The hardest one is probably the order types. I¹ve done them in terms of an army of ants with number tags instead of name tags; their normal formation, of course, is 0 1 2 3 ..., marching to the left, but we soon have them performing all sorts of evolutions. The biggest hurdle is the idea that we don¹t need a physical realization of an abstract order; the problem arises the moment I suggest having Ant 0 fall out and fall back in at the rear of the formation. There¹s also an army of fancy ants, with separate numbers on head and thorax; their basic formation is (0,0) (0,1) (0,2) ... (1,0) (1,1) (1,2) ... (2,0) (2,1) (2,2) ... . . . . . . . . . We eventually discover how to parachute the plain ants onto this formation so that each fancy ant gets exactly one rider, and even how to Þgure out which plain ant lands on ant (p, q). I¹ve even had a couple of students successfully answer all of the following question on a take-home Þnal exam, and quite a few answer the Þrst two parts, even though the formation and ideas involved have not been discussed at all: Suppose that I try to form up the fancy ants according to the following rule: Ant (p, q) is in front of Ant (m, n) if pn < qm. Thus, Ant (3,5) is in front of Ant (4, 6) because 3 * 6 < 5 * 4. (a) Unfortunately, this rule doesnt deÞne a legitimate formation. Find two ants that according to this rule ought to be standing in the same place in the formation. (That is, neither of the ants is in front of the other according to the rule.) (b) To get around the difÞculty mentioned in part (a), well allow ants that ought to be in the same position to stand on one anothers backs. Explain how to Þnd inÞnitely many ants who will have to stand on the back of Ant (1, 2). (c) Ant (11, 25) is in front of Ant (4, 9); Þnd an ant who is behind Ant (11,25) but in front of Ant (4, 9). >> or gives the student a better picture of how mathematics >> Þts into the modern world by looking at (the elementary >> aspects of) some modern applications that are light on >> algebraic prerequisites. > I¹d be interested in what sorts of examples are used there too. Electoral methods, apportionment, fair division; Euler and Hamilton circuits and paths, minimal spanning trees, precedence graphs and scheduling algorithms; population growth (exponential model, discrete logistic model); square-cube law and growth; symmetry, especially frieze and wallpaper patterns. [...] >>> But in the educational arena, a large amount of >>> unfairness is inevitable as long as a main function >>> of the system is to promote social stratiÞcation -- >>> the separation of winners from losers in a manner >>> that the losers see as it being their own fault that >>> they are losers (and thus not making a huge stink about >>> it). >>Fortunately, quite a few of us who teach have no patience >>with the notion that this is the main function of an >>educational system. > I carefully said *a* main, not *the* main. Hm; I thought that I had used Œa¹ as well. Apparently my Þngers got ahead of my brain. [...] Brian === Subject: Re: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9q14709; >Can someone tell me how to solve the integral: >S dx/[(x^2)*ln(x)] ??? Substitution y = ln(x) dy = dx/x leads to S 1/[x^2*ln(x)] dx = S e^(-y)/y dy I do not believe this can be expressed by elementary functions. It is possible to express e^(-y) by a power series and then intergate term by term: inf S e^(-y)/y dy = Sum S (-1)^n*y^(n-1)/n! dy = n=0 inf = ln(|y|) + Sum (-y)^n/(n*n!) + C = n=1 inf = ln[|ln(x)|] + Sum (-1)^n*[-ln(x)]^n/(n*n!) + C n=1 Convergence is fast. For an accuracy better than 10^(-6), use about 10 terms for 0.1 < x < 10, 20 terms for 10^(-3) < x < 10^3, 30 terms for 10^(-5) < x < 10^5. === Subject: Re: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UMBOX28844; S 1/[x^2*ln(x)] dx = inf = ln[|ln(x)|] + Sum [-ln(x)]^n/(n*n!) + C n=1 inf lim { ln[|ln(x)|] + Sum [-ln(x)]^n/(n*n!) } = -g ~= -0.57721567 x->inf n=1 where g is Euler¹s constant. === Subject: Re: Need help with some integral! > > > Can someone tell me how to solve the integral: > > > S dx/[(x^2)*ln(x)] ??? > > > > > x = e^u, then by parts. > Details? integral e^2u u e^u du = integral u.e^3u du = (1/9) integral v.e^v dv thence by parts === Subject: Re: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9S14696; >You all ignored the / sign. >The integrand is: 1/[(x^2)*ln(x)] and not just: (x^2)*ln(x). >I still need help... If you are looking for an elementary expression for the indeÞnite integral, the problem looks hopeless. There might be hope of computing certain deÞnite integrals having the integrand above. Where did this problem come from? Todd Trimble === Subject: Re: Need help with some integral! > > > > > > > Can someone tell me how to solve the integral: > > > > S dx/[(x^2)*ln(x)] ??? > > > > > > > x = e^u, then by parts. > > Details? > integral e^2u u e^u du > = integral u.e^3u du > = (1/9) integral v.e^v dv > thence by parts I think one of us has misread the problem. Note the /. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Need help with some integral! > > > > > > > > > Can someone tell me how to solve the integral: > > > > > S dx/[(x^2)*ln(x)] ??? > > > > > > > > > x = e^u, then by parts. > I think one of us has misread the problem. Note the /. integral 1/(x^2 log x) dx = integral e^-2u / u * e^u du = integral 1/(u.e^u) du = integral (e^v / v) dv then by parts from your local atom smasher. ;-) === Subject: Re: proving trig identities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9G14672; >I am having a lot of trouble with this problem: >1+tanx/1+cotx=tanx >I have changed the tanx to sinx/cosx, and cotx to cosx/sinx >then divided them by multiplying the reciprocal. I have tried many >different ways on how to solve this, but I just get stuck. Can you >help? The identity that you write is not really a trig identity. It is easily show that (1+x)/(1+1/x) = x, for all x /= 0 To wit: (1+x)/(1+1/x) = [x(1+x)]/[x(1+1/x)] = [x(1+x)]/[x + 1] = x The only trig in your version of this identity is tanx = 1/cotx - MO === Subject: Inverse Matrices 1.03 for Windows! Inverse Matrices 1.03 for Windows now available! The program provides detailed, step-by-step solution in a tutorial-like format to the following problem: Given a 2x2 matrix, or a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse matrix by using the Gauss-Jordan elimination method. The check of the solution is given. The program is designed for university students and professors. Price: $15. Product homepage: http://www.dekovsoft.com/inverse_matrices.htm http://www.dekovsoft.com/ddekov/ === Subject: Re: Inverse Matrices 1.03 for Windows! First: You are using a freeware version of ClickTeam Install Creator in your trial versions. Do your paid versions use a non-freeware version of ClickTeam Install Creator - since you are asking for money? Second: You do not allow for decimal entry. Third: You do not allow for complex numbers. Fourth: You simply provide a display of all the steps - like you would get in a book. That is there are no animations (Flash or Java) or sounds that would help make it stick in a student¹s mind - i.e. a student who has challenges with learning from a book is still going to have the same challenges with your so called helpful programs! Fifth: You are charging $15 USD? for each little segment - are there not calculator programs (e.g. for the TI-89 or equivalent) that one can download free (with a one-time purchase of a calculator) that will show the steps of a matrix inversion, etc? Plus, at the end of it all, one still has a calculator to show for it. >Inverse Matrices 1.03 for Windows now available! >The program provides detailed, step-by-step solution in a >tutorial-like format to the following problem: Given a 2x2 matrix, or >a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse >matrix by using the Gauss-Jordan elimination method. The check of the >solution is given. The program is designed for university students and >professors. >Price: $15. >Product homepage: >http://www.dekovsoft.com/inverse_matrices.htm >http://www.dekovsoft.com/ddekov/ -- Casey === Subject: Re: Inverse Matrices 1.03 for Windows! Program¹s description: The program provides detailed, step-by-step solution in a tutorial-like format to the following problem: Given a 2x2 matrix, or a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse matrix by using the Gauss-Jordan elimination method. The check of the solution is given. The program is designed for university students and professors. The above description shows the aim of the program and its scope. Suppose a student wants to see the solution (or the check) in a tutorial-like format. The student cannot Þnd the solution in a textbook. Or suppose the student has some error and cannot see the error in the solution. Or suppose the student wants to see lots of examples to better understand the Gauss-Jordan elimination method. Or suppose a professor wants to give a quiz/homework to students and he/she needs a lots of problems (if the step-by-step solutions are given, the professor can save some time during the check of the quizes). Then the student/professor can use the program. The real question about such a program is as follows: How close is the solution produced by the program to the solutions given in the textbooks? Will the solution produced by the program be better, or worse, than the solutions, given in the textbooks? The authors of textbooks do not follow the strict Gauss-Jordan algorithm. Every author uses a set of deviations. E.g., if the augmented matix is as follows: |3 ... | |1 ... | | ... | in order to make the pivot entry 3 equal to 1, almost all authors prefer to exchange the Þrst and the second row. (Accordingly the strict Gauss-Jordan algorithm we have to divide the Þrst row by 3). Every author uses its own set of deviations. The program uses 11 deviations. The set of deviations used by the program covers almost all sets of deviations used in the textbooks. If we compare the solutions produced by the program with the corresponding solutions given in the textbooks, we can see that in approximately 95% of the cases the solution given by the program coincides with the solution given in the textbooks. (Or it is better - better or worse - from the point of view of the sets of deviations, because if we use the strict Gauss-Jordan algorithm, we can speak only about correct or non-correct solution). Every lecturer uses its own set of deviations. If some university wants the set of the deviations used by the program to coincide with the set of deviations used in the university, a custom build of the program can be produced. Really low price - one can see the offer soon at the web page of the program: http://www.dekovsoft.com/inverse_matrices.htm No doubt, the program will be useful for a large number of students and it will help them to better understand the Gauss-Jordan elimination method and to improve their skills. Dr.Dekov http://www.dekovsoft.com/ddekov/ > First: > You are using a freeware version of ClickTeam Install Creator in your > trial versions. > Do your paid versions use a non-freeware version of ClickTeam Install > Creator - since you are asking for money? > Second: > You do not allow for decimal entry. > Third: > You do not allow for complex numbers. > Fourth: > You simply provide a display of all the steps - like you would get in > a book. > That is there are no animations (Flash or Java) or sounds that would > help make it stick in a student¹s mind - i.e. a student who has > challenges with learning from a book is still going to have the same > challenges with your so called helpful programs! > Fifth: > You are charging $15 USD? for each little segment - are there not > calculator programs (e.g. for the TI-89 or equivalent) that one can > download free (with a one-time purchase of a calculator) that will > show the steps of a matrix inversion, etc? > Plus, at the end of it all, one still has a calculator to show for it. > >Inverse Matrices 1.03 for Windows now available! > >The program provides detailed, step-by-step solution in a > >tutorial-like format to the following problem: Given a 2x2 matrix, or > >a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse > >matrix by using the Gauss-Jordan elimination method. The check of the > >solution is given. The program is designed for university students and > >professors. > >Price: $15. > >Product homepage: > >http://www.dekovsoft.com/inverse_matrices.htm > >http://www.dekovsoft.com/ddekov/ === Subject: Uniqueness of physical objects in the physical universe. I¹ve changed some things around, made some changes, polished up the proof a little, and think that there might be something of interest here. Any serious feedback would be most appreciated. In our last episode I attempted to prove that No two objects in the physical universe are identical. Apparently, this may be reducible to a tautology. I am not sure that being a tautology matters much, because we are talking about physical properties of real objects in the universe. If you demonstrate a physical property of an object in the universe, then it dosent matter how you did it - tautology or not. A physical property is a physical property, and it matters not how you arrive at the proof of that property, as long as the demonstration is valid. Tautologies are trivial within the framework or abstract logical systems. If I say that It is a fast photon because it is a fast photon - then at least you know that you have a fast photon. Tautologies are not neccesarily trivial when you¹re talking about real objects. However, it seems that the original statement might be reducible to a question of uniqueness. So, I have the following statement to work with : ----------------------------------------------- Every object in the physical universe is unique ----------------------------------------------- So, I would like to prove that statement. It is not possible to compare every single physical object in the universe in a physics lab, so it must be proved mathematically. DeÞnitions: Physical object. Any object in the physical universe which exists. This can be a person, place or thing. A region of space/time is an object. A region of empty space is an object. Locations are therefore objects. Events are objects, as per relativity theory. If it exists in the physical universe then it is an object. Unique A physical property of an object in the universe such that if an object is unique, then there is no other object which is identical to that object. There is a physical difference between objects which are distinct, and there are no physical differences between objects which are identical. Most uniqueness proofs require 2 things, Þrst you demonstrate existence, and then you demonstrate uniqueness. However, in this case, I cannot prove that a physical object exists, it must be assumed (possibly via an axiom). So, lets assume that objects really exist in the physical universe, and try something like this- ----------------------------------------------------- Every object in the physical universe is unique Proof Suppose not Let O1 and O2 be distinct objects in the physical universe which are identical. There are 2 possible cases, 1) O1 and O2 are in separate locations 2) O1 and O2 are in in the exact same location Case 1) If O1 and O2 are in two separate locations, then they are not identical, and therefore they are both unique. Case 2) O1 and O2 are in the same location and they are also identical in every possible physical respect. They cannot be distinct, because either O1 or O2 is trivial and one of them does not really exist. If you can have O1 and O2 in the same exact location, doing the same exact thing, then let O3, O4, O(n) be identical to O1 and all in the same exact location. You now have an inÞnite number of identical physical objects in the exact same spot, which is obviously absurd. A contradiction. QED ----------------------------------------------------- Is this science, or have I Þnally cracked ? WillieK === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. Are electrons non-identical? Are atoms of the same isotope of the same element non-identical? Are molecules of the same chemical composition of the same isotopes non-identical? At some level of complexity uniqueness becomes almost inevitable on merely statistical grounds, but that is only a statistical conclusion. === Subject: Re: Uniqueness of physical objects in the physical universe. > > In our last episode I attempted to prove that No two objects in the > > physical universe are identical. > Are electrons non-identical? > Are atoms of the same isotope of the same element non-identical? > Are molecules of the same chemical composition of the same isotopes > non-identical? > At some level of complexity uniqueness becomes almost inevitable on > merely statistical grounds, but that is only a statistical conclusion. Two electrons which must be in two different locations. The location of an object which is situated in the universe is a physical property of that object. If you have two electrons in two separate location, then they cannot be identical. If you have two electrons which are both in the same exact location, then one of them must be trivial. Otherwise, you have two electrons in the same exact spot which are identical to the Þrst, you might as well add a third, fourth, and so on, even inÞnitely many if you wish. Clearly impossible. Two electrons are always unique. Consider two separate regions of space/time which appear to be identical. If the regions are in separate locations then they are fundemantally different by virtue of their locations. They are not in the same location. They cannot be identical. They are unique. All objects in the universe are unique. No two atoms are identical. No two molecules. No two houses. No two people. No two chunks of space/time are identical. No two events. Abstract atoms are identical. Real ones are not. Physicists are always taking things as identical because it makes their models purr, but this is clearly impossible. === Subject: Re: Uniqueness of physical objects in the physical universe. X-RFC2646: Format=Flowed; Original >> > In our last episode I attempted to prove that No two objects in the >> > physical universe are identical. >> Are electrons non-identical? >> Are atoms of the same isotope of the same element non-identical? >> Are molecules of the same chemical composition of the same isotopes >> non-identical? >> At some level of complexity uniqueness becomes almost inevitable on >> merely statistical grounds, but that is only a statistical conclusion. > Two electrons which must be in two different locations. The location of an > object which is situated in the universe is a physical property of that > object. > If you have two electrons in two separate location, then they cannot be > identical. > If you have two electrons which are both in the same exact location, then > one of them must be trivial. Otherwise, you have two electrons in the same > exact spot which are identical to the Þrst, you might as well add a > third, > fourth, and so on, even inÞnitely many if you wish. Clearly impossible. > Two electrons are always unique. > Consider two separate regions of space/time which appear to be identical. > If > the regions are in separate locations then they are fundemantally > different > by virtue of their locations. They are not in the same location. They > cannot > be identical. They are unique. All objects in the universe are unique. > No two atoms are identical. > No two molecules. > No two houses. > No two people. > No two chunks of space/time are identical. > No two events. > Abstract atoms are identical. Real ones are not. > Physicists are always taking things as identical because it makes their > models purr, but this is clearly impossible. Have you ever heard about bosons? -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Uniqueness of physical objects in the physical universe. > >> > In our last episode I attempted to prove that No two objects in the > >> > physical universe are identical. > >> > >> Are electrons non-identical? > >> > >> Are atoms of the same isotope of the same element non-identical? > >> > >> Are molecules of the same chemical composition of the same isotopes > >> non-identical? > >> > >> At some level of complexity uniqueness becomes almost inevitable on > >> merely statistical grounds, but that is only a statistical conclusion. > > Two electrons which must be in two different locations. The location of an > > object which is situated in the universe is a physical property of that > > object. > > If you have two electrons in two separate location, then they cannot be > > identical. > > If you have two electrons which are both in the same exact location, then > > one of them must be trivial. Otherwise, you have two electrons in the same > > exact spot which are identical to the Þrst, you might as well add a > > third, > > fourth, and so on, even inÞnitely many if you wish. Clearly impossible. > > Two electrons are always unique. > > Consider two separate regions of space/time which appear to be identical. > > If > > the regions are in separate locations then they are fundemantally > > different > > by virtue of their locations. They are not in the same location. They > > cannot > > be identical. They are unique. All objects in the universe are unique. > > No two atoms are identical. > > No two molecules. > > No two houses. > > No two people. > > No two chunks of space/time are identical. > > No two events. > > Abstract atoms are identical. Real ones are not. > > Physicists are always taking things as identical because it makes their > > models purr, but this is clearly impossible. > Have you ever heard about bosons? I have. It¹s a problem. Apparently, certain results from QM seem to indicate some strange things which I simply cant accept. If two bosons are in the same location at the same time, doing the exact same thing, then one of them is trivial. Otherwise, the combination of the two creates a third which is identical to either of the Þrst. You might as well add yet another, and another, until you have inÞnitely many bosons in thye same exact spot. This is clearly absurd. The theory is missing something. I simply cannot accept it. I¹m not going to make very many friends by saying that, but it is simply not possible. Thinking abstractly, yes - you can imagine cubes or spheres superimposed on top of each other in R3, inÞnitely many of them if you wish. No problem for a mathematician to do this. But, if we are talking about bricks, you simply cannot jam 2 identical bricks into the same location at the same time without one of them becoming trivial. === Subject: Re: Uniqueness of physical objects in the physical universe. > Every object in the physical universe is unique (Ax)(E!y)(x = y) for all x, there exists a unique y such that x = y > So, I would like to prove that statement. It is not possible to compare > every single physical object in the universe in a physics lab, so it must be > proved mathematically. Prove it in FOL from the axiom (Ax)(x = x) and the deÞnion of (E!y).P(y). > DeÞnitions: > Physical object. > Any object in the physical universe which exists. This can be a person, > place or thing. A region of space/time is an object. A region of empty space > is an object. Locations are therefore objects. Events are objects, as per > relativity theory. If it exists in the physical universe then it is an > object. I object, said the cute little object, to being objectiÞed! > Unique (E!y).P(y) is deÞned as (Ex).P(x) & (Ax)(Ay)(P(x)&P(y) -> x = y) > A physical property of an object in the universe such that if an object > is unique, then there is no other object which is identical to that object. > There is a physical difference between objects which are distinct, and there > are no physical differences between objects which are identical. Nobody is like nobody. -- > Every object in the physical universe is unique > Proof > Suppose not > Let O1 and O2 be distinct objects in the physical universe which are > identical. O1 /= O2 & O1 = O2. That¹s a contradiction. QED > There are 2 possible cases, > 1) O1 and O2 are in separate locations > 2) O1 and O2 are in in the exact same location > Case 1) > If O1 and O2 are in two separate locations, then they are not identical, > and therefore they are both unique. > Case 2) > O1 and O2 are in the same location and they are also identical in every > possible physical respect. They cannot be distinct, because either O1 or O2 > is trivial and one of them does not really exist. > If you can have O1 and O2 in the same exact location, doing the same > exact thing, then let O3, O4, O(n) be identical to O1 and all in the same > exact location. You now have an inÞnite number of identical physical > objects in the exact same spot, which is obviously absurd. A contradiction. > Is this science, or have I Þnally cracked ? No, it¹s philosophy which is never what it¹s cracked up to be. ;-) === Subject: Re: Uniqueness of physical objects in the physical universe. > > Every object in the physical universe is unique > (Ax)(E!y)(x = y) > for all x, there exists a unique y such that x = y Actually, For all objects O1 in the physical universe, there is no O2 such that O1 = O2. > > So, I would like to prove that statement. It is not possible to compare > > every single physical object in the universe in a physics lab, so it must be > > proved mathematically. > Prove it in FOL from the axiom (Ax)(x = x) and the deÞnion of (E!y).P(y). I¹m no logician, > > DeÞnitions: > > Physical object. > > Any object in the physical universe which exists. This can be a person, > > place or thing. A region of space/time is an object. A region of empty space > > is an object. Locations are therefore objects. Events are objects, as per > > relativity theory. If it exists in the physical universe then it is an > > object. > I object, said the cute little object, to being objectiÞed! objection sustained ? > > Unique > (E!y).P(y) is deÞned as > (Ex).P(x) & (Ax)(Ay)(P(x)&P(y) -> x = y) > > A physical property of an object in the universe such that if an object > > is unique, then there is no other object which is identical to that object. > > There is a physical difference between objects which are distinct, and there > > are no physical differences between objects which are identical. > Nobody is like nobody. > -- > > Every object in the physical universe is unique > > Proof > > Suppose not > > Let O1 and O2 be distinct objects in the physical universe which are > > identical. > O1 /= O2 & O1 = O2. That¹s a contradiction. QED My proving has always been skewed, Cause I¹m a ridiculous dude, And this proof may need work, I say with a smirk, And I¹ll try not to come too unglued. > > There are 2 possible cases, > > 1) O1 and O2 are in separate locations > > 2) O1 and O2 are in in the exact same location > > Case 1) > > If O1 and O2 are in two separate locations, then they are not identical, > > and therefore they are both unique. > > Case 2) > > O1 and O2 are in the same location and they are also identical in every > > possible physical respect. They cannot be distinct, because either O1 or O2 > > is trivial and one of them does not really exist. > > If you can have O1 and O2 in the same exact location, doing the same > > exact thing, then let O3, O4, O(n) be identical to O1 and all in the same > > exact location. You now have an inÞnite number of identical physical > > objects in the exact same spot, which is obviously absurd. A contradiction. > > Is this science, or have I Þnally cracked ? > No, it¹s philosophy which is never what it¹s cracked up to be. ;-) I¹ve þipped over items unique, They tell me I¹m just a math geek, But physics to warn, A notion is born, And I should get a kiss on the cheek. === Subject: Re: Uniqueness of physical objects in the physical universe. > > > Every object in the physical universe is unique > > (Ax)(E!y)(x = y) > > for all x, there exists a unique y such that x = y > Actually, For all objects O1 in the physical universe, there is no O2 such > that O1 = O2. When O1 exists, there¹s the object O1 such that O1 = O1. So there! === Subject: Re: Uniqueness of physical objects in the physical universe. > > > > > > > Every object in the physical universe is unique > > > (Ax)(E!y)(x = y) > > > for all x, there exists a unique y such that x = y > > Actually, For all objects O1 in the physical universe, there is no O2 such > > that O1 = O2. > When O1 exists, there¹s the object O1 such that O1 = O1. So there! > Baloney, even an object is an abstraction. .....billlllllly billy billy billy billy ......... I¹m talking about real objects in the physical universe in general. These are not abstractions. The word object _can_ be deployed as a name for an abstraction, if you are considering abstractions. But I¹m not. Thats the whole problem. These two worlds are so easily confused, I have a very hard time keep it straight myself. However, the coffee cup on my desk is not an abstraction, it is physical. I say it is unique among all other objects in the whole universe, and I have proved it. > When O1 exists, there¹s the object O1 such that O1 = O1. So there! Yes - you are exactly correct!! A physical object is exactly identical to itself. This is true, and it is trivial, but it is also very, very useful to know. Being trivial does not make it useless. This fact is as solid as a block of granite. It is a physical property of all physical objects in the real universe. === Subject: Re: Uniqueness of physical objects in the physical universe. Consider two objects in the physical universe, O1 and O2. I would like to prove that if O1 and O2 are oriented in Þxed positions with respect to one another, that this relationship is invariant with respect to how they are oriented with respect to the rest of the universe. Maybe it matters - maybe it dosent - and maybe I need to put it down before I snap, again. === Subject: Re: Uniqueness of physical objects in the physical universe. > > Prove it in FOL from the axiom (Ax)(x = x) and the deÞnion of (E!y).P(y). If I prove it using FOL, then I have proved an abstraction. You have no physical objects in a FOL. Physical objects exist only in the universe, not abstract space. That is the disconnect. You need a fresh start. I dont think it can be done purely with FOL. === Subject: Re: Uniqueness of physical objects in the physical universe. > > > Prove it in FOL from the axiom (Ax)(x = x) and the deÞnion of > (E!y).P(y). > If I prove it using FOL, then I have proved an abstraction. Baloney, even an object is an abstraction. === Subject: question of this system of linear congruences I have following difÞcult(to me) problem. Would you please give me some detailed solution or explanation. _________________________________________________________ Find solution of following system of linear congruences. x=11 (mod 36) x=7 (mod 40) x=32 (mod 75) === Subject: Re: question of this system of linear congruences > I have following difÞcult(to me) problem. > Would you please give me some detailed solution or explanation. > _________________________________________________________ > Find solution of following system of linear congruences. > x=11 (mod 36) > x=7 (mod 40) > x=32 (mod 75) There may be better ways but . . . Look at the Þrst equation. It says x = 11 + 36*r. Now, look at the last equation. It says x = 32 + 75*s. Put the two together and simplify a little to get 36*r - 75*s = 21. Divide by 3 to get 25*s + 7 = 12*r. Since s can¹t be even write s = 2*n + 1, substitute and simplify to get 32 + 50*n = 12*r. Since 32 and 12 are divisible by 4 but 50 isn¹t, n = 2*m. Divide by 4 to get 8 + 25*m = 3*s. Note 8 + 25 = 3*11. Let m = 1 backtrack to get s = 5 and x = 407. Check - a _must_: 407 - 11 = 11*36; 407 - 7 = 10*40 407 - 32 = 5*75 -- Paul Sperry Columbia, SC (USA) === Subject: Math is a hard discipline You can look over my work for yourself, and see that the problem I¹ve found in algebraic number theory is real. So why won¹t top math professors acknowledge my results? How is it that a paper of mine was so easily censored? (Note: I didn¹t withraw it, the editors yanked it when some sci.math¹ers sent some disparaging emails!) The simple answer has to do with your youth and their age. Know how when you were a little kid you could dream of being an astronaut, or of winning a gold medal at the Olympics? When you were a kid, your options were wide. Now for most of you, though still in many ways kids, those dreams are reasonably dead. The older you get, the fewer are your options. You as undergraduates have a bigger potential future than any math professor who is much older, who by now probably knows about where they¹ll end up in the pack. Now consider people supposedly at the top of the heap like Taylor, Wiles or Ribet. My work undermines all of their work, and throws them down into the pack with others. So it¹s far worse for people who feel accomplished in algebraic number theory, as they not only have to look what they can no longer accomplish, but be burdened by what they thought they had accomplished, but didn¹t. Many of you will think it¹s easier to go along with the old folks who¹s futures are lost, as what else can you do? Well, you are losing your own future as well, if you go into algebraic number theory, as you can learn all you want about elliptic curves and Galois Theory, and all these difÞcult subjects that older people will keep teaching you because they have nowhere to go. You¹ll expend all of that energy, and maybe get out your own work, or even get you own notoriety TODAY but the clock will be ticking. It could happen in a few months, or a few years or a few decades. You could be smiling holding your grandkids, with your pride in a history of notable accomplishment when the news comes out, and you haven nothing. It¹s math. Your fantasies, hope and dreams don¹t change it. Eventually the truth will come out, and what if it¹s a hundred years from now, before you¹re a joke? Is it worth it to you to have congratulations from old men who have nothing except the ability to use their social power to hide the truth for them to smile on you? Andrew Wiles didn¹t prove Fermat¹s Last Theorem, most of the supposed accomplishments of algebraic number theory over the last hundred years plus are nonesuch. Math is a hard discipline. Little mistakes grow, when not addressed. Over a hundred years ago some people focused on roots of monic polynomials with integer coefÞcients, and others built a lot of work on this very simple idea that has weird problems. Your professors will keep trying to ignore my work, as what future do they have? Does Andrew Wiles want to go home to his wife and kids and tell them he¹s no longer some great hero to the math world? That he accomplished nothing? That¹s what¹s brutal about mathematics! When you¹re wrong, you can have spent years, and lots of effort, and come out at the end with nothing. They¹re old men. You¹re young. Why shouldn¹t they sacriÞce your future so that they can stay comfortable? Isn¹t that the way it has always been? James Harris === Subject: Re: Math is a hard discipline Much too hard and much too disciplined for James S Harris. === Subject: Re: Math is a hard discipline alt.math.undergrad: >Much too hard and much too disciplined for James S Harris. Galdor of the Havens, who sat near by, overheard him. ŒYou speak for me also,¹ he cried. -- /The Lord of the Rings, Bk II Chapter 2 -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you¹re afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Math is a hard discipline > (Note: I didn¹t withraw it, the editors yanked it when some sci.math¹ers > sent some disparaging emails!) No, what they sent were simple counterexamples that showed that the result of the paper was false. You¹ve went on to claim that the counterexamples don¹t apply, because your paper isn¹t actually about what it said it was about. However, that also is fatal to the paper, because you do not *deÞne* what you are talking about in the paper. ... > My work undermines all of their work, and throws them down into the pack > with others. So far, you work has fallen into three categories: 1. Results that are wrong. 2. Results that only you can understand, because you haven¹t the foggiest notion how to write mathematics. 3. Results that are true, but routine or trivial. If there is anything worthwhile lurking in category #2, then the only way you will make any headway at getting anyone to notice is by LEARNING HOW TO WRITE MATHEMATICS. -- --Tim Smith === Subject: Refuting argument, algebraic number theory The following post steps through the math without much commentary, until the end. The math proof is very basic with mostly simple algebraic manipulations, and the bedrock concepts--like constants are constant--are basic as well. It is enough for those who rely on what can be mathematically proven to be correct. But those unlike them, the truth is never enough. ___JSH ------------------------------------------------------------- --------- I. Factorization What follows is in a commutative ring. Let P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where f is nonzero and note that at m=0, P(0) = u^2 f^2(3x + uf), which, of course, does not vary as m varies. So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) which requires that at least two of the a¹s must equal 0 at m=0, while one equals 3. Now let f be coprime to 3 and x so that 3x + uf is coprime to f. Since, at m=0, two of the a¹s must equal 0, it¹s convenient to just arbitrarily select a_1 and a_2 as those two. Then you have uf for the Þrst, uf for the second and 3x + uf for the third as terms that do not vary when m varies. I¹m going to emphasize that point because it¹s important. Now then, if m=1, what are the *constant* terms? They are uf, for the Þrst, uf for the second, and 3x + uf for the third. That¹s logical because they do not vary with m, so if m=1003909273, what are the constant terms? They are uf, for the Þrst, uf for the second, and 3x + uf for the third. Now divide f^2 from both sides, which gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you note that P(0)/f^2 = u^2(3x + uf), which means that now your constant terms are u for the Þrst, u for the second and 3x + uf for the third. Now then, if m=1, what are the constant terms now? They are u for the Þrst, u for the second, and 3x + uf for the third. If m = 2938479378, what are the constant terms now? They are u for the Þrst, u for the second, and 3x + uf for the third. How can the constant terms of the Þrst two go from uf to u? They must be divided by f. Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided from P(m), it is u; therefore, a_1 x + uf is divided by f, and you have a_1 x/f + u and the constant term of a_2 x + uf is uf, but when f^2 is divided from P(m), it is u; therefore, a_2 x + uf is divided by f, and you have a_2 x/f + u while the constant term of a_3 x + uf is 3x + uf, and after f^2 is divided off, it is 3x + uf, so you have a_3 x + uf so, dividing P(m) by f^2 gives P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). II. Algebraic integers Now take P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside by f^2(a_1 a_2 a_3) and you have P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). Now consider the case that m, f, and u are algebraic integers, then I have the ratios of algebraic integers: uf/a_1, uf/a_2, and uf/a_3, and now let v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 where the v¹s and w¹s are algebraic integers in each case coprime to each other. Making the substitutions I have P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). And I have from before that P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f so (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f as that is the last coefÞcient from the last term u^3 f, which proves that (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 then I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) but I still have that P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) without contradiction. III. Galois Theory Now in the ring of algebraic integers consider the possibility that a_1/f is not an algebraic integer to see if that leads to a contradiction. First, if a_1/f is not an algebraic integer and w_1 is, they obviously cannot be equal. But I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) and P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) so far simultaneously true without contradiction, so there must exist z_1, z_2, and z_3 such that w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 and z_1 z_2 z_3 = 1, where algebraically the z¹s are units, but z_1, z_2 and z_3 are not units in the ring of algebraic integers. But, if z_1, z_2 and z_3 are units algebraically but are not algebraic integers, what does that mean? It means they are not algebraic integers. Nothing more. Consider properties that cover rings like the ring of algebraic integers, and the ring of integers itself, and it can be shown that two are required: 1. No rational in the ring except 1 and -1 is a unit. 2. No non-unit number within the ring is a factor of any two integers that are coprime to each other in the ring of integers. So there are two basic factor properties of rings like the ring of integers and the ring of algebraic integers. Therefore, it sufÞces for z_1, z_2, and z_3 to be in a ring where properties 1. and 2. hold, which includes the ring of algebraic integers, and I have named that ring, the ring of objects. Since they may be in such a ring, there is no contradiction with previous results. However, at least some believe that Galois Theory shows how factors can distribute within roots, and these results contradict with that belief. Unfortunately much of modern math in the area of algebraic number theory relies on the false belief, so that major works thought to be true, are in fact false. But given the simplicity of the argument presented here, the conclusions follow without room for doubt for the mathematician. Given what I¹ve seen up to this time, it seems unlikely that the math community will just suddenly give in to what is true, but some of you still may be saved by knowing it. Those not mathematicians, of course, may just ignore the truth. But if any of you are, you will not. James Harris === Subject: Re: Refuting argument, algebraic number theory The following post steps through the math without much commentary, until the end. The math proof is very basic with mostly simple algebraic manipulations, and the bedrock concepts--like constants are constant--are basic as well. It is enough for those who rely on what can be mathematically proven to be correct. But for too many others, the truth is never enough. ___JSH ------------------------------------------------------------- --------- I. Factorization What follows is in a commutative ring. Let P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where f is nonzero and note that at m=0, P(0) = u^2 f^2(3x + uf), which, of course, does not vary as m varies. So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) which requires that at least two of the a¹s must equal 0 at m=0, while one equals 3. Now let f be coprime to 3 and x so that 3x + uf is coprime to f. Since, at m=0, two of the a¹s must equal 0, it¹s convenient to just arbitrarily select a_1 and a_2 as those two. Then you have uf for the Þrst, uf for the second and 3x + uf for the third as terms that do not vary when m varies. I¹m going to emphasize that point because it¹s important. Now then, if m=1, what are the *constant* terms? They are uf, for the Þrst, uf for the second, and 3x + uf for the third. That¹s logical because they do not vary with m, so if m=1003909273, what are the constant terms? They are uf, for the Þrst, uf for the second, and 3x + uf for the third. Now divide f^2 from both sides, which gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you note that P(0)/f^2 = u^2(3x + uf), which means that now your constant terms are u for the Þrst, u for the second and 3x + uf for the third. Now then, if m=1, what are the constant terms now? They are u for the Þrst, u for the second, and 3x + uf for the third. If m = 2938479378, what are the constant terms now? They are u for the Þrst, u for the second, and 3x + uf for the third. How can the constant terms of the Þrst two go from uf to u? They must be divided by f. Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided from P(m), it is u; therefore, a_1 x + uf is divided by f, and you have a_1 x/f + u and the constant term of a_2 x + uf is uf, but when f^2 is divided from P(m), it is u; therefore, a_2 x + uf is divided by f, and you have a_2 x/f + u while the constant term of a_3 x + uf is 3x + uf, and after f^2 is divided off, it is 3x + uf, so you have a_3 x + uf so, dividing P(m) by f^2 gives P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). II. Algebraic integers Now take P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside by f^2(a_1 a_2 a_3) and you have P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). Now consider the case that m, f, and u are algebraic integers, then I have the ratios of algebraic integers: uf/a_1, uf/a_2, and uf/a_3, and now let v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 where the v¹s and w¹s are algebraic integers in each case coprime to each other. Making the substitutions I have P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). And I have from before that P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f so (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f as that is the last coefÞcient from the last term u^3 f, which proves that (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 then I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) but I still have that P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) without contradiction. III. Galois Theory Now in the ring of algebraic integers consider the possibility that a_1/f is not an algebraic integer to see if that leads to a contradiction. First, if a_1/f is not an algebraic integer and w_1 is, they obviously cannot be equal. But I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) and P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) so far simultaneously true without contradiction, so there must exist z_1, z_2, and z_3 such that w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 and z_1 z_2 z_3 = 1, where algebraically the z¹s are units, but z_1, z_2 and z_3 are not units in the ring of algebraic integers. But, if z_1, z_2 and z_3 are units algebraically but are not algebraic integers, what does that mean? It means they are not algebraic integers. Nothing more. Consider properties that cover rings like the ring of algebraic integers, and the ring of integers itself, and it can be shown that two are required: 1. No rational in the ring except 1 and -1 is a unit. 2. No non-unit number within the ring is a factor of any two integers that are coprime to each other in the ring of integers. So there are two basic factor properties of rings like the ring of integers and the ring of algebraic integers. Therefore, it sufÞces for z_1, z_2, and z_3 to be in a ring where properties 1. and 2. hold, which includes the ring of algebraic integers, and I have named that ring, the ring of objects. Since they may be in such a ring, there is no contradiction with previous results. However, at least some believe that Galois Theory shows how factors can distribute within roots, and these results contradict with that belief. Unfortunately much of modern math in the area of algebraic number theory relies on the false belief, so that major works thought to be true, are in fact false. But given the simplicity of the argument presented here, the conclusions follow without room for doubt for the mathematician. Given what I¹ve seen up to this time, it seems unlikely that the math community will just suddenly give in to what is true, but some of you still may be saved by knowing it. Those not mathematicians, of course, may just ignore the truth. But if any of you are, you will not. James Harris === Subject: Re: Refuting argument, algebraic number theory >The following post steps through the math Oh crap. Time to update my kill Þle again. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you¹re afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Refuting argument, algebraic number theory > I. Factorization > What follows is in a commutative ring. > Let > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) Is m an element of the ring, so that you are talking about P(m) as an element of your commutative ring, or are you talking about P(m) as an element in the ring of polynomials that is associated with your commutative ring? Are f, m, and x arbitrary elements of the ring? See...right here, in the Þrst freaking section of your argument, you go off without saying what the you are doing. That¹s not how to write mathematics. -- --Tim Smith === Subject: Re: Refuting argument, algebraic number theory > The following post steps through the math without much commentary, > until the end. The math proof is very basic with mostly simple > algebraic manipulations, and the bedrock concepts--like constants are > constant--are basic as well. > It is enough for those who rely on what can be mathematically proven > to be correct. > But those unlike them, the truth is never enough. ___JSH > ------------------------------------------------------------- --------- > I. Factorization > What follows is in a commutative ring. I¹m not sure what that means. Can you please deÞne ring. And by the way, do you mean that the coefÞcents are in some particular commutative ring? Or that the roots of your polynomial lie in some ring? === Subject: Implicit Diff. X-RFC2646: Format=Flowed; Original I¹m probably going to have a couple of questions today, but here¹s the one currently getting me: A person walks along a straight sidewalk at a constant speed of 8 feet/second. A spotlight is trained on the person from across the street, which is 20 feet wide. How fast is the spotlight turning (in radians/second) when the distance between the person and the spotlight is 52 feet? I¹ve done this one a few times over, and below is the version that I feel probably came closest to getting it right - I felt conÞdent with the math: A = 20. dA = 0. C = 52. dC = 7.3846. B = 38. dB = 8. c^2 = a ^ 2 + b^ 2 2704 = 400 + b^2 48 = b c^2 = a^2 + b^2 2CdC = 2AdA + 2BdB 104dC = 16B dC = .1538461538B dC = 7.384615385 My diagram, as best I can recreate in ASCII, looks like this: Man | Side | Side C. 52 feet. B | ----- Spotlight Side A, 20 ft. the angle of the spotlight). Going for implicit diff, I get: (CdB - BdC) / C^2 = cosTheta dTheta 52(8) - 48(7.3846) / 2704 = cosTheta dTheta .0227585799 = cosThetadTheta cosTheta = A/C = 20/52 = .3846153846 .0227585799 = .3846153846 dTheta .0591723077 = dTheta And that¹s the right answer. You know, typing it out carefully helped me see the mistake I¹d been making. I was going to cancel this post, but then, I Þgure maybe it¹ll be of help to someone else doing a similar problem, so I¹m letting it go through. === Subject: Re: Implicit Diff. > ... > My diagram, as best I can recreate in ASCII, looks like this: > Man > | > Side | Side C. 52 feet. > B | > ----- Spotlight > Side A, 20 ft. You should do that sort of thing in a constant width font. === Subject: This one, I actually can¹t solve. X-RFC2646: Format=Flowed; Original The circumference of a sphere was measured to be 83 cm with a possible error of 0.7 cm. Use differentials to estimate the maximum error in the calculated surface area. Estimate the relative error in the calculated surface area. Okay, so, I /did/ Þgure out that the maximum error in the calculated surface area is 36.98744. dA / dR approx. deltaA/deltaR 8PiR = deltaA/deltaR 8PiR = deltaA / .111408 [ It took me a while to Þgure out that I should divide the .7 by 2Pi, if I was going to be dividing the circum by 2 Pi to get the radius] 8Pi(13.2098)(.111408) = deltaA 36.98744 = deltaA. However, I¹m making no headway solving for the relative error. The relative error, IIRC, is f¹(x)/f(x). So far, I¹ve tried: 2/r (as 8PiR / 4PiR^2 reduces to 2/R), I¹ve tried 13.2098 / (13.2098 +.111408), 13.2098^2 (13.2098 +.111408)^2, 8Pi(13.2098) / 4Pi(13.2098+.111408)^2. The most straightforward one - 2/R - is the one that I presumed would actually be correct, since it Þts the deÞnition of f¹/f exactly, but none of the above have yielded the correct answer. Any pointers in the right direction would be appreciated. === Subject: Re: This one, I actually can¹t solve. alt.math.undergrad: >The circumference of a sphere was measured to be 83 cm with a possible error >of 0.7 cm. ...Estimate the relative error in the calculated surface area. >... I¹m making no headway solving for the relative error. The relative >error, IIRC, is f¹(x)/f(x). Correct. Well, actually not quite. You want df(x)/f(x) or f¹(x)dx/f(x) to be precise -- the differential of the function divided by the value of the function. And it¹s usually easier to Þnd the relative error if you stay in letters rather than numbers until the very end. C = 2*pi*r S = 4*pi*r^2 = (1/pi)C^2 dS = [2C/pi] dC dS/S = (2C/pi) * dC / [(1/pi)C^2] = 2C*dC/C^2 = 2*dC/C = 2*0.7/83 = about 1.69% >So far, I¹ve tried: 2/r (as 8PiR / 4PiR^2 reduces to 2/R), I¹ve tried >13.2098 / (13.2098 +.111408), 13.2098^2 (13.2098 +.111408)^2, 8Pi(13.2098) / >4Pi(13.2098+.111408)^2. I haven¹t looked thoroughly into your algebra, but I¹ll bet there¹s a mistake in converting between radius and circumference. Remember that what is measured is not radius but circumference, so your relative error is relative to the (surface area as function of the) circumference and not the radius. (That¹s true in general. In this case it shouldn¹t matter, since C is proportional to r.) >The most straightforward one - 2/R - is the one that I presumed would >actually be correct, since it Þts the deÞnition of f¹/f exactly, but >none of the above have yielded the correct answer. Any pointers in the right >direction would be appreciated. Hmm ... let¹s see if I can do this in terms of r: S = 4*pi*r^2; dS = 8*pi*r*dr; dS/S = 2*dr/r But now you need dr/r in terms of dC/C. Since C is a linear function of r, they¹re the same, 0.7/83. 2 times that gives the Þgure I mentioned above. Does your textbook disagree? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you¹re afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: This one, I actually can¹t solve. days. My association with the Department is that of an alumnus. >The circumference of a sphere was measured to be 83 cm with a possible error >of 0.7 cm. Use differentials to estimate the maximum error in the calculated >surface area. >Estimate the relative error in the calculated surface area. >Okay, so, I /did/ Þgure out that the maximum error in the calculated >surface area is 36.98744. >dA / dR approx. deltaA/deltaR >8PiR = deltaA/deltaR >8PiR = deltaA / .111408 [ It took me a while to Þgure out that I should >divide the .7 by 2Pi, if I was going to be dividing the circum by 2 Pi to >get the radius] >8Pi(13.2098)(.111408) = deltaA >36.98744 = deltaA. I¹m going to rework the problem, because I¹m having trouble Þguring out what you did. Notice that I explain what I¹m doing to make it easier for others to Þgure out what I was thinking as I was doing it. The area in terms of the radius is A(r) = 4pi*r^2; but we want the area have that f(r)^2 = 4pi^2*r^2, so A(f) = f(r)^2/pi. That is: if we square the circumference and divide by 4pi, we get the area. So we have A(c) = c^2/pi as a function that gives the area A in terms of the circumference c. Now: the error in A is Delta(A), which is approximately equal to the differencia dA. The differential is dA = A¹(c)*dc, where dc is the differencial in c. Here, A¹(c) = 2c/pi, and -0.7 <= dc <= 0.7. So A¹(83)*(-0.7) <= dA <= A¹(83)*0.7. -166.2*0.7/pi <= dA <= 166.2*0.7/pi -116.2/pi <= dA <= 116.2/pi So the total error is bounded by +/- (116.2)/pi which is approximately +/- 36.9876. (the discrepancy between your solution and mine is that you are approximating twice: once when you approximate .7/2pi by .111408 (a bit too small), and then again when you approximate at the end). >However, I¹m making no headway solving for the relative error. The relative >error, IIRC, is f¹(x)/f(x). The relative error is approximately equal to the absolute error divided by the quantity calculated. That is: you have found that the absolute error is +/- 36.9876 square centimeters. If you use the value of 83 centimeters that you found to begin with to calculate the area, you obtain A(83) = (83)^2/pi = 6889/pi. So we have (-116.2/pi)/(6889/pi) <= dA/A <= (116.2/pi)/(6889/pi) -116.3/6889 <= dA/A <= 116.2/6889 Since 116.3/6889 is approximately 0.01687, this means that the relative error is bounded approximately by -0.01687 <= dA/A <= 0.01687 or a percentage error of about plus or minus 1.687%. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: This one, I actually can¹t solve. X-RFC2646: Format=Flowed; Response > The circumference of a sphere was measured to be 83 cm with a possible > error of 0.7 cm. Use differentials to estimate the maximum error in the > calculated surface area. Nm. I came back to it after doing some other problems, and I Þnally managed to get it. === Subject: Units, and algebraic integers Consider x^2 + 2^{sqrt(2)}x + 1 = 0. It doesn¹t have algebraic integer roots because 2^{sqrt(2)} is transcendental, so the roots can¹t be roots of a monic polynomial with integer coefÞcients. So they are not algebraic integers. But despite not being algebraic integer, they are properly units, i.e. factors of 1. That example bugs me because it highlights how weird things are in algebraic number theory that the units which are the roots of that equation are just ignored. Either they¹re tossed into the grab bag of complex numbers, or the other of transcendentals. Gauss started the ball rolling considering numbers of the form a+bi, where Œa¹ and Œb¹ are integers, now called gaussian integers in his honor. with integer coefÞcients, but then stopped. Why? Some of you may turn to complex numbers but Gauss new of complex numbers and transcendentals when he started studying gaussian integers. So why did he freaking bother? Why did he screw around with those numbers? Why didn¹t progress continue such that the roots of x^2 + 2^{sqrt(2)}x + 1 = 0 would properly be included in some integer-type category? My guess is that the ring of algebraic integers--because of the problems I¹ve highlighted--were like a pleasant drug. It seemed easy to work in them, and big results seemed within reach, except they were wrong. People settled for what was easy because it appeared to work. They didn¹t really progress the Þeld, but claimed they did. By focusing on roots of monic polynomials with integer coefÞcients, you can think you¹re proving a lot, when you make assumptions about numbers not roots of monic polynomials with integer coefÞcients, but what about the roots of x^2 + 2^{sqrt(2)}x + 1 = 0? My work shows the limitations with the arbitrary focus on roots of monic polynomials with integer coefÞcients. It is such a bad focus that the problems can be shown with basic algebra. It¹s so easy to show that I¹ve posted it yet again. The proof almost looks like some algebra homework problem from high school. It¹s that easy. Ignoring it just makes you a fool, if you¹re young. But it has been around for so long that a lot of older people who have no future starting over have good reason to block my work and act like there is no problem. So they break their own rules. I put a paper through formal peer review--at least Southwest Journal of Pure and Applied Mathematics claims they do formal peer review--and some sci.math¹ers manage to get it withdrawn immediately with a few emails. Think that¹s normal? Do you really think math editors care a lot about what sci.math¹ers say? Older people like Wiles or Ribet no longer have a future. They¹re beyond learning new things, getting over the hearbreak of error, and starting over. But many of you are just learning. You have the future ahead of you, and the possibility of real brilliance in mathematics. The older men in the club you wish to join clearly see themselves as having nothing to gain from the truth. Like yeah, sure, Andrew Wiles wants to face the world, or even Ribet or Taylor, if the truth is fully known. Right. Who would? Would you? Why should they tell you the truth? What¹s in it for them? James Harris === Subject: Re: Units, and algebraic integers Consider x^2 + 2^{sqrt(2)}x + 1 = 0. It doesn¹t have algebraic integer roots because 2^{sqrt(2)} is transcendental, so the roots can¹t be roots of a monic polynomial with integer coefÞcients. So they are not algebraic integers. But despite not being algebraic integers, they are properly units, i.e. factors of 1. Yet notice how they¹re categorized. Either they¹re tossed into the grab bag of complex numbers, or the other of transcendentals. Think about it. Gauss started the ball rolling considering numbers of the form a+bi, where Œa¹ and Œb¹ are integers, now called gaussian integers in his honor. He knew of both complex numbers and transcendentals, so why bother? with integer coefÞcients, but then progress stopped so that I can give you this example over a hundred years later: x^2 + 2^{sqrt(2)}x + 1 = 0 Why? Why didn¹t progress continue such that the roots of x^2 + 2^{sqrt(2)}x + 1 = 0 would properly be included in some integer-type category? My guess is that the ring of algebraic integers--because of the problems I¹ve highlighted--was like a pleasant drug. It seemed easy to work in the ring, and get big results. Trouble is, they¹re not right. People settled for what was easy because it appeared to work. They didn¹t really progress the Þeld, but claimed they did. There was no real progress in this area since Gauss. That¹s over a hundred years of stagnation with false claims of results. By focusing on roots of monic polynomials with integer coefÞcients, you can think you¹re proving a lot, when you make assumptions about numbers not roots of monic polynomials with integer coefÞcients, but what about the roots of x^2 + 2^{sqrt(2)}x + 1 = 0? My work shows the limitations with the arbitrary focus on roots of monic polynomials with integer coefÞcients. It is such a bad focus that the problems can be shown with basic algebra. So people argue with me, not because I¹m wrong--after all, you can check the math for yourself--but because I¹m right, and they don¹t like what¹s mathematically correct because of the social implications. But what¹s mathematically correct is so easy to show that I¹ve posted it yet again. The proof almost looks like some algebra homework problem from high school. It¹s that easy. Ignoring it just makes you a fool, if you¹re young, but might seem smart if you¹re older, and either a grad student with a lot invested or even more so if you¹re an actual professor. Besides, it¹s a problems that¹s freaking over a hundred years old! Maybe professors feel they aren¹t responsible! And starting over is so hard. Can you imagine? Having to unlearn and then re-learn, as the mathematical foundations are set aright? Easier to block and act like there is no problem. So they break their own rules. I put a paper through formal peer review--Southwest Journal of Pure and Applied Mathematics claims they do formal peer review--and some sci.math¹ers managed to get it withdrawn immediately with a few emails. Think that¹s normal? Do you really think math editors typically care a lot about what sci.math¹ers say? Older people like Wiles or Ribet probably don¹t have a future. They¹re beyond learning new things, getting over the hearbreak of error, and starting over. It¹s hard to start over, to get over the emotions, and accept truths that don¹t sit well, even in mathematics. But you are different. Many of you are just learning. You have the future ahead of you, and the possibility of real brilliance in mathematics. The older men in the club you wish to join clearly see themselves as having nothing to gain from the truth. Like yeah, sure, Andrew Wiles wants to face the world, or even Ribet or Taylor, if the truth is fully known. Right. Who would? Would you? Why should they tell you the truth? What¹s in it for them? James Harris === Subject: Frieze Patterns I have the following problem: Consider the inÞnite strip pattern: ... TTTTTT ... Prove that every element of the symmetry group of the above pattern has the form t^n or (t^n)r where t is a suitable translation and r is some suitable reþection. I can see clearly why every element of the symmetry group has either of those forms, when you take t to be the translation Œone unit¹ (or one ŒT¹) to the right or left and then r as a reþection through the vertical line passing through any one of the T¹s, but how on earth do I go about proving this rigorously? Can any one help with proving this properly? Gary. === Subject: Re: Is this system causal? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9VD09i32277; >x[n] --> [system] -->y[n] >input outpu >System: y[n] + y[n+1] = x[n] >A system is causal if for every choice of n0, the output sequence >(y[n]) value at the index n=n0, depends only on the input sequence >(x[n]) values for n<=n0 >I think it¹s caual, but just want to corroborate. Any help is >Luis Rewriting your system gives y[n+1]=x[n]- y[n] Compare n+1 left hand ,n right hand ; so you¹re right to think it is causal system (time oriented), === Subject: NEED HELP -- Þnd a basis of a subspace! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9VD0BV32375; Can anybody help me to solve the following question: Show that the plane deÞned by x-2y=0 is a 2-dimensional subspace of R^3(R cubic). Give a basis for this subspace,and extend it to a basis for R^3 === Subject: Re: Please HELP me with this riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UMBNF28788; maybe he used an icerope to kill himself === Subject: graphing caculator by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UMBbq28915; i¹m in algebra and we do a lot of stuff with graphing the only problem is I don¹t have one so is there a good online graphing caculator!!!!??????? === Subject: Any general idea on proving a topology that is a discrete topology? X-RFC2646: Format=Flowed; Original === Subject: Re: Any general idea on proving a topology that is a discrete topology? days. My association with the Department is that of an alumnus. Prove that every singleton is open? -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: math software/games for primary kids whats a good (freeware if possible) software to teach maths for year 5-6? im useless at math and Þnd it hard when my kids ask me questions about maths homework :( ive heard a program called mathamagic (spell) advertised is there a demo of that somewhere, or a torrent Þle ;) === Subject: Looking for paper X-RFC2646: Format=Flowed; Original Hi all, I am looking for a paper by Ashcraft C.C titled A Vector Implementation of the Multifrontal Method for Large Sparse, Symmetric Positive DeÞnite Computer Services Technical Report. No luck in Þnding the actual paper though. Wonder if anyone has any suggestion as where I can Þnd the paper. Thuan Seah === Subject: proof by induction X-RFC2646: Format=Flowed; Original how the hell do you do it? bill === Subject: Re: proof by induction X-RFC2646: Format=Flowed; Response > how the hell do you do it? > bill You show its true for 1. Then you show that if it is true for n, it is true for n+1. Therefore it is true for all numbers. Proof: You have shown it is true for 1. You have shown that if it is true for n, then it is true for n+1. So if its true for 1 (which it is), it must be true for 2. If its true for 2, it must be true for 3. If its true for 3, it must be true for 4. And so on ... === Subject: Re: proof by induction X-RFC2646: Format=Flowed; Response >> how the hell do you do it? >> bill > You show its true for 1. > Then you show that if it is true for n, it is true for n+1. > Therefore it is true for all numbers. > Proof: > You have shown it is true for 1. > You have shown that if it is true for n, then it is true for n+1. > So if its true for 1 (which it is), it must be true for 2. > If its true for 2, it must be true for 3. > If its true for 3, it must be true for 4. > And so on ... how do you show it¹s true for n? === Subject: Re: proof by induction >>> how the hell do you do it? >>> bill >> You show its true for 1. >> Then you show that if it is true for n, it is true for n+1. >> Therefore it is true for all numbers. >> Proof: >> You have shown it is true for 1. >> You have shown that if it is true for n, then it is true for n+1. >> So if its true for 1 (which it is), it must be true for 2. >> If its true for 2, it must be true for 3. >> If its true for 3, it must be true for 4. >> And so on ... > how do you show it¹s true for n? You dont. You show that IF it works for n, it ALSO works for n+1. In other words you end up with two axioms: True_for(1). True_for(n) True_for(n+1). See http://mathworld.wolfram.com/ PrincipleofMathematicalInduction.html gtoomey === Subject: Question about Gomory-Cut for an application I have to Þnd an integer solution. I use the simplex-algorithm and the Gomory-cut. It works Þne, but it is to slowly. There are multiply opportunities to make a Gomory-cut. Which Gomory-cuts should I take, to get a solution in shortest time ? Where can I get information about that (in the internet) ? Ulrich === Subject: Permutations question Can anyone suggest a simple method to calculate the number of permutations of 4 letters chosen from the word MUDGEE? === Subject: Re: Permutations question X-RFC2646: Format=Flowed; Original > Can anyone suggest a simple method to calculate the number of permutations > of 4 letters chosen from the word MUDGEE? Two steps: First, calculate the number of permutations of 4 letters selected from MUDGE (ie with one E) Second, calculate the number of permutations of 4 letters selected from MUDGEE, but containing two Es. Add them together. You do step 1 - the easier step - and what you can of step 2, and I will help you with the rest. === Subject: permutations Can anyone suggest a simple method to calculate the number of permutationsof 4 letters chosen from the word MUDGEE? === Subject: Re: Differential Geometry with Maple 9 And we can have the same thing for free with GRTensor. I¹ve been using for a long time :) http://grtensor.phy.queensu.ca/ > atlas - powerful Maple package for differential geometry calculations > is now available for Maple 9 at > http://www.graphtree.com/Maple/atlas/index.htm > The package works with manifolds, mappings, embeddings, submersions, > p-forms, tensor Þelds etc. > The user may concentrate on geometrical problem not on the > programming. > Standard mathematical notations accepted in modern differential > geometry are used in the package. > All calculations are as coordinate free as possible. > Some calculations with non-numerical dimension are available. > Detailed help Þles, templates and examples are available. > Any differential geometry problem can be solved by one and the same > simple solving scheme. > Besides that new differential geometry powerful tool (atlas 2D/3D > Wizard) is available for Maple 9 at > http://www.graphtree.com/Maple/atlasWizard/index.htm > With this Wizard you can solve typical differential geometry problems > in any 2D or 3D coordinate system: > - plane curves > - space curves > - surface geometry > -- > Norb Bobrin, Graph Tree Ltd. > Maple & Mathematica Packages, Programs & Math Arts > http://www.graphtree.com/ === Subject: Re: question on Mathematica package loading. > I didn¹t try to follow everything that you did because it¹s a lot of > work. But you always need a usage message for any function that is > to be exported from a package. No, this is only a typical way to introduce a symbol in the non-private context. Marcus -- Can an object be called a work of art if it can also be used to clean the stove? -- W. Allen === Subject: Re: question on Mathematica package loading. > MathGroup was not updated for several days during that period. I don¹t see > why your posting wouldn¹t have been accepted and using bug wouldn¹t have > made any difference. news group. That is about 6 days after I posted it. I am sure it was censored and rejected for some reason. Only reason I can think of is that I put the word Œbug?¹ in the subject line. I wish at least I was told why that is. It is difÞcult to have a discussion on a newsgroup where posts take days to show up, if they do, and it seems that only one person is controlling that group and controlling what people are allowed to say. That is why many people are starting to post Mathematica questions here instead. I do not see why a Mathematica discussion should be censored at all. > When working with packages I always work with a package in the .nb format, > use initialization cells and use the Create Auto Save Package option when > Þrst saving the notebook. It is easier to edit a notebook than a .m Þle. > Then after changing the package I always 1) save it, 2) Quit the kernel and > 3)reload the package. I never run into any difÞculties that way. > David Park > djmp@earthlink.net > http://home.earthlink.net/~djmp/ and I am now following steps you outlined. It seems to work much better now. === Subject: Re: question on Mathematica package loading. > I have posted this question 3 days ago (oct 27) to the Mathematica > news group, and as of today oct 30, it still has not shown up, I assume > it was not allowed on that group. > It might be possible becuase I put the word Œbug?¹ in the subject. > So I am removing the word Œbug¹ from the subject and trying this > newsgroup. > I am posting the question here in the hope someone can help. > cheers, > Bill > ----- post ------- > I am having hard time understanding why Mathematica is doing > the following. > To summarize, I create a simple package with one > function test3[] in it. Load the package, and can call > the function ok. I removed the Œusage¹ line, and try to call > the function again, but it will not work AS expected. But this > results in a new symbol added by the same name of the function, > again as expected. > But I then put the Œusage¹ line back into the package, > clean the environment again, and reload the package, > but I still can not access the package function again, it > is still calling a fake symbol instead. > ----- package Þle strange.m ---- > BeginPackage[strange`] > test3::usage=test3 <----- (1) > Begin[`Private`] > test3[x_]:=x^3; > End[] > EndPackage[]; > ---- now test the package, we see we can call strange`test3[] > < Names[strange`*] > {test3} > strange`test3[4] > 64 > ---------------------- > Now edit the Þle strange.m, and remove the line > marked by (1), so we now get > BeginPackage[strange`] > > Begin[`Private`] > test3[x_]:=x^3; > End[] > EndPackage[]; > ---- Now remove all symbols, and reload, we see the > ---- name test3[] is not visible > Remove[Global`*]; Remove[strange`*]; Names[strange`*] > {} > Now load the packge again and try to call test3 > Get[strange.m]; Names[strange`*] > {} > You see, NO names in package. Now try to call strange`test3[], > this will cause a new symbol to be created instead as expected: > strange`test3[4] > test3(4) > Ok, now it seems to Þx this, all what I have to do is > to remove all the names in the strange context, and add the > usage line back to the package, and reload, and then > it should work. However, it still does not work. > Added line(1) back, SAVED the package to disk again (Þle->save) > then did > Remove[Global`*]; Remove[strange`*]; Names[strange`*] > {} > < Names[strange`*] > {test3} > -- so the name is back. OK. Now I would expect to be able > -- to call it again and it should work as before. BUT it > -- does not! > strange`test3[4] > test3(4) <----- This should come back as 64 > -- Why has not Mathematica call the correct function again?? > -- checked that there is NO global test3 function anywhere: > Names[Global`test3] > {} > -- So what is going on?? Let us see how your Þrst package deÞnition works: BeginPackage[strange`] test3::usage=test3 Begin[`Private`] test3[x_]:=x^3; End[] EndPackage[] First BeginPackage makes strange` the current context; the usage line isn¹t necessary per se, but its purpose is to create the symbol test3 in the current context strange` -- that is, it creates the symbol with the full name strange`test3. Next we enter the context strange`Private` and evaluate the expression test3[x_]:=x^3. The parser sees two symbols: test3 and x. It searches for test3 and successfully Þnds it, because strange` is on the context search path ($ContextPath) now, which is one of the functions of BeginPackage. This means that you can access strange`test3 simply as test3. Therefore, no new symbol strange`Private`test3 is created (but the parser does create the new symbol strange`Private`x). On the next step, when you remove the usage line, the symbol strange`test3 isn¹t created; instead, when the evaluator reaches the test3[x_]:=x^3 line, it cannot Þnd the test3 symbol and creates the new symbol strange`Private`test3. After the End[] statement you cannot access this symbol directly as test3, because strange`Private` is not put on the context search path (it¹s not the task of Begin/End) but you can still access it by its full name, e.g. as strange`Private`test3[3]. Finally, when you load the package with the usage line added again, the symbol strange`test3 will be created once more, but what will happen when we reach the test3[x_]:=x^3 line? The key point is that the current context is always searched before the contexts on $ContextPath. Therefore, the evaluator Þnds the symbol strange`Private`test3 (left from the previous execution of the package) and redeÞnes it! In other words, that line will still look like strange`Private`test3[strange`Private`x_]:=strange`Private`x^3 just as during the second run (with the usage line removed) -- the deÞnition isn¹t attached to the strange`test3 symbol. The bottom line is that all three examples work as expected and to get the desired behaviour you should do Remove[strange`Private`test3]. Maxim Rytin m.r@inbox.ru === Subject: Re: question on Mathematica package loading. Please see below. > The bottom line is that all three examples work as expected and to get > the desired behaviour you should do Remove[strange`Private`test3]. The problem is that I did do the following: Remove[strange`*]; Names[strange`*] {} I thought when I did that, I was deleting all references to that package. You see, I get that there are no strange` symbols left. Now I understand from you that some strange` references are still there (the ones with strange`private` context). This really bothers me for 2 reasons. 1. I was using a wild card to delete everything below strange`, so this logically should delete everything below strange` context 2. If I am able to type Remove[strange`Private`test3], then this means I have access to the private symbols of a package. This defeats the whole puprose of having a private part of a package. I think Mathematica package scoping design is not well thought. I think for large applications this can introduce many subtle problems. In addition, Mathemtica should have a command (and a button) that one can use to reinitialize the whole environment to the state it was in when one loaded Mathematica initially, without having to manually remove symbols from every context, and now from the private parts of packages as well to clean things out, Similar to the excellent Maple command called Œrestart¹. Currently I seem to shut down the application and restart it when I am not sure about something. Takes too much time. Why does not Mathematica have such a command? === Subject: Re: question on Mathematica package loading. > In addition, Mathemtica should have a command (and a button) that > one can use to reinitialize the whole environment to the state it > was in when one loaded Mathematica initially [...] > Why does not Mathematica have such a command? Hi Bill, Kernel -> Quit Kernel? Marcus -- All women become like their mothers. That is their tragedy. No man does. That¹s his. -- Oscar Wilde === Subject: Re: MuPad 3.0 Calculus of Variations - syntax > Here is a simple Calculus of Variations problem.... > Which curves can give an extremum of the functional > v(y(x)) = integral from 0 to 1 of ((y¹)^2+12xy)dx, y(0)=0, y(1)=1 ? > The answer is y=x^3 > Let¹s try to solve this.... > L := (diff(y(x),x))^2 + 12*x*y(x); > res := detools::euler(L, [x], y); > This gives the Euler equation 2.y([x,x])-12x (equation #1) > Now to check x^3 is the solution to equation #1 > y := x -> x^3; > op(res,1)+op(res,2); > This gives Error: Illegal operand [_power] during evaluation of y > 2*diff(y(x),x,x)-12*x; //gives 0 as expected. > This is a simple case, but I would like to be able to do this type of > veriÞcation of the Euler equation in more complicated cases without having > to type in the output from detools::euler. (werner.seiler@iwr.uni-heidelberg.de) the developer of the detools library (he has no news access). He asked me to post this reply: The reason for the described problem is very simple. The notation `y([x,x])¹ appearing in the output is a shorthand for derivatives that is only used by the methods in the DETools library. The MuPAD kernel knows nothing about it and hence substituting some function for y will lead to the described error. In principle, a simple option in the call of `detools::euler¹ should have the effect that the output is not in this abbreviated form but in the standard `diff¹ notation of MuPAD. Unfortunately I had to discover that currently there is a bug preventing the proper recognition of this option. This will be corrected in the next release of MuPAD (and the option will be described in the manual). Thus for the moment the only solution of the problem is a little hack using some internal data of the DETools library. If the following procedure is applied to the output of `detools::euler¹ (before any other call to a method in the library is performed!), then the output is transformed in the standard `diff¹ format and the veriÞcation works as the user intends. makeDiff := proc(r) local DF; begin DF := detools::data[DFout]; DF::convert_to(DF(r),diff); end_proc; Applied on your example: >> L := (diff(y(x),x))^2 + 12*x*y(x); 2 12 x y(x) + diff(y(x), x) >> res := makeDiff(detools::euler(L, [x], y)); 2 diff(y(x), x, x) - 12 x >> y := x -> x^3; x -> x^3 >> res 0 -- *---* MuPAD -- The Open Computer Algebra System *---*| |*--|* Ralf Hillebrand (tonner@mupad.de) *---* === Subject: Symbolic eigenvectors X-RFC2646: Format=Flowed; Original I understand that Mathematica calculates the eigenvectors of a matrix containing symbolic elements using polynomial interpolation. Could someone kindly point me to the algorithms used for this sort of thing? Jerry. === Subject: Why doesn¹t MuPAD do...? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA3FuLF10215; tell me why!!!! function=-abs(x-2)/(x+1)^2-1/(x+1); solve(function=0,x); MuPAD doesn¹t give me the output....i¹m still waiting ...