except solver) and formular in excel could find the a , b & c thanks a lot! 1.258333 a + 1.516667 b + 1.688889 c = 21200 1.775a + b + c = 21200 a + 2.55b + c = 21200 a + b + 3.066667c = 21200 ==== >Could anyone help, any functions (except solver) and formular in excel could >find the a , b & c thanks a lot! 1.258333 a + 1.516667 b + 1.688889 c = 21200 >1.775a + b + c = 21200 >a + 2.55b + c = 21200 >a + b + 3.066667c = 21200 Can't be done without more specification on your part. You have 3 variables, a, b and c, but 4 euqations, so your system of equations is overspecified. If any 3 of these are linearly independent, you could use them to determine a, b and c. For instance, put the coefficients for a, b and c from the 2nd through 4th equations into a square matrix, i.e., a 3-by-3 range of cells like A1:C3. 1.775000 1.000000 1.000000 1.000000 2.550000 1.000000 1.000000 1.000000 3.066667 and put 21200 in a 3-row, 1-column range like D1:D3. Then enter the following array formula in a 3-row, 1-column range like E1:E3. =MMULT(MINVERSE(A1:C3),D1:D3) -- 1. Don't attach files to postings in this newsgroup. 2. Snip unnecessary text from quoted text. Indiscriminate quoting is wasteful. 3. Excel 97 & later provides 65,536 rows & 256 columns per worksheet. There are no add-ins or patches that increase them. Need more? Use something else. ==== Harlan, Harlan Grove find the a , b & c thanks a lot! 1.258333 a + 1.516667 b + 1.688889 c = 21200 >1.775a + b + c = 21200 >a + 2.55b + c = 21200 >a + b + 3.066667c = 21200 Can't be done without more specification on your part. You have 3 variables, a, > b and c, but 4 euqations, so your system of equations is overspecified. If any 3 > of these are linearly independent, you could use them to determine a, b and c. > For instance, put the coefficients for a, b and c from the 2nd through 4th > equations into a square matrix, i.e., a 3-by-3 range of cells like A1:C3. 1.775000 1.000000 1.000000 > 1.000000 2.550000 1.000000 > 1.000000 1.000000 3.066667 and put 21200 in a 3-row, 1-column range like D1:D3. Then enter the following > array formula in a 3-row, 1-column range like E1:E3. =MMULT(MINVERSE(A1:C3),D1:D3) -- > 1. Don't attach files to postings in this newsgroup. > 2. Snip unnecessary text from quoted text. Indiscriminate quoting is wasteful. > 3. Excel 97 & later provides 65,536 rows & 256 columns per worksheet. There are > no add-ins or patches that increase them. Need more? Use something else. ==== For those who are interested, I recently installed kazaa (kazaa lite actually http://www.kazaalite.tk/ ) and typed in a search for mathematica. The search returned a free older version of mathematica, version 4.1.0.0, provided by the developer Wolfram Research Inc.. (Current version is 5.0). I downloaded the 48MB file and it runs well on Windows XP. There appears to be versions of Maple as well (provided by Waterloo Mapie Inc) gtoomey ==== Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't distribute it - and the same is to be said for Maple and any other program. is concerned, you are stealing. But did you really belived that kazaa users share legal things? ==== Maybe you are right. It came with a password generator which should have raised my suspicions. gtoomey ==== > Maybe you are right. It came with a password generator which should have > raised my suspicions. gtoomey I suggest you check your computer for any unauthorised / unexpected files that it might also have installed. I am not familiar with kazaa, but some other similar types of software are known to add spyware without telling you. Bevan ==== >Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't distribute >it - and the same is to be said for Maple and any other program. >is concerned, you are stealing. But did you really belived that kazaa users share legal things? Speaking of which, what are the options for the casual tinkerer that wants to play with special functions, graphing, a programming environment and things, but doesn't want to pony up $600+ for one of the usual math packages? -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== >Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't distribute >it - and the same is to be said for Maple and any other program. law >is concerned, you are stealing. But did you really belived that kazaa users share legal things? > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? > Assuming that the hypothetical tinkerer wants to do only legal things, he can look for cheaper software or a trial package, or write his own. Or maybe find someone who has the program and will let him use it on the computer it is already installed on. If a casual driver wants to take a ride, his legal options don't include borrowing one without the owners permission. Why do some people think intellectual property is fair game? ==== Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? Gregory, You want to browse around GAMS, Guide to Available Math software, http://gams.nist.gov/ and The Netlib http://www.netlib.org/index.html Lotsa stuff, much of it free, open source, etc. Google on {GNU, math} gets 158,000 hits, so you can refine on down, and there's lotsa places to go. Best, -dlj. ==== >The developer is Wolfram for Wolfram _made_ it, but they don't distribute >it - and the same is to be said for Maple and any other program. >is concerned, you are stealing. No, as far as the law is concerned you are violating copyright. Gareth ==== >> Speaking of which, what are the options for the casual tinkerer that wants >> to play with special functions, graphing, a programming environment and >> things, but doesn't want to pony up $600+ for one of the usual math >> packages? > >Assuming that the hypothetical tinkerer wants to do only legal things, he >can look for cheaper software or a trial package, or write his own. Or >maybe find someone who has the program and will let him use it on the >computer it is already installed on. If a casual driver wants to take a ride, his legal options don't include >borrowing one without the owners permission. Why do some people think >intellectual property is fair game? Good grief - get off your high horse - dare I suggest that he was asking about cheaper (or free) software packages. Gareth ==== >> >> Speaking of which, what are the options for the casual tinkerer that wants >> to play with special functions, graphing, a programming environment and >> things, but doesn't want to pony up $600+ for one of the usual math >> packages? Gregory, You want to browse around GAMS, Guide to Available Math software, >http://gams.nist.gov/ and The Netlib http://www.netlib.org/index.html Wow, that's a pretty intimidating list. I didn't know the list existed. -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== > >Speaking of which, what are the options for the casual tinkerer that wants >to play with special functions, graphing, a programming environment and >things, but doesn't want to pony up $600+ for one of the usual math >packages? >>Gregory, >>You want to browse around GAMS, Guide to Available Math software, >>http://gams.nist.gov/ and The Netlib http://www.netlib.org/index.html > Wow, that's a pretty intimidating list. I didn't know the list existed. > Gregory, page on SourceForge. It's here: http://scilinux.sourceforge.net/mathpack.html Only a couple of dozen programs -- but what a couple of dozen! Also almost all of these have their own built-in graphics or else relate pretty directly to Gnome or X-windows. Some of them are also available more Micrsoft Windows, and of course if you have a compiler you can always roll your own, since they're Open Source. Best, -dlj. ==== > >Speaking of which, what are the options for the casual tinkerer that wants >to play with special functions, graphing, a programming environment and >things, but doesn't want to pony up $600+ for one of the usual math >packages? >>Gregory, >>You want to browse around GAMS, Guide to Available Math software, >>http://gams.nist.gov/ and The Netlib http://www.netlib.org/index.html > Wow, that's a pretty intimidating list. I didn't know the list existed. > Gregory, page on SourceForge. It's here: http://scilinux.sourceforge.net/mathpack.html Only a couple of dozen programs -- but what a couple of dozen! Also almost all of these have their own built-in graphics or else relate pretty directly to Gnome or X-windows. Some of them are also available for Micrsoft Windows, and of course if you have a compiler you can always roll your own, since they're Open Source. Best, -dlj. ==== > >Speaking of which, what are the options for the casual tinkerer that wants >to play with special functions, graphing, a programming environment and >things, but doesn't want to pony up $600+ for one of the usual math >packages? >>Gregory, >>You want to browse around GAMS, Guide to Available Math software, >>http://gams.nist.gov/ and The Netlib http://www.netlib.org/index.html > Wow, that's a pretty intimidating list. I didn't know the list existed. > Gregory, page on SourceForge. It's here: http://scilinux.sourceforge.net/mathpack.html Only a dozen programs -- but what a dozen! Almost all of these have their own built-in graphics or else relate pretty directly to Gnome or X-windows, which means they're great for just mathnoodling around. Some of them are also available for Micrsoft Windows, and of course if you have a compiler you can always roll your own, since they're Open Source. Best, -dlj. ==== Octave is a free high-level language that has support for things like > matrices, complex numbers, calculus and the like. Available for > Unixesque systems and Windows. Other than that, I'm not aware of > anything. There exist modules for Python for doing this sort of thing, > but it's nowhere near as slick as using octave. > Frodo, looking at Octave. Here's one page for it and a bunch of associated frontends, utilities, and fun tools: http://freshmeat.net/search/?q=Octave§ion=projects&x=0&y=0 You may I hve to take out the free Freshmeat registration before this goes through directly for you. I dunno. -dlj. ==== > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? Already mentioned by others, but I'll add my recommendation, too. For numerical stuff, Octave. Free, and almost MATLAB compatible. Should run on almost all machines (but don't know how well graphing will go on almost any machine, since it uses gnuplot). For symbolic stuff, there are some free packages, but I've never heard them suggested as Mathematica replacements. -- Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html ==== >Why do some people think > intellectual property is fair game? Some people see intellectual property as a fiction which has no place in a free society. Then again, some people see taxes as such a fiction. ==== >For symbolic stuff, there are some free packages, but I've never heard >them suggested as Mathematica replacements. I thought pari looked promising, although I have not done much with it. ==== Octave is a free high-level language that has support for things like > matrices, complex numbers, calculus and the like. Available for > Unixesque systems and Windows. Other than that, I'm not aware of > anything. There exist modules for Python for doing this sort of thing, > but it's nowhere near as slick as using octave. > Frodo, looking at Octave. Here's one page for it and a bunch of associated frontends, > utilities, and fun tools: > http://freshmeat.net/search/?q=Octave§ion=projects&x=0&y=0 You may I hve to take out the free Freshmeat registration before > this goes through directly for you. I dunno. -dlj. There appears to be a small project group trying to keep Macsyma alive. Their product is called Maxima, and it's free. It is a continuation of a much earlier version of Macsyma, not of the commercial version which went down the tubes a few years ago. Here's the site and a blurb from the site: ------------------------------------------------ http://maxima.sourceforge.net/index.shtml Maxima is a descendant of DOE Macsyma, which had its origins in the late 1960s at MIT. It is the only system based on that effort still publicly available and with an active user community, thanks to its open source nature. Macsyma was the first of a new breed of computer algebra systems, leading the way for programs such as Maple and Mathematica. This particular variant of Macsyma was maintained by William Schelter from 1982 until he passed away in 2001. In 1998 he obtained permission to release the source code under GPL. It was his efforts and skill which have made the survival of Maxima possible, and we are very grateful to him for volunteering his time and skill to keep the original Macsyma code alive and well. Since his passing a group of users and developers has formed to keep Maxima alive and kicking. Maxima itself is reasonably feature complete at this stage, with abilities such as symbolic integration, 3D plotting, and an ODE solver, but there is a lot of work yet to be done in terms of bug fixing, cleanup, and documentation. This is not to say there will be no new features, but there is much work to be done before that stage will be reached, and for now new features are not likely to be our focus. ------------------------------------------------ The Windows download is about 9.5 megabytes. You have to have Lisp as well, but it is not clear to me whether the Windows version has Lisp built in. I intend to download it when I get back to a faster connection. I started using Maple in 1989, switched to Mathematica in 1991 at a friend's suggestion, and was for years a Mathematica cheerleader. I still use Mathematica, but no longer have positive feelings about it, mainly because of what I perceive as Wolfram's greed. I would love to see this Maxima project succeed, or at least survive. ==== >> Speaking of which, what are the options for the casual tinkerer that wants >> to play with special functions, graphing, a programming environment and >> things, but doesn't want to pony up $600+ for one of the usual math >> packages? Already mentioned by others, but I'll add my recommendation, too. For >numerical stuff, Octave. Free, and almost MATLAB compatible. Should run on >almost all machines (but don't know how well graphing will go on almost >any machine, since it uses gnuplot). For symbolic stuff, there are some free packages, but I've never heard >them suggested as Mathematica replacements. I'd recommend MuPAD, which is not open-source but is free for personal use (you do have to ask them for a license key to unlock some memory limit but the procedure is pretty painless). The syntax is highly similar to Maple's (Pascal-like), only MuPAD's is a bit cleaner and better thought-out. No doubt Maple's function library is much richer, but MuPAD has a few unique tricks like continued fraction arithmetic. -- Erick ==== I have been using Macsyma for almost 10 years, little bit less time spent with different versions of MAPLE ( up to ver. 6 ), Mathematica, MathCAD. As far as pure and applied mathematics, and physics are concerned, the Macsyma (LISP) is the most professional environment. For example, calculating improper integrals with 'residuals (which is #1 procedure in modern physics) sucks in MAPLE, way behind the Macsymas's capabilities. Since the Macsyma development was suspended for some period of time, its interface, or interaction with other environments looks outdated. However, it is a quite feasible task to catch up with the state of the art requirements: extensibility, embedding, automated generation of the LaTeX image of the Macsyma (Maxima) Notebook, adding new useful packages, etc. Internet resources for Macsyma (Maxima) have always been close to 0, and I don't know why. Andrew Isaverdian ==== > The Windows download is about 9.5 megabytes. You have to have Lisp as well, > but it is not clear to me whether the Windows version has Lisp built in. I > intend to download it when I get back to a faster connection. The Windows Maxima installer has everything you need to use Maxima. Mike Thomas. ==== > better thought-out. No doubt Maple's function library is much richer, > but MuPAD has a few unique tricks like continued fraction arithmetic. These tricks actually come almost for free since MuPAD uses on OO language, so we could implement continued fractions in a way a user could, in principle, do this as well. (There are users out there doing advanced things like this, see for example mupad-combinat.sf.net) Oh, and it works for things like series, too. -- +--+ +--+| |+-|+ Christopher Creutzig (ccr@mupad.de) ==== >> >Speaking of which, what are the options for the casual tinkerer that wants >>to play with special functions, graphing, a programming environment and >>things, but doesn't want to pony up $600+ for one of the usual math >>packages? Gregory, You want to browse around GAMS, Guide to Available Math software, >http://gams.nist.gov/ and The Netlib http://www.netlib.org/index.html >> >> >> Wow, that's a pretty intimidating list. I didn't know the list existed. >> Gregory, page on SourceForge. It's here: >http://scilinux.sourceforge.net/mathpack.html Only a dozen programs -- but what a dozen! Almost all of these have >their own built-in graphics or else relate pretty directly to Gnome >or X-windows, which means they're great for just mathnoodling around. Some of them are also available for Micrsoft Windows, and of course >if you have a compiler you can always roll your own, since they're >Open Source. I actually have an eight year old Mac that can't run the latest system software, so it can't even run the Carbonized GNUPlot, which would have been half of what I need. Maybe some day I'll just roll my own, and then borrow Numerical Recipies in C as needed. It's just kind of annoying to program at the OS level, for the graphics and things. -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== I actually have an eight year old Mac that can't run the latest system > software, so it can't even run the Carbonized GNUPlot, which would have > been half of what I need. Maybe some day I'll just roll my own, and then > borrow Numerical Recipies in C as needed. It's just kind of annoying to > program at the OS level, for the graphics and things. > Yeah. Right. Remember when all of Unix was around 68K. You could carry the whole thing around on a couple of ten inch floppies. -dlj. ==== >Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't distribute >it - and the same is to be said for Maple and any other program. >is concerned, you are stealing. But did you really belived that kazaa users share legal things? > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? Netlib! (www.netlib.org) free compilers, etc). Numerical Recipes in C (and Fortran) are available ONLINE here: http://www.library.cornell.edu/nr/nr_index.cgi Have fun! Mike ==== Gave us: >> >> I actually have an eight year old Mac that can't run the latest system >> software, so it can't even run the Carbonized GNUPlot, which would have >> been half of what I need. Maybe some day I'll just roll my own, and then >> borrow Numerical Recipies in C as needed. It's just kind of annoying to >> program at the OS level, for the graphics and things. >> Yeah. Right. Remember when all of Unix was around 68K. You could >carry the whole thing around on a couple of ten inch floppies. -dlj. > There was no such thing. The form factor for the floppy you describe was/is eight inches. 3.5 5.25 8 12 Nearly the same form factors used for chip epitaxy. :-] ==== >Yeah. Right. Remember when all of Unix was around 68K. You could >>carry the whole thing around on a couple of ten inch floppies. >> There was no such thing. The form factor for the floppy you > describe was/is eight inches. 3.5 > 5.25 > 8 > 12 Nearly the same form factors used for chip epitaxy. :-] Dark, I'm sure you're right. I was going to say 12, but then thought, Naw, that can't be true. That's absurd. At one point I learned to type in Chinese on a Fuji-Xerox J-Star, a $28,000 machine with roughly the power of a mid-range Amiga a couple of years later -- and it handled everything, including your fonts, on a series of floppies that were big enough to flop around, maybe your 12's, I forget. -dlj. ==== >Yeah. Right. Remember when all of Unix was around 68K. You could >>carry the whole thing around on a couple of ten inch floppies. >> There was no such thing. The form factor for the floppy you > describe was/is eight inches. 3.5 > 5.25 > 8 > 12 Nearly the same form factors used for chip epitaxy. :-] Dark, I'm sure you're right. I was going to say 12, but then thought, Naw, that can't be true. That's absurd. At one point I learned to type in Chinese on a Fuji-Xerox J-Star, a $28,000 machine with roughly the power of a mid-range Amiga a couple of years later -- and it handled everything, including your fonts, on a series of floppies that were big enough to flop around, maybe your 12's, I forget. -dlj. ==== >> >> I actually have an eight year old Mac that can't run the latest system >> software, so it can't even run the Carbonized GNUPlot, which would have >> been half of what I need. Maybe some day I'll just roll my own, and then >> borrow Numerical Recipies in C as needed. It's just kind of annoying to >> program at the OS level, for the graphics and things. >> Yeah. Right. Remember when all of Unix was around 68K. You could >carry the whole thing around on a couple of ten inch floppies. -dlj. I think memory = exp(features). -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't > distribute >it - and the same is to be said for Maple and any other program. > law >is concerned, you are stealing. But did you really belived that kazaa users share legal things? > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? > Assuming that the hypothetical tinkerer wants to do only legal things, he > can look for cheaper software or a trial package, or write his own. Or > maybe find someone who has the program and will let him use it on the > computer it is already installed on. If a casual driver wants to take a ride, his legal options don't include > borrowing one without the owners permission. Why do some people think > intellectual property is fair game? Marvin Do you regard the extension of the copyright laws (the Sonny Bono act) as a theft of the commons? (After x amount of years the work could be legally been extended). That is, by 'some people' do you mean, that you think that publishers think that it is fair game to steal from the public domain? Who's stealing from whom here http://bayes.wustl.edu/ ? Darren. ==== > I actually have an eight year old Mac that can't run the latest system > software, so it can't even run the Carbonized GNUPlot, which would have > been half of what I need. Maybe some day I'll just roll my own, and then > borrow Numerical Recipies in C as needed. It's just kind of annoying to > program at the OS level, for the graphics and things. The free macanova package at: http://www.stat.umn.edu/macanova/ will run on old macs. It has plotting and a high level language for numerical and statistical computation. (It will also run on old DOS machines, Windows and UNIX.) ==== >> I actually have an eight year old Mac that can't run the latest system >> software, so it can't even run the Carbonized GNUPlot, which would have >> been half of what I need. Maybe some day I'll just roll my own, and then >> borrow Numerical Recipies in C as needed. It's just kind of annoying to >> program at the OS level, for the graphics and things. The free macanova package at: http://www.stat.umn.edu/macanova/ will run on old macs. It has plotting and a high level language for >numerical and statistical computation. (It will also run on old DOS >machines, Windows >and UNIX.) -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== Go to any university book store and buy the student version. 1/10 the price and works practically the same. They just remove a few features which i could care less. >Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't distribute >it - and the same is to be said for Maple and any other program. law >is concerned, you are stealing. But did you really belived that kazaa users share legal things? > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? -- > Is that plutonium on your gums? > Shut up and kiss me! > -- Marge and Homer Simpson ==== < Why do some people think >intellectual property is fair game? Because it dosent have physical mass. I did consulting with this company, freelance, sorta, and two years later they called me in for a meeting and shoved some legal bullshit in my face and said that if I wqnt to do any more work for them than Id better sign it. Basically a nondisclosure agreement with a few added lines of code: Any and all information, inventions, processes, intellectual knowledge, etc, that was learned as a result of working for that company, is solely owned by said company or something close. To make a long story short If I signed that agreement I would have prevented myself from working for anyone else in the field. Because everything i had learned could only be used for their gain an not mine or someone elses lalalalal. Has anyone ever heard of this before? As a result I sent a lettter saying I would not sign unless they gave me 30% ownership in company. Now I work for competitor and have taken away 20% of their business in less than 1 yr. (water treatment) and I get 50% of the profit. Funny how some people get motivated by greed and fame and popularity and others are motivated by revenge, lol Gregory Toomey ha scritto nel messaggio The developer is Wolfram for Wolfram _made_ it, but they don't > distribute >it - and the same is to be said for Maple and any other program. > law >is concerned, you are stealing. But did you really belived that kazaa users share legal things? > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? > Assuming that the hypothetical tinkerer wants to do only legal things, he > can look for cheaper software or a trial package, or write his own. Or > maybe find someone who has the program and will let him use it on the > computer it is already installed on. If a casual driver wants to take a ride, his legal options don't include > borrowing one without the owners permission. Why do some people think > intellectual property is fair game? > ==== > Go to any university book store and buy the student version. 1/10 the price > and works practically the same. They just remove a few features which i > could care less. > Not totally legal: the bookstore is supposed to check that you are actually a student. I haven't read the fine print, so I don't know whether they are ever trying to force you to claim you're a student. -dlj. ==== >> Go to any university book store and buy the student version. 1/10 the price >> and works practically the same. They just remove a few features which i >> could care less. >> Not totally legal: the bookstore is supposed to check that you are >actually a student. I haven't read the fine print, so I don't know >whether they are ever trying to force you to claim you're a student. -dlj. > So sign up for a class at a local community college and become a legal student. -- http://www.math.fsu.edu/~bellenot bellenot math.fsu.edu +1.850.644.7189 (4053fax) ==== >> Go to any university book store and buy the student version. 1/10 the >> price >> and works practically the same. They just remove a few features which i >> could care less. >> Not totally legal: the bookstore is supposed to check that you are >actually a student. I haven't read the fine print, so I don't know >whether they are ever trying to force you to claim you're a student. -dlj. > So sign up for a class at a local community college and become a legal > student. The Mathematica for Students license is specifically limited to use WHILE YOU ARE A STUDENT. Usually, being a student means being a FULL-TIME student, but I'm not sure whether WRI requires this. Also note: you have to apply for a password after purchasing your copy. Proof of student status may be required at that time. (I've heard one horror story from a valid student at USC.) Also, Wolfram claims there's no difference between the student version and the professional version. YMMV. --Ron Bruck ==== >> Go to any university book store and buy the student version. 1/10 the price >> and works practically the same. They just remove a few features which i >> could care less. > >Not totally legal: the bookstore is supposed to check that you are >actually a student. I haven't read the fine print, so I don't know >whether they are ever trying to force you to claim you're a student. -dlj. > Depends on the company. The student edition of Macromedia Studio MX, has a clause in the contract. for example, that gives the company the right to audit your computer any time they feel like it to make sure you are following the restrictions. I guess that means they could break down your door at 2 in the morning and raid your computer. So sign up for a class at a local community college and become a legal > student. > -- > http://www.math.fsu.edu/~bellenot > bellenot math.fsu.edu > +1.850.644.7189 (4053fax) ==== > >> Octave is a free high-level language that has support for things like >> matrices, complex numbers, calculus and the like. Available for >> Unixesque systems and Windows. Other than that, I'm not aware of >> anything. There exist modules for Python for doing this sort of >> thing, but it's nowhere near as slick as using octave. >> Frodo, looking at Octave. Here's one page for it and a bunch of associated frontends, utilities, > and fun tools: > http://freshmeat.net/search/?q=Octave§ion=projects&x=0&y=0 > Of course YMMV, but depending on your OS you may want to find a precompiled version of Octave (if you can). I've got it working a treat even consider trying to resolve them. Mainly stuff to do with fortran compilers, which I avoid like the plague ;-) -- Frodo Morris http://users.ox.ac.uk/~wadh1342 All your bast are belong to us AKA Graham Lee, Wadham College SpectrumSofts currently on show at URL/speccy/: Speccy@Home SETI Client Also the home of iloveyou.bas, the first PC virus ported to the ZX82!!! ==== >< Why do some people think >>intellectual property is fair game? Because it dosent have physical mass. I did consulting with this company, freelance, sorta, and two years later >they called me in for a meeting and shoved some legal bullshit in my face >and said that if I wqnt to do any more work for them than Id better sign it. Basically a nondisclosure agreement with a few added lines of code: Any and all information, inventions, processes, intellectual knowledge, >etc, that was learned as a result of working for that company, is solely >owned by said company or something close. To make a long story short If I signed that agreement I would have prevented >myself from working for anyone else in the field. Because everything i had >learned could only be used for their gain an not mine or someone elses >lalalalal. >Has anyone ever heard of this before? Yes. It is rather normal, at least in general terms. (It is odd that you were asked to sign this after working for them for two years. Should be done up front.) But there are constraints. While the company can protect their investment, they do not own you. Non-compete agreements, for example, can not unreasonably restrict you from working. Obviously there are differences of opinion here, but the legal system has dealt with this for a long time, and there are lots of guidelines. I suspect this is largely governed by state law, so if you have specific concerns, you should check locally. bob ==== >>Gregory Toomey ha scritto nel messaggio > Speaking of which, what are the options for the casual tinkerer that wants > to play with special functions, graphing, a programming environment and > things, but doesn't want to pony up $600+ for one of the usual math > packages? Have you tried Maxima (symbolic computations) and octave (numerical computations)? They are $free$, free and open-source afaik. Paulo Matos ==== > is concerned, you are stealing. As far as law is concerned one is committing a copyright infringement, not theft. -- Ville ==== I have already asked this question a few weeks ago, but I was away and it seems like the post is no longer on the server. Please have a look at the following: Find b, such that a*b - 1 = 0 modulo c (which means that a*b-1 can be divided by c, without any remainder) I have a feeling the answer is rather trivial, but it is avoiding me at this time. Sparky ==== I have already asked this question a few weeks ago, but I was away and it >seems like the post is no longer on the server. Please have a look at the >following: Find b, such that a*b - 1 = 0 modulo c (which means that a*b-1 can be >divided by c, without any remainder) > Look up multiplicative inverse of a modulo c. For the calculation, try http://www.cs.ucsb.edu/~kirbysdl/cs178/multinv.html for an illustration. ==== >URGENT PLEASE READ >Our friend recently suffered a heart attack at 46 years old. >He is a regularly working contracted employee (self employed) who will be unable to work for 6 or 7 weeks from the time of the heart attack with absolutely no income. >He has paid taxes all of his life but is not eligible for government assistance >of any type as his wife is working earning $1400.00 monthly. This obviously does not support their family including their 2 kids but does exclude them from any agency help. >Mrs. Sheffer was even told that if she were a crack head or unemployed, they could then offer assistance. >They have now missed their rent, had to give up their car, had to add $300 monthly in medications and the quality of their life has sunken to new lows. >This is a kind and considerate hard working family who has fallen on temporary hard times and could sure use a kindly hand before it's too late. >We are trying to prove that people DO still care even if our government doesn't. >We have set up a donation PO Box at the address below and could sure use your help! Donna Sheffer, 3216 Eglinton ave. east, Scarborough, Ontario, Canada, M1J 2H6 >Your generous donation $1, $2, $5, $10, $20 whatever would be greatly appreciated. >Help restore a little faith and save a real family. >God Bless >See the angioplasty image here. http://www.otwebdesign.com/apimage.jpg This is no scam >Please help!! I faxed you a $100 bill ! Do you really think we're all idiots? ==== i'd like to know something about homogeneous coordinates and parametric representation of a line. if i have two points p1(x, y, z, w) and p2(x, y, z, w) with w != 1; i want to avoid dividing coordinates of this two points by w before calculating the parametric equation of the line (p1p2), and i'd like to know if this is a possible result : x = p1x + k * (p2x - p1x) y = ..... ...... w = p1w + k * (p2w - p1w) ? the last line is the one i'm not sure. all kind of help is welcome :) -- Lucas Montes ==== > i'd like to know something about homogeneous coordinates and parametric > representation of a line. if i have two points p1(x, y, z, w) and p2(x, y, z, w) with w != 1; but (x,y,z,w) = (x/w, y/w, z/w, 1) in homogeneous coordinates for w /= 0 > i want to avoid dividing coordinates of this two points by w before > calculating the parametric equation of the line (p1p2), and i'd like to > know if this is a possible result : x = p1x + k * (p2x - p1x) > y = ..... > ...... > w = p1w + k * (p2w - p1w) ? the last line is the one i'm not sure. I fault them not. w = 0, when k = -p1w/(p2w - p1w) Now from p1 to p2, 0 <= k <= 1. So an easy way to avoid w = 0, 'between' p1 and p2 is to assure p2w > p1w Not sure why you want to avoid w = 1, avoiding w = 0 as I recall assures the point is finite, ie not at infinity, not on the plane at infinity. To avoid w = 1 between p1 and p2 when k = -(1 - p1w)/(p2w - p1w) isn't as easy. ==== Dear Mathematicians... I am in great need of a single, good book for someone with an Advanced Year 10 knowledge of mathematics to take me up to a standard ready for first year mathematics, the kind that would be studied in a Bachelor of Science majoring in mathematics. The book should be suited to self study, and have self check exercises. Does anyone have any suggestions for me? I am really quite desperate now... Robert White ==== > Dear Mathematicians... I am in great need of a single, good book for someone with an Advanced Year > 10 knowledge of mathematics to take me up to a standard ready for first > year mathematics, the kind that would be studied in a Bachelor of Science > majoring in mathematics. The book should be suited to self study, and have > self check exercises. Does anyone have any suggestions for me? I am really quite desperate now... To be well prepared for first year in university it is essential that you have very good working knowledge of algebra and geometry, which is studyied at high-school (gymnasium) level. Best way to learn this is to solve lots and lots of problems, that is the only way to learn mathematics, by doing it. One book I could suggest and that is also very cheap is Schaum series, for example Schaum's Outline of Precalculus by Fred Safier. A preview of this book can be found att www.amazon.com. There is also avaible previews of other Schaum series books which can be relevant to you. Do the whole book, and if you still have some time over then I can suggest you some more books. One fascinating book about higer mathematics is What Is Mathematics?: An Elementary Approach to Ideas and Methods by R. Courant & Herbert Robbins > Robert White /Marko ==== Can anyone help me to solve this problem: I have the unknown variables H and F and I dont'd know wich way to go. 12.100 + 27.000F = 200H 22.960 + 40H = 90.000F The solution should be F=0,3 and H=101 please help me!!! ==== You must solve a system of linear equations, wich can be done with or without matrix. I'll present you the resolution without matrix. 12.100 + 27.000F = 200H 22.960 + 40H = 90.000F 200H - 27000F = 12100 -40H + 90000F = 22900 Dividing the first equation by 200 and the second one by 40, we obtain H - 135F = 121/20 -H + 2250F = 1145/2 Adding the two equations (and substituing the second one by that sum), we get H - 135F = 121/20 2115F = 11571/20 H - 135F = 121/20 F = 11571/42300 (Note that 20*2115=42300) H = 121/20 + 135*(11571/42300) F = 11571/42300 H = 2020/47 F = 3857/14100 H = 42,98734... F = 0,2735460993... So, H approx. 43 F approx 0,3 markus p.9ahler escreveu na mensagem > Can anyone help me to solve this problem: > I have the unknown variables H and F and I dont'd know wich way to go. 12.100 + 27.000F = 200H > 22.960 + 40H = 90.000F The solution should be F=0,3 and H=101 please help me!!! ==== I did a mistake in the previous answer. Sorry... You must solve a system of linear equations, wich can be done with or without matrix. I'll present you the resolution without matrix. 12.100 + 27.000F = 200H 22.960 + 40H = 90.000F 200H - 27000F = 12100 -40H + 90000F = 22900 Dividing the first equation by 200 and the second one by 40, we obtain H - 135F = 121/2 (note: here was the mistake!) -H + 2250F = 1145/2 Adding the two equations (and substituing the second one by that sum), we get H - 135F = 121/2 2115F = 633 H - 135F = 121/2 F = 211/705 (Note that 633/2115=211/705) H = 121/2 + 135*(211/705) F = 211/705 H = 9485/94 F = 211/705 H = 100,9042553...... F = 0,2992907801.... So, as you wanted, H approx. 101 F approx 0,3 markus p.9ahler escreveu na mensagem > Can anyone help me to solve this problem: > I have the unknown variables H and F and I dont'd know wich way to go. 12.100 + 27.000F = 200H > 22.960 + 40H = 90.000F The solution should be F=0,3 and H=101 please help me!!! ==== try going to directory.google.com and searching for 'number theory' - its under maths or science or something. I found some interesting stuff about one type of cryptography called RCA or RSA or something. If you're looking to get into cryptography or cryptanalysis (the breaking of codes), it's really interesting, just make sure you take things slowly and read all the definitions of things like coprimes. - T >Hey, I'm looking to learn information theory. Does anyone know where I can >get more info? I am in year 11 in Hobart, maths is my strong point, >and i want to do some detailed examinations of cryptography. for fun. >so, anyone know anywhere i cna get this kind of info? >Also, if you happen to know any references (pref. web) on other >interesting areas of maths I'd be greatful, I'm bored ;). >mjec ==== -- > Hey, I'm looking to learn information theory. Does anyone know where I can > get more info? you want to do Info T but you can not find the info ?? I doubt if I am the only one that sees the irony :) gra I am in year 11 in Hobart, maths is my strong point, > and i want to do some detailed examinations of cryptography. for fun. > so, anyone know anywhere i cna get this kind of info? > Also, if you happen to know any references (pref. web) on other > interesting areas of maths I'd be greatful, I'm bored ;). > mjec --- ==== Central European Science Journals The publisher of electronic journals is looking for candidates interested in INTERNSHIP FOR STUDENTS REQUIREMENTS: Education: student of mathematics - effective knowledge of the internet environment - very good knowledge of English (work language) POSITION DESCRIPTION: - the internet research processes support - information research - data analysis WE OFFER: - on-line work - possibility of cooperation in developing international science project - references letter Please note that we do not offer financial profits. If you are interested in our offer please send us your CV and cover letter both in English with the subject Internship-student of mathematics at: CONTACT INFORMATION: rekrutacja@cesj.com ==== I need for the problems of Kangourou of Mathematic since 1978 in english or italian language ==== 6AQNa.149821$Ny5.4234886@twister2.libero.it... > I need for the problems of Kangourou of Mathematic since 1978 in english or > italian language woohoo! -- Julien Santini, CMI Technop.99le de Ch.89teau-Gombert, France ==== If d(xy) = x dy + y dx am I right to assume: d(xy + tz) = (x dy + t dz) + (y dx + z dt)? Rudy ==== > If > d(xy) = x dy + y dx > am I right to assume: > d(xy + tz) = (x dy + t dz) + (y dx + z dt)? > Rudy > Yes. ==== Rudy von Massow typed: > If > d(xy) = x dy + y dx > am I right to assume: > d(xy + tz) = (x dy + t dz) + (y dx + z dt)? d(xy + tz) = (x dy + t dz) + (y dx + z dt) [1] Yes, it is correct. But I would suggest you to write it as a less ambiguous answer than the once you have typed in the following way: d(xy) + d(tz) = x.dy + y.dx + t.dz + z.dt unless this is not what you were looking for. At any rate, if I have understood properly, equation [1] looks stilted at least to me. -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. ==== I got me a really simple question..and if anyone could explain it, it would be great. Got a system of 3 linear equations as follows: x + y + 2z = 1 x + z = 2 y + z = 0 Once you solve it, you get the following: (1 0 1 | 0) (0 1 1 | 0) (0 0 0 | 1) I agree with the solution and with the fact that the last row (0=1) makes no sense and as such the system has no solution. But if I understand correctly, a solution to the above system is any point or set of points where all three planes meet. Does this mean that in the above example, there is no single point (or set of points) where all the three planes meet? Thanx, Doug ==== > I got me a really simple question..and if anyone could explain it, it > would be great. Got a system of 3 linear equations as follows: > x + y + 2z = 1 > x + z = 2 > y + z = 0 Once you solve it, you get the following: (1 0 1 | 0) > (0 1 1 | 0) > (0 0 0 | 1) I agree with the solution and with the fact that the last row (0=1) makes > no > sense and as such the system has no solution. But if I understand > correctly, a solution to the above system is any point or set of points > where all three planes meet. Does this mean that in the above example, > there is no single point (or set of points) where all the three planes > meet? Basically yes, there is no point in common to these three planes. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > .... > Got a system of 3 linear equations as follows: > x + y + 2z = 1 > x + z = 2 > y + z = 0 I agree .... the system has no solution. But if I understand > correctly, a solution to the above system is any point or set of points > where all three planes meet. Does this mean that in the above example, > there is no single point (or set of points) where all the three planes meet? > .... Yes. If you'd like a geometrical picture, think of your planes as the three faces of a triangular prism (without ends). Each pair of faces intersects along a line, but those three lines are all parallel. Ken Pledger. ==== I'm hopefully going to university next year and there are 6 books on my recommended list, any comments would be good, seeming as I cant get the books in the local bookstores without ordering them, and therefore I have to keep them 1) What is Mathematics? By Richard Courant ISBN 0195105192 2) The pleasures of counting ISBN 0521568234 3) The Book of numbers ISBN 038797993X 4) Calculus Gems ISBN 0070575665 5) The Mathematical Experience ISBN 0817637397 6) The Foundations of Mathematics ISBN 0198531656 At the moment I'm thinking book 6, what I've heard of it, after scowering the web! Any comments whether good or bad would be very helpful Figurt ==== A good list, but also consider... Advanced Engineering Mathematics by Erwin Kryszig (sp?) Without a doubt, the best advanced maths book I've ever read. Extremely readable, with clear examples, my dad used it when he did electrical engineering in the 60s, I used it at Uni as well (somewhat newer edtion). It's not all that engineering-specific, although it does tend towards applied maths rather than pure maths / number theory etc. I found it was good for many 1st, 2nd and 3rd year maths subjects. - Daniel -- **************************************************************************** ** * Daniel Franklin - Postdoctoral research fellow, TITR Institute * University of Wollongong, NSW, Australia * d.franklin@ieee.org **************************************************************************** ** ==== A good list, but also consider... Advanced Engineering Mathematics by Erwin Kryszig (sp?) Without a doubt, the best advanced maths book I've ever read. Extremely > readable, with clear examples, my dad used it when he did electrical > engineering in the 60s, I used it at Uni as well (somewhat newer edtion). > It's not all that engineering-specific, although it does tend towards > applied maths rather than pure maths / number theory etc. I found it was > good for many 1st, 2nd and 3rd year maths subjects. - Daniel > -- > **************************************************************************** ** > * Daniel Franklin - Postdoctoral research fellow, TITR Institute > * University of Wollongong, NSW, Australia * d.franklin@ieee.org > **************************************************************************** ** I agree. I'm using it still after 20 years, eventhough the current edition has grown considerable over the decades. gtoomey ==== Some of the Schaum's outline series are good too. gtoomey ==== I have the 3rd edition (1972) of An Introduction to the Theory of Numbers by Niven and Zuckerman. Seems (problem 20 on p88) that they assert that 16 has 4 prime factors (2,4,8,16) and 72 has 5 (2,4,8,3,9). Inasmuch as 4,8,16,and 9 are composite, how can they be called prime factors? Or did they just botch the statement of the problem? The problem reads: For any integer n >1 define v(n) as (-1)^j, where j is the total number of prime factors of n. For example if n=16, j=4; if n=72, j=5. Also define v(1)=1. Prove that v(n) is a totally multiplicative function and that sum {v(d): d|n} = 1 if n is a perfect square and 0 otherwise. ==== > I have the 3rd edition (1972) of An Introduction to the Theory of > Numbers > by Niven and Zuckerman. Seems (problem 20 on p88) that they assert that > 16 > has 4 prime factors (2,4,8,16) and 72 has 5 (2,4,8,3,9). Inasmuch as > 4,8,16,and 9 are composite, how can they be called prime factors? Who called these primes? > The problem reads: For any integer n >1 define v(n) as (-1)^j, where j > is > the total number of prime factors of n. For example if n=16, j=4; if > n=72, > j=5. 16 = 2 x 2 x 2 x 2: four primes there, and 72 = 2 x 2 x 2 x 3 x 3: I make that five primes :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== The text uses the term prime factors, as if p^n were a prime factor for n>1. Seems like they meant powers of prime factors. I have the 3rd edition (1972) of An Introduction to the Theory of > Numbers > by Niven and Zuckerman. Seems (problem 20 on p88) that they assert that > 16 > has 4 prime factors (2,4,8,16) and 72 has 5 (2,4,8,3,9). Inasmuch as > 4,8,16,and 9 are composite, how can they be called prime factors? Who called these primes? > The problem reads: For any integer n >1 define v(n) as (-1)^j, where j > is > the total number of prime factors of n. For example if n=16, j=4; if > n=72, > j=5. 16 = 2 x 2 x 2 x 2: four primes there, and > 72 = 2 x 2 x 2 x 3 x 3: I make that five primes :-) -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > The League of Gentlemen ==== At my son's school, each year has about 90 children, in 3 classes of 30 children each. At the end of the summer term, the teachers havee the difficult task of deciding how to group the 90 children into 3 (possibly different) classes for next year. They ask each child to list 4 friends they would like to be in the same class with next year (the next autumn term). The child ranks each friend according to strength of friendship. Children do not necessarily agree - just because Tom lists Dick and Harry, does not mean that Dick or Harry list Tom, or that they agree with how he ranks them. The teachers have the problem of how to split the 90 children into new classes for next year, so that all children are as happy as possible. This sounds like a straight-forward problem in optimization (but with 90x4 variables?). Unfortunately my mathematics is very rusty, and I do not know what algorithm to use to get a solution. Once armed with the right algorithm, I am happy to create a test database of preferences per child, and implement the algorithm in a language such as Perl. I would appreciate any advice on the algorithm. Clyde Ingram P.S. Apologies if cross-posting (to 3 News Groups) is considered bad form. ==== > At my son's school, each year has about 90 children, in 3 classes of 30 > children each. > At the end of the summer term, the teachers havee the difficult task of > deciding how to group the 90 children into 3 (possibly different) classes > for next year. They ask each child to list 4 friends they would like to be in the same > class with next year (the next autumn term). > The child ranks each friend according to strength of friendship. Children > do not necessarily agree - just because Tom lists Dick and Harry, does not > mean that Dick or Harry list Tom, or that they agree with how he ranks them. The teachers have the problem of how to split the 90 children into new > classes for next year, so that all children are as happy as possible. This sounds like a straight-forward problem in optimization (but with 90x4 > variables?). Unfortunately my mathematics is very rusty, and I do not know > what algorithm to use to get a solution. It sounds like an integer program. IPs are notoriously difficult to solve. (Lots of NP-complete problems can be modeled as IPs.) You may want to set up a digraph D where the vertices are the children, and (u,v) is an edge of D if u wants to be in the same class as v. Then you are looking for a partition (A,B,C) of the vertices of D where A, B, and C have (about) 30 elements each. [A partition means that the union of A, B, and C is the vertex set of D, and no vertex is in at least two of A, B, and C.) You may be lucky enough to get a reasonable answer by looking for a minimum cut (E, F) [E and F are a partition of the vertices of D, and the number of edges with one end in each of E and F is as small as possible]. If E (or F) happens to have (about) 30 elements, throw out the vertices in E (or F), and look for a minimum cut of the smaller digraph. Again, if you're lucky, this cut will consist of two sets with about 30 elements in each. (There's a lot of terminology here, but it's all standard.) This would be a quick way to do this. You could also write a program to generate all sets (A,B) with exactly 30 elements in A and B, and where A and B have no elements in common. Then count how many edges have ends in A and B, or A and V-A-B, or B and V-A-B, and choose the sets (A,B) to minimize this number of edges. > P.S. Apologies if cross-posting (to 3 News Groups) is considered bad form. Not in this case. -- Christopher Heckman ==== > At the end of the summer term, the teachers have the difficult task of > deciding how to group the 90 children into 3 (possibly different) classes > for next year. They ask each child to list 4 friends they would like to be in the same > class with next year (the next autumn term). > The child ranks each friend according to strength of friendship. Children > do not necessarily agree - just because Tom lists Dick and Harry, does not > mean that Dick or Harry list Tom, or that they agree with how he ranks them. The teachers have the problem of how to split the 90 children into new > classes for next year, so that all children are as happy as possible. > If that's their motivation they're going to have a lot more problems with happy uneducated kids attending social club. They should be concerned about placing kids of similar skills together, so those with skill can properly develop and those without can be fully helped as best as can. ==== > At my son's school, each year has about 90 children, in 3 classes of 30 > children each. > At the end of the summer term, the teachers havee the difficult task of > deciding how to group the 90 children into 3 (possibly different) classes > for next year. They ask each child to list 4 friends they would like to be in the same > class with next year (the next autumn term). > The child ranks each friend according to strength of friendship. Children > do not necessarily agree - just because Tom lists Dick and Harry, does not > mean that Dick or Harry list Tom, or that they agree with how he ranks them. The teachers have the problem of how to split the 90 children into new > classes for next year, so that all children are as happy as possible. This sounds like a straight-forward problem in optimization (but with 90x4 > variables?). Unfortunately my mathematics is very rusty, and I do not know > what algorithm to use to get a solution. Once armed with the right algorithm, I am happy to create a test database of > preferences per child, and implement the algorithm in a language such as > Perl. I would appreciate any advice on the algorithm. Clyde Ingram P.S. Apologies if cross-posting (to 3 News Groups) is considered bad form. One approach is to use the Greedy Algorithm. Place each student in the class that will maximize his happiness. Of course, before you can place any students, you have to place all his friends first. That's where it gets tricky. It's easy towards the end when most students are already placed, but how do you place the first student if none have been placed already? I whipped up a test database in Access. In it I have 90 students named a0 through i9 and set up a table: Me Friend Rank Class a0 a6 1 a0 a4 2 a0 a1 3 a0 a7 4 a1 a0 1 a1 a3 2 a1 a7 3 a1 a9 4 a2 a4 1 . . . Here each student is listed 4 times, once for each friend along with his rating (4=highest) and the class to which he is assigned (initially blank). a6 0301 0403 0101 a7 0304 0203 0402 0101 a8 0203 a9 0304 0102 c3 0302 0404 c4 0203 c5 0404 a9 Other cliques found elswhere: d1-e0 e2-d2-d4-e3 g0-g7-h2-g6 g1-h6 g2-h3 g3-h0 g4-g9 i4 / | i3-i0-i1 |/ i2 (the last one was not semi-random like the others, I deliberately constructed it) A good starting point is to try to keep the cliques intact and then place whatever students remain individually using the Greedy Algorithm. So I put each of the big cliques in a seperate class (by entering a 1 in the [Class] field of the table) and spread the rest about as follows: All the a? and c? went to Class 1 All the g? and h? went to Class 2 All the i? d? and e? went to Class 3 That placed about 35 students, so now we run a query to list which unplaced students match those already placed (summing the [rank] will tell us which class he would be happiest in): Me SumOfRank Class b0 10 1 b1 10 1 b2 10 1 b3 10 1 b4 10 1 b5 10 1 b6 10 1 b7 10 1 b8 10 1 b9 10 1 c2 4 1 c8 3 1 d5 1 1 d5 2 3 . . . g5 10 2 g8 5 2 h1 7 2 h4 6 2 h5 10 2 h7 10 2 h8 10 2 Note that a student's maximum happiness cannot exceed 10, so we can't do any better than putting the b? students in Class 1 and g5, h5, h7, h8 in Class 2. Note that d5 has friends in both Class 1 and Class 3, so we'll defer placing him (or any other student whose SumOfRank is less than 10) until we get the high scorers placed (by updating the table). After placing the students selected by the first pass of the unplaced student query, the query is run a second time. With the placed students growing and the unplaced students shrinking, each pass of the query will be different from the previous pass. In each pass, we always are greedy, we only place the highest scores, defering the lower scores to the next pass to see if the newly placed students improve their score. In the example, at pass 4, I got Me SumOfRank Class c0 10 1 d0 2 1 d3 1 1 d5 4 1 d5 2 3 . . . but there was only one slot left open in Class 1, so once c0 is placed, we have to stop considering Class 1 for the rest of the students. Here d5 loses the chance to be in the class with his best friend, but he'll still have to wait to see if he'll get in Class 2 or 3 as a couple of his friends have not been placed yet. Finally, on the 9th pass (with only one slot remaining in Class 2) Me SumOfRank Class d0 4 2 . . . f0 7 2 f2 1 3 . . . f0 grabs the last spot in Class 2 with his high score of 7 forcing all remaining unplaced students into Class 3. How well did this work? Did I maximize happiness? If every student in a class were perfectly happy (all 4 of his friends are in the same class), each would have a score of 10. So a perfect score for the class would be 300. For my sample data (admittedly unrealistic) the scores came out Class 1 300 Class 2 257 Class 3 194 Compare that to the case where the students are assigned to the classes randomly: Class 1 98 Class 2 106 Class 3 98 This may not be the best that can be done, but it ended up pretty good. Note that I knew ahead of time (from how the list of semi-random friends were chosen) that all a? b? and c? should be in the same class, but I just followed the script of placing the cliques (which include none of b?) and let the Greedy Algorithm do the rest. Had I chosen to place the small 3-person clique c3-c4-c5 in a class other than 1, Class 1 would never have gotten a perfect score. With real data, you may not get any perfect scores, but any total score over 300 means you did better than guessing. ==== I'd use a genetic algorithm: http://www-2.cs.cmu.edu/Groups/AI/html/faqs/ai/genetic/part3/faq-doc-1.html gtoomey ==== > I'd use a genetic algorithm: > http://www-2.cs.cmu.edu/Groups/AI/html/faqs/ai/genetic/part3/faq-doc-1.html That's interesting, but would you trust the advice of someone who can't put together a decent web site? gtoomey ==== [...] > That's interesting, but would you trust the advice of someone who > can't put together a decent web site? [...] If we're going to be superficial, it's better to be logically superficial. Recalling the small rural town with only 2 barbers -- one with a very messy shop, dark cirlces under his eyes, and looking like he's been dragged through a fence backwards; the other with a clean, neat shop and immaculate hair. Which one do you have your hair cut at? ==== > [...] > One approach is to use the Greedy Algorithm. Place each student in the class > that will maximize his happiness. Of course, before you can place any > students, you have to place all his friends first. That's where it gets > tricky. It's easy towards the end when most students are already placed, > but how do you place the first student if none have been placed already? Actually, it doesn't matter which class the first student goes into. There is complete symmetry at that point. -- Christopher Heckman ==== > [...] > That's interesting, but would you trust the advice of someone who > can't put together a decent web site? > [...] If we're going to be superficial, it's better to be logically superficial. I suppose that not being able to navigate upwards past the start of part 3 making it impossible to get to the other parts is a trivial complaint. Maybe the web page designer figured that you would just use your back button to get to the table of contents page. Of course, the assumption that one is actually starting from the table of contents is not warranted in light of the link posted in this thread which takes the user directly to part 3. That's careless indifference to the needs of the user on the part of both the web page desinger and the link poster. And speaking of careless, why was that link pointing to part 3? Part 3 out of context makes no sense at all. A better link would have been to one of the prior parts where genetic algorithms are described. Or how about a link to the table of contents, so the user has the opportunity to read the entire section? There is no place for carelessness in science. Recalling the small rural town with only 2 barbers -- one with a very messy > shop, dark cirlces under his eyes, and looking like he's been > dragged through a fence backwards; the other with a clean, neat > shop and immaculate hair. Which one do you have your hair cut at? Here's a better question: would you wnat to take a ride on the Space Shuttle? ==== > [...] > One approach is to use the Greedy Algorithm. Place each student in the class > that will maximize his happiness. Of course, before you can place any > students, you have to place all his friends first. That's where it gets > tricky. It's easy towards the end when most students are already placed, > but how do you place the first student if none have been placed already? Actually, it doesn't matter which class the first student goes into. There > is complete symmetry at that point. > -- Christopher Heckman Placing a single student into an arbitrary empty class would attract his friends if the Greddy Algorithm were applied immediately. I was thinking that you should have an attractor in each of the three classes and by using cliques, there is less likelyhood of seperating friends. But I could easily find three individuals who have no friends in common that could serve as attractors. After placing all the students and achieving an aggregate score of 751 out of 900, I made another query showing which class each student was in and which class he would be happiest in. Of the 90 students, 79 were in their optimum class. Of the reamining 11, 4 would be happier in class 1, but class 1 is completely populated with scores of 10, so on one would be willing to transfer out. That left 3 in class 2 that would be happier in class 3, 3 in class 3 that would be happier in class 2 and 1 in class 2 that would have equal happiness in class 3. Of course, transfering a pair of students could adversely affect the overall class scores and this did indeed happen when I switched all 6 students (the aggregate dropped from 751 to 746). So I switched them back and then tried swapping all possible pairs one at a time to see which pairs, if any, result in an increase in the aggregate score. I ended up with a net gain of 7 points switching a single pair for a final tally of 758 out of 900. And it might be possible to tease out another couple points by repeating the process after making a new list of transfer candidates, but I didn't take it any further. ==== > [...] > One approach is to use the Greedy Algorithm. Place each student in the class > that will maximize his happiness. Of course, before you can place any > students, you have to place all his friends first. That's where it gets > tricky. It's easy towards the end when most students are already placed, > but how do you place the first student if none have been placed already? Actually, it doesn't matter which class the first student goes into. There > is complete symmetry at that point. > -- Christopher Heckman Placing a single student into an arbitrary empty class would attract > his friends if the Greddy Algorithm were applied immediately. If you are using the greedly algorithm to put students in one at a time, then yes, the student's friends would be the first ones in, by definition. In this case, there's more than one possible greedy algorithm. -- Christopher Heckman ==== > [...] > There is no place for carelessness in science. > Recalling the small rural town with only 2 barbers -- one with a very messy > shop, dark cirlces under his eyes, and looking like he's been > dragged through a fence backwards; the other with a clean, neat > shop and immaculate hair. Which one do you have your hair cut at? Here's a better question: would you wnat to take a ride on the Space > Shuttle? Nope. I don't even like flying. -- Christopher Heckman ==== [...] > There is no place for carelessness in science. [...] As clever as planned innovation. ==== Does anyone know of a great web site (or web document) with a whole bunch of polynomial tricks? I mean, things one wouldn't normally learn in highschool when solving polynomials of degree > 2 (especially) Thanx Doug ==== > Does anyone know of a great web site (or web document) with a whole bunch of > polynomial tricks? > Approximate solutions can be found using Newton's method. There is no general analytic method for finding roots of degree>5. (Proof is left as an exercise to the reader). For polynomials of degree<=5, you can use a package like Maple or Mathematica to find an analytic solution, or looks at the following: http://mathworld.wolfram.com/Polynomial.html http://mathworld.wolfram.com/BairstowsMethod.html http://mathworld.wolfram.com/GraeffesMethod.html gtoomey ==== How can I calculate the various probabilities in a lotto game. Take , for example, the type 6/49 . The calculation for 6 from 49 is obviously easy but how do you calculate the chances of having 5 of the selected numbers amongst the 6 drawn. This, for me, gets even more complicated when a bonus ball is also drawn. I've seen various formulae but they do not always agree -e.g. the chances of getting 3 from amongst the 6 selected and the 6 drawn vary from about 57% to 61% There must be a firm formula for this but I cannot fathom it. Help please. TIA Jack H ==== > How can I calculate the various probabilities in a lotto game. Take , for example, the type 6/49 . The calculation for 6 from 49 is obviously easy but how do you calculate the > chances of having 5 of the selected numbers amongst the 6 drawn. This, for me, gets even more complicated when a bonus ball is also drawn. I've seen various formulae but they do not always agree -e.g. the chances > of getting 3 from amongst the 6 selected and the 6 drawn vary from about > 57% to 61% > There must be a firm formula for this but I cannot fathom it. Help please. TIA > Jack H For a start have a look at: combinations http://mathworld.wolfram.com/Combination.html permutations http://mathworld.wolfram.com/Permutation.html the binomial distributionhttp://mathworld.wolfram.com/BinomialDistribution.html gtoomey ==== Investment on TAB horse racing. This software utilises Internet technology and expert system to track all the professional activity of the racing experts and actions of the professionals when they invest. www.racePredictor.com ==== Yes, this works for integer powers, but what about (particularly) irrational powers, can you just say (eg:) works for p=2, p=3, therefore it works for p=(pi)? This was the point I got stuck on. Rick > (a + b)^2 = a^2 + b^2 + 2ab > = a^2 + b^2 > (a + (b+c))^2 >= a^2 + (b+c)^2 >= a^2 + b^2 + c^2 > ... and so on by induction (a + b)^p = a^p + b^p + stuff >= a^p + b^p > (a + (b + c)) >= a^p + b^p + c^p > ... > I have a problem which is more broad than this, but a proof of the above > would provide a great starting point: > I am trying to prove that for a = {a1, a2, a3, a, ...} (infinite sequence > of > non-negative numbers) > SUM(a) <= [SUM(a^p)]^(1/p) > where 0 (a1 + a2 + a3 + a4 + a5 +...)^p <= a1^p + a2^p + a3^p + a4^p + a5^p +... > Now a proof of p = 0.5 may just get me started. (Nay, would likely help me > out enourmously) > It is intuitively true, but I just can't seem to formulate a proof. > It seems reminiscent of the Triangle Inequality, but ...not. > Rick > will o @ netspaceDOTnet.au =:-) > ==== T H O M A S 20 8 15 13 1 19 = 76 I found Sarah hanging out on Broadway Ave., beside the penis poster poles. 186 Sarah 20 2 87 51/314 10960 Sarah 47 Auralia 63 Thomas 76 Sarah was born on the 20th, her given initials add to 20 and her last initial is a T (20). Her given names add to 110 (Leviticus 20). Sarah was born on day 51, her vowels add to 51 (17+17+17), her prime valued letters add to 51 (17+17+17). Her given names span 17. Her last name adds to 76 (17 plus the 17th prime), corresponding to Exodus 26 (17th non-prime). Her last two names add together for 139 (17+17th prime). Her middle name adds to 134.04% of her first name (chapter 134 is Numbers 17). Her 51st (17+17+17th) day of birth adds with her 186 valued name for 237 (26+49+70+92=237, it's the 17th non-prime plus the 34th non-prime plus the 51st non-prime plus the 68th non-prime). Her names average 62, if she took my 62 valued last name, then her name could add to 172. See books by Bonnie Gaunt for discussions on the number 186. Sarah's common name adds to 123, corresponding to Numbers 6, it is 41+41+41 (or three times the 6th prime in prime position. Her last name is 6 lettered and adds to 76. Her first 6 letters add to 48 (6x8). Her last two names differ in value by 13 and add together for 139 (13 is the 6th prime). Her middle name adds to 63 (Exodus 13), her last name adds to 76 (Exodus 13+13). Her names have an average value of 62, the 658 verses of Book 6 and the 404 verses of Book 66 add together for 1062, and New Testament Book 6 terminates at chapter 1062, it is 666 plus 6x66. Her 10 (6th non-prime) different letters add to 136. Her consonants exceed her vowels by 84, and also her even valued letters add to 84 (the first 13 primes minus the first 13 non-primes). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Sarah's day, month and year of birth adds to 109 (29th prime). She was born in 87 (29+29+29). Her first and last names differ in value by 29. Sarah's given names average 55 while her last name adds to 76 (the 55th non-prime). Her primes, squares and cubes add together for 76 (55th non-prime). Her first name is 5 lettered and adds to the 5+5+5th prime. Primes Non-Primes Lucas 2 1 1 3 4 3 5 6 4 7 8 7 11 9 11 13 10 18 17 12 29 19 14 47 23 15 76 29 16 123 31 18 199 37 20 322 41 21 521 43 22 843 47 <-15th-> 24 <-15th-> 1364 <-Jeremiah is Book 24 with 1364 verses Her first name adds to the 15th prime, her middle name adds to the 15+15+15th non-prime. Her names add to the 15th prime, 45th and 55th non-primes, together for 115. Her consonants add to 135 (9x15). The 15th letter of her name is the 15th letter of the alphabet. Sarah provides the stats on the 24th (15th non-prime), Bible Book 24 contains 1364 verses (the 15th Lucas). Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== So I was doing an elementary rate of change on x with respect to time given the change in y w.r.t time. I also know that x and y are the two sides of the triangle, connected by the constant hypotenous h. Now we all know the following: h^2 = x^2 + y^2 so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations, constants are left out and... -x^2 (proportional to) y^2 #thus x (proportional to) y so since the h^2 is constant, the change in x should reflect the change in y since x^2 is proportional toy^2. However, this is not the case, meaning that if we have a ladder standing on the floor and leaned against a wall, if it begins to slide, it will not slide on x proportional to y. So the question is why? what am I missing....and perhaps I should stop reviewing math late at night? Thanx Doug ==== > So I was doing an elementary rate of change on x with respect to time > given the change in y w.r.t time. I also know that x and y are the > two sides of the triangle, connected by the constant hypotenous h. > Now we all know the following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing > relations, constants are left out and... > -x^2 (proportional to) y^2 #thus > x (proportional to) y so since the h^2 is constant, the change in x should reflect the > change in y since x^2 is proportional toy^2. However, this is not the > case, meaning that if we have a ladder standing on the floor and > leaned against a wall, if it begins to slide, it will not slide on x > proportional to y. So the question is why? what am I missing....and perhaps I should stop > reviewing math late at night? Thanx > Doug G'day Doug, Perhaps too late at night, let's see if I can still get this right (too early in the day? ) x^2 = h^2 - y^2 doesn't imply x proportional to y since x= sqrt (h^2 - y^2) rate of change of x with respect to y: dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy = {-1/2[sqrt (h^2 - y^2)]} . (-2y) = y/sqrt (h^2 - y^2) so x is not proportional to y Julian -- Local IT Bloke CSIRO, Forestry and Forest Products Ph: +61 8 8721 8118 ==== > So I was doing an elementary rate of change on x with respect to time given > the change in y w.r.t time. I also know that x and y are the two sides of > the triangle, connected by the constant hypotenous h. Now we all know the > following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations, x^2 = h^2 - y^2 is neater style > constants are left out and... > -x^2 (proportional to) y^2 #thus > x (proportional to) y > I beg your pardon, but you're learning rules instead of math. y proportional to x when there's some constant c such that y = c * x Thinking thus from the basics, you can see that from x^2 = h^2 - y^2 x^2 is _not_ proportionally to y^2. Now if y^2 is proportional to x^2, then for some constant c y^2 = c * x^2 Hence y = (sqr c) * x and thus as you wished, y is proportional to x and in this case the constant of proportionallity is sqr c. ==== >h^2 = x^2 + y^2 >so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations, >constants are left out and... >-x^2 (proportional to) y^2 #thus Nope. You can leave out constants in the original equation when you're describing _rates_of_change_, but not when you're describing relations of the original variables. In this case the constant h is (by definition of a right triangle) larger than either x or y. x and y are never proportional for constant h. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== Julian Mattay typed: > Perhaps too late at night, let's see if I can still get this right (too > early in the day? ) x^2 = h^2 - y^2 doesn't imply x proportional to y since > x= sqrt (h^2 - y^2) > rate of change of x with respect to y: > dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy > = {-1/2[sqrt (h^2 - y^2)]} . (-2y) > = y/sqrt (h^2 - y^2) > so x is not proportional to y Julain, if you directly differentiate the following equation, x^2 = h^2 - y^2 where _h_ is a constant term, and both _x_ and _y_ variables, then you would end up with the answer, dy/dx = -x/y which if I have interpreted correctly defines clearly the proportional relation of _x_ and _y_. -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. ==== > Julain, if you directly differentiate the following equation, x^2 = h^2 - y^2 where _h_ is a constant term, and both _x_ and _y_ variables, then you > would end up with the answer, dy/dx = -x/y which if I have interpreted correctly defines clearly the proportional > relation of _x_ and _y_. x and y would only be proportional in that equation if dy/dx were constant. It is not and so they are not. -- Rich Carreiro rlcarr@animato.arlington.ma.us ==== exercise. I'm still a newbie so forgive and correct me on any errors. If h is the second hand on a clock, and is 2cm long, the rate at which y (the height) is decreasing when h is at say 2 o'clock can be found using the following method. ---------------------- In a right angled triangle YHX where the opposite sides are labelled yhx, then at 2 o'clock angle Y=30 at the centre, H=90 near 3 on the clock and X=60 near 2 on the clock. y/sin y = h/sin h y = (hsin y)/sin h y = 1 x = sqrt(h^2 - y^2) = sqrt(3) Y is changing at the rate of (pi/30)/second so dY/dt = pi/30 We know that y = (hsin y)/sin H and since h and sin H are constant at 2 and 1 respectively y = 2sin Y and dy/dY = 2cos Y To find the rate at which y is decreasing we look for dy/dt. dy/dt = (dy/dY)(dY/dt) dy/dt = 2cos(Y)(pi/30) At 2 o'clock dy/dt = 2cos(pi/6)(pi/30) = 0.1814 y is decreasing at a rate of 0.1814 cm/s ---------------------- To find the rate at which x is increasing dx/dt = (dx/dy)(dy/dt) We know x = sqrt(h^2 - y^2) dx/dy = -y/[sqrt(h^2 - y^2)] So the rate at which x is increasing is dx/dt = -[y/(sqrt(h^2 - y^2))](0.1814) = -0.1047 x is increasing at a rate of 0.1047 cm/s. ---------------------- x is not proportional to y because the fraction x/y should remain unchanged for all values, which is similar to what William said. Dave. ==== >h^2 = x^2 + y^2 >so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations, >constants are left out and... >-x^2 (proportional to) y^2 #thus Nope. You can leave out constants in the original equation when > you're describing _rates_of_change_, but not when you're describing > relations of the original variables. What do you mean by this? Any rate of change equations is also one that links the two variables whose proportionality we're trying to discern. For example: The Volume of a sphere is: V = 4/3(pi)r^3 so V is related to r by the above equation, and V is proportional to r^3. In this case the constant h is (by definition of a right triangle) > larger than either x or y. x and y are never proportional for > constant h. I must say that I don't understand this statement as well. Of course x and y will not be proportional to constant h as they change, since its a constant and it doesn't change as the other variables change. -- > Stan Brown, Oak Road Systems, Cortland County, New York, USA > http://OakRoadSystems.com/ > You find yourself amusing, Blackadder. > I try not to fly in the face of public opinion. Doug ==== This actually makes the most sense, but can you come up with a quick proof ? You would think that as the rate of change of x or y increases to a significant amount, the constant added on would start to become negligable, and thus allow for what I'm trying to do, no? Doug So I was doing an elementary rate of change on x with respect to time given > the change in y w.r.t time. I also know that x and y are the two sides of > the triangle, connected by the constant hypotenous h. Now we all know the > following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations, x^2 = h^2 - y^2 is neater style constants are left out and... > -x^2 (proportional to) y^2 #thus > x (proportional to) y I beg your pardon, but you're learning rules instead of math. > y proportional to x when there's some constant c such that > y = c * x Thinking thus from the basics, you can see that from > x^2 = h^2 - y^2 > x^2 is _not_ proportionally to y^2. Now if y^2 is proportional to x^2, then for some constant c > y^2 = c * x^2 > Hence y = (sqr c) * x and thus as you wished, y is proportional to x > and in this case the constant of proportionallity is sqr c. > ==== So seeing how you were waking up when me was hitting the sack, (and by the G'Day), I'll assume you're from down-under...so thanx for replying AND for bringing us up here in the north the Crocodile Hunter - one crazy bloke ... contradict with a counter example. I'm guessing that you're assuming x will only be proportional to y if dy/dx = 1? for example as in: x^2=y^2 #where we all agree that x is proportional to y. x=y d/dy(x) = d/dy(y) x' = 1 But now consider adding a constant to one side of the equation: x^2=3y^2 #so that x = (3y^2)^(1/2) and if you find the derivative of d/dy...this time you'll get...(and I'll let you do the math) dx/dy = (3y)/((3y^2)^(1/2)) which certainly no longer appears to be proportional > So I was doing an elementary rate of change on x with respect to time > given the change in y w.r.t time. I also know that x and y are the > two sides of the triangle, connected by the constant hypotenous h. > Now we all know the following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing > relations, constants are left out and... > -x^2 (proportional to) y^2 #thus > x (proportional to) y so since the h^2 is constant, the change in x should reflect the > change in y since x^2 is proportional toy^2. However, this is not the > case, meaning that if we have a ladder standing on the floor and > leaned against a wall, if it begins to slide, it will not slide on x > proportional to y. So the question is why? what am I missing....and perhaps I should stop > reviewing math late at night? Thanx > Doug G'day Doug, Perhaps too late at night, let's see if I can still get this right (too > early in the day? ) x^2 = h^2 - y^2 doesn't imply x proportional to y since > x= sqrt (h^2 - y^2) > rate of change of x with respect to y: > dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy > = {-1/2[sqrt (h^2 - y^2)]} . (-2y) > = y/sqrt (h^2 - y^2) > so x is not proportional to y Julian > -- > Local IT Bloke > CSIRO, Forestry and Forest Products Ph: +61 8 8721 8118 ==== Yeah, I don't think this holds either...go over the example I just put up to Julian Mattay's comment and tell me if I'm still lost in the woods on this? Thanx doug Julain, if you directly differentiate the following equation, x^2 = h^2 - y^2 where _h_ is a constant term, and both _x_ and _y_ variables, then you > would end up with the answer, dy/dx = -x/y which if I have interpreted correctly defines clearly the proportional > relation of _x_ and _y_. x and y would only be proportional in that equation if dy/dx > were constant. It is not and so they are not. -- > Rich Carreiro rlcarr@animato.arlington.ma.us ==== Pardon, fergot to simplify, which would give us: dx/dy = (3y)/((3y^2)^(1/2)) = 3/(3^(1/2)) = constant and now that I think about it, what Rich said makes complete and utter sense with respect to proportionality equations having constants when ones change is compared to another. And I know that taking the derivative of d/dy or d/dx will not yield a constant for y^2 + x^2 = h^2, but why? Simply because we have a constant that is being added to both? Doug > So seeing how you were waking up when me was hitting the sack, (and by the > G'Day), I'll assume > you're from down-under...so thanx for replying AND for bringing us up here > in the north the > Crocodile Hunter - one crazy bloke ... contradict with a counter example. > I'm guessing that you're assuming x will only be proportional to y if dy/dx > = 1? for example as in: > x^2=y^2 #where we all agree that x is proportional to y. > x=y > d/dy(x) = d/dy(y) > x' = 1 But now consider adding a constant to one side of the equation: > x^2=3y^2 #so that > x = (3y^2)^(1/2) > and if you find the derivative of d/dy...this time you'll get...(and I'll > let you do the math) > dx/dy = (3y)/((3y^2)^(1/2)) > which certainly no longer appears to be proportional So I was doing an elementary rate of change on x with respect to time > given the change in y w.r.t time. I also know that x and y are the > two sides of the triangle, connected by the constant hypotenous h. > Now we all know the following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing > relations, constants are left out and... > -x^2 (proportional to) y^2 #thus > x (proportional to) y so since the h^2 is constant, the change in x should reflect the > change in y since x^2 is proportional toy^2. However, this is not the > case, meaning that if we have a ladder standing on the floor and > leaned against a wall, if it begins to slide, it will not slide on x > proportional to y. So the question is why? what am I missing....and perhaps I should stop > reviewing math late at night? Thanx > Doug G'day Doug, Perhaps too late at night, let's see if I can still get this right (too > early in the day? ) x^2 = h^2 - y^2 doesn't imply x proportional to y since > x= sqrt (h^2 - y^2) > rate of change of x with respect to y: > dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy > = {-1/2[sqrt (h^2 - y^2)]} . (-2y) > = y/sqrt (h^2 - y^2) > so x is not proportional to y Julian > -- > Local IT Bloke > CSIRO, Forestry and Forest Products Ph: +61 8 8721 8118 ==== So I was doing an elementary rate of change on x with respect to time > given the change in y w.r.t time. I also know that x and y are the > two sides of the triangle, connected by the constant hypotenous h. > Now we all know the following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing Since you know the change in y with respect to time and want the change in x with respect to time, differentiate the entire equation with respect to time: x^2 + y^2 = h^2 d/dt (x^2 + y^2 = h^2) 2x dx/dt + 2 y dy/dt = 0 x dx/dy + y dy/dt = 0 You said you were given dy/dt, so solve for dx/dt in terms of everything else: x dx/dt = - y dy/dt dx/dt = -(y/x) dy/dt or dx/dt = -(sqrt(h^2 - x^2))/x dy/dt or dy/dt = -y/sqrt(h^2 - y^2) dy/dt Assuming you know either of x or y along with dy/dt, you thus know dx/dt. But to tie into some earlier statements, nothing in any of those equations implies x and y are proportional. Heck, even in something as simple as x + by = h (b, h are non-zero constants) x and y are not proportional. Remember, dy/dx = constant is a necessary *but not sufficient* condition for x and y to be proporational. -- Rich Carreiro rlcarr@animato.arlington.ma.us ==== I got the problem right, and I know that x and y in pythagoras are not working out to be proportional, what's bothering me is that I can't see why the wouldn't be proportional. The answer seems to be simply because there's a constant that is added on to one side (the h), and this seems to through everything off balance. With respect to the simple example you gave below (y = mx +b) you see, y and x ARE proportional if b = 0. Meaning, if we transform the cartesian plane by making sure that our segment goes through the origin, we'll be fine, and able to calculate the proportionality of x w.r.t y. And this further clerifies (and perhaps finally makes me see) why in the pythagoras eqtn x and y are not proportional....it seems, you can't simply eliminate constants that aren't related via a multiplication to the variable Doug So I was doing an elementary rate of change on x with respect to time > given the change in y w.r.t time. I also know that x and y are the > two sides of the triangle, connected by the constant hypotenous h. > Now we all know the following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing Since you know the change in y with respect to time and > want the change in x with respect to time, differentiate > the entire equation with respect to time: > x^2 + y^2 = h^2 > d/dt (x^2 + y^2 = h^2) > 2x dx/dt + 2 y dy/dt = 0 > x dx/dy + y dy/dt = 0 You said you were given dy/dt, so solve for dx/dt in > terms of everything else: > x dx/dt = - y dy/dt > dx/dt = -(y/x) dy/dt > or > dx/dt = -(sqrt(h^2 - x^2))/x dy/dt > or > dy/dt = -y/sqrt(h^2 - y^2) dy/dt Assuming you know either of x or y along with dy/dt, you > thus know dx/dt. But to tie into some earlier statements, nothing in > any of those equations implies x and y are proportional. Heck, even in something as simple as > x + by = h (b, h are non-zero constants) > x and y are not proportional. Remember, dy/dx = constant is a necessary *but not > sufficient* condition for x and y to be proporational. -- > Rich Carreiro rlcarr@animato.arlington.ma.us ==== >h^2 = x^2 + y^2 >>so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing >relations, >>constants are left out and... >>-x^2 (proportional to) y^2 #thus >> Nope. You can leave out constants in the original equation when >> you're describing _rates_of_change_, but not when you're describing >> relations of the original variables. What do you mean by this? Any rate of change equations is also one that >links the two >variables whose proportionality we're trying to discern. He was talking about an _additive_ constant. Since the derivative of a constant is 0, you can ignore additive constants when taking the derivative. >The Volume of a sphere is: V = 4/3(pi)r^3 >so V is related to r by the above equation, and V is proportional to r^3. Sure, but it's not proportional to r. >> In this case the constant h is (by definition of a right triangle) >> larger than either x or y. x and y are never proportional for >> constant h. I must say that I don't understand this statement as well. Of course x and >y will not be proportional to constant h as they change, since its a constant >and it doesn't change as the other variables change. You misread the preposition. I did not say x and y were non- proportional _to_ constant h, but _for_ (i.e., given or under the circumstances of) constant h. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== Following your style, I too will top post. A proof of what? If y = cx, then when x = 0, y = 0. Tho y isn't proportional to x, it's of the same order of magnitude as x. > This actually makes the most sense, but can you come up with a quick proof ? > You would think > that as the rate of change of x or y increases to a significant amount, the > constant added on would start > to become negligable, and thus allow for what I'm trying to do, no? Doug So I was doing an elementary rate of change on x with respect to time > given > the change in y w.r.t time. I also know that x and y are the two sides > of > the triangle, connected by the constant hypotenous h. Now we all know > the > following: > h^2 = x^2 + y^2 > so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing > relations, x^2 = h^2 - y^2 is neater style constants are left out and... > -x^2 (proportional to) y^2 #thus > x (proportional to) y I beg your pardon, but you're learning rules instead of math. > y proportional to x when there's some constant c such that > y = c * x Thinking thus from the basics, you can see that from > x^2 = h^2 - y^2 > x^2 is _not_ proportionally to y^2. Now if y^2 is proportional to x^2, then for some constant c > y^2 = c * x^2 > Hence y = (sqr c) * x and thus as you wished, y is proportional to x > and in this case the constant of proportionallity is sqr c. ==== Rich Carreiro typed: > x and y would only be proportional in that equation if dy/dx > were constant. It is not and so they are not. Upon feeding arbitrary values of _x_ in the equation, and finding with the help of the given equation the values of _y_, I find that _x_ and _y_ and indeed proportional, but only inversely. -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. ==== >Upon feeding arbitrary values of _x_ in the equation, and finding with >the help of the given equation the values of _y_, I find that _x_ and >_y_ and indeed proportional, but only inversely. Sorry, but no. If the original equation is x^2 + y^2 = h^2, which I believe it was, then y = +/- sqrt(h^2 - x^2), and there's no way x and y are inversely proportional. x and y would be inversely proportional if and only if xy=const. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== The solution to the Behrens-Fisher problem (BFP) renders techniques such as analysis of variance, Welch test, two-sample t test, two-sample z-test where variances are in fact being approximated, tests for functional relationships which assume homogeneity of variance, discriminant analysis - in fact other k sample tests, obsolete. Adverse consequences invalidating the results of these outdated tests can be found in: Statist. Med. 14 (1995), 101 J. Clin. Epidemiol. 50 (1997), 631 J. Epidemiol. Community Health 52 (1998), 764 Cancer Epidemiol. Biomark. Prev. 10 (2001), 569 American Journal of Kidney Diseases 40 (2002), 873 Eur. J. Gastroenterology Hepatol 14 (2002), 811 Ophthalmic Epudemiology 9 (2002), 347 The statistical software GSP, which uses the publications of Tsakok in Metron 36, p79 & p105, provides an easy way of implementing the solution of the BFP in all its aspects. Please contact me for copies of GSP. ==== L O N G 12 15 14 7 = 48 Angie Tysseland was playing piano on the NW corner of 21st Street and 4th Ave., I stopped to listen and to collect her stats. Angie was teamed up with Teresa Long, Teresa was providing vocals and doing a very good job of it, I was pretty well astounded by Terri Long's singing ability. This was an incredibly good free concert at the Saskatoon Jazz Festival. 213 Terri 4 5 59 124/241 806 Teresa 68 Lorrelle 97 Long 48 Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Teresa was born in 59, it's the 17th prime while 17 in turn is the 7th prime, while the primes up to 7 add to 17 (59 is the 7th prime in prime position). She was born on day 124 (Numbers 7). She was born 117 days closer to the beginning of the year than to the end of the year. Her first name adds to 68 (4x17 and is the 7x7th non-prime). Her middle name adds to 97 (Leviticus 7). Her first and last names average 58 (the 7 primes up to 17). Her 68 (4x17 and is the 7x7th non-prime) valued first name is 58 (the 7 primes to 17) short of her 124 (Numbers 7) day of birth. Her given names differ in value by 29, it's the 7th prime (17) plus the 7th non-prime (12), or simply 7p+7np. She has 7 different letters in her given names. The vowels in her given names add to 36 (7+7p+7np). She is missing 17 letters from her full name. Her vowels add to 51 (17+17+17). Her odd valued letters add to 77 and are in positions adding to 84 (7 times the 7th non-prime), it is a difference of 7. Her even valued letters add to 136 (8x17) and are in positions adding to 87, it's a difference of 49 (7x7 and is the 17+17th non-prime). Her even valued letters add to 136 (8x17) and exceed her odd valued letters by 59 (the 7th prime in prime position and is her year of birth). Her even valued letters add to 136, it's the 104th non-prime while 104 in turn is the 77th non-prime (136 is the 77th non-prime in non-prime position). Her Fibonacci valued letters add to the 21 (7+7+7) chapters of Bible Book 7 and are in positions adding to 36 (7+7p+7np). Her Lucas valued letters add with their positions for the 108 verses of Bible Book 59 (the 17th prime, her year of birth). Her unrepeated letters add with their positions for the 108 verses of Bible Book 59 (the 17th prime, her year of birth). Her names add to the 49th non-prime, 25th prime and to the 33rd non-prime, together for 107 (the perfect 28th prime while the numbers 1 through 7 add to 28). Her 213 valued name is 171.77% of her 124th (Numbers 7) day of birth. Her repeating letters are in positions adding to 124 (Numbers 7, her day of birth). Her last 7 letters add to 77. Her last two names differ in value by 49 (7x7 and is the 7x7th non-prime). She abbreviates her first name to Terri (70). She is commonly known as Terri (70) Long (48), these names average 59 (her year of birth). Non-Primes 1 27 50 72 4 28 51 74 6 30 52 75 8 32 54 76 9 33 55 77 10 34 56 78 12 35 57 80 14 36 58 81 15 38 60 82 16 39 62 84 18 40 63 85 20 42 64 86 21 44 65 87 22 45 66 88 24 46 68 90 25 48 69 91 26 49 70 92 Teresa adds to 68, her day, month and year of birth adds to 68. She has four E's and four L's, these most frequently repeated letters add together for 68. Her 17 missing letters add to 240 (68.37% of the possible total). Her 213 valued name adds with her 124th day of birth for 337 (the 68th prime). She was born on day 124 (Numbers 7) while her first name adds to 68 (the 7x7th non-prime). Her first name adds to 68 (49th non-prime) while her last two names differ in value by 49. Terri was born on day 124, corresponding to Numbers 7, the chapter contains 89 verses. Her 213 valued name exceeds her 124th day of birth by 89. The Gospels are Bible Books 40, 41, 42 and 43, together for 166. Terri's name adds to 213, it's the 166th non-prime while 166 is the 128th non-prime, while 128 is 2 to the 7th. Chapter 124 is Numbers 7, it is the 7x7th prime (227) short of the numbers up to the 17th non-prime (351). Genesis 7, Exodus 7, Leviticus 7 and Numbers 7 are chapters 7, 57, 97 and 124, together for 285 (the 7x7th chapter of The Samuels). Numbers 7 and Deuteronomy 7 are chapters 124 and 160, together for 284 (Second Samuel 17, the 7th prime). First Samuel 7 and Second Samuel 7 are chapters 243 and 274, together for 517 (17 is the 7th prime, the primes up to 7 add to 17). First Kings 7 and Second Kings 7 are chapters 298 and 320, together for 618 (the number of verses in Bible Book 7). First Chronicles 7 and Second Chronicles 7 are chapters 345 and 374, together for the 719 verses of Bible Book 12 (7th non-prime), this 719 is the 128th (2 to the 7th) prime. First Samuel 7 and Second Samuel 7 are chapters 243 and 274, together for 517, while First Chronicles 7 and Second Chronicles 7 are chapters 345 and 374, together for 719, it's an average of the 618 verses of Bible Book 7. 213 Terri 4 5 59 124/241 806 Teresa 68 Lorrelle 97 Long 48 189 Angie 27 2 61 58/307 1471 Angie 36 Grace 34 Tysseland 119 Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Terri Long and Angie Tysseland are teamed up and producing a CD (7), they were born in months adding to 7. Terri was born on day 124 (Numbers 7), Angie on day 58 (the 7 primes up to 17). Their 6 names have an average value of 67 (Exodus 17). Angie's names add to 36 (7+7p+7np), 34 (17+17) and 119 (7x17), together for 189 (the first 17 primes minus the first 17 non-primes). Their first names add together for 104 (77th non-prime). Their last names add together for the 167 verses of Bible Book 17 (the 7 primes in prime positions up to the 17th prime add together for the 167 verses of Book 17). Esther becomes Queen in Book 17 and Q is the 17th letter of the alphabet. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 Terri and Angie (6x6) are separated by 665 days (Ecclesiastes 6). Teresa (6 letters) was born 66 days further into the year than Angie (6x6). They were together born 366 days closer to the beginning of their years than to the end of their years. Terri's full name adds to 213 (166th non-prime). Their given names add together for 235, it is the 184th non-prime, pretty as the 184th prime (1097) and the 184th non-prime (235) averages 666. Their initials have an average value of 36 (6x6, and 1 through 36 adds to 666). Terri was born on the 4th (Numbers with 36 chapters). I showed them both gems and neither had thanx for showing them evidence that their very names are a gift from God. My math was used repeatedly as an excuse to arrest and chemically lobotomize me (torture me) in psychiatric facilities, and whenever I meet with people and show them patterns that please them, they never have the decency to offer to buy me a measly cookie for my work nor send me a cheap letter expressing thanx. Man oh man, you people are compassionless turds, you are the shit of the earth, soon God will spread you out over the surface of the earth like the dung that you are, and in this I rejoice. 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== I think you're ill mannered, boring and part of the reason why many people are not interested in religion. Dave. ==== > I think you're ill mannered, boring and part of the reason why many > people are not interested in religion. Dave. It's numerology, a religion Plonk-a-dork!!! RJP ==== Can anyone out there spare some time to help me expand an algorithm. I know the answer I want from it and have a starting point. Unfotunately I'm just too dumb to sort it myself. -- Ian Turnbull ==== I have found an algorithm written in pascal - on the internet. I had a hex file which I dis-assembled and found to my surprise to be similar in nature to the pascal one. The data acted upon by these similar algorithms is a 14 digit number. However, I am also trying to do some reverse engineering on a device I have which also takes a 14 digit number and 3 other values. I am therefore trying to emulate this device I have and thus [rightly or wrongly] am assuming the pascal & hex are an earlier version - hence the expand the algorithm statement But, then, I'm probably not making any sense...?! > Can anyone out there spare some time to help me expand an algorithm. > I know the answer I want from it and have a starting point. Unfotunately I'm > just too > dumb to sort it myself. > -- > Ian Turnbull ==== > But, then, I'm probably not making any sense...?! That bit I understand. ==== > Can anyone out there spare some time to help me expand an > algorithm. I know the answer I want from it and have a > starting point. Unfotunately I'm just too dumb to sort it > myself. I have found an algorithm written in pascal - on the internet. > I had a hex file which I dis-assembled and found to my > surprise to be similar in nature to the pascal one. The data > acted upon by these similar algorithms is a 14 digit number. However, I am also trying to do some reverse engineering on > a device I have which also takes a 14 digit number and 3 > other values. I am therefore trying to emulate this device I have and thus > [rightly or wrongly] am assuming the pascal & hex are an > earlier version - hence the expand the algorithm statement But, then, I'm probably not making any sense...?! Why don't you just post the Pascal code together with an explanation of what you are trying to achieve? Then you're more likely to get some useful replies. Daniel ==== V O G T 22 15 7 20 = 64 In the late afternoon I went to Starbucks and met Roger Bristow, soon he was joined by a friend of his, Victoria Vogt, both are medical students at the U of S. 64+ Dad 15 3 /291 64+ Mom 4 8 /149 183 Victoria 29 2 76 60/306 6951 Victoria 97 Lee 22 Vogt 64 64+ Sis 8 3 78 67/298 7689 64+ Bro 29 6 83 180/185 9628 Victoria was born on the 60th day of the year, it is 11 plus the 11th prime (31) plus the 11th non-prime (18), or simply 11+11p+11np. Her first and last names both begin with the 22nd (11+11th) letter of the alphabet. Her first two letters add to 31 (11p). Her middle name adds to the 22 (11+11) chapters of Bible Book 11. Her given names add to 119. Her first and last names differ in value by 33 (3x11). Her names have an average value of 61 (Exodus 11), it is the 18th prime while 18 in turn is the 11th non-prime), it is the 11th prime in non-prime position. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 <-11th-> 18 <-11th-> 55 <-11th-> 199 --- --- --- --- 160 113 143 518 The parents were born in months adding to 11 and also the kids were born in months adding to 11, together for the 22 chapters of Bible Book 11. The kids were born on days of the month adding to 66 (6x11) and in years adding to 237, it's the opening chapter of The Samuels, pretty as The Samuels contain 55 (5x11) chapters. The kids were born on days 29, 8 and 29, these Bible Books contain an average of 77 (7x11) verses. The males were born on days of the month adding to 44 (4x11). The males were born on days of the month averaging the 22 (11+11) chapters of Bible Book 11. The sisters were born on days of the year adding to 127 (the 31st prime while 31 in turn is the 11th prime), it is the 11th prime in prime position. The males were likely born on days of the year averaging 127 (the 11th prime in prime position). Generally the parents have their birthdays 150 days closer to the beginning of their years than to the end of their years, corresponding to Deuteronomy 33 (3x11). Generally the parents have their birthdays 150 days closer to the beginning of their years than to the end of their years, pretty as there are 150 chapters in Bible Book 19 while the parents were born on days of the month adding to 19. Mom was born 67 days closer to the end of the year than to the beginning of the year (19th prime), keeping in mind that the 2460 verses of Book 19 is 7x19x19 minus the 19th prime (67). If the parents were both born in non-leap years, then the family was born on days of the year adding to 597 (Psalm 119), pretty that Victoria's first name adds to 97 while her given names would add together for 119. I meet Victoria on the 19th. Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 <-Bible Book 8 contains 11 4 chapters, pretty as 13 8 is the 4th non-prime 17 19 -- 77 <-the first 8 primes plus 8 more adds to the 85 verses of Bible Book 8 Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 18 61 19 67 <- 67 <-the 8th prime in prime position R U T H <-Book 8 18 21 20 8 = 67 The family was born on days of the month adding to the 85 verses of Bible Book 8, it is a combination of the first 8 primes (up to 19) plus 8 more. The little sister was born on the 8th day of the month and on the 67th (19th prime or the 8th prime in prime position) day of the year, Bible Book 8 is Ruth (67). Generally the parents have their birthdays on days of the year adding to 290. Dad was born with 291 days remaining in the year. The little sister was born with 298 days remaining in the year. The family was together born with 1229 days remaining in their years. The first and the last kids were both born on the 29th day of the month. There are 29 chapters in Bible Book 13 (6th prime) and 29 verses in chapter 666 (Ecclesiastes 7), pretty as 29 is a combination of 6 plus the 6th prime (13) plus the 6th non-prime (10). Dad generally has his birthday on the 74th day of the year (a factor of 666). Mom generally has her birthday on the 216th (6x6x6) day of the year. The kids were born on days of the month adding to 66. 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) The first two letters in Vogt add to 37, the first pair of kids were born on days of the month adding to 37 and also the last two kids were born on days of the month adding to 37. The Vo(37)gt kids were born in years adding to 237 (the opening chapter of The Samuels). Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Because Victoria was born on the 29th I asked to see her pennies (copper, the 29th element), although I was attracted to her and really wanted to see her panties, and explained that her money was a gift from God. Anyway, she had 4 pennies, they were dated 89, 90, 96 and 02, together for 277 (the 59th prime or the 17th prime in prime position). The last two kids are separated by 1939 (7x277) days, or exactly 277 weeks. The kids are together separated by 2677 days. The kids were born on days 29, 8 and 29, these Bible Books contain an average of 77 verses. The kids were born on days and in months adding to 77 and in years adding to 237. The sisters were born on days of the month adding to 37. The kids were born on days of the year adding to 307. The kids were born on days of the month adding to the 66 Books of the Bible, it is 7x7+17 or 7 squared plus the 7th prime, or the 17th prime plus 7 more. Primes 2 73 179 3 79 181 5 83 191 7 89 193 11 97 197 13 101 199 17 103 211 19 107 223 23 109 227 29 113 229 31 127 233 37 131 239 41 137 241 43 139 251 47 149 257 53 151 263 59 157 269 61 163 271 67 167 277 <-59th 71 173 281 Mom was born on the 4th day of the 8th month, and see that she was born with 149 days remaining in the year (the number of verses in Bible Book 48). The little sister was born with 298 (149+149) days remaining in the year. The last two kids are separated by 1939 (7x277) days, or exactly 277 weeks. Victoria's pennies were together 35 years old, pretty as 149 is the 35th prime. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 <-17 is the 7th prime 5 11 <- 11 while the primes up 6 13 to 7 add to 17 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 <-the 7th prime in --- prime position 167 Esther Book 17 <-the 7th prime Leviticus begins with 17 verses and terminates at chapter 117 with 17+17 verses. There are 17 verses at chapters 1 and 3, and 59 (the 17 prime) verses at chapter 13, so the 17's and the 17th prime are at chapter numbers adding to 17 (1+3+13=17). The first 17 versed chapters in the Bible are at chapters 91 and 93, together for 184, or the 167 verses of Book 17 plus 17 more. Leviticus contains 859 verses, it ends in 59 (the 17th prime). The first 17's in the Bible surround chapter 92 (the 4x17th non-prime): Leviticus --------- 91 1 17 92 2 <-68th (4x17th) non-prime 93 3 17 94 4 95 5 96 6 97 7 98 8 99 9 100 10 101 11 102 12 103 13 59 <-17th prime 104 14 105 15 106 16 107 17 108 18 109 19 110 20 111 21 112 22 113 23 114 24 115 25 116 26 117 27 34 <-17+17 The family was born on days of the month averaging 17. The kids were born in years adding to 237, it's a combination of the 17th, 17+17th, 17+17+17th and the 17+17+17+17th non-prime numbers (26, 49, 70 and 92). The last two kids are separated by exactly 277 weeks, Victoria's pennies had years adding to 277 (the 17th prime in prime position). It's another true story. Non-Primes 1 27 50 72 4 28 51 74 6 30 52 75 8 32 54 76 9 33 55 77 10 34 56 78 12 35 57 80 14 36 58 81 15 38 60 82 16 39 62 84 18 40 63 85 20 42 64 86 21 44 65 87 22 45 66 88 24 46 68 90 25 48 69 91 26 49 70 92 <-the 17th level Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== Stand back, everyone! Dar is winding up again. The pitch should come in the next day or so. -- It takes a village to raise an idiot. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== ... > http://www.crbond.com A nice site - simple & informative. gtoomey ==== S T E V E N S O N 10 20 5 22 5 14 19 15 14 = 133 In the afternoon I went to the food court in the Wildwood Mall and met Buddy, he is the 6th of 8 kids. 133+ Dad 4 10 14 277/88 +15477 133+ Mom 9 4 16 100/266 +14924 278 Buddy 10 6 57 161/204 113 William 79 Lyon 66 Stevenson 133 The parents were together born 23 days closer to the end of their years than to the beginning of their years. Buddy and his parents were born on days of the month adding to 23, Buddy was born exactly 23 23 with 66 chapters. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 <-14th-> 22 --- --- 281 176 Dad was born in 14. The parents were born in months adding to 14. The parents were born in 14 and 16, these Bible Books contain an average of 614 verses. Buddy was born 43 days closer to the beginning of the year than to the end of the year (14th prime). He was born in to 79, it is the 22nd prime while 22 in turn is the 14th non-prime (79 parents are separated by exactly 79 weeks while his first name adds to name exceeds his 161st day of birth by the 117 verses of Bible Book 22 (14th non-prime). Dad and Buddy were born on days of the month adding to 14. My birthday was 14 days ago while Buddy and I had our birthdays an average of 140 days ago. Mom was born in 16 and married Stevenson, pretty as the name adds to 133 while Bible chapter 133 is Numbers 16. This 133 (Numbers 16) exceeds it's 101st non-prime position by 32 (16+16). Buddy has 16 letters in his first and last names, his last name adds to 133 born on days 4, 9 and 10, together these Bible Books contain 96 (6x16) chapters. Buddy was born in 57, Bible chapter 57 and Bible Book 57 both contain 25 verses (the 16th non-prime). The parents were born in years adding to 30 (20th non-prime). Buddy and his parents were born in months adding to 20. Buddy has 20 letters. Dad and Buddy were born in years adding to 71 (20th prime). Buddy and his parents were together born 20 days closer to the beginning of their years than to the end of their years. Buddy was born on day 161 (Deuteronomy 8 with 20 verses). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Dad was born on day 277, it's the 59th prime while 59 in turn is the 17th prime (277 is the 17th prime in prime position). The parents were born on days 277 and 100, these are the 59th prime and the 75th non-prime, together for 134 (Numbers 17). Dad was born 189 days closer to the end of the year than to the beginning of the year (the first 17 primes minus the first 17 non-primes). Dad was born 177 days further into the year than mom (3 times the 17th prime). Buddy was born with 204 days remaining in the year (Joshua 17), it is 12x17, or the 7th non-prime times the 7th prime, and is twice 17 plus twice the 17th 161 day of birth by 117. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations The parents were born on days of the month adding to 13 (6th prime). Dad was born in 14, Bible Book 14 contains 36 chapters (6x6 while 1 through 36 adds to 666). Mom was born in 16, Bible Book 16 contains 13 chapters (6th prime). Buddy's 278 valued name exceeds his letters in his given names add to 91 (1 through 13), the repeating letters in his given names add to 54 (13 plus the 13th prime), it is a difference of 37 (37 chapters in the Bible contain the length of 13 verses). I am 113 days older than Buddy, there are 113 verses in Bible Book 54 while his initials add to 54 (13 plus the 13th prime). He is the 6th of the kids, his first 6 letters add to 66, his middle name prime and the 184th non-prime averages 666). Buddy was born on the 10th (6th non-prime) day of the 6th month. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61 27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107 40 109 <-29th-> 42 --- --- 1480 665 Buddy's first name adds to 79 (Exodus 29), his given names add together for 145 (5x29). In his given names, his consonants plus their positions exceed his vowels plus their positions by 29. The primes and squares in his given names add together for 79 (Exodus 29). Buddy and his parents were born in years averaging 29. Mom and Buddy were born in years adding to the 73 verses of Bible Book 29 (and is the Lucas numbers up to 29). Dad and Buddy were born on days of the year adding to 438 (6 times the 73 verses of Book 29). Dad and Buddy were together born with 292 days remaining in their years (4 times the 73 verses of Book 29). Mom and Buddy were together born 209 days closer to the beginning of their years than to the end of their years (29.85 weeks). Buddy and I were born on days of the year adding to 209 (29.85 weeks). Buddy was born with 204 days remaining in the year (29.14 weeks). Mom and Buddy were born on days of the century adding to 26923 (13x19x109), it is both a multiple of 13 (29 chapters in Bible Book 13) and a multiple of 109 (the 29th prime). Dad was 42 (29th non-prime) years old when Buddy was born. We meet when I am 16815 days old, my age ends in 815 (the first 29 primes minus the first 29 non-primes). Bible Book 13 (the 6th prime) contains 29 chapters and Bible chapter 666 (Ecclesiastes 7) contains 29 verses, pretty as 29 is 6 plus the 6th prime (13) plus the 6th non-prime (10). Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Buddy is coming out of a family of 10, his dad was born in the 10th month while mom was born on the 10x10th day of the year. Buddy was born on the 10th day of the month. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 <-11th-> 18 <-11th-> 55 <-11th-> 199 --- --- --- --- 160 113 143 518 Buddy's first name adds to 79 (the 11+11th prime), his middle name initials add to 31 (11th prime). He has 11 letters in his given names and his given names add to 145, corresponding to Numbers 28 with 31 primes, squares and cubes together add with their positions for 211. exceed his vowels by 110. In his given names, his odd valued letters exceed his even valued letters by 45 (the 11th non-prime in prime position). In his given names, his unrepresented letters exceed his represented letters by 127 (the 11th prime in prime position). Buddy was born 62 days after mom's birthday (twice the 11th prime). He was born 249 days after dad's birthday (First Samuel 11). He was born 311 days after his parent's birthdays. Mom was born on the 9th, corresponding to First Samuel with 31 chapters (11th prime). Buddy and his parents were born on days of the year adding to 538, corresponding to Psalm 60 (11+11p+11np). I meet Buddy on the 62nd day of the year (twice the 11th prime). I am 113 days older than him (the first 11 non-primes). 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter parents were born on days and in months and years adding to 57. Dad and Buddy were born on days of the year adding to 438 (62.57 weeks). Daniel and Revelation are the major Books of end-times prophecy (they are in part about 666), they contain 357 verses and 404 (57+57p+57np) verses. He and his parents were born on days of the month adding to 23 (Isaiah with an average of 19.575757... verses per chapter). If I could convince Buddy that the God of the Bible provided him with his name, it would only result in him giving money to a church that has an Egyptian penis on it's roof. Perhaps Buddy would only attend a church in late December, and he would reluctantly do so in order to please his wife and her parents. While at that church he would see their decorated evergreen trees, comment on their beauty, and give the church money. The churches teach the sheep to turn evergreen trees into decorated idols, the evergreen tree was worshipped for centuries as a fertility symbol for it remains green throughout the year. The Old Testament documents and condemns how pagans surrounding and opposed to ancient Israel worshipped the evergreen trees. The Old Testament also documents and condemns the obelisks, some version refer to these symbolic penises as the Towers of Bethshemesh. Like the evergreen trees, the penis was worshipped as a fertility symbol. Or if I could convince Buddy that the God of the Bible provided him with his name, he would seek out a priest that has a fish head hat, and give that person money. The ancient priests of fish god Dagon dressed up in fish outfits, over the years their costumes evolved so that all that now remains is the fish head hat, again the fish is being worshipped as a fertility symbol due to their large number of eggs. Sticking a penis on the roof of your church (or a fish head hat on the head of your priest) is a violation of God's Second Commandment. Turning trees into idol, bowing to decorated trees and worshipping trees, penises and fish is a violation of God's First Commandment. By December 25th the sun is visibly returning from the south, calling this pagan sunwhoreshipping holiday Christmas is a violation of God's Third Commandment (it is a pagan mass and not facilities in an attempt to make me shut up about the false traditions in your churches is nothing less than a violation of God's Sixth Commandment. They tortured me for years, I begged for years for assistance to get out of the country to no avail, and all you people can do is give money to the churches that teach you to violate Commandments. You people collectively spent millions of dollars having me tortured, then annually you people collectively spend billions of dollars in turning trees into idols. My math was used as an excuse to repeatedly arrest and torture me, and then when I manage to meet with you people in restaurants and show you evidence that your name is a gift from God, you are so cheap and ignorant that you don't even have taking the time out of my life to show you such. You are an incompassionate turd, you are the shit of the earth, and I am on my knees begging God to honor Exodus 20:5 and Hosea 4:6 as promises, and terminate your life, the lives of your siblinks, and the lives of your children. Buddy, if I find you or your family members in the obituaries, I will cheer, it will be in accordance to Scripture (Psalm 137:9), and I will post these stats again. All you are good for is to have your stats posted on the usenet and be used as an example to others, and look, here you are!!! And it should be mentioned that Buddy Stevenson is a native Indian, Joe Munroe once informed me that the Indians on his reserve will flip flop between traditional aboriginal and Christian beliefs, depending upon whom they are trying to get into their beds. Your only compassion is for your filthy can try and let your filthy traditions save you. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!