mm-1067 >Let f(z)=(z+i)/(z-2i). Prove that |Int( f(z)) | <= sqrt(2) where the >integral is taken over the contour C which is the line z=t , t in [-1,1]. >So I used the estimate lemma (also called fundamental lemma). >So we first bound |f(z)|. >We notice that |z|=|t|<=1 so |z|<=1 , which implies -|z|>= -1. >|z+i|<=1+1 = 2. and |z-2i|>= |2i|-|z| >= 2-1 = 1. >so |z+i|/|z-2i| <= 2, i.e |f(z) | <=2. >The arclength of z=t in [-1,1] is 2. >So using the estimate lemma we get |int(f(z) dz) | <= 2* arclength = 2* 2 = 4. >I can't seem to improve to lower the bound to sqrt(2), any ideas? You can get sharper bounds than the triangle inequality in both cases. Draw a picture and look at |z + i| for z in [-1,1]. Ditto for |z- 2i|. Other than that, you are on the right track. --Lynn === Subject: Re: Prove this bound on integral (Lynn) So following your hint, we write z as z=x+iy, since z is in [-1,1] then z is real so z=x with x in [-1,1]. Now we consider |z+i|. |z+i|=|x+i|= sqrt(x^2 + 1), this is maximum when x=1 so we get |z+i|<= sqrt(1+1)= sqrt(2). Now |z-2i|= |x-2i| = sqrt(x^2+4), this is minimm when x=0 so we get |z-2i| >= 2. Then we apply estimate lemma and we're done. Correct? === Subject: Re: Prove this bound on integral (Lynn) >So following your hint, we write z as z=x+iy, since z is in [-1,1] >then z is real so z=x with x in [-1,1]. Now we consider |z+i|. >|z+i|=|x+i|= sqrt(x^2 + 1), this is maximum when x=1 so we get >|z+i|<= sqrt(1+1)= sqrt(2). >Now |z-2i|= |x-2i| = sqrt(x^2+4), this is minimm when x=0 so we get >|z-2i| >= 2. Then we apply estimate lemma and we're done. >Correct? I think you know it is correct. :-) --Lynn === Subject: Re: Bible Codes Resurrected? > As a follow up, I notified Art Levitt, the author of the website that > I pointed to, of your objection. He contacted Haralick and it appears > that Haralick indeed was unaware of the other lists that you > mentioned. > He said, I will take a look to see if there are any additional > web site. > This is a good example of the quality of information that gets around > on this subject. > Of course Haralick is aware of the other lists because they are > described in my paper published in Statistical Science. He also > http://cs.anu.edu.au/~bdm/dilugim/StatSci/emanuel_rep.html > So what you report is nonsense. > I directly quoted an email that Haralick cc'ed me. It appears to me > that he only made a mistake by forgetting about those extra lists > he already used. I wouldn't infer from this that he deliberately > ignored these lists because he knew that they would produce bad > results - after all, what purpose would that serve him, as it would > only be a matter of time when you or someone else would try it out and > disprove him. And maybe they produce good results? I too would like to > see him perform the experiment with the lists on your website. > They will only provide a few extra appellations. Anyway, note that > I was responding to a specific claim and did not say that Haralick > should have used or not used any particular data. Rather, the whole > idea of his experiment is broken. A little on this below. > More interestingly, Haralick has even collected appellations himself > that are not in the two lists he used. > ones? > Other way around. Haralick knows lots of valid appellations that are > not used in his experiment. These include appellations he has collected > himself for his own (other) experiments. They also include many that > everyone knows. For example, the original experiment used appellations > of the form Rabbi Yakov, but not Rav Yakov. As you know, the latter > is an extremely common expression; why is it not used? There are many > more examples. Why is the much smaller and more irregular set of > appellations that work in War and Peace the correct set of additional > appellations? And why is it correct to only add appellations when > (according to our hypothesis) a major problem with the original data was > the selective inclusion of doubtful appellations? (There are even two > that nobody ever found a single example of in the literature.) That is for certain true that there are many more possible appellations, at least twice as many as you pointed out in the Rabbi, Rav example. > And that's only the start of the problems with this appallingly bad > paper of Haralick. > What are some other problems? I appreciate the work that you have done > the subjectivity of the lists. If there are flaws in the Haralick > I'll summarise what Haralick did. There are two sets of data and > a third constructed from them. > D1: data used by WRR for Genesis > D2: data used by BMMK (McKay & al.) for War and Peace > D3. the union of D1 and D2 > Note that D1 and D2 have a very large overlap. > D1 works well in Genesis and very poorly in War and Peace. > D2 works well in War and Peace and fairly poorly in Genesis > The question is: how does D3 behave? > Using the analysis method of WRR, D3 works worse in Genesis > than D1 did, and worse in War and Peace than D2 did. I did not know this. I wish Haralick would have mentioned this in his paper. I got the impression from reading his paper that the reason that he used a different metric was because it was more accurate, not that it did not work on the combined lists. This fact doesn't only prove that the experimental methodology of WRR was flawed as you showed in MBBK; it shows that the only way to argue the thesis of WRR is to do what Witztum does on his website, which is to try to convince everyone that the WRR list is superior than your list, which is completely subjective. > Somehow or other, Haralick found a different method of analysis > for which D3 works about the same or slightly better in Genesis > than D1 did, whereas D3 still works worse in War and Peace > than D2 did. (Actually he found 4 new methods of analysis, some > of which behave this way and some of which do not.) > Haralick claims this proves something. I think it only proves that > wishful thinking is strong medicine. > The fact is that the result is extremely sensitive to both the data > and the method of analysis. Seemingly tiny changes (like adding > or deleting a single word, or replacing a mathematical expression > by another with the same properties) can make the result jump > up or down dramatically. Factors of 10 or more are nothing. > I know of tiny changes to the analysis method (far far more minor > than the changes between WRR and Haralick) that make the > result for D2 in War and Peace 100 times better than the result > for D1 in Genesis. What does this prove? Nothing. What does > it prove if Haralick can come up with a method showing some > other behaviour? Nothing. > Besides this, Haralick has not established a significance level for > his results. How likely is it for his observations to occur by chance? > He doesn't even ask the question. (I have reason to believe that > they are not very unlikely.) And let's not get started on the errors > in his new methods. > Brendan. realize how much the choice of the metric has such a big effect on the results of the experiment, as I have never tested it out myself. This fact and the fact that there are probably infinitely many reasonable compactness measures casts a lot of doubt on the methodology of Haralick's experiment, implying that there still is wiggle room in the choice of a metric - even though Haralick only tested 32 metrics on list one, someone else doing the experiment might test another 32 possible metrics on list one and get drastically different results. Haralick would have to give an explanation for the choice of 32 metrics that he tested - and let's not go down that road again as we've been there before regarding the choice of the appellations. Regardless of these critisisms, I would still like to give Professor Haralick the courtesy of a chance to respond to your critisisms (assuming that he is reading this thread) and would be interested in hearing the results of his same experiment adding all of the lists posted on your website (that Avi Norowitz mentioned before). Anyway, I'm sure that you also have criticisms of the other paper (only 4 pages) regarding the phrase Cursed is bin Laden and revenge belongs to the Messiah found in an ELS in the Bible? I and probably others reading this thread would probably be interested in hearing them. Craig === Subject: Partial products It is possible that the following problem has been solved already by someone, but I couldn't find any refence with Google. Given is a set of n=2^m numbers {x(0),...x(n-1)}. It is required to compute all partial products P(0) to P(n-1), such that each includes n-1 numbers from the set: P(0) = 1 *x(1)*...*x(n-1) P(1) =x(0)* 1 *...*x(n-1) ... P(n-1)=x(0)*x(1)*...* 1 The only operation permited is multiplication. I need an efficient algorithm to generate these partial products using minimum number of multiplications. My use of the above is in a factorization algorithm. A.D. === Subject: Re: Partial products > It is possible that the following problem has been solved already by > someone, but I couldn't find any refence with Google. > Given is a set of n=2^m numbers {x(0),...x(n-1)}. It is required > to compute all partial products P(0) to P(n-1), such that each > includes n-1 numbers from the set: > P(0) = 1 *x(1)*...*x(n-1) > P(1) =x(0)* 1 *...*x(n-1) > ... > P(n-1)=x(0)*x(1)*...* 1 > The only operation permited is multiplication. I need an efficient > algorithm to generate these partial products using minimum number > of multiplications. > My use of the above is in a factorization algorithm. > A.D. I hoped that this problem would attract some attention from the readers of this group. In the meantime I made some progress on solving it. Obviously, naive direct method requires n*(n-1) multiplication operations. Desired approach will pre-compute some values and combine them to obtain the products P(i) using minimum number of operations. The pre-computed values can be stored in a 2-dimensional matrix. Then partial products P(i) can be obtained with minimum number of additional operations by cleverly combining the matrix elements. Let pre-computed value be stored in a matrix d(m,2^m) in the following manner: for i=0 to n-1 d(0,i)=x(i) next i ' for j=1 to m-1 k=2^(m-j)-1 for i=0 to k d(j,i)=d(j-1,2*i)*d(j-1,2*i+1) next i next j The filling up of the matrix d(j,i) needs n-2 operations. The following step will combine the elements from the matrix into the products P(i). For example take m=4, n=16. One gets the following: P(0)=d(0,1)*d(1,1)*d(2,1)*d(3,1) which uses 3 operations. However P(1) needs only 1 operation if d(0,1) only is replaced in the above product with d(0,0). It can be shown that P(2) will need 2 operations, P(3) again 1 operation, P(4) 3 operations, and so on ... After counting all the operations I found that the total is 3*(16-2) including those needed to calculate d(j,i) products. It looks that optimum method needs 3*(n-2) operations, versus n*(n-1) operation needed by direct inefficient method. For n=2^7 the difference is quite significant, 378 versus 16256 operations! What I still need is an algorithm to combine the elements from matrix d(m,2^m) to obtain P(i)'s. Any suggestion is appreciated. A.D. === Subject: Re: Partial products This best way to deal with this is the Forward-Backward Algorithm. we construct to arrays: 1st array is: x(0) x(0)*x(1) x(0)*x(1)*x(2), ......... 2st array is: x(n-1) x(n-1)*x(n-2) x(n-1)*x(n-2)*x(n-3), ........... It takes O(n) to construct these 2 arrays. every partial product is the product from 2 number in 1st and 2st array. and it takes constant time to locate the 2 numbers, so the whole process takes O(n). BTW, what are you going to make use of this algorithm? I would like to know it. === Subject: Re: infinitesimal calculation ? > I also understand that because the function 1/0 is not defined in the > standard |R, It cannot be defined in the extended |R if we assume > true the standardisation axiom. Therefore, what does happen if we are > not taking this axiom true? > Does anybody knows if the non standard analysis is self consistent > without the standardisation axiom? (It helps me to understand the > embedding constraints of the standard sets/properties into the new > one) It must be -- there obviously exist models in which the axiom holds, and since it is a non-logical axiom (i.e. not a tautology or first-order validity) there must be models in which it doesn't. This, of course, is assuming that it is independent from the other axioms (in the sense that you can't derive it from the other two). > However, I keep learning on the extended algebra and the possible > representations we can build on these new (for me ; ) sets. May be > some one has a pointer towards a condensed lecture on this aspect? > (more condensed than keisler?). You may want to look at Abraham Robinson's book on the topic, or Handbook of Analysis and Its Foundations by Eric Schechter if condensed is what you're looking for. > One of the questions concerning the representations is if there exists > an isomorphism to the elements of the representation or a class of > elements ? e.g. infinite sequence of standard real numbers. Well, you can't have an order-isomorphism, since the non-standard reals are non-Archimedean. For the same reason, you can't have an (in my cursory examinations), the non-standard reals are something like the direct product of three continuum sized groups with some relations and a multiplicative structure thrown in. So the best I can think of is *maybe* a set theoretic isomorphism, which only requires the cardinalities of the involved sets to be equal. I don't know what the cardinality of the set of non-standard reals is, but I suspect it is that of the continuum. 'cid 'ooh === Subject: Re: JSH: Understanding polynomials ... [context] > > > which is a simple example of non-polynomial factorization as you > > > have factored x-1 into non-polynomial factors. > > > > You have factored a polynomial in x^{1/2} into two polynomials in > > x^{1/2}. > > So that is a polynomial factorisation. > Harris stated that x^{1/2} + 1 is not a polynomial, which is obviously Is it false? Depends. I clearly stated (see the explanation below) that it is a polynomial in some contexts, but not in other contexts. > Your explanation: It is a polynomial in x^{1/2}. Which is true, but silly > and irrelevant. And I added that it is not a polynomial in x. But that is the point. In the context it is *not* irrelevant. In a way (x - 1) = (x^{1/2} + 1)(x^{1/2} - 1) is a polynomial factorisation. When you can provide a polynomial factorisation in such a way (where the polynomial variable is 0 for x = 0), you can visibly inspect the constant terms. When you can *not* do that, you can *not* make such visible inspections. (And I may note in addition that constant term is used normally in the context of polynomials.) And JSH tried to explain why in his factorisation of P(x) into: (5 a1(x) + 7)(5 a2(x) + 7)(5 a3(x) + 7) there was no clear visible connection between the constant term of P(x) and the 7's in the factorisation, by arguing that there are fractional exponents involved in the a_i(x), and so it was not a polynomial factorisation. The explanation was wrong for two reasons. First, in the a_i(x) there are no fractional exponents of x involved. Second, if there had (positive) fractional exponents been involved here, there *would* have been a clear visible connection. If the a_i(x) could be written as polynomials in some x^{1/p} for some natural p, the constant term of P(x) must be 343. This is reminiscent to an error JSH made quite a few years ago when where ... stands for a (possibly infinite) series of a_rs.x^r.y^s with rational r and s. -- 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: what determines upper bound on eigenvalues of a covariance matrix? >>Suppose M is an nxn covariance matrix of n normally distributed random >>variables. >>Thus, M is symmetric and positive semidefinite. >>Let z be the largest eigenvalue of M. In the limit that the size n of >>M goes to infinity, >>what are necessary and sufficient conditions such that: >> i) z/n -> 0? >> ii) 0 < z/n < infinity? >> >> That will always be true except in the trivial case where M = 0. >> >> I'm afraid not so, Robert. Let e be a random variable with mean zero >> and unit variance. >> Let X(n) be the n-tuplet whose every component equals e. >> Then let M(n) = var(X(n)). In this case, M is the nxn square matrix >> whose every >> element is 1. The largest eigenvalue of M in this case is z=n. Thus, >> z/n=1 is strictly >> non-zero for all n. >> >> How is that not so? Last time I looked, 0 < 1 < infinity. >(ii) is true, but (i) is not so. >Roderick The that I was referring to was (ii), not (i). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: induction vs Cantor II Sorry, forgot sci.math > Summary from the first thread: > L_1 is a list of reals. This creates an implied > mapping (injection) F_1 from the naturals to the reals. > D_n is the Cantor number, CN (anti-diagonal number), > that can be formed using the mapping F_n. > Let L_n+1 be a list of reals by inserting D_n into L_n > at row 2n. > Row i in L_n+1 = Row i in L_n if i < 2n. > Row i in L_n+1 = Row i-1 in L_n if i > 2n. > This process is clearly an inductive process. > It creates a new mapping for each natural > number n. > For all j > n, D_n is in L_j at row 2n. > The union of all the L_n, (a countable set) contains > all the D_n. Let's call the union CR (Cantor reals). > There's one problem. Someone is bound to come along > and claim a bijection exists between N and CR and they > are bound to claim that there is no CN to prove that > it's not a bijection N<->R. > We can stop them by acting like 3-year-old babies and > call them names, we can deny its true by utilizing the > results of Cantor's proof, or we can just prove that > they are wrong by coming up with the CN. I think the > best way is to actually support Cantor's proof and > show how to construct the CN. > Someone is bound to claim this is a bijection: > Each n can be associated with a unique real by using > the real it is mapped to with F_n. This is the same > real as the real its maps to with F_i for all i > n/2. > They will then try to claim that the CN associated > with that bijection exists in that bijection. They > are going to try and claim that their bijection > includes it's own CN! > I'm not sure how to construct the CN in this case. > If you can help me, You can post a response. I don't > like a bunch of garbage in my e-mail but if you think > you know me, e-mail me. > GMN-75-23335. > P.S. I'm not looking for anything but the CN so I can > prove those anti-Cantor people wrong. === Subject: Re: induction vs Cantor II Poker Joker says... >Sorry, forgot sci.math >> Summary from the first thread: >> L_1 is a list of reals. This creates an implied >> mapping (injection) F_1 from the naturals to the reals. >> D_n is the Cantor number, CN (anti-diagonal number), >> that can be formed using the mapping F_n. >> Let L_n+1 be a list of reals by inserting D_n into L_n >> at row 2n. >> Row i in L_n+1 = Row i in L_n if i < 2n. >> Row i in L_n+1 = Row i-1 in L_n if i > 2n. Okay. So the original list L_1 is r_1 r_2 r_3 . . . L_2 is the list (where d_1 is the anti-diagonal number of L_1) r_1 d_1 r_2 r_3 . . . L_3 is the list (where d_2 is the anti-diagonal number of L_2) r_1 d_1 r_2 d_2 r_3 . . . In the limit, we have L_infinity which will look like the following r_1 d_1 r_2 d_2 r_3 d_3 r_4 d_4 r_5 d_5 . . . This list contains all the elements of L_1 plus all the anti-diagonal numbers d_1, d_2, etc. If you use Cantor's diagonalization procedure on *this* list, you get a new number d_infinity that is not equal to r_1, nor d_1, nor r_2, nor d_2, etc. d_infinity is guaranteed to be different from r_1 in the first decimal place, different from d_1 in the second decimal place, different from r_2 in the third place, different from d_2 in the fourth place, etc. -- Daryl McCullough Ithaca, NY === Subject: Re: induction vs Cantor II Poker Joker says... >> Summary from the first thread: >> L_1 is a list of reals. This creates an implied >> mapping (injection) F_1 from the naturals to the reals. >> D_n is the Cantor number, CN (anti-diagonal number), >> that can be formed using the mapping F_n. >> Let L_n+1 be a list of reals by inserting D_n into L_n >> at row 2n. >> Row i in L_n+1 = Row i in L_n if i < 2n. >> Row i in L_n+1 = Row i-1 in L_n if i > 2n. >> This process is clearly an inductive process. >> It creates a new mapping for each natural >> number n. >> For all j > n, D_n is in L_j at row 2n. >> The union of all the L_n, (a countable set) contains >> all the D_n. Let's call the union CR (Cantor reals). >> There's one problem. Someone is bound to come along >> and claim a bijection exists between N and CR and they >> are bound to claim that there is no CN to prove that >> it's not a bijection N<->R. Yes, there is a bijection between N and CR, if by CR you mean the set of all reals r such that r is in L_n for some natural number n. You define a merged list L_merged as follows: 1. Let positions 1, 3, 5, 7, etc. in L_merged be filled in with elements of L_1. 2. Let positions 2, 6, 10, 14, etc. in L_merged be filled in with elements of L_2. 3. Let positions 4, 12, 20, 28, etc. in L_merged be filled in with elements of L_3. In general, real number i of list L_j will appear at position 2^{j-1} * (2i - 1) of list L_merged. For example, real number 1 of list L_1 will appear at position 2^0 * (2-1) = 1. Real number 2 of list L_1 will appear at position 2^0 * (4-1) = 3. Real number 3 of list L_3 will appear at position 2^2 * (6-1) = 20. etc. Then you perform the regular Cantor diagonalization procedure to come up with a real that is not in the list L_merged (and therefore, it is not in any of the lists L_1, L_2, etc.) -- Daryl McCullough Ithaca, NY === Subject: Re: induction vs Cantor II >> D_n is the Cantor number, CN (anti-diagonal number), >> that can be formed using the mapping F_n. Some additional explanation is in order here. D_n is really any number that can be generated by any method as long as the construction gaurantees that D_n is not already in L_n. For example, D_n may match one of the reals in the list along the diagonal. That's okay as long as the construction gaurantees a unique real. Note that the construction itself can change at each iteration. The only real restriction is that at each iteration, we describe the construction that we are using and that it generates a unique real. I don't see anything that restricts any given real from showing up on the list. === Subject: Re: induction vs Cantor II Poker Joker says... > D_n is the Cantor number, CN (anti-diagonal number), > that can be formed using the mapping F_n. >Some additional explanation is in order here. >D_n is really any number that can be generated by >any method as long as the construction gaurantees >that D_n is not already in L_n. For example, D_n >may match one of the reals in the list along the >diagonal. That's okay as long as the construction >gaurantees a unique real. Okay, so let's say that a function f is a diagonalizing function if for any infinite list of reals L, f(L) returns a real that is not on list L. One such diagonalizing function is the one Cantor described: f(L) = that real whose nth digit is equal to d+1 (mod 10) where d = the nth digit of the nth real of list L. If you pick one fixed diagonalizing function f, then we can let L_1 be our original list, and then let L_2 be the result of adding f(L_1) to L_1, and let L_3 be the result of adding f(L_2) to L_2, etc. Then if we let L_merged be the merged list of L_1, L_2, L_3, etc. then f(L_merged) will produce a real that is not on any of the lists. -- Daryl McCullough Ithaca, NY === Subject: (image processing) how to match the perceived energy for two images... Hi all, This question is regarding two images. One image is the original image; another image is obtained by proforming downsampling, upsampling, filtering, color-space conversion, gamma conversion, etc, on the original image. After all of these steps, I want the two images to have same perceived energy... (or maybe I should say that I want the two images to appear similar viewing from several meters away... ) currently the processed image is a lot darker than the original image. I should defintely scale up the second image in order to make it brighter and perceived similar to the original one... it is just a question about the scaling factor, after so many linear and non-linear steps, I have already lost track of the scaling factor... Now in order to match the two images and make them look similar when viewing from several meters away, I think I want to be able to take any N x N block from both images Red plane, compare the the mean of two blocks, and make the second one scale-up to the original image with the same mean. The problem with my scheme is that if I choose N larger, then two adjacent different blocks may have two different scaling-factor, and hence it has visible artifacts... If I choose N small, say N=1, then it is in fact I am cheating by making the second image point by point the same as the first one, then I lose all information of my processing of the second image... So are there any good schemes that can let me retain the processed information of the second image, while scale up its brightness and make their perceived energy the same when viewing at several meters distance? === Subject: Re: Operator ambiguity, Escultura > As this example shows, in the domain of complex numbers, > the square root of the product is not equal to the product of the > square roots and the same for quotients. And how do you define the square root in the domain of complex numbers? > As a consequence of the Fundamental Theorem of Algebra, > every complex number has exactly two complex square roots. In any field, every element has at most two square roots. Besides, you don't need the Fundamental Theorem of Algebra in order to prove that every complex complex number has a square root. Jose Carlos Santos === Subject: Re: Operator ambiguity, Escultura >> As this example shows, in the domain of complex numbers, >> the square root of the product is not equal to the product of the >> square roots and the same for quotients. > And how do you define the square root in the domain of complex > numbers? A square root of a complex number a+bi is a solution of the equation z^2 = a+ib. >> As a consequence of the Fundamental Theorem of Algebra, >> every complex number has exactly two complex square roots. > In any field, every element has at most two square roots. Besides, > you don't need the Fundamental Theorem of Algebra in order to prove > that every complex complex number has a square root. === Subject: Re: Operator ambiguity, Escultura >As this example shows, in the domain of complex numbers, >the square root of the product is not equal to the product of the >square roots and the same for quotients. >>And how do you define the square root in the domain of complex >>numbers? > A square root of a complex number a+bi is a solution of the equation > z^2 = a+ib. reproduced above) something about the square root. Now, I know very well what _a_ square root is, thank you very much, but when you use the expression the square root you are making a choice: you are choosing one among the two square roots of a complex number (different from 0). Only after that you can state that in the domain of complex numbers, the square root of the product is not equal to the product of the square roots. So, I repeat my question: how do you define the square root in the domain of complex numbers? Jose Carlos Santos === Subject: Re: Operator ambiguity, Escultura Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Only after that you can state that in the domain of complex numbers, >the square root of the product is not equal to the product of the square >roots. >So, I repeat my question: how do you define the square root in the >domain of complex numbers? Any definition of which root is *the* root will have the problem stated, so the statement is true regardless of which definition you use. -- Richard === Subject: Re: Operator ambiguity, Escultura >>Only after that you can state that in the domain of complex numbers, >>the square root of the product is not equal to the product of the square >>roots. >>So, I repeat my question: how do you define the square root in the >>domain of complex numbers? > Any definition of which root is *the* root will have the problem > stated, so the statement is true regardless of which definition you > use. I know that. I just wanted to know which definition was N. Silver the square root of the product is not equal to the product of the square roots. Note that he did not add regardless of which definition you use. Jose Carlos Santos === Subject: Re: Operator ambiguity, Escultura I think that the error is in the following step: sqrt(1/-1) = 1/i Its a bit like saying sqrt((-3)(-3))=-3 And then proclaiming that you have taken the positive square root. Clearly this is incorrect. > I would like to pull out and highlight something interesting that E. > E. Escultura posted a few days ago, which I'd guess he's probably > talked about many times before, but I just noticed it and think it's > neat. > First some more preamble as *by convention* as has been noted when I > brought up the subject of operator ambiguity before, sqrt(x) is taken > to be positive. > So, by the convention, sqrt(4) = 2, and that's good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. > Naively then, you may believe that you can just say, take the positive > of the square root but as Escultura showed, that doesn't work: > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction. > You see, the ambiguity in the square root operator still remains, > despite the convention. > It doesn't work to just try and always take the positive as > Escultura's example shows so clearly. > Who has the resolution? I'm curious as to whether or not any of you > think you can answer. > James Harris === Subject: Re: Operator ambiguity, Escultura >I would like to pull out and highlight something interesting > at E.E. Escultura posted a few days ago, which I'd guess > he's probably talked about many times before, but I just > noticed it and think it's neat. You just ache to show mathematics is not on firm foundations, don't you?! Because you are stumped at times does not mean that mathematics is contradictory. On the contrary, it means you are not working hard enough, do not know enough, or are not mathematically sophisticated enough to overcome your mathematical difficulties. This elementary stuff has been worked over many, many times before by those much smarter than you or I. It would be prudent to give up the ghost now. You have not a snowball's chance of ever winning in this forum. === Subject: Re: Kick classic in shins, win prizes > Misner, Thorne and Wheeler have undoubtedly each accomplished more > than I might hope to should I live another 50 years. (Ob kowtow). > But one thing they had _not_ accomplished as of 1973 was to edit their > massive textbook free of casual carelessness. > My complaint is with Box 2.3 -- Differentials. > In this box we are made acquainted with the exterior derivative or > gradient Df (boldface small-d italic-f), which is contrasted with > the bad old differential df (italic-d italic-f). The new model is > superior to the old in every way, and comes with a lifetime powertrain > warranty! Let's see what it can do... > A symbol v (bold-v) represents a particular infinitesimally long > displacement of the point P (script-P), which is the argument of the > function f = f(P). An inner product is formed and set equal to > a symbol @_v f (partial d subscript bold-v, italic-f), which is the > change of f in going from the tail of v to its tip. Mighty like a > differential, no? In fact we can identify the old notation with the > new term-by-term: > = @_v f == = df > Each slot on the rhs and lhs is occupied by the same object, but with > a different symbol (grad f is the gradient, dX the infinitesimal > displacement of the argument, df the infinitesimal change in the > function). Notice in particular that df and Df are _not_ mapped > to each other, but are distinct objects. > Ok. Now let's go back and catalogue the sophistry enlisted in a > casual effort to calumniate the old notation: > Df ... is a more rigorous version of the elementary concept of > differential i.e., df. > So they claim. Too bad the two symbols simply represent different > objects in relabled but otherwise identical equations; neither more or > less elementary or rigorous than the other. > In elementary textbooks... > Nice touch. ;-) > ...one is presented with differential df as representing 'an > infinitesimal change in the function f(P)' associated with some > infinitesimal displacment of the point P; but one will recall that the > displacement of P is left arbitrary, albeit infinitesimal. Thus df > represents a change in f in some unspecified direction. > Heavens! This means that df is itself a variable or a function of > its argument (the change in P), as _well_ as an infinitesimal! What > are we to do! > But this is precisely what the exterior derivative Df represents... > Nope. Df represents an _engine_ to _calculate_ such a thing, given an > infinitesimal displacement as input, same as the gradient. In fact, > it _is_ the gradient, as they just finished telling us. Seems they > forgot. > Thus Df, before it has been pierced forming the inner product> to give a number, represents the change in > f in an unspecified direction. > Same comment. Df represents a linear machine (a tensor, as they well > should know) for calculating the change in an arbitrary direction, > given an input. Nor is it not some brave new object never before > seen, it is the gradient. > And now get ready for the coup-de-grace.. > The only failing of the textbook presentation... > Forgot to add elementary to exclude the book we are reading? > ...then, was its suggestion that Df was a scalar or a number; the > explicit recognition of the need for specifying a directiong v to > reduce Df to a number shows that in fact Df is a 1-form, the > gradient of f. > It takes time to appreciate such confusion. It really must be savored: > and particularly the interesting claim that the textbook > presentation suggests that Df a scalar or number. How on > earth could it! _Df_ (bold-d f) is part of a _new_ notation -- > which in truth looks confusingly similar to a part of the old notation > (df) which it does not map to. > The bad old textbook could only have been making a suggestion about > _df_ , not Df, and it would have been perfectly correct to do so: df > is indeed a scalar, albeit a variable one -- a concept we might > perhaps recall -- and is equivalent to the weird @_v f symbol in the > new notation, not to Df. This would possibly account for df acting > more like a number or scalar, and Df acting more like a 1-form or > tensor? And we haven't made the variable infinitesimal any less > variable by relabeling it to include an argument in its symbol, > anymore than relabeling y as y(x) makes y a definite number. > The symbols Df and df represent distinct objects in different notation > schemes, and attempting to identify them leads to the unsettling idea > that somebody must have been confused somewhere -- surely it must have > been the feeble, unrigorous (and conveniently absent) authors of this > bad old notation? But the confusion instead lies in the mind > attempting to make the identification. LOL, MTW's dual prose is certainly interesting, had to try it once. While component based notation may have some fault's, I can always get out my ruler and clock, and figure out where the I am. Understandly, mathematical physicists seek notational concistion, especially to eliminate the unimportant distractions, and that's good. I guess it's up to the new students to determine what fits best, maybe differential algebra will evolve legs, and out run tensor analysis. Ken S. Tucker === Subject: Re: Kick classic in shins, win prizes <... LOL, MTW's dual prose is certainly interesting, > had to try it once. > While component based notation may have some > fault's, I can always get out my ruler and clock, > and figure out where the I am. > Understandly, mathematical physicists seek notational > concistion, especially to eliminate the unimportant > distractions, and that's good. > I guess it's up to the new students to determine what > fits best, maybe differential algebra will evolve legs, > and out run tensor analysis. Huh! I was almost afraid to go back and look at this. I expected at least a few how dare you criticize MTW, you lickspittle's. But nothing... === Subject: Orthogonal complement in L^2(M) Hi all, Let M be a measure space and consider the space L^2(M) of all square-integrable functions from M into the reals. A proof from a book that I was reading uses the following fact implicitely: if E is a closed subspace of L^2(M), then its orthogonal complement is different form {0}, unless, of course, E = L^2(M). My question is: is this obvious? I know that L^2(M) is a Hilbert space and that in any Hilbert space that statement is true, but the authors of the book make no mention at all to Hilbert spaces (they don't appear at the subject index). So, is the statement obvious for Hilbert spaces of the form L^2(M)? I know that every Hilbert space is isomorphic to a Hilbert space of that type, but there could be an easier prove in this context. Jose Carlos Santos === Subject: Re: Orthogonal complement in L^2(M) >Let M be a measure space and consider the space L^2(M) of all >square-integrable functions from M into the reals. A proof from a book >that I was reading uses the following fact implicitely: if E is a closed >subspace of L^2(M), then its orthogonal complement is different form >{0}, unless, of course, E = L^2(M). My question is: is this obvious? >I know that L^2(M) is a Hilbert space and that in any Hilbert space >that statement is true, but the authors of the book make no mention at >all to Hilbert spaces (they don't appear at the subject index). >So, is the statement obvious for Hilbert spaces of the form L^2(M)? >I know that every Hilbert space is isomorphic to a Hilbert space of >that type, but there could be an easier prove in this context. I can't think of anything easier than this, which doesn't use the fact that it's L^2(M). Let y be a member of the Hilbert space not in E, and d = dist(y,E). There is a sequence x_n in E with ||x_n - y|| -> d. Use the parallelogram law to show that x_n is a Cauchy sequence, so by completeness it has a limit x with ||x - y|| = d, and x is in E. Then x - y will be in the orthogonal complement of E. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Orthogonal complement in L^2(M) >>Let M be a measure space and consider the space L^2(M) of all >>square-integrable functions from M into the reals. A proof from a book >>that I was reading uses the following fact implicitely: if E is a closed >>subspace of L^2(M), then its orthogonal complement is different form >>{0}, unless, of course, E = L^2(M). My question is: is this obvious? >>I know that L^2(M) is a Hilbert space and that in any Hilbert space >>that statement is true, but the authors of the book make no mention at >>all to Hilbert spaces (they don't appear at the subject index). >>So, is the statement obvious for Hilbert spaces of the form L^2(M)? >>I know that every Hilbert space is isomorphic to a Hilbert space of >>that type, but there could be an easier prove in this context. > I can't think of anything easier than this, which doesn't use > the fact that it's L^2(M). > Let y be a member of the Hilbert space not in E, and d = dist(y,E). > There is a sequence x_n in E with ||x_n - y|| -> d. Use the parallelogram > law to show that x_n is a Cauchy sequence, so by completeness it > has a limit x with ||x - y|| = d, and x is in E. Then x - y will be in > the orthogonal complement of E. that there was something missing here. Jose Carlos Santos === Subject: Re: Orthogonal complement in L^2(M) >Let M be a measure space and consider the space L^2(M) of all >square-integrable functions from M into the reals. A proof from a book >that I was reading uses the following fact implicitely: if E is a closed >subspace of L^2(M), then its orthogonal complement is different form >{0}, unless, of course, E = L^2(M). My question is: is this obvious? >I know that L^2(M) is a Hilbert space and that in any Hilbert space >that statement is true, but the authors of the book make no mention at >all to Hilbert spaces (they don't appear at the subject index). >So, is the statement obvious for Hilbert spaces of the form L^2(M)? >I know that every Hilbert space is isomorphic to a Hilbert space of >that type, but there could be an easier prove in this context. >> I can't think of anything easier than this, which doesn't use >> the fact that it's L^2(M). >> Let y be a member of the Hilbert space not in E, and d = dist(y,E). >> There is a sequence x_n in E with ||x_n - y|| -> d. Use the parallelogram >> law to show that x_n is a Cauchy sequence, so by completeness it >> has a limit x with ||x - y|| = d, and x is in E. Then x - y will be in >> the orthogonal complement of E. >that there was something missing here. No, surely it's as Chapman suggests - the authors of a book titled Representations of Compact Lie Groups are just assuming the reader already knows a little basic real analysis. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: Orthogonal complement in L^2(M) > Hi all, > Let M be a measure space and consider the space L^2(M) of all > square-integrable functions from M into the reals. A proof from a book > that I was reading uses the following fact implicitely: if E is a closed > subspace of L^2(M), then its orthogonal complement is different form > {0}, unless, of course, E = L^2(M). My question is: is this obvious? It's true: whether it's obvious is a matter of opinion. > I know that L^2(M) is a Hilbert space and that in any Hilbert space > that statement is true, but the authors of the book make no mention at > all to Hilbert spaces (they don't appear at the subject index). > So, is the statement obvious for Hilbert spaces of the form L^2(M)? > I know that every Hilbert space is isomorphic to a Hilbert space of > that type, but there could be an easier prove in this context. Seems unlikely. Which book is it? Is it a physics book? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Orthogonal complement in L^2(M) >>Let M be a measure space and consider the space L^2(M) of all >>square-integrable functions from M into the reals. A proof from a book >>that I was reading uses the following fact implicitely: if E is a closed >>subspace of L^2(M), then its orthogonal complement is different form >>{0}, unless, of course, E = L^2(M). My question is: is this obvious? > It's true: whether it's obvious is a matter of opinion. >>I know that L^2(M) is a Hilbert space and that in any Hilbert space >>that statement is true, but the authors of the book make no mention at >>all to Hilbert spaces (they don't appear at the subject index). >>So, is the statement obvious for Hilbert spaces of the form L^2(M)? >>I know that every Hilbert space is isomorphic to a Hilbert space of >>that type, but there could be an easier prove in this context. > Seems unlikely. > Which book is it? Is it a physics book? Not at all. It's Representations of Compact Lie Groups by Theodor Broecker and Tammo tom Dieck. It appears at the end of the proof of the theorem 2.6 (chapter III), which states that given a compact self-adjoint automorphism of L^2(M), the direct sum of the eigenspaces is dense and, furthermore, given r > 0, the direct sum of those eigenspaces whose eigenvalue has absolute value greater or equal than r is finite-dimensional (in other words, the spectral theorem for compact self-adjoint operators). Jose Carlos Santos === Subject: Re: Orthogonal complement in L^2(M) Originator: grubb@lola >Let M be a measure space and consider the space L^2(M) of all >square-integrable functions from M into the reals. A proof from a book >that I was reading uses the following fact implicitely: if E is a closed >subspace of L^2(M), then its orthogonal complement is different form >{0}, unless, of course, E = L^2(M). My question is: is this obvious? >> It's true: whether it's obvious is a matter of opinion. >I know that L^2(M) is a Hilbert space and that in any Hilbert space >that statement is true, but the authors of the book make no mention at >all to Hilbert spaces (they don't appear at the subject index). >So, is the statement obvious for Hilbert spaces of the form L^2(M)? >I know that every Hilbert space is isomorphic to a Hilbert space of >that type, but there could be an easier prove in this context. >> Seems unlikely. >> Which book is it? Is it a physics book? >Not at all. It's Representations of Compact Lie Groups by Theodor >Broecker and Tammo tom Dieck. It appears at the end of the proof of >the theorem 2.6 (chapter III), which states that given a compact >self-adjoint automorphism of L^2(M), the direct sum of the eigenspaces >is dense and, furthermore, given r > 0, the direct sum of those >eigenspaces whose eigenvalue has absolute value greater or equal than r >is finite-dimensional (in other words, the spectral theorem for compact >self-adjoint operators). I would assume that knowledge of Hilbert spaces and the fact that L^2(M) is a Hilbert space would be a prerequisite for such a book. In fact, I would assume that elementary facts about Hilbert spaces would be used almost without comment. --Dan Grubb === Subject: Re: Orthogonal complement in L^2(M) >Let M be a measure space and consider the space L^2(M) of all >square-integrable functions from M into the reals. A proof from a book >that I was reading uses the following fact implicitely: if E is a closed >subspace of L^2(M), then its orthogonal complement is different form >{0}, unless, of course, E = L^2(M). My question is: is this obvious? >> It's true: whether it's obvious is a matter of opinion. >I know that L^2(M) is a Hilbert space and that in any Hilbert space >that statement is true, but the authors of the book make no mention at >all to Hilbert spaces (they don't appear at the subject index). >So, is the statement obvious for Hilbert spaces of the form L^2(M)? >I know that every Hilbert space is isomorphic to a Hilbert space of >that type, but there could be an easier prove in this context. >> Seems unlikely. >> Which book is it? Is it a physics book? > Not at all. It's Representations of Compact Lie Groups by Theodor > Broecker and Tammo tom Dieck. It appears at the end of the proof of > the theorem 2.6 (chapter III), which states that given a compact > self-adjoint automorphism of L^2(M), the direct sum of the eigenspaces > is dense and, furthermore, given r > 0, the direct sum of those > eigenspaces whose eigenvalue has absolute value greater or equal than r > is finite-dimensional (in other words, the spectral theorem for compact > self-adjoint operators). I'm sure these guys regard that as assumed knowledge. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Orthogonal complement in L^2(M) >Which book is it? Is it a physics book? >>Not at all. It's Representations of Compact Lie Groups by Theodor >>Broecker and Tammo tom Dieck. It appears at the end of the proof of >>the theorem 2.6 (chapter III), which states that given a compact >>self-adjoint automorphism of L^2(M), the direct sum of the eigenspaces >>is dense and, furthermore, given r > 0, the direct sum of those >>eigenspaces whose eigenvalue has absolute value greater or equal than r >>is finite-dimensional (in other words, the spectral theorem for compact >>self-adjoint operators). > I'm sure these guys regard that as assumed knowledge. I did not reply yesterday because I did not have the book at hand then, but now that I have it I can confirm that you're right (together with David C. Ullrich and Daniel Grubb): the authors are assuming that knowledge. In fact, a few pages before that theorem they state (concerning the space H of all continuous functions from a compact Lie group G into the complex numbers, equiped with the norm derived from the standard scalar product) that [c]ompletion of H for this norm yelds the Hilbert space H^ = L^2(G) of square integrable functions. So, some basic facts concerning HIlbert spaces are assumed here. Jose Carlos Santos === Subject: Re: The-State-of-the-Art in Mathematics Hey, Escultura, I disagree. When you say dichotomy, do you mean trichotomy? What are your vague numbers? What is your definition of a real number? Do you consider the nonstandard constructions of the surreal numbers, or hyperreals or colonies of germs and germinal functions? You appear to promote geometry. I'm getting somewhat defensive about all this. I would like to not share with you. Please go off about the real numbers. You think they are decimals? Write the decimal form of square root of two. You claim to solve all paradoxes? What are they? Where are paradoxes these days? The flaw is axiomatization, but not the root probabilistic flaw. What do you think about iota? Not you, Ullrich. Ullrich is a troll. Please explain Banach-Tarski, and the bleeding points on the line. The real numbers are complete. The state of the art in mathematics is symplectic integration and geometric algebra, and not manifold theory, although there are some fantastic combinatorial results out in the polynomial realm. Mathematics tends to lead physics by some time. I encourage you to go off about the real numbers. Please explain your opinion of what is wrong with contemporary standard treatments of the real numbers. This sci.math group is full of reviewers. The real number is a scalar. Ross Finlayson === Subject: The function x^x Hi. Does anyone know what the graph of the function f(x) = Imag(x^x) for x < 0, that is, the imaginary part of x^x (e^(x ln(x)) with principal branch of ln(x)), looks like? I haven't really been able to get a good idea of it's behavior from looking at a few points. === Subject: Re: The function x^x > Does anyone know what the graph of the function f(x) = Imag(x^x) for > x < 0, that is, the imaginary part of x^x (e^(x ln(x)) with principal > branch of ln(x)), looks like? Mark Meyerson has written something about the graph of x^x. Have a look at http://www.usna.edu/MathDept/mdm/homepage.html -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The function x^x > Hi. > Does anyone know what the graph of the function f(x) = Imag(x^x) for x > < 0, that is, the imaginary part of x^x (e^(x ln(x)) with principal > branch of > ln(x)), looks like? I haven't really been able to get a good idea of > it's behavior from looking at a few points. The negative axis, taken as a branch cut for the principal branch of the complex function log, can be parametrized as: x=r*exp(i*pi), 0<=r<+inf. Plug x in exp(x*log(x)) to get: exp(x*log(x))=exp(r*(-1)*(r+i*pi))= exp(-r^2)*exp(-i*r*pi) Separate real and imaginary parts, Im(x^x)=exp(-r^2)*sin(-r*pi) Now graph that for 0<=r<+inf. -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: The function x^x > Hi. > Does anyone know what the graph of the function f(x) = Imag(x^x) for x > < 0, that is, the imaginary part of x^x (e^(x ln(x)) with principal > branch of > ln(x)), looks like? I haven't really been able to get a good idea of > it's behavior from looking at a few points. > The negative axis, taken as a branch cut for the principal branch of the > complex function log, can be parametrized as: > x=r*exp(i*pi), 0<=r<+inf. > Plug x in exp(x*log(x)) to get: > exp(x*log(x))=exp(r*(-1)*(r+i*pi))= I suppose that, instead of (r+i*pi), you meant to have (log(r) + i*pi). etc. > exp(-r^2)*exp(-i*r*pi) > Separate real and imaginary parts, > Im(x^x)=exp(-r^2)*sin(-r*pi) > Now graph that for 0<=r<+inf. With the change noted above, you would have gotten r^(-r)*sin(-r*pi), which one could graph, as you said, for 0<=r<+inf. I suggest writing the answer as Im(x^x) = (-x)^x * sin(pi*x) for x <= 0. David === Subject: Re: The function x^x >Hi. >Does anyone know what the graph of the function f(x) = Imag(x^x) for x >< 0, that is, the imaginary part of x^x (e^(x ln(x)) with principal >branch of >ln(x)), looks like? I haven't really been able to get a good idea of >it's behavior from looking at a few points. >>The negative axis, taken as a branch cut for the principal branch of the >>complex function log, can be parametrized as: >>x=r*exp(i*pi), 0<=r<+inf. >>Plug x in exp(x*log(x)) to get: >>exp(x*log(x))=exp(r*(-1)*(r+i*pi))= > I suppose that, instead of (r+i*pi), you meant to have (log(r) + i*pi). > etc. and my head wasn't working properly :-) >>exp(-r^2)*exp(-i*r*pi) >>Separate real and imaginary parts, >>Im(x^x)=exp(-r^2)*sin(-r*pi) >>Now graph that for 0<=r<+inf. > With the change noted above, you would have gotten r^(-r)*sin(-r*pi), > which one could graph, as you said, for 0<=r<+inf. I suggest writing > the answer as > Im(x^x) = (-x)^x * sin(pi*x) for x <= 0. > David -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Infinite number of football teams single-elimination tournament. All even numbered teams play an odd number >team in the first round. The winners eliminate the losers then they in turn >play one another and eliminate another half. This goes on hundreds, if not >thousands, of times. When is there less than an infinite amount of teams? If >there is always an infinite amount of teams no matter how many rounds are >played, how do we know since an infinite amount has been eliminated? > Suppose that the higher-numbered team always wins. Then the teams in each > round go like this: > (1) 1 2 3 4 5 6 7 8 9 10 11 12 ... > (2) 2 4 6 8 10 12 ... > (3) 4 8 12 ... > ... Let m_0 = 1 and m_j be the smallest numbered undefeated team of the j-th match. The winner of the tournament is lim(j->oo) m_j unless (m_j)_j has no limit, as in the example above, when the tournament is a draw. A limited option giving the tax payers a big break for not having to fund infinitely many stadiums, is to let m_0 = 1 and m_(j+1) be the first team with team number > m_j to defeat m_j. Again lim(j->oo) m_j is the winner unless (m_j)_j has no limit resulting in a drawn tournament. === Subject: Discontinuous everywhere functions in Maple? Is there anyway to define an expression or function in Maple that is discontinuous everywhere, say the characteristic of Q? I looked in the index for some way to input notation x in R and x in R Q but couldn't find it. I'd imagine there would be extreme computational difficulties in determining irrationality of numbers. I'm basically trying to see if it can do Lebesgue integration and how far it will actually go. Jason === Subject: Re: Discontinuous everywhere functions in Maple? > Is there anyway to define an expression or function in Maple that is > discontinuous everywhere, say the characteristic of Q? I looked in the index > for some way to input notation x in R and x in R Q but couldn't find > it. I'd imagine there would be extreme computational difficulties in > determining irrationality of numbers. Try this: f:=x->if type(x,rational) then 1 else 0 end if; OTOH, the comp.soft-sys.math.maple newsgroup is the appropriate place for questions concerning Maple. Jose Carlos Santos === Subject: Re: Discontinuous everywhere functions in Maple? > Try this: > f:=x->if type(x,rational) then 1 else 0 end if; Hmm, I've encountered this thing before ... Question: who is the inventor of this function. And for what purpose? Han de Bruijn === Subject: Re: Discontinuous everywhere functions in Maple? >> Try this: >> f:=x->if type(x,rational) then 1 else 0 end if; > Hmm, I've encountered this thing before ... > Question: who is the inventor of this function. And for what purpose? It's called the Dirichlet function. I don't know its original purpose, but it is frequently encountered as a standard example of a function that is Lebesgue integrable but not Riemann integrable. In fact, it's an example of what is called a simple function in measure theory, since it takes on only finitely many values, each on a measurable set. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Discontinuous everywhere functions in Maple? >Try this: >f:=x->if type(x,rational) then 1 else 0 end if; >>Hmm, I've encountered this thing before ... >>Question: who is the inventor of this function. And for what purpose? > It's called the Dirichlet function. [ ... ] > is Lebesgue integrable but not Riemann integrable. Maybe you're in the position then to answer another question: > Can somebody enumerate all different integration techniques? > I've heard now about Lebesgue integration, Stieltjes integral .. And Riemann integrable too. > As a physicist, I always calculated integrals the common way, > by employing the main theorem of calculus. Or numerically . > What's the beef with these definitions? And how many are there? I posted this in a thread about Cantor's Theory but got no answer. Han de Bruijn === Subject: Re: Discontinuous everywhere functions in Maple? >>Try this: >>f:=x->if type(x,rational) then 1 else 0 end if; >Hmm, I've encountered this thing before ... >Question: who is the inventor of this function. And for what purpose? >> It's called the Dirichlet function. [ ... ] >> is Lebesgue integrable but not Riemann integrable. This is too trivial an example; it can be changed on a set of measure 0 to become Riemann integrable. One can produce a function on the unit interval which only takes on the values 0 and 1, which is Lebesgue measurable, and for which the Lebesgue integral can be easily computed, which cannot be so modified, as there is no interval where it does not take on both values on a set of positive measure. >Maybe you're in the position then to answer another question: >> Can somebody enumerate all different integration techniques? >> I've heard now about Lebesgue integration, Stieltjes integral .. There are limits of these, such as Cauchy and Denjoy. And there is the Riemann-Hellinger integral, which has many applications. I doubt if there is an easy answer to this; for probability, usually what is wanted is the Lebesgue-type integral for an arbitrary probability measure. But there are such things as the somewhat Riemann-Stieltjes integrals with respect to white noise, which are much more irregular, and have more stringent conditions. There are extensions of this as well. >And Riemann integrable too. >> As a physicist, I always calculated integrals the common way, >> by employing the main theorem of calculus. Or numerically . As to how to calculate an integral, use anything that works. But the Fundamental Theorem of Calculus states that the Riemann integral, which does not involve the notion of derivative, can be calculated by antidifferentiation. >> What's the beef with these definitions? And how many are there? The main idea of integral is limit of sum. There are may ways of passing to the limit. The integral sign itself is an old s. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Discontinuous everywhere functions in Maple? > [ ... snip ... ] But there are such things > as the somewhat Riemann-Stieltjes integrals with respect to > white noise, which are much more irregular, and have more > stringent conditions. There are extensions of this as well. How about smoothing the irregular (possibly discrete) signal, by convoluting it with Gauss distributions and then integrating the result. It seems that I'll just need a Riemann integration, with no significant deviations from results obtained otherwise. I could do it via the Fourier domain as well. Any remarks on that kind of procedure? Han de Bruijn === Subject: Re: Discontinuous everywhere functions in Maple? >> [ ... snip ... ] But there are such things >> as the somewhat Riemann-Stieltjes integrals with respect to >> white noise, which are much more irregular, and have more >> stringent conditions. There are extensions of this as well. >How about smoothing the irregular (possibly discrete) signal, >by convoluting it with Gauss distributions and then integrating >the result. It seems that I'll just need a Riemann integration, >with no significant deviations from results obtained otherwise. >I could do it via the Fourier domain as well. If this could be done, it would have been. As it is, it makes a difference which point is used to multiply the length of the interval. If W is the Wiener process, do dW is white noise, the Ito integral from a to b, int_a^b W(t) dW(t), is (W^2(b) - W^2(a) - b + a)/2, while the Rubin-Fisk-Stratonovich integral is what one would expect. This is because the Riemann type integral int_a^b (dW(t))^2 = b-a. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Discontinuous everywhere functions in Maple? >> It's called the Dirichlet function. [ ... ] >> is Lebesgue integrable but not Riemann integrable. > Maybe you're in the position then to answer another question: >> Can somebody enumerate all different integration techniques? >> I've heard now about Lebesgue integration, Stieltjes integral .. > And Riemann integrable too. In addition to the references that Dik T. Winter provided, you might look at . The Lebesgue integral is integration with respect to Lebesgue measure. The concept of a measure can be generalized, as explained on that page, and when you consider integration with respect to arbitrary measures you get pretty much all the types of integrals. A little further generalization is also possible. See . -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Discontinuous everywhere functions in Maple? ... > Can somebody enumerate all different integration techniques? > I've heard now about Lebesgue integration, Stieltjes integral .. > And Riemann integrable too. See , especially the section see also at the bottom, and the quote in the first paragraph. It will also answer your other questions. -- 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: Discontinuous everywhere functions in Maple? > See , especially the section > see also at the bottom, and the quote in the first paragraph. It will > also answer your other questions. Indeed! Quoted from this page: it appears that cases where these methods [i.e., generalizations of the Riemann integral] are applicable and Riemann's [definition of the integral] is not are too rare in physics to repay the extra difficulty. Well, that explains why, as a physicist, I didn't need anything but Han de Bruijn === Subject: Re: Discontinuous everywhere functions in Maple? >> Is there anyway to define an expression or function in Maple that is >> discontinuous everywhere, say the characteristic of Q? I looked in the index >> for some way to input notation x in R and x in R Q but couldn't find >> it. I'd imagine there would be extreme computational difficulties in >> determining irrationality of numbers. >Try this: >f:=x->if type(x,rational) then 1 else 0 end if; Yes, if you only give f numeric inputs. But type only looks at the surface, so for example type((sqrt(2)+1)^2 + (sqrt(2)-1)^2), rational) will return false. Somewhat better is to use is, which will make some effort (I'm not sure how much) to see whether the expression can be simplified to a rational: is((sqrt(2)+1)^2 + (sqrt(2)-1)^2), rational) will return true. On the other hand, even is is not infallible: is((arctan(1/3)+arctan(1/2))/Pi, rational) will return false although (arctan(1/3)+arctan(1/2))/Pi simplifies to 1/4. >OTOH, the comp.soft-sys.math.maple newsgroup is the appropriate place >for questions concerning Maple. Yes. >>I'm basically trying to see if it can do Lebesgue integration and how far it >>will actually go. No, it can't do Lebesgue integration. All the integration methods are meant for functions that are at least piecewise smooth. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A simple proof for the following? /Carl > Is there for the following statement a simple proof (or a way to show > that it is plausible)? > Statement: Any 8-bit word can be stored as a 14-bit word, with the > 14-bit word having at least two zeros between every one. > Carl === Subject: Re: topology...~! > let X be a countably infinite set and > T be a finite-complement topology on X. > (b) show that every continuous real valued function on X > is constant > I thought it might be amusing to use the finite-complement topology > more directly, so here's an alternate proof. > Let f: X ---> R be continuous and non-constant. > Suppose a < b are in f(X), and let c = (a+b)/2. > Actually, this appears to prove that you can't even have > a non-constant continuous map into the rationals. Here's ultimate generalization. A space is hyper-connected when for all open nonnul U,V, U/V /= nulset (all open nonnul sets intersect) Notice cofinite and cocountable spaces are hyper-connected. hyper-connected X, Hausdorff Y, f in C(X,Y) ==> constant. Otherwise some x,y with f(x) /= f(y) some open U,V separate f(x), f(y) open f^-1(U), f^-1(V) separate x,y which contradicts X is hyper-connected open U,V separate x,y when U,V disjoint, x in U, y in V Thus for example a continuous function from a cocountable space into the complex rationals Q[i], is constant. === Subject: Re: topology...~! > let X be a countably infinite set and > T be a finite-complement topology on X. (b) show that every continuous real valued function on X > is constant > Actually, this appears to prove that you can't even have > a non-constant continuous map into the rationals. Minor correction made below regarding this generalization. > A space is hyper-connected when > for all open nonnul U,V, U/V /= nulset > (all open nonnul sets intersect) > Notice cofinite and cocountable spaces are hyper-connected. > hyper-connected X, Hausdorff Y, f in C(X,Y) ==> f is constant. > Otherwise some x,y with f(x) /= f(y) > some open U,V separate f(x), f(y) > open f^-1(U), f^-1(V) separate x,y > which contradicts X is hyper-connected > open U,V separate x,y when U,V disjoint, x in U, y in V > Thus for example a continuous function from a cocountable space > into the complex rationals Q[i], is constant. === Subject: Re: topology....... > hello.....doctor~ > let X be a countably infinite set and > T be a finite-complement topology on X. > (a) show that X is compact and connected > (b) show that every continuous real valued function on X > is constant > ----------------------------------------------------- > i can show (a). > but i can't show (b). > so, i need your advice. Let F:X->R be continuous. Then F(X) is connected. But X is countable, so F(X) is likewise countable, and no non-singleton countable subset of R is connected. F(X) is therefore a singleton, which is to say, F is constant. === Subject: Re: Discontinuous convex function >Has somebody got an example of a discontinuous convex functional over a >Banach space? Depends on what it means to have an example - you can show such things exist using the Axiom of Choice. >Pereger ************************ David C. Ullrich === Subject: Re: Discontinuous convex function > Has somebody got an example of a discontinuous convex functional over a > Banach space? > Pereger A discontinuous linear functional will do, right? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? There are a number of authors and a number of contributors to this NG who claim that sampling of analogue signals is represented by multiplying the incoming waveform f(t) by a comb of Diracian Delta Functions of the form d(t - T) and who then go on to claim that each such sample gives rise to a contribution to the spectrum of the sampled signal of f(T).e^(-sT). As this claim, this apparently faulty meme, ought to be such a fundamental part of DSP, its foundation stone in fact, it should be possible to prove this assertion by appeal to Dirac's properties of his Delta Function, and to the Laplace transform. I think that such a proof should be a simple thing to be provided by that body of authors and contributors if the claim were to be true. In our training of the Laplacian method, we are presented with a whole range of such derivations, the spectra for d(t), u(t), t, sin(t), rect(t), sinc(t), etc etc etc. Why not a similar derivation for f(t).d(t - T)? It cannot be a difficult matter for those who make the claim that sampling is so represented! As that body of authors and contributors seem unable to provide such a proof and resort to side issues and rather silly and childish ad hominem attacks when challenged upon the matter the conclusion that I reach is that the claim is false, and that that body of authors and contributors hold a religious-like stance to the matter and respond with the emotional maladjustment which is the mark of all those who are the religious loonies of the world today. (11/9 and the resultant war brought on by the religionists Bush, Blair and Windsor being prime examples of catastrophes brought on by emotional maladjustment) === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? > There are a number of authors and a number of contributors > to this NG who claim that sampling of analogue > signals is represented by multiplying the incoming waveform > f(t) by a comb of Diracian Delta Functions of the form > d(t - T) and who then go on to claim that each such sample > gives rise to a contribution to the spectrum of the sampled > signal of f(T).e^(-sT). > As this claim, this apparently faulty meme, ought to be such a > fundamental part of DSP, its foundation stone in fact, it should > be possible to prove this assertion by appeal to Dirac's properties > of his Delta Function, and to the Laplace transform. I think that such > a proof should be a simple thing to be provided by that > body of authors and contributors if the claim were to be true. > In our training of the Laplacian method, we are presented > with a whole range of such derivations, the spectra for > d(t), u(t), t, sin(t), rect(t), sinc(t), etc etc etc. Why not > a similar derivation for f(t).d(t - T)? It cannot be a difficult > matter for those who make the claim that sampling > is so represented! It appears that when you say spectrum you mean Laplace Transform. Assuming that's what you mean, then this is indeed very easy to show. The definition of the LT is (i) L(f)(s) = int_0^infinity f(t) e^(-st) dt. And the definition of delta(t-T) is that if g is continuous (and T > 0) then (ii) int_0^infinity g(t) delta(t-T) dt = g(T). (I'm calling it delta instead of d to avoid confusion with the d in dt.)) Now (i) says that L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T) e^(-st) dt, and applying (i) with g(t) = f(t) e^(-st) shows that int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT). > As that body of authors and contributors seem unable > to provide such a proof and resort to side issues and > rather silly and childish ad hominem attacks when > challenged upon the matter the conclusion that I reach > is that the claim is false, and that that body of authors > and contributors hold a religious-like stance to the matter > and respond with the emotional maladjustment which is > the mark of all those who are the religious loonies of the > world today. (11/9 and the resultant war brought on > by the religionists Bush, Blair and Windsor being prime > examples of catastrophes brought on by emotional > maladjustment) Uh, right. Who is it who's been unable to give the proof above? Are you sure it's unable and not just unwilling? I mean this is all covered in every book on differential equations I've ever seen... ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Up to the point quoted below, I agree with input. But. to evaluate the (unilateral) Laplace transform, we either have to be able to present f(t) as a function of exponentials, in which case simple integration applies together with the adding of exponents, or we have to apply Integration by Parts. (or else for a time domain product we have to evaluate an s domain convolution) You simply cannot make a simple suggestion that g(t) = f(t).d(t - T) as you have done below. Integration by parts..... int(UV) = U.int(v) - int[ dU.int(V) ] which becomes even more unwieldy when it needs to be int(UVW) as in evaluating the LT of f(t).d(t - T) > Now (i) says that > L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T) e^(-st) dt, > and applying (i) with g(t) = f(t) e^(-st) shows that > int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT). === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >Up to the point quoted below, I agree with >input. >But. to evaluate the (unilateral) Laplace >transform, we either have to be able to >present f(t) as a function of exponentials, >in which case simple integration applies >together with the adding of exponents, >or we have to apply Integration by Parts. >(or else for a time domain product we >have to evaluate an s domain convolution) >You simply cannot make a simple suggestion >that g(t) = f(t).d(t - T) as you have done >below. Uh, right. What I gave was a complete and correct proof (well, it was extremely informal by mathematical standards). You just saying I can't do what I did doesn't make the proof invalid, sorry. The fact that you don't understand something doesn't make it wrong. >Integration by parts..... >int(UV) = U.int(v) - int[ dU.int(V) ] >which becomes even more unwieldy when >it needs to be int(UVW) as in evaluating >the LT of f(t).d(t - T) >> Now (i) says that >> L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T) e^(-st) dt, >> and applying (i) with g(t) = f(t) e^(-st) shows that >> int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT). ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? The fact that you cannot do what you did _DOES_ make the proof invalid. I understand fully what I am proposing. If there is to be any criticism that there is a lack of understanding then it seems to arise in your own contribution. Nevertheless, thank-you for your attempt at assistance, wrong though it was. > Uh, right. What I gave was a complete and correct > proof (well, it was extremely informal by mathematical > standards). You just saying I can't do what I did > doesn't make the proof invalid, sorry. The fact > that you don't understand something doesn't make > it wrong. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >The fact that you cannot do what you did _DOES_ >make the proof invalid. It would if you gave an explanation for _why_ I can't do that, instead of just asserting it. >I understand fully what I am proposing. >If there is to be any criticism that there is a lack >of understanding then it seems to arise in your >own contribution. >Nevertheless, thank-you for your attempt at >assistance, wrong though it was. I begin to understand the comments that you called ad hominem a few posts up, although I haven't seen them. If you were interested in understanding this you'd ask me to explain why the steps in the proof were correct instead of just asserting they're invalid. But you're somehow convinced that you must be right, even though we're talking about standard mathematical facts that you can read in undergraduate textbooks. The idea that you can be so certain you're right and all those books are wrong is simply kooky. >> Uh, right. What I gave was a complete and correct >> proof (well, it was extremely informal by mathematical >> standards). You just saying I can't do what I did >> doesn't make the proof invalid, sorry. The fact >> that you don't understand something doesn't make >> it wrong. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? I did give you an explanation, suggesting that you must either integrate by parts or else must resort to frequency-domain convolution. >The fact that you cannot do what you did _DOES_ >make the proof invalid. > It would if you gave an explanation for _why_ I > can't do that, instead of just asserting it. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? > I did give you an explanation, suggesting that > you must either integrate by parts or else > must resort to frequency-domain convolution. OK, let's integrate by parts: L[f(t).delta(t-T)] = int_0^infinity exp(-st).f(t).delta(t-T).dt using int_a^b udv = (uv)|_a^b - int_a^b v.du choose u = exp(-st) => du = -s.exp(-st)dt and, dv = f(t).delta(t-T) dt => v = int_0^t x(u) delta(u-T)du = 0 if t < T, or x(T) if t >= T so, L[f(t).delta(t-T)] = [exp(-st).v(t)]_0^inf - int_0^inf -s.exp(-st).v(t)dt = 0 + s.int_T^inf x(T)exp(-st)dt = [-x(T).exp(-st)]_T^inf = x(T).exp(-sT) hopefully that is satisfactory for you, excuse the sloppiness Richard === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? The point at which I have stopped quoting you is the point at which you seem to have made a simple change of variable from t to u which does not seem to support your change of limits from a^b to 0^t (assuming that I have understood your notation for expressing limits) How have you calculated your change of limits? > OK, let's integrate by parts: > L[f(t).delta(t-T)] = int_0^infinity exp(-st).f(t).delta(t-T).dt > using int_a^b udv = (uv)|_a^b - int_a^b v.du > choose u = exp(-st) => du = -s.exp(-st)dt > and, > dv = f(t).delta(t-T) dt > => v = int_0^t x(u) delta(u-T)du > = 0 if t < T, or x(T) if t >= T === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >I did give you an explanation, suggesting that >you must either integrate by parts or else >must resort to frequency-domain convolution. That's exactly the part that you simply _stated_ without proof. It's not true, by the way. >>The fact that you cannot do what you did _DOES_ >>make the proof invalid. >> It would if you gave an explanation for _why_ I >> can't do that, instead of just asserting it. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? .....[Pantomime mode ON]..... Oh, Yes! It is! .....[Pantomime mode ON]..... > That's exactly the part that you simply _stated_ > without proof. It's not true, by the way. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >.....[Pantomime mode ON]..... >Oh, Yes! It is! >.....[Pantomime mode ON]..... Uh, right. Now why don't you give us a _proof_ of your assertion that the only way to evaluate a Laplace transform is integration by parts or convolution? It's a funny thing. I've been a mathematician for over 20 years, and this is the first time I've ever seen anyone insist that X is the _only_ way to accomplish Y. >> That's exactly the part that you simply _stated_ >> without proof. It's not true, by the way. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? What other way is there to evaluate it that is not based upon the two methods I described? Your reputation as a debater is not enhanced by your wont to sneer but without producing counter-argument. I have never asserted that X is the _only_ way to accomplish Y >.....[Pantomime mode ON]..... >Oh, Yes! It is! >.....[Pantomime mode OFF]..... > Uh, right. Now why don't you give us a _proof_ of your > assertion that the only way to evaluate a Laplace transform > is integration by parts or convolution? > It's a funny thing. I've been a mathematician for over 20 > years, and this is the first time I've ever seen anyone > insist that X is the _only_ way to accomplish Y. >> That's exactly the part that you simply _stated_ >> without proof. It's not true, by the way. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >What other way is there to evaluate it that is not >based upon the two methods I described? I explained exactly how to evaluate this Laplace transform correctly in my first post on the topic. (I also just posted a more complicated proof, using the fact that the delta function is in some sense a limit of ordinary functions.) >Your reputation as a debater is not enhanced by >your wont to sneer but without producing >counter-argument. Guffaw. >I have never asserted that X is the _only_ way >to accomplish Y Uh, yes you did. In your reply to my first post you said in part But. to evaluate the (unilateral) Laplace transform, we either have to be able to present f(t) as a function of exponentials, in which case simple integration applies together with the adding of exponents, or we have to apply Integration by Parts. (or else for a time domain product we have to evaluate an s domain convolution) which does say that this and that are the only way to do something. >>.....[Pantomime mode ON]..... >>Oh, Yes! It is! >>.....[Pantomime mode OFF]..... >> Uh, right. Now why don't you give us a _proof_ of your >> assertion that the only way to evaluate a Laplace transform >> is integration by parts or convolution? >> It's a funny thing. I've been a mathematician for over 20 >> years, and this is the first time I've ever seen anyone >> insist that X is the _only_ way to accomplish Y. > That's exactly the part that you simply _stated_ > without proof. It's not true, by the way. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? So first of all you say that I said, X is the _only_ way to accomplish Y Then you said that I said, this and that are the only way to do something. You're not consistent. In any case, I suggested three ways of proceeding and certainly did not state that X is the _only_ way to accomplish Y nor did I state this and that are the only way to do something.. >I have never asserted that X is the _only_ way >to accomplish Y > Uh, yes you did. In your reply to my first post > you said in part > which does say that this and that are the > only way to do something. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? .....[Pantomime mode ON]..... Oh, no! You didn't! .....[Pantomime mode OFF]..... >What other way is there to evaluate it that is not >based upon the two methods I described? > I explained exactly how to evaluate this Laplace > transform correctly in my first post on the topic. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Is not the same thing true of you, and therefore an invalid comment to make? > But you're somehow convinced that you must be right, === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >Is not the same thing true of you, and therefore an >invalid comment to make? No, because I gave an actual proof that I was right, starting from the definitions. >> But you're somehow convinced that you must be right, ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? If you say, No to the question posed below, then you are saying that you are not convinced that you must be right. >Is not the same thing true of you, and therefore an >invalid comment to make? > No, because I gave an actual proof that I was right, > starting from the definitions. >> But you're somehow convinced that you must be right, > ************************ > David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >If you say, No to the question posed below, then you >are saying that you are not convinced that you >must be right. Uh, sorry, I didn't read carefully, in particular I didn't realize that you were now quoting less than complete sentences. Yes, I am convinced I'm right. The complete sentence that I thought the is not the same true of you was asking about was this: But you're somehow convinced that you must be right, even though we're talking about standard mathematical facts that you can read in undergraduate textbooks. Yes, I'm convinced I'm right. No, it's not true that I'm convinced I'm right _in spite of_ the fact that I'm contradicting what's in hundreds of textbooks. The person who's somehow convinced he's right and all those books are wrong is you. >>Is not the same thing true of you, and therefore an >>invalid comment to make? >> No, because I gave an actual proof that I was right, >> starting from the definitions. > But you're somehow convinced that you must be right, >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Therefore it is also true to say of you But you're somehow convinced that you must be right, and therefore it was an inappropriate thing for you to introduce. Furthermore, if something that you read in one of your textbooks is your reply to a point of debate, then it is a simple matter for you to type it out. Far better than sneering at your correspondent! >If you say, No to the question posed below, then you >are saying that you are not convinced that you >must be right. > Yes, I am convinced I'm right. The complete sentence > that I thought the is not the same true of you was > asking about was this: > But you're somehow convinced that you must be right, > even though we're talking about standard mathematical > facts that you can read in undergraduate textbooks. > Yes, I'm convinced I'm right. No, it's not true that > I'm convinced I'm right _in spite of_ the fact that > I'm contradicting what's in hundreds of textbooks. > The person who's somehow convinced he's right and > all those books are wrong is you. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? You just saying that it was a complete and correct proof does not make the proof valid, sorry. The fact that you don't understand something doesn't make it right. > Uh, right. What I gave was a complete and correct > proof (well, it was extremely informal by mathematical > standards). You just saying I can't do what I did > doesn't make the proof invalid, sorry. The fact > that you don't understand something doesn't make > it wrong. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Hi there, Im still a student but I recall studying the effect of delta function sampling and I recall the proof by David. The way I understood the proof of it was that f(t).delta(t-T) = f(T) where T is often a constant delay or shift. You can say that f(t) has been sampled at point T. This occurs because the delta function is zero everywhere except at T so if you multiply the two, every point is zero except for the one at T (where delta(t-T) = 1). This is a basic property of a delta function and can be proved but Its fairly obvious so im not going to. Now you can evaluate f(t) at T, and use that in the laplace transform like: integral(from 0 to inf) of ( f(T).e^-(st) ) dt if f(T) is a constant you can pull it out the front. If you periodically sample a signal, e.g with a delta comb: f(t).(sigma (from n=0 to inf) of (delta( t - nT))) function you get an array of values, such that sampled f(t) = f(0) + f(T) + f(2T)........ since it is not continuous now, you can represent the signal as f(kT). Now you are looking at a starred laplace transform which is no longer an integral because it is not continuous. Instead it is a summation which is multiplied by a complex variable: F*(s) = sigma(k=0 to inf) of ( f(kT).e^-(T.s.k) ) To complete the transform, apply the proof of a geometric sum to get a neat ratio of polynomials (in basic problems). The z-transform is more commonly used and is strongly related to the starred laplace transform. I hope that is what you were looking for and also that it is all correct (my memory sometimes plays tricks on me) Marc Now that you >Up to the point quoted below, I agree with >input. >But. to evaluate the (unilateral) Laplace >transform, we either have to be able to >present f(t) as a function of exponentials, >in which case simple integration applies >together with the adding of exponents, >or we have to apply Integration by Parts. >(or else for a time domain product we >have to evaluate an s domain convolution) >You simply cannot make a simple suggestion >that g(t) = f(t).d(t - T) as you have done >below. > Uh, right. What I gave was a complete and correct > proof (well, it was extremely informal by mathematical > standards). You just saying I can't do what I did > doesn't make the proof invalid, sorry. The fact > that you don't understand something doesn't make > it wrong. >Integration by parts..... >int(UV) = U.int(v) - int[ dU.int(V) ] >which becomes even more unwieldy when >it needs to be int(UVW) as in evaluating >the LT of f(t).d(t - T) >> Now (i) says that >> L(f(t)delta(t - T))(s) = int_0^infinity f(t)delta(t-T) e^(-st) dt, >> and applying (i) with g(t) = f(t) e^(-st) shows that >> int_0^infinity f(t)delta(t-T) e^(-st) dt = f(T) e^(-sT). > ************************ > David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >Hi there, >Im still a student but I recall studying the effect of delta function >sampling >and I recall the proof by David. The way I understood the proof of it was >that f(t).delta(t-T) = f(T) That's not true. What's true is that f(t).delta(t-T) = f(T).delta(t-T). > where T is often a constant delay or shift. >You can say that >f(t) has been sampled at point T. >This occurs because the delta function is zero everywhere except at T so >if you multiply the two, every point is zero except for the one at T (where >delta(t-T) = 1). delta(t-T) = 1 when t = T is not true either. Actually there's no such thing as the value of delta(t-T) when t = T. The delta function is not a function. What it is is a generalized function. What that means is a slightly long story, but the simplest way to think of it is probably this: There's no such thing as delta(t-T) _except_ when it appears under an integral sign. And if g is a continuous function then (*) int_-infinity^infinity g(t)delta(t-T) dt = g(T). Really - (*) is a version of the _definition_ of what the delta function really is. That's the way it has to be to make things like the Laplace transform come out right. If you set g(t) = e^(-st) in (*) you see that the Laplace transform of delta(t-T) is e^(-Ts), which is what we want it to be. If on the other hand it were true that delta(t-T) = 0 when t <> T and 1 when t = T then that Laplace transform would be the integral of a function that vanishes except at one point, and that integral is _zero_. (If we were talking about discrete variables instead of continuous variables then what you say about delta(t-T) would be exactly right.) ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? I have been doing more DSP lately (without sampling) and had somehow managed to confuse the real-time version of the delta with the discrete one ( delta(n) = { 1 if n = 0 { 0 otherwise ) After reading both of your explanations I realised that my working was incorrect as it was a weird hybrid. Now that I have revised the correct continuous time version, I understand David's original working and also remember trying to do it the way Airy is doing it when I first learnt about laplace transforms. It wasn't fun lol. Just to clarify to make sure im not off the rails again: Airy, I dont understand why you can't call f(t).delta(t-T) = g(t)? I personally cannot see anything wrong with it. g(t) itself will be undefined at T and zero at t = not T. But since it is being integrated, the sifting theorem should still work shouldn't it? More importantly you should be able to go: g(t) = e^-(st).f(t) now you have: int(from 0 to inf) of ( g(t).delta(t-T) ) dt which is a basic sifting theorem problem which David illustrated previously. Marc >Hi there, >Im still a student but I recall studying the effect of delta function >sampling >and I recall the proof by David. The way I understood the proof of it was >that f(t).delta(t-T) = f(T) > That's not true. What's true is that > f(t).delta(t-T) = f(T).delta(t-T). > where T is often a constant delay or shift. >You can say that >f(t) has been sampled at point T. >This occurs because the delta function is zero everywhere except at T so >if you multiply the two, every point is zero except for the one at T (where >delta(t-T) = 1). > delta(t-T) = 1 when t = T is not true either. > Actually there's no such thing as the value of delta(t-T) when > t = T. The delta function is not a function. What it is is a > generalized function. What that means is a slightly long story, > but the simplest way to think of it is probably this: There's > no such thing as delta(t-T) _except_ when it appears under > an integral sign. And if g is a continuous function then > (*) int_-infinity^infinity g(t)delta(t-T) dt = g(T). > Really - (*) is a version of the _definition_ of what the > delta function really is. > That's the way it has to be to make things like the Laplace > transform come out right. If you set g(t) = e^(-st) in (*) > you see that the Laplace transform of delta(t-T) is e^(-Ts), > which is what we want it to be. If on the other hand > it were true that delta(t-T) = 0 when t <> T and 1 when > t = T then that Laplace transform would be the integral > of a function that vanishes except at one point, and that > integral is _zero_. > (If we were talking about discrete variables instead of > continuous variables then what you say about delta(t-T) > would be exactly right.) > ************************ > David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? In the evaluation of integral transforms, the sifting theory will work if it is present as one of the integrals. > Airy, I dont understand why you can't call f(t).delta(t-T) = g(t)? I > personally cannot see > anything wrong with it. g(t) itself will be undefined at T and zero at t = > not T. But since it > is being integrated, the sifting theorem should still work shouldn't it? === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? It doesn't agree with integration by parts. > Airy, I dont understand why you can't call f(t).delta(t-T) = g(t)? I > personally cannot see > anything wrong with it. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >It doesn't agree with integration by parts. The integration by parts that you're doing is _wrong_. The fact that integration by parts works is a theorem. Theorems have hypotheses - whatever delta(t-T) is, it's not a continuous function. >> Airy, I dont understand why you can't call f(t).delta(t-T) = g(t)? I >> personally cannot see >> anything wrong with it. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Show where it is wrong, or else resort to infantile sneering. Oh!!! You're doing that already! >It doesn't agree with integration by parts. > The integration by parts that you're doing is _wrong_. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >Show where it is wrong, or else resort >to infantile sneering. Ok. Took me a minute to find the place where you showed us the integration by parts: Certainly, if you integrate by parts, you would choose int(f(t).d(t-T)) as the integrated bit to yield f(T), but when I try this, I get 0!...... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-sT) as U ..... f(T).e^(-st) - int(e^(-st)/-s . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. Your first step, f(T).e^(-st) - int(e^(-st)/-s . f(T)) is already wrong. My guess is because of a confusion over definite integrals versus antiderivatives, ie indefinite integrals. The (definite) integral from 0 to infinity of f(t).d(t-T) is indeed f(T). But what we need here is an antiderivative. An antiderivative of f(t).d(t-T) is given by F(t), where F(t) = 0 for t < T and f(T) for t > T. So that line should be F(t).e^(-st) - int(e^(-st)/-s . F(t)). Golly, I assumed that you had some idea how to do simple calculus - looking at this I realize you took the derivative wrong. What we really get is this: F(t).e^(-st) - int(e^(-st)(-s) . F(t)). Now when we put in the limits t = 0 to t = infinity the first term vanishes since F(0) = 0 and e^(-st) tends to 0 as t -> infinty (at least if s > 0; if s < 0 this integration by parts is not going to work.) So out Laplace transform becomes - int_0^infinity (e^(-st)/-s . F(t)) Since F(t) = 0 for t < T and f(T) for t > T this equals f(T) int_T^infinity (e^(-st)s ). Again, it's easy to evaluate the last integral; int_T^infinity (e^(-st)s ) = e^(-sT), so we finally get f(T)e^(-sT) for the Laplace transform. >Oh!!! You're doing that already! So now we've found _two_ mistakes in your integration by parts, confusing antiderivatives and definite integrals and taking a very simple derivative incorrectly. >>It doesn't agree with integration by parts. >> The integration by parts that you're doing is _wrong_. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Dirac did not define his function a as generalised function because the concept did not exist in 1930. The introduction of generalised functions and the Theory Of Distributions is entirely irrelevant when no mathematics resulting from that theory is applied. The product of a generalised function is another generalised function. No generalised functions appear in our DSPs. > Actually there's no such thing as the value of delta(t-T) when > t = T. The delta function is not a function. What it is is a > generalized function. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >Dirac did not define his function a as generalised function because >the concept did not exist in 1930. I didn't say that Dirac used the term generalized function. The delta function _is_ a generalized function, aka distribution. Dirac's delta function was one of the first examples of the _concept_, which was named and put on a rigorous mathematical basis by Schwarz years later. >The introduction of generalised functions and the Theory Of >Distributions is entirely irrelevant when no mathematics >resulting from that theory is applied. >The product of a generalised function is another generalised >function. No generalised functions appear in our DSPs. >> Actually there's no such thing as the value of delta(t-T) when >> t = T. The delta function is not a function. What it is is a >> generalized function. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? The Delta function was on a rigorous mathematical basis when Dirac defined it, although he spoke to the contrary. The idea of a function being the limit of another function has always been entirely respectable and is found in the evaluation of every derivative. The introduction of generalised functions and the Theory Of Distributions is entirely irrelevant when no mathematics resulting from that theory is applied. >Dirac did not define his function a as generalised function because >the concept did not exist in 1930. > I didn't say that Dirac used the term generalized function. > The delta function _is_ a generalized function, aka distribution. > Dirac's delta function was one of the first examples of the > _concept_, which was named and put on a rigorous mathematical > basis by Schwarz years later. >The introduction of generalised functions and the Theory Of >Distributions is entirely irrelevant when no mathematics >resulting from that theory is applied. >The product of a generalised function is another generalised >function. No generalised functions appear in our DSPs. >> Actually there's no such thing as the value of delta(t-T) when >> t = T. The delta function is not a function. What it is is a >> generalized function. > ************************ > David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >The Delta function was on a rigorous mathematical basis >when Dirac defined it, although he spoke to the contrary. >The idea of a function being the limit of another function >has always been entirely respectable and is found in >the evaluation of every derivative. Yes, but the delta function is _not_ the limit of a sequence of ordinary functions, at least not in the sense that one speaks of limits in calculus. It's not a function at all. But it is true that one can define the _action_ of a delta function as a limit - that is, the linear functional defined by the delta function is indeed a limit of the linear functionals defined by a sequence of ordinary functions. If you want to look at it that way it's very easy to show from that point of view that the Laplace transform is what it is: Fix T, and say f_n(t) = n for T < t < T + 1/n, f_n(t) = 0 for other t. Then the f_n _do_ converge to delta(t-T) in a certain sense - in particular the Laplace transform of delta(t-T) _is_ the limit of the Laplace transform of f_n as n -> infinity. Let's say F_n is the LT of f_n. Then F_n(s) = n int_T^{T+1/n} e^{-st} dt = n (e^{-s(T+1/n)) - e^{-sT}) / (-s) = e^{-sT} (1 - e^{-s/n}) / (s/n). It's an easy calculus exercise to show that (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity. So the limit of F_n(s) is e^{-sT}. >The introduction of generalised functions and the Theory Of >Distributions is entirely irrelevant when no mathematics >resulting from that theory is applied. >>Dirac did not define his function a as generalised function because >>the concept did not exist in 1930. >> I didn't say that Dirac used the term generalized function. >> The delta function _is_ a generalized function, aka distribution. >> Dirac's delta function was one of the first examples of the >> _concept_, which was named and put on a rigorous mathematical >> basis by Schwarz years later. >>The introduction of generalised functions and the Theory Of >>Distributions is entirely irrelevant when no mathematics >>resulting from that theory is applied. >>The product of a generalised function is another generalised >>function. No generalised functions appear in our DSPs. > Actually there's no such thing as the value of delta(t-T) when > t = T. The delta function is not a function. What it is is a > generalized function. >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Make your mind up - first you say that it is not a function, then you say that it is the delta function. . > It's not a function at all. > But it is true that one can define the _action_ > of a delta function as a limit === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? So you have argued a derivation, without using the concept of generalised functions which were in any case irrelevant, to arrive at a result that wasn't in dispute. I wonder why? > It's an easy calculus exercise to show that > (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity. > So the limit of F_n(s) is e^{-sT}. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >So you have argued a derivation, without using the >concept of generalised functions which were in any case >irrelevant, to arrive at a result that wasn't in dispute. The result I arrived at was that the Laplace transform of delta(t-T) is e^{-sT}. That wasn't in dispute? >I wonder why? Because you're too stupid or stubborn to agree with the much simpler argument proving the same thing. >> It's an easy calculus exercise to show that >> (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity. >> So the limit of F_n(s) is e^{-sT}. ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? .....[Pantomime mode ON]..... Oh, yes! It _IS_ a function. .....[Pantomime mode OFF]..... And I have never defined it as a limit of a sequence of functions any more than I would define a derivative as such a limit. The limit employed is very much the same limit as used in calculus; t > 0. >The Delta function was on a rigorous mathematical basis >when Dirac defined it, although he spoke to the contrary. >The idea of a function being the limit of another function >has always been entirely respectable and is found in >the evaluation of every derivative. > Yes, but the delta function is _not_ the limit of > a sequence of ordinary functions, at least not in > the sense that one speaks of limits in calculus. > It's not a function at all. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? No, that's not true and is not supported in any way by the properties of the Diracian Delta Function. There are no simple multiplication properties unless expressed under an integral sign. What is true is that int(+/- inf)(f(t).d(t-T)) is f(T), but the Delta is no longer present having been integrated out. > That's not true. What's true is that > f(t).delta(t-T) = f(T).delta(t-T). === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >No, that's not true and is not supported in any way >by the properties of the Diracian Delta Function. >There are no simple multiplication properties >unless expressed under an integral sign. You really don't know nearly as much about any of this as you think you do. delta is a _measure_. The product of a measure and a continuous function is a standard thing; by _definition_ saying that f(t).delta(t-T) = f(T).delta(t-T) means that if g is a continuous function then (*) int g(t).f(t).delta(t-T) = int g(t).f(T).delta(t-T) (where int is the integral from -infinity to infinity in general, or from 0 to infinity here). And (*) is true, because both sides equal g(T)f(T). >What is true is that int(+/- inf)(f(t).d(t-T)) is f(T), >but the Delta is no longer present having been >integrated out. Yes, that's true. Doesn't imply that f(t).delta(t-T) = f(T).delta(t-T) is false. >> That's not true. What's true is that >> f(t).delta(t-T) = f(T).delta(t-T). ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? You cannot prove something merely by defining it to be correct. > delta is a _measure_. The product of a measure > and a continuous function is a standard thing; > by _definition_ saying that f(t).delta(t-T) = f(T).delta(t-T) > means that if g is a continuous function then === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? To use your own argument, just saying it, does not prove it one way or the other. > delta is a _measure_. The product of a measure > and a continuous function is a standard thing; > by _definition_ saying that f(t).delta(t-T) = f(T).delta(t-T) > means that if g is a continuous function then > (*) int g(t).f(t).delta(t-T) = int g(t).f(T).delta(t-T) > (where int is the integral from -infinity to infinity > in general, or from 0 to infinity here). And (*) is > true, because both sides equal g(T)f(T). === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >To use your own argument, just saying it, does >not prove it one way or the other. Of course not. Here the question was about a _definition_. I didn't claim to have proved I was giving a correct definition. The definition I gave _is_ correct, and perfectly standard, your ignorance of the matter notwithstanding. >> delta is a _measure_. The product of a measure >> and a continuous function is a standard thing; >> by _definition_ saying that f(t).delta(t-T) = f(T).delta(t-T) >> means that if g is a continuous function then >> (*) int g(t).f(t).delta(t-T) = int g(t).f(T).delta(t-T) >> (where int is the integral from -infinity to infinity >> in general, or from 0 to infinity here). And (*) is >> true, because both sides equal g(T)f(T). ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Your reputation will not be enhanced by your tendency to resort to ad hominem remarks. > The definition I gave _is_ correct, and perfectly > standard, your ignorance of the matter notwithstanding. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Looking at the work done by Dirac, his Delta function is a function defined as a limit. No problem there. All derivatives are so defined. >No, that's not true and is not supported in any way >by the properties of the Diracian Delta Function. >There are no simple multiplication properties >unless expressed under an integral sign. > You really don't know nearly as much about any > of this as you think you do. > delta is a _measure_. The product of a measure > and a continuous function is a standard thing; === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Shame on you for making personal remarks. The conclusion about your university-style email address is that you are a first-year man. >No, that's not true and is not supported in any way >by the properties of the Diracian Delta Function. >There are no simple multiplication properties >unless expressed under an integral sign. > You really don't know nearly as much about any > of this as you think you do. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? > No, that's not true and is not supported in any way > by the properties of the Diracian Delta Function. > There are no simple multiplication properties > unless expressed under an integral sign. > What is true is that int(+/- inf)(f(t).d(t-T)) is f(T), > but the Delta is no longer present having been > integrated out. > That's not true. What's true is that > f(t).delta(t-T) = f(T).delta(t-T). Perhaps you should try to understand it a bit more before saying that someone is making false statements. The above means that (ie is equivalent with): int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo f(T).delta(t-T) dt The left handside simplifies to f(T). The right handside simplies to f(T). If you don't see the latter, define h(t)=1 (for all t). Now int_{-oo}^oo f(T).delta(t-T) dt = int_{-oo}^oo f(T).h(t).delta(t-T) dt = f(T) . int_{-oo}^oo h(t).delta(t-T) dt = f(T).h(T) = f(T) since h(T)=1. Hence they are indeed equal. Wilbert === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Multiplying one or both sides of an identity by 1 does not value to your argument. >If you don't see the latter, define h(t)=1 (for all t). Now > int_{-oo}^oo f(T).delta(t-T) dt = int_{-oo}^oo f(T).h(t).delta(t-T) dt > = f(T) . int_{-oo}^oo h(t).delta(t-T) dt = f(T).h(T) = f(T) since > h(T)=1. > Hence they are indeed equal. > Wilbert === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Is it necessarily true that if two integrals are equal, then the functions under those integrals are equal? > No, that's not true and is not supported in any way > by the properties of the Diracian Delta Function. > There are no simple multiplication properties > unless expressed under an integral sign. > What is true is that int(+/- inf)(f(t).d(t-T)) is f(T), > but the Delta is no longer present having been > integrated out. > That's not true. What's true is that > f(t).delta(t-T) = f(T).delta(t-T). > The above means that (ie is equivalent with): > int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo f(T).delta(t-T) dt === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >Is it necessarily true that if two integrals are >equal, then the functions under those integrals >are equal? Of course not. >> No, that's not true and is not supported in any way >> by the properties of the Diracian Delta Function. >> There are no simple multiplication properties >> unless expressed under an integral sign. >> What is true is that int(+/- inf)(f(t).d(t-T)) is f(T), >> but the Delta is no longer present having been >> integrated out. >> That's not true. What's true is that >> f(t).delta(t-T) = f(T).delta(t-T). >> The above means that (ie is equivalent with): >> int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo f(T).delta(t-T) dt ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Perhaps you should pay attention to the discussion rather than resorting in the first instance to personal remarks? > Perhaps you should try to understand it a bit more > before saying that someone is making false statements. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >> No, that's not true and is not supported in any way >> by the properties of the Diracian Delta Function. >> There are no simple multiplication properties >> unless expressed under an integral sign. >> What is true is that int(+/- inf)(f(t).d(t-T)) is f(T), >> but the Delta is no longer present having been >> integrated out. >> That's not true. What's true is that >> f(t).delta(t-T) = f(T).delta(t-T). >Perhaps you should try to understand it a bit more >before saying that someone is making false statements. >The above means that (ie is equivalent with): >int_{-oo}^oo f(t).delta(t-T) dt = int_{-oo}^oo f(T).delta(t-T) dt Well actually no, that's not what it means. See my reply to Airy. >The left handside simplifies to f(T). The right handside simplies >to f(T). If you don't see the latter, define h(t)=1 (for all t). Now >int_{-oo}^oo f(T).delta(t-T) dt = int_{-oo}^oo f(T).h(t).delta(t-T) dt >= f(T) . int_{-oo}^oo h(t).delta(t-T) dt = f(T).h(T) = f(T) since >h(T)=1. >Hence they are indeed equal. >Wilbert ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? What you say below is not supported by the properties of the Diracian Delta function. It has to be under an integral sign to get the f(T). > Im still a student but I recall studying the effect of delta function > sampling > and I recall the proof by David. The way I understood the proof of it was > that f(t).delta(t-T) = f(T) where T is often a constant delay or shift. > You can say that > f(t) has been sampled at point T. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? with as follows..... You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to determine your result. You simply cannot say that f(t).d(t-T) yields f(T) unless you do so under an integral. This arises from the fundamental properties of the Diracian Delta function . Certainly, if you integrate by parts, you would choose int(f(t).d(t-T)) as the integrated bit to yield f(T), but when I try this, I get 0!...... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-sT) as U ..... f(T).e^(-st) - int(e^(-st)/-s . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. What I seek is a sound mathematical proof of the claim that f(t).d(t-T) gives rise to a spectrum contribution of f(T).e^-(st), and claiming that f(t).d(t-T) equals f(T) is not sound. We use the properties of the Diracian Delta Function in so many other aspects of signal processing that it is just not right to pull-a-fast-one. > Now you can evaluate f(t) at T, and use that in the laplace transform like: > integral(from 0 to inf) of ( f(T).e^-(st) ) dt > if f(T) is a constant you can pull it out the front. === Subject: Re: Infantile authours degrading this NG. It is interesting the size of the amount of posts. If I am wrong in what I say, it would be a simple matter to ignore me and what I posit. That so many people, people who would seek to represent themselves as authorities, respond with emotional postings that do not address the points raised, seems to suggest that I am not wrong, and that they are embarrassed about a basic failing in their professed knowledge and so feel threatened. the amount of posts, posts that are non-technical and of an undesirable ad hominem style has reached such a scale that I have binned them all this morning. This leaves just one point that has not yet been addressed, and that is, that any mathematical analysis of a real system must, to be respectable, deal with measurements of that system. There is no system of which I am aware that has sampling pulses whose measurements match those of the Diracian Impulse... 1. Their amplitudes do not approach infinity. 2. Their areas do not approach unity. 3. In any case, the operation of f(t).d(t - T) is not defined unless under an integral sign, and cannot be evaluated unless under that integral sign. 4. The evaluation of the spectrum of the Diracian relies on it being over all time, -oo^+oo. It is improper, therefore, to attempt to evaluate other operations using the Diracian with a reduced domain, and yet still rely on the spectrum derivation. > of posts that it has generated, it stands to reason that you are > unlikely to get a response suitable to you from them. === Subject: Re: Infantile authours degrading this NG. > It is interesting the size of the amount of posts. Airy The expression Bollocks springs to mind as far as your diatribes are concerned. Marco --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. Why don't you and James Harris collaborate and solve all the major outstanding problems. >It is interesting the size of the amount of posts. >If I am wrong in what I say, it would be a simple >matter to ignore me and what I posit. >That so many people, people who would seek to >represent themselves as authorities, respond with >emotional postings that do not address the >points raised, seems to suggest that I am not wrong, and that >they are embarrassed about a basic failing in their >professed knowledge and so feel threatened. >the amount of posts, posts that are non-technical and >of an undesirable ad hominem style has reached such a >scale that I have binned them all this morning. >This leaves just one point that has not yet been addressed, and >that is, that any mathematical analysis of a real system must, >to be respectable, deal with measurements of that system. >There is no system of which I am aware that has sampling >pulses whose measurements match those of the Diracian >Impulse... >1. Their amplitudes do not approach infinity. >2. Their areas do not approach unity. >3. In any case, the operation of f(t).d(t - T) is not >defined unless under an integral sign, and cannot be >evaluated unless under that integral sign. >4. The evaluation of the spectrum of the Diracian relies >on it being over all time, -oo^+oo. It is improper, therefore, >to attempt to evaluate other operations using the Diracian >with a reduced domain, and yet still rely on the spectrum >derivation. >> of posts that it has generated, it stands to reason that you are >> unlikely to get a response suitable to you from them. === Subject: Re: Infantile authours degrading this NG. Classic crackpot. If nobody says you're wrong you must be right. But if a large number of people say you're wrong it follows you must be right as well. Things must be very pleasant in your little world. Btw, since you don't seem to have noticed my post where I reply to your request to show _where_ your integration by parts fails: When I said that integration by parts does not apply because delta is not a continuous I was assuming that you were making one sort of moderately subtle error. Hadn't looked at the actual argument you gave. In case you didn't see that post, in your integration by parts you simply do the basic calculus wrong. >It is interesting the size of the amount of posts. >If I am wrong in what I say, it would be a simple >matter to ignore me and what I posit. >That so many people, people who would seek to >represent themselves as authorities, respond with >emotional postings that do not address the >points raised, seems to suggest that I am not wrong, and that >they are embarrassed about a basic failing in their >professed knowledge and so feel threatened. >the amount of posts, posts that are non-technical and >of an undesirable ad hominem style has reached such a >scale that I have binned them all this morning. >This leaves just one point that has not yet been addressed, and >that is, that any mathematical analysis of a real system must, >to be respectable, deal with measurements of that system. >There is no system of which I am aware that has sampling >pulses whose measurements match those of the Diracian >Impulse... >1. Their amplitudes do not approach infinity. >2. Their areas do not approach unity. >3. In any case, the operation of f(t).d(t - T) is not >defined unless under an integral sign, and cannot be >evaluated unless under that integral sign. >4. The evaluation of the spectrum of the Diracian relies >on it being over all time, -oo^+oo. It is improper, therefore, >to attempt to evaluate other operations using the Diracian >with a reduced domain, and yet still rely on the spectrum >derivation. >> of posts that it has generated, it stands to reason that you are >> unlikely to get a response suitable to you from them. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. (1/root(2.PI.N)).e^(t^2/n^2) is pretty continuous. : When I said that > integration by parts does not apply because delta > is not a continuous === Subject: Re: Infantile authours degrading this NG. You claim to be a mathematician of 20 years' standing. I suggest that you do not try to be a teacher of mathematics. Mathematics is an abstract subject for which your students must be relaxed in mind in order to be able to take on abstractions. Your emotive and insulting posts will disrupt such a state of mind. In my time as a part-time tutor of mathematics to adults I came across this time and time again - people who have been put off mathematics, not by the subject per se, but by the aggression and personal attacks addressed to them by their teachers. Your infantile tirade below illustrates well the persons proscribed Shame on you. (I notice that you do not address the mathematical points below) > Classic crackpot. If nobody says you're wrong you must > be right. But if a large number of people say you're wrong > it follows you must be right as well. > Things must be very pleasant in your little world. > Btw, since you don't seem to have noticed my post > where I reply to your request to show _where_ your > integration by parts fails: When I said that > integration by parts does not apply because delta > is not a continuous I was assuming that you > were making one sort of moderately subtle error. > Hadn't looked at the actual argument you gave. > In case you didn't see that post, in your > integration by parts you simply do the basic > calculus wrong. >It is interesting the size of the amount of posts. >If I am wrong in what I say, it would be a simple >matter to ignore me and what I posit. >That so many people, people who would seek to >represent themselves as authorities, respond with >emotional postings that do not address the >points raised, seems to suggest that I am not wrong, and that >they are embarrassed about a basic failing in their >professed knowledge and so feel threatened. >the amount of posts, posts that are non-technical and >of an undesirable ad hominem style has reached such a >scale that I have binned them all this morning. >This leaves just one point that has not yet been addressed, and >that is, that any mathematical analysis of a real system must, >to be respectable, deal with measurements of that system. >There is no system of which I am aware that has sampling >pulses whose measurements match those of the Diracian >Impulse... >1. Their amplitudes do not approach infinity. >2. Their areas do not approach unity. >3. In any case, the operation of f(t).d(t - T) is not >defined unless under an integral sign, and cannot be >evaluated unless under that integral sign. >4. The evaluation of the spectrum of the Diracian relies >on it being over all time, -oo^+oo. It is improper, therefore, >to attempt to evaluate other operations using the Diracian >with a reduced domain, and yet still rely on the spectrum >derivation. >> of posts that it has generated, it stands to reason that you are >> unlikely to get a response suitable to you from them. > ************************ > David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > In my time as a part-time tutor of mathematics to adults the more of this thread I read the more I pity those poor souls === Subject: Re: Infantile authours degrading this NG. > You claim to be a mathematician of 20 years' standing. > I suggest that you do not try to be a teacher of mathematics. > Mathematics is an abstract subject for which your students > must be relaxed in mind in order to be able to take on > abstractions. What qualifies a failed software engineer (remember Westinghouse) to make such a statement? What are YOUR qualifications? Did you ever actually get a degree? === Subject: Re: Infantile authours degrading this NG. > You claim to be a mathematician of 20 years' standing. > I suggest that you do not try to be a teacher of mathematics. > Mathematics is an abstract subject for which your students > must be relaxed in mind in order to be able to take on > abstractions. Your emotive and insulting posts > will disrupt such a state of mind. In my time as a part-time > tutor of mathematics to adults Sounds a bit far fetched to me. Was that before or after the Westinghouse affair? === Subject: Re: Infantile authours degrading this NG. > You claim to be a mathematician of 20 years' standing. Yet, he posted a detailed analysis of your errors in your 'integration by parts' of the posited function. If you stay focused on the problem at hand, you would see that your mistakes have been pointed out to you. Then you can learn from your mistakes. daestrom === Subject: Re: Infantile authours degrading this NG. Unfortunately, although he claimed to have posted such, it was masked by his over-riding urge to be insulting and was therefore not seen by me. personal remarks, then I would be happy to respond. I have challenged a number of posers in this respect, but all have declined to reply without their spleen a'venting, which leads me to question that they had a genuine contribution to offer in the first place. Sic transit gloria Mundi. have anything to do with childish outbursts. There were no personal attacks for you to snip. Defence in asserting the right of reply, maybe, but no attacks. > You claim to be a mathematician of 20 years' standing. > parts' of the posited function. > If you stay focused on the problem at hand, you would see that your mistakes > have been pointed out to you. Then you can learn from your mistakes. === Subject: Re: Infantile authours degrading this NG. that... > Unfortunately, although he claimed to have posted such, it was > masked by his over-riding urge to be insulting and was therefore > not seen by me. Because of your over-riding urge to be insulted. -- Checkmate all rights reserved === Subject: Re: Infantile authours degrading this NG. > Unfortunately, although he claimed to have posted such, it was > masked by his over-riding urge to be insulting and was therefore > not seen by me. Well, here is what David C. Ullrich posted.... (lines with >> are from Airy R. Bean's previous post, those preceded by '>' are from David C. Ullrich's pose, my comments are embedded with '<*<') >>Show where it is wrong, or else resort >>to infantile sneering. >Ok. Took me a minute to find the place >where you showed us the integration >by parts: >>Certainly, if you integrate by parts, you would choose >>int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >>I try this, I get 0!...... >>int(UV) = U.int(V) - int[dU.int(V)] giving..... >> int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >>with f(t).d(t-T) as V and e^(-sT) as U ..... >>f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >>f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >>f(T).e^(-sT) - f(T).e^(-sT).... >>0. >Your first step, > f(T).e^(-st) - int(e^(-st)/-s . f(T)) >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). <**Now when we put in the limits t = 0 to t = infinity >the first term vanishes since F(0) = 0 and e^(-st) >tends to 0 as t -> infinty (at least if s > 0; if s < 0 >this integration by parts is not going to work.) >So out Laplace transform becomes > - int_0^infinity (e^(-st)/-s . F(t)) >Since F(t) = 0 for t < T and f(T) for t > T >this equals > f(T) int_T^infinity (e^(-st)s ). <*Again, it's easy to evaluate the last integral; >int_T^infinity (e^(-st)s ) = e^(-sT), so we >finally get f(T)e^(-sT) for the Laplace transform. <**> daestrom === Subject: Re: Infantile authours degrading this NG. Where do you get that from? The definite integral int -oo^+oo f(t).d(t-T) gives us f(T), but there is no information given to us about the indefinite integral nor of the anti-derivative. Indeed, the definite integral gives us f(T), which is a constant function having no aspect of t in its evaluation and which graphs as a horizontal line from -oo to +oo. This comes from the basic definitions of the Diracian. Your claimed anti-derivative has a strong dependence on t, and therefore has not come from anything formally defined. Are you making this up as you go along? Well, here is what David C. Ullrich posted.. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be === Subject: Re: Infantile authours degrading this NG. >Where do you get that from? Just explained that in a different post. >The definite integral int -oo^+oo f(t).d(t-T) >gives us f(T), but there is no information >given to us about the indefinite integral >nor of the anti-derivative. >Indeed, the definite integral gives us f(T), which >is a constant function having no aspect of t in its >evaluation and which graphs as a horizontal >line from -oo to +oo. This comes from the >basic definitions of the Diracian. True, and of no relevance whatever to the question of what an antiderivative is. Hint: You're _really_ showing over and over that you have no understanding of _basic_ calculus. You keep kindly informing me that I'm making a laughingstock of myself, presumably you'd want to know how utterly ignorant you're revealing yourself to be. PS: Look up the definition of ad hominem. I'm not saying you must be wrong because you're stupid, I'm saying you're looking very stupid because the things you're saying are wrong, at such a basic level. >Your claimed anti-derivative has a strong >dependence on t, and therefore has not >come from anything formally defined. >Are you making this up as you go along? >Well, here is what David C. Ullrich posted.. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. >>Where do you get that from? >Just explained that in a different post. >>The definite integral int -oo^+oo f(t).d(t-T) >>gives us f(T), but there is no information >>given to us about the indefinite integral >>nor of the anti-derivative. >>Indeed, the definite integral gives us f(T), which >>is a constant function having no aspect of t in its >>evaluation and which graphs as a horizontal >>line from -oo to +oo. This comes from the >>basic definitions of the Diracian. >True, and of no relevance whatever to the question >of what an antiderivative is. >Hint: You're _really_ showing over and over that >you have no understanding of _basic_ calculus. >You keep kindly informing me that I'm making >a laughingstock of myself, presumably you'd >want to know how utterly ignorant you're >revealing yourself to be. I tangled with him in another newsgroup. He is untrainable and is proud of that fact. /BAH /BAH Subtract a hundred and four for e-mail. === Subject: Re: Infantile authours degrading this NG. An ad hominem attack from you of no value to the discussion. Shame on you. You should no better. I do not keep kindly informing you that you're making a laughingstock of myself. I have suggested that once and once only. Please get your facts correct. You say that the things quoted below from me are true and then you say that I'm utterly ignorant. You are inconsistent. >Where do you get that from? > Just explained that in a different post. >The definite integral int -oo^+oo f(t).d(t-T) >gives us f(T), but there is no information >given to us about the indefinite integral >nor of the anti-derivative. >Indeed, the definite integral gives us f(T), which >is a constant function having no aspect of t in its >evaluation and which graphs as a horizontal >line from -oo to +oo. This comes from the >basic definitions of the Diracian. > True, and of no relevance whatever to the question > of what an antiderivative is. > Hint: You're _really_ showing over and over that > you have no understanding of _basic_ calculus. > You keep kindly informing me that I'm making > a laughingstock of myself, presumably you'd > want to know how utterly ignorant you're > revealing yourself to be. > PS: Look up the definition of ad hominem. > I'm not saying you must be wrong because you're > stupid, I'm saying you're looking very stupid > because the things you're saying are wrong, > at such a basic level. >Your claimed anti-derivative has a strong >dependence on t, and therefore has not >come from anything formally defined. >Are you making this up as you go along? >Well, here is what David C. Ullrich posted.. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be === Subject: Re: Infantile authours degrading this NG. >An ad hominem attack from you of no value to the >discussion. >Shame on you. >You should no better. I suppose it should be no surprise that you can't follow the definition of a term like ad hominem any better than you follow definitions in mathematics. Look it up. Google ad hominem - the first hit is http://www.nizkor.org/features/fallacies/ad-hominem.html That's a correct definition, and nothing I've said here fits. (If you don't understand why not, I gave an explanation in the PS below.) >I do not keep kindly informing you that you're making > a laughingstock of myself. I have suggested that once >and once only. Uh, no. >Please get your facts correct. >You say that the things quoted below from me are true >and then you say that I'm utterly ignorant. >You are inconsistent. No. What you say about the value of the definite integral is correct - your idea that the integral from -infinity to infinity can be used in place of the antiderivative in an integration by parts shows amazing ignorance. >>Where do you get that from? >> Just explained that in a different post. >>The definite integral int -oo^+oo f(t).d(t-T) >>gives us f(T), but there is no information >>given to us about the indefinite integral >>nor of the anti-derivative. >>Indeed, the definite integral gives us f(T), which >>is a constant function having no aspect of t in its >>evaluation and which graphs as a horizontal >>line from -oo to +oo. This comes from the >>basic definitions of the Diracian. >> True, and of no relevance whatever to the question >> of what an antiderivative is. >> Hint: You're _really_ showing over and over that >> you have no understanding of _basic_ calculus. >> You keep kindly informing me that I'm making >> a laughingstock of myself, presumably you'd >> want to know how utterly ignorant you're >> revealing yourself to be. >> PS: Look up the definition of ad hominem. >> I'm not saying you must be wrong because you're >> stupid, I'm saying you're looking very stupid >> because the things you're saying are wrong, >> at such a basic level. >>Your claimed anti-derivative has a strong >>dependence on t, and therefore has not >>come from anything formally defined. >>Are you making this up as you go along? >>Well, here is what David C. Ullrich posted.. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > An ad hominem attack from you of no value to the > discussion. > Shame on you. > You should no better. > I do not keep kindly informing you that you're making > a laughingstock of myself. I have suggested that once > and once only. > Please get your facts correct. > You say that the things quoted below from me are true > and then you say that I'm utterly ignorant. > You are inconsistent. >>Where do you get that from? >> Just explained that in a different post. >>The definite integral int -oo^+oo f(t).d(t-T) >>gives us f(T), but there is no information >>given to us about the indefinite integral >>nor of the anti-derivative. >>Indeed, the definite integral gives us f(T), which >>is a constant function having no aspect of t in its >>evaluation and which graphs as a horizontal >>line from -oo to +oo. This comes from the >>basic definitions of the Diracian. >> True, and of no relevance whatever to the question >> of what an antiderivative is. >> Hint: You're _really_ showing over and over that >> you have no understanding of _basic_ calculus. >> You keep kindly informing me that I'm making >> a laughingstock of myself, presumably you'd >> want to know how utterly ignorant you're >> revealing yourself to be. >> PS: Look up the definition of ad hominem. >> I'm not saying you must be wrong because you're >> stupid, I'm saying you're looking very stupid >> because the things you're saying are wrong, >> at such a basic level. >>Your claimed anti-derivative has a strong >>dependence on t, and therefore has not >>come from anything formally defined. >>Are you making this up as you go along? >>Well, here is what David C. Ullrich posted.. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. Hang on a moment... you say that I cannot use the properties of a definite integral.....and then you go on and do so yourself! If I cannot rely on the property of a definite integral that int -oo^+oo f(t).d(t-T) is f(T) then neither can you! Well, let us suppose that our f(t) is sin(wt)...how does your claimed anti-derivative then differentiate to become sin(wt).d(t-T)? Or, let us suppose that our f(t) is e^t...how does your claimed anti-derivative then differentiate to become e^t.d(t-T)? So, how does your anti-derivative differentiate to become two different products of functions? Remember, until we actually evaluate the definite integral that we're dealing in whole functions over all t in symbolic maths. > Well, here is what David C. Ullrich posted > My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). === Subject: Re: Infantile authours degrading this NG. >Hang on a moment... you say that I cannot use the properties >of a definite integral.....and then you go on and do so yourself! >If I cannot rely on the property of a definite integral that >int -oo^+oo f(t).d(t-T) is f(T) then neither can you! I didn't say any such thing. I said that you need an antiderivative in the integration by parts formula. That formula you quote is _why_ the antiderivative is what I say it is. By definition an antiderivative for f(t).d(t-T) would be F, where F(x) = int_0^x f(t).d(t-T). If x < T then that gives F(x) = 0, while if x > T that gives F(x) = f(T), precisely because of that formula above. >Well, let us suppose that our f(t) is sin(wt)...how >does your claimed anti-derivative then differentiate >to become sin(wt).d(t-T)? Impossible to answer this question since you insist generalized functions have nothing to do with it - these antiderivatives are not differentiable in the classical sense, they have derivatives in the sense of distributions. >Or, let us suppose that our f(t) is e^t...how >does your claimed anti-derivative then differentiate >to become e^t.d(t-T)? >So, how does your anti-derivative differentiate to >become two different products of functions? >Remember, until we actually evaluate the definite >integral that we're dealing in whole functions over >all t in symbolic maths. >> Well, here is what David C. Ullrich posted >> My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. The (definite) integral from >>0 to infinity of f(t).d(t-T) is indeed f(T). But >>what we need here is an antiderivative. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be >> F(t).e^(-st) - int(e^(-st)/-s . F(t)). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. Apart from throwing in the term generalised functions, how does the presented mathematics for your derivatives differ from those of the classical sense? Remember that any one function in your class of generalised functions represents the properties of the whole class. >Well, let us suppose that our f(t) is sin(wt)...how >does your claimed anti-derivative then differentiate >to become sin(wt).d(t-T)? > Impossible to answer this question since you insist > generalized functions have nothing to do with it - > these antiderivatives are not differentiable in > the classical sense, they have derivatives in the > sense of distributions. === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > Apart from throwing in the term generalised functions, > how does the presented mathematics for your > derivatives differ from those of the classical sense? > Remember that any one function in your class of > generalised functions represents the properties of the > whole class. >>Well, let us suppose that our f(t) is sin(wt)...how >>does your claimed anti-derivative then differentiate >>to become sin(wt).d(t-T)? >> Impossible to answer this question since you insist >> generalized functions have nothing to do with it - >> these antiderivatives are not differentiable in >> the classical sense, they have derivatives in the >> sense of distributions. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. >Apart from throwing in the term generalised functions, >how does the presented mathematics for your >derivatives differ from those of the classical sense? The antiderivative is not even continuous, much less differentiable in the classical sense. >Remember that any one function in your class of >generalised functions represents the properties of the >whole class. >>Well, let us suppose that our f(t) is sin(wt)...how >>does your claimed anti-derivative then differentiate >>to become sin(wt).d(t-T)? >> Impossible to answer this question since you insist >> generalized functions have nothing to do with it - >> these antiderivatives are not differentiable in >> the classical sense, they have derivatives in the >> sense of distributions. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. The anti-derivative of (sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2)) is pretty continuous in the classical sense, and indeed, in any sense you care to present. > The antiderivative is not even continuous, much less > differentiable in the classical sense. === Subject: Re: Infantile authours degrading this NG. >The anti-derivative of >(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2)) >is pretty continuous in the classical sense, and >indeed, in any sense you care to present. Uh, yes it is. If that function were a delta function you'd have a point. >> The antiderivative is not even continuous, much less >> differentiable in the classical sense. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. Where does that assertion come from? Remember - you proscribe the definite integral int -oo^+oo f(t).d(t-T) > By definition an antiderivative for f(t).d(t-T) > would be F, where > F(x) = int_0^x f(t).d(t-T). > If x < T then that gives F(x) = 0, while if x > T > that gives F(x) = f(T), precisely because of that > formula above. === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > Where does that assertion come from? > Remember - you proscribe the definite > integral int -oo^+oo f(t).d(t-T) >> By definition an antiderivative for f(t).d(t-T) >> would be F, where >> F(x) = int_0^x f(t).d(t-T). >> If x < T then that gives F(x) = 0, while if x > T >> that gives F(x) = f(T), precisely because of that >> formula above. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. By what definition? - You have presented here a definite integral and not an anti-derivative which contradicts earlier assertions by you. Indeed, in the only related definition of which I am aware, the limits of integration should be +/- oo and not 0 to x. To which definition in your By definition do you refer? > By definition an antiderivative for f(t).d(t-T) > would be F, where > F(x) = int_0^x f(t).d(t-T). === Subject: Re: Infantile authours degrading this NG. get lost bean dr. x > By what definition? - You have presented here a > definite integral and not an anti-derivative which > contradicts earlier assertions by you. > Indeed, in the only related definition of which I > am aware, the limits of integration should be +/- oo > and not 0 to x. To which definition in your By definition > do you refer? >> By definition an antiderivative for f(t).d(t-T) >> would be F, where >> F(x) = int_0^x f(t).d(t-T). --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. >By what definition? - You have presented here a >definite integral and not an anti-derivative which >contradicts earlier assertions by you. >Indeed, in the only related definition of which I >am aware, the limits of integration should be +/- oo >and not 0 to x. To which definition in your By definition >do you refer? You really need to find a calculus book somewhere. If F(x) = int_-infinty^x f(t) then F' = f; this is a thing called the Fundamental Theorem of Calculus. (In the present context int_0^x is the same, since we're assuming that f(t) = 0 for t < 0). >> By definition an antiderivative for f(t).d(t-T) >> would be F, where >> F(x) = int_0^x f(t).d(t-T). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. You really need to grow up. This is an international forum and not the staff room of the infants's school in which you apparently teach. You're changing your tune, once again. The discussion was about int f(t).d(t-T), now your equivocating about int f(t) only. Integration and differentiation of products is not done by simple substitution. > You really need to find a calculus book somewhere. > If F(x) = int_-infinty^x f(t) then F' = f; this > is a thing called the Fundamental Theorem of Calculus. > (In the present context int_0^x is the same, since > we're assuming that f(t) = 0 for t < 0). === Subject: Re: Infantile authours degrading this NG. >You really need to grow up. >This is an international forum and not >the staff room of the infants's school in >which you apparently teach. >You're changing your tune, once again. >The discussion was about int f(t).d(t-T), >now your equivocating about int f(t) only. It's hard to decide. You're just trolling, or you're really unable to follow simple arguments. Hint: The f in the statement from me you quote below is not the same as the f that we've been talking about. If I'd realized it was going to cause confusion I would have used a different letter... >Integration and differentiation of products >is not done by simple substitution. >> You really need to find a calculus book somewhere. >> If F(x) = int_-infinty^x f(t) then F' = f; this >> is a thing called the Fundamental Theorem of Calculus. >> (In the present context int_0^x is the same, since >> we're assuming that f(t) = 0 for t < 0). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > You're changing your tune, once again. Quack? Honk? === Subject: Re: Infantile authours degrading this NG. Yes - I corrected that in a later post. However, the differentiation and integration was -s/-s giving unity. > <* is apparent the '-' was taken out front. But since the beginning, dU should > have been '-e^(-st)s' with the 's' in the numerator, not denominator. I > suspect this error came from ARB's original attempt at the integration by === Subject: Re: Infantile authours degrading this NG. You may well have a point here, if my error is in not evaluating the definite integral over the whole term. However, other errors in what you propose nullify your argument, (see other posts) > Well, here is what David C. Ullrich posted >So that line should be > F(t).e^(-st) ...... === Subject: Re: Infantile authours degrading this NG. > You may well have a point here, if my error is in > not evaluating the definite integral over the whole > term. > However, other errors in what you propose nullify your > argument, (see other posts) The 'other errors' are on your part, not David's. (see other replies) daestrom === Subject: Re: Infantile authours degrading this NG. Your argument falls down when T = 0, giving t -T as just t, for then F(0) = f(0), and not 0. > Well, here is what David C. Ullrich posted. >Now when we put in the limits t = 0 to t = infinity >the first term vanishes since F(0) = 0 === Subject: Re: Infantile authours degrading this NG. > Your argument falls down when T = 0, > giving t -T as just t, for then F(0) = f(0), > and not 0. Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) f(0). And the Dirac delta function evaluated at (0-T) is zero, regardless of the value of f(0). (assuming that T > 0) daestrom === Subject: Re: Infantile authours degrading this NG. if F(t) = f(t).d(t-T), then with T equal to zero, this comes down to f(t).d(t). This might be undefined because it is not expressed under an integral sign, but it is (probably) not 0. If it were to be expressed under an integral sign, then it would be f(0) as I stated. > Your argument falls down when T = 0, > giving t -T as just t, for then F(0) = f(0), > and not 0. > Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) f(0). > And the Dirac delta function evaluated at (0-T) is zero, regardless of the > value of f(0). (assuming that T > 0) === Subject: Re: Infantile authours degrading this NG. nope, you've got it all wrong bean. obviously you're incompetent. well we knew that from the start. need better than that if you're going to stay here. dr. x > if F(t) = f(t).d(t-T), then with T equal to zero, > this comes down to f(t).d(t). > This might be undefined because it is not expressed > under an integral sign, but it is (probably) not 0. > If it were to be expressed under an integral sign, then > it would be f(0) as I stated. >> Your argument falls down when T = 0, >> giving t -T as just t, for then F(0) = f(0), >> and not 0. >> Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) > f(0). >> And the Dirac delta function evaluated at (0-T) is zero, regardless of >> the >> value of f(0). (assuming that T > 0) --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. The evaluation yields f(T) which is a constant and is a graphed as a horizontal straight line having no expression of time in its evaluation. This is fundamental to the definition of the Diracian, where no other properties are specified. This property is used in other aspects of DSP and so it must be used consistently here. > Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. === Subject: Re: Infantile authours degrading this NG. > The evaluation yields f(T) which is a constant > and is a graphed as a horizontal straight line > having no expression of time in its evaluation. > This is fundamental to the definition of the Diracian, > where no other properties are specified. This > property is used in other aspects of DSP and so > it must be used consistently here. >> Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous function of t. Where t T it has a value of f(T) (note carefully the case of each statement). It has two possible values depending on whether you are evaluating it at tT. That is not the same thing as what you're saying, ... is a constant... daestrom === Subject: Re: Infantile authours degrading this NG. That does not agree with the definition with which I work, which is int -oo^+oo f(t).d(t-T) = f(T). In that evaluation, all dependence upon t disappears. What definition are you using to claim discontinuity? > The evaluation yields f(T) which is a constant > and is a graphed as a horizontal straight line > having no expression of time in its evaluation. > This is fundamental to the definition of the Diracian, > where no other properties are specified. This > property is used in other aspects of DSP and so > it must be used consistently here. >> Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. > No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous > function of t. === Subject: Re: Infantile authours degrading this NG. >That does not agree with the definition with which >I work, which is int -oo^+oo f(t).d(t-T) = f(T). >In that evaluation, all dependence upon t disappears. That evaluation of the _antiderivative_ is simply _wrong_. >What definition are you using to claim discontinuity? >> The evaluation yields f(T) which is a constant >> and is a graphed as a horizontal straight line >> having no expression of time in its evaluation. >> This is fundamental to the definition of the Diracian, >> where no other properties are specified. This >> property is used in other aspects of DSP and so >> it must be used consistently here. > Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >> No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous >> function of t. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. That evaluation is the one by which all electronic engineers are taught. >That does not agree with the definition with which >I work, which is int -oo^+oo f(t).d(t-T) = f(T). >In that evaluation, all dependence upon t disappears. > That evaluation of the _antiderivative_ is simply > _wrong_. === Subject: Re: Infantile authours degrading this NG. But the gives a different result if the integration is done in the opposite direction, and therefore cannot be valid. Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. === Subject: Re: Infantile authours degrading this NG. > But the gives a different result if the integration > is done in the opposite direction, and therefore > cannot be valid. > Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. Evaluating F(t) where tT you have a value of f(t). If evaluated in the reverse order (i.e. over the interval +inf to zero), you get -f(t) instead of f(t). This is exactly as it should be. daestrom === Subject: Re: Infantile authours degrading this NG. If you evaluate the anti-derivative given by Mr.Ullrich (without accepting it as a correct formula, BTW) then you would get the properties where F(t) = 0 for t > T and f(T) for t < T. > But the gives a different result if the integration > is done in the opposite direction, and therefore > cannot be valid. > Well, here is what David C. Ullrich posted.... >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. > Evaluating F(t) where tT you have a value of f(t). > If evaluated in the reverse order (i.e. over the interval +inf to zero), you > get -f(t) instead of f(t). This is exactly as it should be. > daestrom === Subject: Re: Infantile authours degrading this NG. >> But the gives a different result if the integration >> is done in the opposite direction, and therefore >> cannot be valid. >> Well, here is what David C. Ullrich posted.... >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. > Evaluating F(t) where tT you have a value of > f(t). Oops, where t> T you have F(t) = f(T), not f(t). My mistake. > If evaluated in the reverse order (i.e. over the interval +inf to zero), > you get -f(t) instead of f(t). This is exactly as it should be. That should be -f(T) instead of f(T), again, my mistake. daestrom === Subject: Re: Infantile authours degrading this NG. I disagree that it is wrong. The operation being undertaken is Integration By Parts and therefore Integration is required and not Anti-Differentiation. My f(T).e^(-st) is the result of evaluating U.int(V), and, as it is part of a sum, evaluting the integral between the limits of 0^oo to save further tedious writing. > Unfortunately, although he claimed to have posted such, it was > masked by his over-riding urge to be insulting and was therefore > not seen by me. >>Certainly, if you integrate by parts, you would choose >>int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >>I try this, I get 0!...... >>int(UV) = U.int(V) - int[dU.int(V)] giving..... >> int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >>with f(t).d(t-T) as V and e^(-sT) as U ..... >>f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >Your first step, > f(T).e^(-st) - int(e^(-st)/-s . f(T)) >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). === Subject: Re: Infantile authours degrading this NG. >I disagree that it is wrong. The operation being >undertaken is Integration By Parts and >therefore Integration is required and not >Anti-Differentiation. Hilarious. I think you might want to take Calc 101 again before lecturing us on all this stuff. >My f(T).e^(-st) is the result of evaluating >U.int(V), If int means int_0^infinty then yes it is, that's exactly whu it's wrong. > and, as it is part of a sum, evaluting >the integral between the limits of 0^oo to save >further tedious writing. Making less and less sense - if you'd done that evaluation already then there would be no t appearing at this point. It's really good of you to warn me about what a laughingstock I'm making of myself, btw. >> Unfortunately, although he claimed to have posted such, it was >> masked by his over-riding urge to be insulting and was therefore >> not seen by me. >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >>Your first step, >> f(T).e^(-st) - int(e^(-st)/-s . f(T)) >>is already wrong. My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. The (definite) integral from >>0 to infinity of f(t).d(t-T) is indeed f(T). But >>what we need here is an antiderivative. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be >> F(t).e^(-st) - int(e^(-st)/-s . F(t)). ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. >> I disagree that it is wrong. The operation being >> undertaken is Integration By Parts and >> therefore Integration is required and not >> Anti-Differentiation. > Hilarious. I think you might want to take > Calc 101 again before lecturing us on all > this stuff. >> My f(T).e^(-st) is the result of evaluating >> U.int(V), > If int means int_0^infinty then yes it is, that's > exactly why it's wrong. i just realized now how far this is cross-posted. i suspect that David is hanging out on sci.math (but i dunno). David, he really doesn't get it. and he doesn't want to get it. if it were up to Beanie, we would rewrite all the calculus textbooks. let's not feed the troll. maybe if we stop feeding it, it will shrivel up and blow away. r b-j === Subject: Re: Infantile authours degrading this NG. > I disagree that it is wrong. The operation being > undertaken is Integration By Parts and > therefore Integration is required and not > Anti-Differentiation. >> Hilarious. I think you might want to take >> Calc 101 again before lecturing us on all >> this stuff. > My f(T).e^(-st) is the result of evaluating > U.int(V), >> If int means int_0^infinty then yes it is, that's >> exactly why it's wrong. >i just realized now how far this is cross-posted. i suspect that David is >hanging out on sci.math (but i dunno). Yes. > David, he really doesn't get it. >and he doesn't want to get it. up. Was actually trying to help the guy with the math in my first one or two replies, but since then it's all been just for the fun of watching the way the guy's technique. He's pretty good, has things like being insulting and then whining when people reply insultingly, ignoring refutations and then claiming they don't exist, claiming that the fact that there have been a lot of replies proves he must be right, etc down pat. We've got better here on sci.math but he's pretty good. I can't decide whether he's just trolling or really believes he's right... > if it were up to Beanie, we would rewrite >all the calculus textbooks. >let's not feed the troll. maybe if we stop feeding it, it will shrivel up >and blow away. What fun would that be? >r b-j ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. I see that you cannot resist the temptation to lower the tone. > Hilarious. I think you might want to take > Calc 101 again before lecturing us on all > this stuff. > It's really good of you to warn me about what a > laughingstock I'm making of myself, btw. === Subject: Re: Infantile authours degrading this NG. >I see that you cannot resist the temptation to >lower the tone. If you had any sense you'd stop making an idiot of yourself over the math, stop being so utterly hilarious complaining about the tone of people's comments, and just take a calculus class. >> Hilarious. I think you might want to take >> Calc 101 again before lecturing us on all >> this stuff. >> It's really good of you to warn me about what a >> laughingstock I'm making of myself, btw. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. I see that you cannot resist the temptation to lower the tone. >I see that you cannot resist the temptation to >lower the tone. > If you had any sense you'd stop making an idiot of > yourself over the math, stop being so utterly > hilarious complaining about the tone of people's > comments, and just take a calculus class. >> Hilarious. I think you might want to take >> Calc 101 again before lecturing us on all >> this stuff. >> It's really good of you to warn me about what a >> laughingstock I'm making of myself, btw. === Subject: Re: Infantile authours degrading this NG. >I see that you cannot resist the temptation to > lower the tone. In an earlier post David pointed out legitimate problems with the math behind your argument. Rather than read, and learn from, that post, you accidentally deleted it. I suggest you find the post on Google and read it. You may learn something. Charles Perry P.E. === Subject: Re: Infantile authours degrading this NG. I suggest that you read what is written before you open your mouth and come out with untruths. You may learn something. I did not accidentally delete anything. In response to a barrage of quite unnecessary gratuitous and insulting remarks by Mr.Ullrich, I _DELIBERATELY_ deleted things. If Mr.Ullrich had anything valuable to offer, then it was lost in the fog of his childish outbursts. There is no concept of legitimacy. Law does not come into it. >I see that you cannot resist the temptation to > lower the tone. > In an earlier post David pointed out legitimate problems with the math > behind your argument. Rather than read, and learn from, that post, you > accidentally deleted it. I suggest you find the post on Google and read > it. You may learn something. === Subject: Re: Infantile authours degrading this NG. >I suggest that you read what is written before you open > your mouth and come out with untruths. > You may learn something. > I did not accidentally delete anything. In response to > a barrage of quite unnecessary gratuitous and insulting > remarks by Mr.Ullrich, I _DELIBERATELY_ deleted > things. One of the posts you _DELIBERATELY_ deleted is the one I quoted earlier, where David C. Ullrich, started out quite politely showing you where you made your mistakes. And in his more recent post, he states: If you take f(T) exp(-st), (assume s > 0), evaluate at t = infinity, t = 0 and subtract you get -f(t). If F is an antiderivative of f(t) delta(t-T), so F(t) = 0 for t < T, F(t) = f(T) for t > T, and you do the same evaluate-subtract with F(t)e^{-st} you get 0. This was clear from the difference in your calculation and my correct version of the same calculuation, btw. Not much 'unnecessary gatuitous and insulting remarks' in that unless you consider the last sentence such. And if you do, then you obviously have a much thinner 'skin' than the way you treat others. daestrom === Subject: Re: Infantile authours degrading this NG. let's ask crystal, shall we bean? dr. x >I suggest that you read what is written before you open > your mouth and come out with untruths. > You may learn something. > I did not accidentally delete anything. In response to > a barrage of quite unnecessary gratuitous and insulting > remarks by Mr.Ullrich, I _DELIBERATELY_ deleted > things. > If Mr.Ullrich had anything valuable to offer, then it was lost > in the fog of his childish outbursts. > There is no concept of legitimacy. Law does not come into it. >>I see that you cannot resist the temptation to >> lower the tone. >> In an earlier post David pointed out legitimate problems with the math >> behind your argument. Rather than read, and learn from, that post, you >> accidentally deleted it. I suggest you find the post on Google and >> read >> it. You may learn something. --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Infantile authours degrading this NG. How does your Indefinite Integral or (Antiderivative) differ from a Definite Integral when the limits of integration are imposed? > Well, here is what David C. Ullrich posted >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. === Subject: Re: Infantile authours degrading this NG. >How does your Indefinite Integral or (Antiderivative) >differ from a Definite Integral when the limits of >integration are imposed? If you take f(T) exp(-st), (assume s > 0), evaluate at t = infinity, t = 0 and subtract you get -f(t). If F is an antiderivative of f(t) delta(t-T), so F(t) = 0 for t < T, F(t) = f(T) for t > T, and you do the same evaluate-subtract with F(t)e^{-st} you get 0. This was clear from the difference in your calculation and my correct version of the same calculuation, btw. >> Well, here is what David C. Ullrich posted >>is already wrong. My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. No you don't. You get -f(T). >How does your Indefinite Integral or (Antiderivative) >differ from a Definite Integral when the limits of >integration are imposed? > If you take f(T) exp(-st), (assume s > 0), evaluate > at t = infinity, t = 0 and subtract you get -f(t). === Subject: Re: Infantile authours degrading this NG. >> Unfortunately, although he claimed to have posted such, it was >> masked by his over-riding urge to be insulting and was therefore >> not seen by me. >Well, here is what David C. Ullrich posted.... (lines with >> are from Airy >R. Bean's previous post, those preceded by '>' are from David C. Ullrich's >pose, my comments are embedded with '<*<') >Show where it is wrong, or else resort >to infantile sneering. >>Ok. Took me a minute to find the place >>where you showed us the integration >>by parts: >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. >>Your first step, >> f(T).e^(-st) - int(e^(-st)/-s . f(T)) >>is already wrong. My guess is because of a confusion >>over definite integrals versus antiderivatives, ie >>indefinite integrals. The (definite) integral from >>0 to infinity of f(t).d(t-T) is indeed f(T). But >>what we need here is an antiderivative. >>An antiderivative of f(t).d(t-T) is given by >>F(t), where F(t) = 0 for t < T and f(T) for t > T. >>So that line should be >> F(t).e^(-st) - int(e^(-st)/-s . F(t)). ><**>Now when we put in the limits t = 0 to t = infinity >>the first term vanishes since F(0) = 0 and e^(-st) >>tends to 0 as t -> infinty (at least if s > 0; if s < 0 >>this integration by parts is not going to work.) >>So out Laplace transform becomes >> - int_0^infinity (e^(-st)/-s . F(t)) >>Since F(t) = 0 for t < T and f(T) for t > T >>this equals >> f(T) int_T^infinity (e^(-st)s ). ><*is apparent the '-' was taken out front. But since the beginning, dU should >have been '-e^(-st)s' with the 's' in the numerator, not denominator. I >suspect this error came from ARB's original attempt at the integration by I pointed out explicitly, in lines you snipped, that Airy had written e^{-st}/s where it should have been e^{-st} s. I thought that I'd fixed the error at that point, it's possible that some erroneous cut&pasted lines were not edited properly. (Also pointed out at least twice so far that this is somewhat amusing. Before looking at it I knew that his integration by parts must be wrong, since it gave the wrong answer, but I assumed that the problem was with something subtle about delta functions. Turns out that in addition to not getting the antiderivative right his amazing conclusion is based in part on simply differentiating an exponential incorrectly! It's easy to see why this was the point at which he suddenly decided he needed to announce that he was not reading any more replies... Although announcing that he'd deleted the replies containing math and then also complaining that the replies he was reading didn't contain any math was a very bold step.) >>Again, it's easy to evaluate the last integral; >>int_T^infinity (e^(-st)s ) = e^(-sT), so we >>finally get f(T)e^(-sT) for the Laplace transform. ><**daestrom ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > Unfortunately, although he claimed to have posted such, it was > masked by his over-riding urge to be insulting and was therefore > not seen by me. >>Well, here is what David C. Ullrich posted.... (lines with >> are from >>Airy >>R. Bean's previous post, those preceded by '>' are from David C. Ullrich's >>pose, my comments are embedded with '<*<') >>Show where it is wrong, or else resort >>to infantile sneering. >Ok. Took me a minute to find the place >where you showed us the integration >by parts: >Certainly, if you integrate by parts, you would choose >>int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >>I try this, I get 0!...... >>int(UV) = U.int(V) - int[dU.int(V)] giving..... >> int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >>with f(t).d(t-T) as V and e^(-sT) as U ..... >>f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >>f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >>f(T).e^(-sT) - f(T).e^(-sT).... >>0. Your first step, > f(T).e^(-st) - int(e^(-st)/-s . f(T)) >is already wrong. My guess is because of a confusion >over definite integrals versus antiderivatives, ie >indefinite integrals. The (definite) integral from >0 to infinity of f(t).d(t-T) is indeed f(T). But >what we need here is an antiderivative. >An antiderivative of f(t).d(t-T) is given by >F(t), where F(t) = 0 for t < T and f(T) for t > T. >So that line should be > F(t).e^(-st) - int(e^(-st)/-s . F(t)). >><**Now when we put in the limits t = 0 to t = infinity >the first term vanishes since F(0) = 0 and e^(-st) >tends to 0 as t -> infinty (at least if s > 0; if s < 0 >this integration by parts is not going to work.) >So out Laplace transform becomes > - int_0^infinity (e^(-st)/-s . F(t)) >Since F(t) = 0 for t < T and f(T) for t > T >this equals > f(T) int_T^infinity (e^(-st)s ). >><*>It >>is apparent the '-' was taken out front. But since the beginning, dU >>should >>have been '-e^(-st)s' with the 's' in the numerator, not denominator. I >>suspect this error came from ARB's original attempt at the integration by > I pointed out explicitly, in lines you snipped, that Airy had > written e^{-st}/s where it should have been e^{-st} s. I thought > that I'd fixed the error at that point, it's possible that some > erroneous cut&pasted lines were not edited properly. > (Also pointed out at least twice so far that this is somewhat > amusing. Before looking at it I knew that his integration by > parts must be wrong, since it gave the wrong answer, but I > assumed that the problem was with something subtle about > delta functions. Turns out that in addition to not getting > the antiderivative right his amazing conclusion is based > in part on simply differentiating an exponential incorrectly! Yes, I was not trying to point the finger at you for the original error, just pointing out that at that particular point you corrected your explanation (for those that may have wondered how the 's' term suddenly moved from denominator to numerator). And that, as I said, I suspect this error came from ARB's original... posting, not you per se. Interesting bit of calculus, I may have to go visit sci.math to see what other interesting things are bandy'd about. daestrom === Subject: Re: Infantile authours degrading this NG. > Interesting bit of calculus, I may have to go visit sci.math to see what > other interesting things are bandy'd about. > daestrom Take ARB with you. === Subject: Re: Infantile authours degrading this NG. The bold step that I took was to make a stand against infantile ad hominem attacks as is your wont. You continue with your motivation to insult in your posting below. If you have anything of value to offer, then offer it in a mature post. Shame on you. You should know better. (If you adopt the same arrogant stance to your pupils as you do here, and if any of those pupils are reading this NG, you can be sure that you are the laughing stock in your school (if not so already)) > It's easy to see why this was the point at which he suddenly > decided he needed to announce that he was not reading any > more replies... Although announcing that he'd deleted the > replies containing math and then also complaining that > the replies he was reading didn't contain any math was > a very bold step.) === Subject: Re: Infantile authours degrading this NG. <310ft1F35b2uqU4@uni-berlin.de> <3132s3F30bqd5U1@uni-berlin.de> <315hu5F3491fhU4@individual.net> <315nt5F372fc6U1@individual.net> <316s92F37f4hjU1@individual.net> <38ttq0pn4ujphmaesmpmjfhj8m767apgos@4ax.com> <318bvsF35ggl4U15@individual.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > The bold step that I took was to make a stand > against infantile ad hominem attacks as is your wont. > You continue with your motivation to insult in your > posting below. > If you have anything of value to offer, then offer it > in a mature post. > Shame on you. You should know better. The irony of it, the irony. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Infantile authours degrading this NG. >> The bold step that I took was to make a stand >> against infantile ad hominem attacks as is your wont. >> You continue with your motivation to insult in your >> posting below. >> If you have anything of value to offer, then offer it >> in a mature post. >> Shame on you. You should know better. >The irony of it, the irony. Indeed. But you left out the best bit, irony-wise: >(If you adopt the same arrogant stance to your >pupils as you do here, and if any of those pupils are >reading this NG, you can be sure that you are the >laughing stock in your school (if not so already)) ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. > The bold step that I took was to make a stand > against infantile ad hominem attacks as is your wont. You continue with your motivation to insult in your > posting below. If you have anything of value to offer, then offer it > in a mature post. Shame on you. You should know better. >> The irony of it, the irony. > Indeed. But you left out the best bit, irony-wise: >> (If you adopt the same arrogant stance to your >> pupils as you do here, and if any of those pupils are >> reading this NG, you can be sure that you are the >> laughing stock in your school (if not so already)) Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie (real name Gareth) is not competent enough in the subject matter to be teaching it to anyone. despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002.shtml Even kids would be able to figure out quickly that they cannot learn (except by rote, decidedly a bad way to learn math) from Beanie because what he says makes no sense because it is false. even more so with adults. you cannot systemically learn underlying concepts from misconceptions. and you cannot learn math without learning conceptually. i imagine that Beanie is unemployed and (due to his character) unemployable. r b-j === Subject: Re: Infantile authours degrading this NG. > Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie > (real name Gareth) is not competent enough in the subject matter to be > teaching it to anyone. > despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002.shtml Wrong Gareth The Gareth that is giving you the run around, and that is exactly what he is doing, is in his 50s. He attended Essex University, although it appears there little evidence he graduated, and has an varied employment history. He has admitted being dismissed from on job. All the evidence is he screwed up big time. Other posts by him suggest that this was not an isolated case. He trolls varies newsgroups with his wind-ups on a range of topics, DSP is one his favorite topics- his theories on sampling first came to light several years ago. They have been refuted many times. Other topics have included a justification for the Soham murders (for US readers, two teenage girls were murdered in Soham a few years ago). He appears to be unemployable and becomes fixated on those who have won out arguements against him. Do some googling and you will find evidence of all the above. If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. === Subject: Re: Infantile authours degrading this NG. : He appears to be unemployable and becomes fixated on those who have won out : arguements against him. said NIMROD hiding in the shadows thowing stones from a safe distance : : If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. said NIMROD hiding in the shadows thowing stones from a safe distance === Subject: Re: Infantile authours degrading this NG. >> Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie >> (real name Gareth) is not competent enough in the subject matter to be >> teaching it to anyone. >> despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002.shtml > Wrong Gareth > The Gareth that is giving you the run around, and that is exactly what he is > doing, is in his 50s. i found a few old postings where *he* says he's class of '72 (not 2002). same middle name, same first name, same last name. education in mathematics. lot'sa coincidences for someone on this side of the pond. i'm sure you UKers know him better. > He attended Essex University, although it appears there little evidence he > graduated, and has an varied employment history. He has admitted being > dismissed from on job. All the evidence is he screwed up big time. i picked that up. > Other posts by him suggest that this was not an isolated case. > He trolls varies newsgroups with his wind-ups on a range of topics, DSP is > one his favorite topics- his theories on sampling first came to light > several years ago. They have been refuted many times. i know that. another reason i thought it was the young guy in the photo is that his arguments are really neophyte (his integration-by-parts and his concepts surrounding bandlimited sampling are indicative of someone who hasn't really done it) and his means of argument are very immature. not reflective of someone in their 50s. > Other topics have included a justification for the Soham murders (for US > readers, two teenage girls were murdered in Soham a few years ago). > He appears to be unemployable and becomes fixated on those who have won out > arguements against him. > Do some googling and you will find evidence of all the above. it's exactly what i did. > If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL. i agree. r b-j === Subject: Re: Infantile authours degrading this NG. >same middle name, same first name, same last name. education in >mathematics. lot'sa coincidences for someone on this side of the pond. i'm >sure you UKers know him better. There are some historical consistencies with the email addresses with which he's been associated that put him as an experienced ringer (a person who, with other folks, pulls the ropes to toll peals and carols from church bell towers), with email traffic to related lists back to at least 1998. He was upsetting some of those folks with his views, too, and his account of his historical geographic locations is consistent with where he is now. So I think the young guy in the website you found is unlikely to be Airy. The employment history issues also seem to indicate the same. He seems to be interested in and experienced with music, which is something he has in common with you and I, though. ;) Apologies to those of you who've dealt with this for way too long. He's still a little new to us at comp.dsp, but we've had very similar experiences with other people. Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org === Subject: Re: Infantile authours degrading this NG. > So I think the young guy in the website you found is unlikely to be > Airy. The employment history issues also seem to indicate the same. i've come around to that conclusion. it's just that with an identical first, middle, and last name and some formal study in electrical engineering and math and being in the UK, that just seemed to be too many coincidences for me initially. > Apologies to those of you who've dealt with this for way too long. > He's still a little new to us at comp.dsp, but we've had very similar > experiences with other people. *very* similar? i think Beanie is in a league of his own. his trolling mastery *way* outstrips eBob. r b-j === Subject: Re: Infantile authours degrading this NG. Nimrod for a while caused the following to be typed: > If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE > TROLL. Best advice I've heard on this subject so far. -- Radio glossary #21 Packet: An item which causes your XYL to not speak to you for five days when it arrives along with a Waters and Stanton invoice. === Subject: Re: Infantile authours degrading this NG. briefly between bottles, to write: >He appears to be unemployable and becomes fixated on those who have won out >arguements against him. Which let's be honest, is pretty much everybody. Nick. === Subject: Re: Infantile authours degrading this NG. >Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie >(real name Gareth) is not competent enough in the subject matter to be >teaching it to anyone. >despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002.shtml ... >i imagine that Beanie is unemployed and (due to his character) unemployable. >r b-j I'm don't think that's the same Gareth Evans (way too young). Maybe he can enlighten us. It was evident to me yesterday that that's apparently not an uncommon name in the UK (or Australia). Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org === Subject: Re: Infantile authours degrading this NG. >I'm don't think that's the same Gareth Evans (way too young). Maybe >he can enlighten us. It was evident to me yesterday that that's >apparently not an uncommon name in the UK (or Australia). Gareth is a pretty common Welsh name. Evans is very common Welsh name. Put them together and you get a fairly common, and very Welsh name :-) Steve === Subject: Re: Infantile authours degrading this NG. > Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie >(real name Gareth) is not competent enough in the subject matter to be >teaching it to anyone. >despite http://www.informatics.bangor.ac.uk/public/news/prizes_2002.shtml >Even kids would be able to figure out quickly that they cannot learn (except >by rote, decidedly a bad way to learn math) from Beanie because what he says >makes no sense because it is false. even more so with adults. you cannot >systemically learn underlying concepts from misconceptions. and you cannot >learn math without learning conceptually. >i imagine that Beanie is unemployed and (due to his character) unemployable. >r b-j Not too sure about that logic. When I was doing A-level maths (A-levels are the UK exams at 18, before entering university) we had a really bad maths teacher. Everyone knew she was really bad, so we largely ignored her. We worked extra hard on our own from books, and did lots of exercising with the published past papers. The final result was we all achieved really excellent grades in which is a bit sad, but the bottom line is her incompetance actually achieved good results. :-) Steve === Subject: Re: Infantile authours degrading this NG. >> Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie >> (real name Gareth) is not competent enough in the subject matter to be >> teaching it to anyone. ... >> Even kids would be able to figure out quickly that they cannot learn (except >> by rote, decidedly a bad way to learn math) from Beanie because what he says >> makes no sense because it is false. ... > Not too sure about that logic. > When I was doing A-level maths (A-levels are the UK exams at 18, before > entering university) we had a really bad maths teacher. Everyone knew > she was really bad, so we largely ignored her. makes sense. i was saying that they cannot learn ... *from* Beanie.... > We worked extra hard on > our own from books, and did lots of exercising with the published past > papers. The final result was we all achieved really excellent grades in > which is a bit sad, but the bottom line is her incompetance actually > achieved good results. :-) just because of two outcomes that might be correlated, it does not mean that one was the cause (ty teacher) and the other was the effect (good test results). you do not know how well you'ld have done with a really good teacher. perhaps it is that you did so well despite the poor teacher not because of the poor teacher. i am an advocate of good teaching. good, effective, teachers have to be, at the very least, competent in the subject and in related subjects that they are teaching. my $0.02 . r b-j === Subject: Re: Infantile authours degrading this NG. >You claim to be a mathematician of 20 years' standing. >I suggest that you do not try to be a teacher of mathematics. >Mathematics is an abstract subject for which your students >must be relaxed in mind in order to be able to take on >abstractions. Your emotive and insulting posts >will disrupt such a state of mind. In my time as a part-time >tutor of mathematics to adults I came across this time and >time again - people who have been put off mathematics, not >by the subject per se, but by the aggression and personal >attacks addressed to them by their teachers. Guffaw. Every once in a while I get advice like this here on sci.math. Curiously, it always comes from someone who's totally wrong about some bit of mathematics, who's insisting he's right and refusing to accept simple explanations of why he's wrong. Been teaching math for 20 years, and no situation even remotely like that has ever come up in the classroom. >Your infantile tirade below illustrates well the persons proscribed >Shame on you. >(I notice that you do not address the mathematical points below) Guffaw again. The mathematical points have been addressed quite completely. Below I refer to a post where I do address a new mathematical point (that you're simply doing the basic calculus wrong), which post you _say_ that you deleted unread. Which is of course another classic crackpot technique: ignore substantive objections and then claim they don't exist (or repeatedly ignore substantive objections and then whine about the fact that they're not repeated in every reply.) >> Classic crackpot. If nobody says you're wrong you must >> be right. But if a large number of people say you're wrong >> it follows you must be right as well. >> Things must be very pleasant in your little world. >> Btw, since you don't seem to have noticed my post >> where I reply to your request to show _where_ your >> integration by parts fails: When I said that >> integration by parts does not apply because delta >> is not a continuous I was assuming that you >> were making one sort of moderately subtle error. >> Hadn't looked at the actual argument you gave. >> In case you didn't see that post, in your >> integration by parts you simply do the basic >> calculus wrong. >>It is interesting the size of the amount of posts. >>If I am wrong in what I say, it would be a simple >>matter to ignore me and what I posit. >>That so many people, people who would seek to >>represent themselves as authorities, respond with >>emotional postings that do not address the >>points raised, seems to suggest that I am not wrong, and that >>they are embarrassed about a basic failing in their >>professed knowledge and so feel threatened. >>the amount of posts, posts that are non-technical and >>of an undesirable ad hominem style has reached such a >>scale that I have binned them all this morning. >>This leaves just one point that has not yet been addressed, and >>that is, that any mathematical analysis of a real system must, >>to be respectable, deal with measurements of that system. >>There is no system of which I am aware that has sampling >>pulses whose measurements match those of the Diracian >>Impulse... >>1. Their amplitudes do not approach infinity. >>2. Their areas do not approach unity. >>3. In any case, the operation of f(t).d(t - T) is not >>defined unless under an integral sign, and cannot be >>evaluated unless under that integral sign. >>4. The evaluation of the spectrum of the Diracian relies >>on it being over all time, -oo^+oo. It is improper, therefore, >>to attempt to evaluate other operations using the Diracian >>with a reduced domain, and yet still rely on the spectrum >>derivation. > of posts that it has generated, it stands to reason that you are > unlikely to get a response suitable to you from them. >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Infantile authours degrading this NG. >Which is of course another classic crackpot technique: ignore >substantive objections and then claim they don't exist (or >repeatedly ignore substantive objections and then whine >about the fact that they're not repeated in every reply.) At one time on uk.radio.amateur, Bean, in one of his previous sock-puppets, had requested help with a technical problem with an off-the-shelf amateur transceiver he had bought. He received a number of helpful, technically-based replies and suggestions. It came as little surprise that it was at that precise moment he suffered an ISP failure, and never saw the replies. He was the only one on the ng who used that particular ISP who reported such a failure..... FUs set to remove uk.radio.amateur. -- from Aero Spike === Subject: Re: Infantile authours degrading this NG. There is nothing amiss about deleting posts that are a thinly-disguised vehicles for delivering infantile abuse, such as yours below. Shame on you - you should know better if you claim 20 years experience as an expert. No expert needs to hind behind abusive and infantile tirades. > Guffaw again. The mathematical points have been addressed > quite completely. Below I refer to a post where I do address > a new mathematical point (that you're simply doing the basic > calculus wrong), which post you _say_ that you deleted > unread. > Which is of course another classic crackpot technique: ignore > substantive objections and then claim they don't exist (or > repeatedly ignore substantive objections and then whine > about the fact that they're not repeated in every reply.) === Subject: Re: Infantile authours degrading this NG. As with all escapees from the infants' school playground, I invite you to repost any technically relevant posts, but without your childish and unnecessary infantile outpourings - then I will deal with the points raised. I assume that okstate.edu is some form of infants' school and that you are copying the behaviour of your pupils? > There is nothing amiss about deleting posts that > are a thinly-disguised vehicles for delivering > infantile abuse, such as yours below. > Shame on you - you should know better if you > claim 20 years experience as an expert. > No expert needs to hind behind abusive and > infantile tirades. > Guffaw again. The mathematical points have been addressed > quite completely. Below I refer to a post where I do address > a new mathematical point (that you're simply doing the basic > calculus wrong), which post you _say_ that you deleted > unread. > Which is of course another classic crackpot technique: ignore > substantive objections and then claim they don't exist (or > repeatedly ignore substantive objections and then whine > about the fact that they're not repeated in every reply.) === Subject: Re: Infantile authours degrading this NG. I have deleted posts which are excuses for gratuitous abuse. No point reading them further. > Guffaw again. The mathematical points have been addressed > quite completely. Below I refer to a post where I do address > a new mathematical point (that you're simply doing the basic > calculus wrong), which post you _say_ that you deleted > unread. === Subject: Re: Infantile authours degrading this NG. >Mathematics is an abstract subject for which your students >must be relaxed in mind in order to be able to take on >abstractions. >In my time as a part-time >tutor of mathematics to adults I came across this time and >time again - people who have been put off mathematics, not >by the subject per se, but by the aggression and personal >attacks addressed to them by their teachers. In my time I came across classes in which the range of mathematical abilities varied from the incompetent to the brilliant. I can't recall a case of anyone being put off by the teacher - but plenty that were put off by teaching methods. You may, of course, be referring to your own case - which is not necessarily generally true. FUs set to remove uk.radio.amateur. -- from Aero Spike === Subject: Re: Infantile authours degrading this NG. I suggest that if you wish to gain any respect and be seen as a potential mathematics professional that you drop your wont for making personal remarks and that you concentrate on the matters in hand. You seem to have no experience of acadaemic debate, and I can tell you that Argumentum Ad Hominem, which is your wont, has no place in reputable acadaemic institutions. Do us all a favour, Mr.Ullrich and grow up, or else leave this forum. > You claim to be a mathematician of 20 years' standing. > I suggest that you do not try to be a teacher of mathematics. > Mathematics is an abstract subject for which your students > must be relaxed in mind in order to be able to take on > abstractions. Your emotive and insulting posts > will disrupt such a state of mind. In my time as a part-time > tutor of mathematics to adults I came across this time and > time again - people who have been put off mathematics, not > by the subject per se, but by the aggression and personal > attacks addressed to them by their teachers. > Your infantile tirade below illustrates well the persons proscribed > Shame on you. > (I notice that you do not address the mathematical points below) > Classic crackpot. === Subject: Re: Infantile authours degrading this NG. I suggest that if you wish to gain anyone's respect in this forum, you learn to quit top posting like a newbie, quit trying to impress everyone with your amazingly brilliant but trivial technobabble, and learn not to get all pissy the first time someone begs to differ with you. > I suggest that if you wish to gain any respect and be > seen as a potential mathematics professional that you > drop your wont for making personal remarks and > that you concentrate on the matters in hand. > You seem to have no experience of acadaemic debate, and > I can tell you that Argumentum Ad Hominem, which is your > wont, has no place in reputable acadaemic institutions. > Do us all a favour, Mr.Ullrich and grow up, or else leave > this forum. > You claim to be a mathematician of 20 years' standing. > I suggest that you do not try to be a teacher of mathematics. > Mathematics is an abstract subject for which your students > must be relaxed in mind in order to be able to take on > abstractions. Your emotive and insulting posts > will disrupt such a state of mind. In my time as a part-time > tutor of mathematics to adults I came across this time and > time again - people who have been put off mathematics, not > by the subject per se, but by the aggression and personal > attacks addressed to them by their teachers. > Your infantile tirade below illustrates well the persons proscribed > Shame on you. > (I notice that you do not address the mathematical points below) > Classic crackpot. -- Fairy A. Bean === Subject: Re: Infantile authours degrading this NG. >That so many people, people who would seek to >represent themselves as authorities, respond with >emotional postings that do not address the >points raised. But you yourself have deleted whole posts - some of which were long and detailed and which dealt with the points you raised - because you perceived something you defined as 'childish' early on. If you believe that to be a mature approach to debate, then you are in a very small minority. You have been repeatedly told that the way to deal with (what you perceive as) 'childishness' is to ignore it and deal with the substance. This (positive) behaviour reinforces the message that negativity is not the means to achieve a satisfactory outcome. Unfortunately, your approach merely reinforces negative behaviour and is ultimately self-defeating. FUs set to remove uk.radio.amateur, where Bean's approach to discussion has been discussed ad nauseum. -- from Aero Spike === Subject: Re: Infantile authours degrading this NG. > I suggest that if you wish to gain any respect and be > seen as a potential mathematics professional that you > drop your wont for making personal remarks and > that you concentrate on the matters in hand. Alas, you are confused. And what is worse, given to top-posting. === Subject: Re: Infantile authours degrading this NG. authors. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Infantile authours degrading this NG. >It is interesting the size of the amount of posts. Please remove uk.radio.amateur from your FUs. TIA -- from Aero Spike === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? A > Certainly, if you integrate by parts, you would choose > int(f(t).d(t-T)) as the integrated bit to yield f(T), but when > I try this, I get 0!...... > int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... > with f(t).d(t-T) as V and e^(-sT) as U ..... > f(T).e^(-st) - int(e^(-st)/-s . f(T))..... > f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... > f(T).e^(-sT) - f(T).e^(-sT).... > 0. let's integrate by parts correctly: L[x(t).delta(t-T)] = int_0^infinity exp(-st).x(t).delta(t-T).dt using int_a^b udv = (uv)|_a^b - int_a^b v.du as our formulation of parts, choose: u = exp(-st) => du = -s.exp(-st)dt dv = x(t).delta(t-T) dt => v = int_0^t x(u) delta(u-T)du = 0 if t < T, or x(T) if t >= T so, L[x(t).delta(t-T)] = [exp(-st).v(t)]_0^inf - int_0^inf -s.exp(-st).v(t)dt = 0 + s.int_T^inf x(T)exp(-st)dt = [-x(T).exp(-st)]_T^inf = x(T).exp(-sT) hopefully that is satisfactory for you, I think the sloppiness is warranted. Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? hey Beanie, i have proposal for you. would you promise to shut up and go away (or, at least, just shut up) if i show you where your flaw is in your integration by parts? it's amazing why you don't just accept the simple proof (applying the definition of the dirac impulse function) that +inf integral{ x(t) d(t-T) e^(-st) dt} ( d(t) = dirac impulse function ) -inf +inf = integral{ [x(t) e^(-st)] d(t-T) dt} -inf +inf = integral{ x1(t) d(t-T) dt} ( where x1(t) = x(t) e^(-st) ) -inf +inf = integral{ x1(u+T) d(u) du} ( substitute u = t-T ) -inf +inf = integral{ x2(u) d(u) du} ( where x2(t) = x1(t+T) ) -inf = x2(0) ( by definition of the dirac impulse ) = x1(T) = x(T) e^(-sT) now that's how normal people without mental illness will evaluate that bilateral Laplace Transform integral. (people *with* mental illness may want to use a less direct and more difficult method so that they will more likely screw up and get the wrong answer.) for some reason Beanie wants to evaluate it by parts and he insists he gets a different answer (zero). setting aside the lesson of Occam's Razor ( http://en.wikipedia.org/wiki/Occam%27s_razor the simplest explanation should be used), the answer *if* you integrate by parts *correctly* (which Beanie can't seem to do correctly), the integral turns out to be the same. Laplace{ x(t) d(t-T) } = x(T) e^(-sT) not zero as Beanie claims Beanie, if you promise to behave yourself and stop trolling this newsgroup, i'll show your flaw in your integration by parts. both integrating by parts and the straight forward integration comes out with the same answer (as if any of us suspected otherwise). but i'm gonna use the *textbook* definition of integrating by parts: b | b b integral{ u dv} = (uv)| - integral{ v du} a | a a (Beanie, you'll need a mono-spaced font to read this ASCII math well.) Beanie wants (if he agrees to conventional notation) u = e^(-st) and dv = x(t) d(t-T) dt Beanie, if you agree to those (very reasonable) terms, i will show you (and everyone else reading it) the flaw in your derivation that erroneously says the result is zero. r b-j > with as follows..... > You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to > determine your result. > You simply cannot say that f(t).d(t-T) yields f(T) unless > you do so under an integral. This arises from the fundamental > properties of the Diracian Delta function . > Certainly, if you integrate by parts, you would choose > int(f(t).d(t-T)) as the integrated bit to yield f(T), but when > I try this, I get 0!...... that's because your knowledge and insight is far less than you think and you're also quite mentally ill. (perhaps only the latter is true if you're the master troll Jerry makes you out to be, but both are true if you're simply the obnoxious crackpot and jerk that is more ostensible.) > int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... > with f(t).d(t-T) as V and e^(-sT) as U ..... > f(T).e^(-st) - int(e^(-st)/-s . f(T))..... > f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... > f(T).e^(-sT) - f(T).e^(-sT).... > 0. > What I seek is a sound mathematical proof of the claim > that f(t).d(t-T) gives rise to a spectrum contribution > of f(T).e^-(st), and claiming that f(t).d(t-T) equals > f(T) is not sound. We use the properties of the Diracian > Delta Function in so many other aspects of signal > processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? Because of your childish behaviour quoted below, something of value to contribute to the discussion, then please present it in a style more appropriate to an international forum. > hey Beanie, > i have proposal for you. would you promise to shut up and go away (or, at > least, just shut up) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? are you afraid to see your mistake explicitly exposed? cluck, cluck r b-j > Because of your childish behaviour quoted below, > something of value to contribute to the discussion, > then please present it in a style more appropriate to > an international forum. >> hey Beanie, >> i have proposal for you. would you promise to shut up and go away (or, at >> least, just shut up) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? I am not afraid of anything, be that anything you, or the truth. It is simply that I refuse to deal with you when you indulge yourself in infantile outbursts. Shame on you. You should know better, including not indulging in silly yah-boo-sucks tirades as in your latest rant below. You do nothing for your reputation by so behaving. > are you afraid to see your mistake explicitly exposed? > cluck, cluck > Because of your childish behaviour quoted below, > something of value to contribute to the discussion, > then please present it in a style more appropriate to > an international forum. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > What I seek is a sound mathematical proof of the claim > that f(t).d(t-T) gives rise to a spectrum contribution > of f(T).e^-(st) For what it's worth, tongue firmly in cheek, and with apologies to David Ulrich, who did it properly: You seem to have a problem with the Dirac Delta definition. Anyway, presumably you are happy with the Laplace transform L[f(t).delta(t-T)] = int_0^infinity f(t).delta(t-T).exp(-st) dt Maybe the notion that the Delta function is, loosely speaking, zero nearly everywhere is sort-of vaguely familiar to you, so you might be convinced that something like this is true: L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt And maybe considering f(t) and exp(-st) might get you ruminating on their continuity, and a daring insight like this may leap out: L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt Forgive me if I'm getting carried away, but isn't the Riemann Integral linear? I think it might be, so wouldn't that mean that L[f(t).delta(t-T)] = lim M->0 f(T).exp(-sT) int_(T-M)^(T+M) delta(t-T)dt I really think that I might be onto something here, but I need to work out how to integrate that pesky Dirac Delta ... any takers? === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? In your second equation, how come you've still got t in your expression of the Diracian, but you've replaced it by T in the other two terms? It would seem to me that if a principle is to be applied, then it must be applied everywhere, so that you must end up with d(T -T) = d(0). > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt > And maybe considering f(t) and exp(-st) might get you ruminating on > their continuity, and a daring insight like this may leap out: > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> What I seek is a sound mathematical proof of the claim >> that f(t).d(t-T) gives rise to a spectrum contribution >> of f(T).e^-(st) ... > I really think that I might be onto something here, but I need to work > out how to integrate that pesky Dirac Delta ... any takers? it's actually pretty straight-forward to do it right (it's on the other post i just made). Beanie prefers to use integration-by-parts, and then makes a mistake (he actually doesn't do it right from the beginning) and comes out with zero so then he claims that all accumulated knowledge since Heaviside and Nyquist about it is in error. don't fall for his bull. he really doesn't know what he's typing about. r b-j === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >>I really think that I might be onto something here, but I need to work >>out how to integrate that pesky Dirac Delta ... any takers? > it's actually pretty straight-forward to do it right (it's on the other post > i just made). Beanie prefers to use integration-by-parts, and then makes a > mistake (he actually doesn't do it right from the beginning) and comes out > with zero so then he claims that all accumulated knowledge since Heaviside > and Nyquist about it is in error. don't fall for his bull. he really > doesn't know what he's typing about. > r b-j Disclaimer, my previous post was not to be taken very seriously - the maths was intentionally hand-wavy and tongue-in-cheek. David Ullrich did it the standard way further up. I just integrated it by parts for Mr Bean, so he is hopefully now satisfied ... Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > I just integrated it by parts for Mr Bean, so he is hopefully now > satisfied ... but shucks! you gave it away! for free! i wanted to extract a promise (that the asshole would probably not keep) before telling him what he did wrong. oh well, it's public knowledge anyway. r b-j === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > For what it's worth, tongue firmly in cheek, and with apologies to David > Ulrich, who did it properly: Even more apologies to David Ullrich for spelling his name wrong === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> For what it's worth, tongue firmly in cheek, and with apologies to David >> Ulrich, who did it properly: >Even more apologies to David Ullrich for spelling his name wrong No problem - I assumed you were talking about someone else, wondered why I didn't see any posts from him... (the spelling with one l is much more common.) ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? Can you either stop intentially re-arranging the letters and spacings in the subject line of these posts ... or if it's unintentional, fix your usenet posting application so it stops doing so? It's a little disorientating to have a thread breaking off into 5 or 6 different permutations within the group ( appears to be localized to your replies ). It's making a mess in Google as well. >with as follows..... >You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to >determine your result. >You simply cannot say that f(t).d(t-T) yields f(T) unless >you do so under an integral. This arises from the fundamental >properties of the Diracian Delta function . >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >What I seek is a sound mathematical proof of the claim >that f(t).d(t-T) gives rise to a spectrum contribution >of f(T).e^-(st), and claiming that f(t).d(t-T) equals >f(T) is not sound. We use the properties of the Diracian >Delta Function in so many other aspects of signal >processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform >like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. ( modify address for return mail ) www.numbersusa.com www.americanpatrol.com === Subject: Deviration Of The Spectrom Due To f(t).d(t - T)????? No. Sorry. Good reason to do so. > Can you either stop intentially re-arranging the letters and spacings > in the subject line of these posts ... or if it's unintentional, fix > your usenet posting application so it stops doing so? It's a little > disorientating to have a thread breaking off into 5 or 6 different > permutations within the group ( appears to be localized to your > replies ). It's making a mess in Google as well. >with as follows..... >You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to >determine your result. >You simply cannot say that f(t).d(t-T) yields f(T) unless >you do so under an integral. This arises from the fundamental >properties of the Diracian Delta function . >Certainly, if you integrate by parts, you would choose >int(f(t).d(t-T)) as the integrated bit to yield f(T), but when >I try this, I get 0!...... >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. >What I seek is a sound mathematical proof of the claim >that f(t).d(t-T) gives rise to a spectrum contribution >of f(T).e^-(st), and claiming that f(t).d(t-T) equals >f(T) is not sound. We use the properties of the Diracian >Delta Function in so many other aspects of signal >processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform >like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. > ( modify address for return mail ) > www.numbersusa.com > www.americanpatrol.com Author-Supplied-Address: Anonymous-Remailer See Comment Header === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? Comments: This message did not originate from the above address. It was remailed by two or more anonymous mail services. Send complaints of REAL abuse ONLY to remaileradmin at optonline dot n Mail-To-News-Contact: abuse@dizum.com > No. Sorry. Good reason to do so. Your desperate need for attention is a good reason only in the mind of selfish arseholes like yourself. -=- This message was posted via two or more anonymous remailing services. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? Stupid boy. I see that you don't have the strength of character Stupid cowardly boy. > No. Sorry. Good reason to do so. > Your desperate need for attention is a good reason only in the mind of > selfish arseholes like yourself. > -=- > This message was posted via two or more anonymous remailing services. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? And what would that good reason be? BTW, MS Outlook Express, despite its failings, does group things by thread even if the subject line has been changed. > No. Sorry. Good reason to do so. > Can you either stop intentially re-arranging the letters and spacings > in the subject line of these posts ... or if it's unintentional, fix > your usenet posting application so it stops doing so? It's a little > disorientating to have a thread breaking off into 5 or 6 different > permutations within the group ( appears to be localized to your > replies ). It's making a mess in Google as well. === Subject: Re: Deviration Of Teh Spectrom Due To f(t).d(t - T)????? Confidential, sorry. > And what would that good reason be? BTW, MS Outlook Express, despite its > failings, does group things by thread even if the subject line has been changed. > No. Sorry. Good reason to do so. > Can you either stop intentially re-arranging the letters and spacings > in the subject line of these posts ... or if it's unintentional, fix > your usenet posting application so it stops doing so? It's a little > disorientating to have a thread breaking off into 5 or 6 different > permutations within the group ( appears to be localized to your > replies ). It's making a mess in Google as well. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? > No. Sorry. Good reason to do so. >int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-sT) as U ..... f(T).e^(-st) - int(e^(-st)/-s . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. Your dU, ie dU/dt, ie d/dt e^(-sT) equals zero. Thus int[dU.int(V)] = int(0.f(T)) = 0. and U.int(V) = e^(-sT) . f(T), which is equal to the left handside. Wilbert === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? Yes it's a constant. Where did I cock up? Let's see..... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-sT) as U ..... f(T).e^(-st) - int(e^(-st)/-s . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. Ah yes, typo, should read..... with f(t).d(t-T) as V and e^(-st) as U ..... also the next line is wrong, should read..... f(T).e^(-st) - int(-s.e^(-st) . f(T))..... However the effect of a consecutive differentiation followed by an integral is still to give -s/-s. Giving.... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with f(t).d(t-T) as V and e^(-st) as U ..... f(T).e^(-st) - int(-s.e^(-st) . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT).... 0. > No. Sorry. Good reason to do so. >int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... >with f(t).d(t-T) as V and e^(-sT) as U ..... >f(T).e^(-st) - int(e^(-st)/-s . f(T))..... >f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... >f(T).e^(-sT) - f(T).e^(-sT).... >0. > Your dU, ie dU/dt, ie d/dt e^(-sT) equals zero. === Subject: Re: Deviration Of The Spectrom Due To f(t).d(t - T)????? > No. Sorry. Good reason to do so. Do what? === Subject: Deviration Of The Spectrom Deu To f(t).d(t - T)????? Do what was complained of. > No. Sorry. Good reason to do so. > Do what? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > Do what was complained of. Where? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? In a previous post. > Do what was complained of. > Where? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > In a previous post. Top-posting makes the whole thing impractical. === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? Nothing wrong with top-posting. It is the preferred method, because you don't have to page down through much repeated and already-seen history. skipped over unread. > In a previous post. > Top-posting makes the whole thing impractical. === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? that... > Nothing wrong with top-posting. That's what clueless newbies always say. > It is the preferred method, because you don't have to > page down through much repeated and already-seen > history. > skipped over unread. > In a previous post. > Top-posting makes the whole thing impractical. -- Checkmate all rights reserved === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > Nothing wrong with top-posting. Clueless! > It is the preferred method, because you don't have to Really? And what exactly is your supporting evidence? A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > A: Maybe because some people are too annoyed by top-posting. It is precisely because of this that he does it. His pathology requires that he annoy people. Bob -- Things should be described as simply as possible, but no simpler. A. Einstein === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? > Nothing wrong with top-posting. > It is the preferred method, because you don't have to > page down through much repeated and already-seen > history. More properly, It is preferred by some, because they don't have to page down through much repeated and already-seen history. It is not universally 'preferred' by all use-net users. In fact, there are several use-nettiquite > skipped over unread. ... by those that don't want to scroll through the previous context. Others, that don't mind reading the postings in context don't mind at all. In fact, many prefer in-context posting as it is much clearer what the poster is replying to. In the end, in-context posting, with snippage of unrelated, repeatedly quoted material is 'preferred' by many. You don't speak for the entire use-net community, so don't pretend that you do. daestrom === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? >> Nothing wrong with top-posting. >> It is the preferred method, because you don't have to >> page down through much repeated and already-seen >> history. > More properly, It is preferred by some, because they don't have to page > down through much repeated and already-seen history. It is not > universally > 'preferred' by all use-net users. I'd say it was preferred by wankers, but that's just my humble opinion. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Longest Thread Ever In de.sci.mathematik Thread: Hat Cantor doch geirrt? Actually 3095 postings === Subject: Re: Longest Thread Ever words floating about: >In >de.sci.mathematik >Thread: Hat Cantor doch geirrt? >Actually 3095 postings Dot's nice, still no pun for discussion? === Subject: Re: Longest Thread Ever isn't > In > de.sci.mathematik > Thread: Hat Cantor doch geirrt? > Actually 3095 postings The Greatest Man Ever hit 20,000 postings last April. === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Just because the factors are divisible by 7 when x=0, it doesn't > follow that the factors are divisible by 7 for all values of x. > It follows from the distributive property. James Harris === Subject: Re: JSH:Understanding constant terms ... > Just because the factors are divisible by 7 when x=0, it doesn't > follow that the factors are divisible by 7 for all values of x. ... > It follows from the distributive property. Back on that again James? No, it does *not* follow from the distributive property. You have r_3(x) + 22 which is not divisible by any factor of 7 when x = 0. You claim that it is not divisible by any factor of 7 when x != 0 (and base that on the distributive property). The claim you made quite some time ago was that if for some x, and some factor f of 7, (r_3(x) + 22)/f was an algebraic integer that than (due to the distributive property) in r_3(x)/f + 22/f both terms should be algebaric integers. But that is *not* the distributive property, and it is false. Consider x + 1 in the algebraic integers. By your interpretation for all x (x + 1)/2 would not be an algebraic integer. On the other hand, you have r_1(x) + 7, which is divisible when x = 0, but is not necessarily divisible by 7 when x != 0. And I do not know how you intend to show such using the distributive property, because this is also obviously false. x + 2 is divisible by 2 when x = 0, but not for many other values of x. -- 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: JSH:Understanding constant terms > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Just because the factors are divisible by 7 when x=0, it doesn't > follow that the factors are divisible by 7 for all values of x. > James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Just because the factors are divisible by 7 when x=0, it doesn't > follow that the factors are divisible by 7 for all values of x. > James Harris How? === Subject: Re: JSH:Understanding constant terms >> which means that 7 is divided from the constant terms as well, and >> only two of the constant terms, c_1 and c_2 have 7 as a factor, so >> necessarily 7 is divided from the factors where the constant terms are >> 7. >> Just because the factors are divisible by 7 when x=0, it doesn't >> follow that the factors are divisible by 7 for all values of x. >> > Just because the factors are divisible by 7 when x=0, it doesn't >> follow that the factors are divisible by 7 for all values of x. >> Just because the factors are divisible by 7 when x=0, it doesn't >> follow that the factors are divisible by 7 for all values of x. > Care to step through that argument for us slowpokes? Sure. The factor g_1(x) has *two* parts, where one of them is the constant term, which is constant as it is in fact, 7, and 7 is constant. The other varies as x varies. Now the constant term goes from 7 to 1, which means that it is divided by 7. Understand? Well then, by the distributive property, the other term must be divided by 7 as well, as if you have two parts, then you can't get to one without going through the other. For example to help you understand the concept consider z = x + 7 versus z/7 = x/7 + 1 and notice that by the distributive property, you can't divide 7 from the second piece without going through the first! That's how the distributive property applies. For other readers, note that I can give detail as necessary with the argument, and also notice the *level* of detail that posters wish to request! That makes the argument in my original post one of the most worked out in math history with a level of detail that few ever bother to try in today's math world. It is a perfect argument, in every detail, as a proof must be. James Harris === Subject: Re: JSH:Understanding constant terms <87llcmoxk6.fsf@phiwumbda.org> Discussion, linux) > Sure. > The factor g_1(x) has *two* parts, where one of them is the constant > term, which is constant as it is in fact, 7, and 7 is constant. > The other varies as x varies. > Now the constant term goes from 7 to 1, which means that it is divided > by 7. > Understand? > Well then, by the distributive property, the other term must be > divided by 7 as well, as if you have two parts, then you can't get to > one without going through the other. > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 Let me try, too. Let g(x) = 17 / (x+1). What's the constant term? If I understand you correctly, it is g(0), i.e., 17. Right? Now, 17 divides 17. Right? So I should conclude (by the distributive property) that 17 divides g(x) for all x, right? In particular, I guess that 17 divides g(16). But g(16) = 1, so therefore, 17 divides 1, right? Where did I go wrong? -- Jesse F. Hughes I love Mathematics[...] She doesn't care about my feelings, or my pride or how I so wish to get out and travel. Truth is truth, and anything else is just plain wrong. -- James Harris, /A Love Story/ === Subject: Re: JSH:Understanding constant terms > Where did I go wrong? I can help here. You went wrong when you tried to explain some mathematics to JSH. This is a simple, but common, error. === Subject: Re: JSH:Understanding constant terms <87llcmoxk6.fsf@phiwumbda.org> Discussion, linux) > > Just because the factors are divisible by 7 when x=0, it doesn't > follow that the factors are divisible by 7 for all values of x. > > It follows from the distributive property. >> Care to step through that argument for us slowpokes? > Sure. > The factor g_1(x) has *two* parts, where one of them is the constant > term, which is constant as it is in fact, 7, and 7 is constant. Er, I guess. Trivially, we can write g_1(x) = (g_1(x) - g_1(0)) + g_1(0). ^^^^^^^^^^^^^^^^^ ^^^^^^ so-called constant bit ^^^^^^^^^^^^^^^^^ evidently the non-constant bit > The other varies as x varies. > Now the constant term goes from 7 to 1, which means that it is divided > by 7. > Understand? > Well then, by the distributive property, the other term must be > divided by 7 as well, as if you have two parts, then you can't get to > one without going through the other. What the distributive property says is (a + b)/c = a/c + b/c. You have g_1(0)/7 is an algebraic integer (or something). You want to conclude that for all x, g_1(x)/7 is an algebraic integer. All the distributive property seems to give you is: g_1(0) / 7 = (g_1(0) / 7 - g_1(0) / 7) + g_1(0) / 7 or, if you prefer, g_1(x) / 7 = (g_1(x) / 7 - g_1(0) / 7) + g_1(0) / 7 Neither of these equations proves that g_1(x) is divisible by 7. I don't see how I go from that to the claim that 7 divides g_1(x) - g_1(0) for all x. Maybe a little more detail? I still don't get it. -- I don't want to wine and dine and date you once or twice. I want to hold you now. I just want to spend the night. You tell me a better plan. Baby, I'm not a patient man. -- Jimmy Lafave, the romantic troubadour. === Subject: Re: JSH:Understanding constant terms > Just because the factors are divisible by 7 when x=0, it doesn't >> follow that the factors are divisible by 7 for all values of x. > It follows from the distributive property. Care to step through that argument for us slowpokes? > Sure. > The factor g_1(x) has *two* parts, where one of them is the constant > term, which is constant as it is in fact, 7, and 7 is constant. > The other varies as x varies. > Now the constant term goes from 7 to 1, which means that it is divided > by 7. > Understand? > Well then, by the distributive property, the other term must be > divided by 7 as well, as if you have two parts, then you can't get to > one without going through the other. > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 > and notice that by the distributive property, you can't divide 7 from > the second piece without going through the first! > That's how the distributive property applies. > For other readers, note that I can give detail as necessary with the > argument, and also notice the *level* of detail that posters wish to > request! > That makes the argument in my original post one of the most worked out > in math history with a level of detail that few ever bother to try in > today's math world. > It is a perfect argument, in every detail, as a proof must be. > James Harris You haven't proved what you said you were going to prove. You said you were going to prove that the factor was divisible by 7 for all values of x. You haven't proved that. === Subject: Re: JSH:Understanding constant terms > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 > and notice that by the distributive property, you can't divide 7 from > the second piece without going through the first! Ok, you've shown that when z (which is better written z(x) since it's a function of x) is divided by 7, both terms are divided by 7. That's fine, and obviously true by the distributive law, as you claim. But what you were supposed to show was that if z(0) is divisible by 7, then z(x) is divisible by 7 for all x. Clearly that's NOT true, since in your example, z(0)=7 is divisible by 7, but z(1)=8 is not divisible by 7. --Mark === Subject: Re: JSH:Understanding constant terms > For example to help you understand the concept consider > z = x + 7 > versus > z/7 = x/7 + 1 > and notice that by the distributive property, you can't divide 7 from > the second piece without going through the first! > Ok, you've shown that when z (which is better written z(x) since it's a > function of x) is divided by 7, both terms are divided by 7. That's fine, > and obviously true by the distributive law, as you claim. But what you were > supposed to show was that if z(0) is divisible by 7, then z(x) is divisible > by 7 for all x. Clearly that's NOT true, since in your example, z(0)=7 is > divisible by 7, but z(1)=8 is not divisible by 7. > --Mark Ok, the actual expression from my post is g_1(x) = 5a_1(x) + 7 where 7 is a constant term as a_1(0) = 0. And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 as a multiple. Dividing P(x) by 49 gives a contant term of 22. That means that when P(x) is divided by 49, g_1(x) must be divided by 7, by the distributive property. Or do you disagree? James Harris === Subject: Re: JSH:Understanding constant terms <87llcmoxk6.fsf@phiwumbda.org> Discussion, linux) > Ok, the actual expression from my post is > g_1(x) = 5a_1(x) + 7 > where 7 is a constant term as a_1(0) = 0. > And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 > as a multiple. > Dividing P(x) by 49 gives a contant term of 22. > That means that when P(x) is divided by 49, g_1(x) must be divided by > 7, by the distributive property. > Or do you disagree? I don't see how you're appealing to the distributive property here. Can you be more explicit? The distributive property says a(b + c) = ab + ac. What is a, b and c? I guess that a is 1/y, b is 5 a_1(x) and c is 7, right? But so what? So g_1(x)/7 = 5 a_1(x) / 7 + 1. If I knew that g_1(x) was divisible by 7 (in what ring? algebraic integers?), then I could conclude that 5 a_1(x) is also divisible by 7. And if I knew that 5 a_1(x) was divisible by 7, then I would know that g_1(x) is, too. But I don't know either of these things, do I? -- I have to break the code of how [mere humans] work, and I have made a lot of progress over years of effort, and I feel like I am close to figuring out all the inner details of human wiring. -- James S. Harris on the extra problems of conveying his research === Subject: Re: JSH:Understanding constant terms >> Ok, you've shown that when z (which is better written z(x) since >> it's a function of x) is divided by 7, both terms are divided by 7. >> That's fine, and obviously true by the distributive law, as you >> claim. But what you were supposed to show was that if z(0) is >> divisible by 7, then z(x) is divisible by 7 for all x. Clearly >> that's NOT true, since in your example, z(0)=7 is divisible by 7, >> but z(1)=8 is not divisible by 7. > Ok, the actual expression from my post is > g_1(x) = 5a_1(x) + 7 > where 7 is a constant term as a_1(0) = 0. > And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 > as a multiple. > Dividing P(x) by 49 gives a contant term of 22. > That means that when P(x) is divided by 49, g_1(x) must be divided by > 7, by the distributive property. > Or do you disagree? Before going back to your original posting, I'd like to clear up this point first. Several people have made the claim that this statement, which I'll slightly generalize and call S, is false: S: If g(0) is divisible by k, then g(x) must be divisible by k for all x. You disputed these claims, seeming to be saying that you believe that S is true. Do you now agree that S is false? --Mark === Subject: Re: JSH:Understanding constant terms > Ok, the actual expression from my post is > g_1(x) = 5a_1(x) + 7 > where 7 is a constant term as a_1(0) = 0. > And g_1(x) is a factor of P(x), where P(0) = 7(7)(22), and P(x) has 49 > as a multiple. > Dividing P(x) by 49 gives a contant term of 22. > That means that when P(x) is divided by 49, g_1(x) must be divided by > 7, by the distributive property. > Or do you disagree? You're distributing different quantities when 'x' is not equal to 0. Your original decomposition into factors applied to the x = 0 case, not to the general case. Go ahead and substitute x = 1 into the equation, follow through with it and see what you get. (By the way, that information was given to you months ago, but you ignored it!) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, > Up to this point you are fine. > so > necessarily 7 is divided from the factors where the constant terms are > 7. > No. What you are saying here is that if 7 divides g(0) then > 7 must divide g(x) for all x. This is simply not true. It follows algebraically. If you need me to explain in further detail then I can, but simple denial is not going to work. > The rest of your post in nonsense. > - William Hughes I can explain the algebra in detail to you, but if you simply choose to claim it's nonsense, then no progress can be made, right? Now then, if you will accept algebra, and point out an area of confusion then I can explain to you. But if you are simply going to just deny then there's little point. Which is it? Will you follow algebra, and give a mathematical area for me to explain, or are you just going to deny? James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, > Up to this point you are fine. > so > necessarily 7 is divided from the factors where the constant terms are > 7. > No. What you are saying here is that if 7 divides g(0) then > 7 must divide g(x) for all x. This is simply not true. > It follows algebraically. If you need me to explain in further detail > then I can, but simple denial is not going to work. Actually it does not follow, algebraically or otherwise. Simple assertion is not going to work. I have explained in great detail, why you are wrong. You have ignored these posts. > The rest of your post in nonsense. > - William Hughes > I can explain the algebra in detail to you, but if you simply choose > to claim it's nonsense, then no progress can be made, right? Good, start by explaining all the counterexamples I have provided. You have quite a backlog. As an easy first step explain why, with a(0) = 0 and w1(0) =1 the constant terms of (a(x) + 7) and (a(x)/w1(x) + 7/w1(x)) are different. > Now then, if you will accept algebra, and point out an area of > confusion then I can explain to you. But if you are simply going to > just deny then there's little point. > Which is it? Will you follow algebra, and give a mathematical area > for me to explain, or are you just going to deny? Which is it? Will you reply to the above and give mathematical arguments, or are you just going to ignore? > James Harris === Subject: Re: JSH:Understanding constant terms > Where am I going wron ? I'll explain. > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > OK, (x-1)(x-2) = x^2 -3x +2 > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > No idea what you mean by algebraic integr functions Well, just in case people didn't understand, I explained in my next paragraph. I'll skip past your next and go to that next. > Possibly: > A) (x-1)=0 and (x-2)=0 are both monic polys > wuth integer coefficients so their roots are > algebraic integers. > B) y = (x-1), Z = (x-2): if x is an algebraic integer > then y and z both are. > Which one of these do you mean or do you mean something else ? > If you mean that the product of the g's gives you P(x) > and that the root or roots of g_r(x) =0 are algebraic integers > then the g's must be monic and so must P(x). > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. And that's what I mean, as I went ahead and explained what I mean. An algebraic function f(x), will give an algebraic integer result for any algebraic integer x. Hopefully that clears things up. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > Yes this follows from Vieta's theorem > Forn any poly x^n + B_1x^(n-1) + ... +B_n =0 > B_n = product of roots x(-1)^n > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > If and only if the g's are monic polys No. The g's are algebraic integer functions. Here's an example: f(x) = 2x is an algebraic integer function, as for any algebraic integer x, it gives an algebraic integer, but notice it's not a monic polynomial. > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > r_1(x) = x-1 -1 > r_2(x) = x-2 -2 > => (x-1)(x-2) = x^2 -3x +2 > So you have gone round in a circle. Well, try it out on x^2 + 3x + 2 = (x+2)(x+1) with g_1(x) = x+2, and g_2(x) = x+1 where I'm deliberately using polynomial factors, and you'll find that c_1 = 2, r_1(x) = 2, and c_2 = 1, r_2(x) = x, so (r_1(x) + c_1)(r_2(x) + c_2) = (x+2)(x+1) and it's trivial there, but that should give you the general idea. What I've done is generalize from the basic to something a little more complicated where instead of polynomial factors I can use non-polynomial ones. > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > P(X) is not monic so it cannot be the P(X) > you describe above . Why not? Consider 2x^2 + 3x + 1 = (2x + 1)(x + 1) where here g_1(x) = 2x + 1 and it's an algebraic integer function. > So I'm confused - was your previous explabation irelevant > to what comes ? > Perhaps you should go through the steps as if you were > writing a computer program defining precisely > what each term means, how one step follows from the next. > Otherwise, you have no hope of communicating your valuable insights > to anyone. A computer would be easier, and I know as I can program computers. People are quirky, as they can choose to ignore information. You can lead a horse to water... > I am none the wiser as to what you are trying to prove. > Perhaps you should state the theorem you are trying to > prove clearly and then outline the strategy > of your proof with clear,precise definitions of all the terms > What criteria do the g's have to meet ? They are algebraic integer functions and factors of P(x). > Are they monic polys ? - presumably so if their roots > are to be albegraic integers. Perhaps you > are not sure yourself ? I am sure. And no they are not necessarily monic polynomials. > Are they irreducible or does this not matter ? It doesn't matter. > Does the factorisation have to be unique ? No. James Harris === Subject: Re: JSH:Understanding constant terms > Where am I going wron ? > I'll explain. Yes,I think I understand you now. To recap A g_i(x) is any function that produces an algebraic integer when x is an algebraic integer The roots r1,r2 of x^2 -2x +2 are akgebraic integers The roots, s1,s2,s3 of x^3 -7x +5 are likewise algebraic integers so g_1(x)= r1*x^(1/2) + s2*x + 2 g_2(x)= s2*x^(1/2) + r3 give algebaric integers when x is an algebraic integer. Their product P(x)is a polynomial in x^(1/2) and g_1(0)*g_2(0) = 2 r3 is an algebraic integer. So where does that get us ? === Subject: Re: JSH:Understanding constant terms >If and only if the g's are monic polys > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). Don't the g_i, polys in x, also have to be algebraic integers whenever x is an algebraic integer ? So, say 6x^2 +(pi)x + 2 couldn't be a g_1 even though its constant term is a rational integer. === Subject: Re: JSH:Understanding constant terms >>If and only if the g's are monic polys >> No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Don't the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? I'm sorry. Was that English? You were assuming that the g_i(x) were polynomials, and that the product of the g_i(x) was equal to P(x). Then your statement if and only if the g's are monic polys was about the statement that each g_i(0) divides, in the algebraic integers, P(0). But that is false. For each g_i(0) to divide P(0) in the ring of all algebraic integers, it is necessary and sufficient for each of the values g_i(0) to be algebraic integers. For certainly we need each g_i(0) to be an algebraic integer. And if they all are, then clearly each of them divides P(0) in the ring of all algebraic integers. You don't need the g_i(0) to be monic; you seemed to be led astray by the fact that g_i(0) is plus or minus the product of the roots of g_i(x). Certainly if g_i(x) is a monic polynomial with algebraic integer coefficients, then g_i(0) will be an algebraic integer; but it is possible for g_i(0) to be an algebraic integer without all roots of g_i(x) to be algebraic integer. > So, say 6x^2 +(pi)x + 2 couldn't >be a g_1 even though its constant term >is a rational integer. Doesn't matter. Your assertion was about the g_i(0) dividing P(0) in the algebraic integers. For that statement, you don't need the g_i(x) to be monic (note that P(x) was not monic either). -- 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: JSH:Understanding constant terms >>If and only if the g's are monic polys > No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Don't the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? > I'm sorry. Was that English? I can't say that I've found your prose particularly pristine. Are you a non-native English speaker ? > You were assuming that the g_i(x) were polynomials, and that the > product of the g_i(x) was equal to P(x). Then your statement if and > only if the g's are monic polys was about the statement that each > g_i(0) divides, in the algebraic integers, P(0). According to Harris, the g_i(x) are algebraic integers whenever x is ( his algebraic integer functions). Clearly g_i(0) has to be an algebraic integer and so the constant term of g_i(x)has to be an algerbaic integer. You say that the gi can be ANY polynominal. This is clearly not the case. For example pi*x +1. Are you saying that p1*1 +1 is an alegebraic integer ? > But that is false. For each g_i(0) to divide P(0) in the ring of all > algebraic integers, it is necessary and sufficient for each of the > values g_i(0) to be algebraic integers. Yes, > For certainly we need each > g_i(0) to be an algebraic integer. And if they all are, then clearly > each of them divides P(0) in the ring of all algebraic integers. > You don't need the g_i(0) to be monic; you seemed to be led astray by > the fact that g_i(0) is plus or minus the product of the roots of > g_i(x). No, that was not the problem. Certainly if g_i(x) is a monic polynomial with algebraic > integer coefficients, then g_i(0) will be an algebraic integer; but it > is possible for g_i(0) to be an algebraic integer without all roots of > g_i(x) to be algebraic integer. Yes. > So, say 6x^2 +(pi)x + 2 couldn't >be a g_1 even though its constant term >is a rational integer. > Doesn't matter. I'm afraid it does matter if you are using the g_i(x) as defined by Harris. === Subject: Re: JSH:Understanding constant terms days. My association with the Department is that of an alumnus. >If and only if the g's are monic polys > > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). >>Don't the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? >> I'm sorry. Was that English? >I can't say that I've found your >prose particularly pristine. I do try to avoid the use of cute abbreviations like polys, though. >Are you a non-native English speaker ? Indeed. English is, alas, my fourth language. It often shows. >> You were assuming that the g_i(x) were polynomials, and that the >> product of the g_i(x) was equal to P(x). Then your statement if and >> only if the g's are monic polys was about the statement that each >> g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >whenever x is ( his algebraic integer functions). >Clearly g_i(0) has to be an algebraic >integer and so the constant term of g_i(x)has to be an >algerbaic integer. >You say that the gi can be ANY polynominal. Please note that I was not addressing James's argument. I was addressing your restriction to the g_i being polynomials, and your assertion, ->under that further restriction<-, that each g_i(0) would divide P(0) in the ring of all algebraic integers if and only if the g_i(x) were monic. ->That<- assertion is false. >This is clearly not the case. For example pi*x +1. >Are you saying that p1*1 +1 is an alegebraic integer ? No. I am saying that if g(x)=pi*x + 1, then it is entirely possible for g(0) to divide P(0) in the ring of all algebraic integers, EVEN THOUGH g(x) is not a monic polynomial. Hell: take g1(x) = 2x+1 and g2(x) = x-1. Then g1(x)*g2(x) = P(x) = 2x^2 - x - 1. g1(0) and g2(0) both divide P(0) in the ring of all algebraic integers, EVEN THOUGH g1(x) is not monic. You claimed the divisibility conclusion would hold if and only if the g's are monic poly[nomial]s. The if is true. The only if is false. -- 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: JSH:Understanding constant terms >If and only if the g's are monic polys > No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). >>Don't the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? > I'm sorry. Was that English? >I can't say that I've found your >prose particularly pristine. > I do try to avoid the use of cute > abbreviations like polys, though. Cute ? Poly used to be a widespr. abb. for polytechnic. I've never heard it reviled on the basis of cuteness before. But then you learn many strange things on sci.math. >Are you a non-native English speaker ? > Indeed. English is, alas, my fourth language. It often shows. >> You were assuming that the g_i(x) were polynomials, and that the >> product of the g_i(x) was equal to P(x). Then your statement if and >> only if the g's are monic polys was about the statement that each >> g_i(0) divides, in the algebraic integers, P(0). >According to Harris, the g_i(x) are algebraic integers >whenever x is ( his algebraic integer functions). >Clearly g_i(0) has to be an algebraic >integer and so the constant term of g_i(x)has to be an >algerbaic integer. >You say that the gi can be ANY polynominal. > Please note that I was not addressing James's argument. I was > addressing your restriction to the g_i being polynomials, and your > assertion, ->under that further restriction<-, that each g_i(0) would > divide P(0) in the ring of all algebraic integers if and only if the > g_i(x) were monic. ->That<- assertion is false. Yes, I agree >This is clearly not the case. For example pi*x +1. >Are you saying that p1*1 +1 is an alegebraic integer ? > No. I am saying that if g(x)=pi*x + 1, then it is entirely possible > for g(0) to divide P(0) in the ring of all algebraic integers, EVEN > THOUGH g(x) is not a monic polynomial. Yes, but we now know that the g_i(x) must meet another condition. What I said prev. was incorrect, but your statement that ANY polyno. would do is crassly passe in the light of new revelations. === Subject: Re: JSH:Understanding constant terms days. My association with the Department is that of an alumnus. [.snip.] > You were assuming that the g_i(x) were polynomials, and that the > product of the g_i(x) was equal to P(x). Then your statement if and > only if the g's are monic polys was about the statement that each > g_i(0) divides, in the algebraic integers, P(0). >>According to Harris, the g_i(x) are algebraic integers >>whenever x is ( his algebraic integer functions). >>Clearly g_i(0) has to be an algebraic >>integer and so the constant term of g_i(x)has to be an >>algerbaic integer. >>You say that the gi can be ANY polynominal. >> Please note that I was not addressing James's argument. I was >> addressing your restriction to the g_i being polynomials, and your >> assertion, ->under that further restriction<-, that each g_i(0) would >> divide P(0) in the ring of all algebraic integers if and only if the >> g_i(x) were monic. ->That<- assertion is false. >Yes, I agree >>This is clearly not the case. For example pi*x +1. >>Are you saying that p1*1 +1 is an alegebraic integer ? >> No. I am saying that if g(x)=pi*x + 1, then it is entirely possible >> for g(0) to divide P(0) in the ring of all algebraic integers, EVEN >> THOUGH g(x) is not a monic polynomial. >Yes, but we now know that the g_i(x) >must meet another condition. >What I said prev. was incorrect, >but your statement that ANY polyno. >would do is crassly passe in the light >of new revelations. So... I addressed your comment as written. Since that comment, you have now changed your mind about what the g_i(x) should be; based on that change of heart, you complain that my correction was wrong, because it failed to meet your newfound and not-previously asserted conditions on the g_i(x). How... interesting... -- 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: JSH:Understanding constant terms > [.snip.] > You were assuming that the g_i(x) were polynomials, and that the > product of the g_i(x) was equal to P(x). Then your statement if and > only if the g's are monic polys was about the statement that each > g_i(0) divides, in the algebraic integers, P(0). >>According to Harris, the g_i(x) are algebraic integers >>whenever x is ( his algebraic integer functions). >>Clearly g_i(0) has to be an algebraic >>integer and so the constant term of g_i(x)has to be an >>algerbaic integer. >>You say that the gi can be ANY polynominal. > Please note that I was not addressing James's argument. I was >> addressing your restriction to the g_i being polynomials, and your >> assertion, ->under that further restriction<-, that each g_i(0) would >> divide P(0) in the ring of all algebraic integers if and only if the >> g_i(x) were monic. ->That<- assertion is false. >Yes, I agree >>This is clearly not the case. For example pi*x +1. >>Are you saying that p1*1 +1 is an alegebraic integer ? > No. I am saying that if g(x)=pi*x + 1, then it is entirely possible >> for g(0) to divide P(0) in the ring of all algebraic integers, EVEN >> THOUGH g(x) is not a monic polynomial. >Yes, but we now know that the g_i(x) >must meet another condition. >What I said prev. was incorrect, >but your statement that ANY polyno. >would do is crassly passe in the light >of new revelations. > So... I addressed your comment as written. Since that comment, you > have now changed your mind about what the g_i(x) should be; based on > that change of heart, you complain that my correction was wrong, > because it failed to meet your newfound and not-previously asserted > conditions on the g_i(x). > How... interesting... The disingenuousness of your attempted exculpation is breathtaking. I am unversed in exegesising the pronouncements of the Supreme Coprolocutor. You, however, analyse His productions with monomaniacal intensity. Therefore, to claim that the conditions on the g_i(x), although new to me, were unknown to you is sophistry of the most egregious kind. A simple admission of your trivial error would bring the catharsis and ultimate peace of mind you yearn for. === Subject: Re: JSH:Understanding constant terms days. My association with the Department is that of an alumnus. [.snip.] >> So... I addressed your comment as written. Since that comment, you >> have now changed your mind about what the g_i(x) should be; based on >> that change of heart, you complain that my correction was wrong, >> because it failed to meet your newfound and not-previously asserted >> conditions on the g_i(x). >> How... interesting... >The disingenuousness of your attempted >exculpation is breathtaking. >I am unversed in exegesising the >pronouncements of the Supreme Coprolocutor. >You, however, analyse His productions with >monomaniacal intensity. Therefore, >to claim that the conditions on the g_i(x), >although new to me, were unknown to you >is sophistry of the most egregious kind. Sigh. I addressed YOUR comment. I noted an error in YOUR comment, based on YOUR hypothesis. You acknowledge that error, but complain that I noted it. I did not address anything the original poster said. Presumably, you are either joking, or dense. I prefer to assume the former, though it seems hard to ignore the latter possibility. -- 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: JSH:Understanding constant terms > [.snip.] >> So... I addressed your comment as written. Since that comment, you >> have now changed your mind about what the g_i(x) should be; based on >> that change of heart, you complain that my correction was wrong, >> because it failed to meet your newfound and not-previously asserted >> conditions on the g_i(x). > How... interesting... >The disingenuousness of your attempted >exculpation is breathtaking. >I am unversed in exegesising the >pronouncements of the Supreme Coprolocutor. >You, however, analyse His productions with >monomaniacal intensity. Therefore, >to claim that the conditions on the g_i(x), >although new to me, were unknown to you >is sophistry of the most egregious kind. > Sigh. I addressed YOUR comment. I noted an error in YOUR comment, > based on YOUR hypothesis. You acknowledge that error, but complain > that I noted it. I did not address anything the original poster said. Typical No, I wasn't complaining that you noted it at all. My soul was suffused with the light that you brought unto me. I wasn't saying that you explicitly addressed anything the OP said, but that you must have known implicitly what the OP said about the g_i(x) due to your regular enlightening intercourse with the OP. Perhaps a fleeting fit of insanity overcame you as you pondered these deep matters causing you to cast aside your usual intellectual caution and make that rash and impetuous statement about ANY polynomial. > Presumably, you are either joking, or dense. Joking ? Gay persiflage ? Badinage of an ironic nature ? Egad Sir,I would rather ritually disembowel myself than make light of your forensic quest for Truth. > I prefer to assume the former, though it seems hard to ignore the > latter possibility. Your sharpness of mind is matched only by your generosity of spirit. === Subject: Re: JSH:Understanding constant terms Discussion, linux) > No, I wasn't complaining that you noted it at all. > My soul was suffused with the light that you > brought unto me. Jesus, give it a rest and put away the thesaurus. We're already ing impressed. At least I know I am. -- Jesse F. Hughes To be honest, I don't have enough interest in math to spend the time it would take to clean up the mess that I believe has been created in the past 100 or so years. -- Curt Welch lets the world down. === Subject: Re: JSH:Understanding constant terms >If and only if the g's are monic polys No. If and only if the values g_i(0) are algebraic integers. This will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for example, > if the g_i are ANY polynomials with integer constant terms). >>Don't the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? >> I'm sorry. Was that English? >I can't say that I've found your >prose particularly pristine. > I do try to avoid the use of cute > abbreviations like polys, though. > Cute ? Poly used to be a widespr. abb. Um Davidson, widespr. and abb. are not words in the English language. If you think they are acceptable abbreviations in the English language you are clearly not qualified to be asking others rudely if they are a non-native English speaker. > for polytechnic. I've never heard So you were saying Don't the g_i, polytechnics in x, also have to be ...? KeithK > it reviled on the basis of cuteness before. > But then you learn many strange things on sci.math. >Are you a non-native English speaker ? > Indeed. English is, alas, my fourth language. It often shows. >> You were assuming that the g_i(x) were polynomials, and that the >> product of the g_i(x) was equal to P(x). Then your statement if and >> only if the g's are monic polys was about the statement that each >> g_i(0) divides, in the algebraic integers, P(0). According to Harris, the g_i(x) are algebraic integers >whenever x is ( his algebraic integer functions). >Clearly g_i(0) has to be an algebraic >integer and so the constant term of g_i(x)has to be an >algerbaic integer. You say that the gi can be ANY polynominal. > Please note that I was not addressing James's argument. I was > addressing your restriction to the g_i being polynomials, and your > assertion, ->under that further restriction<-, that each g_i(0) would > divide P(0) in the ring of all algebraic integers if and only if the > g_i(x) were monic. ->That<- assertion is false. > Yes, I agree >This is clearly not the case. For example pi*x +1. >Are you saying that p1*1 +1 is an alegebraic integer ? > No. I am saying that if g(x)=pi*x + 1, then it is entirely possible > for g(0) to divide P(0) in the ring of all algebraic integers, EVEN > THOUGH g(x) is not a monic polynomial. > Yes, but we now know that the g_i(x) > must meet another condition. > What I said prev. was incorrect, > but your statement that ANY polyno. > would do is crassly passe in the light > of new revelations. === Subject: Re: JSH:Understanding constant terms >If and only if the g's are monic polys No. If and only if the values g_i(0) are algebraic integers. This > will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for > example, > if the g_i are ANY polynomials with integer constant terms). >>Don't the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? >> I'm sorry. Was that English? >I can't say that I've found your >prose particularly pristine. I do try to avoid the use of cute > abbreviations like polys, though. > Cute ? Poly used to be a widespr. abb. > Um Davidson, widespr. and abb. are not words in the English language. Yes, that's wherefore me chosed they. > If you think they are acceptable abbreviations in the English language you > are clearly not qualified to be asking others rudely if they are a > non-native English speaker. Me not it were which the one say first bad bad things about my spoke Brit-talk was. You cogitate big deep, you get it much quickly. === Subject: Re: JSH:Understanding constant terms >If and only if the g's are monic polys No. If and only if the values g_i(0) are algebraic integers. This > will > certainly happen if the g_i are monic polynomials with integer > coefficients, but it can happen in other cases as well (for > example, > if the g_i are ANY polynomials with integer constant terms). >>Don't the g_i, polys in x, also have to be >>algebraic integers whenever x is an algebraic >>integer ? >> I'm sorry. Was that English? >I can't say that I've found your >prose particularly pristine. I do try to avoid the use of cute > abbreviations like polys, though. Cute ? Poly used to be a widespr. abb. > Um Davidson, widespr. and abb. are not words > in the English language. > Yes, that's wherefore me chosed they. > If you think they are acceptable abbreviations in the English language you > are clearly not qualified to be asking others rudely if they are a > non-native English speaker. > Me not it were which the one say > first bad bad things about my spoke Brit-talk was. > You cogitate big deep, you get it much quickly. Me humbled muchly your by delightf. reply. :) KeithK === Subject: Re: JSH:Understanding constant terms If and only if the g's are monic polys >> No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Don't the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? So, say 6x^2 +(pi)x + 2 couldn't >be a g_1 even though its constant term >is a rational integer. The g_i 's are not necessarily polynomials. In the specific case JSH considers later in his posting, the a_i(x) are the roots of a cubic whose coefficients are functions of x. They could be found explicitly using Cardano's formula: complex expressions involving cube- and square-roots of polynomials in x. (They are algebraic integers whenever x is, because the cubic involved is monic.) -- David Hartley === Subject: Re: JSH:Understanding constant terms >>If and only if the g's are monic polys >> No. If and only if the values g_i(0) are algebraic integers. This will >> certainly happen if the g_i are monic polynomials with integer >> coefficients, but it can happen in other cases as well (for example, >> if the g_i are ANY polynomials with integer constant terms). >Don't the g_i, polys in x, also have to be >algebraic integers whenever x is an algebraic >integer ? So, say 6x^2 +(pi)x + 2 couldn't >be a g_1 even though its constant term >is a rational integer. > The g_i 's are not necessarily polynomials. In the specific case JSH > considers later in his posting, the a_i(x) are the roots of a cubic > whose coefficients are functions of x. They could be found explicitly > using Cardano's formula: complex expressions involving cube- and > square-roots of polynomials in x. (They are algebraic integers whenever > x is, because the cubic involved is monic.) Apparently, all that is required of the g_i(x) is that when x is an algebraic integer g_1(x) is also an algebraic integer. So something of the form A*log(x) + pi + tan(x) would not do. As the sums, products and roots of algebraic integers are algebraic integers then, say, g_1(x) A x^(3/5) + Bx^(1/2) + C, where A,B,C are algebraic integers would be an algebaraic integer if x is an algebraic integer. But this is still a polynomial in fractional powers of x. Non-polynomial seems a bit misleading. === Subject: Re: JSH:Understanding constant terms ... > Apparently, all that is required of the g_i(x) > is that when x is an algebraic integer > g_1(x) is also an algebraic integer. > So something of the form A*log(x) + pi + tan(x) > would not do. > As the sums, products and roots of algebraic > integers are algebraic integers then, say, g_1(x) > A x^(3/5) + Bx^(1/2) + C, where A,B,C > are algebraic integers would be an algebaraic integer > if x is an algebraic integer. But this is > still a polynomial in fractional powers of x. > Non-polynomial seems a bit misleading. What if g_i(x) are the roots of g^2 + x.g + 1? I would certainly call that non-polynomial. And what do you think of sqrt(x + 1)? -- 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: JSH:Understanding constant terms > ... > > Apparently, all that is required of the g_i(x) > > is that when x is an algebraic integer > > g_1(x) is also an algebraic integer. > > So something of the form A*log(x) + pi + tan(x) > > would not do. > > As the sums, products and roots of algebraic > > integers are algebraic integers then, say, g_1(x) > > A x^(3/5) + Bx^(1/2) + C, where A,B,C > > are algebraic integers would be an algebaraic integer > > if x is an algebraic integer. But this is > > still a polynomial in fractional powers of x. > > Non-polynomial seems a bit misleading. > What if g_i(x) are the roots of g^2 + x.g + 1? I would certainly > call that non-polynomial. And what do you think of sqrt(x + 1)? I don't know what your definition of polynomial or non-polynomial might be. So I can only guess as to why you think your first example is a non-polynomial. A polynomial is finite sum of terms of the form Ax^k, where A is an algebraic integer, x a variable and k a positive rational number How are you defining polynomial and non-polynomial ? === Subject: Re: JSH:Understanding constant terms ... > > Non-polynomial seems a bit misleading. > > What if g_i(x) are the roots of g^2 + x.g + 1? I would certainly > call that non-polynomial. And what do you think of sqrt(x + 1)? > I don't know what your definition of polynomial or non-polynomial > might be. > So I can only guess as to why you think your first example > is a non-polynomial. In wat way is -x + sqrt(x^2 - 4) a polynomial? > A polynomial is finite sum of terms of the form Ax^k, where A > is an algebraic integer, x a variable and k a positive > rational number > How are you defining polynomial and non-polynomial ? The same. -- 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: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > Algebraic integers are roots of monic polynomials with integer > coefficients. > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the a's is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. Which is, of course, an out-and-out contradiction, right? So unless mathematics is inconsistent, there must be something wrong. What is it? It could be that algebraic number theory is wrong. Or possibly Galois theory. Or both. Or it could be that you have a logical error somewhere in what you have said above. You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly seems OK. I don't disagree with it. Then you noted that 49 = 7*7 must be factored out of the expression (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). Let's look in detail at that troublesome third term, r_3(x) + 22. The objective here is to examine whether I can divide a nonunit factor of 7 out of this expression, with the result after the factoring being an algebraic integer. Let w be a nonunit factor of 7. You note that 22 is coprime to 7. Therefore 22/w cannot be an algebraic integer. So that is potentially a problem. But recall that r_3(x) + 22 = 5 a_3(x) + 7. Suppose I divide the latter expression through by w: I get 5 a_3(x)/w + 7/w. The term 7/w is not a problem here because I was assuming that w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. Now, if such a w can be chosen so that a_3(x)/w is also an algebraic integer, I conclude that in the expression 5 a_3(x)/w + 7/w all the coefficients are algebraic integers, and the sum is an algebraic integer. Can such a w be defined ? If so, it is going to have to be a function of x: because when x = 0, w has to be 1. Whereas, for most other integers x, the polynomial that you mention above is irreducible, and w is a nonunit factor of 7. Here is how w_1(x), w_2(x), and w_3(x) need to be defined: w_1(x) = GCD(a_1(x), 7) w_2(x) = GCD(a_2(x), 7) w_3(x) = GCD(a_3(x), 7). Here GCD denotes the 'greatest common divisor' function. It is defined in the ring of algebraic integers by a theorem of Dedekind. Obviously these definitions imply that all of (5 a_1(x) + 7)/w_1(x), (5 a_2(x) + 7)/w_2(x), and (5 a_3(x) + 7)/w_3(x) are algebraic integers. Now the crucial thing to show is that w_1(x) * w_2(x) * w_3(x) = 49. Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general not by any additional factors of 7. Why? Because 2401 x^3 - 147 x^2 + 3x is coprime to 7 [except when x itself is divisible by 7], and -49*(2401 x^3 - 147 x^2 + 3x) is the constant term of the polynomial that the a's satisfy, as you note above. Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product must equal 49, as desired. What this shows is that it IS possible to divide 49 out of (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) in such a way that each factor is an algebraic integer. But what about the constant terms ? It is *not required* that the constant terms be respected. That is, even though it is true that (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, (as you can immediately check), there is NO REASON to require that each of these three terms be algebraic integers; and of course, 22/w_3(x) is not. Again, why? Isn't the product of the constant terms equal to the constant term of the product ??? Yes. But here is the last key fact. The constant term of (r_2(x) + 22)/w_3(x) is NOT what you think it is. You think it must be 22/w_3(x). It isn't. You need to remember YOUR OWN DEFINITION of constant term. It is instead 22/w_3(0). So then you note that w_3(0) = 1, and you have no contradiction. Everything hangs together. Again: you defined 'constant term' in a perfectly reasonable way, and you should have stuck with your definition. Instead you got confused and assumed that whatever is in the POSITION of the constant term *is* the constant term. But here, 22/w_3(x) is not even constant, because by its definition, w_3(x) is not constant. Nora B. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > Algebraic integers are roots of monic polynomials with integer > coefficients. > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the a's is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. > Which is, of course, an out-and-out contradiction, right? There is the appearance of contradiction, which is readily resolved by not giving the ring of algebraic integers a special position. So then, just because the roots do not have 7 as a factor in the ring of algebraic integers, it's not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? If you overrate the ring of algebraic integers, then you can make arguments that are wrong. Essentially, you have to understand that the requirement that a number be the root of a monic polynomial with integer coefficients is a meaningless technicality, with no real mathematical importance. There is no weight to the requirement mathematically that a number be the root of a monic polynomial with integer coefficients. That's what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. It turns out to be a problem in algebraic number theory, which has lead to a mis-use of Galois Theory. It's not even really complicated, but there are reasons for people to get emotional over the issue, as it's a mistake at the foundations of the discipline of mathematics. It's an error in core. > Or it could be that you have a logical error somewhere in what > you have said above. You can continue with that assumption. I have no problem defending the math I outlined in my original post. Ultimately, it boils down to constants being constant. Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I don't disagree with it. Good. > Then you noted that 49 = 7*7 must be factored out of the expression > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > Let's look in detail at that troublesome third term, It's not troublesome. It's asymmetrical with regard to the rest. The asymmetry is what blocks you out of the ring of algebraic integers, which requires a lot of symmetry. > r_3(x) + 22. > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. Now you're going to cheat, and I'm going to explain how you cheat before you do it, so readers can see how it works. A number cannot be an algebraic integer if it is not the root of some monic polynomial with integer coefficients. You say nonunit above as if that applies globally, when you're going to rely on non-unit status in the ring of algebraic integers. So consider a unit function u(x), which is not a unit in the ring of algebraic integers. Now you can multiply with that unit, and then assert that you have a non-unit in the ring of algebraic integers, to claim that you can do that multiplication. But it's only a non-unit in the ring of algebraic integers because it's not the root of some monic polynomial with integer coefficients. Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. Notice that the poster didn't say, in the ring of algebraic integers. It's important that you note that the claim is in the ring of algebraic integers. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. The poster is apparently acknowledging that 22 properly is coprime to 7. And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > Suppose I divide the latter expression through by w: I get > 5 a_3(x)/w + 7/w. > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. Remember readers, non-unit in the ring of algebraic integers! > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > 5 a_3(x)/w + 7/w > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > Can such a w be defined ? > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > w_1(x) = GCD(a_1(x), 7) > > w_2(x) = GCD(a_2(x), 7) > w_3(x) = GCD(a_3(x), 7). > Here GCD denotes the 'greatest common divisor' function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > Obviously these definitions imply that all of > (5 a_1(x) + 7)/w_1(x), > (5 a_2(x) + 7)/w_2(x), and > (5 a_3(x) + 7)/w_3(x) > are algebraic integers. Notice also that you have the implication that the factors do not have a constant term, which has to be made explicit in a bit. Essentially the w's the poster is trying to use are unit factors, but are not units in the ring of algebraic integers by a technicality that they are not roots of a monic polynomial with integer coefficients. It's a neat trick when you think about it that requires relying on the very problem that has been outlined to try and use the ring of algebraic integers to disprove that there is a problem with that ring! > Now the crucial thing to show is that > w_1(x) * w_2(x) * w_3(x) = 49. > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > 2401 x^3 - 147 x^2 + 3x > is coprime to 7 [except when x itself is divisible by 7], and > -49*(2401 x^3 - 147 x^2 + 3x) > is the constant term of the polynomial that the a's satisfy, as > you note above. > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > What this shows is that it IS possible to divide 49 out of > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > in such a way that each factor is an algebraic integer. Well, yes, it's possible. > But what about the constant terms ? > It is *not required* that the constant terms be respected. That Notice the poster doesn't even try at this point, simply resorting to hand-waving by asserting that the constant terms don't need to be respected! The problem is that if you have a constant term that is 7, another that is 7, and one that's 22, then to get rid of the 7's, you have to divide out by 7, but that's algebra that's inconvenient to the poster! So instead the poster just TELLS you that the math doesn't care. Let's watch to see what else this poster tries on you. > is, even though it is true that > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > > Again, why? Isn't the product of the constant terms equal to the > constant term of the product ??? > Yes. But here is the last key fact. The constant term of > (r_2(x) + 22)/w_3(x) > is NOT what you think it is. You think it must be > 22/w_3(x). > It isn't. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > 22/w_3(0). > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. Notice that the poster has asserted that the constant term is constant at x=0, but every where else it's actually a function of x. So the full assertion is that the constant term is NOT CONSTANT, but is instead a function of x. Remember, when people try to fight mathematics they have to at some point rely on something that is just wacky. Here you can see that ultimately the poster is trying to get you to believe that a constant is in fact a function of x. But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) = P(0), and notice, you don't have any functions of x, as x has been set to 0. So the poster is using unit factors, which by a technicality are not units in the ring of algebraic integers, unless you do wish to believe that mathematics is inconsistent, and so what? Your belief would be wrong. > Again: you defined 'constant term' in a perfectly reasonable way, > and you should have stuck with your definition. Instead you got > confused and assumed that whatever is in the POSITION of the constant > term *is* the constant term. But here, 22/w_3(x) is not even constant, > because by its definition, w_3(x) is not constant. > Nora B. Well, I say you're using unit factors, which because they are not roots of a monic polynomial with integer coefficients are not units in the ring of algebraic integers. How do you answer? James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that g_1(x) g_2(x)...g_n(x) = P(x) and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. Now consider setting x=0, and notice that gives g_1(0) g_2(0)...g_n(0) = P(0) and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have c_1 c_2...c_n = P(0) and the c's are necessarily algebraic integers and factors of the > constant term P(0). Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. Then I can substitute and have (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). Now let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 then P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) when the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so let g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). Setting x = 0 with the cubic defining the a's gives a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. Since indices are arbitrary let the first two equal 0, and I have c_1 = 7, c_2 = 7, and c_3 = 22 which is consistent with the constant term of P(x), which is 1078. But dividing P(x) by 49, gives P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. Algebraic integers are roots of monic polynomials with integer > coefficients. For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the a's is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. Which is, of course, an out-and-out contradiction, right? > There is the appearance of contradiction, which is readily resolved by > not giving the ring of algebraic integers a special position. > So then, just because the roots do not have 7 as a factor in the ring > of algebraic integers, it's not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. > That's what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > It's not even really complicated, but there are reasons for people to > get emotional over the issue, as it's a mistake at the foundations of > the discipline of mathematics. > It's an error in core. > Or it could be that you have a logical error somewhere in what > you have said above. > You can continue with that assumption. I have no problem defending > the math I outlined in my original post. > Ultimately, it boils down to constants being constant. > Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I don't disagree with it. > Good. > Then you noted that 49 = 7*7 must be factored out of the expression > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > Let's look in detail at that troublesome third term, > It's not troublesome. It's asymmetrical with regard to the rest. > The asymmetry is what blocks you out of the ring of algebraic > integers, which requires a lot of symmetry. > r_3(x) + 22. > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. > Now you're going to cheat, and I'm going to explain how you cheat > before you do it, so readers can see how it works. > A number cannot be an algebraic integer if it is not the root of some > monic polynomial with integer coefficients. > You say nonunit above as if that applies globally, when you're going > to rely on non-unit status in the ring of algebraic integers. > So consider a unit function u(x), which is not a unit in the ring of > algebraic integers. > Now you can multiply with that unit, and then assert that you have a > non-unit in the ring of algebraic integers, to claim that you can do > that multiplication. > But it's only a non-unit in the ring of algebraic integers because > it's not the root of some monic polynomial with integer coefficients. > Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. > Notice that the poster didn't say, in the ring of algebraic integers. > It's important that you note that the claim is in the ring of > algebraic integers. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > The poster is apparently acknowledging that 22 properly is coprime to > 7. And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > Suppose I divide the latter expression through by w: I get > 5 a_3(x)/w + 7/w. > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. > Remember readers, non-unit in the ring of algebraic integers! > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > 5 a_3(x)/w + 7/w > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > Can such a w be defined ? > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > w_1(x) = GCD(a_1(x), 7) > w_2(x) = GCD(a_2(x), 7) > w_3(x) = GCD(a_3(x), 7). > Here GCD denotes the 'greatest common divisor' function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > Obviously these definitions imply that all of > (5 a_1(x) + 7)/w_1(x), > (5 a_2(x) + 7)/w_2(x), and > (5 a_3(x) + 7)/w_3(x) > are algebraic integers. > Notice also that you have the implication that the factors do not have > a constant term, which has to be made explicit in a bit. > Essentially the w's the poster is trying to use are unit factors, but > are not units in the ring of algebraic integers by a technicality that > they are not roots of a monic polynomial with integer coefficients. > It's a neat trick when you think about it that requires relying on the > very problem that has been outlined to try and use the ring of > algebraic integers to disprove that there is a problem with that ring! > Now the crucial thing to show is that > w_1(x) * w_2(x) * w_3(x) = 49. > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > 2401 x^3 - 147 x^2 + 3x > is coprime to 7 [except when x itself is divisible by 7], and > -49*(2401 x^3 - 147 x^2 + 3x) > is the constant term of the polynomial that the a's satisfy, as > you note above. > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > What this shows is that it IS possible to divide 49 out of > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > in such a way that each factor is an algebraic integer. > Well, yes, it's possible. > But what about the constant terms ? > It is *not required* that the constant terms be respected. That > Notice the poster doesn't even try at this point, simply resorting to > hand-waving by asserting that the constant terms don't need to be > respected! > The problem is that if you have a constant term that is 7, another > that is 7, and one that's 22, then to get rid of the 7's, you have to > divide out by 7, but that's algebra that's inconvenient to the poster! > So instead the poster just TELLS you that the math doesn't care. > Let's watch to see what else this poster tries on you. > is, even though it is true that > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > Again, why? Isn't the product of the constant terms equal to the > constant term of the product ??? > Yes. But here is the last key fact. The constant term of > (r_2(x) + 22)/w_3(x) > is NOT what you think it is. You think it must be > 22/w_3(x). > It isn't. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > 22/w_3(0). > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. > Notice that the poster has asserted that the constant term is constant > at x=0, but every where else it's actually a function of x. What nonsense! Nora Baron said no such thing!! She said YOU fail to see that 22/w_3(x) is NOT the constant term of (r_2(x) + 22)/w_3(x). Why? because w_3(x) is a function of x. > So the full assertion is that the constant term is NOT CONSTANT, but > is instead a function of x. The full assertion is that 22/w_3(x) is NOT CONSTANT, but is instead a function of x. (DUH) KeithK > Remember, when people try to fight mathematics they have to at some > point rely on something that is just wacky. Here you can see that > ultimately the poster is trying to get you to believe that a constant > is in fact a function of x. > But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are > algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) = > P(0), and notice, you don't have any functions of x, as x has been set > to 0. > So the poster is using unit factors, which by a technicality are not > units in the ring of algebraic integers, unless you do wish to believe > that mathematics is inconsistent, and so what? Your belief would be > wrong. > Again: you defined 'constant term' in a perfectly reasonable way, > and you should have stuck with your definition. Instead you got > confused and assumed that whatever is in the POSITION of the constant > term *is* the constant term. But here, 22/w_3(x) is not even constant, > because by its definition, w_3(x) is not constant. > Nora B. > Well, I say you're using unit factors, which because they are not > roots of a monic polynomial with integer coefficients are not units in > the ring of algebraic integers. > How do you answer? > James Harris === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. > That is, for an algebraic integer x, and natural number 'a' where > 1<=a<=n, g_a(x) is an algebraic integer. > Now consider setting x=0, and notice that gives > g_1(0) g_2(0)...g_n(0) = P(0) > and notice there is no dependency on x, as the constant term is > defined by terms where x is equal to 0, and that's not a trick. > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have > c_1 c_2...c_n = P(0) > and the c's are necessarily algebraic integers and factors of the > constant term P(0). > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on. > Then I can substitute and have > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x). > Now let > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > then > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so let > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7 > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0). > Setting x = 0 with the cubic defining the a's gives > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3. > Since indices are arbitrary let the first two equal 0, and I have > c_1 = 7, c_2 = 7, and c_3 = 22 > which is consistent with the constant term of P(x), which is 1078. > But dividing P(x) by 49, gives > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22 > which means that 7 is divided from the constant terms as well, and > only two of the constant terms, c_1 and c_2 have 7 as a factor, so > necessarily 7 is divided from the factors where the constant terms are > 7. > Now if you continue the analysis the conclusion that two of the > factors of P(x) have 7 as a factor leads to the conclusion that two of > the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > have 7 as a factor, but you can rather easily prove that with integer > x, if the cubic is irreducible over rationals, then that will not be > true in the ring of algebraic integers. > Algebraic integers are roots of monic polynomials with integer > coefficients. > For a number to be an algebraic integer, it must be the root of some > monic polynomial that has integer coefficients. > Therefore, you can conclude from the mathematics that for a given x, > when the cubic defining the a's is irreducible over rationals then its > roots do not have 7 as a factor in the ring of algebraic integers. > Which is, of course, an out-and-out contradiction, right? > There is the appearance of contradiction, which is readily resolved by > not giving the ring of algebraic integers a special position. > So then, just because the roots do not have 7 as a factor in the ring > of algebraic integers, it's not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. > That's what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > It's not even really complicated, but there are reasons for people to > get emotional over the issue, as it's a mistake at the foundations of > the discipline of mathematics. > It's an error in core. > Or it could be that you have a logical error somewhere in what > you have said above. > You can continue with that assumption. I have no problem defending > the math I outlined in my original post. > Ultimately, it boils down to constants being constant. > Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I don't disagree with it. > Good. > Then you noted that 49 = 7*7 must be factored out of the expression > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > Let's look in detail at that troublesome third term, > It's not troublesome. It's asymmetrical with regard to the rest. > The asymmetry is what blocks you out of the ring of algebraic > integers, which requires a lot of symmetry. > r_3(x) + 22. > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. > Now you're going to cheat, and I'm going to explain how you cheat > before you do it, so readers can see how it works. > A number cannot be an algebraic integer if it is not the root of some > monic polynomial with integer coefficients. > You say nonunit above as if that applies globally, when you're going > to rely on non-unit status in the ring of algebraic integers. > So consider a unit function u(x), which is not a unit in the ring of > algebraic integers. > Now you can multiply with that unit, and then assert that you have a > non-unit in the ring of algebraic integers, to claim that you can do > that multiplication. > But it's only a non-unit in the ring of algebraic integers because > it's not the root of some monic polynomial with integer coefficients. > Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. > Notice that the poster didn't say, in the ring of algebraic integers. > It's important that you note that the claim is in the ring of > algebraic integers. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > The poster is apparently acknowledging that 22 properly is coprime to > 7. And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > Suppose I divide the latter expression through by w: I get > 5 a_3(x)/w + 7/w. > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. > > Remember readers, non-unit in the ring of algebraic integers! > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > 5 a_3(x)/w + 7/w > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > Can such a w be defined ? > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > w_1(x) = GCD(a_1(x), 7) > > w_2(x) = GCD(a_2(x), 7) > w_3(x) = GCD(a_3(x), 7). > Here GCD denotes the 'greatest common divisor' function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > Obviously these definitions imply that all of > (5 a_1(x) + 7)/w_1(x), > (5 a_2(x) + 7)/w_2(x), and > (5 a_3(x) + 7)/w_3(x) > are algebraic integers. > Notice also that you have the implication that the factors do not have > a constant term, which has to be made explicit in a bit. > Essentially the w's the poster is trying to use are unit factors, but > are not units in the ring of algebraic integers by a technicality that > they are not roots of a monic polynomial with integer coefficients. > It's a neat trick when you think about it that requires relying on the > very problem that has been outlined to try and use the ring of > algebraic integers to disprove that there is a problem with that ring! > Now the crucial thing to show is that > w_1(x) * w_2(x) * w_3(x) = 49. > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > 2401 x^3 - 147 x^2 + 3x > is coprime to 7 [except when x itself is divisible by 7], and > -49*(2401 x^3 - 147 x^2 + 3x) > is the constant term of the polynomial that the a's satisfy, as > you note above. > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > What this shows is that it IS possible to divide 49 out of > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > in such a way that each factor is an algebraic integer. > Well, yes, it's possible. > But what about the constant terms ? > It is *not required* that the constant terms be respected. That > Notice the poster doesn't even try at this point, simply resorting to > hand-waving by asserting that the constant terms don't need to be > respected! > The problem is that if you have a constant term that is 7, another > that is 7, and one that's 22, then to get rid of the 7's, you have to > divide out by 7, but that's algebra that's inconvenient to the poster! > So instead the poster just TELLS you that the math doesn't care. > Let's watch to see what else this poster tries on you. > is, even though it is true that > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > > Again, why? Isn't the product of the constant terms equal to the > constant term of the product ??? > Yes. But here is the last key fact. The constant term of > (r_2(x) + 22)/w_3(x) > is NOT what you think it is. You think it must be > 22/w_3(x). > It isn't. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > 22/w_3(0). > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. > Notice that the poster has asserted that the constant term is constant > at x=0, but every where else it's actually a function of x. The poster has asserted that the contant term is 22/w_3(0). The poster has asserted explicitely that the constant term is not 22/w_3(x). 22/w_3(0) is not does not vary with x, it has the same value if x=0, x=10, x=pi x = whatever. -William Hughes === Subject: Re: JSH:Understanding constant terms > Consider a polynomial P(x), with n factors such that > g_1(x) g_2(x)...g_n(x) = P(x) > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions. [snip to save space ...] > Which is, of course, an out-and-out contradiction, right? > There is the appearance of contradiction, which is readily resolved by > not giving the ring of algebraic integers a special position. No, it's not the *appearance* of a contradiction. It is just an outright contradiction - unless someone, somewhere has made a mistake. > So then, just because the roots do not have 7 as a factor in the ring > of algebraic integers, it's not significant. > So unless mathematics is inconsistent, there must be something > wrong. What is it? > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. Overrate ??? The set of algebraic integers has a clear, unambiguous definition. There is a textbook proof that it forms a ring. It is not a field; it is simply a natural extension of the integers. It has some key properties: for example, every algebraic number can be written in the form a/n, where a is an algebraic integer and n is an ordinary integer. And it has the property that any two elements have a greatest common divisor. Unlike the integers, it does NOT have the property of factorization into primes. Which of these characteristics would you describe as overrating the ring of algebraic integers? > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. I disagree. If you throw that out, you lose some of the interesting properties that the ring has. Dedekind devoted intense study to it for good reason. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. A matter of opinion, clearly, rather than a matter of fact. > That's what follows. > It could be that algebraic number theory is wrong. Or possibly > Galois theory. Or both. > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > It's not even really complicated, but there are reasons for people to > get emotional over the issue, as it's a mistake at the foundations of > the discipline of mathematics. > It's an error in core. The error is in your non-rigorous application of your own definition of constant term, as shown below. > Or it could be that you have a logical error somewhere in what > you have said above. > You can continue with that assumption. I have no problem defending > the math I outlined in my original post. > Ultimately, it boils down to constants being constant. > Specifically 7 and 22 are constants, and behave like constants. > You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly > seems OK. I don't disagree with it. > Good. > Then you noted that 49 = 7*7 must be factored out of the expression > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22). > Let's look in detail at that troublesome third term, > It's not troublesome. It's asymmetrical with regard to the rest. Spare me the semantics. > The asymmetry is what blocks you out of the ring of algebraic > integers, which requires a lot of symmetry. > r_3(x) + 22. > The objective here is to examine whether I can divide a nonunit > factor of 7 out of this expression, with the result after the > factoring being an algebraic integer. > Now you're going to cheat, and I'm going to explain how you cheat > before you do it, so readers can see how it works. > A number cannot be an algebraic integer if it is not the root of some > monic polynomial with integer coefficients. No disagreement there. That's the *definition*. > You say nonunit above as if that applies globally, when you're going > to rely on non-unit status in the ring of algebraic integers. Sorry that I did not qualify every statement with in the algebraic integers. I thought it was clearly implied. In any case, just to be definite, what I meant was: nonunit *in the ring of algebraic integers*. > So consider a unit function u(x), which is not a unit in the ring of > algebraic integers. Now it's your turn to be imprecise, apparently. What is a unit function ??? > Now you can multiply with that unit, A unit in what ring? Any nonzero complex number is a unit in SOME ring; many are nonunits in others. The word unit in the context you are using it here is not well-defined. You must specify: in what ring? > and then assert that you have a > non-unit in the ring of algebraic integers, to claim that you can do > that multiplication. > But it's only a non-unit in the ring of algebraic integers because > it's not the root of some monic polynomial with integer coefficients. > Ok readers, here we go, pay careful attention... > Let w be a nonunit factor of 7. > Notice that the poster didn't say, in the ring of algebraic integers. OK, I'll say it: Let w be a nonunit factor of 7 in the ring of algebraic integers. I think you know that is what I meant in the first place. > It's important that you note that the claim is in the ring of > algebraic integers. Sure. > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > The poster is apparently acknowledging that 22 properly is coprime to > 7. *In the ring of algebraic integers*. That is clearly implied. > And is in fact coprime to 7 in the the ring of algebraic integers. > But recall that r_3(x) + 22 = 5 a_3(x) + 7. > Suppose I divide the latter expression through by w: I get > 5 a_3(x)/w + 7/w. > The term 7/w is not a problem here because I was assuming that > w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer. > > Remember readers, non-unit in the ring of algebraic integers! No quarrel with that. Do tell us when you find where I start cheating. > Now, if such a w can be chosen so that a_3(x)/w is also an > algebraic integer, I conclude that in the expression > 5 a_3(x)/w + 7/w > all the coefficients are algebraic integers, and the sum is > an algebraic integer. > Can such a w be defined ? > If so, it is going to have to be a function of x: because when > x = 0, w has to be 1. Whereas, for most other integers x, the > polynomial that you mention above is irreducible, and w is a nonunit > factor of 7. > Here is how w_1(x), w_2(x), and w_3(x) need to be defined: > w_1(x) = GCD(a_1(x), 7) > > w_2(x) = GCD(a_2(x), 7) > w_3(x) = GCD(a_3(x), 7). > Here GCD denotes the 'greatest common divisor' function. It > is defined in the ring of algebraic integers by a theorem of > Dedekind. > Obviously these definitions imply that all of > (5 a_1(x) + 7)/w_1(x), > (5 a_2(x) + 7)/w_2(x), and > (5 a_3(x) + 7)/w_3(x) > are algebraic integers. > Notice also that you have the implication that the factors do not have > a constant term, which has to be made explicit in a bit. Wrong! Here is where you make your mistake! You define constant term of a function G(x) to be G(0). So here, the constant term of, for example, (r_3(x) + 22)/w_3(x), is (r_3(0) + 22)/w_3(0) = r_3(0)/w_3(0) + 22/w_3(0). No problem at all with *existence* of a constant term !!! YOUR problem is, the constant term is not what you think it is, or want it to be! > Essentially the w's the poster is trying to use are unit factors, They most certainly are not: not in the ring of algebraic integers. Is THIS where you think I am cheating? Dividing algebraic integers by other (nonunit) algebraic integers? Is that cheating? Remember, I am NOT claiming that 22/w_3(x) is an algebraic integer (unless x = 0). > but > are not units in the ring of algebraic integers by a technicality that > they are not roots of a monic polynomial with integer coefficients. It's not a technicality, unless you regard the very phrase that defines the ring to be a technicality ! > It's a neat trick when you think about it that requires relying on the > very problem that has been outlined to try and use the ring of > algebraic integers to disprove that there is a problem with that ring! The ring has certain properties that I described above, one of which is the existence of greatest-common-divisors of any two of its elements. It is NOT a neat trick or cheating in any way to make use of known properties! The integers, for example, have the property of unique factorization into primes. Would it be *cheating* to use that property to prove a theorem ??? > Now the crucial thing to show is that > w_1(x) * w_2(x) * w_3(x) = 49. > Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general > not by any additional factors of 7. Why? Because > 2401 x^3 - 147 x^2 + 3x > is coprime to 7 [except when x itself is divisible by 7], and > -49*(2401 x^3 - 147 x^2 + 3x) > is the constant term of the polynomial that the a's satisfy, as > you note above. > Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors > of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product > must equal 49, as desired. > What this shows is that it IS possible to divide 49 out of > (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22) > in such a way that each factor is an algebraic integer. > Well, yes, it's possible. concede everything. > But what about the constant terms ? > It is *not required* that the constant terms be respected. That > Notice the poster doesn't even try at this point, simply resorting to > hand-waving by asserting that the constant terms don't need to be > respected! Keep reading. I explain very clearly what I meant. > The problem is that if you have a constant term that is 7, another > that is 7, and one that's 22, then to get rid of the 7's, you have to > divide out by 7, but that's algebra that's inconvenient to the poster! > So instead the poster just TELLS you that the math doesn't care. > Let's watch to see what else this poster tries on you. > is, even though it is true that > (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22, > (as you can immediately check), there is NO REASON to require that each > of these three terms be algebraic integers; and of course, 22/w_3(x) > is not. > > Again, why? Isn't the product of the constant terms equal to the > constant term of the product ??? > Yes. But here is the last key fact. The constant term of > (r_2(x) + 22)/w_3(x) > is NOT what you think it is. You think it must be > 22/w_3(x). > It isn't. You need to remember YOUR OWN DEFINITION of constant > term. It is instead > 22/w_3(0). > So then you note that w_3(0) = 1, and you have no contradiction. > Everything hangs together. > Notice that the poster has asserted that the constant term is constant > at x=0, but every where else it's actually a function of x. Read the following very, very carefully, and try to absorb something. I am using YOUR definition of constant term. The constant term of r_3(x)/w_3(x) + 22/w_3(x) is NOT 22/w_3(x). By YOUR OWN DEFINITION, it is 22/w_3(0). Got that? > So the full assertion is that the constant term is NOT CONSTANT, but > is instead a function of x. You are missing the point so incredibly thoroughly. The constant term in question is 22/w_3(0). It IS CONSTANT. Yes, 22/w_3(x) is NOT CONSTANT. But also, by your definition, it is NOT THE CONSTANT TERM. You are deeply confused on this point. As William Hughes has pointed out in another thread, your thinking is based on the idea that you can look at a function, and BY INSPECTION tell what the constant term is. You say, the constant term of r_3(x) + 22 is obviously 22. That's correct. Then you consider r_3(x)/w_3(x) + 22/w_3(x) and you think: hey, 22/w_3(x) is in the POSITION where the constant term ought to be, so it must BE the constant term. In other words, you just throw YOUR OWN DEFINITION out the window, and declare what you believe to be the constant term BY INSPECTION. And it's not. You should have used your own definition consistently and RIGOROUSLY throughout. When you deviated from it, you got into trouble and led yourself down the path to this erroneous conclusion. THIS is what has led you to think there is a contradiction, which in your addled logic leads you to believe there is some kind of problem with the ring of algebraic integers. Again, just to be really, really clear about this. The constant term of (r_3(x) + 22)/w_3(x) is NOT what you think it is, i.e., it is NOT 22/w_3(x). No, by YOUR OWN DEFINITION, the constant term is 22/w_3(0). What you need to do is RIGOROUSLY USE that definition. It does not lead you down the path to the wrong conclusion which you have followed previously. > Remember, when people try to fight mathematics they have to at some > point rely on something that is just wacky. Here you can see that > ultimately the poster is trying to get you to believe that a constant > is in fact a function of x. > But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are > algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) = > P(0), and notice, you don't have any functions of x, as x has been set > to 0. > So the poster is using unit factors, because you know that 22 and w_3(x) are coprime in the ring of algebraic integers. That is a CORRECT FACT. However, it is not important. The important thing here is that, when you divide the WHOLE EXPRESSION (r_3(x) + 22) by w_3(x), the result IS an algebraic integer. That is what you require in the first place. There is no reason that the individual terms must be algebraic integers. Here's an analogy in integers. Note that 15 = 13 + 2. Divide both sides by 3. On the left, I get 5, an integer. On the right I get 13/3 + 2/3. Neither of these is an integer. So what? Do I conclude that the right side is not divisible by 3? No!!! The SUM of the two parts still equals 5, an integer. The fact that 2/3 is not an integer is NOT A PROBLEM. You can add two nonintegers and get an integer. Similarly, you can add two non-algebraic-integers and get an algebraic integer. Not a problem at all. > which by a technicality are not > units in the ring of algebraic integers, Saying the DEFINING PROPERTY of the ring is a technicality is utterly absurd. You must see that at least. > unless you do wish to believe > that mathematics is inconsistent, and so what? Your belief would be > wrong. I agree, or at least I agree that what we have discussed here does not imply that mathematics is inconsistent. If you were right, it would. But because you have failed to use YOUR OWN DEFINITION of constant term, you are not right, and mathematics is still safe. > Again: you defined 'constant term' in a perfectly reasonable way, > and you should have stuck with your definition. Instead you got > confused and assumed that whatever is in the POSITION of the constant > term *is* the constant term. But here, 22/w_3(x) is not even constant, > because by its definition, w_3(x) is not constant. > Nora B. > Well, I say you're using unit factors, which because they are not > roots of a monic polynomial with integer coefficients are not units in > the ring of algebraic integers. > How do you answer? As above: 22/w_3(x) is not an algebraic integer. I don't claim it is, and there is no reason it SHOULD be. It is NOT the constant term of (r_3(x) + 22)/w_3(x), unless x = 0. Or do you actually still think 22/w_3(x) IS the constant term, in spite of your own definition ?? Try it - use YOUR OWN DEFINITION of constant term - tell me what you get! Nora B. > James Harris === Subject: Re: JSH:Understanding constant terms ... > You note that 22 is coprime to 7. Therefore 22/w cannot be an > algebraic integer. So that is potentially a problem. > > > The poster is apparently acknowledging that 22 properly is coprime to > 7. > *In the ring of algebraic integers*. That is clearly implied. Actually, with most reasonable definitions of coprime, in any ring that contains both 7 and 22, they are coprime. Standard definition: a and b are coprime if p and q in the ring can be found such that p*a + q*b = 1. Alternative definition: a and b are coprime if the have no common non-unit factor in the ring. With both definitions 7 and 22 are coprime regardless of the ring involved. I have no idea what James means with properly coprime. -- 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: JSH:Understanding constant terms > If you overrate the ring of algebraic integers, then you can make > arguments that are wrong. And *that* is precisely what you have done. You have imposed demands on the ring of algebraic integers which are inconsistent with their properties. > Essentially, you have to understand that the requirement that a number > be the root of a monic polynomial with integer coefficients is a > meaningless technicality, with no real mathematical importance. That requirement defines a perfectly valid ring. Whether it is important or not depends on whether it is useful in some mathematical context or not. In your application it is useless. That doesn't speak to the interests of others. > There is no weight to the requirement mathematically that a number be > the root of a monic polynomial with integer coefficients. No, of course not. Rational numbers may not meet that requirement. If your requirements are not satisfied by algebraic integers, don't use them. [snip irrelevant and redundant exposition of fallacious arguments] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH:Understanding constant terms Discussion, linux) > It turns out to be a problem in algebraic number theory, which has > lead to a mis-use of Galois Theory. > It's not even really complicated, but there are reasons for people to > get emotional over the issue, as it's a mistake at the foundations of > the discipline of mathematics. > It's an error in core. Er, when did algebraic number theory or Galois theory become part of the foundations of mathematics? I missed that memo. Probably, they dropped me from the mailing list when I moved to a philosophy section that's part of a management department in a technical university. -- I've been thinking about my problems with getting any kind of admission that my math arguments showing the core error in mathematics are correct, so I've gone to marketing books. -- James S. Harris, on when mathematics isn't enough === Subject: Re: final proof Einstein 1905 is crap. > Nice one, Mike. > Androcles I tried hard to incorporate every known formal/informal fallacy exactly like the originator of the thread has done. :) But speaking about fallacies, I think thw 'non sequitur' is the most serious one, that is the conclusion does not follow from the premises, a disconected idea. Mike === Subject: Re: final proof Einstein 1905 is crap. >> Nice one, Mike. >> Androcles > I tried hard to incorporate every known formal/informal fallacy > exactly like the originator of the thread has done. :) > But speaking about fallacies, I think thw 'non sequitur' is the most > serious one, that is the conclusion does not follow from the premises, > a disconected idea. > Mike Actually the commonest fallacy in Special Relativity is to call a conclusion a postulate via the fallacy of circularity, and that is pretty serious. The velocity of light is c in all inertial frames is bandied about as if it were common knowledge, but stems from V = (c+w)/(1+ w/c), and that was a conclusion Einstein drew in section 5 with the help of the equations of transformation developed in section 3 (i.e. the Lorentz transformations) from which he extracts sqrt(1-v^2/c^2). Ref. http://www.fourmilab.ch/etexts/einstein/specrel/www/ The idiots claim to be able to derive the LTs from a conclusion that was itself derived from the LT and circularity is thus established. In fact the original postulate ( which by definition of postulate it cannot be) is light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. The reason it cannot be a postulate is that it is not intuitive, it is testable, and it has no experimental data to support it. It is an hypothesis. That it reaches absurd conclusions indicates it to be a failed hypothesis. Reading Einstein's paper carefully, it becomes obvious he was perpetrating a hoax. He begins with Galilean Relativity, using a magnet and a conductor as an example, carefully avoiding any explicit description but vaguely referring to laws that hold good, claims that although it is irreconcilable with his second hypothesis this is only apparent , further claims that the velocity of light in our theory plays the part, physically, of an infinitely great velocity and produces a mathematical fiction, xi = x'/sqrt(1-v^2/c^2) ..... (recall that x' = x-vt) The reason Einstein so carefully avoided stating the PoR is that he makes use of it when he says But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v... so he must be careful not to be too explicit in the introduction and preceding sections. To conclude It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain V = (c+w)/(1+w/c) = c. denies him the use of the ray moves [at] c-v... without which he cannot derive the Lorentz transforms. Shubert makes the same fallacy of circularity in his rather long-winded but actually quite simple minded derivation, although he now admits pulling sqrt(1-v^2/c^2) from thin air. He is nothingimportant, of course, although he likes to think he is everythingimportant. Egomania is treatable, I understand, although it may take the services of a trained psychiatrist. Androcles === Subject: Re: final proof Einstein 1905 is crap. > Actually the commonest fallacy in Special Relativity is to call a > conclusion > a postulate via the fallacy of circularity, and that is pretty serious. Well, you are hetting into really dark domains here. Before getting as dvanced as SR, the use of circularity was enforced by those who missinterpreted Newton, or actually desired to present Newton's metaphysics as a science. To be explicit: Newton used a modified version of Kepler's third law in order to derive the law of universal gravitation. Please note that without such law, which was purely empirical, the derivation of LUG is impossible. Then, in mechanics books you see the famous Kepler problem and the derivation of Kepler's third law using the LUG as given! But of course, if the conclusion is true, all premised in the deduction process must be true otherwise the deduction was unsound in the first place. The same of course hold with Einstein and some claims the constancy of c can be derived , etc. Let me tell you what it boild down to: Math infidels with no understanding of the principles of logic the Greeks established and Aristotle put in a massive organized works called the Organon, have taken over and their misunderstanding and paranoia has resulted in the chaos we see in science todat with inductions presented as deductions, deductions as induction, abductions as inductions and deductions. Question of the day: how to you distinguish between paranoid and genious? Often, both sound as equally ntelligent. My answer: Paranoids are easy to spot is you understand formal and informal fallacies and the ways they can be concealed in what appears to be a sound inference. Watch for: Non sequitur Circularity affirmation of the consequent denial of the antecedant false analogy red herring faulty causality faulty generalization Straw man etc. I belive Einstein has concealed all of the above fallacies in what is a masterpiece of paranoid thinking to result in a circular theory where all predictions will match observations but no prediction will result that will alter the fundamental basis of the theory which is: paranoia. Mike === Subject: Re: final proof Einstein 1905 is crap. Mike: >Newton used a modified version of Kepler's third law in order to >derive the law of universal gravitation. Please note that without such >law, which was purely empirical, the derivation of LUG is impossible. Newton did not derive his law of gravitation, nor can it be derived from kepler's laws. Newton deduced a general principle from kepler's laws and if that principle was to stand a chance of being correct, the first requirement is that kepler's laws must be derivable from that principle. >Then, in mechanics books you see the famous Kepler problem and the >derivation of Kepler's third law using the LUG as given! Kepler's laws _can_ be derived from newton's law of gravity. It's called consistency, not circularity. Newton's law if gravity _cannot_ be derived from kepler's laws. Kepler's laws say nothing about forces or potentials or even why the planetary motion should be described by his laws. By contrast, the 1/r potential newton proposed can be used to show that stable orbits only occur for forces of the form r^n, where n = +/- 2. >But of course, if the conclusion is true, all premised in the >deduction process must be true otherwise the deduction was unsound in >the first place. >The same of course hold with Einstein and some claims the constancy of >c can be derived , etc. The assertion that `c' is a constant cannot be derived, since obviously it's possible to create a self-consistent theory of mechanics in which c -> infty, called galilean relativity. Distinguishing between special relativity and galilean relativity is an experimental issue and since galilean relativity makes predictions that are incompatible with what is observed, experiment has determined that there exists some velocity, `c', which is finite. Electromagnetic theory predicts that `c' and the speed of light must be the same. [..] >I belive Einstein has concealed all of the above fallacies in what is >a masterpiece of paranoid thinking to result in a circular theory >where all predictions will match observations but no prediction will >result that will alter the fundamental basis of the theory which is: >paranoia. Special relativity has been used to make quite a number of predictions. For example, the standard model is based on special relativity. If relativity was wrong, the standard model would have had zero chance of correctly predicting the mass ratio between the W and Z or the myriad of other predictions it make correctly. While there is more to the standard model than relativity, relativity is the foundation upon which the standard model is constructed. === Subject: Re: final proof Einstein 1905 is crap. >> Actually the commonest fallacy in Special Relativity is to call a >> conclusion >> a postulate via the fallacy of circularity, and that is pretty >> serious. > Well, you are hetting into really dark domains here. Before getting as > dvanced as SR, the use of circularity was enforced by those who > missinterpreted Newton, or actually desired to present Newton's > metaphysics as a science. To be explicit: > Newton used a modified version of Kepler's third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Kepler's third law using the LUG as given! > But of course, if the conclusion is true, all premised in the > deduction process must be true otherwise the deduction was unsound in > the first place. Hold on a moment. Kepler's third law is empirical, you've said. That's originating in or based on observation or experience. Kepler's third law (however modified) is The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes: (P1/P2)^2 = (R1/R2)^3 Are you saying this is NOT what is observed? All I see here is that a simple equation has been determined to describe observation. > The same of course hold with Einstein and some claims the constancy of > c can be derived , etc. That is a very different matter. Nobody has ever observed light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. There is no empirical foundation to it. In fact the empirical data contradicts it, aside from any logical deduction. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. You'll get no argument from me on that. > Question of the day: how to you distinguish between paranoid and > genious? Often, both sound as equally ntelligent. > My answer: Paranoids are easy to spot is you understand formal and > informal fallacies and the ways they can be concealed in what appears > to be a sound inference. Watch for: > Non sequitur > Circularity > affirmation of the consequent > denial of the antecedant > false analogy > red herring > faulty causality > faulty generalization > Straw man > etc. > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. To you he was paranoid. You are looking into the cause of his ill-concieved ideas and attributing it to a form of mental illness. To me he was a huckster. Without getting into psychology or psychiatry, fields in which I do not claim expertise, for maybe thieves and murderers suffer from mental abberations and it can all be blamed on an abusive childhood, nevertheless I want them locked away in a correctional institution or be taken out of society if it can be justly determined that they are guilty of crime. It is too late to correct Einstein the man, he's lived and died. We are still obligated to eradicate the consequences of his fraud, whatever the cause. Therefore I have to say your analysis is actually --- Non sequitur. But then, so is mine. Whatever the cause, it is our duty to promote freedom of thought in the next generation of budding young physicists and to dissuade them from the indoctrinators that infest this newsgroup. Androcles. > Mike === Subject: Re: final proof Einstein 1905 is crap. >> Actually the commonest fallacy in Special Relativity is to call a >> conclusion >> a postulate via the fallacy of circularity, and that is pretty >> serious. > Well, you are hetting into really dark domains here. Before getting as > dvanced as SR, the use of circularity was enforced by those who > missinterpreted Newton, or actually desired to present Newton's > metaphysics as a science. To be explicit: > Newton used a modified version of Kepler's third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Kepler's third law using the LUG as given! > But of course, if the conclusion is true, all premised in the > deduction process must be true otherwise the deduction was unsound in > the first place. > Hold on a moment. Kepler's third law is empirical, you've said. > That's originating in or based on observation or experience. > Kepler's third law (however modified) is > The ratio of the squares of the revolutionary periods for two planets > is equal to the ratio of the cubes of their semimajor axes: > (P1/P2)^2 = (R1/R2)^3 > Are you saying this is NOT what is observed? > All I see here is that a simple equation has been determined to describe > observation. > The same of course hold with Einstein and some claims the constancy of > c can be derived , etc. > That is a very different matter. Nobody has ever observed light is > always propagated in empty space with a definite velocity c which is > independent of the state of motion of the emitting body. > There is no empirical foundation to it. > In fact the empirical data contradicts it, aside from any logical > deduction. Not that hard to do! Use the standard apparatus for testing light velocity, the two, toothed wheels, form which c can be calculated from the difference in their rotation and separation when a beam is shone through. Now stick it on a rocket, and test the sunlight speed when the rocket is travelling away/towards, the sun. Probably cost a few million on the next interplanetary run- FAR too expensive for the DHR budget! (make that too likely to embarass) > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. > You'll get no argument from me on that. > Question of the day: how to you distinguish between paranoid and > genious? Often, both sound as equally ntelligent. > My answer: Paranoids are easy to spot is you understand formal and > informal fallacies and the ways they can be concealed in what appears > to be a sound inference. Watch for: > Non sequitur > Circularity > affirmation of the consequent > denial of the antecedant > false analogy > red herring > faulty causality > faulty generalization > Straw man > etc. > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. > To you he was paranoid. > You are looking into the cause of his ill-concieved ideas and > attributing > it to a form of mental illness. > To me he was a huckster. > Without getting into psychology or psychiatry, fields in which I do not > claim expertise, for maybe thieves and murderers suffer from mental > abberations and it can all be blamed on an abusive childhood, > nevertheless > I want them locked away in a correctional institution or be taken > out of society if it can be justly determined that they are guilty of > crime. > It is too late to correct Einstein the man, he's lived and died. > We are still obligated to eradicate the consequences of his fraud, > whatever the cause. > Therefore I have to say your analysis is actually --- Non sequitur. > But then, so is mine. Whatever the cause, it is our duty to promote > freedom of thought in the next generation of budding young physicists > and to dissuade them from the indoctrinators that infest this newsgroup. > Androcles. > Mike === Subject: Re: final proof Einstein 1905 is crap. > > Actually the commonest fallacy in Special Relativity is to call a > conclusion > a postulate via the fallacy of circularity, and that is pretty > serious. > Well, you are hetting into really dark domains here. Before getting >> as >> dvanced as SR, the use of circularity was enforced by those who >> missinterpreted Newton, or actually desired to present Newton's >> metaphysics as a science. To be explicit: >> Newton used a modified version of Kepler's third law in order to >> derive the law of universal gravitation. Please note that without >> such >> law, which was purely empirical, the derivation of LUG is >> impossible. >> Then, in mechanics books you see the famous Kepler problem and the >> derivation of Kepler's third law using the LUG as given! >> But of course, if the conclusion is true, all premised in the >> deduction process must be true otherwise the deduction was unsound >> in >> the first place. >> Hold on a moment. Kepler's third law is empirical, you've said. >> That's originating in or based on observation or experience. >> Kepler's third law (however modified) is >> The ratio of the squares of the revolutionary periods for two >> planets >> is equal to the ratio of the cubes of their semimajor axes: >> (P1/P2)^2 = (R1/R2)^3 >> Are you saying this is NOT what is observed? >> All I see here is that a simple equation has been determined to >> describe >> observation. >> The same of course hold with Einstein and some claims the constancy >> of >> c can be derived , etc. >> That is a very different matter. Nobody has ever observed light is >> always propagated in empty space with a definite velocity c which is >> independent of the state of motion of the emitting body. >> There is no empirical foundation to it. >> In fact the empirical data contradicts it, aside from any logical >> deduction. > Not that hard to do! > Use the standard apparatus for testing light velocity, the two, > toothed wheels, form which c can be calculated from the difference in > their rotation and separation when a beam is shone through. Now stick > it on a rocket, and test the sunlight speed when the rocket is > travelling away/towards, the sun. > Probably cost a few million on the next interplanetary run- FAR too > expensive for the DHR budget! (make that too likely to embarass) The easiest way is Roemer's method and Cassini. All it takes is data gathering. The DHRs would say time was increasing and decreasing during the orbit, but it wouldn't be all that convincing. Androcles >> Let me tell you what it boild down to: Math infidels with no >> understanding of the principles of logic the Greeks established and >> Aristotle put in a massive organized works called the Organon, have >> taken over and their misunderstanding and paranoia has resulted in >> the >> chaos we see in science todat with inductions presented as >> deductions, >> deductions as induction, abductions as inductions and deductions. >> You'll get no argument from me on that. >> Question of the day: how to you distinguish between paranoid and >> genious? Often, both sound as equally ntelligent. >> My answer: Paranoids are easy to spot is you understand formal and >> informal fallacies and the ways they can be concealed in what >> appears >> to be a sound inference. Watch for: >> Non sequitur >> Circularity >> affirmation of the consequent >> denial of the antecedant >> false analogy >> red herring >> faulty causality >> faulty generalization >> Straw man >> etc. >> I belive Einstein has concealed all of the above fallacies in what >> is >> a masterpiece of paranoid thinking to result in a circular theory >> where all predictions will match observations but no prediction >> will >> result that will alter the fundamental basis of the theory which >> is: >> paranoia. >> To you he was paranoid. >> You are looking into the cause of his ill-concieved ideas and >> attributing >> it to a form of mental illness. >> To me he was a huckster. >> Without getting into psychology or psychiatry, fields in which I do >> not >> claim expertise, for maybe thieves and murderers suffer from mental >> abberations and it can all be blamed on an abusive childhood, >> nevertheless >> I want them locked away in a correctional institution or be taken >> out of society if it can be justly determined that they are guilty of >> crime. >> It is too late to correct Einstein the man, he's lived and died. >> We are still obligated to eradicate the consequences of his fraud, >> whatever the cause. >> Therefore I have to say your analysis is actually --- Non sequitur. >> But then, so is mine. Whatever the cause, it is our duty to promote >> freedom of thought in the next generation of budding young physicists >> and to dissuade them from the indoctrinators that infest this >> newsgroup. >> Androcles. >> Mike === Subject: Re: final proof Einstein 1905 is crap. > The same of course hold with Einstein and some claims the constancy of > c can be derived , etc. > That is a very different matter. Nobody has ever observed light is > always propagated in empty space with a definite velocity c which is > independent of the state of motion of the emitting body. Androcles is being a retard, as per usual. > There is no empirical foundation to it. More of Androcles not reading the literature, as per usual. > In fact the empirical data contradicts it, aside from any logical > deduction. More of Androcles mistakening his suppositions for reality, as per usual. > Not that hard to do! > Use the standard apparatus for testing light velocity, the two, > toothed wheels, form which c can be calculated from the difference in > their rotation and separation when a beam is shone through. Now stick > it on a rocket, and test the sunlight speed when the rocket is > travelling away/towards, the sun. > Probably cost a few million on the next interplanetary run- FAR too > expensive for the DHR budget! (make that too likely to embarass) Since you have no idea what goes into designing a mission package, such ignorant statements come as no surprise. Why don't you read the literature? Have you ever considered the possibility that you are wrong? === Subject: Re: final proof Einstein 1905 is crap. Jim Greenfield: >Not that hard to do! >Use the standard apparatus for testing light velocity, the two, >toothed wheels, form which c can be calculated from the difference in >their rotation and separation when a beam is shone through. Now stick >it on a rocket, and test the sunlight speed when the rocket is >travelling away/towards, the sun. >Probably cost a few million on the next interplanetary run- FAR too >expensive for the DHR budget! (make that too likely to embarass) I'll be happy to perform the experiment if you pay for it. My guess is that you aren't willing to back up your beliefs with cash. === Subject: Re: final proof Einstein 1905 is crap. [snip > Hold on a moment. Kepler's third law is empirical, you've said. > That's originating in or based on observation or experience. > Kepler's third law (however modified) is > The ratio of the squares of the revolutionary periods for two planets > is equal to the ratio of the cubes of their semimajor axes: > (P1/P2)^2 = (R1/R2)^3 > Are you saying this is NOT what is observed? > All I see here is that a simple equation has been determined to describe > observation. Actually that's the observed law. Newton did not use that because it will gets you nowhere. Newton used kmT^2 = R^3 where k is a constant. This is not mentioned in most books. The constant k ends up part of G in the LUG. This is what Newton did, he assumed that in the two-body problem: k(m1+m2)T^2 = (R1+R2)^3=R^3 He then assumed that m1>m2 to get: k1m1T^2 ~ R^3 and he assumed again that this hold for all planets so that: km1T1^2 = R1^3 km2T2^2 = R2^3 divide and you get Kepler's observed law! This series of assumptions by Newton hides the hypothesis that one of the bodies is fixed in absolute space and is immovable, and that is the SUN. He actually states that hypothesis I: HYPOTHESIS I. That the center of the system of the world is immovable. This is acknowledged by all, while some contend that the earth, others that the sun is fixed in that center. Let us see what may from hence follow. PROPOSITION XI. THEOREM XI. That the common center of gravity of the earth, the sun, and all the planets, is immovable. LOL Without such hypothesis, derivation of the LUG is impossible. Now, math infidels such as Newton and Einstein did not realize that unless bodies trajectories are fixed in space by a magical 'spirit' (actually Newton believed in such spirit) their gravitational dynamics lead to a collapse of the universe in a single point over time. This is why Einstein wanted to include the cosmological constant to prevent that. Yet, everybody is deceived to believe that the derived theories explain gravitation but is deprived of the underline assumptions that basically invoke magic. Most books attempt to circumvent the above using hand waiving arguments and Kolker eats them for breakfast. Yet, what is, is. Mike === Subject: Re: final proof Einstein 1905 is crap. > [snip >> Hold on a moment. Kepler's third law is empirical, you've said. >> That's originating in or based on observation or experience. >> Kepler's third law (however modified) is >> The ratio of the squares of the revolutionary periods for two >> planets >> is equal to the ratio of the cubes of their semimajor axes: >> (P1/P2)^2 = (R1/R2)^3 >> Are you saying this is NOT what is observed? >> All I see here is that a simple equation has been determined to >> describe >> observation. > Actually that's the observed law. Newton did not use that because it > will gets you nowhere. Newton used kmT^2 = R^3 where k is a constant. > This is not mentioned in most books. The constant k ends up part of G > in the LUG. This is what Newton did, he assumed that in the two-body > problem: > k(m1+m2)T^2 = (R1+R2)^3=R^3 > He then assumed that m1>m2 to get: > k1m1T^2 ~ R^3 > and he assumed again that this hold for all planets so that: > km1T1^2 = R1^3 > km2T2^2 = R2^3 > divide and you get Kepler's observed law! > This series of assumptions by Newton hides the hypothesis that one of > the bodies is fixed in absolute space and is immovable, and that is > the SUN. He actually states that hypothesis I: > HYPOTHESIS I. > That the center of the system of the world is immovable. > This is acknowledged by all, while some contend that the earth, others > that the sun is fixed in that center. Let us see what may from hence > follow. Einstein does the same thing, calling it the stationary system. However, Newton is careful to call an hypothesis by its name, but Einstein calls his own light speed hypothesis a postulate. > PROPOSITION XI. THEOREM XI. > That the common center of gravity of the earth, the sun, and all the > planets, is immovable. > LOL > Without such hypothesis, derivation of the LUG is impossible. Granted that Newton envisaged a heliocentric universe, nevertheless the solar system can be treated as having a common centre of gravity. He didn't know of the existence of Pluto, the orbit of which is Sol dependent. Moreover, any perturbation of the orbits of the planets by the nearest star are neglible. There is a considerable difference between Pluto being circa 4 light hours from Sol and circa 4 light years from Proxima Centauri, particularly when R^3 is taken into account. (24 * 365.25 )^3 ~= 5,000,000. > Now, math infidels such as Newton and Einstein did not realize that > unless bodies trajectories are fixed in space by a magical 'spirit' > (actually Newton believed in such spirit) their gravitational dynamics > lead to a collapse of the universe in a single point over time. A trajectory cannot be fixed in space by definition of trajectory, which implies motion. I 've missed the point you are attempting to make. Until such times as our solar system enters the vicinity of a massive body such as another star, I'll go along with Newton's description, it is an adequate approximation. > This > is why Einstein wanted to include the cosmological constant to prevent > that. > Yet, everybody is deceived to believe that the derived theories > explain gravitation but is deprived of the underline assumptions that > basically invoke magic. Action at a distance IS magic, in that we do not understand it. However, we do observe it and Newton gives a good approximation that will adequately predict the future positions of the planets, even though it will ultimately be chaotic. > Most books attempt to circumvent the above using hand waiving > arguments and Kolker eats them for breakfast. Yet, what is, is. Why bother with Kolker? He's just another parrot, repeating what he's read. He's not capable of actually thinking, few are. There are three distinct categories of people that write to this newsgroup. 1) Parrots that can only repeat what they've read. 2) Egocentrics that want to promote their own pet theory. 3) Rational thinkers. There are three mutually exclusive concepts that compete. a) the velocity of light is medium dependent b) the velocity of light is source dependent c) the velocity of light is observer dependent Kolker is a 1c) Androcles. > Mike === Subject: Re: final proof Einstein 1905 is crap. > Newton used a modified version of Kepler's third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. Not true. A central force with a 1/r potential will give you all of the Keplarian laws. It is quite true that Keplers third law -suggest- to Newton that the inverse square law is correct, but Kepler independent ways of deriving the famous inversesquare law plus all of Keplers empirical laws exist as I have indicated. Refere to Goldstein's book on Mechanics for the mathematical details. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Kepler's third law using the LUG as given! Not true, as I have shown. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. Aristotle's logic is a restricted subset of first order logic. Furthermore the syllogism is not the ideal form of mathematical proof structure. The Stoics* developed a form of logic** that is much closer to first order logic than Aristotle. Unfortunately, most of their works in logic have not survived. For details refer to -The Development of Logic- by Kneale and Kneale (hom et ux). > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. Utter nonsense. SR is potentially falsifiable. Later ways of expressing SR by means of Minkowski geometry show how stragight forward SR is. The group of transformation (Lorentz Group) which is the bases of SR were established first and independently by Lorentz and Poincare so there are no mathematical mysteries here. Bob Kolker * especially Chrysippus ** modus ponens > Mike === Subject: Re: final proof Einstein 1905 is crap. > Newton used a modified version of Kepler's third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Not true. A central force with a 1/r potential will give you all of the > Keplarian laws. It is quite true that Keplers third law -suggest- to > Newton that the inverse square law is correct, but Kepler independent > ways of deriving the famous inversesquare law plus all of Keplers > empirical laws exist as I have indicated. Refere to Goldstein's book on > Mechanics for the mathematical details. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Kepler's third law using the LUG as given! > Not true, as I have shown. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. > Aristotle's logic is a restricted subset of first order logic. > Furthermore the syllogism is not the ideal form of mathematical proof > structure. The Stoics* developed a form of logic** that is much closer > to first order logic than Aristotle. Unfortunately, most of their works > in logic have not survived. For details refer to -The Development of > Logic- by Kneale and Kneale (hom et ux). > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. > Utter nonsense. SR is potentially falsifiable. Later ways of expressing > SR by means of Minkowski geometry show how stragight forward SR is. The > group of transformation (Lorentz Group) which is the bases of SR were > established first and independently by Lorentz and Poincare so there are > no mathematical mysteries here. Just the biggie! HOW does photon A, emitted from a source stationary ref us, KNOW how fast B, emitted from a moving source at that location, is travelling????????????? Jim G c'=c+v > Bob Kolker > * especially Chrysippus > ** modus ponens > Mike === Subject: Re: final proof Einstein 1905 is crap. Jim Greenfield: >Just the biggie! >HOW does photon A, emitted from a source stationary ref us, KNOW how >fast B, emitted from a moving source at that location, is >travelling????????????? Why would a photon have to know that? === Subject: Re: final proof Einstein 1905 is crap. > Newton used a modified version of Kepler's third law in order to > derive the law of universal gravitation. Please note that without such > law, which was purely empirical, the derivation of LUG is impossible. > Not true. A central force with a 1/r potential will give you all of the > Keplarian laws. It is quite true that Keplers third law -suggest- to > Newton that the inverse square law is correct, but Kepler independent > ways of deriving the famous inversesquare law plus all of Keplers > empirical laws exist as I have indicated. Refere to Goldstein's book on > Mechanics for the mathematical details. > Then, in mechanics books you see the famous Kepler problem and the > derivation of Kepler's third law using the LUG as given! > Not true, as I have shown. > Let me tell you what it boild down to: Math infidels with no > understanding of the principles of logic the Greeks established and > Aristotle put in a massive organized works called the Organon, have > taken over and their misunderstanding and paranoia has resulted in the > chaos we see in science todat with inductions presented as deductions, > deductions as induction, abductions as inductions and deductions. > Aristotle's logic is a restricted subset of first order logic. > Furthermore the syllogism is not the ideal form of mathematical proof > structure. The Stoics* developed a form of logic** that is much closer > to first order logic than Aristotle. Unfortunately, most of their works > in logic have not survived. For details refer to -The Development of > Logic- by Kneale and Kneale (hom et ux). > I belive Einstein has concealed all of the above fallacies in what is > a masterpiece of paranoid thinking to result in a circular theory > where all predictions will match observations but no prediction will > result that will alter the fundamental basis of the theory which is: > paranoia. > Utter nonsense. SR is potentially falsifiable. Not for religious fanatics like Mike (aka Bill Smith, Undeniable, Eleatis etc...) > Later ways of expressing > SR by means of Minkowski geometry show how stragight forward SR is. The > group of transformation (Lorentz Group) which is the bases of SR were > established first and independently by Lorentz and Poincare so there are > no mathematical mysteries here. You are wasting your time with this idiot. Dirk Vdm === Subject: Re: final proof Einstein 1905 is crap. > Shubert makes the same fallacy of circularity in his rather long-winded > but actually quite simple minded derivation, although he now admits > pulling sqrt(1-v^2/c^2) from thin air. I missed his derivation but have noted here that Tom Robert's derivation involves hand-waving when he reaches his penultimate step: the evaluation of a criterion, value one for Newton, over or under for SR and ... polar coordinates? (Don't remember because I didn't actually read that part.) The Newton part was automatic, the SR part's logic trail disappeared and he just said, well make it this particular value and we have SR. SR-cult cretins here have agreed Einstein's 1916/1962 (let's hear it from the idiots) did really idiotic things. AE 1905 was just as bad but less obvious. I've seen a number of so-called derivations that started with the conclusion. eleaticus He is nothingimportant, of > course, although he likes to think he is everythingimportant. Egomania > is treatable, I understand, although it may take the services of a > trained psychiatrist. > Androcles === Subject: Re: final proof Einstein 1905 is crap. >> Shubert makes the same fallacy of circularity in his rather >> long-winded >> but actually quite simple minded derivation, although he now admits >> pulling sqrt(1-v^2/c^2) from thin air. > I missed his derivation but have noted here that Tom Robert's > derivation > involves hand-waving when he reaches his penultimate step: the > evaluation of > a criterion, value one for Newton, over or under for SR and ... polar > coordinates? (Don't remember because I didn't actually read that > part.) The > Newton part was automatic, the SR part's logic trail disappeared and > he just > said, well make it this particular value and we have SR. I've carved Roberts up big time in the thread Roberts considers Einstein paper to be irrelevant verbiage. He really is a sad case of fanaticism. Androcles > SR-cult cretins here have agreed Einstein's 1916/1962 (let's hear it > from > the idiots) did really idiotic things. AE 1905 was just as bad but > less > obvious. > I've seen a number of so-called derivations that started with the > conclusion. > eleaticus > He is nothingimportant, of >> course, although he likes to think he is everythingimportant. >> Egomania >> is treatable, I understand, although it may take the services of a >> trained psychiatrist. >> Androcles === Subject: Re: final proof Einstein 1905 is crap. eleaticus: >I've seen a number of so-called derivations that started with the >conclusion. on a regular basis.