mm-3024 === Subject: Order relation between two radicals....(Ramanujan ) Which is the order relations between $ A, B $ where A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . === Subject: Re: Order relation between two radicals....(Ramanujan ) Other interesting/magic/ pairs (A(i),B(i)), to be compared, are [S.Ramanujan] : A(1):= sqrt[3]{sqrt[3]{2}-1} B(1):= sqrt[3]{1/9} - sqrt[3]{2/9}+sqrt[3]{4/9} A(2):= 3*sqrt{sqrt[3]{5}-sqrt[3]{4}} B(2):= sqrt[3]{2} + sqrt[3]{20}- sqrt[3]{25} A(i)=B(i) ? , i in {1,2} . === Subject: Re: Order relation between two radicals....(Ramanujan ) > Other interesting/magic/ pairs > (A(i),B(i)), to be compared, are [S.Ramanujan] : > A(1):= sqrt[3]{sqrt[3]{2}-1} > B(1):= sqrt[3]{1/9} - sqrt[3]{2/9}+sqrt[3]{4/9} > A(2):= 3*sqrt{sqrt[3]{5}-sqrt[3]{4}} > B(2):= sqrt[3]{2} + sqrt[3]{20}- sqrt[3]{25} > A(i)=B(i) ? , i in {1,2} . If you are going to keep this up. using curt(x), or (x)^(1/3) for the cube root of x is a lot more intelligible than sqrt[3]{x} === Subject: Re: Order relation between two radicals....(Ramanujan ) > Which is the order relations between $ A, B $ where > A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . > I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . They are equal. Let 'curt(x) ' denote the cube root of x, then A^6 = 7*curt(20) - 19, trivially, and B^6 = (curt(5/3) - curt(2/3))^6 = 7*curt(20) - 7, though it is a good deal less trivial to do the simplification. === Subject: Re: Order relation between two radicals....(Ramanujan ) > Which is the order relations between $ A, B $ where > A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . A = Howisbeingused? B = (5/3)^(1/3) - (2/3)^(1/3) > I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . Bah. Even unrelated to square root. Can you come up with a better presentation? === Subject: Re: Order relation between two radicals....(Ramanujan ) William Elliot skribis: >> Which is the order relations between $ A, B $ where >> A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . > A = Howisbeingused? > B = (5/3)^(1/3) - (2/3)^(1/3) >> I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . > Bah. Even unrelated to square root. > Can you come up with a better presentation? The presentation gives absolutely correct LaTeX if you set the in front of each sqrt. First I suspected what the problem is: Grab a calculator, I thought. But then I used different calculators myself: TI-66: A = 0.312050634263 B = 0.312050636755 => A < B TI-89: A = 0.31205063679722 B = 0.3120506367604 => A > B Being a physicists I would claim them equal, though ;-) s ---------------------------------------------------- ( 2+2=5 for small values of 5 and big values of 2) === Subject: Re: Order relation between two radicals....(Ramanujan ) skribis: >> Which is the order relations between $ A, B $ where >> A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . >> I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . > First I suspected what the problem is: > Grab a calculator, I thought. But then I used different calculators myself: > TI-66: > A = 0.312050634263 > B = 0.312050636755 => A < B > TI-89: > A = 0.31205063679722 > B = 0.3120506367604 => A > B > Being a physicists I would claim them equal, though ;-) > s Which, of course, they are. Interesting that the TI-89's CAS gets it wrong unless you give it a hint: (7*20^(1/3)-19)^(1/6)=(5/3)^(1/3)-(2/3)^(1/3) returns false. 7*20^(1/3)-19=((5/3)^(1/3)-(2/3)^(1/3))^6 returns true. === Subject: Re: Order relation between two radicals....(Ramanujan ) On Tue, 21 Jun 2005 15:37:57 +0200, s >Being a physicist I would claim them equal, though ;-) You could take the sixth power of each and find the answer quite easily. === Subject: Re: Order relation between two radicals....(Ramanujan ) > Which is the order relations between $ A, B $ where > A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . > I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . Yicks, how cryptic. B = sqr 3? === Subject: Re: Order relation between two radicals....(Ramanujan ) >> Which is the order relations between $ A, B $ where >> A=sqrt[6]{7sqrt[3]{20}-19} and B= sqrt[3]{5/3} -sqrt[3]{2/3} . >> I have denoted sqrt[p]{X}:= X^{1/p} , p>= 2 . >Yicks, how cryptic. B = sqr 3? No; A is the sixth root of 7(20)^{1/3}-19, and B is the difference between the cubic roots of 5/3 and of 2/3. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Searching for simple drawing program with coordinate input XPident: Unknown > Could anyone direct me to a simple drawing program, preferably shareware > or freeware, that can draw figures containing lines, circles and > possibly ellipses - not with a pen or a brush but from coordinate > input, something like Circle (3.0 ,2.8, r=2.5) or Line (-2.3, 3.6) - > (4.4, 2.7), with an option to export to a popular graphical format like > GIF? Nothing fancy, not even plotting graphs. (Colour filling would be > nice though.) All programs I've seen so far do a lot more than I need, > and work with some kind of pen or pencil. > I'd like to be able to make simple pictures of solutions to mathematical > puzzles. I use MetaPost that comes as part of the TeX typesetting package. To do a single diagram to illustrate your examples, beginfig(1) draw fullcircle scaled 2.5cm shifted (3.0cm, 2.8cm); draw (-2.4cm, 3.6cm) -- (4.4cm, 2.7cm); endfig; end; When processed by mpost, this generates a PostScript file. The downside is that MetaPost is really a programming language and unless you are a programmer accustomed to the code-compile-test cycle you are unlikely to find using it very intuitive. Glen -- Glen Prideaux (My email address is really glen at.) === Subject: Re: Searching for simple drawing program with coordinate input > I use MetaPost that comes as part of the TeX typesetting package. To do > a single diagram to illustrate your examples, > beginfig(1) > draw fullcircle scaled 2.5cm shifted (3.0cm, 2.8cm); > draw (-2.4cm, 3.6cm) -- (4.4cm, 2.7cm); > endfig; > end; > When processed by mpost, this generates a PostScript file. > The downside is that MetaPost is really a programming language and > unless you are a programmer accustomed to the code-compile-test cycle > you are unlikely to find using it very intuitive. I'm familiar with the fundamentals of Pascal and C++, so that shouldn't be a problem. The code that you cited seems rather === Subject: The grandest Math Ocean Liner The world's largest, longest, tallest, grandest Math Ocean Liner ever! The most famous Math Ocean Liner in the world! Up to 5,000,000 corporate & individual passengers. Absolutely safe! Please fill out the form below http://webstore.maplesoft.com/ to tell us about your math cruise vacation preferences. A Maplesoft personal consultant will then contact you to assist you with the details and to help you finalize your math vacation plans. You can expect a quick response during our regular hours of operation: Please find the first 16 Mb of the spectrum of entertainment stuff we offer you on route to the final point of our voyage http://maple.bug-list.org/MapleCrisis-Review-01.pdf Bon voyage! === Subject: Re: The grandest Math Ocean Liner Nothing coherent. +------------------+ /| /| | | || || | Please do not | / O O | feed the Troll | / | | / ------------------+ / | || / | | | | / || / | | |/ | || / / | | || / | | | --| | | | | --| * | | | | | | -/ *-- -- | || / | / `´ * / /- | | | * c c c C/ C c c c -- the shadow === Subject: Re: The grandest Math Ocean Liner Sirs, the careful investigation has shown, that VB has no serious mathematical education, academic degrees and any publications on CAS during 1994 - 2005, excluding his amusing posts in non-moderated forums similar to that. === Subject: Re: The grandest Math Ocean Liner It may be true, but (from my point of view) his CAS bug hunting is most useful, and I don't care whether he is a megalomaniac or not. -- Valeri Astanoff === Subject: Re: An exact simplification challenge - 4 (beat all world's CASs, again) <11b5utt7kvvm805@corp.supernews.com> <11b6717m4dtnf00@corp.supernews.com> Fri, Jun 17 2005 2:54 pm Hello Jon, First let me tell you that I am delighted to find such a dauntless kindred soul like you! ;-) The centerpiece of my life is, The devil is not so black as he is painted. Please find the detals below. > on Fri, Jun 17 2005 12:35 pm VB> An excellent question. You hit the nail by the head. JS> Since you are a genius at CAS, VB> Woow! Really? How much nice hearing such an estimation! :) VB> To get serious, in our not large Simferopol, just 300,000 VB> inhabitants, where I know in person about 300 programmers, VB> about 5 of them claim that they are geniueses (what an VB> overpopulation ;) VB> But I am competent in a couple of fields, and to be competent VB> means to realize the limitations. This is why I am not and VB> was not ever among those 5 programmers who claim they are VB> geniuses. VB> What I claim however, is that I am gifted, but this is VB> something very much different from 'genius'. JS> why don't you create the ultimate system? VB> It would be silly of mine never think about such a feat, VB> and about a decade ago I tried, just for fun, to see, what VB> is necessary... and in my opinion such a miraculous CAS - VB> which efficiently deals with all the entities at say VB> http://mathworld.wolfram.com VB> http://www.math-atlas.org/welcome.html VB> could require up to 100,000,000 lines of code and data 8-( VB> Compare, Mathematica and Maple are of about 1,000,000 lines VB> of code each... after about 20 years of development. JS> The fundamental concepts of these systems are very simple. JS> The complexity tends to come in the implementation and JS> growth of the systems. I fully agree with you. JS> I once thought about doing a CAS before that would have JS> some features that are much better than what is currently JS> available. Your approach is definitely a very promising. To concieve and implement a prototype, then test it carefully, see the actual and potential problems and bottlenecks, and if all works nice and there is no solid reasons to expect some tangible quality degradation in the process of the expansion of such a system then reinforce it step by step. If unfortunately the given prototype's performance is not adequate (sigh) then drink coffee, think-think-think, and from overnight insomnia concieve a new prototype, and the cycle repeat itself... ;) My impression is that maybe within just 10 cycles at most a viable tech solution can be discovered. Now please note that as you perfectly realise this, this exploit would be of truly enormous industrial and educational impact! JS> (I personally think that maple and matlab are poor excuses JS> for a computer program) When I think of Maple design, the more the VM machine drills down to new and new layers of distinct Maple bugs, the more often I find that Maple is kinda a crash project. As one of the most insightful, competent, and wittiest engineers of the 20th century, Designer-in-Chief of the Saturn V launch vehicle, the most powerful rocket ever successfully flown, Wernher von Braun has it: Crash programs fail because they are based on theory that, with nine women pregnant, you can get a baby a month. ;-) VB> But the toughest point might be the overall design and VB> dozens of other very essential points. JS> Well, I think the main issue is to design a coherent system JS> that is efficient and elegant. Again, I fully agree with you! Personally, I am much impressed with Stephan Wolfram's overall design of Mathematica, and his main concept, list. A huge stock of revellations, a real treasure trove is also both RJF' published papers (well, you know) http://www.cs.berkeley.edu/~fateman/ and his most geneous *thousands* remarks over the forums... and an amazing amount of comments and remarks on MAXIMA http://www.math.utexas.edu/pipermail/maxima/ all his remarks :) JS> It is not an easy task of course, but maybe not as hard as JS> you think Actually, while mentioning 100,000,000 lines of code/data I was speaking of deploying a full-fledged CAS. JS> (but if you don't start you will never learn anything and JS> never accomplish anything) Why sure! Voting with 2 hands! ;) VB> In other words, in a nutshell the answer to you question VB> sounds, By the same reason that the best world's powerlifter VB> cot lift a 1x1x1 m cube made of osmium - its mass is VB> 22 tons and 610 kg. JS> Well, if he never tries then he will never know... the JS> reason he would know he couldn't because he tried... JS> then he might say that he can't but if could if he works JS> out harder and builds more muscle... JS> the worse thing is a defeatist attitude before you are JS> even defeated. Just an example at hand. :-) Had I only think of the VM machine and how it is difficult to implement it, thus intimidating myself, I'd never reached the level of its implementation we have now achieved. VB> But still Cyber Tester and me can bring into such a building VB> site an integral part of its success, something without which VB> the whole project - it it will be lauched ever - will fail. VB> And this is a point I'd definitely prefer to explain in detail VB> later, even more, it has been pled to do this within about VB> the next 12 months. VB> But even if such a formidable, majestic, stupendous project VB> would be never launched - which would be a great pity! - even VB> in a reasonably near future Cyber Tester will bring talking VB> improvement in CASs quality. VB> And - on universe - I am speaking just about this stuff - VB> and the less I am distracted the sooner all of you will VB> see what exactly Cyber Tester offers. JS> Well, you can contribute also by mapping out your design goals. JS> If they are worth their weight then someone will come along JS> and use them. complaining about other systems will not get you JS> where you want. JS> I hear your complaining loud and clear VB> Yep, this is a first step - to show how the things are not VB> black but pitch-dark at least with 1 computer algebra system... JS> I think everyone knows this Let me tell you here, yes and no. Yes, a hell of a lot of persons realize that Maple is, to put it mildly, not quite perfect. JS> but most people are to lazy to do anything about it. Maple JS> and Mathematica are sufficient for most peoples needs (and JS> are actually amazingly good at some of the things they do). JS> Its similar to bitching about the gov... everyone knows its JS> broke but no one wants to go through the trouble to fix it. :) JS> but yet I don't see you doing anything to fix the problem VB> Firstly, you cot see some stuff because it is our trade VB> secret. Secondly, the events happen about which you still VB> do not know because we find it to be premature to inform VB> the public until certain results are achieved. JS> Ok, I will accept that. But for the last several months JS> I have seen you post many flaws about maple which is JS> somewhat irrelevant to your project... that is, your project JS> should be independent of how maple performs except to know JS> that it performs badly and that you can do better (or at JS> least think that you can). JS> (if you think complaining will help then you are wrong). VB> This is generally right. But with Maplesoft, the core is that VB> the owners of this company seem do not want (or did not? and VB> my fears are in the past?) to pay their most serious attention VB> to stabilizing their development process instead of adding - VB> on the basis of painstaking research over a long string of VB> Maple versions it is possible to claim safely - inadequately VB> tested - news chunks of code. JS> Man, if you don't know about this stuff now then have you JS> been living in a cave? ROTFL... a zero cool comparison! JS> Almost all big businesses are like this. There only goal is JS> to continue to make money. They have no reason to expend JS> money on something to improve there product if they do not JS> get 10x the return. Most of those guys that were initially JS> with the creation of those projects are living it up now. JS> I recently have had very similar experience where a company JS> has refused to act on their rules and guidelines on a JS> customer who has violated them (actually it was my father's JS> situation... and the act was fraud). The company has no JS> desire to pursue this because it will cost them money and JS> that is the bottom line (unfortunately). They do not care JS> if they can make a better enviroment for the customers unless JS> it will increase profit. So, my advice, if I can offer any, JS> is that you should spend your time focusing on designing a JS> better CAS than wasting it on anti-maple stuff. You will JS> get farther and faster if you did. VB> You will see much more within the counted days, VB> just wait a bit. JS> Ok, I hope so ;) and I really do hope that you can do JS> something better than maple. Hot dog! This is one of the highest appraisals I ever hear over my lifetime. Up to this comment of yours, I was considering in my mind, for many times, how the things with a new, novel, more powerful CAS could be done. But this was mostly just for fun, for my own amusement... Now with these warm words of yours, I am feeling like something intangible, oversubtle has changed in me. I promise nothing but... at least I will think about this topic more often ;-) A devilishly pretty goal! JS> What would be even nicer is if you could put them out JS> of business (by offering a superior product as a very JS> reasonable cost). JS> Though, chances are that if you developed a product like JS> maple then you would become just like them and maybe 10-15 JS> years down the line there would be a someone trying to JS> oust your product. ;-) Maybe at this point someone would sign, c'est la vie. But for me this sounds this way: This is why I adore life and people. Go beyond all the limits imaginable. Best, Vladimir -- Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: Maple's fd crash code, and FORTRAN in Maple Sat, Jun 18 2005 12:24 pm First of all, let me thank you much for a really non-trivial message! It certainly offers several nice ideas to discuss. It is remarkable that you have introduced a notion Bug Ombudsman ;-) CC> Bug Ombudsman Mr V. Bondarenko All computer algebra systems have bugs, but with an *eerie* amount of distinct bug manifestations in Maple 9.5.2, with the *minimalistic* estimation of 20,000+ and presumably, even worser situation with Maple 10 - we will come to this soon - your notion seems to be really important for the whole symbolic computation community. I am going to come back to this later, and as for the time being let me please confine myself with just a short remark about the crashes you refer to CC> Maybe there are these 3 possibilities: CC> (1) Mr Bondarenko saw more than 1 memory address CC> but it got omitted. address. Before reporting publicly, I have a habit to check the reported stuff about 10 times, rebooting each time. To cross check myself, these days, I produced yet another line of 10 rebooting and found that again, always the same address has been reported by Maple to the user, 0x00245713. Best wishes, Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf [16 Mb] VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: Bondarenko denunciations continue <48f73fb4a8fa4655bde217934e4ea6fa.122484@mygate.mailgate.org> Hello Jay, JB> i think he could have done something useful with what he has, JB> like writing a book on sorted Maple bugs which could've been JB> useful for Maple users, I find your idea to be really excellent! Also, I believe that it would be very nice to add, as an integral part of such a book, a list of possible workarounds whenever possible, to help the user in case he or she encountered a bug. As you know, a huge stock of workarounds can be found at the Other Bugs Links link at the Maple Bugs Encyclopeadia, for example Give and receive help for the mathematical software Maple. Questions at all levels are welcome. http://groups.yahoo.com/group/maple-assist/ Discussing bugs in Maple 9, suggestions for improvement, and differences with earlier versions of Maple. This group is NOT for general questions about Maple. http://groups.yahoo.com/group/maple-new/ The Maple User Group (MUG) was an electronic mailing list designed to give Maple users an opportunity to discuss applications, problems and issues with other users. Monthly as text files from http://www.scg.uwaterloo.ca/~maple_gr/Digests/ Discussing bugs in Maple 8, suggestions for improvement, and differences with earlier versions of Maple. This group is NOT for general questions about Maple. http://groups.yahoo.com/group/maple8/ The Maple Advisor Database is a project of the Symbolic Computation Group of the University of British Columbia, under the direction of Robert Israel. http://www.math.ubc.ca/~israel/advisor/advisor.html where the brightest world's Maple wits, the Maple experts http://webpages.shepherd.edu/amihailo/maple.asp showed their invariably workable, at times, extremely witty solutions. In fact, during the last 3 years I have already thought many times about writing such a Maple supporting book, and the biggest problem here seems to be the following. Next Maple versions should be developed and released, and be competitive enough that such a book would have real sense (in the context of Maple syntax/specifics) Unfortunately, our multi-year detailed reaserch of the Maple development process, on the basis of bug propagation details, reveals an increasing level of its instability (you could observe, an unaided eye, with a relatively recent really user- unfriendly move of Maplesoft, I'd rather say, a gesture of the Maplesoft's top officers despair, the releasing an unstable, crash-prone, BETA version under the brand of a regular commercial release) Maple 9.02 for MacOSX : Always a Shame !! 1) The recovery of my former files (Maple 7) is practically impossible [details...] 2) Printing is catastrophic : [details...] To summarize : On Macintosh OSX platforms, Maple 9.0.2 is unusable in the current release. Former files are irretrievable (text input and standard math). Paper publishing is unusable. Maplesoft is ing around with Macintosh users. ---> Are the engineers of Maplesoft as stupid as they let it appear ?? ---> Either do they make deliberately so that everybody goes away to Mathematica ?? ---> I am ashamed for them ... ---> Don't buy Maple 9 (neither 9.0.1, neither 9.0.2) for MacOSX !! Carl DeVore: http://groups.yahoo.com/group/maple8/message/111 Some users are getting very frustrated with the lack of public information about bugs. I believe that if nothing is done to address this issue, then Maple will quickly become a system which is only useful for teaching calculus and other low-level courses. It will be useless for research. Prof Dr Rafal Ablamowicz http://www.math.tntech.edu/rafal/ http://groups.yahoo.com/group/maple8/message/183 I strongly support the idea of having a site listing Maple bugs as posted by Vladimir Bondarenko and commenetd on by Bertfried Fauser Harald Pleym http://www.hpleym.no/ Sorry to say, but I think Maplesofts days are numbered. Maple bugs: the next freeze frame calculated by GEMM Maple bugs: A dangerous, misleading ad statement at the main Maplesoft's site And Cyber Tester, at the given span of time and later on, is focused not on Maple whatsoever. This day, you only see our Maple stuff, but we work painstakingly across both free and commercial computer algebra systems. Thus, at the given time, I am afraid that the task of preparation to the streamlined improvement of CASs quality and gradual transforming the whole CAS field is so burning & urgent than I simply could not afford writing anything like the above-said Maple Bug Book 8-( In part, you can find quite a number of workarounds at the Maple Bugs Encyclopaedia, JB> and publishing something about his method, if they are any JB> good. We aleady has provided the symbolic computational community with fully free Maple Bugs Encyclopedia - and ever much more various fully free stuff is coming... But... Let me please quote here the wording I like very much by Herr Christopher Creutzig AZ> MuPAD is great (but it would be even better if you made it AZ> open source) CC> Give us a realistic way of earning a living from that. CC> Sorry, but we all like to have something to eat. ;) JB> instead, he goes down the wrong road, Here, maybe I'd politely diasgree a bit with you ;) as, in fact, the business plan behind Cyber Tester has been designed very diligently, with genuine care. JB> opting to become megalomaniacal instead. I fully agree with you that, for a sector of persons here, what I, the leader and the senior owner of the Cyber Tester, am doing (by design) might really look like a behavior of a person who exaggerrates (and, surprisingly, this is also a component of our business plan... let me stop here ;) As a matter of fact, however, the especial exquisiteness and uniqueness of the whole case is that my statements are true, they are based on multi-years experimentation, my mathematical, engineering, and managing powers and multi-year experience, Extended Resume http://www.cas-testing.org/index.php?list=3 Opinions (much outdated, selected of selected only, all this cries for an upgrade!) http://www.cas-testing.org/index.php?list=7 Credits http://www.cas-testing.org/index.php?list=4 the power of the VM machine, our gifted & dedicated programmers (the average programming experience of 17.5 years), the already calculated Maple - not only Maple! - bugs stocks, to count just a part of our actual strengths. Both you and the other folks will see this with your own eyes within just the next 12 months, within the framework of the JB> at times i feel pity for him. I am thankful you for showing your sincere feelings. Nowadays, alas, persons seem to be kind of bit ashamed to show their real feelings, this is why I truly appreciate your confession. I hope, upon seeing our next moves, by the end of 2006 you will feel sheer light, and maybe even effervescent proudness for our achievements ;) Best of luck, Vladimir -- Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: simple query I have DCF = 1/(1.08)^2 = 0.8573 I tried this on a calculator and got a different result. Can anyone tell me how I can get the answer above. === Subject: Re: simple query >I have DCF = 1/(1.08)^2 = 0.8573 >I tried this on a calculator and got a different result. Can anyone >tell me how I can get the answer above. Enter 1.08 press x^2 button press 1/x button. --Lynn === Subject: For the time being Recent results of experimental evidence integral quibble dt = waste(t) can be summarized as quibble = dwaste(t)/dt Let T(t) be whatever it was that'll do for the time being at time t. Can anyone give an estimate for integral(prehistory,today) T(t) dt the accumulated offerings to the time being? Totalitarianism Society is partially ordered. Can society be well ordered? By use of the axiom of choice, a partial order can be extended to a total order. Therefore society can be totally ordered. However do recall, a total order may not be well ordered. ---- === Subject: Novel factorizations, episode IV Hi! > Begin Factors.cpp <<< #include // Rudimentary code! Number-to-factorize is hard-coded as 'N': #define N 3*5*7*11*13 // 3*3*3*3*3*3*3*3*3*3 // 1001 // 1009 int Expand( int n ) { int bit = 1; while( n > 4*bit ) { bit = bit * 2; } return bit * n; } int Oddify( int n ) { while( (n&1) == 0 ) { n = n/2; } return n; } void FindFactors( int x,int y,int b ) { if( Expand(N)/y <= x || Expand(N)/(y+2*b) > (x+6*b) || b == 0 || x > 2*y ) { if( Expand(N) == x*y ) { cout << N << = << Oddify(x) << * << Oddify(y) << endl; // Throw anything here to quit at first successful factorization // Comment out or remove the line below to find all factorizations throw Done.; } return; } FindFactors( x,y,b/2 ); FindFactors( x+b,y,b/2 ); FindFactors( x+b/2,y+b/2,b/2 ); FindFactors( x+3*b/2,y+b/2,b/2 ); } int main() { int n; for( n=0x00008000;n!=0;n=n/2 ) { if( n*n < Expand(N) ) { try { FindFactors( n,n,n ); } catch( ... ) { break; } } } return 0; } > End Factors.cpp <<< Loosely based on the idea presented in... Go ahead, try it out, analyze, get ideas, enjoy! - Risto - === Subject: Re: Inequalities: x,y,z with x+y+z=0 On 14 Jun 2005 20:41:32 -0700, alexandru.lupas@ulbsibiu.ro >Suppose that x , y, z are real numbers such that x+y+z=0 . >Prove or disprove the inequalities [...] >(3.2) x^{11}+y^{11}+z^{11}=< > =<(11/18)*|x^3+y^3+z^3|^3(x^2+y^2+z^2) if x^3+y^3+z^3 =< 0 . Alex, this last one looks strange. Since all odd power sums have the same sign, the absolute value makes the inequality trivial, and without the conversion to absolute value the best constant seems to be 341/216 instead of 11/18. Leon === Subject: Re: Inequalities: x,y,z with x+y+z=0 Let us denote S(j)=x^j+y^j+z^j with x+y+z=0 . Then S(2)*S(3)=A*S(5) , A:=6/5 , S(7)^{2}=B*S(4)*S(5)^{2} , B:= 49/50 , Also S(11)=C_1*S(3)*S(8) - C_2*S(3)^{3} *S(2) , C_1=11/6 ,C_2=11/18 . If S(3)=x^3+y^3+z^3= - K, K>=0, |S(3)|=K , then S(11)=- C_1*K*S(8) +C_2*|S(3)|^{3} *S(2) =< C_2*|S(3)|^{3} *S(2), i.e the last inequality. Of course the best constant C(*) in the inequality |S(11)| =< k* |S(3)|^{3} *S(2) it's very interesting. === Subject: Re: Inequalities: x,y,z with x+y+z=0 On 22 Jun 2005 20:37:48 -0700, Alex. Lupas >Let us denote S(j)=x^j+y^j+z^j with x+y+z=0 . Then [...] >Of course the best constant C(*) in the inequality > |S(11)| =< k* |S(3)|^{3} *S(2) it's very interesting. A few more details: Let F(t) = (t-x)(t-y)(t-z). Newton-Girard formulas show that F(t) = t^3 - 1/2 S(2) t - 1/3 S(3). This gives the recursion formula S(k+3) = 1/2 S(2) S(k+1) + 1/3 S(3) S(k). So each S(k) is a linear combination, with non-negative coefficients, of products S(2)^m S(3)^n with 2 m + 3 n = k. For odd k this means that n can only be odd, so S(k) < 0 if and only if S(3) < 0. Patiently applying the recursion formula again and again finally results in S(11) = 11/48 S(2)^4 S(3) + 11/54 S(2) S(3)^3. Because F(t) has three real zeroes, its discriminant is non-negative, which proves to be equivalent to the condition that 6 S(3)^2 =< S(2)^3 For S(3) >= 0 this means that S(11) >= (11/6 + 11/54) S(2) S(3)^3. Leon === Subject: Re: Inequalities: x,y,z with x+y+z=0 >Patiently applying the recursion formula again and again finally >results in S(11) = 11/48 S(2)^4 S(3) + 11/54 S(2) S(3)^3. >Because F(t) has three real zeroes, its discriminant is non-negative, >which proves to be equivalent to the condition that 6 S(3)^2 =< S(2)^3 >For S(3) >= 0 this means that S(11) >= (11/6 + 11/54) S(2) S(3)^3. Oops, make that S(11) >= (11/8 + 11/54) S(2) S(3)^3. Leon === Subject: Boot Strap === Subject: Re: Boot Strap A confidential message from mathedman has admitted to WE all that he's wanting a boot straping. === Subject: Re: Boot Strap That's the most intelligent thing you've ever posted! === Subject: Re: Boot Strap > That's the most intelligent thing you've ever posted! Please excuse, it wasn't intended to go over your head. === Subject: International Symposiums on Symbolic and Algebraic Computation http://www.mmrc.iss.ac.cn/~issac2005/ ISSAC is the yearly premier international symposium in Symbolic and Algebraic Computation that provides an opportunity to learn of new developments and to present original research results in all areas of symbolic mathematical computation. http://www.mmrc.iss.ac.cn/~issac2005/callposter.htm The Poster Sessions are an ideal avenue for presenting recent research results or reports on ongoing research projects that might not yet be complete, but whose preliminary results are already of potential interest to the community. ISSAC'2005 Chinese Academy of Sciences, Beijing, China http://www.mmrc.iss.ac.cn/~issac2005/ ISSAC 2002 Laboratoire d'Informatique Fondamentale de Lille, Lille, France http://www.lifl.fr/issac2002 ISSAC 2001 University of Western Ontario, London, Canada http://www.orcca.on.ca/issac2001 ISSAC 2000 St Andrews University, St Andrews, Scotland http://www-groups.dcs.st-and.ac.uk/issac2000/ ISSAC'99 Simon Fraser University, Vancouver, Canada http://www.cecm.sfu.ca/ISSAC99/ ISSAC'98 University of Rostock, Rostock, Germany http://wwwteo.informatik.uni-rostock.de/ISSAC98/ ISSAC'97 Maui, Hawaii, USA http://www.sigmod.org/dblp/db/conf/issac/issac97.html ISSAC'96 Swiss Federal Institute of Technology in Zurich Zurich, Switzerland http://www.sigmod.org/dblp/db/conf/issac/issac96.html ISSAC'95 Concordia University, Montreal, Canada http://www.sigmod.org/dblp/db/conf/issac/issac95.html ISSAC'94 St Catherine's College, Oxford, UK http://www.sigmod.org/dblp/db/conf/issac/issac94.html ISSAC'93 Glushkov Institute of Cybernetics, Kiev, Ukraine http://www.sigmod.org/dblp/db/conf/issac/issac93.html ISSAC'92 Berkeley, California, USA http://www.sigmod.org/dblp/db/conf/issac/issac92.html ISSAC'91 Bonn, Germany http://www.sigmod.org/dblp/db/conf/issac/issac91.html ISSAC'90 Tokyo, Japan http://www.sigmod.org/dblp/db/conf/issac/issac90.html ISSAC'89 Portland, Oregon, USA http://www.sigmod.org/dblp/db/conf/issac/issac89.html ISSAC'88 Rome, Italy http://www.sigmod.org/dblp/db/conf/issac/issac88.html Predecessor: EUROCAL SYMSAC 1986 Waterloo, Canada SYMSAC 1981 Snowbird, UT, USA SYMSAC 1976 Yorktown Heights, NY, USA SYMSAC 1971 Los Angelles, CA, USA SYMSAC 1966 Washington, DC, USA === Subject: Vladimir Bondarenko - Refereed Publication # 1 - International Symposium on Symbolic and Algebraic Computation - 1995 - Status: Accepted: Dr Vilmar Trevisan: Poster Session Chair - ISSAC'95 Received: from SMTP by EXPLORER (Mercury 1.13); Fri, 16 Jun 95 17:42:52 +0300 Return-path: issac95@mat.ufrgs.br Received: from spider.cris.crimea.ua by expl.cris.crimea.ua (Mercury 1.13) with ESMTP; Fri, 16 Jun 95 17:41:59 +0300 Received: from mat.ufrgs.br (euler.mat.ufrgs.br [143.54.24.1]) by spider.cris.crimea.ua (8.6.10/8.6.10.1) with SMTP id RAA17996 for SVA@expl.cris.crimea.ua ; Fri, 16 Jun 1995 17:38:13 +0400 Received: by mat.ufrgs.br (4.1/941007) id AA04164; Fri, 16 Jun 95 10:39:04-030 === Subject: poster at ISSAC'95 Cc: issac95@mat.ufrgs.br I am pleased to inform you that your poster submission for the ISSAC'95 has been accepted for presentation. Enclosed you will find comments from the referees. I ask that you carefully consider them for preparing your presentation. By mid-June you will receive information on the schedule and poster format. Vilmar Trevisan Poster Session Chair - ISSAC'95 ------------------------------------------------------------------ Referee 1. The author seems to have a heuristic which is able to solve multiple integrals with an arbitrary number of variables. Although no theoretical substance is evidenced by the abstract, I recommend that the poster be accepted. Referee 2. In the revision the author has made clear what has actually been accomplished, thus eliminating my original objections. Referee 3. In the previous refereeing, we recommended the following three for revision. (i) stating explicitly the derived formulas (ii) emphasizing the motivation for considering these particular functions (iii) explaining the subproblems where the author needed CAS showing what features of a CAS one needs. For (i) and (iii), the revised paper is much improved, and for (ii), examples are chosen for a certain demonstration of his non-algorithmic/heuristic approach to find exact and easily computable expressions for definite multiple integrations. Thus, the goal of the paper becomes clear. In this time, I am not sure the correctness of his formulations (1),(2) and ability/effectivity of his proposing approach for mathematical study supported by Computer Algebra System. There are uncertain points in his explanation of his general approach. Moreover, there is no description on correctness/precision on evaluation of the sums with floating arithmetics. However, it is very interesting and valuable to discuss on ability/effectivity of such an approach as computer aided mathematical research. In this point, it seems nice to give the author an opportunity to present his ideas in more details and discuss with mathematicians/experts on computational mathematics. Referee 4. The English of the paper should be improved, especially the use of emphasizing should have been taken seriously. The author should use just one way of emphasizing {em } for example. What we have now is really disturbing and it makes the paper difficult to read. Motivation: We still do not know where the formulas are coming from.(!!) It would be necessary to consider a formula of the considered type from the literature which others tried to proof. I mean integrals like the Mehta integral (see I.G. MacDonald, pp. 988-1007). Title: I would suggest to use Computer Algebra System Aided evaluation of some definite n-fold integrals with undetermined n Keywords: Keep just the only one definite multiple integral The others are either irrelevant or make no sense. Now I am going to consider the paper paragraph by paragraph. 1. Sentence 1 replace defined by given. Sentence 2 definition of --> knowing Sentence 3 cancel , if there exist at all. 3. Here (1)-(2) appear without having presented the corresponding formulas before. First the formulas should appear!! Reformulate the underlined part of Sent. 1 by saying that in the given references the formulas do not appear. Reformulate the last two sequences of this paragraph. 4. Sent. 1 these are yet unknown at these point. Sent. 2 segment -> interval Sent. 3 omit are right. The function name term is not nice. I would propose rank instead of it. S_n should be explained in a more elegant way for example by considering linear polynomials in a_i with coefficients 0 and 1. 7. Avoid the word subheuristic! 8. lucky --> successful given equation but not in general for relations that are not displayed, hence one does not see which is left and which is right. Omit the Note column form the table (No information for the reader). 11. polinomial -> polynomial (Use spell checkers!) 12. Reformulate. Let just the integral stay on the lhs. 14. Sent. 1 said --> noted 3) closed-form in which sense? As a sum it is by no means closed. 15. Thirdly?? Reformulate this paragraph. As far as the author can handle summation problem like the one here with the parameter l I would not speak about method. Bring the reference list to a uniform style. === Subject: Vladimir Bondarenko - Refereed Publication # 1 - International Symposium on Symbolic and Algebraic Computation - 1995 - Scheduled Presentation Time: Dr Vilmar Trevisan: Poster Session Chair - ISSAC'95 ................................................................ First world's man+machine based Maple review http://maple.bug-list.org/MapleCrisis-Review-01.pdf [16 Mb] Maple Bugs Encyclopaedia [Select options & click the Go button] http://maple.bug-list.org/ Extended Resume http://www.cas-testing.org/index.php?list=3 ................................................................ Download PDF [291 Kb] http://www.cas-testing.org/ARTICLES/VBsampl3.pdf Download ZIPped Postscript [72.3 Kb] http://www.cas-testing.org/ARTICLES/VBsampl3.zip Received: from SMTP by EXPLORER (Mercury 1.13); Tue, 20 Jun 95 20:41:28 +0300 Return-path: issac95@mat.ufrgs.br Received: from spider.cris.crimea.ua by expl.cris.crimea.ua (Mercury 1.13) with ESMTP; Tue, 20 Jun 95 20:41:17 +0300 Received: from mat.ufrgs.br ([143.54.24.1]) by spider.cris.crimea.ua (8.6.10/8.6.10.1) with SMTP id UAA09839; Tue, 20 Jun 1995 20:36:27 +0400 Received: by mat.ufrgs.br (4.1/941007) id AA18421; Tue, 20 Jun 95 13:36:21-030 === Subject: ISSAC poster Cc: issac95@mat.ufrgs.br (ISSAC 95) Content-Length: 1127 Your poster Definite Multiple Integrals... has been scheduled for presentation on Monday, July 10 from 4 to 5 pm The location of the room will be ounced at the conference. In the poster room, you will be given a poster board of about 1.5m by 2m. On this poster board, you can fasten sheets of paper with diagrams and text that explain your main ideas. You should plan to arrive at the poster room at least 45 minutes early in order to get set up. You may bring your own tape, pins etc. in order to attach your displays to the board, or we can provide you with these. Please begin to give careful thought about how to arrange your poster, so that it is interesting. Keep in mind that since the poster session is only an hour, many people will have only a few minutes to look at each poster. Therefore, it is best not to make the poster too busy. You are also encouraged to keep a few copies of your extended abstract for people to take. In the case of joint authors, at least one of the authors must be present. Vilmar Trevisan Poster Session Chair ................................................................ First world's man+machine based Maple review http://maple.bug-list.org/MapleCrisis-Review-01.pdf [16 Mb] Maple Bugs Encyclopaedia [Select options & click the Go button] http://maple.bug-list.org/ Extended Resume http://www.cas-testing.org/index.php?list=3 ................................................................ === Subject: Vladimir Bondarenko - Refereed Publication # 2 - International Symposium on Symbolic and Algebraic Computation - 1996 - Status: Accepted: Dr Wolfgang Kuechlin: Poster Session Chair - ISSAC'96 ................................................................ International Symposia on Symbolic and Algebraic Computation ................................................................ Received: from SMTP by EXPLORER (Mercury 1.13); Tue, 4 Jun 96 21:55:06 +0300 Return-path: issac96@informatik.uni-tuebingen.de Received: from spider.cris.crimea.ua by expl.cris.crimea.ua (Mercury 1.13) with ESMTP; Tue, 4 Jun 96 21:54:39 +0300 Received: from macon.informatik.uni-tuebingen.de (macon.Informatik. Uni-Tuebingen.De [134.2.12.17]) by spider.cris.crimea.ua (8.6.10/8.6.10.1) with SMTP id VAA09410 for sva@expl. cris.crimea.ua; Tue, 4 Jun 1996 21:46:42 +0400 Received: from utu.Informatik.Uni-Tuebingen.De by macon.informatik. uni-tuebingen.de (AIX 3.2/UCB 5.64/4.03) id AA16644; Tue, 4 Jun 1996 19:45:00 +0200 Received: by utu.informatik.uni-tuebingen.de (4.1/SMI-4.1) id AA01831; Tue, 4 Jun 96 19:46:27 +0200 (SVA@expl.cris.crimea.ua) === Subject: Re: ISSAC'96 poster session submission (V. Bondarenko) based on the referee reports, the poster session program committee has decided to accept your poster for presentation at ISSAC'96 in Zurich. Below you find the grades and comments (if any) we have received from the referees. Note that the referee process was less formal than for the conference part, and some reports came in by phone. We also include our refereeing guidelines for your information. Please take a minute and use the guidelines to decode the Grades given to your paper by the referees. Please try to improve your poster accordingly, if at all possible. During the ISSAC'96 conference, each poster will be displayed in hardcopy on a stand in the ETH main building. (You should plan for about 6 pages in A3 format; the exact dimensions of the space available will follow) During the Poster Session times, an author is expected to be present near the poster for explanations to anyone interested. There will be no talks or slide presentations for posters. We are also pling to print the posters as an internal report of ETH Zurich. The report will be distributed to the ISSAC participants, but it is not a part of the ISSAC proceedings. Each poster will be limited to a maximum of 4 pages A4. See formatting guidelines below. Please submit the final version of your poster to us by June 22, or notify us if you want it printed unchanged. Otherwise we will not print your poster. I hope to meet you at ISSAC'96. The conference starts on Wednesday, July 24, and ends on Friday, July 26. To register for ISSAC'96 fill in the registration form http://www.inf.ethz.ch/ISSAC96/registration.html and send it by e-mail, air mail or fax to the address mentioned there. Should you not be able to come and present the poster, please inform us as soon as possible. Wolfgang Kuechlin ISSAC'96 Poster Session Chair ................................................................ Prof. Dr. Wolfgang Kuechlin http://www-sr.informatik.uni-tuebingen.de/pages/personen/wolfgang_kuechlin.h tml ................................................................ === Subject: Vladimir Bondarenko - Refereed Publication # 2 - International Symposium on Symbolic and Algebraic Computation - 1996 - Poster Session - ISSAC'96 ................................................................ First world's man+machine based Maple review http://maple.bug-list.org/MapleCrisis-Review-01.pdf [16 Mb] Maple Bugs Encyclopaedia [Select options & click the Go button] http://maple.bug-list.org/ ................................................................ Computer Algebra System As A Tool: Exact Evaluation Of Definite n-Dimensional Integrals With Undetermined A Priory Integer n, International Symposium on Symbolic and Algebraic Computation, 1996, Poster Session, Zurich, Switzerland Download PDF [132 Kb] http://www.cas-testing.org/ARTICLES/VBsampl1.pdf Download ZIPped Postscript [43.3 Kb] http://www.cas-testing.org/ARTICLES/VBsampl1.zip ................................................................ Kelly B. Roach http://www.kellyroach.com/ This note discusses a generalization of some formulas in A Heuristic for Exact Calculation of n-Dimensional (n+1)- Parametric Integrals of Some Elementary and Special Functions by Vladimir V. Bondarenko. http://www.planetquantum.com/Notes/Ndim.pdf http://www.planetquantum.com/News/050601.html ................................................................ Poster Session V. Balakirsky Computational method for solving the systems of recurrent equations R. Barrere Functional aspects of some approximate methods V. Bondarenko CA system as a tool: exact evaluation of definite n-fold integrals with undetermined a priory integer n F. Boulier Differential algebra with Maple L. Brenig, M. Codutti, A. Figueiredo Numerical integration of dynamical systems using Taylor series M. Caboara, M. Silvestri Compatible module orderings M. Cipu Replicable functions: a computational approach K. Drescher Domains of computation in the computer algebra system Mupad B. Dupee A: An expert system for choosing and applying numerical library routines I. Gaal Algorithms for the computation of power integral bases in algebraic number fields G. Greuel, G. Pfister, H. Schoenem SINGULAR: A computer algebra system for singularity theory, algebraic geometry and commutative algebra I. Heitlager, N. van den Berg, J. van Veelen, V. Goldman, J. van Hulzen An abstract datatype for simplified numerical code generation in REDUCE S. Hessinger Determining the Galois group of an order four linear differential equation A. Heuerding, G. Jager, S. Schwendim LWB --- The logics workbench 1.0 G. Hotz, E. Rohnert, E. Schomer Graphical interface for geometric theorem proving O. Ivanov, S. Ivanov Module design of computer algebra interfaces and windows technology for PC P. Kenderov, M. Spiridonova, L. Alamanov Implementation of convex analysis operations using Maple A. Kondratyev On zero divisor recognition J. Llovet, R. Martinez, B. Castano Multivariate polynomial matrices using straight-line programs V. Min, I. Poloskov Random effects analysis with computer algebra systems L. Miller Algorithms for computing in subalgebras of polynomial algebras over a ring H. Park Grobner Bases and Signal Processing M. Pesch Left and right Grobner bases in Ore extensions of polynomial rings E. Pflugel On the latest version of DESIR-II A. Popov Symbolic computation of potential energy functions E. Prisman derSymb: simplifies derivatives with symbols K. Shirayanagi, M. Sweedler Automatic algorithm stabilization W. Steeb, T. Kiat Shi Object-oriented design and symbolic computation R. Stobbe FACTORY: A C++ Class library for multivariate polynomial arithmetic S. Tertichniy, I. Obukhova GRC-EC: Computer algebra system for applications in gravitation theory E. Volcheck Testing torsion divisors for symbolic integration S. Wetzel, W. Backes, T. Lauer The lattice class in LiDIA M. Wolff, C. Richard, O. Gloor Analysis alive - An interactive system for learning Analysis P. Zimmerm, L. Bernardin, M. Monagan Polynomial factorization challenges O. Zhukov Highly effective computing algebra ------------------------------------------------------------------- [Up] ------------------------------------------------------------------- Last update: 23 Jul 1996 Comments to mhart@inf.ethz.ch ................................................................ First world's man+machine based Maple review http://maple.bug-list.org/MapleCrisis-Review-01.pdf [16 Mb] Maple Bugs Encyclopaedia [Select options & click the Go button] http://maple.bug-list.org/ ................................................................ === Subject: Re: Vladimir Bondarenko - Refereed Publication # 2 - International Symposium on Symbolic and Algebraic Computation - 1996 - Poster Session - ISSAC'96 Received: from SMTP by EXPLORER (Mercury 1.13); Thu, 13 Jun 96 22:54:10 +0300 Return-path: kuechlin@informatik.uni-tuebingen.de Received: from spider.cris.crimea.ua by expl.cris.crimea.ua (Mercury 1.13) with ESMTP; Thu, 13 Jun 96 22:53:38 +0300 Received: from macon.informatik.uni-tuebingen.de (macon.Informatik. Uni-Tuebingen.De [134.2.12.17]) by spider.cris.crimea.ua (8.6.10/8.6.10.1) with SMTP id WAA25437 for sva@expl. cris.crimea.ua ; Thu, 13 Jun 1996 22:34:53 +0400 Received: from greina.Informatik.Uni-Tuebingen.De by macon. informatik.uni-tuebingen.de (AIX 3.2/UCB 5.64/4.03) id AA22097; Thu, 13 Jun 1996 20:33:54 +0200 Received: by greina.wsi (SMI-8.6/SMI-SVR4) id UAA07066; Thu, 13 Jun 1996 20:33:09 +0200 === Subject: Re: Two problems (Vladimir Bondarenko)] Cc: kuechlin@informatik.uni-tuebingen.de it is ok if you submit a Tex file, as long as we manage to print it. We will try our best, but if we can not print it it will not go into the report. Probably everything will be ok. > The second problem is that I absolutely need a fax or a formal > letter of invitation (with your signature) to get some money from > a local foundation to attend ISSAC. And it takes some (not > necessarily infinitesimal) time. My secretary will send a fax on the morning of Friday June 14. If the letter is not sufficient for you, or if you need additional letters for obtaining a Swiss visa, you may contact the local organizer in Zurich, Michael Kalkbrener, mkalk@math.ethz.ch -wolfgang kuechlin ______________________________________________________________________ Wolfgang Kuechlin University of Tuebingen Associate Professor W. Schickard Institute for Informatics Tel.: (+49) 7071-29.7047 Sand 13 Fax: (+49) 7071-67540 D-72076 Tuebingen, Germany e-mail: kuechlin@informatik.uni-tuebingen.de URL: http://www-sr.informatik.uni-tuebingen.de/ NEW PHONE NUMBER: (+49) 7071-29.7.7047 AS OF 24 JUNE 1996 ______________________________________________________________________ === Subject: Decode this mathematical equation? Hey all, I found this while surfing and was hoping someone could translate this into english. Looks like trig but I haven't taken a math class in quite http://mysite.verizon.net/jsaputo/images/trig.jpg === Subject: Re: Decode this mathematical equation? Another possible interpretation ? Maths is less than love . === Subject: Re: Decode this mathematical equation? > Hey all, ... It's not an equation, it's an inequality. === Subject: Re: Decode this mathematical equation? > Hey all, I found this while surfing and was hoping someone could translate > this into english. Looks like trig but I haven't taken a math class in > http://mysite.verizon.net/jsaputo/images/trig.jpg thats a Laplace transform - nothing to do with trigonometry === Subject: Re: Decode this mathematical equation? What's it mean? What does it have to do with love? >> Hey all, I found this while surfing and was hoping someone could >> translate this into english. Looks like trig but I haven't taken a math >> http://mysite.verizon.net/jsaputo/images/trig.jpg > thats a Laplace transform - nothing to do with trigonometry === Subject: Re: Decode this mathematical equation? > What's it mean? What does it have to do with love? well laplace transforms are typically used to solve certain classes of differential equations, and make some indefinite integrals easier. as what they are to do with love , thats a pretty meaningless question... very little at a rough guess > Hey all, I found this while surfing and was hoping someone could > translate this into english. Looks like trig but I haven't taken a math > http://mysite.verizon.net/jsaputo/images/trig.jpg >> thats a Laplace transform - nothing to do with trigonometry === Subject: Re: Decode this mathematical equation? So someone just felt like writing a laplace transform is less than a heart....? I would venture to say there is a deeper meaning. >> What's it mean? What does it have to do with love? > well laplace transforms are typically used to solve certain classes of > differential equations, and make some indefinite integrals easier. as > what they are to do with love , thats a pretty meaningless question... > very little at a rough guess >> Hey all, I found this while surfing and was hoping someone could >> translate this into english. Looks like trig but I haven't taken a math >> http://mysite.verizon.net/jsaputo/images/trig.jpg > thats a Laplace transform - nothing to do with trigonometry >> === Subject: Re: Decode this mathematical equation? > So someone just felt like writing a laplace transform is less than a > heart....? I would venture to say there is a deeper meaning. Where did you get it from? The context might hint at a meaning. === Subject: Re: Decode this mathematical equation? It was an anonymous message, the only other thing he said was: I'd rather be popular and dumb than lonely and a genius then it was signed with the aforementioned math equation. >> So someone just felt like writing a laplace transform is less than a >> heart....? I would venture to say there is a deeper meaning. > Where did you get it from? The context might hint at a meaning. === Subject: Re: Decode this mathematical equation? > It was an anonymous message, the only other thing he said was: I'd rather > be popular and dumb than lonely and a genius then it was signed with the > aforementioned math equation. at a guess then i'd say its an 18yr old under graduate who has to do some applied maths on whatever course he's taking, and hes having a little difficulty with laplace transforms.... hence his underwhelmed statement that he less than loves them.... > So someone just felt like writing a laplace transform is less than a > heart....? I would venture to say there is a deeper meaning. >> Where did you get it from? The context might hint at a meaning. === Subject: Re: Decode this mathematical equation? > So someone just felt like writing a laplace transform is less than a > heart....? I would venture to say there is a deeper meaning. It says: L of f of t is less than heart. A transformation is short of being love. === Subject: Re: Decode this mathematical equation? > So someone just felt like writing a laplace transform is less than a > heart....? I would venture to say there is a deeper meaning. > It says: > L of f of t is less than heart. > A transformation is short of being love. Thou hast forgot the u. The transform is of the function f(x)u(x) of x. === Subject: Re: Decode this mathematical equation? >> So someone just felt like writing a laplace transform is less than a >> heart....? I would venture to say there is a deeper meaning. >> It says: >> L of f of t is less than heart. >> A transformation is short of being love. > Thou hast forgot the u. The transform is of the function f(x)u(x) of x. Nay, sir. The limits of integration for a real valued Laplace transform are from zero to infinity. The function u(t) turns off the f (makes it zero) when t is negative. === Subject: Re: Decode this mathematical equation? >> So someone just felt like writing a laplace transform is less than a >> heart....? I would venture to say there is a deeper meaning. >> It says: >> L of f of t is less than heart. >> A transformation is short of being love. > Thou hast forgot the u. The transform is of the function f(x)u(x) of x. > Nay, sir. The limits of integration for a real valued Laplace transform > are from zero to infinity. The function u(t) turns off the f (makes it zero) > when t is negative. Oh! I had not realized that u had a specific meaning. It's a step function is it? The one that I know of as H after Heaviside? I thought it was just there so one could read the function x |-> f(x)u(x) as eff you which is less than loving, I suppose. === Subject: Re: Decode this mathematical equation? Interesting. Do L, f, or t have any values to them? Or are they not known? >> So someone just felt like writing a laplace transform is less than a >> heart....? I would venture to say there is a deeper meaning. > It says: > L of f of t is less than heart. > A transformation is short of being love. === Subject: An exact 1-D limit challenge - 16 (Maple only; Mathematica succeeds ;) Is there a person who can calculate using Maple the exact value of the following limit limit( (1 - exp(z/2))/sqrt(exp(2*z) - 1) - sqrt(2)*EllipticF(sqrt(exp(z) + 1), 1/sqrt(2)), z= infinity); ? Best wishes, Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: An exact 1-D limit challenge - 16 (Maple only; Mathematica succeeds ;) I'd say 1/2 1/2 2 2 EllipticK(----) I 2 > Is there a person who can calculate using Maple > the exact value of the following limit > limit( > (1 - exp(z/2))/sqrt(exp(2*z) - 1) - > sqrt(2)*EllipticF(sqrt(exp(z) + 1), 1/sqrt(2)), > z= infinity); > Best wishes, > Vladimir Bondarenko > http://maple.bug-list.org/MapleCrisis-Review-01.pdf > VM and GEMM architect > Co-founder, CEO, Mathematical Director > Cyber Tester, LLC > 13 Dekabristov Str, Simferopol > Crimea 95000, Ukraine > tel: +38-(0652)-447325 > tel: +38-(0652)-230243 > tel: +38-(0652)-523144 > fax: +38-(0652)-510700 > http://www.cybertester.com/ > http://maple.bug-list.org/ > http://www.CAS-testing.org/ === Subject: simple query part 2 sorry about the original simple query it didn't have all the required information. I have a sum, its related to accounting. If you have a pound and you want to find it value in a year's time, you could write a simple equation like so. (1 + 0.10)^4 = (1.1)^4 = 1.4641 where 0.10 is the interest rate (10%), and ^4 the number of years. My problem is that recreating this on a calculator. (1.1)^4, do I times this by 4 or do something else with it. find the percentage? and if it is the latter how do I do this on a calculator, I'm not versed in finding percentages. === Subject: Re: simple query part 2 > sorry about the original simple query it didn't have all the required > information. > I have a sum, its related to accounting. > If you have a pound and you want to find it value in a year's time, you > could write a simple equation like so. > (1 + 0.10)^4 > = (1.1)^4 > = 1.4641 > where 0.10 is the interest rate (10%), and ^4 the number of years. > My problem is that recreating this on a calculator. > (1.1)^4, do I times this by 4 No. Multiply 1.1 by itself 4 times: (1.1)^4 = 1.1 * 1.1 * 1.1 * 1.1 Your calculator may have a key marked x superscript y (i.e. x^y) on it. Key in 1.1, then press that key, then press 4. That is read 1.1 to the power 4. > or do something else with it. find the > percentage? and if it is the latter how do I do this on a calculator, > I'm not versed in finding percentages. I don't understand. In your problem you are _given_ the percentage, you don't have to find it. === Subject: Re: simple query part 2 could write a simple equation like so. > (1 + 0.10)^4 > = (1.1)^4 > = 1.4641 > where 0.10 is the interest rate (10%), and ^4 the number of years. > My problem is that recreating this on a calculator. > (1.1)^4, do I times this by 4 or do something else with it. find the > percentage? and if it is the latter how do I do this on a calculator, On most calculators the function you want is accessed with a key marked x^y -- that is, x with a y superscript. The example above would be keyed 1.1 [x^y] 4 [=] . On a calculator with an automatic constant function the sequence 1.1 [*] 1.1 [=] [=] [=] will give the same result, where [*] represents the multiplication key. The second entry of 1.1 would be unnecessary on some models. Multiplying by 4 is something else altogether: one way or another you need to multiply the number by itself, four times in all. -- Odysseus === Subject: Statistical test of Pick-4 What kind of statistical test might one perform on the last 500 draws of a Pick-4 lottery game to determine if there is a bias? Stig Holmquist === Subject: Re: Statistical test of Pick-4 > What kind of statistical test might one perform on the last 500 > draws of a Pick-4 lottery game to determine if there is a bias? > Stig Holmquist Stig, A few weeks back, a friend at work claimed that when a number hasn't been picked for a while, it is due, so you can increase your odds of winning by playing recently unused numbers. We downloaded about 130 Pick-3 drawing winners from the Maryland lottery and tried to develop a scheme to break the bank. We played around with the numbers, taking under-used numbers for a month and seeing if they popped up in the next week, but didn't find any solid patterns that would guarantee a winner or even give that much of an edge. The historical numbers for each month are at http://www.mdlottery.com/ for both Pick-3 and Pick-4. You can copy the numbers and paste them on Excel, then do a Data - Text To Columns to parse the numbers into columns. Or, if you like, I can email my Excel worksheet to you to save you some effort. I think I did the Pick-4 as well but the data is all on the work computer. Paul === Subject: Re: Statistical test of Pick-4 draws of a Pick-4 lottery game to determine if there is a bias? > Stig Holmquist > Stig, > A few weeks back, a friend at work claimed that when a number hasn't been > picked for a while, it is due, so you can increase your odds of winning by > playing recently unused numbers. We downloaded about 130 Pick-3 drawing > winners from the Maryland lottery and tried to develop a scheme to break the > bank. > We played around with the numbers, taking under-used numbers for a month and > seeing if they popped up in the next week, but didn't find any solid > patterns that would guarantee a winner or even give that much of an edge. Unsurprising. The idea that you can improve your odds in games of pure chance by analysing what's happened previously is known as the gambler's fallacy. It is as old as time, and is completely and utterly wrong. === Subject: Re: Statistical test of Pick-4 >> What kind of statistical test might one perform on the last 500 >> draws of a Pick-4 lottery game to determine if there is a bias? >> Stig Holmquist >> Stig, >> A few weeks back, a friend at work claimed that when a number hasn't been >> picked for a while, it is due, so you can increase your odds of winning >> by >> playing recently unused numbers. We downloaded about 130 Pick-3 drawing >> winners from the Maryland lottery and tried to develop a scheme to break >> the >> bank. >> We played around with the numbers, taking under-used numbers for a month >> and >> seeing if they popped up in the next week, but didn't find any solid >> patterns that would guarantee a winner or even give that much of an edge. > Unsurprising. The idea that you can improve your odds in games of pure > chance by analysing what's happened previously is known as the > gambler's fallacy. It is as old as time, and is completely and > utterly wrong. That's what I told my friend but would he believe me? No. Since things were a bit slow around the shop that morning, we decided to have some fun by downloading over a hundred sequential lottery results and looking for patterns. (Yes, we are number geeks.) What surprised me is that even after five months worth of drawings, two of the ten digits had occurred significantly more often than expected and one had occurred significantly less than expected, the other seven being within a standard deviation of the expected 10%. That is an indication of the long-term nature of a number being due. Paul === Subject: Re: Statistical test of Pick-4 What kind of statistical test might one perform on the last 500 >> draws of a Pick-4 lottery game to determine if there is a bias? >> Stig Holmquist >> Stig, >> A few weeks back, a friend at work claimed that when a number hasn't been >> picked for a while, it is due, so you can increase your odds of winning >> by >> playing recently unused numbers. We downloaded about 130 Pick-3 drawing >> winners from the Maryland lottery and tried to develop a scheme to break >> the >> bank. >> We played around with the numbers, taking under-used numbers for a month >> and >> seeing if they popped up in the next week, but didn't find any solid >> patterns that would guarantee a winner or even give that much of an edge. > Unsurprising. The idea that you can improve your odds in games of pure > chance by analysing what's happened previously is known as the > gambler's fallacy. It is as old as time, and is completely and > utterly wrong. But it's not a fallacy if the machines have true bias, which is less likely now that the lotteries have more experience. The Illinois Lotto machine was biased when it first started in 1984, and I hit a couple times with my statistical analysis (never anything big) before they corrected it. > That's what I told my friend but would he believe me? No. Since things were > a bit slow around the shop that morning, we decided to have some fun by > downloading over a hundred sequential lottery results and looking for > patterns. (Yes, we are number geeks.) > What surprised me is that even after five months worth of drawings, two of > the ten digits had occurred significantly more often than expected and one > had occurred significantly less than expected, the other seven being within > a standard deviation of the expected 10%. That is an indication of the > long-term nature of a number being due. But given there is true bias, the less frequently occuring numbers are NOT due. They are not occuring because of the bias that favors the most frequently occuring numbers. Betting on the numbers that are due is the stupidest thing you can do. Your only chance is the slim hope that maybe the bias shows up in the stats. I notice that in the UK, all the lottery machines and the ball sets are named and published along with the results so that the number geeks can track the bias of specific combinations of machine/ball set. Of course, the bias is likely just be an illusion, but it sells tickets. > Paul === Subject: f=polynomial , f(1)= ? Let f(X)=X^7-14*X^6+63*X^5-A*X^4+B*X^3-C*X^2+D*X-E . 1) Prove that there exists such a polynomial having only real roots. Further suppose that f(X) has only real roots, namely x_1 =< x_2=< ...=< x_7 . 2) If x_n - x_1 =< 7*sqrt(2)/2 , find f(1) . It's true that f(1) is in the set {- 245.75+236.5*sqrt(2), 245.75+236.5*sqrt(2)} ? === Hello the world, Before me, on the left, a monography is placed. Yu. A. Shevljakov, Matrix algorithms in the theory of elasticity of heterogeneous media, 1977, Kiev-Odessa, 215 pp Prof. Dr. Shevljakov's monography treats a string of theoretical and practical issues in building mechanics. Unlike its main competitor, Wolfram Research, Inc, since at Waterloo Maple/Maplesoft was NEVER able to tackle a wide range of the integrals occurring in this book. Below come just the simplest examples, *partial* cases of the original integrals which involve up to 3 parametres and where the Maplesoft's slogan, Command the brilliance of thousand mathematicians, even more sounds like rave cries of a bombed teenager at the disco hall. TEST CASE: int(BesselK(0,z)/sqrt(z^2-1), z= 1..infinity); EXPECTED: 1/2*BesselK(0,1/2)^2 .4272753096 CHECKUP: evalf(Int(BesselK(0,z)/sqrt(z^2-1),z=1..infinity)); .4272753096 int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (2002) Maple 8 ---------------------------- int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (2001) Maple 7 ---------------------------- int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (2000) Maple 6 ---------------------------- int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (1997) Maple V Rel 5 ---------------------- int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (1995) Maple V Rel 4 ---------------------- int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (1994) Maple V Rel 3 ---------------------- int(BesselK(0,z)/(z^2-1)^(1/2),z = 1 .. infinity) ------------------------------------------------------------------ Now ask the same Mathematica 5.1 and get an instant answer. Integrate[BesselK[0, z]/Sqrt[z^2 - 1], {z, 1, Infinity}] BesselK[0, 1/2]^2/2 0.427275 TEST CASE: int(z^2*BesselK(1,z)/sqrt(z^2-1), z= 1..infinity); EXPECTED: Pi/exp(1) 1.155727350 CHECKUP: evalf(Int(z^2*BesselK(1,z)/sqrt(z^2-1), z= 1..infinity)); 1.155727350 int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (2002) Maple 8 --------------------------- int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (2001) Maple 7 --------------------------- int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (2000) Maple 6 --------------------------- int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (1997) Maple V Rel 5 --------------------- int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (1995) Maple V Rel 4 --------------------- int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) --------------------- (1994) Maple V Rel 3 --------------------- int(z^2*BesselK(1,z)/(z^2-1)^(1/2),z = 1 .. infinity) ----------------------------------------------------------------- Again, ask the same Mathematica 5.1 and get an instant answer. Integrate[z^2 BesselK[1, z]/Sqrt[z^2 - 1], {z, 1, Infinity}] Pi/E The same Maple bug manifestation (NOT a weakness, a term so much adored by some Maple experts, because it is not possible, over at least 11 years, NOT to see for the Maplesoft's owner/top managers that Maple fails miserably here) with int(z*BesselK(0,z)/sqrt(z^2-1), z= 1..infinity); int(BesselK(1,z)/sqrt(z^2-1), z= 1..infinity); int(z*BesselK(1,z)/sqrt(z^2-1), z= 1..infinity); int(z^2*BesselK(1,z)/sqrt(z^2-1), z= 1..infinity); int(BesselK(2,z)/sqrt(z^2-1), z= 1..infinity); int(z*BesselK(2,z)/sqrt(z^2-1), z= 1..infinity); int(z^2*BesselK(2,z)/sqrt(z^2-1), z= 1..infinity); int(BesselK(3,z)/sqrt(z^2-1), z= 1..infinity); int(z*BesselK(3,z)/sqrt(z^2-1), z= 1..infinity); int(z^2*BesselK(3,z)/sqrt(z^2-1), z= 1..infinity); int(BesselY(0,z)/sqrt(z^2-1), z= 1..infinity); int(BesselY(1,z)/sqrt(z^2-1), z= 1..infinity); int(BesselY(2,z)/sqrt(z^2-1), z= 1..infinity); int(z*BesselY(0,z)/sqrt(z^2-1), z= 1..infinity); int(z*BesselY(1,z)/sqrt(z^2-1), z= 1..infinity); int(z*BesselY(2,z)/sqrt(z^2-1), z= 1..infinity); int(z^2*BesselY(0,z)/sqrt(z^2-1), z= 1..infinity); int(z^2*BesselY(1,z)/sqrt(z^2-1), z= 1..infinity); int(z^2*BesselY(2,z)/sqrt(z^2-1), z= 1..infinity); Any comments on Maple 10 behaviour? Best wishes, Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Originator: richard@cogsci.ed.ac.uk (Richard Tobin) [deleted] Sorry, we don't care. Take your private feud somewhere else. And even if we might care, this isn't the way to make us. -- Richard === rich...@cogsci.ed.ac.uk (Richard Tobin) RT> private feud To start with, if an issue involves 3,000,000 Maple customers all over the world, it is amazing to hear that it can be, at any rate, interpreted as private feud. Next, let me please pass the word to one of our supporters: === Subject: Re: Maple bugs: Backward compatibility - 1 F> I find Vladimir posts the most interesting and educational F> posts. F> By sharing with us his findings, we get to learn more. F> If you do not find them interesting for you, then no one F> is forcing you to read them. warm wording!) Also, the stuff Cyber Tester posts is about a new approach to CAS development, about making CASs, so to say, more intelligent, faster, and - in the long run - hopefully - less expensive. Last but not least, all this is also a point of fighting for regular customer's rights (our own included) to enjoy nice CAS for our money. Also, please do not name yourself we. You are not the King/an Omnipotent Ruler of sci.math.symbolic etc These forums are read by many many thousands persons, and up to now about a dozen only made some fuss about writing/feud/not writing - among them 80% are just trivial slanderous persons. I hope that you have absolutely nothing to do with them. Best wishes, Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ................................................................ === elasticity cot be calculated - Family 1 > Unlike its main competitor, Wolfram Research, Inc, since at > Waterloo Maple/Maplesoft was NEVER able to tackle a wide range > of the integrals occurring in this book. Maple project started in november 1980. Stephen Wolfram first worked on SMP, then created Wolfram Research in 1987 to develop Mathematica. Dunno where you found 1994. === Jean-Claude Arbaut JCA> where you found 1994. Actually, of cause I realize what is the date of the Maple project launch as I am a Maple fan since 1993 :) to the Maple versions we acquired or have indirect access to via our helpers. These versions are as follows 2002 Maple 8 2001 Maple 7 2000 Maple 6 1997 Maple V Release 5 1995 Maple V Release 4 1994 Maple V Release 3 Sure, the same bug might be present in earlier versions to which we unfortunately this moment have no access and even an idea where to find a person who could test our results using these versions. (If, by lucky strike, you think you could be such a person, please do not hesitate to contact us at v b @ cybertester . com - the 5 spaces should be deleted, they are inserted to fool the Google email-hiding stuff) Please also note that over 1-2-3 successive versions maybe the described should NOT be interpreted as a bug but rather as a mentioned earlier fav Maple expert's sweet, weakness ;) But over a decade, and with the main competitor cracking it instantly, it surely turns into a bug. Actually, one of novelties which Cyber Tester has introduced is careful, diligent, painstaking research of bug manifestations evolution. This is important, because it can give you an idea of how these (at least) dozens thousands Maple bugs behave themselves. This is one of keys to failure prediction analysis, to a streamlined development process, to satisfied customers. === Subject: Try your luck at the Maple roulette! A sample of 1000s Hello Maple 10 potential customers, Why don't try your luck at the roulette named Maple? Really, why don't put your success at stake via making your career depending on chances? If you would lose, Maplesoft loses nothing: they already collected your money, and there is no refund. Below are, detected by the VM machine, just a couple of similar examples out of thousands distinct ones of Maple growing instability calculated automatically. You may wish to find more in the beta 0.1 of the first world's Maple review written in a close cooperation between a human being and a computational environment http://maple.bug-list.org/MapleCrisis-Review-01.pdf Please note the 3rd type of answer, Error, (in property/AndProp/+) too many levels of recursion in Maple 9. Pattern: 2 distinct outcomes at random 2 distinct outcomes at random 3 distinct outcomes at random -------------------- (2002) Maple 8 -------------------------- mathematically, always the same -------------------- (2001) Maple 7 -------------------------- 1 outcome, always -------------------- (2000) Maple 6 -------------------------- mathematically, always the same -------------------- (1997) Maple V Rel 5 -------------------- 1 outcome, always -------------------- (1995) Maple V Rel 4 -------------------- 1 outcome, always -------------------- (1994) Maple V Rel 3 -------------------- 1 outcome, always (unevaluated integral) Details: restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); # 2 distinct outcomes at random 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) # 2 distinct outcomes at random int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) # 3 distinct outcomes at random ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) Error, (in property/AndProp/+) too many levels of recursion int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) -------------------- (2002) Maple 8 -------------------------- # mathematically, always the same 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) -------------------- (2001) Maple 7 -------------------------- # 1 outcome, always 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) -------------------- (2000) Maple 6 -------------------------- # mathematically, always the same -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) -------------------- (1997) Maple V Rel 5 -------------------- # 1 outcome, always 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) -------------------- (1995) Maple V Rel 4 -------------------- # 1 outcome, always 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) 3^(1/2)-2*Re(arctanh(3^(1/2))) -------------------- (1994) Maple V Rel 3 -------------------- # 1 outcome, always (unevaluated integral) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) --------------------------------------------------------------- restart; f := abs(z)+2: int(1/f^2/ln(f), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) -------------------- (2002) Maple 8 -------------------------- int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) -------------------- (2001) Maple 7 -------------------------- int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) -------------------- (2000) Maple 6 -------------------------- int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) -------------------- (1997) Maple V Rel 5 -------------------- 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) -------------------- (1995) Maple V Rel 4 -------------------- 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) 2*Ei(1,ln(2)) -------------------- (1994) Maple V Rel 3 -------------------- int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) --------------------------------------------------------------- Best wishes, Vladimir Bondarenko http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s Sirs, really for you it is still not clear, that mr. Bondarenko simply the loser and the nervously sick person? === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s Mr. Bondarenko, in my opinion, your strange posts (politely saying) and behaviour seriously disturb normal work of the forum and other forums. You act in a role of spammer discreditable the Maple products. Meanwhile, in spite of all shortcomings, Maple, in my opinion, is the best CAS for today. Patience at many people at the breaking point. I recommend you to create the forum or on your odious site, or on Google. There you can debate on scientific themes interesting for you with yourselves and with other people similar to you. === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s > Mr. Bondarenko, in my opinion, your strange posts (politely saying) and > behaviour seriously disturb normal work of the forum and other forums. > You act in a role of spammer discreditable the Maple products. > Meanwhile, in spite of all shortcomings, Maple, in my opinion, is the > best CAS for today. Patience at many people at the breaking point. I > recommend you to create the forum or on your odious site, or on Google. > There you can debate on scientific themes interesting for you with > yourselves and with other people similar to you. The problem is, he would be alone ;-) === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s It is very reasonable advice! === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s > Hello Maple 10 potential customers, > Why don't try your luck at the roulette named Maple? +------------------+ /| /| | | || || | Please do not | / O O | feed the Troll | / | | / ------------------+ / | || / | | | | / || / | | |/ | || / / | | || / | | | --| | | | | --| * | | | | | | -/ *-- -- | || / | / `´ * / /- | | | * c c c C/ C c c c -- the shadow === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s Maybe you are seeing that your attempts to create some vacuum around Maple bugs seem do not bring you success? A piece of hit. A cardinal question here: Is VB really a troll? And if you think IS, what exactly makes you thinking so. What are the necessary features of a troll? Do I really have all (or at least 1 of them)? (This troll statement is not true whatsoever, but I will come back to this later) === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s > Is VB really a troll? > And if you think IS, what exactly makes you thinking so. > What are the necessary features of a troll? > Do I really have all (or at least 1 of them)? +------------------+ /| /| | | ||__|| | Please do not | / O O__ | feed the Troll | / | | / _ ------------------+ / |____ || / | | | |____/ || / |_|_|/ | _|| / / |____| || / | | | --| | | | |____ --| * _ | |_|_|_| | -/ *-- _-- _ | || / _ | / `´ * / _ /- | | | * ___ c_c_c_C/ C_c_c_c____________ === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s Brad Cooper === Subject: Re: Maple bugs: Backward compatibility - 1 BK> I find it wonderful that Vladimir is persuing excellence BK> in CAS for the user community. I find it shocking that BK> there appears to be so little regression testing of Maple. BK> Alternatively, if there is regression testing, then it is BK> just as shocking that it is so ineffective. BK> When, a few weeks ago, Vladimir toasted the birth of the BK> first reported bug in Maple 15 years ago and then showed BK> that it is STILL THERE I was shocked. BK> How can that be? BK> I use a CAS almost daily in my work and it is a fabulous BK> tool. With Vladimir beavering away at the task of CAS BK> testing, these systems will ultimately be better and we BK> all benefit. Vladimir, I salute you. BK> Brad === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s Just a question, as I'm rather naive: why not look for bugs or problems in other CAS ? Like Mathematica, MuPAD, Maxima, etc... The guys at Maplesoft refused a job to you ? It has become rather obvious it's a personnal problem, so why would we bother reading all your posts ? > Hello Maple 10 potential customers, > Why don't try your luck at the roulette named Maple? > Really, why don't put your success at stake via making > your career depending on chances? If you would lose, > Maplesoft loses nothing: they already collected your > money, and there is no refund. > Below are, detected by the VM machine, just a couple > of similar examples out of thousands distinct ones > of Maple growing instability calculated automatically. > You may wish to find more in the beta 0.1 of the first > world's Maple review written in a close cooperation > between a human being and a computational environment > http://maple.bug-list.org/MapleCrisis-Review-01.pdf > Please note the 3rd type of answer, > Error, (in property/AndProp/+) too many levels of recursion > in Maple 9. > Pattern: > 2 distinct outcomes at random > 2 distinct outcomes at random > 3 distinct outcomes at random > -------------------- (2002) Maple 8 -------------------------- > mathematically, always the same > -------------------- (2001) Maple 7 -------------------------- > 1 outcome, always > -------------------- (2000) Maple 6 -------------------------- > mathematically, always the same > -------------------- (1997) Maple V Rel 5 -------------------- > 1 outcome, always > -------------------- (1995) Maple V Rel 4 -------------------- > 1 outcome, always > -------------------- (1994) Maple V Rel 3 -------------------- > 1 outcome, always (unevaluated integral) > Details: > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > restart; int(1/((abs(z)+2)^2*(sqrt(abs(z)+3))), z=-infinity..infinity); > # 2 distinct outcomes at random > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > # 2 distinct outcomes at random > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > # 3 distinct outcomes at random > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > Error, (in property/AndProp/+) too many levels of recursion > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > -------------------- (2002) Maple 8 -------------------------- > # mathematically, always the same > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) > -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > 3^(1/2)+ln(3^(1/2)-1)-ln(1+3^(1/2)) > -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) > -------------------- (2001) Maple 7 -------------------------- > # 1 outcome, always > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > -------------------- (2000) Maple 6 -------------------------- > # mathematically, always the same > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > ln(3^(1/2)-1)-ln(1+3^(1/2))+3^(1/2) > 3^(1/2)-ln(1+3^(1/2))+ln(3^(1/2)-1) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > -ln(1+3^(1/2))+ln(3^(1/2)-1)+3^(1/2) > ln(3^(1/2)-1)+3^(1/2)-ln(1+3^(1/2)) > -ln(1+3^(1/2))+3^(1/2)+ln(3^(1/2)-1) > -------------------- (1997) Maple V Rel 5 -------------------- > # 1 outcome, always > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > -------------------- (1995) Maple V Rel 4 -------------------- > # 1 outcome, always > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > 3^(1/2)-2*Re(arctanh(3^(1/2))) > -------------------- (1994) Maple V Rel 3 -------------------- > # 1 outcome, always (unevaluated integral) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/(abs(z)+3)^(1/2),z = -infinity .. infinity) > --------------------------------------------------------------- > restart; f := abs(z)+2: int(1/f^2/ln(f), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > restart; f := abs(z)+2: int(1/(f^2*ln(f)), z= -infinity..infinity); > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > -------------------- (2002) Maple 8 -------------------------- > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > -------------------- (2001) Maple 7 -------------------------- > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > -------------------- (2000) Maple 6 -------------------------- > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > -------------------- (1997) Maple V Rel 5 -------------------- > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > -------------------- (1995) Maple V Rel 4 -------------------- > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > 2*Ei(1,ln(2)) > -------------------- (1994) Maple V Rel 3 -------------------- > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > int(1/(abs(z)+2)^2/ln(abs(z)+2),z = -infinity .. infinity) > --------------------------------------------------------------- > Best wishes, > Vladimir Bondarenko > http://maple.bug-list.org/MapleCrisis-Review-01.pdf > VM and GEMM architect > Co-founder, CEO, Mathematical Director > Cyber Tester, LLC > 13 Dekabristov Str, Simferopol > Crimea 95000, Ukraine > tel: +38-(0652)-447325 > tel: +38-(0652)-230243 > tel: +38-(0652)-523144 > fax: +38-(0652)-510700 > http://www.cybertester.com/ > http://maple.bug-list.org/ > http://www.CAS-testing.org/ === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s Jean-Claude Arbaut JCA> Just a question, as I'm rather naive: You are a novice in this group and just did not see all the case. JCA> why not look for bugs or problems in other CAS ? 1) Over years, I already reported publicly numerous bugs in Mathematica, MuPAD, Derive, AXIOM... just search over Invernet... for example, a *part* of bugs in Mathematica http://forums.wolfram.com/mathgroup/search/?q=Vladimir+Bondarenko&restrict=M athGroup&x=12&y=11 Next, I have reported more then 500 MuPAD bugs here, go and count for vvb@mail.strace.net http://www.mupad.de/bugs.shtml Stefan Wehmeier, Feb 24, 2005: SW> Actually, Vladimir has contributed many bugs to SW> www.mupad.de/BUGS SW> and even got an award for being our best beta tester. Really. Just look here http://www.cas-testing.org/SciFace.phtml 2) I have 24 hours a day, only. 3) I have 1 head and 2 hands only. Unfortunately, as of today, I cot type in with the toes ;-) 4) Please have a look here at the list of CASs I already helped to the manufacturers to improve http://www.cas-testing.org/index.php?list=3 5) Why don't consider reading opinions about quality & scope of my work produced by experts in the field of computer algebra systems, CAS manufacturers and top CAS folks since 1993. Find a few selected of selected opinions from them placed here http://www.cas-testing.org/index.php?list=7 ? which belong, respectively, to COMMENT 1 Albert D. Rich http://www.derive.com/ COMMENT 2 Prof Dr Oleg Marichev (WRI) www.functions.wolfram.com COMMENT 3 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 4 Prof Dr Walter Oevel (SciFace GmbH) http://www.math.uni-paderborn.de/~walter/ COMMENT 5 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 6 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 7 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 8 Dr Michael Wester www.math.unm.edu/~wester COMMENT 9 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 10 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 11 Stefan Wehmeier (SciFace GmbH) http://www-math.uni-paderborn.de/~stefanw/ COMMENT 12 Dr Anwar Shiekh (WRI) Test Development Group Supervisor COMMENT 13 Kelly Roach http://www.kellyroach.com/ COMMENT 14 Dr Anwar Shiekh (WRI) Test Development Group Supervisor 6) As of today, Maple is the buggiest system. MOST non-ergonomic CAS. So it is necessary to first handle the biggest trouble before going to easier ones. 7) If you want us to identify more problems in other systems, this is easy, just send us money so we can acquire more computers to test all these CAS automatically - simultaneously. 8) Or write to Maplesoft, let them send us money :) JCA> Like Mathematica, MuPAD, Maxima, etc... See the above. JCA> The guys at Maplesoft refused a job to you ? The guys at Maplesoft paid you to ask me this? JCA> It has become rather obvious it's a personnal problem, It has become rather obvious you have no software testing experience. Otherwise, you would understand that Maplesoft is going to go bunkrupt within a visible span of time if they would be unable to fight their competitor, Mathematica. Now they keep sleeping and I am trying to kick them of out this dangerous hibernation and ostrich policy. JCA> so why would we bother reading all your posts ? Who we, on earth? You are NOT the King here. NOT a ruler. NOT my superior (ha-ha, I am clever, I have no superiors now, because I am the principal owner of Cyber Tester ;-) If you do not like reading my posts, don't victimize yourself, and just ignore them. === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s > JCA> It has become rather obvious it's a personnal problem, > It has become rather obvious you have no software testing > experience. In my software testing experience, I have never trolled a newsgroup. It's probably a gap. > Otherwise, you would understand that Maplesoft is going to > go bunkrupt within a visible span of time if they would be > unable to fight their competitor, Mathematica. > Now they keep sleeping and I am trying to kick them of out > this dangerous hibernation and ostrich policy. No, you are trying to prevent people from buying Maple, which is not your job, unless you are paid for that. > JCA> so why would we bother reading all your posts ? > Who we, on earth? When several people are involved, they use we, it's a tradition. > If you do not like reading my posts, don't victimize yourself, > and just ignore them. Difficult to forget they are in my box. === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s JCA> No, you are trying to prevent people from buying Maple, Why SURE, AND this too! Over 2 last years I have wasted LOTS of time fighting with Maple. For all who read this: ***** * Think twice before buying Maple 10 * ***** Ask yourself: If 3 companies propose you a trial version to test over weeks, and only Maplesoft - not - WHY this? Why Maplesoft does not give you access to Maple 10 trial version? At this, nobody cot say that Mathematica 5.1 is essentially weaker - in any sense - than Maple - but all the same Wolfram Research gives you the chance to make YOUR decision. Try for free - and buy. Try for free - and don't buy. Whatever is your impression of the tried product. JCA> which is not your job, Oh you seem to feel to be my superior? If so, this is completely false, however. JCA> unless you are paid for that. Woow! Oh so you are like a unique live X-ray machine!? Why you don't report about your phenomenal, extraordinaire vision qualities into the Guinness Record Book? You will collect the world's fame soon. JCA> unless you are paid for that. Then I only can surmise that you are paid too. Because you pick with me already several times, within these days. JCA> When several people are involved Several slanderous persons, you mean. Who claimed that I have no refereed publications, post only non-moderate groups http://forums.wolfram.com/mathgroup/search/?q=Vladimir+Bondarenko&restrict=M athGroup&x=12&y=11 etc I feel sorry that by saying we you seem to going to associate yourself with those persons? But hopefully I am wrong here. JCA> Difficult to forget they are in my box. Filter then. The best, write to Maplesoft than they improve Maple ;) (1) if they want to do this and 2) if they CAN do this. Then, hopefully, I will have no incentive to post the Maple bugs publicly whatsoever. === Subject: Re: Try your luck at the Maple roulette! A sample of 1000s 1] Indeterminate You may wish to get a fully functional, 15-day, save-disabled trial version of Mathematica 5.1.1 at http://www.wolfram.com/products/mathematica/trial.cgi .................................................................. === Subject: Re: The Pentium bug - 1 - The puissance of the customers community ... > As a result, Intel gave up in December 1994 and offered to > replace all faulty Pentium chips and had to stop the complete > production of that particular chip, even though it was right > that only very few were affected. All chips were affected, but not all people were affected. It was a bug in the mask used for the Pentium, which was corrected fast enough in the succeeding batches. The production of that particular chip was not stopped, the production of that particular chip with that particular mask was stopped. (If I remember right there were two bits wrong in the constant table used by the division operation, that could lead to wrong decisions, and so to wrong results.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The Pentium bug - 1 - The puissance of the customers community > ... > As a result, Intel gave up in December 1994 and offered to > replace all faulty Pentium chips and had to stop the complete > production of that particular chip, even though it was right > that only very few were affected. > All chips were affected, but not all people were affected. It was > a bug in the mask used for the Pentium, which was corrected fast > enough in the succeeding batches. The production of that particular > chip was not stopped, the production of that particular chip with > that particular mask was stopped. (If I remember right there were > two bits wrong in the constant table used by the division operation, > that could lead to wrong decisions, and so to wrong results.) According to Cleve Moler, there were five missing entries in a lookup table. Take a look at the bottom of the second column of page 2 of A Tale of Two http://www.mathworks.com/company/newsletters/news_notes/clevescorner/ Cleve also put a collection of documents about this bug up here: http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=166 6&objectType=file# -- Steve Lord slord@mathworks.com === Subject: Re: The Pentium bug - 1 - The puissance of the customers community > ... I'd like to congratulate you on your use of the word puissance. === Subject: Re: The Pentium bug - 1 - The puissance of the customers community Well, English is not my mother tongue but what I have just found http://www.cogsci.princeton.edu/cgi-bin/webwn2.1?s=puissance puissance: power to influence or coerce the puissance of the labor vote Well, there is also such a definition :-) http://www.horsescanada.com/glossary.htm puissance: a non-timed jumping competition where the fences are fewer but higher until only one combination of horse and rider succeed in completing the course === Subject: Re: The Pentium bug - 1 - The puissance of the customers community How many Pentuim engineers does it take to change a light bulb? Three: one to hold the ladder and one to turn the bulb. === Subject: Re: The Pentium bug - 1 - The puissance of the customers community > How many Pentuim engineers does it take to change a light bulb? Three: > one to hold the ladder and one to turn the bulb. Intel: Quality is Job 0.999986 -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: The Pentium bug - 1 - The puissance of the customers community > How many Pentuim engineers does it take to change a light bulb? Three: > one to hold the ladder and one to turn the bulb. Shouldn't that be: one to hold the bulb and one to turn the ladder? === === Subject: A Proposal to mr. Bondarenko Mr. Bondarenko, in my opinion, your strange posts (politely saying) and behaviour seriously disturb normal work of the forum and other forums. You act in a role of spammer discreditable the Maple products. Meanwhile, in spite of all shortcomings, Maple, in my opinion, is the best CAS for today. Patience at many people at the breaking point. I recommend you to create the forum or on your odious site, or on Google. There you can debate on scientific themes interesting for you with yourselves and with other people similar to you. === Subject: Re: A Proposal to mr. Bondarenko > Mr. Bondarenko, in my opinion, your strange posts (politely saying) and > behaviour seriously disturb normal work of the forum and other forums. > You act in a role of spammer discreditable the Maple products. > Meanwhile, in spite of all shortcomings, Maple, in my opinion, is the > best CAS for today. Patience at many people at the breaking point. I > recommend you to create the forum or on your odious site, or on Google. > There you can debate on scientific themes interesting for you with > yourselves and with other people similar to you. And what about George Ghiata ? I am quite new to sci.math, but it seems to me that this guy is not very serious... (or should not be taken seriously ;-)). === Subject: Re: A Proposal to mr. Bondarenko they have something in common: they are deeply entrenched in the delusion of their self-greatness. === Subject: Re: A Proposal to mr. Bondarenko I completely agree with the given offer and the presented opinions on such visitors of forums. Only at such forums they can show one's worth. V.V. === Subject: Re: A Proposal to mr. Bondarenko Hi all V.B. dispraisers, sth like that? This becomes more oying than V.B.'s posts. -- Peter Pein Berlin === Subject: Re: A Proposal to mr. Bondarenko It is very reasonable advice! === Subject: q-diophantic equations ? For q real, q =/= 1 , let us denote [x]:=[x]_q=(q^x - 1)/(q-1}). It may be observed that lim_{q --->1}[x]_q = x . These q-numbers arise in Quantum Physics, q-Calculus or ,,Quantum Calculus. By a ,, q-integer let us mean numbers as ..., [-3] , [-2],[-1], [0] , [1] ,[2],[3],...,[n],.... ; . == i) There are studied in q-integers different equations, like Pell-equation ? ii) Does there exists q , q>0 , q=/=1 , for which equation [b]^{[a]}= [a]^{[b]} + 1 , ( q-Cat equation ??) , has solutions in q-integers ? === Subject: OT: spam question for math fans I first posted an on-topic (math-related) question to this newsgroup in April. Since that time I've been inundated with email spam urging me to buy any number of different stocks. Before my first post to this newsgroup I had never received a single spam about the stock market. So -- you intellectual giants -- what conclusions can we draw from these data? Making the *strong* assumption that spammers do their homework before actually spamming a particular newsgroup, I conclude that the readers of this group *are* indeed interested in the stock market. Is my assumption too strong? As an aside (for those of you who might be interested in markets) I noted that I saw a huge number of emails in April from spammers urgting me to buy oil-related stocks just as crude oil prices peaked in early April. I've seen *no* such spams this week as crude oil prices exceeded the April top. === Subject: Re: OT: spam question for math fans >> Is my assumption too strong? << Probably. It is just a matter of what spammer harvested what set of email addresses one week. I get porno website ads in Chinese and mail-order drug offers. I have also figured out why Africa is so poor. If all those emails I get from Nigeria are true, the total amount of money in unclaimed bank accounts is almost twice the Gross Domestic Product of all of sub-Sharian Africa. === Subject: Re: spam question for math fans >I first posted an on-topic (math-related) question to this newsgroup > in April. Since that time I've been inundated with email spam urging > me to buy any number of different stocks. > Before my first post to this newsgroup I had never received a single > spam about the stock market. > So -- you intellectual giants -- what conclusions can we draw from these > data? Try munging your address a bit more. Making it more difficult of the bots to pick the address up will help reduce the spam levels. Do not attempt to use something like nobody@noname.com, as that is already pretty well flagged as a troll address. Michael === Subject: Re: spam question for math fans > Try munging your address a bit more. Making it more difficult of the bots > to pick the address up will help reduce the spam levels. > Do not attempt to use something like nobody@noname.com, as that is already > pretty well flagged as a troll address. > Michael What exactly is the problem with nobody@noname.com? If this stops spambots, then that's fine, no? I know you shouldn't use an address which acutally exists, because someone, somewhere (ISP?) may suffer the consequence. That's exactly why I took the domain of a spammer and put it as my email address. I did try to contact these spammers before using their domain, but never got a reply (I wonder why...) === Subject: Re: spam question for math fans to pick the address up will help reduce the spam levels. > Do not attempt to use something like nobody@noname.com, as that is already > pretty well flagged as a troll address. > Michael > What exactly is the problem with nobody@noname.com? If this stops > spambots, then that's fine, no? > I know you shouldn't use an address which acutally exists, because > someone, somewhere (ISP?) may suffer the consequence. That's exactly why > I took the domain of a spammer and put it as my email address. > I did try to contact these spammers before using their domain, but never > got a reply (I wonder why...) Unfortunately if you use Google Groups you HAVE to give a real address in order to register (they send you a magic link in an email which activates your account). At least, that's the way it seemed when I registered a while ago... unless anyone knows any different. It's SHOCKING that they give no warning on the registration screen about the abuse that the email address will suffer. === Subject: Re: spam question for math fans : >>I first posted an on-topic (math-related) question to this newsgroup >> in April. Since that time I've been inundated with email spam urging >> me to buy any number of different stocks. >> Before my first post to this newsgroup I had never received a single >> spam about the stock market. >> So -- you intellectual giants -- what conclusions can we draw from these >> data? > Try munging your address a bit more. Making it more difficult of the bots > to pick the address up will help reduce the spam levels. Making the unmunging instructions too clear is not to be recommended - spambots can understand replace foo with bar to reply and do just that. See my example below, which (so far) no spambot has been able to figure out. -- A couple of questions. How do I stop the wires short-circuiting, and what's this nylon washer for? Interchange the alphabetic letter groups to reply === Subject: Re: spam question for math fans > Making the unmunging instructions too clear is not to be recommended - > spambots can understand replace foo with bar to reply and do just that. > See my example below, which (so far) no spambot has been able to figure > out. Because this entire thread is off-topic, I intended to send you an email instead of replying to this newsgroup -- but I was unable to decode your instructions :o) I tried various combinations/permutations of abc and zyx, but ultimately I was defeated. Has anyone ever sent you an email by 'interchanging alphabetic letter-groups' ? === Subject: Re: spam question for math fans >I first posted an on-topic (math-related) question to this newsgroup > in April. Since that time I've been inundated with email spam urging > me to buy any number of different stocks. > Before my first post to this newsgroup I had never received a single > spam about the stock market. > So -- you intellectual giants -- what conclusions can we draw from these > data? > Making the *strong* assumption that spammers do their homework before > actually spamming a particular newsgroup, I conclude that the readers > of this group *are* indeed interested in the stock market. > Is my assumption too strong? > As an aside (for those of you who might be interested in markets) I > noted that I saw a huge number of emails in April from spammers urgting > me to buy oil-related stocks just as crude oil prices peaked in early > April. I've seen *no* such spams this week as crude oil prices exceeded > the April top. === Subject: Re: OT: spam question for math fans > I first posted an on-topic (math-related) question to this newsgroup > in April. Since that time I've been inundated with email spam urging > me to buy any number of different stocks. > Before my first post to this newsgroup I had never received a single > spam about the stock market. > So -- you intellectual giants -- what conclusions can we draw from these > data? > Making the *strong* assumption that spammers do their homework before > actually spamming a particular newsgroup, I conclude that the readers > of this group *are* indeed interested in the stock market. > Is my assumption too strong? > As an aside (for those of you who might be interested in markets) I > noted that I saw a huge number of emails in April from spammers urgting > me to buy oil-related stocks just as crude oil prices peaked in early > April. I've seen *no* such spams this week as crude oil prices exceeded > the April top. Speaking, if I may, as an intellectual giant, I think that any unisguised email address posted to a newsgroup is immediately bombarded with crap. I think mostly the addresses are harvested by automated programs, which then send the same mail to all addresses without too much human intervention or intelligence applied. The solution is to either disguise your email address, or (like me) use a disposable address purely for newsgroups so that all the spam is separated from your proper email. I often post to math(s) newsgroups but I must say I have never noticed a bias towards math-related spam. For me it's just the usual urgent security patch from Microsoft, problems with your eBay/PayPal account, congratulations, you have won the Nigerian lottery, legal matter requiring your immediate attention etc. etc. etc. === Subject: Re: OT: spam question for math fans >I first posted an on-topic (math-related) question to this newsgroup >in April. Since that time I've been inundated with email spam urging >me to buy any number of different stocks. >Before my first post to this newsgroup I had never received a single >spam about the stock market. >So -- you intellectual giants -- what conclusions can we draw from these >data? My guess is that you made the mistake of putting your correct email address in the header. If wa1ter@myrealbox.com is your actual email address, that might explain it. Otherwise, who knows? Many usenet posters try to spam-proof their posts. An example is my email address in this message. --Lynn === Subject: Re: OT: spam question for math fans > ... I conclude that the readers > of this group *are* indeed interested in the stock market. I am a reader of this newsgroup. I am not interested in the stock market. (Is there is only one?) -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: OT: spam question for math fans > Making the *strong* assumption that spammers do their homework before > actually spamming a particular newsgroup, I conclude that the readers > of this group *are* indeed interested in the stock market. > Is my assumption too strong? Yes, by far. - Tim === Subject: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? For the third time I must attract your attention to the fact that some Jerzy Karczmarczuk http://www.info.unicaen.fr/~karczma/ of l'Universite' de Caen Basse-Normandie, a complete ignoramus in Software Testing field, who has thrown in packs variegated insults to a person he even NEVER met before these weeks, me, still continues defaming your organization via sending - wilfully, by design - even more insults as well as other completely false information about me. Some *selected* examples only: a clinical idiot Bondarenko visibly suffers from a heavy mental disorder needing a competent psychiatrist he will be compelled by somebody to undergo a medical treatment a person far from mental equilibrium And I used such words, or even stronger not capable of understanding the simplest logic despite lack of *ANY* refereed publications Bondarenko investigations Beware, everybody. *Stalinist denunciations* V. Bondarenko continues to spread his hatred of any critics Perhaps you believe also in this team of programmers under the rulership of Mr. Bondarenko? I am afraid we must ask some Mudjaheddin for rescue. sent two denunciatory letters [<- I think this is a gem! - VB] ^^^^^^^^^^^^ Please find the details below. sci.math.symbolic, comp.soft-sys.math.maple JK> Mr. Bondarenko was busy answering two other people on the JK> subject of his expertise, despite lack of *ANY* refereed JK> publications Here are details about my refereed publications accepted at the yearly premier international symposium, ISSAC, the International Symposium on Symbolic and Algebraic Computation: JK> I tried to discourage people from discussing with somebody JK> so far from mental equilibrium. And I used such words, or JK> even stronger. No comments. In other words, a worker of your Computer Science department, l'Universite' de Caen Basse-Normandie, France, your worker, Jerzy Karczmarczuk PUBLICLY confirmed about himself that he IS a slanderer. AND EVEN THIS IS NOT ALL! JK> Mr. Bondarenko sent two denunciatory letters to the ^^^^^^^^^^^^ (! - VB) A WILFUL SLANDERER, Jerzy Karczmarczuk, COMPLAINS THAT THE PERSON ABOUT WHOM HE THROWS LIBELS - REPORTS ABOUT THESE ABUSES?! JK> administration of my University. WHY SURE! BUT AND EVEN THIS IS NOT ALL! Now - after your, Professor Bretto's inaction - a progress from systematic wilful libel to pestering my acquaintances, references and persons who expressed their opinion about quality of my work, has been achieved by your worker, Jerzy Karczmarczuk, and thus, a part of their most valuable time was eaten up to read libels sent willfully by your worker. Jun 18 2005 JK> I sent a copy of the newsgroup of a particularly succulent JK> posting by V. Bondarenko to people whom he considers his JK> supporters, since I was not sure whether they know that JK> their names are related to his; and this they might not JK> like... Professor, why don't read these opinions, and ask yourself, If such top-notch computer experts value this persons so much, maybe I, Professor Bretto, am relly doing something incorrect allowing a slanderous person to behave so? http://www.cas-testing.org/index.php?list=7 Albert D. Rich, http://www.derive.com/ Prof Dr Oleg Marichev, WRI http://www.functions.wolfram.com/ Dr Anwar Shiekh, WRI Test Development Group Supervisor Prof Dr Walter Oevel (SciFace GmbH) Dr Michael Wester, http://www.math.unm.edu/~wester Stefan Wehmeier, SciFace GmbH Kelly Roach, http://www.kellyroach.com/ EVEN THIS IS NOT ALL! Sat, 18 Jun 2005 JK> I send a copy of this posting also to other people mentioned JK> as references on V. Bondarenko personal pages, since they JK> deserve also to be informed. Since I informed already Michael JK> Wester, and the Paderborn group does not need further warning, JK> I had some contact with them, they know what is going on, JK> the people concerned are JK> Prof. Dr. Anatoly Svidzinsky JK> Prof. Dr. Alexander Letichevsky JK> Dr. Anwar Shiekh BUT EVEN THIS IS NOT ALL!!! How do you like, Professor Bretto, a recent message from your worker Jerzy Karczmarczuk? ................................................................. JK> Beware, everybody. You are potential targets as well. ................................................................. AND EVEN DIRECT THREATS TO MY LIFE! ................................................................. JK> I am afraid we must ask some Mudjaheddin for rescue... ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ................................................................. President of l'Universite' de Caen Basse-Normandie asking for her MOST urgent and decisive support againt a wilful slanderer working at your university? Or you, dear Professor Bretto, will stop this wilful libeler, Jerzy Karczmarczuk, ONCE AND FOR ALL, by yourself? Best wishes, Vladimir Bondarenko vb@cybertester.com http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ P.S. How a student willing to enter l'Universite' de Caen Basse- Normandie will feel about learning - maybe even at the computer/informatics department where you teach - from a person who systematically both shows his sheer computer incompetence in the field of software testing and throws in packs variegated insults - to a person you even NEVER met before these weeks? Last but not least. if he is obviously cot foresee even the nearest impact of his actions - the impact of his wilful libel and scandalous character on his further career? ------------------------------------------------------------------- Letter # 1 - Jun 6, 2005 === Subject: Jerzy Karczmarczuk defames the l'Universite' de Caen Basse-Normandie ------------------------------------------------------------------- Local: Mon,Jun 6 2005 3:43 pm === Subject: Jerzy Karczmarczuk defames the l'Universite' de Caen Basse-Normandie alain.bre...@wanadoo.fr, Nicole LE QUERLER nlequ...@crisco.unicaen.fr, Patrick DUBOIS dubo...@admin.unicaen.fr, v...@cybertester.com A leading world's CAS researcher, the inventor of the LIFT and CYCLE technologies to be demonstrated in the near future, I am sorry to inform you that a worker of l'Universite' de Caen Basse- Normandie, some Jerzy Karczmarczuk, defames your organization via systematic sending publicly the insulting messages. I am too busy with research work; but now I am feeling that the time has come to stop the improper behaviour of Jerzy Karczmarczuk and ask for your resolute help in modifying the behavior of this person. Here you can find some samples of his statements about me, a person whom he NEVER even met. JK> person, whose ego reached the stage needing a competent JK> psychiatrist. JK> the insulting slogans, as taken directly from an Ancien JK> Regime era... JK> This fellow Bondarenko visibly suffers from a heavy mental JK> disorder. JK> Insane people usually deserve some compassion JK> perhaps he will be compelled by somebody to undergo a medical JK> treatment... JK> a clinical idiot JK> V. Bondarenko is not capable of understanding the JK> simplest logic. JK> a person far from mental equilibrium and more, but I hope the quoted links will do. Also please note that, at any rate, Jerzy Karczmarczuk goes easily beyond his competence, a big defect for a research worker as well as an instructor. Had you need any kind of support to drive the nail home please don't hesitate to let me know and I guarantee you the strongest backstop imaginable to stop the above demonstrated disorderly conduct. Best wishes, Vladimir Bondarenko vb@cybertester.com http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ------------------------------------------------------------------- ------------------------------------------------------------------- Letter # 2 - Jun 8, 2005 === Subject: Jerzy Karczmarczuk defames the l'Universite' de Caen Basse-Normandie ------------------------------------------------------------------- === Subject: Numerous, systematic public intentional insults, Or, Jerzy Karczmarczuk keeps defaming the l'Universite' de Caen B asse-Normandie Alain Bretto alain.bre...@info.unicaen.fr, alain.bre...@wanadoo.fr, Patrick DUBOIS dubo...@admin.unicaen.fr, v...@cybertester.com A leading world's CAS researcher, whose services had been used by the biggest market players such as Texas Instruments, Inc, and Wolfram Research, Inc http://www.cas-testing.org/index.php?list=3 the MuPAD Beta Contest First Prize winner launched by SciFace GmbH http://www.cas-testing.org/SciFace.phtml the inventor of the LIFT and CYCLE technologies to be demonstrated in the near future, http://maple.bug-list.org/MapleCrisis-Review-01.pdf I must inform you sadly that, alas, a worker of l'Universite' de Caen Basse-Normandie, some Jerzy Karczmarczuk, continues defaming your organization. The day before yesterday, on Monday, 6 Jun 2005 12:19:18 I've sent you a carbon copy of a report on improper behaviour of this person. The main copy had been addressed to Professor Bretto from whom I really expected help in stopping the insulting messages the above-said Jerzy Karczmarczuk posts into Google Groups. Being too much immersed into research activity, I did not visit the Google Groups yesterday. However, today, on June 8, 2005, with much surprise, I discovered the following remark from Jerzy Karczmarczuk showing his full disrespect to me, to you, to the whole administration of your University, and thus, to the whole l'Universite' de Caen Basse-Normandie: JK> Until now I was the only one in my department who knew that JK> we have a mentally sick person in this newsgroup, now there JK> are more. Imagine the explosion of laughter when one of the JK> receivers of this text of Bondarenko: > A leading world's CAS researcher, the inventor of the LIFT and > CYCLE technologies to be demonstrated in the near future, I am > sorry to inform you [...] JK> tried, being a serious person, to verify on the Internet the JK> case of this leading world's CAS researcher, to find his JK> publications, software, or *whatever*. My detailed answer to this person you can find here: The behavior of Jerzy Karczmarczuk who even does not acquainted with me personally, and who does not have any relation to the field I am working in, automated software testing, cot be described in other way as a wilful trespass, an wilful illegal act against my rights. Being a linguist, I hope that upon reading the statements made upon me, you certainly feel that I am right. I kindly request from you that you stop at last the described improper actions of a worker of l'Universite' de Caen Basse- Normandie, Jerzy Karczmarczuk wrt to me. Having too many a task on software testing automation, I hope that I will not be forced to get involved in further handling of this case. Just a detail, if the libel sent by Jerzy Karczmarczuk and the scandal would continue further, what would think your potential students about joining your University where their rights could be broken so easily? Best wishes, Vladimir Bondarenko vb@cybertester.com VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://maple.bug-list.org/MapleCrisis-Review-01.pdf http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ------------------------------------------------------------------- === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? HI GROUP, I propose to rename this group to *sci.math.spam* or *spam.math.symbolic* REGARDS, Marcus === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? > For the third time I must attract your attention to the fact that > some Jerzy Karczmarczuk > http://www.info.unicaen.fr/~karczma/ > of l'Universite' de Caen Basse-Normandie, a complete ignoramus in > Software Testing field Not to denigrate the competence of my friends in the software testing business; but ignoramii in that field are not entirely unheard of. === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? I think, all people already could be convinced, that VB is ordinary pathological and pettifogging type trying somehow to attach oneself to science without any grounds. === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? Indeed, Mr. Bondarenko evokes very unpleasant feelings. J.R. === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? CYCLE technologies, among really leading experts on CAS I have not found mr. Bondarenko. As far as I have understood, your scientific successes are limited only by the strange enough posts on forums and by a few poster papers on the CAS conferences of mean level . First, it is necessary to present results of the researches, instead of promise for the future, and then the scientific community will estimate them in essence. I think, the forum not a place for your creative work and wrangle with your critics. === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? i think in the future you should replace worker with employee or academic staff, i don't think worker is the appropriate word here :D but really you should stop writing these silly letters. or is all this just intentional comedy? you sure got me then..... === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? > a clinical idiot > Bondarenko visibly suffers from a heavy mental disorder > needing a competent psychiatrist > he will be compelled by somebody to undergo a medical treatment > a person far from mental equilibrium ... Don't be upset. Go take your medicine. -- the shadow === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? > For the third time I must attract your attention to the fact that > some Jerzy Karczmarczuk > http://www.info.unicaen.fr/~karczma/ > of l'Universite' de Caen Basse-Normandie, a complete ignoramus in > Software Testing field, who has thrown in packs variegated insults > to a person he even NEVER met before these weeks, me, still continues > defaming your organization via sending - wilfully, by design - even > more insults as well as other completely false information about me. [snip lengthy diatribe on defaming, insults, etc.] Pot... kettle... say, that packs variegated coinage is new to me! === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? > For the third time I must attract your attention > to the fact that some Jerzy Karczmarczuk > http://www.info.unicaen.fr/~ka[CapitalEth]rczma/ > of l'Universite' de Caen Basse-Normandie, > a complete ignoramus in Software Testing > field, who has thrown in packs variegated > insults to a person he even NEVER met before > these weeks, me, still continues defaming > your organization via sending - wilfully, > by design - even more insults as well as > other completely false information about me. [snip rest] on my drive home this evening I need to stop by a grocery store and get a gallon of milk. Dave L. Renfro === Subject: Cubic lattices Can anyone explain the differences between simple, face-centred and body-centred cubic lattices? I can find dozens of references to them but no definition of what they are. AJ === Subject: Re: Cubic lattices > Can anyone explain the differences between simple, face-centred and > body-centred cubic lattices? > I can find dozens of references to them but no definition of what they are. > AJ For the simple cubic lattice, consider points in R^3 whose coordinates are integers. For the face-centered cubic lattice, add an additional point on each face of each cube in this lattice, eg. points with one integer coordinate and two coordinates that differ from an integer by 1/2. For the body-centered cubic lattice, start once again with the simple cubic lattice and add points whose coordinates are all different from an integer by 1/2. Note that if the points from both the face-centered and body-centered cubic lattices are combined, one returns to a simple cubic lattice with nearest neighbor spacings half of what appears in the all integer coordinate lattice. === Subject: Re: John Thomas Walton - October 8th 1946 > Tuesday June 28th 2005 179/186 17663 > W A L T O N > 23 1 12 20 15 14 = 85 > Oh dear. You said it... Anyway, here's my proof that Menai Bridge, a village on Anglesey, is Hell: Menai Bridge. OK, let's start with Bridge. A bridge is something that takes you from one side of a river, ravine, etc to the other, or more abstractly from one state of being to another. Hold that thought for now. Menai. M=1000, I=1. ENA obviously means enable negative arithmetic, or subtract. 1000-1=999; use the bridge to mirror that number horizontally, that gives: 666. Therefore Menai Bridge is Hell. === Subject: Re: John Thomas Walton - October 8th 1946 > 1000-1=999; use the bridge to mirror that number > horizontally, that gives: > 666. Alas, 666 is not a horizontal reflection of 999... try it! - Risto - === Subject: Re: John Thomas Walton - October 8th 1946 >> 1000-1=999; use the bridge to mirror that number >> horizontally, that gives: >> 666. > Alas, 666 is not a horizontal reflection of 999... try it! > - Risto - A 180 deg. rotation would have filled the bill. Lee === Subject: This Mail Changed My Life: This Mail Changed My Life: I WAS APPALLED WHEN I SAW, HOW MUCH MONEY ON MY A 100% legal business! I made 10 ? to 78,520 ? within the first 60 days with this marketing plan, To ounce which I am in the concept, cost-free to you. If you decide to seize the following directions and required measures, can enjoy you a similar return! 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Alone this knowledge is already far over 10,-- ? appraise. Imagine once, which gigantic internet businesses (entirely equal, whether characteristic web pages, Partner - Sites or pure enamel advertising) open themselves you therewith for you in the future! THERE should have been already people, know forked out have that for such a much cable transmitter. You get this knowledge for 10,-- ?! STEPS 3 As soon as you a payment of 10 Euros to the first address of the list sent have (together with the use purpose (Order ebook - This is VERY IMPORTANT!!!), must do you following process yet terminating. Copy this text and/or this side and send you it at (you can also this text advertise on your characteristic web page, like I it to at least done would have ), 100 people. You retained is the number 100 in the head , 100 a good number of people that one easily can reach, in the internet, hundreds of even thousand could under you forth source. The copy that you send out contains YOUR E-MAILS address at the place no. 4 in the list - You delete set the address at no. 1 in the list and the other you around a position towards the top. The best mer in order to send the mails now, is simple all to marking and then with the right mouse key to copying in order to insert after that the text into the E-MAILS, which you send now at ca. 100 different persons. Do not forget: Each person, of that receive you in the future a payment in the amount of 10.00 ? ; on its PayPal - account, send you also a copy of the ebook! Market your side over different German and American / international Paidmaildienste. There are innumerable ways to reach several people and let have it at this astonishing program part. The most single tasks that you are note must that YOUR E-MAILS address is at no. 4 in the list, as well as let approach , that you really each person, who pays you now 10 , -- ?, also a copy of the ebook! Please you send is no spam, that the most only what could damage this really brilliant system! What itself make stands must all in the eBOOK. Obviously : the address that was previously at no. 1, should have been removed were supposed to have become, and the other E-MAILS addresses a position higher in order to set your E-MAILS address at no. 4. So long you this correctly done have, are your E-MAILS for sending ready! A WORD OF THE WARNING! Do not let yourself mislead to acquire to add your E-MAILS address at position 1 around money quickly! It does not function so! If you do that, you reach ONLY the people whom you send directly the E-MAILS and then your address is removed immediately no. 1 of place again and reach you not thousands of people! Your salary possibility remains otherwise LIMITED. But , if you add your E-MAILS address on position to no. 4, you reach receive and/or send literally 10 thousands of people, that your E-MAILS. THAT THE POWER OF THE INTERNET IS!! As soon as your E-MAILS are ready, you send a minimum of 100 copies the E-MAIL. Through the sending, this E-MAIL and the payment over PayPal is the Therefore it will last only some days until 10 Euro-payments will inundate your PayPal account! THAT WAS ALL! The complete process should last approximately 30 minutes. YOU WOULD BE PREPARED ENTHUSED .... IT FUNCTIONS!! A half hour of the simplest work is required. No head editions, no stamps, no pressure , copies etc. and the draft is allowed 100%. Within 30 days over 30,000 Euros in your PayPal Account. And that many reached already before you. Therefore you can that also. Actually, you expect a considerable number of 10 Euros payments within the first days! Retain a copy of your E-MAILS so that you can use it again, require whenever you more money! Here exactly described like it functioned: If you send your E-MAILS, set your E-MAILS address first at the no.4 in the list. That is the best position that there can be, if you would like to earn correctly much money. The answer-rate for this program is much higher, than each typical E-MAILS marketing-campaign on the basis of a number of reasons, that later is explained. They can expect an answer of approximately 35% of the people , at which you send the E-MAILS. But let us first of all extremely conservatively be and suppose that you have an average talk quota of only 10%. If you send your E-MAILS at 100 different persons , you can expect that you reach at least 10 of those persons, that do exactly that, what you did. (10 %of 100 = 10). To this time is reached your E-MAILS address at no. of 4 the list, and this list now 100 people (1 x 100). Of those 100 persons, you can expect, participate that at least 10 (10% of 100 = 10). Therewith you are sent out become on further 1,000 E-MAIL (10 xs 100), that, with your E-MAILS address on place 3. Of those 1,000 persons, you can expect, participate that at least 100 (10% of 1,000 = 100. Therewith you are sent out become on further 10,000 E-MAIL (100 xs 100), that, with your E-MAILS address on place 2. Through it you are sent out become now on 10,000 E-MAILS (100 xs 100), that, with your E-MAILS address on place 1!! Of those 10,000 persons, you can expect again that 1,000 react. And there you now at NO. 1 in the list are , receive you: 10,000.00 Euro (1,000 x 10 Euro). If your E-MAILS address reaches the position no. 1, you will have within the next 30 days of some thousand people, exactly like you your 20 Euros delivery, sent got money. That means: YOU sold then 1,000 copies of the ebook , which you yourself acquired for only 10,-- ?! Naturally you think is functioned now that to simply as that it. But straight BECAUSE it is so SIMPLE to understand, and because each it the business so simple!! To invest who is ready, a half hour, approximately 30,000 Euro is ready or to received more. In the next days, they will receive ca. 100 payments per day at approximately 30 days. After that the volume of the payments begins to diminish itself while your E-MAILS address clears the position no. 1 again. That is all what you must do! More than 30,000 Euros are through 20 Euro-orders, that you can expect on your PayPal account within the next weeks. Ca. 30,000 Euros for a work of just once 30 minutes! That is real money which you can issue for everything, what you only wish! They can the entire money on your PayPal Account store transfer, or from there on your bank account. Is the not one half hour value for !!?? I think already. Do you remember yourself? That examples 10% above, supposes that 90 of the 100 persons delete your E-MAILS. However, if you correctly follow the plan and send the persons your E-MAILS , is the probability of those persons who will participate on that, ca. 35% Consequently , that examples 10% becomes only as a worst case given. Furthermore the example arranges itself after that that each participants only 100 E-MAIL sends out. Place yourself before what happened would become, if each participants 1,000 E-MAILS sends out in place of of just once 100! You believe do me, many people this and send yet much more! I am even also such an example. Observe following! Over 718 million people surf in the internet every day and that worldwide! More than 200,000 new internets registrations each month! What itself make stands must all in the eBOOK. 10 Euros and 30 minutes of its time invest receive , and for that however a considerable sum within a month. If you uninterruptedly run let would want this business, you should generate regularly apply send let a characteristic web page, how these here , and, or regularly your E-MAILS over different Paidmaildienste. Would be meaningful naturally also polyglot. There is endlessly many German and American / international Paidmaildienste, over that you economically your E-MAILS send, or your characteristic , copied web page, disclose can. You think turn must only once of all the independent advisers in the network-marketing, that permanently expensive newspaper adertisement for them product sale and in order to find colleague, and at the same time ONLY REGIONALLY limited work can. They know certainly some or several of them. Many network-marketings businesses work even WORLDWIDE. You place happen could yourself once before what if you tell... these independent advisers of the eBOOK. The possibilities are endless, and it is so simple! How one exactly makes this, exactly is described in the ebook, which you yourself acquire for 10.00 ?. Euro. Only the first person on the list receives your order of 10 Euros, but each in the list climbs into the position 1. Because it is so simple, the talk speed is VERY HIGH and - FAST. And you begin to see drastic results in less than two weeks! You remember require yourself, you to send only 100 copies, in order to begin therewith. 100 E-MAILS to are received enough around a considerable sum within 30 days. Send it for example at personal contacts and as an answer at people, who send you its programs, that work like it , already in the Net. They know that this program already convinced functioned and are by this system! It does NOT count AS a spam if you react to offer of the people, that send you mails. More examples in the eBOOK. Therefore your E-MAILS (and or generate you itself a characteristic web page as well as this here) send now, and prepare before for a very large influx of the cash within the following 30 days! 5 factors that so successfully make this program. . . 1. EXTREMELY FAST ANSWER 2. EXTREMELY HIGH talk speed 3. UNLIMITED PROFIT - POTENTIAL 4. QUICKLY, SIMPLY AND LOW-PRICED TO BEGINNINGS. A GIGANTIC UTILITY, THAT YOU YOURSELF END THE KNOWLEDGE WHICH IS DESCRIBED IN THE EBOOK, FOR MANY OTHER THINGS IN THE INTERNET PULLING BECOMING! AND THAT FOR UNIQUELY ONLY 10.00 ? 5th ways of the PRACTICAL zero investment (uniquely only 10 Euro), SPEED, and HIGH PROFIT - POTENTIAL, has this program a VERY HIGH talk speed! Most E-MAILS marketing-campaign have an average of 0,5% to 5%. However this certain program produces usually a talk quota of 20% and 35%. Why? Because this program is to be begun so simply , it almost nothing, it costs lasts just 30 minutes, and the results can be seen within the next days. I observed this type of the MLM program over years and am this the simplest and fastest which there is. No stamps, no envelopes, no printing, no copies - just a little effort and a belief!! This program holds it really short and simply! UNLIMITED income potential! This program is structured for each and includes only to send 100 E-MAILS. Do not set however a 100er boundary. You send can so many E-MAILS how you. All 100 E-MAILS have a return of at least 20,000 - 30,000 Euros WITHIN 30 DAYS. Therefore if you 200 E-MAILS, 300 or which amount also always can send, you do it! However you notice yourself that the think to place your E-mail adress does NOT obtain your by the name of on a higher plain of the list the result. You keep in mind made yourself, the people before you the effort and earned it to be at its place. Comply therefore with the guidelines, you would be honest and the money becomes you come. Perhaps you are was yet sceptical, exact like I it originally and superior whether it really functions. Everything, what goes you now through the head, I have behind me and know the feeling. Do we but want to be yet once honest, thought have me I at that time, what I that really to lose? 30 minutes and 10 Euro. Is that not laughable? I am today more than happy that I have invests it. You but comply would ask with the guidelines! This program costs no longer, except 10 Euros and a half hour of your time, and if each complies with the guidelines , each wins! They have now the knowledge which enables you to earn more than 30,000 Euros within the next weeks. The single thing that you can hold now back is a lack of belief or a lack of even belief. All possible doubts that you have at present disappear within some days after they transferred this plan into the practice. Trust me! They not undoubtedly will regret it. Some comments over http://www.paypal.com, That you usefully find can. PayPal Address lets open each with an E-MAIL an account and is worldwide the no. 1 in the Online payment service. PayPal is supposed in more than of 11 million eBAY members n, as well as in an innumerable number of businesses Online. If you send money through PayPal, you can finance your payments with your credit card or Kontokorrentkonto. They don't have to be concerned with your privacy because PayPal certainly leads your statement information. To terminate a transaction with PayPal is far more certainly than the postal service shipment of a check, or to give your credit cards number to a strange. Therefore make use of PayPal already over 63 MILLION person in order to send and to receive money WORLDWIDE and that in real time, the least Paypal-users knowj edoch this eBOOK. You know what means that for you? The opening of a PayPal account is totally cost-free, simple, and it takes only a couple of minutes in claim. Therefore broth you yourself a good cup of tea or coffee up and begin you to transfer that, what began here before long. Finally you can lose nothing, but you stand shortly before that, in the next weeks much to profits! Order the eBOOK yet today and make you the first steps on the way to the financial freedom. Now are YOU At it! === Subject: This Mail Changed My Life: This Mail Changed My Life: I WAS APPALLED WHEN I SAW, HOW MUCH MONEY ON MY A 100% legal business! I made 10 ? to 78,520 ? within the first 60 days with this marketing plan, To ounce which I am in the concept, cost-free to you. If you decide to seize the following directions and required measures, can enjoy you a similar return! Each person that outside of Europe lives can the Euro also into following currencies: US - $$ British pounds Canadian dollars Australian dollars Japanese yens Cost-free convert let and reversed. How? Read further, learn it. If you think, unique 10 Euros are much money , consider you yet once, how much you would have to pay probably for a single impressive advertisement in a newspaper or somewhere in the internet, in order to find visitor / interested parties. EXPANSE more than 10 Euros. In this eBOOK, you find an ENTIRE collection at cost-free advertisement possibilities. You see there fore that the investment is into THIS eBOOK far more value than a 10 Euros. They receive it however for only 10 Euros. The marketing plan shows so favorably is you already why the eBOOK. You imagine have, you an own internet business directly of home from, with which you earn 24 hours all around the clock money. Incomes automatically recurring secured each month, also if you are straight on vacation. Then can be you equal whether your job is jeopardized (each is replaceable). > THAT IS life quality< The developer one of the best known software-manufacturers said once: In this millennium, there will be 2 types of businesses: 1) that in the internet and 2) that that there is not. Think once once of the Online-Shops. Another businessman said also once: Without advertising would be I millionaire, J. P. Getty - Milliard.8ar / Billionaire Were at the same time already from the beginning. With the business that I you into the hand would give: 1) as your own internet-business 2) to the application of your just existing business Do not stand this program sceptically vis-.88-vis. Think about at least some days about that. Otherwise you will throw away cash over 30,000 Euros! This marketing plan has worked for more than 2 years, thousands of people participated already on that and were surprised it by the results in shortest time and that with 10 only unique, -- Euro! It become you also! My experiences: employee of a pressure business, in which I worked. To this time, I lived far over my ratios and was in earnest guilt. The loss of my work released a chain reaction and I lost my car and the house. How you probably can imagine, my outlooks more looked than gloomy. received, how one can earn more over 30,000 Euros and yet much. I ignored this message. Very simply said I was sceptical. However I did not delete these E-MAILS because I thought myself, that any something at this thing at it is. It went me day after day, even for weeks through the head whether it could be possible to earn a such high sum in so short time. In the meantime, my guilt grew always more highly, and I had reached actually the point of the despair. I assessed finally that I had to lose absolutely nothing , if I follow this plan, that was offered me with this E-MAIL. And seen therefrom, I had no more large outlooks for the future anyway. Did therefore I think must undertake myself, I something and what is, if it really functions? Consequently I placed my doubts to the side, made the first step and followed the simple directions that were given me in this E-MAIL. The execution lasted less than 30 minutes and it cost me laughable 10 Euros. family a vacation in Miami and bought I a brand-new audi A8. in the autumn Euros. And the best, I owe no also only one only cent, no one. Until now , I earned over 497,580.00 ?. My accountant set up a Cash-Flow, he predicted would become that I within the following 24 months millionaire and that with only this business plan. If I write you this, I find it remarkably how my life yet so advantageously developed finally. I would not say that you become IMMEDIATELY millionaire , that is also not my task to you. It is your decision whether You this business for itself use would want, or to apply also only around its characteristic business with the contents of the eBOOK. HOWEVER you can be certain that just based on the VERY short LEVEL very many persons begin this business. Even if you earn ONLY 100,000 ? through this business, the 10 ? are overtaken, already a long time. With the cost-free advertisement possibilities in the eBOOK, it is also conceivably simple, also to earn more than 100,000 ?. I ALREADY SHWAS IT, LIKE YOU SEEING ABILITY. I have how most people hard worked and fighted. Then fells me so somewhat laughably simple like a drop into the lap and turns my life completely around. If I think back, at all the similar E-MAILS, that I simply so had deleted previously, I get geese skin because I know now, that it functions. Only if you now expanse harvest, are learned you how it can function also for you! This business plan that I am in the concept to explain you , can be transferred in shortest time. I never transferred less than 20,000 Euros in each offered opportunity. Let me you assure that it is here a LEGAL PERMITTED business opportunity and a faultless money salary business. It does not require that you speak with your friends or family (it would be then, you wish it). Why is this business legal? The legislator calls a system always then as illegal if (even if only in the theory) supposed will must, that one day the last the dogs bite. In the clear text: if the last customer user receives no value more for its financial use or receive can! Each paying customer of this system receives make can however a completely ebook with knowledge inestimable for each internet user, namely, like one cost-free (without paying ) internationally internet-advertising. The user can this knowledge for each of its already existing web page / partner side and or enamel advertising use, around therewith its characteristic products / services or to make confessed settlement businesses in the internet. The user uses this knowledge in addition for this gigantic marketing plan, he has in addition the possibility to obtain the income described here. Therewith once clearly is placed a for everyone, participate in is not illegal that that this marketing plan! Each paying customer receives a value (ebook including copyrighting to resale ) which he can offer now over this marketing plan itself and uses or the knowledge out of the ebook only for itself! Actually you need to not to step once with other people in contact. Anybody, with a characteristic impulse , successfully can become and can construct a fortune with this marketing plan. If you believe, that each dog has its day, you follow the simple, gradually exactly declared plan which is described now. If you do only this alone, you can receive can within the following 30 days over at least 20,000 Euros in cash over your PayPal account. I know that it incredibly must sound for you, very especially if you never in the possession of such a sum were, but believe me, this business works like nothing of other, what you saw in your life previously. With this following, simple 3 - step - plan will plan your life itself within a couple weeks basically change. Please you do not let stand your scepticism your financial success in the way. If you decide for that to use this opportunity not , will have you certainly your reasons for that. I respeam your decision and wish you the very best for your future. I cot say this with absolute sincerity because I have no financial advantage, whether you transfer it or. The life is short, and no one holds off you therefrom to have and to do the financial freedom, what you would like to do. But you must pay the price first for that. The price here is, to transfer the following and is to be become active. It is so simple! I pride say to can that I fulfilled my task vis-.88-vis my wife and the children, in that I secured its future in a very unsteady world. I believe that no price is too high, for this financial freedom. - This the MOST LOW-PRICED, the FASTEST and SIMPLEST way, money is to be earned Online - POINT END!! It gives bring a quantity of firms the one incredible uses in order to earn money. I am ready to explain you now how you can begin IMMEDIATELY. You do not remember ask yourself, I you to send me a single Penny for this lucrative business. They will earn fast and simply cash money with only a copy of this web page as an E-MAIL and the simplest , as well as most popular internet-payment of system! They have probably already 2 20/20 of this project on television in the shipments or Oprah Winfrey seen. Perhaps you read also already in the Wall Street Journal about that. If, here it is not. In detail and gradually. This program is new on no case. It exists in many forms and existed already over several decades. But in the early days, it required much more time and exertions, as well as internet , are the editions now practical ZERO! And in addition comes that the total process is now FASTER ,MORE SIMPLY, and MORE LUCRATIVE , than it been previously GENERALLY ever possible is! If you have already a main business, you leave can operate it at the same time , you this in the future on the side. If not, is this in the story of the internet the FASTEST and SIMPLEST way to earn very much money Online. I guarantee that you so somewhat never saw! This program functions, live equal in which country you, or which currency you have there. It is equal how old or like young you are. And you require undoubtedly no special knowledge or talent. They need run no websites to let, or to answer calls, to generate no photocopies or to send letters through the postal service and also no advertising etc. The single things that you require are: 1) An E-MAIL address 2) On PayPal account on the uniquely 10 Euro deposited are 3) and 30 minutes of your time. P.s. With the PayPal currency computer, you can Euro in US - $$ British pounds Canadian dollars Australian dollars Japanese yens Cost-free convert let and reversed. For the unsteadies: PayPal offers a COST-FREE buyer protection until 400, -- Euro, so there is no risk for you This program lasts a half hour. After that you it started have, have you no longer the slightest work therewith. And yet several thousands Euros are to be earned. Within the next weeks , you will gain through these 30 minutes simplest work of uncy profit! Yes , I know, it sounds to be too well around aware! I thought exactly the same like you now until I was convinced by the opposite! They have gives absolutely NOTHING to lose and it no boundary of the sum that you can earn with this! The fact is: you never will earn more money FASTER, LOW-PRICED and LUCRATIVE. Take yourself the time and read you please all! If you now no time, have retain then you these E-MAILS in your mail box and come you later on that back. Everything what you require, a PayPal account and an E-MAIL address is! You have an E-MAIL address yes Already. Each has already of PayPal belonged (if become not you it soon) and as I this draft got, knew I who would function it, when member of PayPal, I had already already experience with the excellent efficiency. PayPal The simplest method that you saw ever to is received payments Online. Each with an E-MAIL address COST-FREE can register! As soon as you on PayPal Have account, can send and receive you credit cards payments or cash everywhere worldwide! Please you read unconditionally further. If you yourself this half hour will take you this day your life long no longer forget. HERE A PROOF OF 4 participants that themselves decided only 10 Euros and a half hour to invest: What kind of astonishing plan! Before just 3 weeks, I followed your plan although I earned previously no 10 thousands, am happy I over the previous 17,480 Euros. I am absolute of the socks Humphries, Leicester I does not know say should ... what I THANK. MANY ..VIELEN. THANKS TO! I sent 40 this E-MAIL and then, forgot I simply on the thing. If I honestly its believed in should I not really the entire thing. But, when I reviewed my Pay-pal account a week later , more were than 6,000.00 Euro therein! After 30 days, I have now over 26,000.00 Euro! I cot thank you enough! Lisa McDonald, Northampton I was appalled when I saw, how much money flowed into my Pay-pal account. Within 3 weeks, tightly more 20,480.00 Euro climbed my accounts. First I thought is that any type incorrect entry happened! Richard Barrie, Cirencester I participated in order over to see, what one gets for the minimal use and the minimal money, that is required, back. To my amazement, I recovered 36,000 ? in the first 10 weeks and it, more money comes from day to day. Just some months ago some people did the same thing as well as you in this moment. If they decide to follow the following, simple directions , stand there it very quickly financially considerably better. And there is no single reason why you cot participate at this success. YOU have nothing to lose, but to win ALL! Let us begin, you follow represented simply the directions how now and ready you yourself then for a VERY LARGE money influx in the next 30 days before! Here is what you must do. . . STEP 1 If you not already on PayPal Users are, is the very first task that you must do of clicking on the PayPal connection below and an account cost-free To open. It lasts just 2 minutes! Here the connection is: https://www.paypal.com (then choose your country) Eighth you on that that you open a private PREMIUM-ACCOUNT. They must have a private PREMIUM-ACCOUNT (and not only a PERSONAL account)so that you can receive also credit cards payments of the people. For the unsteadies: PayPal offers a COST-FREE buyer protection until 400, -- Euro, so there is no risk for you STEPS 2 It is a law of the universe that we first must give in order to received. In your purchases in businesses, it is also not differently. Consequently the first action that is to be done now is, if you opened your PayPal account: numbers you 10 Euros of your PayPal Account at the FIRST E-MAILS address in the list. As a use purpose of the payment, you write: Order ebook. LEGALLY LEGAL. Directions as well as one a payment undertakes, is to be read under MONEY SENDING on the Paypal side. It is so simple! P.s. With the PayPal currency computer, you can Euro in US - $$ British pounds Canadian dollars Australian dollars Japanese yens Cost-free convert let and reversed. If you send your unique payment of 10 Euros to the first address in the list, you do it with a large large smile on your face , for How you would see, harvest you! Here the present list: 1) netmoney@gmx.net 2) bookmoney24@yahoo.de 3) kingjob24@yahoo.de 4) superlife24@gmx.net After you made a payment of 10 Euros at the first E-MAILS address of the list, something uncy is happened. It gives you an overpowering type of the security, a belief and the conviction at the system. You proved Yourselve just even, functioned that it , for YOU did it, and consequently many other people must give it, who are ready to do exactly that. Now you learned it at the characteristic body, out of first hand that this business really functions! YOU receive now the eBOOK: Internet advertising internationally cost-free like I operate can. Alone this knowledge is already far over 10,-- ? appraise. Imagine once, which gigantic internet businesses (entirely equal, whether characteristic web pages, Partner - Sites or pure enamel advertising) open themselves you therewith for you in the future! THERE should have been already people, know forked out have that for such a much cable transmitter. You get this knowledge for 10,-- ?! STEPS 3 As soon as you a payment of 10 Euros to the first address of the list sent have (together with the use purpose (Order ebook - This is VERY IMPORTANT!!!), must do you following process yet terminating. Copy this text and/or this side and send you it at (you can also this text advertise on your characteristic web page, like I it to at least done would have ), 100 people. You retained is the number 100 in the head , 100 a good number of people that one easily can reach, in the internet, hundreds of even thousand could under you forth source. The copy that you send out contains YOUR E-MAILS address at the place no. 4 in the list - You delete set the address at no. 1 in the list and the other you around a position towards the top. The best mer in order to send the mails now, is simple all to marking and then with the right mouse key to copying in order to insert after that the text into the E-MAILS, which you send now at ca. 100 different persons. Do not forget: Each person, of that receive you in the future a payment in the amount of 10.00 ? ; on its PayPal - account, send you also a copy of the ebook! Market your side over different German and American / international Paidmaildienste. There are innumerable ways to reach several people and let have it at this astonishing program part. The most single tasks that you are note must that YOUR E-MAILS address is at no. 4 in the list, as well as let approach , that you really each person, who pays you now 10 , -- ?, also a copy of the ebook! Please you send is no spam, that the most only what could damage this really brilliant system! What itself make stands must all in the eBOOK. Obviously : the address that was previously at no. 1, should have been removed were supposed to have become, and the other E-MAILS addresses a position higher in order to set your E-MAILS address at no. 4. So long you this correctly done have, are your E-MAILS for sending ready! A WORD OF THE WARNING! Do not let yourself mislead to acquire to add your E-MAILS address at position 1 around money quickly! It does not function so! If you do that, you reach ONLY the people whom you send directly the E-MAILS and then your address is removed immediately no. 1 of place again and reach you not thousands of people! Your salary possibility remains otherwise LIMITED. But , if you add your E-MAILS address on position to no. 4, you reach receive and/or send literally 10 thousands of people, that your E-MAILS. THAT THE POWER OF THE INTERNET IS!! As soon as your E-MAILS are ready, you send a minimum of 100 copies the E-MAIL. Through the sending, this E-MAIL and the payment over PayPal is the Therefore it will last only some days until 10 Euro-payments will inundate your PayPal account! THAT WAS ALL! The complete process should last approximately 30 minutes. YOU WOULD BE PREPARED ENTHUSED .... IT FUNCTIONS!! A half hour of the simplest work is required. No head editions, no stamps, no pressure , copies etc. and the draft is allowed 100%. Within 30 days over 30,000 Euros in your PayPal Account. And that many reached already before you. Therefore you can that also. Actually, you expect a considerable number of 10 Euros payments within the first days! Retain a copy of your E-MAILS so that you can use it again, require whenever you more money! Here exactly described like it functioned: If you send your E-MAILS, set your E-MAILS address first at the no.4 in the list. 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Therefore broth you yourself a good cup of tea or coffee up and begin you to transfer that, what began here before long. Finally you can lose nothing, but you stand shortly before that, in the next weeks much to profits! Order the eBOOK yet today and make you the first steps on the way to the financial freedom. Now are YOU At it! === === Subject: Internal angles of a polygon If polygon has > 2 sides, and is simple and closed, then the sum of it's internal angles = (sides - 2) * 180 degree. Correct? If it is just closed, then if the sum of the internal angles is > (sides - 2) * 180 degrees then the polygon is self-intersecting? If it is closed, is there anyway possible that the sum of the internal angles < (sides - 2) * 180 degrees? === Subject: Re: Internal angles of a polygon On 1-Jul-2005, Hamish 2 sides, and is simple and closed, then the sum of it's > internal angles = (sides - 2) * 180 degree. Correct? Yes. > If it is just closed, then if the sum of the internal angles is > (sides - > 2) * 180 degrees then the polygon is self-intersecting? I think so. > If it is closed, is there anyway possible that the sum of the internal > angles < (sides - 2) * 180 degrees? No. To see this, note that if you 'walk' around the 'outside', the 'turn' angles, i.e., the supplements of the corresponding internal angles, must sum to 360 degrees to leave you facing in the same direction as when you started. -- Jim Heckman === Subject: Re: Internal angles of a polygon >If polygon has > 2 sides, and is simple and closed, then the sum of it's >internal angles = (sides - 2) * 180 degree. Correct? >If it is just closed, then if the sum of the internal angles is > (sides - >2) * 180 degrees then the polygon is self-intersecting? >If it is closed, is there anyway possible that the sum of the internal >angles < (sides - 2) * 180 degrees? Just out of curiosity, what do you think is the sum of the internal angles of the quadrilateral with sides starting at (0,0) going to (1,0) then to (0,1) then (1,1) and finally back to (0,0)? === Subject: Re: Internal angles of a polygon >If polygon has > 2 sides, and is simple and closed, then the sum of it's >internal angles = (sides - 2) * 180 degree. Correct? >If it is just closed, then if the sum of the internal angles is > (sides - >2) * 180 degrees then the polygon is self-intersecting? >If it is closed, is there anyway possible that the sum of the internal >angles < (sides - 2) * 180 degrees? > Just out of curiosity, what do you think is the sum of the internal > angles of the quadrilateral with sides starting at (0,0) going to > (1,0) then to (0,1) then (1,1) and finally back to (0,0)? 720. === Subject: Re: Internal angles of a polygon >>If it is closed, is there anyway possible that the sum of the internal >>angles < (sides - 2) * 180 degrees? >> Just out of curiosity, what do you think is the sum of the internal >> angles of the quadrilateral with sides starting at (0,0) going to >> (1,0) then to (0,1) then (1,1) and finally back to (0,0)? >720. Then the answer is No === Subject: Limit of a sequence , Zeta'(-1) In the following : SUM_n A_k:=A_1+...+A_n , {.} denotes fractional part, F(x):= (6x^3+3x^2-3x-1)/12 , Zeta(s) is the Riem-Zeta function. How it's possible to prove( or disprove) that lim_{n-->infty}(1/n)*SUM_n F({n/k})= - Zeta'(-1) ? === Subject: need help with formula I sell custom made free-form covers. In order to help my customers visualize the result of the datasets they provide in order to spot possible problems before submitting their plot points for production, I would like to be able to take their input and create a scatterplot using excel. This requires that the points be converted to xy coordinates, but I am not sure how to go about doing this. I am aware that this entails the use of a few simple formulas, but, unfortunately, not simple enough for me. It's beem years since I have had any advanced math classes. Points are made around the perimeter at certain intervals, say 1 - 60. A-B would represent the baseline Point A1 and B1 and AB would be used to calculate x1 and y1 Point A2 and B2 and AB would be used to calculate x2 and y2 and so on up to x60 and y60. Each triangle would then consist of Distance between A and B Distance between A and Point-n Distance between B and Point-n Make sense??? If anyone can help I would be most grateful. Please respond directly to this email. === Subject: Re: need help with formula format=flowed; reply-type=response > This requires that the points be converted > to xy coordinates, but I am not sure how to go about > doing this. > Each triangle would then consist of > Distance between A and B <-- c > Distance between A and Point-n <-- a > Distance between B and Point-n <-- b Take A at (0,0) and B at (c,0), and say the data point is at distance a from A and distance b from B, with the angle at A in the range from 0 to pi (inclusive). By the law of cosines, b^2 = c^2 + a^2 - 2 c a cosA, and (a cosA) is just the x-coordinate of point P. So the x- and y-coordinates of P are x = (c^2 + a^2 - b^2) / (2c) and y = sqrt(a^2 - x^2). --r.e.s. === Subject: Re: need help with formula format=flowed; reply-type=response >> This requires that the points be converted >> to xy coordinates, but I am not sure how to go about >> doing this. >> Each triangle would then consist of >> Distance between A and B <-- c >> Distance between A and Point-n <-- a >> Distance between B and Point-n <-- b > x = (c^2 + a^2 - b^2) / (2c) > y = sqrt(a^2 - x^2). For whatever reason, the earlier replies by Lynn Kurtz have only now appeared on my server -- sorry for the repetition. Anyway, this makes for easy Excel plots. --r.e.s. === Subject: Re: need help with formula >I sell custom made free-form covers. >In order to help my customers visualize >the result of the datasets they provide in order >to spot possible problems before submitting their >plot points for production, I would like to be able to take >their input and create a scatterplot using excel. >This requires that the points be converted >to xy coordinates, but I am not sure how to go about >doing this. I am aware that this entails the use >of a few simple formulas, but, unfortunately, >not simple enough for me. It's beem years since >I have had any advanced math classes. >Points are made around the perimeter at certain intervals, >say 1 - 60. >A-B would represent the baseline >Point A1 and B1 and AB would be used to calculate >x1 and y1 >Point A2 and B2 and AB would be used to calculate >x2 and y2 >and so on up to >x60 and y60. >Each triangle would then consist of >Distance between A and B >Distance between A and Point-n >Distance between B and Point-n >Make sense??? No. You need to explain a bit more. For example, where you say: >A-B would represent the baseline >Point A1 and B1 and AB would be used to calculate >x1 and y1 you need to explain what A1 and B1 are, what you are given about A1 and B1, how you use that to get x1 and y1, and for that matter, what x1 and y1 represent. When you say A-B represents the baseline to you mean A is the origin and B is on the positive x axis? You have a lot of meaningless symbols until you explain what they represent. --Lynn === Subject: Re: need help with formula > I sell custom made free-form covers. What's a free form cover? > In order to help my customers visualize > the result of the datasets they provide in order > to spot possible problems before submitting their > plot points for production, I would like to be able to take > their input and create a scatterplot using excel. What are they giving you? In what form is in given and what does it represent? > This requires that the points be converted to xy coordinates Not possible without any information about the points. What you written below makes little sense. Here's what I make of it. You have a closed simple curve with 60 points about the curve p1, p2,.. p60 which are use for a polygonal approximation of the curve. The coordinates of p1 are given (0,0) The coordinates of p2 are given (0,distance from p1 to p2) The coordinates of p3 require knowing the distance from p2 to p3 and the angle between, p1.p2 and p2.p3 or the distance from p1 to p3 and the angle between the base line p1.p2 and p1.p3 This later way may be easier. > Points are made around the perimeter at certain intervals, > say 1 - 60. > A-B would represent the baseline > Point A1 and B1 and AB would be used to calculate > x1 and y1 > Point A2 and B2 and AB would be used to calculate > x2 and y2 > and so on up to > x60 and y60. > Each triangle would then consist of > Distance between A and B > Distance between A and Point-n > Distance between B and Point-n > Make sense??? > If anyone can help I would be most grateful. > Please respond directly to this email. === Subject: Re: need help with formula >>I sell custom made free-form covers. > What's a free form cover? >>In order to help my customers visualize >>the result of the datasets they provide in order >>to spot possible problems before submitting their >>plot points for production, I would like to be able to take >>their input and create a scatterplot using excel. > What are they giving you? > In what form is in given and what does it represent? This is commonly used for land surveying. What they are giving me are the measurements to each point around the circumference, which typically are anywhere from 1-2' apart, from points A and B, (the base). The base itself is a line we establish which is a reference point and is parallel to the longest side of the free form object. It is typically anywhere from 8-15' long, meaning A and B are 8-15' apart. The data is supplied in ft and inches and converted to inches. I basically am trying to find the formulas by which I can triangulate the measurements supplied by my customers and convert them directly into xy coordinates so as to create a graphical representation of the area of the free form via a plotting program, or even a spreasheet. Each point around the circumference should have an equivalent xy coordinate, which can only be obtained by triangulation, correct? Does this help any? >>This requires that the points be converted to xy coordinates > Not possible without any information about the points. > What you written below makes little sense. Here's what I make of it. > You have a closed simple curve with 60 points about the curve > p1, p2,.. p60 which are use for a polygonal approximation of > the curve. The coordinates of p1 are given (0,0) > The coordinates of p2 are given (0,distance from p1 to p2) > The coordinates of p3 require knowing the distance from > p2 to p3 and the angle between, p1.p2 and p2.p3 > or the distance from p1 to p3 and the angle between > the base line p1.p2 and p1.p3 This later way may be easier. >>Points are made around the perimeter at certain intervals, >>say 1 - 60. >>A-B would represent the baseline >>Point A1 and B1 and AB would be used to calculate >>x1 and y1 >>Point A2 and B2 and AB would be used to calculate >>x2 and y2 >>and so on up to >>x60 and y60. >>Each triangle would then consist of >>Distance between A and B >>Distance between A and Point-n >>Distance between B and Point-n >>Make sense??? >>If anyone can help I would be most grateful. >>Please respond directly to this email. === Subject: Re: need help with formula >What they are giving me are the measurements to each >point around the circumference, which typically >are anywhere from 1-2' apart, from points A and B, (the base). >The base itself is a line we establish which is a reference point >and is parallel to the longest side of the free form object. >It is typically anywhere from 8-15' long, meaning A and B are 8-15' apart. >The data is supplied in ft and inches and converted to inches. >I basically am trying to find the formulas by which I can triangulate >the measurements supplied by my customers and convert them directly into >xy coordinates so as to create a graphical representation of the area of >the free form via a plotting program, or even a spreasheet. >Each point around the circumference should have an equivalent xy >coordinate, which can only be obtained by triangulation, correct? >Does this help any? Yes. Say your baseline is b units long, put your origin at A so that the xy coordinates of A are (0, 0) and B are (b, 0). Let's say the distance from A to a point P is m and from B to P is n. Then the point P lies on the intersection of a circle of radius m about A and of radius n about B. The equations of these two circles are: x^2 + y^2 = m^2 and (x - b)^2 + y^2 = n^2 You can eliminate y by subtracting them and solve for x which gives: x = (b^2 + m^2 - n^2) / (2b) Put this in the first equation to get y^2 = m^2 - {(b^2 + m^2 - n^2) / (2b)}^2 and y will be the positive square root of this presuming all the points are on the y side of the baseline. This would be very easy to automate and plot with something like Maple. --Lynn === Subject: Re: need help with formula Hi again Lynn. Will any of this be a problem with -x coordinates? Probably half the points will be on the -x axis. Art >>What they are giving me are the measurements to each >>point around the circumference, which typically >>are anywhere from 1-2' apart, from points A and B, (the base). >>The base itself is a line we establish which is a reference point >>and is parallel to the longest side of the free form object. >>It is typically anywhere from 8-15' long, meaning A and B are 8-15' apart. >>The data is supplied in ft and inches and converted to inches. >>I basically am trying to find the formulas by which I can triangulate >>the measurements supplied by my customers and convert them directly into >>xy coordinates so as to create a graphical representation of the area of >>the free form via a plotting program, or even a spreasheet. >>Each point around the circumference should have an equivalent xy >>coordinate, which can only be obtained by triangulation, correct? >>Does this help any? > Yes. Say your baseline is b units long, put your origin at A so that > the xy coordinates of A are (0, 0) and B are (b, 0). Let's say the > distance from A to a point P is m and from B to P is n. Then the point > P lies on the intersection of a circle of radius m about A and of > radius n about B. The equations of these two circles are: > x^2 + y^2 = m^2 and (x - b)^2 + y^2 = n^2 > You can eliminate y by subtracting them and solve for x which gives: > x = (b^2 + m^2 - n^2) / (2b) > Put this in the first equation to get > y^2 = m^2 - {(b^2 + m^2 - n^2) / (2b)}^2 > and y will be the positive square root of this presuming all the > points are on the y side of the baseline. > This would be very easy to automate and plot with something like > Maple. > --Lynn === Subject: Re: need help with formula >Hi again Lynn. >Will any of this be a problem with >-x coordinates? Probably half the points >will be on the -x axis. >Art No, that wouldn't matter. The only place where care must be taken is if some points are on opposite sides of your baseline, which I presume you can easily prevent by positioning your baseline appropriately. Tell you what, I have a little time on my hands currently, and if you want to send me a set of data points and baseline length, I will try it out in Maple and show you what the results look like. If you want to do that, just email me a text file with your data points. Remove the anti-spam portion of my email address if you email me. --Lynn === Subject: Re: need help with formula Hi Lynn: You have no idea how many people have responded to this request in the past that didn't seem to have any clue what I was asking for, even after several clarifications. I will test this out and see what happens - it looks to be exactly what I was looking for. BTW, what is maple? Would maple by any chance do all this calculating for me? I would ultimately like to have a program that I can import my customers data into and have it calculated automatically. Art >>What they are giving me are the measurements to each >>point around the circumference, which typically >>are anywhere from 1-2' apart, from points A and B, (the base). >>The base itself is a line we establish which is a reference point >>and is parallel to the longest side of the free form object. >>It is typically anywhere from 8-15' long, meaning A and B are 8-15' apart. >>The data is supplied in ft and inches and converted to inches. >>I basically am trying to find the formulas by which I can triangulate >>the measurements supplied by my customers and convert them directly into >>xy coordinates so as to create a graphical representation of the area of >>the free form via a plotting program, or even a spreasheet. >>Each point around the circumference should have an equivalent xy >>coordinate, which can only be obtained by triangulation, correct? >>Does this help any? > Yes. Say your baseline is b units long, put your origin at A so that > the xy coordinates of A are (0, 0) and B are (b, 0). Let's say the > distance from A to a point P is m and from B to P is n. Then the point > P lies on the intersection of a circle of radius m about A and of > radius n about B. The equations of these two circles are: > x^2 + y^2 = m^2 and (x - b)^2 + y^2 = n^2 > You can eliminate y by subtracting them and solve for x which gives: > x = (b^2 + m^2 - n^2) / (2b) > Put this in the first equation to get > y^2 = m^2 - {(b^2 + m^2 - n^2) / (2b)}^2 > and y will be the positive square root of this presuming all the > points are on the y side of the baseline. > This would be very easy to automate and plot with something like > Maple. > --Lynn === Subject: Re: need help with formula >Hi Lynn: >BTW, what is maple? Maple is a comprehensive symbolic mathematics program that can simplify expressions, solve equations, plot curves and surfaces etc. See: http://www.maplesoft.com/ There is a less powerful student edition which would likely be more than adequate for your problem. >Would maple by any chance >do all this calculating for >me? Yes. You could input your set of measurements and receive a plot as output. I am sure there are also other, possibly less expensive, programs that will do it too but I have no experience with them. >I would ultimately >like to have a program that I can >import my customers data into >and have it calculated automatically. That is no problem. --Lynn === === Subject: Representation of Integers Eah positive integer can be uniquely represented by three ordered smaller integers per the following examples: 11 = (2,1,2), 18 = (2,3,1), 31 = (3,2,0). What is the rule? Alex === Subject: Re: Representation of Integers > Eah positive integer can be uniquely represented by three ordered > smaller integers per the following examples: 11 = (2,1,2), 18 = > (2,3,1), 31 = (3,2,0). What is the rule? Although there are infinitely many such rules, the first one that sprang to mind worked: N is represented by (a, b, c), where a = floor(N ^ (1/3)), b = floor((N - a^3) ^ (1/2)), c = N - a^3 - b^2. Or more colloquially, a is the cube root, b is the square root of what's left, and c is what's left after that. The reason why it sprang to mind is that I thought to myself that the most compact such representation would asymptotically have max(a,b,c) ~= N^(1/3). If you take out the largest possible a^3, you'd have a remainder on the order of 3a^2, so taking out a largest possible square would be useful. Then you'd be left with a linear remainder. - Tim === Subject: Re: Representation of Integers > Eah positive integer can be uniquely represented by three ordered > smaller integers per the following examples: 11 = (2,1,2), 18 = > (2,3,1), 31 = (3,2,0). What is the rule? > Although there are infinitely many such rules, the first one that > sprang to mind worked: > N is represented by (a, b, c), where > a = floor(N ^ (1/3)), > b = floor((N - a^3) ^ (1/2)), > c = N - a^3 - b^2. > Or more colloquially, a is the cube root, b is the square root of > what's left, and c is what's left after that. > The reason why it sprang to mind is that I thought to myself that the > most compact such representation would asymptotically have > max(a,b,c) ~= N^(1/3). > If you take out the largest possible a^3, you'd have a remainder on > the order of 3a^2, so taking out a largest possible square would be > useful. Then you'd be left with a linear remainder. > - Tim Yes. How do primes behave in this representation? Alex === Subject: Re: Representation of Integers > Eah positive integer can be uniquely represented by three ordered > smaller integers per the following examples: 11 = (2,1,2), 18 = > (2,3,1), 31 = (3,2,0). What is the rule? > Although there are infinitely many such rules, the first one that > sprang to mind worked: > N is represented by (a, b, c), where > a = floor(N ^ (1/3)), > b = floor((N - a^3) ^ (1/2)), > c = N - a^3 - b^2. > Or more colloquially, a is the cube root, b is the square root of > what's left, and c is what's left after that. > The reason why it sprang to mind is that I thought to myself that the > most compact such representation would asymptotically have > max(a,b,c) ~= N^(1/3). > If you take out the largest possible a^3, you'd have a remainder on > the order of 3a^2, so taking out a largest possible square would be > useful. Then you'd be left with a linear remainder. > - Tim > Yes. How do primes behave in this representation? > Alex What I have in mind here is this: 7 = (1,2,2), 11 = (2,1,2) and 13 = (2,2,1) so that all permutations here yield primes. This is by no means general but one can ask the question thus: Are there any other primes p = (a,b,c) such that all permutations are also primes? Alex === Subject: Collision Hi All, A gamer posed to me the following: Two cubes, of different sizes are tumbling toward, or nearly so, each other. They have different, but known (mathematically specified), mers of tumbling. How do you ascertain, within reasonable computer time, if a collision occurs? How is this problem usually handled? --- Coping out to specifying spheres seems weak. Best wishes, Jim === Subject: Re: Collision > Hi All, > A gamer posed to me the following: > Two cubes, of different sizes are tumbling toward, or nearly so, each other. > They have different, but known (mathematically specified), mers of > tumbling. How do you ascertain, within reasonable computer time, if a > collision occurs? > How is this problem usually handled? --- Coping out to specifying spheres > seems weak. > Best wishes, Jim http://www.geometrictools.com/Documentation/IntersectionRotatingBoxes.pdf which references: http://www.geometrictools.com/Documentation/DynamicCollisionDetection.pdf Carl G. === Subject: Re: Collision > Hi All, > A gamer posed to me the following: > Two cubes, of different sizes are tumbling toward, or nearly so, each other. > They have different, but known (mathematically specified), mers of > tumbling. How do you ascertain, within reasonable computer time, if a > collision occurs? > How is this problem usually handled? --- Coping out to specifying spheres > seems weak. > Best wishes, Jim Hi All, prove interesting reading. Best wishes, Jim === Subject: Re: Collision See Section 3.3 Detecting Collisions of http://cache-www.intel.com/cd/00/00/01/76/17651_coll_det.pdf > Hi All, > A gamer posed to me the following: > Two cubes, of different sizes are tumbling toward, or nearly so, each > other. > They have different, but known (mathematically specified), mers of > tumbling. How do you ascertain, within reasonable computer time, if a > collision occurs? > How is this problem usually handled? --- Coping out to specifying > spheres > seems weak. > Best wishes, Jim === Subject: Re: Collision > Two cubes, of different sizes are tumbling toward, or nearly so, each other. > They have different, but known (mathematically specified), mers of > tumbling. How do you ascertain, within reasonable computer time, if a > collision occurs? > How is this problem usually handled? --- Coping out to specifying spheres > seems weak. > Best wishes, Jim Perhaps one can approach it this way. Consider the, say, larger cube fixed and all the tumbling resides in the smaller one (= the two tumblings combined). Place a grid of, say, 1000 x 1000 x 1000 points in the small cube and then using one of todays fast computers keep checking to see if one of these points ever enters the larger cube. Alex === Subject: Re: Collision > Two cubes, of different sizes are tumbling toward, or nearly so, each other. > They have different, but known (mathematically specified), mers of > tumbling. How do you ascertain, within reasonable computer time, if a > collision occurs? If the centers of mass pass within a distance less than or equal to half the sum of their edge lengths, they must collide. If they pass outside a distance of sqrt(3) times that, they will not. For intermediate distances, it depends upon how they are tumbling. If the tumbling motion is physically reasonable for a rigid cube of uniform density not subject to external torques, then by symmetry it is simply a rotation. This is simpler to solve, and maybe even feasible analytically. Collisions in general can be corner-to-face or edge-to-edge. These may include special cases such as corner-to-edge or edge-to-face, which may need to be treated differently if you intend to handle bounces. Alternatively, the motion may be true tumbling when the assumptions in the previous paragraph are not satisfied. In general, there's not much we can do but numerically model the motion within the distances of interest. Some physically reasonable tumbling motions are actually chaotic, and we will not in general be able to predict many tumbles in advance. For each numerical step, we can analytically interpolate vertex positions (e.g. linearly) to determine whether a collision occurred in the meantime. Linear interpolation gives up the assumption that the cube is perfectly rigid between time steps. Rotational interpolation maintains the rigidity assumption, but is vastly more complex and in some cases may be difficult to determine from the tumbling model. If you want a very simple approximation, you could just skip the interpolation and just check based on position. However, this is subject to fairly gross errors such as a corner that would pass through the other cube, but the time step snapshots only caught it before and after. It also makes it difficult to determine the resulting motion. - Tim === === Subject: I don't believe in global suppression, but ... Hi All, I don't believe in global suppression, but on specific cases, I believe I should be able to suppress. Let me be specific: Some individuals post items I'm not interested in. I would like to suppress posts by these individuals so that they don't appear on MY computer. There is a way to do this, a year or more ago the technique was posted on this web site. Alas, I'm a computer maladroit and have lost the procedure. Would Best wishes, Jim === Subject: Re: I don't believe in global suppression, but ... In Outlook Express Select Message -> Block Sender This will block out that poster, but unfortunately not the replies to his/her posts. > Hi All, > I don't believe in global suppression, but on specific cases, I believe I > should be able to suppress. Let me be specific: > Some individuals post items I'm not interested in. I would like to > suppress > posts by these individuals so that they don't appear on MY computer. > There > is a way to do this, a year or more ago the technique was posted on this > web > site. Alas, I'm a computer maladroit and have lost the procedure. Would > Best wishes, Jim === Subject: Re: I don't believe in global suppression, but ... > I don't believe in global suppression, but on specific cases, I believe I > should be able to suppress. Let me be specific: > Some individuals post items I'm not interested in. I would like to suppress > posts by these individuals so that they don't appear on MY computer. There > is a way to do this, a year or more ago the technique was posted on this web > site. Alas, I'm a computer maladroit and have lost the procedure. Would I'm not familiar with Outlook Express, but I suggest you look for a menu item or preference setting called Message Filters or something like that. Switching to a dedicated newsreader would likely give you many more options, including the ability to assign numerical scores or 'weights' to messages according to various criteria, but setting up a basic killfile doesn't require more than the rudimentary filtering capabilities most e-mail clients provide. -- Odysseus === === Subject: A blackjack probability A puzzle for you readers (I already know the answer): I took a day trip to Atlantic City the other day. A casino I visited there offerered the following side bet at some of its blackjack tables. For each of the player's initial two cards which matches the rank but not the suit of the dealer's face-up card, the payoff is 3 to 1. For each card which matches both rank and suit of the dealer's face-up card, the payoff is 14 to 1. The table uses eight decks. With no knowledge of other cards previously dealt, what is the expected net winnings for this bet? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability > A puzzle for you readers (I already know the answer): > I took a day trip to Atlantic City the other day. A casino I visited > there offerered the following side bet at some of its blackjack tables. > For each of the player's initial two cards which matches the rank but > not the suit of the dealer's face-up card, the payoff is 3 to 1. For > each card which matches both rank and suit of the dealer's face-up card, > the payoff is 14 to 1. The table uses eight decks. With no knowledge of > other cards previously dealt, what is the expected net winnings for this > bet? Here is the promised more elegant approach (which is basically the same as 's). Let I1 and I2 be the indicator variables for the cards' matching the rank of the dealer's card. Let J1 and J2 be the indicator variables for the cards' matching both rank and suit of the dealer's card. Let K = (1-I1) (1-I2) be the indicator variable for neither card matchig the rank. The total payoff W is 3I1 + 3I2 + 11J1 + 11J2 - K. By the linearity of expectation and identical distributions, EW = 6 EI1 + 22 EJ1 - EK = 6(31/511) + 22(7/511) - C(511-31, 2) / C(511, 2) = -18/83 = -.217, approximately, where C(n,m)= n! / [m! (n-m)!] is the number of combinations of n objects taken m at a time. -S.J. Herschkorn Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability > A puzzle for you readers (I already know the answer): > I took a day trip to Atlantic City the other day. A casino I visited > there offerered the following side bet at some of its blackjack tables. > For each of the player's initial two cards which matches the rank but > not the suit of the dealer's face-up card, the payoff is 3 to 1. For > each card which matches both rank and suit of the dealer's face-up card, > the payoff is 14 to 1. The table uses eight decks. With no knowledge of > other cards previously dealt, what is the expected net winnings for this > bet? To add to the mix, this is my answer. In the table below: Cards shows the player's two cards, in order. 0 means no match with the dealer's card; R means match rank but not suit; E means match both rank and suit. P&L shows the player's profit/loss for this outcome for a nominal $1 bet. Probability is the probability of this outcome. Cards P&L Probability ----- --- ---------------------------- 00 -1 384/415 * 383/414 ~ 0.856015 0R +3 384/415 * 24/414 ~ 0.053641 0E +14 384/415 * 7/414 ~ 0.015645 R0 +3 24/415 * 384/414 ~ 0.053641 RR +6 24/415 * 23/414 ~ 0.003213 RE +17 24/415 * 7/414 ~ 0.000978 E0 +14 7/415 * 384/414 ~ 0.015645 ER +17 7/415 * 24/414 ~ 0.000978 EE +28 7/415 * 6/414 ~ 0.000244 The player's expected profit/loss is calculated by summing P&L * Probability over all cases. It comes to -6312/171810, or about -0.0367. Thus, the player can expect to lose about 3.67 cents for each dollar bet. Or, to put it another way, the casino's take is about 3.67% on this play. === Subject: Re: A blackjack probability boundary=----=_NextPart_000_000E_01C583CD.177735B0 --------------------------------------------------------------------- Matt (prisoner number 271892), I suspect your answer must be right, because it yields a plausibly small (3.67%) gain for the casino. But I am puzzled by the wording of the problem, which says that a bet is made on *each* of the player's two cards. If the player's card 1 wins, the player gets either $14 or $3, and if his card 1 loses, the player must pay $1. If his card 2 wins, he gets either $14 or $3, and if his card 2 loses, he must pay $1. So if both cards lose (the case you have labeled 00), the player must pay $2. The player can win on both cards and be paid $28, $17, or $6. What happens if he loses on both cards? Does he lose $1, or $2? Either the wording of the problem is defective, or it uses gambling terminology (payoff is 3 to 1) whose precise meaning is not stated and is unfamiliar to me. -- Mark Spahn > A puzzle for you readers (I already know the answer): > I took a day trip to Atlantic City the other day. A casino I visited > there offerered the following side bet at some of its blackjack tables. > For each of the player's initial two cards which matches the rank but > not the suit of the dealer's face-up card, the payoff is 3 to 1. For > each card which matches both rank and suit of the dealer's face-up card, > the payoff is 14 to 1. The table uses eight decks. With no knowledge of > other cards previously dealt, what is the expected net winnings for this > bet? To add to the mix, this is my answer. In the table below: Cards shows the player's two cards, in order. 0 means no match with the dealer's card; R means match rank but not suit; E means match both rank and suit. P&L shows the player's profit/loss for this outcome for a nominal $1 bet. Probability is the probability of this outcome. Cards P&L Probability ----- --- ---------------------------- 00 -1 384/415 * 383/414 ~ 0.856015 0R +3 384/415 * 24/414 ~ 0.053641 0E +14 384/415 * 7/414 ~ 0.015645 R0 +3 24/415 * 384/414 ~ 0.053641 RR +6 24/415 * 23/414 ~ 0.003213 RE +17 24/415 * 7/414 ~ 0.000978 E0 +14 7/415 * 384/414 ~ 0.015645 ER +17 7/415 * 24/414 ~ 0.000978 EE +28 7/415 * 6/414 ~ 0.000244 The player's expected profit/loss is calculated by summing P&L * Probability over all cases. It comes to -6312/171810, or about -0.0367. Thus, the player can expect to lose about 3.67 cents for each dollar bet. Or, to put it another way, the casino's take is about 3.67% on this play. === Subject: Re: A blackjack probability > Matt (prisoner number 271892), > I suspect your answer must be right, because it yields a > plausibly small (3.67%) gain for the casino. > But I am puzzled by the wording of the problem, > which says that a bet is made on *each* of the player's > two cards. If the player's card 1 wins, the player gets > either $14 or $3, and if his card 1 loses, the player must pay $1. > If his card 2 wins, he gets either $14 or $3, and if > his card 2 loses, he must pay $1. So if both cards > lose (the case you have labeled 00), the player must pay $2. Incorrect. > The player can win on both cards and be paid $28, $17, or $6. > What happens if he loses on both cards? Does he lose $1, or $2? > Either the wording of the problem is defective, or it > uses gambling terminology (payoff is 3 to 1) whose > precise meaning is not stated and is unfamiliar to me. The player may bet any amount from $1 to the amount of his blackjack wager. Assuming s/he bets $1, that $1 is for *both* cards. S/he loses the dollar if and only if neither card matches. Thus, the possible payoffs are -1, 3, 6, 14, 17, and 28. For a numerical answer, I get -9/245, which is approximately -.0367 and which agrees with Matt's answer. Bill and neglect the outcome of losing the bet entirely. However, with the exception of 's, the solution methods posted are quite inelegant - e.g., needlessly laborious. I will post my method later. Before I do so, consider the following generalization. How would you approach this? Suppose a deck has r ranks and s suits (for a total of r*s cards). Suppose there are d decks. The dealer gives him/herself h face-up cards and the player p cards. For each of the p*h comparisons, the player wins a dollars if his/her card matches the rank but not the suit of the dealer's; the player wins a+b dollars for each comparison where both suit and rank match The player loses $1 if none the ranks of his/her cards all differ from all of the ranks of the dealer's cards. What is the expected player's net winnings now? (In the original problem, r = 13, s = 4, d = 8, h = 1, p = 2, a = 3, and b = 11.) We could generalize matters further by allowing payoffs for matching suit but not rank and (to go completely crazy) by considering cards with more than two characteristics (such as rank and suit). -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability On Fri, 08 Jul 2005 18:43:42 -0400, Stephen J. Herschkorn > The player may bet any amount from $1 to the amount of his blackjack > wager. Assuming s/he bets $1, that $1 is for *both* cards. S/he loses > the dollar if and only if neither card matches. Thus, the possible > payoffs are -1, 3, 6, 14, 17, and 28. > For a numerical answer, I get -9/245, which is approximately -.0367 and > which agrees with Matt's answer. Bill and neglect the outcome of > losing the bet entirely. I don't have the original statement of the problem, so I can't be sure of the wording. I worked on the assumption that the $3 I win for a rank-only match includes the dollar I bet and $2 from the house. If I get my dollar back plus $3, I need to recalculate. In that case I get a payout of about $.9633 for my dollar bet, which means that I lose an average of $.0367 of each dollar bet. That matches your result. I included the outcome of losing the bet in calculating the payout because a payout of nothing times the probability of getting that result is zero. Bill Swap first and last parts of username and ISP for address. === Subject: Re: A blackjack probability <11ctj0vn3i1bv60@corp.supernews.com Matt (prisoner number 271892), My prisoner number is actually 271829. > I suspect your answer must be right, because it yields a > plausibly small (3.67%) gain for the casino. > But I am puzzled by the wording of the problem, > which says that a bet is made on *each* of the player's > two cards. If the player's card 1 wins, the player gets > either $14 or $3, and if his card 1 loses, the player must pay $1. > If his card 2 wins, he gets either $14 or $3, and if > his card 2 loses, he must pay $1. So if both cards > lose (the case you have labeled 00), the player must pay $2. The way I understood the problem is that the stake covers BOTH cards. If neither card is a winner then you lose your stake. If either or both cards are winners then you get your stake back plus your combined winnings for the two cards, at the odds stated. So, if you stake $1 and you get my 00 then you lose $1. If you stake $1 and you get my RR then you gain $6. Etc. > The player can win on both cards and be paid $28, $17, or $6. > What happens if he loses on both cards? Does he lose $1, or $2? Well, he loses his stake. If he staked $1 then he loses $1. === Subject: Re: A blackjack probability boundary=----=_NextPart_000_001B_01C582C5.51BC72C0 --------------------------------------------------------------------- I may be misunderstanding the problem, but the situation seem to be that one card of the player is compared with one card of the dealer. If they have the same rank and suit (probability 1/52), the player receives $14. If they have the same rank but are of different suits (probability 3/52), the player receives $3. Otherwise (probability 48/52), the player receives minus $1. So the payoff is (1/52)14 + (3/52)3 + (48/52)(-1) = (14+9-48)/52 = -25/52, which amounts to losing about 48 cents for every dollar bet. -- Mark Spahn A puzzle for you readers (I already know the answer): I took a day trip to Atlantic City the other day. A casino I visited there offerered the following side bet at some of its blackjack tables. For each of the player's initial two cards which matches the rank but not the suit of the dealer's face-up card, the payoff is 3 to 1. For each card which matches both rank and suit of the dealer's face-up card, the payoff is 14 to 1. The table uses eight decks. With no knowledge of other cards previously dealt, what is the expected net winnings for this bet? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability > I may be misunderstanding the problem, but the situation > seem to be that one card of the player is compared with > one card of the dealer. If they have the same rank and suit > (probability 1/52), the player receives $14. If they > have the same rank but are of different suits > (probability 3/52), the player receives $3. Otherwise > (probability 48/52), the player receives minus $1. > So the payoff is > (1/52)14 + (3/52)3 + (48/52)(-1) = (14+9-48)/52 = -25/52, > which amounts to losing about 48 cents for every dollar bet. Nope. As Bill pointed out, your probabilities are wrong. (Bill's arithmetic is also wrong). Plus, you neglect the fact that the player places one bet which applies to both of his/her initial two cards. > Stephen J. Herschkorn A puzzle for you readers (I already know the answer): > I took a day trip to Atlantic City the other day. A casino I visited > there offerered the following side bet at some of its blackjack > tables. > For each of the player's initial two cards which matches the rank but > not the suit of the dealer's face-up card, the payoff is 3 to 1. For > each card which matches both rank and suit of the dealer's face-up > card, > the payoff is 14 to 1. The table uses eight decks. With no > knowledge of > other cards previously dealt, what is the expected net winnings > for this > bet? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability boundary=----=_NextPart_000_0008_01C583A9.9FC27480 --------------------------------------------------------------------- Stephen, So the player puts $2 at risk, betting $1 on each of his two cards, which are each compared with the dealer's one card. Here is my solution, giving the probabilities and payoff... P{player's first card and second card each match the dealer's card in both rank and suit} (pays $14 + $14) = 7 cards of same rank and suit as dealer's card in 8 decks /(8*52-1) cards from which player's first card is selected * 6 cards of same rank and suit as dealer's card /(8*52-1) cards from which player's second card is selected = (7/415)(6/414) P{player's first card matches dealer's card in rank and suit, and player's second card matches dealer's card in rank but not suit} (pays $14 + $3) = (7/415)(8*3 cards in 8 decks of same rank, different suit as dealer's card/414 cards from which player's second card is selected) = (7/415)(8*3/414) P{player's first card matches dealer's card in rank but not suit, and player's second card matches dealer's card in rank and suit} (pays $3 + $14) = (8*3/415)(7/414) P{player's first card matches dealer's card in rank and suit, and player's second card has different rank from dealer's card} (pays $14 - $1) = (7/415)(8*48/414) P{player's first card differs in rank from dealer's card, and player's second card matches dealer's card in rank and suit} (pays -$1 + $14) = (8*48/415)(7/414) P{player's first and second cards each match dealer's card in rank but not suit} (pays $3 + $3) = (8*3/415)((8*3-1)/414) P{player's first card matches dealers' card in rank but not suit, and player's second card differs from dealer's card in rank} (pays $3 - $1) = (8*3/415)(8*48/414) P{player's first card differs from dealer's card in rank, and player's second card matches dealer's card in rank but not suit} (pays -$1 + $3) = (8*48/415)(8*3/414) P{player's first and second card each differ from dealer's card in rank} (pays -$1 - $1) = (8*48/415)((8*48-1)/414) These probabilities cover all the possibilities, and it can be verified that they add up to 1 (because the numerators add up to 7*6+2*7*8*3+2*7*8*24+8*3*(8*3-1)+2*8*3*8*48+8*48*(8*48-1) = 171,810 = 415*414 = denominator). Thus the payoff (expected net winnings) for this dual bet is $28*7*6 + $17*2*7*8*3 + $13*2*7*8*43 + $6*8*3*(8*3-1) + $2*2*8*3*8*48 - $2*8*48*(8*48-1) = -$177,192, divided by the denominator (8*52-1)(8*52-2) = 415*417 = 171,810, which comes to $1.031325301 per $2 risked. This problem specified a dual bet. But would the payoff be different if the player bet on only one of his cards? In that case, the payoff would be $14*P{winning $14} + $3*P{winning $3} - $1*P{losing the bet} = $14*(7/415) + $3*(8*3/415) - $1(8*48/415) = -214/415 dollars gained per dollar bet. This is exactly the same payoff as in the dual bet of the original problem, because (1/2)(-177,192/171,810) = -88,596/171,810 = -214/415. But soft! Is this result plausible? On average, a player loses $1.03 ever time he bets $1 each on his two cards being similar to the dealer's card. Over the long run, for every time you make such a bet, you lose a little more than half the money you put at risk? Is this what is known as a sucker bet? Or is my analysis wrong? -- Mark Spahn (West Seneca, NY) > I may be misunderstanding the problem, but the situation > seem to be that one card of the player is compared with > one card of the dealer. If they have the same rank and suit > (probability 1/52), the player receives $14. If they > have the same rank but are of different suits > (probability 3/52), the player receives $3. Otherwise > (probability 48/52), the player receives minus $1. > So the payoff is > (1/52)14 + (3/52)3 + (48/52)(-1) = (14+9-48)/52 = -25/52, > which amounts to losing about 48 cents for every dollar bet. Nope. As Bill pointed out, your probabilities are wrong. (Bill's arithmetic is also wrong). Plus, you neglect the fact that the player places one bet which applies to both of his/her initial two cards. > Stephen J. Herschkorn A puzzle for you readers (I already know the answer): > I took a day trip to Atlantic City the other day. A casino I visited > there offerered the following side bet at some of its blackjack > tables. > For each of the player's initial two cards which matches the rank but > not the suit of the dealer's face-up card, the payoff is 3 to 1. For > each card which matches both rank and suit of the dealer's face-up > card, > the payoff is 14 to 1. The table uses eight decks. With no > knowledge of > other cards previously dealt, what is the expected net winnings > for this > bet? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability boundary=----=_NextPart_000_00E8_01C582D2.F3F12F60 --------------------------------------------------------------------- I may be misunderstanding the problem, but the situation seem to be that one card of the player is compared with one card of the dealer. If they have the same rank and suit (probability 1/52), the player receives $14. (No the probability would be less than 1/52. you have 7 cards that match from each of the 7 other decks. and a total number of cards to be match against of 52*7 +51 so 7 in 425, If they have the same rank but are of different suits (probability 3/52), the player receives $3. Otherwise (probability 48/52), the player receives minus $1. So the payoff is (1/52)14 + (3/52)3 + (48/52)(-1) = (14+9-48)/52 = -25/52, which amounts to losing about 48 cents for every dollar bet. -- Mark Spahn A puzzle for you readers (I already know the answer): I took a day trip to Atlantic City the other day. A casino I visited there offerered the following side bet at some of its blackjack tables. For each of the player's initial two cards which matches the rank but not the suit of the dealer's face-up card, the payoff is 3 to 1. For each card which matches both rank and suit of the dealer's face-up card, the payoff is 14 to 1. The table uses eight decks. With no knowledge of other cards previously dealt, what is the expected net winnings for this bet? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability boundary=----=_NextPart_000_000A_01C58313.57D21450 --------------------------------------------------------------------- Stephen, I still don't understand the statement of the problem. The player has two cards from 8 decks, and the dealer has one card from the same 8 decks. If one of the player's cards matches the dealer's card in both rank and suit, and the other player's card matches the dealer's card in neither rank nor suit, the player receives $14. If one of the player's cards matches the dealer's card in rank but not suit, and the other player's card matches the dealer's card in neither rank nor suit, the player receives $3. If neither player's card matches the dealer's card in rank, the player must pay $1. But what happens in the other possibilities? If the player has two 4Cs (4s of clubs) and the dealer's card is 4C, does the player still receive only $14? Or does he receive $28? If the player has 4C and JD (jack of diamonds), what happens when the dealer's card is (1) JD, (2) JS, (3) 2H? If the player has 5C and 5H and the dealer has 5D, what happens? (Maybe it would help if I knew what a side bet is. How is a side bet distinguished from a non-side bet? I take it that one does not need to know what blackjack is to answer this problem; that is, that your title A *blackjack* probability is a little joke.) -- Mark Spahn I may be misunderstanding the problem, but the situation seem to be that one card of the player is compared with one card of the dealer. If they have the same rank and suit (probability 1/52), the player receives $14. (No the probability would be less than 1/52. you have 7 cards that match from each of the 7 other decks. and a total number of cards to be match against of 52*7 +51 so 7 in 425, If they have the same rank but are of different suits (probability 3/52), the player receives $3. Otherwise (probability 48/52), the player receives minus $1. So the payoff is (1/52)14 + (3/52)3 + (48/52)(-1) = (14+9-48)/52 = -25/52, which amounts to losing about 48 cents for every dollar bet. -- Mark Spahn A puzzle for you readers (I already know the answer): I took a day trip to Atlantic City the other day. A casino I visited there offerered the following side bet at some of its blackjack tables. For each of the player's initial two cards which matches the rank but not the suit of the dealer's face-up card, the payoff is 3 to 1. For each card which matches both rank and suit of the dealer's face-up card, the payoff is 14 to 1. The table uses eight decks. With no knowledge of other cards previously dealt, what is the expected net winnings for this bet? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A blackjack probability