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 X-RFC2646: Format=Flowed; Response 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 X-RFC2646: Format=Flowed; Response >> 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... X-RFC2646: Format=Flowed; Original 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 X-RFC2646: Format=Flowed; Response >> 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 X-RFC2646: Format=Flowed; Original >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. X-RFC2646: Format=Flowed; Original > 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 === Sub