mm-4489 === Subject: Re: NASA exposes Apollo Hoax > It is not a coincidence that the greater one's scientific education, > the more absurd this drivel about hoaxes seems. G> I say the opposite is true, the more I increase my knowledge the more > G> it appears to be a hoax! Which explains all the material we left up on the moon. > Several Electric vehicles, > Cameras galore, > Landing Stages, > Several US Flags, > And a spoon. > Someone in Bean Counting at NASA stated on paper we left roughly > $409,832,000 worth of material on the Moon. The spoon cost, $0.39 http://home.comcast.net/~anglewyrm/marsphoto.jpg If you can picture a scenario where large amounts of money are required to 'make it happen', then I can picture a scenario where it's done reeeeely cheap. === Subject: teaching notes Hi I am looking for Issues in Accounting Education - Teaching Notes Vol. 22, No. 2 May 2007 Does anyone have it? If yes, what can i do to get it. === Subject: Re: Minimal Oval > Though it may be worth mentioning that if the curvature of concavities > is considered to be negative (which is not an unreasonable thing to > do), then an infinitesimal distortion of the crescent *does* satisfy > the conditions. > Sudden jump from convexity to concavity entails a second order > discontinuity or a sharp inflection point at the entry to the concave > arc which is violative of the conditions. Do you mean a discontinuity in the second order derivative? Yes, there > is an inflection point, but no, there is no need for a sharp > inflection point, or any discontinuity in derivatives. Consider, for > example, y = sin(x), which has inflection points yet all derivatives > are continuous. There is no reason why a similarly well-behaved > transition from convexity to concavity cannot be arranged for this > curve. Yes, sin(x) sort of ovaloids have a smooth inflection points between > the convex and concave parts but they are not allowed.I am referring > to the abrupt spikes of the crescent and also the earlier pictures > wherein the sketch had curvature discontinuity of second > order. I referred to an infinitesimal distortion of the crescent. A smooth curve -- one that satisfies all your differentiability and continuity requirements (though obviously not the original convexity condition) -- can be constructed to approximate the crescent as closely as you want. The curvature at the points of the crescent gets arbitrarily large. By infinitesimal distortion I am referring to such a curve that is arbitrarily close to the crescent. The fact that curves with sharp points are acceptable limiting solutions, in the sense that arbitrarily close smooth curves can be constructed, has been explained over and over and over again, right from the very beginning of this thread. So I am at a loss to understand why again and again you keep raising exactly the same objection. Do you not understand this point? Do you disagree with it? Do you read it, understand it, and agree with it, and then five minutes later completely forget it? > Convexity demands zero or positive sign for curvature continuously and negative sign is not acceptable. ORIGINALLY we were talking about convex curves. Then YOU introduced a concave example, and I mentioned, as a point of interest, that IF you allow curves of this nature THEN the area can be made as small as you like, and I gave the crescent example. OBVIOUSLY the crescent is not a convex curve. OBVIOUSLY it doesn't satisfy the original convexity condition. I KNOW that. === Subject: #236 these triangles; Multiplication on AP-adics is Conservation of Angular Momentum; new textbook: Mathematical Physics (AP-adics primer) for age 6 years onward (snipped) In the case of ....9999 x 2 we have the longitude line going from > North Pole to one unit short > of South Pole and for latitude of 2 we have two units arclength going > west (eastern hemisphere > is imaginaries (pi to 2pi)) This forms a triangle whose legs are .... > 9999 and ...00002 > These triangles are a bit difficult to see for the long and skinny slender ones such as 2 x ....99999. Especially difficult is the triangle ...9999 X ....99999 for it goes from the Greenwich longitude all the way down to the South Pole except stopping short of the South Pole by one unit distance and then it travels westward for a rotation of ....99999 and which the resultant triangle formed is almost the entire Hemisphere. The product of the above is 999......001 Easier triangles to visualize are the ones such as 5000.....0000 X 5000....00000. One leg of this triangle is the equator line and then it goes 90 degrees west. So the three legs of this triangle would be like Greenwich longitude going to the Equator and then 90 degrees westward to the longitude that goes though approx Chicago and lands on the equator. Easier to visualize because they look more like the triangles we are used to, although the legs are concave outward. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: #237 calculus picketfence is more apt or spherical triangles; Multiplication on AP-adics is Conservation of Angular Momentum; new textbook: Mathematical Physics (AP-adics primer) for age 6 years onward > (snipped) In the case of ....9999 x 2 we have the longitude line going from > North Pole to one unit short > of South Pole and for latitude of 2 we have two units arclength going > west (eastern hemisphere > is imaginaries (pi to 2pi)) This forms a triangle whose legs are .... > 9999 and ...00002 > These triangles are a bit difficult to see for the long and skinny > slender > ones such as 2 x ....99999. > Especially difficult is the triangle ...9999 X ....99999 for it goes > from the > Greenwich longitude all the way down to the South Pole except stopping > short of the South Pole by one unit distance and then it travels > westward for a rotation of ....99999 and which the resultant triangle > formed is almost the entire Hemisphere. The product of the above is > 999......001 > Easier triangles to visualize are the ones such as 5000.....0000 X > 5000....00000. One leg of this triangle is the equator line and > then it goes 90 degrees west. So the three legs of this triangle would > be like Greenwich longitude going to the Equator and then 90 degrees > westward to the longitude that goes though approx Chicago and > lands on the equator. Easier to visualize because they look more > like the triangles we are used to, although the legs are concave > outward. > Maybe Dik or someone else knows whether the literature already has terms for what I am constructing. Earlier this morning I thought of them as triangles. Triangles have to explain 3 X ....33333 = 1 X .....999999 So what I have resolved is a better system this evening. I think all the bugs and kinks are out. It is the Calculus of summation of slender line segments that sums up as the area. Call it picketfence calculus or call it spherical-triangles. I guess spherical-triangles is my best term. Whether mathematics already has a term for what I am about to describe, please convey. I am looking at a globe and the hemisphere for the numbers 1 to 999....99999 are on this hemisphere. And I use the Greenwich longitude line in full to mark off this hemisphere where of course the North Pole is 2pi and the South Pole is pi and both are imaginary thus making the Greenwich Longitude imaginary also. So my hemisphere is smaller than the imaginary hemisphere. Now I use the Equator line only 1/2 of it for the other 1/2 is in the imaginary region. And I use another longitude line for reference sake only and it is the 90 degree longitude line which passes through Wisconsin and Illinois and is the 5000.....00000 line. Now any mulitiplication is easy to do in this set up. So if you asked me to multiply 3 X ....33333 which is equal to .....999999 then on the globe I go down the Greenwich longitude of ....3333 which is 60degrees down and be somewhere in Spain and then I go to the equator line and mark off 3 units west thus forming a tiny sliver of a spherical-rectangle whose internal area is .....999999 square units. Or if you wanted the reverse of where I go down the Greenwich Longitude by 3 units and then starting at the Equator go across the Equator by ....33333 units still leaving me a area of ....99999 square units. Now for 1 X ....99999 would be the entire distance down from 1 near the North Pole to ....99999 near the South Pole and the sliver of the spherical rectangle is only 1 unit across and its internal area is ....99999 square units. So here we have a agreement that 3 X ....33333 = 1 X ....999999 Now let us check out 50000.....00000 X 5000....0000 and its product is 2500....0000 but let us see if the spherical triangle matches the arithmetic. So I go halfway down the Greenwich longitude which is in fact the Equator and I sum over all the line segments of the Equator out to 5000...0000 which is that 90 degree longitude. So the model gives us the spherical-triangle for it is the summation of all the line segments from Greenwich to 90 degree longitude. But does it reconcile with the product of 25000....0000. It most certainly does in that it is 1/4 of the hemisphere. Now let me check on another multiplication. That of ...99999 X .... 99999. Here we have the entire Greenwich Longitude except missing the Pole points. And we sum all the longitude lines of the entire hemispheric-Equator. So in essence we sum every line segment in the entire Hemisphere. And what is the product of ....99999 X .....999999 and it is 9999......0001 Now does that final answer of 9999....0001 make any sense as far as the Model? Yes indeed it makes very much sense and even accuracy. Keep in mind that this hemisphere is a abbreviated hemisphere in that the Greenwich Longitude is imaginary. So that one would think that if you multiplied over every line segment in the hemisphere that you would end up with the entire hemisphere. But we must remember that the Greenwich Longitude is imaginary and so the final answer is not ....99999 X .....999999 equals ....99999 but slightly less to compensate for the imaginary lines. Interesting question: now when I go over and play with the imaginary hemisphere, do I pick up those lost lines over there?? Now here is a horrifying prospect that as we stand back and look at this model that we can have the biggest number of ....99999 yet within the smallest of areas and where the triangle of 3 X ....33333 which is a tiny sliver of a triangle yet has more area than 1/4 of the hemisphere embodied in 5000....00000 X 50000.....00000. Question: do we see this sort of stuff in other aspects of Elliptic or Hyperbolic Geometry where we have huge numbers but small size and where we have large size from small numbers? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Implementable Set Theory and Consistency of ZFC >So, even if I don't make use of (5-8), a proof of A from (1-4) is a >proof from (1-8) ? So, even if I say there exists a Foo, then such >a statement is a valid premise for proving that the integral of 1/t >from 1 to x is ln(x) ? Weird .. > No one said anything about valid premise. > Here is the definition (I'm giving it a bit informally, it be can be > made even more precise, but this is good enough for our purposes), > and, for convenience, I'll use Hilbert style proof (it's easy to adapt > this to other systems such as natural deduction): > P is a proof of S from G iff > P is a sequence of formulas in the language such that every entry in > the sequence is either an axiom of logic (with identity if identity > theory is used or needed) or a member of G or follows from previous > entries in the sequence by an inference rule, and the last entry in > the sequence is S. > Then > Theorem: If P is a proof of S from G, and G is a subset of H, then P > is a proof S from H. > Proof: > Suppose P is a proof of S from G, and G is a subset of H. > Then P is a sequence of formulas in the language such that every entry > in the sequence is either an axiom of logic (with identity if identity > theory is used or needed) or a member of H (SINCE EVERY MEMBER OF G IS > A MEMBER OF H) or follows from previous entries in the sequence by an > inference rule, and the last entry in the sequence is S. > Do you get it now? No. Unbelievable! WHAT don't you understand now? MoeBlee There is so much that HdB does not understand that this NG is not large enough to contain it. === Subject: Re: Implementable Set Theory and Consistency of ZFC >If you had a constructive proof, then you'd be able to show a >DERIVATION of any of (5)-(8) from (1)-(4). >Sure. That's what I did. There's not a single mention of any of the axioms (1)-(4) in your > proof of (5). It is obviously not a proof of (5) from (1)-(4). Yes it is. JFH: No it isn't. HdB: Yes it is. JFH: No it isn't. HdB: Yes it is. .. Ad infinitum. Han de Bruijn So there is an infinity after all! === Subject: Re: Implementable Set Theory and Consistency of ZFC >Fine. By ~Infinity, I mean Each set is equinumerous with a set of >the form {0,1,...,n}. Is it your belief that (1)-(4) prove *that* >statement? >The anwser is: yes. Great! Then it is rather easier to show that ZFC is inconsistent. > Just give the proof of ~Infinity using only (1)-(4) and we conclude > that (1)-(4)+(X) is inconsistent. You say it. But we have not proved it. Nor have you. That your model does not allow any other sort of set does not establish that your axiom set does not allow it. === Subject: Re: Implementable Set Theory and Consistency of ZFC >Yes. I've used _only_ Extensionality, Empty set, Pairing and >Union. Then it follows I can make _any_ (finite) set with these. The >other axioms do _not_ add any constructive power to this. Therefore >I conclude they must be theorems, instead of axioms. As long as the >theory is _constructive_, i.e. without the axiom of Infinity. Okay, so you don't know what theorem means. We get it. I don't understand how _you_ can find something that's true, but I don't > consider it to be my problem. We can't understand how _you_ determine what you think is true, but it is not really our problem, either. === Subject: Re: Implementable Set Theory and Consistency of ZFC But ZFC does not consist of ZFC plus the negation of Infinity > plus more axioms (it _is_ equal to ZFC without Infinity plus > more axioms). Let's add to it the response by Jesse F. Hughes, and see if that might > lead to something that resembles agreement between us. [ ... ] It seems to me that your model contains every set required > by axioms (1)-(4) and no sets that are not required and so it is > apparently minimal. If I were to hazard a guess, I would say that > there is a model homomorphism mapping your model into any other model > of (1)-(4). I would further guess that this homomorphism is > one-to-one. What do you think? Han de Bruijn One-to-one (injective or monomorphic) seems likely but onto (surjective or epimorphic) does not. === Subject: Re: Implementable Set Theory and Consistency of ZFC 'I will say it once more. I am typing this slowly, since I don't want >you to miss anything I say. If you have a proof in a theory >consisting of axioms (1)-(4), it is also a proof in the theory >consisting of axioms (1)-(4)+(X). >How do you know that? Has the Pope told you, by dogma, that it is so? No, you royal ignoramus, it's PROVEN as a basic property of the >deductive system. It's the property of monotonicity of deduction. And >we PROVE it. >Yeah, yeah. Then why are (5-9) required in ZFC, once Infinity has become >an axiom of it? Why is e.g. Choice provable in (ZFC-Infinity) and not in >common ZFC? This has been explained several times. If by ZFC-Infinity you mean > ZFC without the axiom of infinity then Choice is _not_ provable > in that syatem. ZFC without the axiom of infinity is not the > same as ZFC plus the negation of the axiom of infinity - choice > _is_ provable in ZFC plus the negation of the axiom of infinity. Sure. (ZFC - Foo) is not the same as (ZFC - Foo + ~Foo). Quite clear. > Tip, hint: how can Foo be denied iff Foo is just plain nonsense in > the first place? It is only in the unreliable opinion of HdB that Foo is nonsense, but that is not the issue here, in finite set terms which even HdB can parse: {a,b} {b} = {a}, but ({a,b} {b}) u {notb} = {a, notb} so they are not the same. === Subject: evaluating limits Problem: Identify a value of delta that corresponds to epsilon = 0.01 given lim_x->-1 (x^2 + 3) = 4 according to the definition of limits. |f(x) - L| < Epsilon |(x^2 + 3) - 4| < .01 |x^2 - 1| < .01 |x + 1||x - 1| < .01 Objective: get the above in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less than 1. Which is reasonable because it should represent an extremely small distance. Why is that? Why can't it represent an extremely large distance? If delta < 1, then |x + 1| < 1 which means that -1 < x + 1 < 1. Subtract 2 from each of those expressions to get -3 < x - 1 < -1. Therefore, |x - 1| < 3 This part, I just don't follow at all. Why subtract 2? What drives us to do this? If we start out with -1 < x + 1 < 1, to say that we have to subtract 2 seems arbitrary. Your goal is to produce the expression |x - a| in the middle of the compound expression. Recall that |x - 1| < 3 and substitute 3 into the inequality. |x - 1||x + 1| < 0.01 3|x + 1| < 0.01 |x + 1| < 0.01/3 Therefore, delta = .01/3 I don't follow why we use |x - 1| < 3 or where it comes from. If we had: -1 < x + 1 < 1, algebra does not say to pull a -2 from thin air and apply it to the expression or to magically produce |x - 1| < 3. I see a whole lot of fluff here and too little reason being applied. I would like to reduce the fluff and increase the reasoning. For example, sure I can recognize that |x - 1| < 3 because originally we had |x - 1||x + 1| < .01 but by that same token, I could just as well say |x - 1| < 4, |x - 1| < 100, |x - 1| < 10^100 etc, what makes 3 so special? Also, he continues from stating that |x - 1| < 3 to substituting 3 for |x - 1|, again, I'm left wondering why? Any clarification appreciated. -- conrad === Subject: Re: evaluating limits > Problem: Identify a value of delta that > corresponds to epsilon = 0.01 given > lim_x->-1 (x^2 + 3) = 4 according to > the definition of limits. |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 > |x + 1||x - 1| < .01 Objective: get the above > in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less > than 1. Which is reasonable > because it should represent > an extremely small distance. Why is that? Why can't it represent > an extremely large distance? You do not need to make delta large, you only need to make it work. And if any positive delta works then all smaller positive deltas will also work. So if you can find a delta > 1 which works so will delta = 1 and so will 0 < delta < 1 === Subject: Re: evaluating limits > Problem: Identify a value of delta that > corresponds to epsilon = 0.01 given > lim_x->-1 (x^2 + 3) = 4 according to > the definition of limits. |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 Another way, that avoids the book's remarks completely, is to note that you want x^2 - 1 to lie between -0.01 and +0.01, so 0.99 <= x^2 <= 1.01, hence sqrt(1-.01) <= x <= sqrt(1+.01). You should be able to show that 1 - 0.01/3 > sqrt(1-.01) and 1 + .01/3 < sqrt(1+.01), so the interval |x-1| < 0.01/3 will suffice. R.G. Vickson > |x + 1||x - 1| < .01 Objective: get the above > in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less > than 1. Which is reasonable > because it should represent > an extremely small distance. Why is that? Why can't it represent > an extremely large distance? If delta < 1, then |x + 1| < 1 > which means that -1 < x + 1 < 1. Subtract 2 from each of those > expressions to get > -3 < x - 1 < -1. Therefore, > |x - 1| < 3 This part, I just don't follow at all. > Why subtract 2? What drives us > to do this? If we start out with > -1 < x + 1 < 1, to say that we have > to subtract 2 seems arbitrary. Your goal is to produce the > expression |x - a| > in the middle of the compound > expression. Recall that |x - 1| < 3 > and substitute 3 into the inequality. |x - 1||x + 1| < 0.01 > 3|x + 1| < 0.01 > |x + 1| < 0.01/3 Therefore, delta = .01/3 I don't follow why we use > |x - 1| < 3 or where it comes from. > If we had: > -1 < x + 1 < 1, algebra does not > say to pull a -2 from thin air and apply > it to the expression or to magically > produce |x - 1| < 3. I see a whole lot > of fluff here and too little reason being > applied. I would like to reduce the fluff > and increase the reasoning. For example, > sure I can recognize that |x - 1| < 3 because > originally we had |x - 1||x + 1| < .01 > but by that same token, I could just as well > say |x - 1| < 4, |x - 1| < 100, |x - 1| < 10^100 > etc, what makes 3 so special? Also, he continues from stating that > |x - 1| < 3 to substituting > 3 for |x - 1|, again, I'm left wondering > why? Any clarification appreciated. -- > conrad === Subject: Re: evaluating limits > Problem: Identify a value of delta that > corresponds to epsilon = 0.01 given > lim_x->-1 (x^2 + 3) = 4 according to > the definition of limits. |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 > |x + 1||x - 1| < .01 Objective: get the above > in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less > than 1. Which is reasonable > because it should represent > an extremely small distance. Why is that? Why can't it represent > an extremely large distance? If delta < 1, then |x + 1| < 1 > which means that -1 < x + 1 < 1. > Subtract 2 from each of those > expressions to get > -3 < x - 1 < -1. Therefore, > |x - 1| < 3 This part, I just don't follow at all. > Why subtract 2? What drives us > to do this? If we start out with > -1 < x + 1 < 1, to say that we have > to subtract 2 seems arbitrary. Your goal is to produce the > expression |x - a| > in the middle of the compound > expression. Recall that |x - 1| < 3 > and substitute 3 into the inequality. |x - 1||x + 1| < 0.01 > 3|x + 1| < 0.01 > |x + 1| < 0.01/3 Therefore, delta = .01/3 I don't follow why we use > |x - 1| < 3 or where it comes from. > If we had: > -1 < x + 1 < 1, algebra does not > say to pull a -2 from thin air and apply > it to the expression or to magically > produce |x - 1| < 3. I see a whole lot > of fluff here and too little reason being > applied. I would like to reduce the fluff > and increase the reasoning. For example, > sure I can recognize that |x - 1| < 3 because > originally we had |x - 1||x + 1| < .01 > but by that same token, I could just as well > say |x - 1| < 4, |x - 1| < 100, |x - 1| < 10^100 > etc, what makes 3 so special? Also, he continues from stating that > |x - 1| < 3 to substituting > 3 for |x - 1|, again, I'm left wondering > why? > Any clarification appreciated. > -- > conrad You are fully justified in questioning the author's comments: I would question them, too (after more than 40 years of teaching calculus). The first observation is: there is no preferred value of delta in the definition: if my neighbor finds a correct delta, and I choose a smaller positive delta (called delta1), mine would be also correct. Why? If |x - a| < delta1 then also |x - a| < delta, and the rest od the definition checking goes well. Next observation: All the manipulations will become clearer if we recast the study to a study of small numbers, so let us rename x - a = h, so we can restore x = a + h. Your a = -1, and the inequality to solve was |x^2 - 1| < 0.01 Change the variable: x + 1 = h, so x = h - 1 |x^2 - 1| = |h^2 - 2*h + 1 - 1| = |h^2 - 2*h| and we need to solve |h^2 - 2*h| < 0.01 in the sense that |h| < delta would do the job for a delta to be found. The inequality is quadratic, and we can (1) either follow through, solving the quadratic inequality, or (2) impose initial restrictions on delta (remember, smaller delta won't hurt) to make the rest linear. The book chose path (2). (So would I.) Their initial restiction was delta < 1. So: Suppose |h| < delta < 1, then back to the quadratic inequality, |h^2 - 2*h| = |h| * |h-2| <= |h| * 3 (because the value of |h-2| is less than |-1-2| = 3) Then: if we solve 3*|h| < 0.01, we will also solve |h^2 - 2*h| < 0.01. That makes |h| < (0.01)/3 as a correct (although not the largest correct) answer: delta can be (0.01)/3. Summary - return to h = x+1: If |x - (-1)| = |x+1| < (0.01)/3 then |f(x) - 4| < 0.01 . Done: an answer is delta = (0.01)/3. ---------------------------------------------- Now for the tough ones who want to solve the quadratic inequality: try to verify that the largest possible delta in this setup is Delta = 0.01 / (1 + (1.01)^(1/2)). In decimals (four valid digits), Delta = 0.04987... delta = 0.03333... (smaller delta doesn't hurt... so Delta may not have been worth the extra effort). Hope it helps, ZVK(Slavek). === Subject: Re: evaluating limits corresponds to epsilon = 0.01 given > lim_x->-1 (x^2 + 3) = 4 according to > the definition of limits. |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 > |x + 1||x - 1| < .01 Objective: get the above > in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less > than 1. Which is reasonable > because it should represent > an extremely small distance. Why is that? Why can't it represent > an extremely large distance? If delta < 1, then |x + 1| < 1 > which means that -1 < x + 1 < 1. Subtract 2 from each of those > expressions to get > -3 < x - 1 < -1. Therefore, > |x - 1| < 3 This part, I just don't follow at all. > Why subtract 2? What drives us > to do this? If we start out with > -1 < x + 1 < 1, to say that we have > to subtract 2 seems arbitrary. Your goal is to produce the > expression |x - a| > in the middle of the compound > expression. Recall that |x - 1| < 3 > and substitute 3 into the inequality. |x - 1||x + 1| < 0.01 > 3|x + 1| < 0.01 > |x + 1| < 0.01/3 Therefore, delta = .01/3 I don't follow why we use > |x - 1| < 3 or where it comes from. > If we had: > -1 < x + 1 < 1, algebra does not > say to pull a -2 from thin air and apply > it to the expression or to magically > produce |x - 1| < 3. I see a whole lot > of fluff here and too little reason being > applied. I would like to reduce the fluff > and increase the reasoning. For example, > sure I can recognize that |x - 1| < 3 because > originally we had |x - 1||x + 1| < .01 > but by that same token, I could just as well > say |x - 1| < 4, |x - 1| < 100, |x - 1| < 10^100 > etc, what makes 3 so special? Also, he continues from stating that > |x - 1| < 3 to substituting > 3 for |x - 1|, again, I'm left wondering > why? Any clarification appreciated. -- > conrad You are fully justified in questioning the author's comments: I would > question them, too (after more than 40 years of teaching calculus). The first observation is: there is no preferred value of delta in the > definition: if my neighbor finds a correct delta, and I choose a smaller > positive delta (called delta1), mine would be also correct. Why? If |x - a| < delta1 then also |x - a| < delta, > and the rest od the definition checking goes well. Next observation: All the manipulations will become clearer if we recast > the study to a study of small numbers, so let us rename x - a = h, so we can restore x = a + h. Your a = -1, and the inequality to solve was |x^2 - 1| < 0.01 Change the variable: x + 1 = h, so x = h - 1 |x^2 - 1| = |h^2 - 2*h + 1 - 1| = |h^2 - 2*h| and we need to solve |h^2 - 2*h| < 0.01 in the sense that |h| < delta would do the job for a delta to be found. The inequality is quadratic, and we can (1) either follow through, solving the quadratic inequality, or (2) impose initial restrictions on delta (remember, smaller delta won't > hurt) to make the rest linear. The book chose path (2). (So would I.) Their initial restiction was > delta < 1. So: Suppose |h| < delta < 1, then back to the quadratic inequality, |h^2 - 2*h| = |h| * |h-2| <= |h| * 3 (because the value of |h-2| is less than |-1-2| = 3) Then: if we solve 3*|h| < 0.01, we will also solve |h^2 - 2*h| < 0.01. That makes |h| < (0.01)/3 as a correct (although not the largest correct) > answer: delta can be (0.01)/3. Summary - return to h = x+1: If |x - (-1)| = |x+1| < (0.01)/3 then |f(x) - 4| < 0.01 . Done: an answer is delta = (0.01)/3. ---------------------------------------------- Now for the tough ones who want to solve the quadratic inequality: try to > verify that the largest possible delta in this setup is Delta = 0.01 / (1 + (1.01)^(1/2)). In decimals (four valid digits), Delta = 0.04987... > delta = 0.03333... (smaller delta doesn't hurt... so Delta may not have been worth the > extra effort). > -- conrad === Subject: Re: evaluating limits > Problem: Identify a value of delta that corresponds to epsilon = 0.01 given > lim_x->-1 (x^2 + 3) = 4 according to the definition of limits. |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 > |x + 1||x - 1| < .01 Objective: get the above in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less than 1. Which is reasonable > because it should represent an extremely small distance. Why is that? Why can't it represent an extremely large distance? It can. But it _should_ represent an extremely small distance because that's the idea: you want to prove that if _x_ is very close to -1 (that is, if the distance from _x_ to -1 is very small), then x^2 + 3 is very close to 4 > If delta < 1, then |x + 1| < 1 which means that -1 < x + 1 < 1. > Subtract 2 from each of those expressions to get > -3 < x - 1 < -1. Therefore, |x - 1| < 3 This part, I just don't follow at all. Why subtract 2? What drives us > to do this? If we start out with -1 < x + 1 < 1, to say that we have > to subtract 2 seems arbitrary. Because saying that x^2 + 3 is very close to 4 is saying that x^2 - 1 is very close to 0. But x^2 - 1 = (x - 1)(x + 1). You already know that x + 1 is small; using the previous product, you will be able to deduce that x^2 - 1 cannot be large. > Your goal is to produce the expression |x - a| > in the middle of the compound expression. Recall that |x - 1| < 3 > and substitute 3 into the inequality. |x - 1||x + 1| < 0.01 > 3|x + 1| < 0.01 > |x + 1| < 0.01/3 Therefore, delta = .01/3 I don't follow why we use |x - 1| < 3 or where it comes from. You explained where it comes from a few lines ago. And since |x + 1| < 0.01 and since |x - 1| < 3, it follows that |x^2 - 1| < 0.03. Jose Carlos Santos === Subject: Re: evaluating limits <5oup3lFn9ocaU1@mid.individual.net > Problem: Identify a value of delta that corresponds to epsilon = 0.01 given > lim x->-1 (x^2 + 3) = 4 according to the definition of limits. |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 > |x + 1||x - 1| < .01 Objective: get the above in the form: |x - a| < delta Below is quoted from my book: Begin by assuming delta is less than 1. Which is reasonable > because it should represent an extremely small distance. Why is that? Why can't it represent an extremely large distance? It can. But it should represent an extremely small distance because > that's the idea: you want to prove that if x is very close to -1 > (that is, if the distance from x to -1 is very small), then x^2 + 3 is > very close to 4 If delta < 1, then |x + 1| < 1 which means that -1 < x + 1 < 1. Subtract 2 from each of those expressions to get > -3 < x - 1 < -1. Therefore, |x - 1| < 3 This part, I just don't follow at all. Why subtract 2? What drives us > to do this? If we start out with -1 < x + 1 < 1, to say that we have > to subtract 2 seems arbitrary. Because saying that x^2 + 3 is very close to 4 is saying that x^2 - 1 is > very close to 0. But x^2 - 1 = (x - 1)(x + 1). You already know that > x + 1 is small; using the previous product, you will be able to deduce > that x^2 - 1 cannot be large. So why subtract 2? Why not 3, 4, 5, or -10? etc. Your goal is to produce the expression |x - a| > in the middle of the compound expression. Recall that |x - 1| < 3 > and substitute 3 into the inequality. |x - 1||x + 1| < 0.01 > 3|x + 1| < 0.01 > |x + 1| < 0.01/3 Therefore, delta = .01/3 I don't follow why we use |x - 1| < 3 or where it comes from. You explained where it comes from a few lines ago. And since > |x + 1| < 0.01 and since |x - 1| < 3, it follows that |x^2 - 1| < 0.03. My book stated -3 < x - 1 < -1 is what you get after subtracting 2. They then say, Therefore, |x - 1| < 3 Which is obviously true if x - 1 < -1, however, why state that |x - 1| is less than 3? Why not less than 100 or 1000, or 10^100? -- conrad === Subject: Re: evaluating limits > Problem: Identify a value of delta that corresponds to epsilon = 0.01 given > lim_x->-1 (x^2 + 3) = 4 according to the definition of limits. > |f(x) - L| < Epsilon > |(x^2 + 3) - 4| < .01 > |x^2 - 1| < .01 > |x + 1||x - 1| < .01 > Objective: get the above in the form: |x - a| < delta > Below is quoted from my book: > Begin by assuming delta is less than 1. Which is reasonable > because it should represent an extremely small distance. > Why is that? Why can't it represent an extremely large distance? > It can. But it _should_ represent an extremely small distance because > that's the idea: you want to prove that if _x_ is very close to -1 > (that is, if the distance from _x_ to -1 is very small), then x^2 + 3 is > very close to 4 > If delta < 1, then |x + 1| < 1 which means that -1 < x + 1 < 1. > Subtract 2 from each of those expressions to get > -3 < x - 1 < -1. Therefore, |x - 1| < 3 > This part, I just don't follow at all. Why subtract 2? What drives us > to do this? If we start out with -1 < x + 1 < 1, to say that we have > to subtract 2 seems arbitrary. > Because saying that x^2 + 3 is very close to 4 is saying that x^2 - 1 is > very close to 0. But x^2 - 1 = (x - 1)(x + 1). You already know that > x + 1 is small; using the previous product, you will be able to deduce > that x^2 - 1 cannot be large. So why subtract 2? Why not 3, 4, 5, or -10? etc. If you subtract 2, you get something about x - 1. That's good, because you're interested in making x^2 - 1 small, and x^2 - 1 = (x + 1)(x - 1). If you want to subtract 5, for instance, you'll get something about x - 4. What will you do with that? > Your goal is to produce the expression |x - a| > in the middle of the compound expression. Recall that |x - 1| < 3 > and substitute 3 into the inequality. > |x - 1||x + 1| < 0.01 > 3|x + 1| < 0.01 > |x + 1| < 0.01/3 > Therefore, delta = .01/3 > I don't follow why we use |x - 1| < 3 or where it comes from. > You explained where it comes from a few lines ago. And since > |x + 1| < 0.01 and since |x - 1| < 3, it follows that |x^2 - 1| < 0.03. My book stated -3 < x - 1 < -1 is what you get > after subtracting 2. They then say, Therefore, |x - 1| < 3 Which is obviously true if x - 1 < -1, Not at all enough. If x = -5, it is still true that x - 1 < -1, but it will *not* be true that |x - 1| < 3. > however, why state > that |x - 1| is less than 3? Why not less than 100 or 1000, or > 10^100? Because |x - 1| < 3 is the best result, under the hypothesis that -3 < x - 1 < 1. But, yes, it also true that |x - 1| < 100, and you can also solve your problem with this inequality. You will get delta = .01/100 then. Jose Carlos Santos === Subject: Re: Equation a*x^2 + b*y^2 = c*z^2 for x;y;z of gcd=1 > <10871419.1193404235300.JavaMail.jakarta@nitrogen.math > forum.org>, <33520934.1193343912165.JavaMail.jakarta@nitrogen.math > forum.org>, > Roman B. Binder Already we now several cases: > 1) For a=b=c=k there is classical Pythagorean > equation solved for let: > x = m^2 -n^2 ; > y = 2mn ; > z = m^2 +n^2 ; > 2) For a=b=1; c=2 > 3) For let a=k once b and c = 1 > etc. > But does it was indicated such equation > for a;b;c of gcd = 1 ? > Some other question, if there could be > possible for fixed x;y;z values for to find > two different solutions ? > for a1;b1;c1 of gcd=1 > and for a2;b2;c2 of gcd=1 > and where at least a1;a2 of gcd = 1 etc. > ( not some proportional case ) For any largish c there will be lots of a and b > such that a 2^2 + b 3^2 = c 5^2. -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) > O.K. > I can imagine as You've find some first set > of a;b;c ... > My second part of the question was if the next > possible set be only proportional or not: > a1 = ka; b1 = kb; c1 = kc and b? > Did you try finding any of them to see whether they > were > proportional? -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) > You are very right: Solutions for such a.2^2 +b.3^2 = c.5^2 are floating not only in proportional area of certain a;b;c . There will be not so easy for to generalize such behaviour... Samples: 16*4 + 4*9 = 4*25 or 4*4 + 1*9 = 1*25 but 55*4 + 20*9 = 16*25 etc. the last simplifies: 11*4 + 4*9 = 16*5 11 + 9 = 20 I am not certain if with the help of the condition: a;b;c;x;y;z of gcd=1 we'll be able for to find only one primitive solution (except of proportional) or may by no one solution at all ? Ro-Bin === Subject: Re: Equation a*x^2 + b*y^2 = c*z^2 for x;y;z of gcd=1 <7439738.1193944993242.JavaMail.jakarta@nitrogen.mathforum.org > Solutions for such a.2^2 +b.3^2 = c.5^2 > are floating not only in proportional > area of certain a;b;c . > There will be not so easy for to > generalize such behaviour... > Samples: > 16*4 + 4*9 = 4*25 or 4*4 + 1*9 = 1*25 > but 55*4 + 20*9 = 16*25 etc. > the last simplifies: > 11*4 + 4*9 = 16*5 > 11 + 9 = 20 Are you actually wondering if sets (a, b, c) satisfying something like a.2^2 +b.3^2 + c.5^2 = 0 (using the negative of your c) are in short supply or difficult to find, even once you find one? Just think of (2^2, 3^2, 5^2) as a vector, and forget the fact that the components happen to be perfect squares. Then a set (a, b, c) can be obtained from _any_ set of integers (p, q, r) as the vector product (2^2, 3^2, 5^2) ^ (p, q, r), explicitly: a, b, c = 9.r - 25.q, 25.p - 4.r, 4.q - 9.p John R Ramsden === Subject: Re: ACA, ACA_0, and their ordinals [Lots of helpful stuff] a.s.a.p. :-) === Subject: Re: JSH: Explaining DMESE, again > IF you cannot understand this story then you will make very dumb > replies that show you have the intellect of a cow. Please try to sound at least quasi-intelligent in reply. James Harris Mr. Harris, I am from California - so I am a happy cow. Sqrt(Cow^2) = +/- Cow So, what is - Cow? Bah ... oops ... Moooooo === Subject: Re: JSH: Why self-delusion? <873avtz9mr.fsf@phiwumbda.org> standard that I've created, would ask for the bought DVD after 30 days > to be re-introduced to the drive, so your friend will need the bought > one again in 30 days if you really, really, really want to help people > make copies illegally. What stops the DVD player from demanding the original every time I attempt to play a backup, after the 30 days? How does it know I presented the original, say, three days ago? Does it keep an internal database of DVDs it has copied? Or does this scheme only work on re-writable DVDs? The whole idea of a backup is in case the original is lost or destroyed. What use is a backup if I need the original anyway? What happens when I buy a new DVD player? Are all my backups useless? === Subject: Re: Help Double Numbersusa membership by calling cspan and mentioning Numbersusa etc > Help Us Double Our Membership: Call C-SPAN > NUSA : October 31 , 2007 -- by NUSA Help us double our membership before the next big battle in Congress > begins. Please call C-SPAN (c-span.org) to tell others about NumbersUSA.com, > and invite the audience to join us to experience the satisfaction of > sending a few faxes (and report cards!) to their Representatives and > Senators. We need your help to get the word out about NumbersUSA.com to hundreds of thousands of these people who are > interested in politics, aware of the need for action, but unaware of > NumbersUSA.com. The hosts WELCOME callers to read out their favorite > WEBSITE ADDRESSES, to make political statements, and to question > political guests who are on the show. Note: Call anytime, but calling during the Open Phones segments is > the best, because any and all subjects are welcome. Open Phones are > generally 7:00 a.m. to about 7:45 a.m. and 9:30 a.m. to 10:00 a.m. > Eastern time. Listeners/viewers can ask Washington Journal guests a question, or > make a comment during Open Phone segments at: Support Republicans: (202) 737-0001 Support Democrats: (202) 737-0002 Support Independents: (202) 628-0205 Outside U.S.: (202) 628-0184 What is C-SPAN? Callers reach a nationwide audience of all political views and > parties. C-SPAN has three television channels available on cable and > satellite television, and is broadcast on XM (channel 132) and Sirius > (Channel 139) satellite radio, as well as on several local radio > stations, and on the internet at:http://c-span.org/ NumbersUSA Comment: > C-SPAN is a unique, non-partisan, non-profit, commercial-free > television and radio network, devoted to covering Congress and > political and cultural issues. C-SPAN features a morning call-in show > called 'Washington Journal which is broadcast seven days a week, > including holidays. It begins at 7:00 a.m. and runs until 10:00 a.m. > Eastern Time (4:00 a.m. to 7:00 a.m. Pacific Time). When Congress is > in session, it often cuts away to the House or Senate floor action > earlier than 10:00 a.m. Listeners call C-SPAN from across America virtually every morning to > complain about the effects of illegal immigration. Most are bewildered > by what is happening to their country due to illegal immigration > numbers out of control. These folks are desperately searching for a > way to effectively express their concerns to Congress. A call to the C-SPAN audience THAT MENTIONS THE NUMBERSUSA.COM WEBSITE > and how easy the website makes it to contact Congress, can get the > word out to hundreds of thousands of Americans who don't get > NumbersUSA.com Action Alerts. Often, the caller can tell the audience > that Congress is about to vote on an immigration related bill, and > invite listeners to visit NumbersUSA.com to find out what they can do > to efficiently and effectively let Congress know what they think. We > can get help from thousands of non-members and recruit many new > members with just one phone call to this bi-partisan audience. NumbersUSA Comment: > Like most Americans, nearly all of these callers have never heard of > NumbersUSA.com, and they have no idea that NumbersUSA.com is available > to them, free of charge, to help them economically leverage their > efforts to make their views known to Congress. Thousands would join > NumbersUSA.com if they could hear of our existence and what we can > do. We appreciate you making this effort! Please tell us about your call! Submit my feedback! === Subject: Re: Help Double Numbersusa membership by calling cspan and mentioning Numbersusa etc > Help Us Double Our Membership: Call C-SPAN > NUSA : October 31 , 2007 -- by NUSA Help us double our membership before the next big battle in Congress > begins. Please call C-SPAN (c-span.org) to tell others about NumbersUSA.com, > and invite the audience to join us to experience the satisfaction of > sending a few faxes (and report cards!) to their Representatives and > Senators. We need your help to get the word out about NumbersUSA.com to hundreds of thousands of these people who are > interested in politics, aware of the need for action, but unaware of > NumbersUSA.com. The hosts WELCOME callers to read out their favorite > WEBSITE ADDRESSES, to make political statements, and to question > political guests who are on the show. Note: Call anytime, but calling during the Open Phones segments is > the best, because any and all subjects are welcome. Open Phones are > generally 7:00 a.m. to about 7:45 a.m. and 9:30 a.m. to 10:00 a.m. > Eastern time. Listeners/viewers can ask Washington Journal guests a question, or > make a comment during Open Phone segments at: Support Republicans: (202) 737-0001 Support Democrats: (202) 737-0002 Support Independents: (202) 628-0205 Outside U.S.: (202) 628-0184 What is C-SPAN? Callers reach a nationwide audience of all political views and > parties. C-SPAN has three television channels available on cable and > satellite television, and is broadcast on XM (channel 132) and Sirius > (Channel 139) satellite radio, as well as on several local radio > stations, and on the internet at:http://c-span.org/ NumbersUSA Comment: > C-SPAN is a unique, non-partisan, non-profit, commercial-free > television and radio network, devoted to covering Congress and > political and cultural issues. C-SPAN features a morning call-in show > called 'Washington Journal which is broadcast seven days a week, > including holidays. It begins at 7:00 a.m. and runs until 10:00 a.m. > Eastern Time (4:00 a.m. to 7:00 a.m. Pacific Time). When Congress is > in session, it often cuts away to the House or Senate floor action > earlier than 10:00 a.m. Listeners call C-SPAN from across America virtually every morning to > complain about the effects of illegal immigration. Most are bewildered > by what is happening to their country due to illegal immigration > numbers out of control. These folks are desperately searching for a > way to effectively express their concerns to Congress. A call to the C-SPAN audience THAT MENTIONS THE NUMBERSUSA.COM WEBSITE > and how easy the website makes it to contact Congress, can get the > word out to hundreds of thousands of Americans who don't get > NumbersUSA.com Action Alerts. Often, the caller can tell the audience > that Congress is about to vote on an immigration related bill, and > invite listeners to visit NumbersUSA.com to find out what they can do > to efficiently and effectively let Congress know what they think. We > can get help from thousands of non-members and recruit many new > members with just one phone call to this bi-partisan audience. NumbersUSA Comment: > Like most Americans, nearly all of these callers have never heard of > NumbersUSA.com, and they have no idea that NumbersUSA.com is available > to them, free of charge, to help them economically leverage their > efforts to make their views known to Congress. Thousands would join > NumbersUSA.com if they could hear of our existence and what we can > do. We appreciate you making this effort! Please tell us about your call! Submit my feedback! One person can't possibly affect the world anymore in this day and age. === Subject: Re: Help Double Numbersusa membership by calling cspan and mentioning Numbersusa etc > Help Us Double Our Membership: Call C-SPAN > NUSA : October 31 , 2007 -- by NUSA Help us double our membership before the next big battle in Congress > begins. Please call C-SPAN (c-span.org) to tell others about NumbersUSA.com, > and invite the audience to join us to experience the satisfaction of > sending a few faxes (and report cards!) to their Representatives and > Senators. We need your help to get the word out about NumbersUSA.com to hundreds of thousands of these people who are > interested in politics, aware of the need for action, but unaware of > NumbersUSA.com. The hosts WELCOME callers to read out their favorite > WEBSITE ADDRESSES, to make political statements, and to question > political guests who are on the show. Note: Call anytime, but calling during the Open Phones segments is > the best, because any and all subjects are welcome. Open Phones are > generally 7:00 a.m. to about 7:45 a.m. and 9:30 a.m. to 10:00 a.m. > Eastern time. Listeners/viewers can ask Washington Journal guests a question, or > make a comment during Open Phone segments at: Support Republicans: (202) 737-0001 Support Democrats: (202) 737-0002 Support Independents: (202) 628-0205 Outside U.S.: (202) 628-0184 What is C-SPAN? Callers reach a nationwide audience of all political views and > parties. C-SPAN has three television channels available on cable and > satellite television, and is broadcast on XM (channel 132) and Sirius > (Channel 139) satellite radio, as well as on several local radio > stations, and on the internet at:http://c-span.org/ NumbersUSA Comment: > C-SPAN is a unique, non-partisan, non-profit, commercial-free > television and radio network, devoted to covering Congress and > political and cultural issues. C-SPAN features a morning call-in show > called 'Washington Journal which is broadcast seven days a week, > including holidays. It begins at 7:00 a.m. and runs until 10:00 a.m. > Eastern Time (4:00 a.m. to 7:00 a.m. Pacific Time). When Congress is > in session, it often cuts away to the House or Senate floor action > earlier than 10:00 a.m. Listeners call C-SPAN from across America virtually every morning to > complain about the effects of illegal immigration. Most are bewildered > by what is happening to their country due to illegal immigration > numbers out of control. These folks are desperately searching for a > way to effectively express their concerns to Congress. A call to the C-SPAN audience THAT MENTIONS THE NUMBERSUSA.COM WEBSITE > and how easy the website makes it to contact Congress, can get the > word out to hundreds of thousands of Americans who don't get > NumbersUSA.com Action Alerts. Often, the caller can tell the audience > that Congress is about to vote on an immigration related bill, and > invite listeners to visit NumbersUSA.com to find out what they can do > to efficiently and effectively let Congress know what they think. We > can get help from thousands of non-members and recruit many new > members with just one phone call to this bi-partisan audience. NumbersUSA Comment: > Like most Americans, nearly all of these callers have never heard of > NumbersUSA.com, and they have no idea that NumbersUSA.com is available > to them, free of charge, to help them economically leverage their > efforts to make their views known to Congress. Thousands would join > NumbersUSA.com if they could hear of our existence and what we can > do. We appreciate you making this effort! Please tell us about your call! Submit my feedback! One person can't possibly affect the world anymore in this day and age. === Subject: Re: Help Double Numbersusa membership by calling cspan and mentioning Numbersusa etc > Help Us Double Our Membership: Call C-SPAN > NUSA : October 31 , 2007 -- by NUSA > Help us double our membership before the next big battle in Congress > begins. > Please call C-SPAN (c-span.org) to tell others about NumbersUSA.com, > and invite the audience to join us to experience the satisfaction of > sending a few faxes (and report cards!) to their Representatives and > Senators. > We need your help to get the word out > about NumbersUSA.com to hundreds of thousands of these people who are > interested in politics, aware of the need for action, but unaware of > NumbersUSA.com. The hosts WELCOME callers to read out their favorite > WEBSITE ADDRESSES, to make political statements, and to question > political guests who are on the show. > Note: Call anytime, but calling during the Open Phones segments is > the best, because any and all subjects are welcome. Open Phones are > generally 7:00 a.m. to about 7:45 a.m. and 9:30 a.m. to 10:00 a.m. > Eastern time. > Listeners/viewers can ask Washington Journal guests a question, or > make a comment during Open Phone segments at: > Support Republicans: (202) 737-0001 > Support Democrats: (202) 737-0002 > Support Independents: (202) 628-0205 > Outside U.S.: (202) 628-0184 What is C-SPAN? > Callers reach a nationwide audience of all political views and > parties. C-SPAN has three television channels available on cable and > satellite television, and is broadcast on XM (channel 132) and Sirius > (Channel 139) satellite radio, as well as on several local radio > stations, and on the internet at:http://c-span.org/ > NumbersUSA Comment: > C-SPAN is a unique, non-partisan, non-profit, commercial-free > television and radio network, devoted to covering Congress and > political and cultural issues. C-SPAN features a morning call-in show > called 'Washington Journal which is broadcast seven days a week, > including holidays. It begins at 7:00 a.m. and runs until 10:00 a.m. > Eastern Time (4:00 a.m. to 7:00 a.m. Pacific Time). When Congress is > in session, it often cuts away to the House or Senate floor action > earlier than 10:00 a.m. > Listeners call C-SPAN from across America virtually every morning to > complain about the effects of illegal immigration. Most are bewildered > by what is happening to their country due to illegal immigration > numbers out of control. These folks are desperately searching for a > way to effectively express their concerns to Congress. > A call to the C-SPAN audience THAT MENTIONS THE NUMBERSUSA.COM WEBSITE > and how easy the website makes it to contact Congress, can get the > word out to hundreds of thousands of Americans who don't get > NumbersUSA.com Action Alerts. Often, the caller can tell the audience > that Congress is about to vote on an immigration related bill, and > invite listeners to visit NumbersUSA.com to find out what they can do > to efficiently and effectively let Congress know what they think. We > can get help from thousands of non-members and recruit many new > members with just one phone call to this bi-partisan audience. > NumbersUSA Comment: > Like most Americans, nearly all of these callers have never heard of > NumbersUSA.com, and they have no idea that NumbersUSA.com is available > to them, free of charge, to help them economically leverage their > efforts to make their views known to Congress. Thousands would join > NumbersUSA.com if they could hear of our existence and what we can > do. > We appreciate you making this effort! > Please tell us about your call! > Submit my feedback! >One person can't possibly affect the world anymore in this day and age. After all, Lee Harvey Oswald only had one vote. One vote can't count... === Subject: Re: Help Double Numbersusa membership by calling cspan and mentioning Numbersusa etc > Help Us Double Our Membership: Call C-SPAN > NUSA : October 31 , 2007 -- by NUSA > Help us double our membership before the next big battle in Congress > begins. > Please call C-SPAN (c-span.org) to tell others about NumbersUSA.com, > and invite the audience to join us to experience the satisfaction of > sending a few faxes (and report cards!) to their Representatives and > Senators. > We need your help to get the word out > about NumbersUSA.com to hundreds of thousands of these people who are > interested in politics, aware of the need for action, but unaware of > NumbersUSA.com. The hosts WELCOME callers to read out their favorite > WEBSITE ADDRESSES, to make political statements, and to question > political guests who are on the show. > Note: Call anytime, but calling during the Open Phones segments is > the best, because any and all subjects are welcome. Open Phones are > generally 7:00 a.m. to about 7:45 a.m. and 9:30 a.m. to 10:00 a.m. > Eastern time. > Listeners/viewers can ask Washington Journal guests a question, or > make a comment during Open Phone segments at: > Support Republicans: (202) 737-0001 > Support Democrats: (202) 737-0002 > Support Independents: (202) 628-0205 > Outside U.S.: (202) 628-0184 What is C-SPAN? > Callers reach a nationwide audience of all political views and > parties. C-SPAN has three television channels available on cable and > satellite television, and is broadcast on XM (channel 132) and Sirius > (Channel 139) satellite radio, as well as on several local radio > stations, and on the internet at:http://c-span.org/ > NumbersUSA Comment: > C-SPAN is a unique, non-partisan, non-profit, commercial-free > television and radio network, devoted to covering Congress and > political and cultural issues. C-SPAN features a morning call-in show > called 'Washington Journal which is broadcast seven days a week, > including holidays. It begins at 7:00 a.m. and runs until 10:00 a.m. > Eastern Time (4:00 a.m. to 7:00 a.m. Pacific Time). When Congress is > in session, it often cuts away to the House or Senate floor action > earlier than 10:00 a.m. > Listeners call C-SPAN from across America virtually every morning to > complain about the effects of illegal immigration. Most are bewildered > by what is happening to their country due to illegal immigration > numbers out of control. These folks are desperately searching for a > way to effectively express their concerns to Congress. > A call to the C-SPAN audience THAT MENTIONS THE NUMBERSUSA.COM WEBSITE > and how easy the website makes it to contact Congress, can get the > word out to hundreds of thousands of Americans who don't get > NumbersUSA.com Action Alerts. Often, the caller can tell the audience > that Congress is about to vote on an immigration related bill, and > invite listeners to visit NumbersUSA.com to find out what they can do > to efficiently and effectively let Congress know what they think. We > can get help from thousands of non-members and recruit many new > members with just one phone call to this bi-partisan audience. > NumbersUSA Comment: > Like most Americans, nearly all of these callers have never heard of > NumbersUSA.com, and they have no idea that NumbersUSA.com is available > to them, free of charge, to help them economically leverage their > efforts to make their views known to Congress. Thousands would join > NumbersUSA.com if they could hear of our existence and what we can > do. > We appreciate you making this effort! > Please tell us about your call! > Submit my feedback! One person can't possibly affect the world anymore in this day and age.- Hide quoted text - - Show quoted text - Tell that to Bill Gates === Subject: Re: Help Double Numbersusa membership by calling cspan and mentioning Numbersusa etc > Help Us Double Our Membership: Call C-SPAN NUSA : October 31 , 2007 > -- by NUSA > Help us double our membership before the next big battle in > Congress begins. > Please call C-SPAN (c-span.org) to tell others about > NumbersUSA.com, and invite the audience to join us to experience > the satisfaction of sending a few faxes (and report cards!) to > their Representatives and Senators. > We need your help to get the word out > about NumbersUSA.com to hundreds of thousands of these people who > are interested in politics, aware of the need for action, but > unaware of NumbersUSA.com. The hosts WELCOME callers to read out > their favorite WEBSITE ADDRESSES, to make political statements, and > to question political guests who are on the show. > Note: Call anytime, but calling during the Open Phones segments > is the best, because any and all subjects are welcome. Open > Phones are generally 7:00 a.m. to about 7:45 a.m. and 9:30 a.m. to > 10:00 a.m. Eastern time. > Listeners/viewers can ask Washington Journal guests a question, or > make a comment during Open Phone segments at: > Support Republicans: (202) 737-0001 > Support Democrats: (202) 737-0002 > Support Independents: (202) 628-0205 > Outside U.S.: (202) 628-0184 What is C-SPAN? > Callers reach a nationwide audience of all political views and > parties. C-SPAN has three television channels available on cable > and satellite television, and is broadcast on XM (channel 132) and > Sirius (Channel 139) satellite radio, as well as on several local > radio stations, and on the internet at:http://c-span.org/ > NumbersUSA Comment: > C-SPAN is a unique, non-partisan, non-profit, commercial-free > television and radio network, devoted to covering Congress and > political and cultural issues. C-SPAN features a morning call-in > show called 'Washington Journal which is broadcast seven days a > week, including holidays. It begins at 7:00 a.m. and runs until > 10:00 a.m. Eastern Time (4:00 a.m. to 7:00 a.m. Pacific Time). When > Congress is in session, it often cuts away to the House or Senate > floor action earlier than 10:00 a.m. > Listeners call C-SPAN from across America virtually every morning > to complain about the effects of illegal immigration. Most are > bewildered by what is happening to their country due to illegal > immigration numbers out of control. These folks are desperately > searching for a way to effectively express their concerns to > Congress. > A call to the C-SPAN audience THAT MENTIONS THE NUMBERSUSA.COM > WEBSITE and how easy the website makes it to contact Congress, can > get the word out to hundreds of thousands of Americans who don't > get NumbersUSA.com Action Alerts. Often, the caller can tell the > audience that Congress is about to vote on an immigration related > bill, and invite listeners to visit NumbersUSA.com to find out what > they can do to efficiently and effectively let Congress know what > they think. We can get help from thousands of non-members and > recruit many new members with just one phone call to this > bi-partisan audience. > NumbersUSA Comment: > Like most Americans, nearly all of these callers have never heard > of NumbersUSA.com, and they have no idea that NumbersUSA.com is > available to them, free of charge, to help them economically > leverage their efforts to make their views known to Congress. > Thousands would join NumbersUSA.com if they could hear of our > existence and what we can do. > We appreciate you making this effort! > Please tell us about your call! > Submit my feedback! > One person can't possibly affect the world anymore in this day and > age.- Hide quoted text - > - Show quoted text - Tell that to Bill Gates replace the word possibly with positively and gates is excluded === Subject: Re: Dinstinct Permutaitons of a Multiset of length N > If M is the number of balls (M>N)number of choices wil be M!/(N!(M- > N)!) This assumes that the arrangement within N is unimportant. If it is > not M!/(N-M)! as there are N! arrangements of the set of N. - Ian Parker No, this is not correct. This would only hold if the bag contained N elements. I was asking about bags with > N elements. === Subject: the n n-th roots of complex exponential map I'd like some help with the following. What are the n-th roots of exp(2 * PI * i * z),where z is a complex variable, and i is imaginary unit? === Subject: Re: the n n-th roots of complex exponential map > I'd like some help with the following. What are the n-th roots of exp(2 * PI * i * z),where z is a complex > variable, and i is imaginary unit? Hint: Let f_0(z) be the most obvious solution. What can you say about f(z)/f_0(z)? -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: I can't find definitions of these things I have come across some notation that I can't find a concrete definition for. I somewhat have a guess what they should be, but I want to make sure. L^p_loc(X) - Would this just be the set of functions f : X --> R such that f restricted to any compact set K is in L^p(K) ? Also, if X is a collection of functions (say R --> R), what is the C^k_loc topology on this set? === Subject: Re: I can't find definitions of these things > I have come across some notation that I can't find a concrete > definition for. I somewhat have a guess what they should be, but I > want to make sure. If it's from a book, the notation should be described somewhere. If it's a paper, they may refer to a book where that notation is defined. > L^p_loc(X) - Would this just be the set of functions f : X --> R such > that f restricted to any compact set K is in L^p(K) ? That would be my guess; the loc could mean locally. --- Christopher Heckman > Also, if X is a collection of functions (say R --> R), what is the > C^k_loc topology on this set? > === Subject: Re: e <4726188e$0$47130$892e7fe2@authen.yellow.readfreenews.n et>, > Is there a simple proof of the identity lim n->oo (1+x/n)^n = e^x ? There is a completely elementary proof in 100 Great Problems of Elementary Mathematics by Heinrich Dorrie (Dover Classics of Science & Mathematics) (Paperback) -- Michael Press === Subject: Convergence!!! Hi I have to prove this but don't know where to begin or how: Prove that if X_n -> X almost surely, and Y_n ->Y almost surely, then X_nY_n -> XY almost surely. Could someone please help me please!!! === Subject: Re: Convergence!!! > Hi I have to prove this but don't know where to begin or how: Prove that if X_n -> X almost surely, and Y_n ->Y almost surely, then > X_nY_n -> XY almost surely. Could someone please help me please!!! Hint: the union of two sets of measure 0 has measure 0. === Subject: Re: Convergence!!! On 2007-11-01 22:42:59 -0400, The World Wide Wade said: ' Hi ' I have to prove this but don't know where to begin or how: ' Prove that if X_n -> X almost surely, and Y_n ->Y almost surely, then > X_nY_n -> XY almost surely. ' Could someone please help me please!!! Hint: the union of two sets of measure 0 has measure 0. Nice way of looking at it. -- -kira === Subject: Re: Convergence!!! > .... > Prove that if X_n -> X almost surely, and Y_n ->Y almost surely, then > X_nY_n -> XY almost surely. > .... To set up the necessary inequalities using epsilon, this little bit of algebra should be helpful: (X-n)(Y-n) - XY = (X-n)(Y-n) - (X-n)Y + (X-n)Y - XY = (X-n)(Y-n - Y) + (X-n - X)Y. Ken Pledger. === Subject: Re: Convergence!!! > Hi I have to prove this but don't know where to begin or how: Prove that if X_n -> X almost surely, and Y_n ->Y almost surely, then > X_nY_n -> XY almost surely. Could someone please help me please!!! Let A be the event that X_n -> X, and B be the event that Y_n -> Y, so P(A)=P(B)=1. Let C be the event that X_n -> X and Y_n -> Y, so C is the intersection of A and B. What can you say about the probability of C? What can you say about the convergence of X_nY_n on the event C? === Subject: Re: The negative dimension Give an example of a space which has -1 dimension. The empty set? Yes. By the small inductive definition of dimension. > Hm, IIRC the dimension of a space is the same by > the various definitions for compact Hausdorff spaces. > No. The empty set equals the number zero, which is represented as a point, actually as a point of reference, on the real line in geometry. Which makes it dimension zero. A space of negative D, f.e of - 2, must reduce the number of D of a space, f.e. of 3 - when united. So the resulting space is of 1 D in this case. > Thus does one conclude the empty set has dimension -1 by > the other definitions including the Hausdorff dimension? Do you have a copy of or recall the author of Dimensional Dementia > wherein all of this consider with multidimensional illustrations? ;-) Yes dimension reducing is like getting old in the brain. Actually with computer tomography one can see, that the brain of an alcoholic has shrunk inside the skull. A very similar process to negative D could be the operation, which removes a set into just it's elements. Start with zero, the empty set. The set, which contains the zero ( so the number one) is reached by an operation. The opposite: one has a set, applies the reducing operation, and get's the elements. F.e. the set of integers can be reduced into just the integers. So try to apply the reducing operation onto the zero, which is the empty set, may be even two or three times. The first application must be something like the smile of a Cheshire Cat. Have fun Hero === Subject: Re: The negative dimension Give an example of a space which has -1 dimension. > The empty set? Yes. By the small inductive definition of dimension. > Hm, IIRC the dimension of a space is the same by > the various definitions for compact Hausdorff spaces. No. The empty set equals the number zero, which is represented as a > point, actually as a point of reference, on the real line in geometry. > Which makes it dimension zero. [...] {0} has dimension zero, but {0} is not the empty set. Try again. --- Christopher Heckman === Subject: Re: JSH: Why self-encryption? I want to talk again about a simple idea I have for copy protection > where when, for instance, you make a DVD of your favorite movie, like, > say Transformers, your DVD burner encrypts the copy, where it can > still read it despite that encryption, but no one else can, without a > key. By your DVD burner, you must mean any DVD burner the reader buys. Otherwise when a DVD burner went belly up the encrypted DVDs it had recorded would become irretrievable coasters. So unless a DVD burner was severely limited in the number of encryption key changes allowed (like those blasted region code chips that allow at most three changes), you could change the key at will and the system would be no more than a handy encryption device which could play anyone's recording, if they provided the encryption key they had used. === Subject: Re: JSH: Why self-encryption? <873avtz9mr.fsf@phiwumbda.org> Obviously if I have an idea that takes off in the entertainment > industry then I would have a lot of power to shape how the world > looks at mathematicians, and you know a lot about what I'd say. > Right. Like that guy who invented CSS. He's super-powerful these > days and he's using his new found power to punish English literature > departments worldwide. > So if he can do it... > So you compare inventing CSS to allowing people to copy their DVD's > without hassle? > And were people getting sued over anything with CSS? > Reminder to people who didn't read my lead post, my idea is that your > DVD burner would encrypt a copy you made of, say, the movie > Transformers, so only it could read it without a key. that makes it platform dependent, how could anyone do a wide distribution ? > A DVD that works on only one DVD burner ? No. You'd buy the DVD from the store like you do now. And when you > took it home you could copy it, freely. Easily. And you'd have no problems, as long as only you used it, but if you > gave it away, no one else could use it as the copy was encrypted--to > your equipment. No. The copy was encrypted to your *computer*. It will not play on the dvd player you have hooked to your TV. It will not play on the dvd player you have in the den. It will not play on the portable dvd player in the van. It will not play on the new computer you buy (the last is not a problem, you never watch DVD's on a computer anyway). - William Hughes === Subject: Re: Fermat's Last Theorem simple proof impossible? I'm wondering. Has it been proven that it is impossible to prove >Fermat's Last Theorem using only the mathematics Fermat would have had >available at his time? No. Did Fermat himself have a real proof like he >claimed? Barring a time machine, it seems unlikely that this question can be > given a final, definitive answer. The general consensus, however, is that it is unlikely. Why? We never saw his claimed proof. So then nobody can know if it's valid or flawed. > Keep in mind > that the claim was a personal note made on the margin of his copy of > the book, a note that he probably never imagined someone else > reading. Years after making the note he did publicly discuss a special > case (he gave the full proof in the case n=4, a rarity for him), and > related problems, but never once even hinted in public at the general > statement. But like you said, his giving proofs was rare. So why expect him to divulge this one? > As such, it is not unreasonable to think that he may have > later realized he was mistaken; why not make a correction then? > Because the original statement was not public; it was a note to > himself in a personal book. There was nobody to correct. Just because we don't know how to do it does not mean it is >impossible to do. Obviously a way to settle one of the questions, >namely that of Fermat having a proof, would be to find what he claimed >to have as a proof and see if it was valid. Unless you discover previously unknown material written by Fermat > himself, this is impossible. That's right. That's exactly what I was talking about. If somehow, somewhere, a note was found from Fermat that outlined the proof, it would clinch the matter. > Even if you found an elementary proof > that used nothing but the machinery that we know existed at the time > of Fermat, you would have no way of knowing whether Fermat had > considered that proof or not. > === Subject: Re: Fermat's Last Theorem simple proof impossible? Did Fermat himself have a real proof like he claimed? > Barring a time machine, it seems unlikely that this question > can be given a final, definitive answer. > The general consensus, however, is that it is unlikely. Why? We never saw his claimed proof. So then nobody can know > if it's valid or flawed. ' Keep in mind > that the claim was a personal note made on the margin of his copy of > the book, a note that he probably never imagined someone else > reading. Years after making the note he did publicly discuss a special > case (he gave the full proof in the case n=4, a rarity for him), and > related problems, but never once even hinted in public at the general > statement. But like you said, his giving proofs was rare. > So why expect him to divulge this one? As Weil notes, there is evidence that Fermat conjectured the general case early in his mathematical career (one can't be sure since there is no way to date Fermat's marginal notes in his copy of Bachet's Diophantus). Never again does Fermat mention the general case in any of his writings -- only the special cases for exponents 3 and 4 (on several occasions). This seems to indicate that Fermat later realized that his methods did not apply to the general case. Moreover, we now know that to be the case -- his techniques are all special cases of general results on conic and elliptic curves; these techniques do not generalize to higher exponents. See my 2 posts in [1] for more, including quotes and references. Keep in mind that the amount of number theory known by Fermat and his contemporaries is _far_ less than one obtains nowadays in a first course in number theory. --Bill Dubuque === Subject: Re: Fermat's Last Theorem simple proof impossible? Hi. I'm wondering. Has it been proven that it is impossible to prove > Fermat's Last Theorem using only the mathematics Fermat would have had > available at his time? No. How could you prove such a thing? The complexity of the Wiles proof has caused most > people to be pessimistic on this question though. > That's what I do not understand. How does the existence of a long and complicated solution to a problem bar the existence of short, simple solutions, anyway? > Did Fermat himself have a real proof like he > claimed? How can we know? We can only make educated guesses, > based on the centuries of unsuccessful efforts. > I would guess that he had a subtle fallacy and did > not have an actual proof. Even Wiles' monster proof > had a subtle gap in its first release. Just because we don't know how to do it does not mean it is > impossible to do. Of course not. Obviously a way to settle one of the questions, > namely that of Fermat having a proof, would be to find what he claimed > to have as a proof and see if it was valid. How do you recommend we do that? > Well, that's the rub, now isn't it? > - Randy === Subject: Re: Fermat's Last Theorem simple proof impossible? > Hi. > I'm wondering. Has it been proven that it is impossible to prove > Fermat's Last Theorem using only the mathematics Fermat would have had > available at his time? No. How could you prove such a thing? The complexity of the Wiles proof has caused most > people to be pessimistic on this question though. That's what I do not understand. How does the existence > of a long and complicated solution to a problem bar > the existence of short, simple solutions, anyway? It doesn't. In fact, the first paper I got published was a short proof of a result which had been originally proved in 1979. But that's a problem which was only 20-25 years old, so it was more likely that there was a shorter proof. However, if there was a short proof of FLT, wouldn't someone (among the thousands who have worked on the problem) have re-discovered it in the past 300 years? As for Fermat not proving other results: He has a reputation of stating things without proof, where a proof was found later. The fact that FLT isn't called Fermat's Conjecture is based on this reputation. --- Christopher Heckman === Subject: Re: Fermat's Last Theorem simple proof impossible? Did Fermat himself have a real proof like he > claimed? to others that they did show something for which he claimed to have a > proof (this clearly shows his background, i.e. not mathematics, he was > just an amateur). Why does one need to be something other than an amateur to have a background (ie. knowledge, experience) in mathematics? > He had pretty good ideas, but there are not many > actual proofs by Fermat known. > -- > 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: Fermat's Last Theorem simple proof impossible? Apart from the other responses... > Did Fermat himself have a real proof like he > claimed? to others that they did show something for which he claimed to have a > proof (this clearly shows his background, i.e. not mathematics, he was > just an amateur). Why does one need to be something other than an amateur to have > a background (ie. knowledge, experience) in mathematics? Knowing what other people have tried is important, so you don't end up reinventing the wheel. --- Christopher Heckman === Subject: Re: Fermat's Last Theorem simple proof impossible? Nntp-Posting-Host: hera.cwi.nl ... > letters to others that they did show something for which he claimed > to have a proof (this clearly shows his background, i.e. not > mathematics, he was just an amateur). ' Why does one need to be something other than an amateur to have > a background (ie. knowledge, experience) in mathematics? Well, he was an amateur (just like me BTW). And you do not need to be something else, but he just was something else. > Knowing what other people have tried is important, so you don't end up > reinventing the wheel. Not publicising what you have found before you can safely tell what you have found (so that recognition comes to the proper place) seems to be quite common in law circles. He has always been very secretive about his proofs (although he did reveal the methods he did use, like infinite descent). -- 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: Fermat's Last Theorem simple proof impossible? letters to others that they did show something for which he claimed > to have a proof (this clearly shows his background, i.e. not > mathematics, he was just an amateur). > Why does one need to be something other than an amateur to have > a background (ie. knowledge, experience) in mathematics? Well, he was an amateur (just like me BTW). And you do not need to be > something else, but he just was something else. > Knowing what other people have tried is important, so you don't end up > reinventing the wheel. Not publicising what you have found before you can safely tell what you > have found (so that recognition comes to the proper place) seems to be > quite common in law circles. He has always been very secretive about his > proofs (although he did reveal the methods he did use, like infinite > descent). Also, back in Fermat's day, mathematicians tended to work alone, the job was competitive, and lots of problems could be solved using a few basic ideas. In that sort of environment, you don't give away your best tricks, especially if they're easy to do. --- Christopher Heckman === Subject: Re: Fermat's Last Theorem simple proof impossible? > Hi. I'm wondering. Has it been proven that it is impossible to prove > Fermat's Last Theorem using only the mathematics Fermat would have had > available at his time? Did Fermat himself have a real proof like he > claimed? Just because we don't know how to do it does not mean it is > impossible to do. Obviously a way to settle one of the questions, > namely that of Fermat having a proof, would be to find what he claimed > to have as a proof and see if it was valid. You're probably aware there can be the world of a difference between simple and elementary. At least two professional mathematicians to my knowledge are on record as not ruling out an elementary solution, Doran Zeilberger and Sir Peter Swinnerton Dyer. But one or both (?) said it would almost certainly be extremely intricate and subtle. Maybe after more of the advanced theory underlying Wiles's proof is discovered and clarified, his proof (perhaps considerably shortened) will be seen as simpler than the elementary solution if one is ever found! John R Ramsden === Subject: Re: Fermat's Last Theorem simple proof impossible? > |Perhaps one can formulate a Liouville theorem for number > theoretical > |conjectures, for example, limiting steps in a proof to finite > |operations of arithmetic and exponential functions, representing > |exponentiation and powers, and thus keeping elliptical functions out. I dimly recall having read that the kind of > descent argument favored by Fermat only works > when a certain kind of cohomological obstruction > doesn't exist, but that the Fermat curve for > some primes has a nontrivial such cohomology. > Take this with a grain of salt. Before Wiles's proof the Fermat conjecture was proven for only a finite number of primes. This was always significant to me. (still is, for that matter.) -- Michael Press === Subject: tommy's division...a halting problem f(n)=/=g(n) for any n f(n) and g(n) nonzero positive integer functions. f(n)'*'g(n) = choose[f(n)/g(n),g(n)/f(n)]if integer > 1. else f(n)*g(n) examples 7'*'14 = 2 14'*'7 = 2 3'*'7 = 21 7'*'3 = 21 1'*'24 = 24 ('*' -> coined tommy's division ) r2(n)= n - largest root of n r2(25)= 25 -25 = 0 r2(24)= 24 - 16 = 8 r2(101) = 101 - 100 = 1 if g(n) < 20 -> output : HALT g(0)= m (positive integer) g(n) = max[r2[f(n)'*'g(n-1)],r2[r2(f(n))'*'g(n-1)]] ( keep in mind g(n) =/= f(n) ! ) wheiter or not g(n) halts for a given f(n) is the halting problem mentioned in the title of the tread , in its generalized form. if f(n) is an irreducable polynomial then g(n) halts. tommy1729 ps : for the ones intrested in dimensional numbers and timothy golden's polysigned ; tommy's division is the key operator in the solution to dividing two numbers of different kind and/or dimension. P2 / P3 = P6 ( 2'*'3 = 6) P2 * P3 = P6 (2'*'3 = 6) note however this is a computation in vector sence ; there might be dimensional reduction so e.g. P6 can be potentially reduced to P4. similarly P3 in this context means 3 algebraicly closed vectors or in other words isomorphic to complex or P3. if you are curious about what the **** polysigned means : www.bandtechnology.com === Subject: Re: tommy's division...a halting problem > f(n)=/=g(n) for any n > f(n) and g(n) nonzero positive integer functions. > f(n)'*'g(n) = > choose[f(n)/g(n),g(n)/f(n)]if integer > 1. > else f(n)*g(n) Whoa! Let's clear that up! f and g are positive integer function such that f(n) is not g(n) for any n. The tommy division ( would be a better symbol, as * suggests multiplication) is a b = a/b if a/b is an integer > 1, a b = b/a if b/a is an integer > 1, a b = a*b otherwise. > r2(n)= n - largest root of n r2(n) = n - floor(sqrt(n))^2. > if g(n) < 20 -> output : HALT g(0)= m (positive integer) g(n) = max[r2[f(n)'*'g(n-1)],r2[r2(f(n))'*'g(n-1)]] Is this a definition, or an equation, or what? > ( keep in mind g(n) =/= f(n) ! ) whether or not g(n) halts for a given f(n) is the halting problem mentioned in the > title of the tread, in its generalized form. if f(n) is an irreducible polynomial then g(n) halts. Have you proved this, or is it just a conjecture? > ps: for the ones interested in dimensional numbers and timothy golden's polysigned ; tommy's division is the key operator in the solution to dividing two numbers of > different kind and/or dimension. P2 / P3 = P6 ( 2'*'3 = 6) > P2 * P3 = P6 (2'*'3 = 6) note however this is a computation in vector sense ; there might be dimensional > reduction so e.g. P6 can be potentially reduced to P4. similarly P3 in this context means 3 algebraicly closed vectors or in other words > isomorphic to complex or P3. if you are curious about what the **** polysigned means : www.bandtechnology.com More like www.badnotation.com. --- Christopher Heckman === Subject: Re: tommy's division...a halting problem > On Nov 1, 3:12 pm, tommy1729 f(n) and g(n) nonzero positive integer functions. > f(n)'*'g(n) = > choose[f(n)/g(n),g(n)/f(n)]if integer > 1. > else f(n)*g(n) Whoa! Let's clear that up! f and g are positive > integer function such > that f(n) is not g(n) for any n. The tommy division ( would be a better symbol, as > * suggests > multiplication) is > a b = a/b if a/b is an integer > 1, > a b = b/a if b/a is an integer > 1, > a b = a*b otherwise. exactly sorry if i defined in C+++++ ;-) you understood what i meant. as for the criticism on my symbol : as the otherwise indicates , it can be seen as a product too ... but if you like that's fine by me ... or '*' or * or even :) the thing is see it somewhat more of a multiplication ... and i forgot was on my keyboard too , i admit ... on the other hand , calling something a kind of division and considering it a product is kind a strange yes :) r2(n)= n - largest root of n r2(n) = n - floor(sqrt(n))^2. i.e. what happens to n, especially for the ones who are not used to working with floor. if g(n) < 20 -> output : HALT g(0)= m (positive integer) g(n) = max[r2[f(n)'*'g(n-1)],r2[r2(f(n))'*'g(n-1)]] Is this a definition, or an equation, or what? both, its a defined equation :) ( keep in mind g(n) =/= f(n) ! ) whether or not g(n) halts for a given f(n) is the > halting problem mentioned in the > title of the tread, in its generalized form. if f(n) is an irreducible polynomial then g(n) > halts. Have you proved this, or is it just a conjecture? ps: for the ones interested in dimensional numbers > and timothy golden's polysigned ; tommy's division is the key operator in the > solution to dividing two numbers of > different kind and/or dimension. P2 / P3 = P6 ( 2'*'3 = 6) > P2 * P3 = P6 (2'*'3 = 6) note however this is a computation in vector sense > ; there might be dimensional > reduction so e.g. P6 can be potentially reduced to > P4. similarly P3 in this context means 3 algebraicly > closed vectors or in other words > isomorphic to complex or P3. if you are curious about what the **** polysigned > means : www.bandtechnology.com More like www.badnotation.com. hahaha srr --- Christopher Heckman > tommy1729 === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I want Engineering Mechanics - Statics 6th edition Meriam === Subject: Re: number sequence , > , > I am recreationally trying to figure out this sequence, but cannot. > Would someone please complete this for me and explain: > 1,9,25,__,81,100 > a) 36 > b) 45 > c) 56 > d) 64 > 42 > You are welcome. > Why 42? It's the answer to the question of Life, > the Universe and Everything. ' It is not one of the answers? That's a joke, son. Not a joke fella, In order to describe an integer sequence it must be > able to be written in formula notation: That may have been the case at one time in mathematical practice. It is no longer. And as for 1,9,25,__,81,100 it is not a sequence. An integer sequence is a map from the natural numbers to the integers. You indicated the presence of six elements, well short of a sequence. > x + (x+2)^2 + ((x+2)^2)^2 .... or some such, i'm still working on this > and i think it has to do with primes. Imploring help from the ' Does anyone have a serious answer? Did you show any of your thinking on the matter? Did you recognize 1,9,25,81,100 to be squares of natural numbers? > 64 The first 3 are squares of consective > odd integers (1**2, 3**2, 5**2), but > the last two are just squares of > consecutive integers (9**2, 10**2). But, by symmetry, there must be three > consecutive integers squared. Thus, the > number that precedes 9**2 must be 8**2 > or 64. -- Michael Press === Subject: Re: number sequence > I am recreationally trying to figure out this > sequence, but cannot. > Would someone please complete this for me and > explain: 1,9,25,__,81,100 a) 36 > b) 45 > c) 56 > d) 64 > 1, 9, 25, r, 81, 100, being r any real number. Fernando. === === Subject: Re: arctan(x)/x^2 > Hello all.. > I have a problem ... II?am suppost to fiind > the > intergral for: > arctan(x) / x^2 > ...please :) > ln(x)-(1/2)ln(x^2+1)-(arctan(x)/x) > Presumably, the antiderivative is to be real, > valid > for nonzero x, and so > I don't give your result full marks. [Note that, > for > x < 0, your result is > not real.] My result is > - ln(1 + 1/x^2) - arctan(x)/x + C > where C is an arbitrary real constant. > David W. Cantrell funny I suppose that by funny you mean strange, as > opposed to humorous. But > it's not strange. It's the same sort of thing as > giving ln|x| + C, rather > than ln(x) + C, in elementary calculus for the > antiderivative of 1/x. David by funny i could also mean 1) you try to correct my integral with a wrong one of yourself... 2) ever heard of cauchy PV ( principal value ) ? you seem to confuse PV with integral ... or can antiderivate mean PV too ? anyways you try to give a PV ( a wrong one ) yet my integral works fine... take the integral from a to b and you will see. so i dont write |x| nor + C another reason i avoid |x| are certain methods of integration and substitution which fail or are undefined so this |x| gives an unworkable part for further computations and d|x| at 0 is .... 3) my integral is correct. *** btw this reminds me of discussions about the + C for certain or all integrals. what is the most logical choice for C if we had to pick one ... quasi gave a nice example where taylor series were used to give another logical value... perhaps the most logical value is the one logicly corresponding to ramanujan's master theorem its makes sence compared to taylor too. since g(n) is the n-th derivate g(-n) is then the n-th integral : with the appropriate C. just a quick idea ...perhaps intresting ... *** ramanujan and i used to investigate functions like arcsin(x)^a * x^b with a real and b negative. i dont know if it got published , but i think so. *** tommy1729 === Subject: Re: arctan(x)/x^2 > Hello all.. > I have a problem ... IIa[Hyphen]I[Hyphen]a m suppost to fiind > the intergral for: > arctan(x) / x^2 > ...please :) > ln(x)-(1/2)ln(x^2+1)-(arctan(x)/x) > Presumably, the antiderivative is to be real, valid > for nonzero x, and so I don't give your result full marks. > [Note that, for x < 0, your result is not real.] My result is > - ln(1 + 1/x^2) - arctan(x)/x + C > where C is an arbitrary real constant. As noted by me previously in this thread, I merely failed to type a division by 2. In fact, my result was actually - ln(1 + 1/x^2)/2 - arctan(x)/x + C > funny I suppose that by funny you mean strange, as opposed to humorous. > But it's not strange. It's the same sort of thing as giving ln|x| + C, > rather than ln(x) + C, in elementary calculus for the antiderivative > of 1/x. by funny i could also mean 1) you try to correct my integral with a wrong one of yourself... True. Yet, in my case, it was wrong only due to a trivial typographical error. > 2) ever heard of cauchy PV ( principal value ) ? Of course. > you seem to confuse PV with integral ... No. > or can antiderivate mean PV too ? No. > anyways you try to give a PV ( a wrong one ) No, I didn't. > yet my integral works fine... I never said it didn't work fine _in any context_. But it seems highly likely that the OP is a student in elementary calculus, and thus, appropriate to his context, I had said Presumably, the antiderivative is to be real, valid for nonzero x. Your antiderivative does not satisfy that condition. By contrast, - ln(1 + 1/x^2)/2 - arctan(x)/x + C does. > take the integral from a to b and you will see. You seem to be confusing a definite integral with an indefinite one. But indeed, your antiderivative can be used to calculate definite integrals via real result. > so i dont write |x| nor + C another reason i avoid |x| are certain methods of integration and > substitution which fail or are undefined so this |x| gives an unworkable > part for further computations and d|x| at 0 is .... I agree that avoiding absolute value is often desirable. Note that my - ln(1 + 1/x^2)/2 - arctan(x)/x + C does not use absolute value. [BTW, if you were thinking of my comment about the antiderivative of 1/x: In elementary calculus, it is normally stated as ln|x| + C. But if you want to give a result which is real for all nonzero x and which also avoids absolute value, then just use ln(x^2)/2 + C.] > 3) my integral is correct. In the context which I had specified, which is appropriate to elemenatary calculus, your result is not fully correct since it is not real for x < 0. David W. Cantrell === Subject: Re: #231 circle or sphere fail to have a Commutative; new textbook: Mathematical-Physics (pi) x (pi) = 180 x 180 = 32,400 degrees = 90 (2pi) = 2pi > since they are all modulo. > Isn't 90 half of 180 (degrees)? > Shouldn't that be: > (pi) * (pi) = 180 x 180 = 32,400 = 90 (mod 2 pi) = 90 = (pi)/2 > (pi) * (pi) = (pi)/2 > So then (pi)*(pi) = (pi)^2 = (pi)/2. > Which also means that: > pi^2 + pi^2 = pi/2 + pi/2 = pi(1/2 + 1/2) = pi I have no complaint with any of the above. Trouble was me, for I still > do not have many matters settled. Under modulo 360 then I agree > with the above. Since you agree with my math above, perhaps you can answer another question for when we expand on those equations. You have (pi)*(pi) = (pi)^2 = (pi)/2, from above. So then (pi)*(pi) = (pi)/2 (pi)*(pi) - (pi)/2 = 0 pi * (pi - 1/2) = 0 Which means that either pi = 0 or pi - 1/2 = 0 pi = 1/2 I don't think you want pi to be 0. So is pi = 1/2? === Subject: #239 circle or sphere fail to have a Commutative; new textbook: Mathematical-Physics '(pi) x (pi) = 180 x 180 = 32,400 degrees = 90 (2pi) = 2pi >since they are all modulo. >Isn't 90 half of 180 (degrees)? >Shouldn't that be: > (pi) * (pi) = 180 x 180 = 32,400 = 90 (mod 2 pi) = 90 = (pi)/2 > (pi) * (pi) = (pi)/2 So then (pi)*(pi) = (pi)^2 = (pi)/2. Which also means that: > pi^2 + pi^2 = pi/2 + pi/2 = pi(1/2 + 1/2) = pi >I have no complaint with any of the above. Trouble was me, for I still >do not have many matters settled. Under modulo 360 then I agree >with the above. > Since you agree with my math above, perhaps you can answer > another question for when we expand on those equations. You have (pi)*(pi) = (pi)^2 = (pi)/2, from above. So then > (pi)*(pi) = (pi)/2 > (pi)*(pi) - (pi)/2 = 0 > pi * (pi - 1/2) = 0 Which means that either > pi = 0 > or > pi - 1/2 = 0 > pi = 1/2 I don't think you want pi to be 0. > So is pi = 1/2? > Okay, I have enough of the multiplication for the realistic side of the sphere model to be able to answer those questions about the imaginary hemisphere. The realistic hemisphere are all the points from 1 to 999...99999 where the Greenwich longitude line is imaginary since the North Pole and South Pole are imaginary points of 2pi and pi respectively. On the realistic hemisphere multiplication ends up as a picketfence summation of all the line segments and where it is closed to multiplication. Where multiplication is spherical-triangles and there area. So now, I impose the same system over on the Imaginary hemisphere, and since pi is a zero point that pi X pi is just pi just as 0 X 0 is 0 Now (pi + 1) X (pi + ....99999) is the same as 1 X ....99999 with a pi stuck in front of it. So that answer is pi + ....99999. Now what is (pi + 3) X (pi + ....3333) and the answer is pi + ....99999. So any multiplication on this Imaginary hemisphere is like doing the multiplying of the Realistic component and then just sticking on a pi in front of it. So that leaves multiplication Closed on the imaginary hemisphere. Except for the issue of pi and 2pi when they are multiplied in various ways. So pi X pi is pi and pi X 2 is pi and pi X 3 is pi so pi is like 0 and also 2pi is 0. But it leaves one minor tricky thing for me to figure out. In that there is a longitude line that is not covered in either the Realistic hemisphere or the Imaginary hemisphere and that is the longitude line that of Greenwich since the Equator line goes from 1 to 999.....9999. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: #231 circle or sphere fail to have a Commutative; new textbook: Mathematical-Physics (pi) x (pi) = 180 x 180 = 32,400 degrees = 90 (2pi) = 2pi > since they are all modulo. > Isn't 90 half of 180 (degrees)? > Shouldn't that be: > (pi) * (pi) = 180 x 180 = 32,400 = 90 (mod 2 pi) = 90 = (pi)/2 > (pi) * (pi) = (pi)/2 > So then (pi)*(pi) = (pi)^2 = (pi)/2. > Which also means that: > pi^2 + pi^2 = pi/2 + pi/2 = pi(1/2 + 1/2) = pi > I have no complaint with any of the above. Trouble was me, for I still > do not have many matters settled. Under modulo 360 then I agree > with the above. Since you agree with my math above, perhaps you can answer > another question for when we expand on those equations. You have (pi)*(pi) = (pi)^2 = (pi)/2, from above. So then > (pi)*(pi) = (pi)/2 > (pi)*(pi) - (pi)/2 = 0 > pi * (pi - 1/2) = 0 Which means that either > pi = 0 > or > pi - 1/2 = 0 > pi = 1/2 I don't think you want pi to be 0. > So is pi = 1/2? I am going to hold off on answering or thinking about the imaginary hemisphere. It makes commonsense to me, anyway, that if I don't have the true-blue side of the globe in order as to multiplication, that it would be silly of me to spend any time over on the imaginary side until I have the other hemisphere in order. === Subject: Re: Can we find the function? > How about f(x) = 1/x if x != 0, > 0 if x = 0 > f(f(x)) = x, certainly a polynomial function. > ... but also a monomial. Right. But, as preface to my example: If f(f(x)) were allowed to be a monomial, I would use simply f(x) = 1/x. > [Note that I'm taking the codomain of f to be R*, the one-point extension > of R, so that we have f(0) = oo and f(oo) = 0.] Then f(f(x)) = x. -------------------------------- Now for the desired example: f(x) = 2/(x + 1)^2 - 1 yields f(f(x)) = (x + 1)^4/2 - 1 Note that again I am taking the codomain to be R*. Thus, f(-1) = oo and > f(oo) = -1. Perhaps I should make a comment for anyone who thinks that using R* is > somehow cheating. If we want to have a _real_ non-polynomial function > f(x) such that f(f(x)) is a polynomial with more than one term, then just > modify my function above slightly to give | { -1 if x = -1, > | f(x) = { > | { 2/(x + 1)^2 - 1 otherwise. Then we still have f(f(x)) = (x + 1)^4/2 - 1. David Or somewhat more generally consider f(x) = b (x-a)^(-n) + a for x <> a, f(a) = a where n is a positive integer. Then f(f(x)) = b^(1-n) (x-a)^(n^2) + a. Similarly, f(x) = b |x-a|^sqrt(n) + a where n is an even positive integer that isn't a square. Then f(f(x)) = b |b|^sqrt(n) (x-a)^n + a. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Can we find the function? > f(x) = b |x-a|^sqrt(n) + a where n is an even positive integer that isn't a square. Then f(f(x)) = b |b|^sqrt(n) (x-a)^n + a. Or if n is an odd positive integer that isn't a square, f(x) = b signum(x-a) |x-a|^sqrt(n) + a and f(f(x)) = |b|^(sqrt(n)+1) (x-a)^n + a -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Can we find the function? Am 01.11.2007 16:15 schrieb matt271829-news@yahoo.co.uk: Well, if f(x) = a*x + b, a <> 1, then a general solution is f^n(x) = a^n*x + b*(a^n - 1)/(a - 1) (where f^n means f iterated n times). Ah, that's nice and concise (and agrees with my approach) - it's so simple;-) Gottfried Helms -- --- Gottfried Helms, Kassel === Subject: Re: #216 in fact a circle does not even obey associativity or commutative addition; new textbook: Mathematical-Physics <2007103120554816807-kquirici@yahoocom> If you read Dik's idea closely, you'll see that the algebra is defined > so that all the sums and products are done modulo 2 pi (360). > So adding two arcs that sum to greater than 2 pi results in an > arc that wraps around the zero point of the unit circle, and > likewise for multiplication. Suppose you've defined a direction from the north pole (there are two > possible directions) as the std 'positive' direction. Suppose going in that direction you add > 65 + 105 + 210 = 380 = 20 mod 360. > a1 a2 a3 Using the other direction as canonical the same 'points' add > 295 + 255 + 150 = 700 = 340 mod 360. > b1 b2 b3 If you mix and match directions, you get for one possibility: > 65+255+210 = 530 = 170 mod 360. > a1 b2 a3 It would seem you need to specify a canonical 'direction' to > measure angles from the origin and ensure consistent > results. Is that right? Yes. For instance, we can assume that adding an arc in the Northern direction is positive, while adding an arc in the Southerly direction is negative. For your math to work out correctly, you must see that +65 = -295, +105 = -255, and +210 = -150 (all mod 360). Don't forget to use the correct sign to indicate the direction. So you have: 65 + 105 + 210 = 380 = 20 mod 360. a1 a2 a3 and -295 + -255 + -150 = -700 = 20 mod 360. b1 b2 b3 It's all very simple, but you have to do the arithmetic operations as they are defined for the ring in order to get correct results. AP doesn't seem to grasp this fact yet. === Subject: Re: #216 in fact a circle does not even obey associativity or commutative addition; new textbook: Mathematical-Physics On 2007-11-01 19:30:11 -0400, David R Tribble said: > If you read Dik's idea closely, you'll see that the algebra is defined > so that all the sums and products are done modulo 2 pi (360). > So adding two arcs that sum to greater than 2 pi results in an > arc that wraps around the zero point of the unit circle, and > likewise for multiplication. ' Suppose you've defined a direction from the north pole (there are two > possible directions) as the std 'positive' direction. ' Suppose going in that direction you add > 65 + 105 + 210 = 380 = 20 mod 360. > a1 a2 a3 ' Using the other direction as canonical the same 'points' add > 295 + 255 + 150 = 700 = 340 mod 360. > b1 b2 b3 ' If you mix and match directions, you get for one possibility: > 65+255+210 = 530 = 170 mod 360. > a1 b2 a3 ' It would seem you need to specify a canonical 'direction' to > measure angles from the origin and ensure consistent > results. Is that right? Yes. For instance, we can assume that adding an arc in the > Northern direction is positive, while adding an arc in the Southerly > direction is negative. For your math to work out correctly, you must see that > +65 = -295, +105 = -255, and +210 = -150 (all mod 360). > Don't forget to use the correct sign to indicate the direction. So you have: > 65 + 105 + 210 = 380 = 20 mod 360. > a1 a2 a3 > and > -295 + -255 + -150 = -700 = 20 mod 360. > b1 b2 b3 It's all very simple, but you have to do the arithmetic operations > as they are defined for the ring in order to get correct results. > AP doesn't seem to grasp this fact yet. that's cool it made me laugh out loud. The sign was the thing === Subject: Re: #216 in fact a circle does not even obey associativity or commutative addition; new textbook: Mathematical-Physics <47202978.7020004@hotmail.com> <47217B9F.3080503@hotmail.com> <4722D1E5.3070008@hotmail.com> <4722FD90.1040002@hotmail.com> <47294A5D.5020701@hotmail.com> I just don't see the problem. David, I condensed this problem to the simple case of Frankfurt to > Zurich. Dik's system does not address the problem that Frankfurt to > Zurich has two choices, either 300 km or 39,700 km. So that system > always has choices, whether you go clockwise or counterclockwise, > whether you use the major arc or the minor arc. So every operation > has this problem of choices. And thus, Dik's system is noncommutative, > nonassociative and nondistributive. And there is noway of removing those choices and thus never a ring or field. Only if you choose to ignore the direction of each arc. Obviously, going South is the opposite of going North, so adding an arc in the Southerly direction is the equivalent of subtracting an arc of the same length in the Northerly direction (and vice versa). Adding A + 300km S (mod 40,000km) is equivalent to A - 300km N (mod 40,000km) A + 39,700km S (mod 40,000km) In other words, traveling 300km South is equivalent to traveling 39,700km North (or -39,700km South) on a 40,000km circle. That's pretty obvious, isn't it? That's how addition and multiplication are defined on the unit circle to make the whole system a ring (and a field). You can't simply ignore the signs. === Subject: Re: distribution of distribution I got n training distribution functions, each of them is a mixture of > two Gaussians. What does mixture mean here? Then I got a new distribution function, which is also a mixture of two > Gaussian. Is there any way to estimate the probability of this new > distribution function? What relevance does the first paragraph have to this question? Since we do not know the underlying distribution of Gaussian mixtures, > it seems to me that this task is pretty hard. > === Subject: Measurable set in R and mean values Let E be a measurable set in R. Must there exist points x,y in E such that (x+y)/2 is also in E ? === Subject: Re: Measurable set in R and mean values <24221068.1193961601716.JavaMail.jakarta@nitrogen.mathforum.org>, > Let E be a measurable set in R. Must there exist points x,y in E such that > (x+y)/2 is also in E ? > Of course you mean distinct x and y. If m(E) = 0 the answer is clearly no. Suppose m(E) > 0. Recall that a.e. point of E is a point of density of E. WLOG, 0 is one of these. Choose h > 0 such that m((-h, h) n E) > 7h/4, where n denotes intersection. Let E' = (0, h) n E. Then m(E') > 3h/4, which implies m(E'/2) > 3h/8. Now check that for every x in (-h, 0), m((x/2 + E'/2) n E) > 0. A lot of those x's also lie in E, so the answer to the question is yes. === Subject: Re: Measurable set in R and mean values On 2007-11-01 22:38:19 -0400, The World Wide Wade said: > <24221068.1193961601716.JavaMail.jakarta@nitrogen.mathforum.org>, ' Let E be a measurable set in R. Must there exist points x,y in E such that > (x+y)/2 is also in E ? ' Of course you mean distinct x and y. If m(E) = 0 the answer is clearly > no. Suppose m(E) > 0. Recall that a.e. point of E is a point of > density of E. WLOG, 0 is one of these. Choose h > 0 such that m((-h, > h) n E) > 7h/4, where n denotes intersection. Let E' = (0, h) n E. > Then m(E') > 3h/4, which implies m(E'/2) > 3h/8. Now check that for > every x in (-h, 0), m((x/2 + E'/2) n E) > 0. Ok... > A lot of those x's also > lie in E, so the answer to the question is yes. You seem to be implying that for all these x's, -x is also in E. How do you see this? Or did I misunderstood your argument? -- -kira === Subject: Re: Measurable set in R and mean values , > <24221068.1193961601716.JavaMail.jakarta@nitrogen.mathforum.org>, Let E be a measurable set in R. Must there exist points x,y in E such that > (x+y)/2 is also in E ? > Of course you mean distinct x and y. If m(E) = 0 the answer is clearly > no. Meaning: if m(E) = 0 is allowed, the answer is no. > Suppose m(E) > 0. Recall that a.e. point of E is a point of > density of E. WLOG, 0 is one of these. Choose h > 0 such that m((-h, > h) n E) > 7h/4, where n denotes intersection. Let E' = (0, h) n E. > Then m(E') > 3h/4, which implies m(E'/2) > 3h/8. Now check that for > every x in (-h, 0), m((x/2 + E'/2) n E) > 0. A lot of those x's also > lie in E, so the answer to the question is yes. === Subject: Re: Measurable set in R and mean values On Nov 1, 9:46 pm, The World Wide Wade , <24221068.1193961601716.JavaMail.jaka...@nitrogen.mathforum.org>, > Let E be a measurable set in R. Must there exist points x,y in E such that > (x+y)/2 is also in E ? > Of course you mean distinct x and y. If m(E) = 0 the answer is clearly > no. Meaning: if m(E) = 0 is allowed, the answer is no. Moreover, if E is merely assumed to have positive outer measure, but its inner measure is allowed to be zero, the answer is no. Namely, assuming the axiom of choice, you can construct a set of real numbers which has full outer measure but does not contain three distinct points in arithmetic progression. === Subject: Re: Measurable set in R and mean values > Let E be a measurable set in R. Must there exist points x,y in E such that (x+y)/2 is also in E ? > No, the empty set is a counterexample. It is the only counterexample: if E is any nonempty subset of R, then there exist points x,y in E such that (x+y)/2 is also in E. Perhaps you meant to ask a different question? E.g., if E is a set of *positive* Lebesgue measure in R, must there exist *distinct* points x,y in E wuch that (x+y)/2 is in E? Seems likely to me. === Subject: Re: Measurable set in R and mean values > Let E be a measurable set in R. Must there exist points x,y in E such > that (x+y)/2 is also in E ? > No, the empty set is a counterexample. It is the only counterexample: > if E is any nonempty subset of R, then there exist points x,y in E > such that (x+y)/2 is also in E. Perhaps you meant to ask a different question? E.g., if E is a set of > *positive* Lebesgue measure in R, must there exist *distinct* points > x,y in E wuch that (x+y)/2 is in E? Seems likely to me. Yes. Hint: Lebesgue Density Theorem. Note that if z is a Lebesgue point for E, then it is also a Lebesgue point for the reflection of E across z. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Measurable set in R and mean values On 2007-11-01 20:50:50 -0400, Butch Malahide said: > Let E be a measurable set in R. Must there exist points x,y in E such > that (x+y)/2 is also in E ? ' No, the empty set is a counterexample. It is the only counterexample: > if E is any nonempty subset of R, then there exist points x,y in E > such that (x+y)/2 is also in E. Perhaps you meant to ask a different question? E.g., if E is a set of > *positive* Lebesgue measure in R, must there exist *distinct* points > x,y in E wuch that (x+y)/2 is in E? Seems likely to me. There are measureable sets that are nowhere dense with positive measure, e.g. some version of the Cantor set. Is it possible some such sets are counterexamples? The construction of the Cantor set keeps on removing an interval involving the middle point of any intervals. So... can this idea be used to construct a counterexample? -- -kira === Subject: Re: Measurable set in R and mean values >On 2007-11-01 20:50:50 -0400, Butch Malahide said: Let E be a measurable set in R. Must there exist points x,y in E such > that (x+y)/2 is also in E ? ' > No, the empty set is a counterexample. It is the only counterexample: > if E is any nonempty subset of R, then there exist points x,y in E > such that (x+y)/2 is also in E. ' Perhaps you meant to ask a different question? E.g., if E is a set of > *positive* Lebesgue measure in R, must there exist *distinct* points > x,y in E wuch that (x+y)/2 is in E? Seems likely to me. There are measureable sets that are nowhere dense with positive >measure, e.g. some version of the Cantor set. Is it possible some such >sets are counterexamples? The construction of the Cantor set keeps on removing an interval >involving the middle point of any intervals. So... can this idea be >used to construct a counterexample? I doubt it. Assuming E is a counterexample, E and the complement of E compete for measure on every interval. That can't happen. There have to be intervals where E dominates and other intervals where the complement of E dominates. In other words, E must have pockets of full measure (and other pockets of zero measure), but the lack of midpoints would make pockets of full measure unachievable. Thus, it's almost certain in my mind that the conjecture is true (the corrected version, as stated by Butch Malahide). In fact, I'd bet on it. quasi === Subject: Re: Measurable set in R and mean values On 2007-11-01 23:44:43 -0400, quasi said: ' On 2007-11-01 20:50:50 -0400, Butch Malahide said: ' Let E be a measurable set in R. Must there exist points x,y in E such > that (x+y)/2 is also in E ? ' No, the empty set is a counterexample. It is the only counterexample: > if E is any nonempty subset of R, then there exist points x,y in E > such that (x+y)/2 is also in E. Perhaps you meant to ask a different question? E.g., if E is a set of > *positive* Lebesgue measure in R, must there exist *distinct* points > x,y in E wuch that (x+y)/2 is in E? Seems likely to me. ' There are measureable sets that are nowhere dense with positive > measure, e.g. some version of the Cantor set. Is it possible some such > sets are counterexamples? ' The construction of the Cantor set keeps on removing an interval > involving the middle point of any intervals. So... can this idea be > used to construct a counterexample? I doubt it. Assuming E is a counterexample, E and the complement of E compete for > measure on every interval. That can't happen. There have to be > intervals where E dominates and other intervals where the complement > of E dominates. In other words, E must have pockets of full measure > (and other pockets of zero measure), Ok... > but the lack of midpoints would > make pockets of full measure unachievable. Can you elaborate a bit more how you concluded this? Thus, it's almost certain in my mind that the conjecture is true (the > corrected version, as stated by Butch Malahide). In fact, I'd bet on it. quasi -- -kira === Subject: Re: Measurable set in R and mean values > Let E be a measurable set in R. Must there exist points x,y in E such that (x+y)/2 is also in E ? Certainly not. (At least, not if you want x and y to be distinct.) You must have in mind some more conditions on E. === Subject: Re: solution manuals, solution manual, solutions > I have solution manuals, solution manual, solutions > manuals, sultion manuel manuals solution in electronic format for > the following textbooks. They > include complete solutions to all the problems in the text, . Payment is through Paypal. Email me lsms9[at] yahoo.com > but please DO NOT POST HERE because I will > not be able to help you, but instead email and ask me for the solution > that you need. Downloads emailed immediately - within 30 minutes! Calculus: Early Transcendentals [Hardcover] by Stewart, James > vector Mechanics for Engineers : E. Russell Johnston, Ferdinand Pierre > niversity Physics with Modern Physics : Hugh D. Young, Roger A. > University Physics : Hugh D. Young, Roger A. Freedman (Hardcover, > Basic Engineering Circuit Analysis : J. David Irwin, R. Mark Nelms > hgo2002 - William H. Hayt, John A. Buck - Engineering > Electromagnetics, 6th Edition + Solutions Manual .pdf > Electric Machinery 6Ed Fitzgerald, Kingsley, Uman - Solutions > Manual.pdf > massey_-_mechanics_of_fluids_-_solutions_manual.pdf > fundamentals.of.digital.logic.with.VHDL.design.solutions.manual.rar > ch6-11 Solution Manual to engineering fluid mechanics 7e.pdf > Solution Manual - Microelectronic circuits by Sedra & Smith 5thEd Computone Corporation: An Instructional Case in Earnings Management and Revenue Recognition H. Lynn Stallworth Robert L. Braun from Issues in Accounting Education | Volume: 22 | Issue: 2 | Pps: 53 - 62 Publisher: American Accounting Association Do you have it? if no, where can I get it? Ruzanna Davtyan === Subject: Re: Solution Manual on Separation Process Principles by Seader and Henley <3879462.1193023496762.JavaMail.jakarta@nitrogen.mathforum.org > Oh god, can I PLEASE have this also? I would be forever grateful if you could send it to tibbyvroom @ yahoo.com If anyone wouldn't mind, could I please get the solution manual of > Separation Process Principles by Seader and Henley, Second Edition, Hi CAn I get the solution manual also. I would be so so grateful. === Subject: Re: #234 in fact a circle does not even obey associativity or commutative Nntp-Posting-Host: hera.cwi.nl ... > And again you fail to see that in my system the distance between Frankfurt > and Zurich is *not* addition, but subtraction. And there you indeed do > have two choices, and subtraction obviously is *not* commutative. ' And there is noway of removing those choices and thus never a ring or > field. ' You just can't read or understand my system. ... > Wrong Dik, it is you who is unable to understand your mistakes. So you > defined the minor arc as addition and the major arc as subtraction. But > now what do you do for podal and antipodal points where minor arc equals > major arc? Do you roll the dice or flip a coin to see which direction or > clockwise or counterclockwise. Indeed, you do not understand my system, because that is *not* the way I defined it. One of the arcs is A - B, the other is B - A. A + B is neither. Again, to repeat the definition, take a circle with radius R and midpoint M. Define an origin on the circle, say O and a direction from O. Let's have A and B two points on the circle. Let alpha be the angle between the line from A to M and the line from O to M in the direction from O to A according to the preferred direction. Similar for beta and B. (Also choose alpha and beta such that they are in [0, 2.pi).) To add A and B we calculate: gamma = alpha + beta mod 2.pi. Take a line from O to M and go in the preferred direction along an angle gamma; that is the point C = A + B. For multiplication we calculate: delta = fraction((A * B) / (4.pi^2)) * 2.pi, where fraction yields the fractional part of a number that is in [0,1). This delta gives D = A * B. Now, what about E = A - B? It would be the number that, when it is added to B yields A. Or in angles, it angle would be the angle from the line through B and M and the line through A and M in the preferred direction. In arcs, it is the arc from B to A in the preferred direction. Similar for B - A. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: #238 in fact a circle does not even obey associativity or commutative And again you fail to see that in my system the distance between Frankfurt > and Zurich is *not* addition, but subtraction. And there you indeed do > have two choices, and subtraction obviously is *not* commutative. > And there is noway of removing those choices and thus never a ring or > field. > You just can't read or understand my system. > ... > Wrong Dik, it is you who is unable to understand your mistakes. So you > defined the minor arc as addition and the major arc as subtraction. But > now what do you do for podal and antipodal points where minor arc equals > major arc? Do you roll the dice or flip a coin to see which direction or > clockwise or counterclockwise. Indeed, you do not understand my system, because that is *not* the way I > defined it. One of the arcs is A - B, the other is B - A. A + B is > neither. > You never considered that your model is all wrong, perhaps you never found anyone to discuss it with. I found over 200,000 search hits talking on commutativity on a sphere and the very first one says it is noncommutative in general. JSTOR: Random Walk on a Sphere and on a Riemannian Manifold - Oct 31 [ 317 ] RANDOM WALK ON A SPHERE AND ON A RIEMANNIAN MANIFOLD BY P. H. ROBERTS ...... It is not known whether any commutative manifold exists which is not ... links.jstor.org/sici?sici=0080-4614(19600331)252%3A1012%3C317%3ARWOASA %3E2.0.CO%3B2-5 - Similar pages - Note this Here, Dik, the first hit in the first paragraphs of the abstract, by P.H.Roberts and H.D. Ursell in a Royal Society journal of Math and Physics state: It is shown that commutability does not hold in general, but that it does hold in completely harmonic spaces and in some others. Dik, those guys don't often make statements like that if they were not sure that a sphere is not Commutative in general. The interesting question, Dik, is what has to be done to your circle in order to make it Commutative, in order to make it Commutative in general. And what has to be done is to cut away your circle so that there is no choice of direction nor is there a choice of major or minor arc. So when you remove 1/2 of your circle you you no longer have a circle but a curved Euclidean line segment and thus you can establish your Ring and Field. And that begs the question of a sphere. How to make it a Field and Ring, and I suspect the answer is cutting away 1/2 and what I suspect the minimum deletion is a hemisphere. So you have to cut away a hemisphere of a globe in order to make it a Field or Ring. In those 200,000 hits where they talk about establishing a Field or Ring over a sphere, what they are doing is merely cutting away alot of the sphere. And it is no different in the perspective that Earth is so large that we see land that looks mostly flat and so if you set up a Model on a small patch of a circle or sphere, then you can get away with the illusion that you have a Field or Ring on that small patch. And that is what yours has come to, Dik, a small patch of the circle that seems to obey your definitions, but not the circle in full. > Again, to repeat the definition, take a circle with radius R and midpoint > M. Define an origin on the circle, say O and a direction from O. Let's > have A and B two points on the circle. Let alpha be the angle between > the line from A to M and the line from O to M in the direction from O to A > according to the preferred direction. Similar for beta and B. (Also The Ring over Reals has no preferred direction. The Field over Reals has no preferred direction. So that should have warned you that your system is doomed to failure. > choose alpha and beta such that they are in [0, 2.pi).) To add A and B we > calculate: gamma = alpha + beta mod 2.pi. Take a line from O to M and go > in the preferred direction along an angle gamma; that is the point > C = A + B. For multiplication we calculate: > delta = fraction((A * B) / (4.pi^2)) * 2.pi, > where fraction yields the fractional part of a number that is in [0,1). > This delta gives D = A * B. Now, what about E = A - B? It would be the number that, when it is added > to B yields A. Or in angles, it angle would be the angle from the line > through B and M and the line through A and M in the preferred direction. > In arcs, it is the arc from B to A in the preferred direction. Similar > for B - A. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ If you deleted half of your circle, then you can get away with your above because there never is two different number answers to a multiplication or addition. If Earth were a hemisphere and not a sphere then the distance between Frankfurt and Zurich would be only 300 km and not a second answer of 39,700 km. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: 2000 Solutions manual My List of Solutions Manual contact me to : newbergh123yahoo.com newbergh123(at)yahoo.com ot to : mattosbw1@gmail.com mattosbw1(at)gmail.com If your wanted solutions manual ins't on this list, also can ask me if is available . These are some only. This same list of tites (not links) is available from : http://rapidshare.com/files/64945514/List of solutions manual.txt - Mechanics, Mechanical Engineering & Aerospace Engineering: Classical mechanics (2nd Ed., Goldstein) Classical Mechanics (Douglas Gregory) + original Ebook Advanced Dynamics (Greenwood) + original Ebook Advanced Engineering Dynamics (2nd Ed., Jerry Ginsberg) + Ebook Classical Dynamics (Jorge V. Jos.8e) + Ebook Impact Mechanics (W.J. 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Thompson) Galin) Object Oriented Software Development Using Java (2nd Ed., Xiaoping Jia) Introduction to the Team Software Process (Watts S. Humphrey) Software Project Management: A Real-World Guide to Success (Joel Henry) Software Engineering (8th Ed., Ian Sommerville) Object-Oriented Programming featuring Graphical Applications in Java (Michael J. Laszlo) Project-Based Software Engineering: An Object-Oriented Approach (Evelyn Stiller & Cathie LeBlanc) Engineering of Software, The: A Technical Guide for the Individual (Dick Hamlet & Joe Maybee) Concepts of Programming Languages (7th Ed., Robert W. Sebesta) Concepts of Programming Languages (8th Ed., Robert W. 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Ngan) contact me to : newbergh123@yahoo.com or mattosbw1@gmail.com Not exchange (for sale) === Subject: Re: The math symbol for Dirac delta function in multidimension. Let us say we have the following boxcar function boxcar(x;a,b) = > 1 if x in [a,b], > 0 otherwise. We define f(x,y) = boxcar(x;a,b) delta(y), where delta is the Dirac > delta function. Then I rotate the coordinate (x,y) -> (x',y'), I end > up with a new function f_rotated(x',y'). The expression of f_rotated(x',y') become complex. I'm wondering if > there is such common practice to denote such function Dirac delta > like function over a line segment, which does not depend on the > coordinate system. For example, I can use delta(r) to denote delta(x)delta(y). But the > notation delta(x)delta(y). does not apply to polar coordinate > system. But delta(r) does. Peng The Russian authors (I am looking at Gelfand-Shilov Vol I, p 222) use delta(P) where P denotes a point in a space of arbitrary dimensions. What is important of course is not delta by itself, but its effect as a continuous linear functional, and its Fourier/Laplace transforms. Anyway, delta(r) , with r=vector, is fine. Russians call those objects (after Sobolev) generalized functions, westerners (after Schwartz) call them distributions. The G-S book translator says that GF.s are more general than distributions but does not explain why. === Subject: Re: The math symbol for Dirac delta function in multidimension. Let us say we have the following boxcar function boxcar(x;a,b) = > 1 if x in [a,b], > 0 otherwise. We define f(x,y) = boxcar(x;a,b) delta(y), where delta is the Dirac > delta function. Then I rotate the coordinate (x,y) -> (x',y'), I end > up with a new function f_rotated(x',y'). The expression of f_rotated(x',y') become complex. I'm wondering if > there is such common practice to denote such function Dirac delta > like function over a line segment, which does not depend on the > coordinate system. For example, I can use delta(r) to denote delta(x)delta(y). But the > notation delta(x)delta(y). does not apply to polar coordinate > system. But delta(r) does. Peng Russian mathematicians (I am looking at Gelfand-Shilov Vol I, p.222) use delta(P) where P is a point in an arbitrary dimension space. So using delta(r), where r is a position vector, should be fine. Terminology twist: Russians call delta(P) a generalized function (after Sobolev), westerners call it a distribution (after Schwartz). The translator of the G-S book says in the preface that a generalized function is more general than a distribution but does not explain why. === Subject: Re: The math symbol for Dirac delta function in multidimension. <011120071334367966%anniel@nym.alias.net.invalid Let us say we have the following boxcar function boxcar(x;a,b) = > 1 if x in [a,b], > 0 otherwise. We define f(x,y) = boxcar(x;a,b) delta(y), where delta is the Dirac > delta function. Then I rotate the coordinate (x,y) -> (x',y'), I end > up with a new function f_rotated(x',y'). The expression of f_rotated(x',y') become complex. I'm wondering if > there is such common practice to denote such function Dirac delta > like function over a line segment, which does not depend on the > coordinate system. For example, I can use delta(r) to denote delta(x)delta(y). But the > notation delta(x)delta(y). does not apply to polar coordinate > system. But delta(r) does. Peng Mathematicians would probably use measures for this. I'm not a mathematician. Would you please elaborate more on this? Peng === Subject: Re: Triangle with more than 180 degrees- Nntp-Posting-Host: hera.cwi.nl ... > These are triangles with curved sides. I can draw triangles with > curved and not straight sides in a planar plane, which have a sum > of > angles exceeding 180 degrees, just as well. ' With that definition you cannot draw a triangle on a sphere. What > you > are doing is to apply Euclidean geometry here, while the question was > about Non-Euclidean geometry. ... > I often read about tangent space and tangent bundle and Riemann math > and i never grasped the essence up to now. > So my question: in hyperbolic geometry of 3D one has tangents, or not? Yes, once you have defined things. But when you want to think about hyperbolic spaces it is best to look at some of the possible models. There are four well known models for the hyperbolic plane. They are explained in . > Are in hyperbolic geometry these tangents straight, is a tangent plane > a plane or can it be sometimes a curved surface? They are not necessarily straight in Euclidean geometry either, although it is most common to think about tangents as being straight lines. But non straight tangents also occur, like the inscribed circle of a triangle, which is tangent to all three sides. > Is the tangent plane a hyperbolic plane or not? > Is the tangent plane a Lobachevsky plane or not? Lobachecsky and hyperbolic are the same. > Is in hyperbolic geometry every geodesic straight? What is straight? Look at the models. Two of the models use (in 2D) the inside of a circle. In one of the models the straight lines are the chords of the bounding circle, in the other the straight lines are the circles orthogonal to the bounding circle plus the diameters of the bounding circle. So, what *is* a geodesic in these models? > Think of hyperbolic paraboloid or the hyperboloid of one sheet, where > there are straight geodesics and not straight ones. Before you even can contemplate such things you need a distance function on the space. But look at where a distance function by Klein for his model is shown. Obviously circles do exist (although I do not particularly wish to visualise them). > PS Actually, are there spheres in hyperbolic geometry, as this seems > to be, when i look at Your answers? Once you have defined some distance function, yes, there are. But what a circle and whatever looks like depends upon the distance function used. But whatever, 3D hyperbolic space is not yet well-understood. -- 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: Epistemology 501: Angular Mechanics > Epistemology 501: Angular Mechanics > ~v~~ > (The following represents a loose collation of topics concerning > angular mechanics and related concepts) > Classical Angular Mechanics - Uhu. How about mental masturbation ad nauseum? -- The most powerful Usenet tool you have ever heard of. NewsMaestro v. 4.0.6 - Dictionary Update/Expert Mode has been released. * Significant improvement in symbol substitution mechanism for verb tense and plurals. * Expert mode. * Miscellaneous improvements and bug fixes. * Templates generator improvements. * Multi-job support. Note: In some previous releases some class files were missing. As a result, the program would not run. Sorry for the inconvenience. Web page: http://newsmaestro.sourceforge.net/ Download page: http://newsmaestro.sourceforge.net/Download_Information.htm Send any feedback, ideas, suggestions, test results to newsmaestroinfo at/ mail.ru. Your personal info will not be released and your privacy will be honored. === Subject: Re: Epistemology 501: Angular Mechanics > Well I provided an example where motion in a straight line results in > infinite L. Yes, but R is also infinite. Why is that a problem? >Huh! If you choose some infinitely distant axis (R=inf), then of >course angular momentum will be infinite. But at normal (finite) >values of R angular momentum will be also normal (finite). It is >impossible to turn a material point using an infinitely long lever! >That'd require an infinite moment of force (torque). Except that r=00 is motion in a straight line. I see no reason for raising exceptions here. Rectilinnear motion is not rotary, so Angular mechanics is not of great use here, though it creates no problems/contradictions with LM anyway. If you see contradiction, show it to us. > The point being that mathematics is not at liberty to describe certain > cases and not others and angular mechanics should apply to all cases. So it does!!! Set forth a situation where it doesn't work, if you can! > Okay. Then how is it you get off with enumerating certain facts like > dancers with their arms in various positions but I don't? Unlike you, I use facts to explain theory. You cite facts on their own, drawing no mathematically correct conclutions therefrom. What is it that follows from your examples that somehow contradicts with reality or with theory? > Except that linear Newtonian mechanics is finite and angular mechanics > isn't. 1. What's finite-ness??? 2. Angular mechanics if based exclusoively on Linear Newtonian Mechanics. All Angular formulas follow from Newton's laws. Angular mechanics is just an application of Linear mechanics to rotatory motion, intended only at faciliating calculations. It doesn't use/introduce any additional axioms/laws. Though I am not going to post here the derivation of Angular mechanics from linear (Newtonian), which is straightforward and can be found in mech. textbooks. I refer you to the books so that you can find fallacies (but I am sure there are no ones). > I never said angular mechanics didn't work. I just said L = r x p > yields infinite values for motion in a straight line. If it works, then what's all jabber about? I agree with you on this. No, it yields infinite values for infinitely distant origins. One infinity causes the other. In Linear mechnics infinite mass will similarly yield infinite momentum. Linear mechanics is infinite ;) An infinite radius is as abstract and impssible as infite mass. So the both kind of mechanics are equal. P.S.: I say: Linear mechanics needs remake because infinite mass yields infinite momentum, infinite force yields infinite acceleration and so on... P.P.S.: Please, are you going to finally provide a _ground_ for your swear contradiction with LM, because I have run out of words trying to explain it informally. So were's you proof? === Subject: Re: Epistemology 501: Angular Mechanics > Epistemology 501: Angular Mechanics > ~v~~ (The following represents a loose collation of topics concerning > angular mechanics and related concepts) Classical Angular Mechanics - The classical definition for angular momentum is L=r x p where r > represents a radius of rotation and p some linear momentum, mv. Linear momentum remains constant resulting straight line motion unless > some transverse acceleration, a, acts to produce rotation of p=mv. But the classical definition for angular momentum is defective since r > and a vary inversely. That is, for a given constant linear momentum > p=mv, angular momentum L=r x p varies from zero at r=0 to infinite > values at r=00. HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA! However, consider what this means: a value of transverse centripetal > acceleration a=0 which results in straight line motion causes infinite > values of angular momentum L=r x p=00 whereas infinite values of > transverse centripetal acceleration a=00 which results in motion at a > point causes zero values of angular momentum L=r x p=0. Clearly this situation is unsatisfactory in mechanical terms and we > must find some alternative definition for angular momentum with > consistent relations among all factors. > Oh, Lester, why don't you try preening at alt.discussions.look.at.me or sci.general.buffoons where there is a greater chance that when you say something it won't be *immediately* obvious how much of a goofball you are. If someone intends to impersonate a surgeon, they generally do it in front of people they think they can fool. They don't go to a surgical conference and attempt it. You are not only an imposter, you're an incompetent imposter, and one with poor judgement in selecting targets as well. PD === Subject: Re: Epistemology 501: Angular Mechanics > In other words I don't think we can understand the physics or expect > to understand the physics until we first understand the mathematics. Hmm, matter of viewpoint, it seems to me. I understand that since >Galileo, it is believed that math comes first and then physical >reality can be understood. I have a different viewpoint. Mathematics is just a language that >we (humans) have invented, and like all languages, it serves to >describe. Horses existed before we invented the word horse to name them >or cheval or Pferd or any other word from all other languages >to name them. Had the horse never existed or been imagined, no >one would have conceived of a word to name it. What I mean is that the object to be described came before >a word could be imagined to name it. My view is that it is the same for math. The inverse square law >of electrostatic interaction has been at play ever since electrons >positrons protons have come into being, presumably billions of >years even before our species came about and only lately invented >the mathematical word to describe it. When I use the word horse, the animal that comes to mind may >be quite different from the one that comes to yours or anybody >else's, with more or less characteristics depending on each >our personal experience with horses, and the word gets all >people who do understand the word conjure up his own subjective >experience of horses, not an objectively identical experience >of horses, unless all start discussing their view and come >to an agreement that they now momentarily come in sync about >it. My view is that it is the same with math. When you pose >such an equation as L=rp, L= r x p is not my definition; it's the classical definition of angular momentum. > what comes to my mind is the sum >I mentioned. To me, the motion of masses is the end result >of all the individual rv relations that a mass is subjected >to in the real physical universe. When a particular rv >relation becomes infinitesimal on account of distance, there >remains all other not infinitesimally active rv s still in >action. The straight L=rp equation to me is just a completely >skinned down hypothetical unit case that only marginally >matches reality. As you probably concluded, the r does not have be directed >normal to the trajectory, it can even be straight in >front or in the back. Can be spherically anywhere about >the mass and the force will be acting the same. Even in this hypothetical case, if r reaches infinity, >then associated velocity factor would reach zero, which >means no motion, thus no momentum. A perfectly acceptable >result in the ideal mathematical representation, but an >impossibility in physical reality that I long ago discarded >as not representing reality. I use the law with physically >possible situations. I don't bother with the impossible. We individually decide how we bend the ideal unbriddled >mathematical word to fit objective reality, for our >own purposes. Well the purpose of epistemology is to define the framework of objective knowledge in terms of logic. The difficulty for math are implications of its formulation with respect to radius of rotation. Infinite radii of rotation do not require zero velocity. Any motion at any finite velocity in a straight line implies an infinite radius of rotation. Just because the radius of rotation is infinite doesn't imply we're looking at the velocity from the origin of rotation or from infinite distance. We could just as easily be looking at the velocity close up and infer the infinite radius of rotation from the properties of the straight line. In any event it isn't necessary to infer an actual infinite radius of rotation from straight lines. All we have to do is note that according to the classical definition, angular momentum would increase from zero at zero radius to unlimited values at unlimited radii, whatever finite nonzero values of v may apply. And the same applies to every form of language used in the context of science whether applied to horses or inverse square laws of motion. It just doesn't matter if the object of the language exists, has ever existed, hasn't existed, or could never have existed. The language and the laws of logic still govern use and application of the language and epistemology explicates those laws including mathematics and science. ~v~~ === Subject: Re: Epistemology 501: Angular Mechanics <7s1ki3da9fi9osu51m9igpd0fee0mqk3dr@4ax.com> In other words I don't think we can understand the physics or expect > to understand the physics until we first understand the mathematics. Hmm, matter of viewpoint, it seems to me. I understand that since >Galileo, it is believed that math comes first and then physical >reality can be understood. I have a different viewpoint. Mathematics is just a language that >we (humans) have invented, and like all languages, it serves to >describe. Horses existed before we invented the word horse to name them >or cheval or Pferd or any other word from all other languages >to name them. Had the horse never existed or been imagined, no >one would have conceived of a word to name it. What I mean is that the object to be described came before >a word could be imagined to name it. My view is that it is the same for math. The inverse square law >of electrostatic interaction has been at play ever since electrons >positrons protons have come into being, presumably billions of >years even before our species came about and only lately invented >the mathematical word to describe it. When I use the word horse, the animal that comes to mind may >be quite different from the one that comes to yours or anybody >else's, with more or less characteristics depending on each >our personal experience with horses, and the word gets all >people who do understand the word conjure up his own subjective >experience of horses, not an objectively identical experience >of horses, unless all start discussing their view and come >to an agreement that they now momentarily come in sync about >it. My view is that it is the same with math. When you pose >such an equation as L=rp, L= r x p is not my definition; it's the classical definition of > angular momentum. what comes to my mind is the sum >I mentioned. To me, the motion of masses is the end result >of all the individual rv relations that a mass is subjected >to in the real physical universe. When a particular rv >relation becomes infinitesimal on account of distance, there >remains all other not infinitesimally active rv s still in >action. The straight L=rp equation to me is just a completely >skinned down hypothetical unit case that only marginally >matches reality. As you probably concluded, the r does not have be directed >normal to the trajectory, it can even be straight in >front or in the back. Can be spherically anywhere about >the mass and the force will be acting the same. Even in this hypothetical case, if r reaches infinity, >then associated velocity factor would reach zero, which >means no motion, thus no momentum. A perfectly acceptable >result in the ideal mathematical representation, but an >impossibility in physical reality that I long ago discarded >as not representing reality. I use the law with physically >possible situations. I don't bother with the impossible. We individually decide how we bend the ideal unbriddled >mathematical word to fit objective reality, for our >own purposes. Well the purpose of epistemology is to define the framework of > objective knowledge in terms of logic. But also of requestioning past conclusions in light of more data gathered after such conclusion is reached, so the conclusion can be more finely tuned if need be. Mandatory it seems to me if epistemology is to completely fulfill its ultimate purpose, which is ultimately to define a universal method for knowing the truth (presumably about objective reality). > The difficulty for math are > implications of its formulation with respect to radius of rotation. > Infinite radii of rotation do not require zero velocity. Barring accumulated momentum from past or other current interactions with force from other spatial origin, it seems to me that yes, it does for the incremental velocity factor due to that particular r, since classical mechanics deals by definition with force acting as the inverse cube of the distance. > Any motion at > any finite velocity in a straight line implies an infinite radius of > rotation. Or that the force being applied was acting axially in line with the direction of motion of the mass maybe. In a universe where there existed only one force interacting from that distance with the mass considered. > Just because the radius of rotation is infinite doesn't > imply we're looking at the velocity from the origin of rotation or > from infinite distance. We could just as easily be looking at the > velocity close up and infer the infinite radius of rotation from the > properties of the straight line. I see what you mean, but how can analyzing how to optimize math in a fictional ideal universe help optimizing a universal method of knowing objective truth (that can concern only the really existing objectively physical reality ? > In any event it isn't necessary to infer an actual infinite radius of > rotation from straight lines. All we have to do is note that according > to the classical definition, angular momentum would increase from zero > at zero radius to unlimited values at unlimited radii, whatever finite > nonzero values of v may apply. If a force is acting on the mass involved, in classical mechanics, it seems to me that the velocity at r tending towards zero would mathematically turn out tending to infinity since the product rv is constant by definition, no ? > And the same applies to every form of language used in the context of > science whether applied to horses or inverse square laws of motion. It > just doesn't matter if the object of the language exists, has ever > existed, hasn't existed, or could never have existed. The language and > the laws of logic still govern use and application of the language and > epistemology explicates those laws including mathematics and science. As it should. There remains to agree on a common definition of epistemology. Andr? Michaud === Subject: Re: Is this checkmate known? > Why not concentrate on a real game such as Reality rather than chess, > which is sextillions of times more simple than it. Really? Where can I download this Reality game from, then? --- Christopher Heckman === Subject: Re: Is this checkmate known? it successfully; of course the first version I used was much longer, > but this is the essential. I give it if black falls victim; white can > also, of course, given that the symmetry of the board. The first 5 moves are the symmetrical Giuoco Pianissimo. Then 6 B-KN5 > O-O? 7 N-Q5 B-K3? - black should have done P-KR3, driving back the > bishop. Other 7th moves are no better, in particular B-KN5 is met by > white's P-KR3 later, moving the bishop back and giving the same > situation. Now the mate goes 8 NxN ch PxN 9 B-KR6 KR-K 10 N-KR4 > anything 11 Q-KB3 anything 12 Q-KN3 ch K-KR (forced) 13 Q-KN7 mate. > Or, alternately, if black had played BxB, 13 B-KN7 ch K-KN (forced) 14 > N-KB5 anything 15 N-KR6 mate is perhaps more elegant in giving mate > with a knight. Black can avoid this mate only by sacrificing his queen > (several ways). How about 10...N-K2 and 11...N-N3 ? Tonyy: tony@mountifield.org - http://tony.mountifield.org Fails to 11. BxB, and if 11. ... PxB 12. Q-N4+ N-N3 12. NxN. After 10. B-KN5 (the Canal Variation), 10. ... 0-0 has long been know as a blunder. The old main line 10. ... P-KR3 11. BxN QxB 12. N-Q5 gives White a little advantage. Korchnoi gave up the Giuoco Pianissimo because of 10 ... N-QR4. === Subject: Re: Is this checkmate known? After 6. B-KN5 (the Canal Variation), 6. ... 0-0 has long been known > as a blunder. The old main line 6. ... P-KR3 7. BxN QxB 8. N-Q5 > gives White a little advantage. Korchnoi gave up the Giuoco Pianissimo > because of 6. ... N-QR4. I corrected your move numbers here, careful. Andrew Usher === Subject: Re: Is this checkmate known? How about 10...N-K2 and 11...N-N3 ? > Tonyy: t...@mountifield.org -http://tony.mountifield.org Fails to 11. BxB, and if 11. ... PxB 12. Q-N4+ N-N3 13. NxN. 13. ... Q-Q2 14. N-KR4 dis ch K-KR, and white's attack is stopped but > he's > up a piece. Yes, I did not think last night before accepting Tony's > proposed > defence, I should surely have seen it. 13. ... Qd7 14. Nf8+ wins the Black Queen. After 6. B-KN5 (the Canal Variation), 6. ... 0-0 has long been known > as a blunder. The old main line 6. ... P-KR3 7. BxN QxB 8. N-Q5 > gives White a little advantage. Korchnoi gave up the Giuoco Pianissimo > because of 6. ... N-QR4. I corrected your move numbers here, careful. Andrew Usher I must echo Ray Gordon's comment that this would be easier if you would learn algebraic notation. As I recall, Edward Lasker analyzed a very similar line arising from the Four Knights' Game in _Chess Strategy_. It's available on Amazon for under $5, and it's even in descriptive notation. Read a few books like that one, and then you'll have the basis for discussion. === Subject: fun puzzle Let the line AB join points A(a,0) and B(0,b), on the x, y axes respectively, with a and b both positive. Let the line pass through the point (8,27). Prove that the minimum distance between A and B is 13.sqrt(13). Do not use calculus in your proof. Hint: what can you say about 8 and 27. === Subject: Re: Third dimension... > If a dot is to one dimension and > a line is to two dimension, > whats to the third dimension? > Jay Bala. Since you start if you might want to look at that first! In what > sense if a dot is to one dimension and a line is to two dimensions > true? Presumably, Jay intended to ask about an analogy: What is to the third dimension as a point is to the first dimension and as > a line is to the second dimension? As I noted in my original response, the answer should be plain to see. David http://kmr.nada.kth.se/files/gok/firstproto/index.php?gallery=Fenomen_och _Begrepp/Pythagoras/Misc&image=Number_related_to_form.jpg Can you get all that into your browser? Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, http://www.geocities.com/forwardintothepast/ Don't be a square or a blockhead; see: http://bfi.org/node/574 http://library.wolfram.com/infocenter/search/?search_results=1;search_per son_id=607 === Subject: Re: Third dimension... <18101346.1193840391486.JavaMail.jakarta@nitrogen.mathforum.org> <20071031102807.843$Gp@newsreader.com> <200710311344138930-kirakun@earthlinknet Fourth? I believe is a curved surface of thinckness zero. How curved, no one is asking tho... but the curvature is a function of of gravity. Higher the graviity, sharper the curve of the plane. And the free-vector, a function of time or anchored by time. Jay Bala. === Subject: Re: Third dimension... <18101346.1193840391486.JavaMail.jakarta@nitrogen.mathforum.org> <20071031102807.843$Gp@newsreader.com> <200710311344138930-kirakun@earthlinknet > On 2007-10-31 12:52:01 -0400, jay1b...@aol.com said: > What is to the third dimension as a point is to the first dimension and as > a line is to the second dimension? > As I noted in my original response, the answer should be plain to see. > David > Well put. Now ... borrowing that... > What is to the fourth dimension > as a point is to the first dimension, > as a line is to the second dimension and > as a plain is to the third dimension? > Jay Bala. The answer is an affine linear subspace of codimension 1 a.k.a. a hyperplane. This answer also works for all your other analogies in this pattern too. -- -kira This, I thought would be the 5th. but a funtion of time. Yes, time is > always one of the aditional dimensions of first, second, third, etc. > Let me hear some thoughts on this. Fourth? I believe is a curved surface of thinckness zero. Jay Bala. === === Subject: Re: 1^2 =3, Discovered > Square of one equals three.-Aiya-Oba(Poet/Philosopher) Thus, 1^2 = 2(3 - 1)/2 + 1 > = 1^2 = (2 x 1) + 1 > = 1^2 = 2 + 1 > = 1^2 = 3 Where every natural number ( n ) is : > n^2 = 2(n^2 - n)/2 + n > Q E D. Sorry, but I'm not impressed. I've been using boolean arithmetic for many years now, so I've known for quite a while that 1 xor 2 = 3. === Subject: Re: 1^2 =3, Discovered <8906053.1193940526660.JavaMail.jakarta@nitrogen.mathforum.org> On Nov 1, 11:08 am, Anthony A. Aiya-Oba Natural Number Isosceles Triple Theorem. > According to this new theorem: > Every natural number is a base of an isosceles triplet, whose total sum is square of the base. Hence, > 1 = (1 + 1 + 1) = 3 = 1^2 2 = (1 + 1 + 2) = 4 = 2^2 3 = (3 + 3 + 3) = 9 = 3^2 . . . Such that, > n^2 = 2(n^2 - n)/2 + n Where n (base), can be any possible natural number, and > 2(n^2 - n)/2, is sum of the other two equal sides. > -Aiya-Oba. Crank. === Subject: rotation of axes formula Having arrived at the formula: x' = x*cos t + y*sin t y' = y*cos t - x*sin t I can supposedly derive the formula in terms of x', y' for x and y Such that I should get: x = x'*cos t - y'*sin t y = x'*sin t + y'*cos t What I did was try and find the inverse of the matrix: [ cos t sin t | 1 0ncos t -sin t | 0 1] The | represents an augmented matrix. The result I got was: [1 0 | 3/(2sin t) 1/(2cos t)n0 1 | 1/(2sin t) -1/(2sin t)] Which is way off. Did I attack this problem in trying to solve for x and y incorrectly? -- conrad === Subject: Re: rotation of axes formula > Having arrived at the formula: > x' = x*cos t + y*sin t > y' = y*cos t - x*sin t > What I did was try and find > the inverse of the matrix: > [ cos t sin t | 1 0ncos t -sin t | 0 1] You started with the wrong matrix. Look at the equation y' = y*cos t - x*sin t again very carefully. I think you should have better luck starting with matrix [ cos t sin t | 1 0n-sin t cos t | 0 1] instead. === Subject: Quick question about vector products I was wondering, is: a x (b x c ) equal to c x ( b x a ) ? === Subject: Re: Quick question about vector products > I was wondering, is: > a x (b x c ) equal to c x ( b x a ) ? > Google for triple vector product. === Subject: Re: Quick question about vector products I was wondering, is: > a x (b x c ) equal to c x ( b x a ) ? No. Try a=b=(1,0,0), c=(0,1,0). === Subject: Re: Quick question about vector products I was wondering, is: > a x (b x c ) equal to c x ( b x a ) ? No. Try a=b=(1,0,0), c=(0,1,0). Actually, I meant to write: a x ( b x c ) equal to ( c x b ) x a Also, because a = c, it boils down to: c x ( b x c ) and (c x b ) x c where b is not equal to c. === Subject: Re: Quick question about vector products > Actually, I meant to write: a x ( b x c ) equal to ( c x b ) x a Also, because a = c, it boils down to: > c x ( b x c ) and (c x b ) x c where b is not equal to c. I've verified this numerically for c = (1, 0, 0) and b = (0, 1, 0), but this not proof that it is true for all vectors. Here's a sketch proof I just dreamed up, using the anti-commutative property of vectors i.e. u x v = -v x u. c x ( b x c ) = -( b x c ) x c = -(-c x b) x c Here's the part I'm not so sure about: -(-c x b) x c = -1.-1(c x b) x c = (c x b) x c Da? === Subject: Re: Quick question about vector products > Actually, I meant to write: a x ( b x c ) equal to ( c x b ) x a ' Also, because a = c, it boils down to: c x ( b x c ) and (c x b ) > x c ' where b is not equal to c. I've verified this numerically for c = (1, 0, 0) and b = (0, 1, 0), > but this not proof that it is true for all vectors. Here's a sketch proof I just dreamed up, using the anti-commutative > property of vectors i.e. u x v = -v x u. c x ( b x c ) = -( b x c ) x c = -(-c x b) x c Here's the part I'm not so sure about: -(-c x b) x c = -1.-1(c x b) x c = (c x b) x c Da? What's the problem? Scalars multiply through trivially. This fact can be verified by noting that the formulas for the cross product are homogeneous of degree 1 in each of the factors, and that multiplication of a vector by a scalar is accomplished by multiplying each component by that scalar. Or maybe you are really asking whether (-1)(-1) = 1? The anti-commutative property is sufficient to answer your (corrected) general question: Is a x ( b x c ) equal to ( c x b ) x a? Note that (c x b) = - (b x c). So we find (c x b) x a = - (b x c) x a and further (b x c) x a = - a x (b x c) so we find (c x b) x a = - (- a x (b x c)) (c x b) x a = a x (b x c). Dale === Subject: Linear algebra with ...T Hello sir~ A, B are 4x4 matrix. Let (A^3)B - 2AB + 3E = 0. T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) Find the dimension of T(R^4). ------------------------------------------------- Sorry. I don't know well. so, I need your advice. === Subject: Re: Linear algebra with ...T > A, B are 4x4 matrix. Let (A^3)B - 2AB + 3E = 0. > What's E about? > T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) > T(B) = 3E. > Find the dimension of T(R^4). > Depends upon A. For example, 0 if A = (sqr 2)I, 4 if A = I. === Subject: Re: Linear algebra with ...T T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) T(B) = 3E. Find the dimension of T(R^4). Depends upon A. For example, > 0 if A = (sqr 2)I, 4 if A = I. If B is nonsingular matrix, I proved that T(R^4) = 4. === Subject: Re: Linear algebra with ...T On 2007-11-01 22:53:48 -0400, mina_world said: > Hello sir~ A, B are 4x4 matrix. Let (A^3)B - 2AB + 3E = 0. T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) Find the dimension of T(R^4). ------------------------------------------------- > Sorry. I don't know well. > so, I need your advice. The equation A^3 B - 2AB + 3 = 0 is equivalent to 3 = B(2A-A^3). So, for any v in R^4, 3v = B T(v). So, BT is full rank. Therefore... -- -kira === Subject: Re: Linear algebra with ...T <2007110123481716807-kirakun@earthlinknet On 2007-11-01 22:53:48 -0400, mina_world said: Hello sir~ A, B are 4x4 matrix. Let (A^3)B - 2AB + 3E = 0. T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) Find the dimension of T(R^4). ------------------------------------------------- > Sorry. I don't know well. > so, I need your advice. The equation A^3 B - 2AB + 3 = 0 is equivalent to > 3 = B(2A-A^3). > So, for any v in R^4, > 3v = B T(v). > So, BT is full rank. Therefore... Maybe, (2A - A^3)B = 3. I think that this problem need a condition with there is B^-1. so, (2A - A^3) = 3.B^-1 so, For any v in R^4, T(v) = (2A - A^3).v = 3.B^-1 T(1,0,0,0) is 1-column of 3.B^-1. T(0,1,0,0) is 2-column of 3.B^-1. T(0,0,1,0) is 3-column of 3.B^-1. T(0,0,0,1) is 4-column of 3.B^-1. Since |B| =/= 0, (r(B) =4) dim = 4. so, T(R^4) = imT = = 4. === Subject: Re: Linear algebra with ...T On 2007-11-02 00:25:07 -0400, mina_world@hanmail.net said: > On 2007-11-01 22:53:48 -0400, mina_world said: >Hello sir~ >A, B are 4x4 matrix. >Let (A^3)B - 2AB + 3E = 0. >T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) >Find the dimension of T(R^4). >------------------------------------------------- > Sorry. I don't know well. > so, I need your advice. ' The equation A^3 B - 2AB + 3 = 0 is equivalent to > 3 = B(2A-A^3). > So, for any v in R^4, > 3v = B T(v). > So, BT is full rank. ' Therefore... Maybe, (2A - A^3)B = 3. I think that this problem need a condition with there is B^-1. No need for B^-1. Suppose V := T(R^4) with dim V < 4. What can you say about dim B(V)? But BT is full rank. -- -kira === Subject: Re: Linear algebra with ...T > On 2007-11-02 00:25:07 -0400, mina_world@hanmail.net said: On 2007-11-01 22:53:48 -0400, mina_world Hello sir~ > A, B are 4x4 matrix. > Let (A^3)B - 2AB + 3E = 0. > T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) > Find the dimension of T(R^4). > ------------------------------------------------- > Sorry. I don't know well. > so, I need your advice. The equation A^3 B - 2AB + 3 = 0 is equivalent to > 3 = B(2A-A^3). > So, for any v in R^4, > 3v = B T(v). > So, BT is full rank. Therefore... > Maybe, (2A - A^3)B = 3. > I think that this problem need a condition with there is B^-1. No need for B^-1. Suppose V := T(R^4) with dim V < 4. What can you say about dim B(V)? But > BT is full rank. If 3E = B(2A-A^3), I can understand your solution. But 3E = (2A - A^3)B. If 3E = B(2A-A^3), then 3E.v = B.T(v) for any v in R^4. Let L = [B]. Namely, B is the matrix of linear transformation L for standard basis. so, 3E.v = (LoT)(v) (LoT)(1,0,0,0) = (3,0,0,0) (LoT)(0,1,0,0) = (0,3,0,0) (LoT)(0,0,1,0) = (0,0,3,0) (LoT)(0,0,0,1) = (0,0,0,3) so, im(LoT) = <(3,0,0,0), (0,3,0,0), (0,0,3,0), (0,0,0,3)> so, r(LoT) = 4. Since r(LoT) <= r(L) and r(LoT) <= r(T), 4 <= r(T). so, r(T) = 4. But 3E =/= B(2A-A^3). === Subject: Re: Linear algebra with ...T On 2007-11-02 01:19:56 -0400, mina_world said: ' On 2007-11-02 00:25:07 -0400, mina_world@hanmail.net said: ' On 2007-11-01 22:53:48 -0400, mina_world said: ' Hello sir~ ' A, B are 4x4 matrix. ' Let (A^3)B - 2AB + 3E = 0. ' T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) ' Find the dimension of T(R^4). ' ------------------------------------------------- > Sorry. I don't know well. > so, I need your advice. ' The equation A^3 B - 2AB + 3 = 0 is equivalent to > 3 = B(2A-A^3). > So, for any v in R^4, > 3v = B T(v). > So, BT is full rank. ' Therefore... Maybe, (2A - A^3)B = 3. I think that this problem need a condition with there is B^-1. ' No need for B^-1. ' Suppose V := T(R^4) with dim V < 4. What can you say about dim B(V)? But > BT is full rank. If 3E = B(2A-A^3), I can understand your solution. > But 3E = (2A - A^3)B. If 3E = B(2A-A^3), then 3E.v = B.T(v) for any v in R^4. Let L = [B]. > Namely, B is the matrix of linear transformation L for standard basis. so, 3E.v = (LoT)(v) > (LoT)(1,0,0,0) = (3,0,0,0) > (LoT)(0,1,0,0) = (0,3,0,0) > (LoT)(0,0,1,0) = (0,0,3,0) > (LoT)(0,0,0,1) = (0,0,0,3) so, im(LoT) = <(3,0,0,0), (0,3,0,0), (0,0,3,0), (0,0,0,3) so, r(LoT) = 4. Since r(LoT) <= r(L) and r(LoT) <= r(T), > 4 <= r(T). > so, r(T) = 4. But 3E =/= B(2A-A^3). Yes, you are right. I made a mistake in the computation. Sorry about that. It should've been (1) 3 = (2A - A^3) B = T B. But we can also argue the same way too. Assume dim T(R^4) < 4. Then no matter what B is, we must have B(R^4) be a subspace of R^4. So, dim T(B(R^4)) < 4, which implies TB is not full rank. However, (1) says TB is full rank. Contradiction. I hope I didn't make a mistake this time. Lately, my brain has been making the most trivials of mistakes. -- -kira === Subject: Re: Linear algebra with ...T <2007110123481716807-kirakun@earthlinknet> <2007110201325411272-kirakun@earthlinknet On 2007-11-02 01:19:56 -0400, mina_world said: > On 2007-11-02 00:25:07 -0400, mina_wo...@hanmail.net said: > On 2007-11-01 22:53:48 -0400, mina_world said: > Hello sir~ > A, B are 4x4 matrix. > Let (A^3)B - 2AB + 3E = 0. > T : R^4 -> R^4, T(X) = (2A - A^3)X (X in R^4) > Find the dimension of T(R^4). > ------------------------------------------------- > Sorry. I don't know well. > so, I need your advice. > The equation A^3 B - 2AB + 3 = 0 is equivalent to > 3 = B(2A-A^3). > So, for any v in R^4, > 3v = B T(v). > So, BT is full rank. > Therefore... Maybe, (2A - A^3)B = 3. I think that this problem need a condition with there is B^-1. > No need for B^-1. > Suppose V := T(R^4) with dim V < 4. What can you say about dim B(V)? But > BT is full rank. If 3E = B(2A-A^3), I can understand your solution. > But 3E = (2A - A^3)B. If 3E = B(2A-A^3), then 3E.v = B.T(v) for any v in R^4. Let L = [B]. > Namely, B is the matrix of linear transformation L for standard basis. so, 3E.v = (LoT)(v) > (LoT)(1,0,0,0) = (3,0,0,0) > (LoT)(0,1,0,0) = (0,3,0,0) > (LoT)(0,0,1,0) = (0,0,3,0) > (LoT)(0,0,0,1) = (0,0,0,3) so, im(LoT) = <(3,0,0,0), (0,3,0,0), (0,0,3,0), (0,0,0,3) so, r(LoT) = 4. Since r(LoT) <= r(L) and r(LoT) <= r(T), > 4 <= r(T). > so, r(T) = 4. But 3E =/= B(2A-A^3). Yes, you are right. I made a mistake in the computation. Sorry about > that. It should've been > (1) 3 = (2A - A^3) B = T B. But we can also argue the same way too. Assume dim T(R^4) < 4. Then no matter what B is, we must have B(R^4) > be a subspace of R^4. So, dim T(B(R^4)) < 4, which implies TB is not > full rank. However, (1) says TB is full rank. Contradiction. I hope I didn't make a mistake this time. Lately, my brain has been > making the most trivials of mistakes. Even if 3 = T B is unbalanced expression, I can understand it as previous post. === Subject: Re: Problem related to a linear regression > Hello The problem I'm facing may arise from trying to do a regression of the > same one parameter set where x is the esimated value and y the actual > observed value. Let me explain step by step what I'm doing; may be you > can recognise from this if the way I proceed is (a) correct and (b) > the direct way, i.e. not performing a detour for obtaining the correct > result. What I'm aiming at is to calculate the actual measured value (Y) from > the results of a fitting function (Xf) calculated with a set of other > parameters in the form: > Y=A+B*Xf > with A and B obtained by regression analysis. In order to obtain the values for A and B a regression has been > calculated with the Xf and Y values by maximising the correlation > (R**2). Depending on the type of function and the parameters used (for > calculating Xf) the resulting points (Xf, Y) are 'somewhere' along the > x-axis around the regression line having a certain slope. Assuming that this regression calculation would already be the final > result, the X-values were scaled with A and B of the regression line. > This shifted the data points (Xf and Y) around the regrression line > which after this transformation corresponded to Y=Xf. It was thought > that this regression line (Y=Xf) would correspond to the final result. > However, looking at the graph suggests that this line doesn't really > correspond to the 'best fit' and that it represents only to an > intermediate result; I'm kind of confused. Initially you found the values of A and B that give the best fit of the line y = A + B*x to the data points (Xf, Y)? Then you applied the transformation Xf' = A + B*Xf, where Xf' is the transformed X-value? (This is what I assume you mean by the X-values were scaled with A and B: you're transforming x so as to take the line y = A + B*x to the line y = x.) And now you're wondering if this choice of A and B actually gives the best fit of the transformed data points (Xf', Y) to the line y = x? Doesn't it amount to exactly the same thing? In the first case you're choosing A and B so as to minimise the sum of (Y - (A + B*Xf))^2, and in the second case you're choosing A and B so as to minimise the sum of (Y - Xf')^2, where Xf' = A + B*Xf? Maybe I misunderstood it... > therefore A+B*Xf is not yet Y but an intermediate > X (Xi): > Xi=A+B*Xf > requiering an additional regression (minimising the sum of dXf**2) for > obtaining Y: > Y=C+D*Xi > This regression line looks now more to correspond to the best fit > between Y (observed values) and Xf (results of fitting function). Question: Is it really necessary to perform 2 regression for this > particular problem, or am I doing a detour for obtaining the correct > result? === Subject: separation process principle I need this one too. plz plz send me! === Subject: GR9768 problems 23, 29, 32, 36, 37, 50, 63 (last set of questions before the test) Would you guys mind giving me your thoughts regarding the following problems? If you prefer to read the actual questions, the pdf for this test is located at [url]http://www.ets.org/Media/Tests/GRE/pdf/ 23. In the euclidean plane, point A is on a circle centered at point O, and O is on a circle centered at A. The circles intersect at points B and C. What is the measure of the angle BAC? ans. 120 degrees 29. Assume that p is a polynomial function on the set of real numbers. If p(0) = p(2) = 3 and p'(0) = p'(2) = -1, then integral(0 to 2) {xp''(x)dx =? ans. -2 32. When 20 children in a classroom line up for lunch, Pat insists on being somewhere ahead of Lynn. If Pat's demand is to be satisfied, in how many ways can the children line up? ans. 20!/2 I know the answer is not 20! 36. For each real number x, let mu(x) be the mean of the numbers 4, 5, 7, 9 and x; and let n(x) be the median of these five numbers. For how many values of x is mu(x) = to n(x)? ans. 3 mu(x) = (25 + x)/5 37. Sum(k=1 to infinity) k^2/k! ans. 2e I used the ratio test and came up with zero. Any thoughts? 50. How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]? ans. four 63. At how many points in the xy-plane do the graphs of y = x^12 and y = 2^x intersect? ans. Three I tried to solve this and get to the point, 12 log x = x log 2. === Subject: Re: GR9768 problems 23, 29, 32, 36, 37, 50, 63 (last set of questions before the test) > 23. > In the euclidean plane, point A is on a circle centered at point O, > and O is on a circle centered at A. The circles intersect at points B > and C. What is the measure of the angle BAC? > ans. 120 degrees > 60 deg. > 29. > Assume that p is a polynomial function on the set of real numbers. If > p(0) = p(2) = 3 and p'(0) = p'(2) = -1, then integral(0 to 2) > {xp''(x)dx =? > ans. -2 > integral(0,2) p'(x) dx = p(2) - p(0) = 0 integral(0,2) p(x) dx = p'(2) - p'(0) = -1 p(x) can be any function in C. === Subject: Re: GR9768 problems 23, 29, 32, 36, 37, 50, 63 (last set of questions before the test) > Would you guys mind giving me your thoughts regarding the following > problems? If you prefer to read the actual questions, the pdf for > this test is located at [url]http://www.ets.org/Media/Tests/GRE/pdf/ 23. > In the euclidean plane, point A is on a circle centered at point O, > and O is on a circle centered at A. The circles intersect at points B > and C. What is the measure of the angle BAC? > ans. 120 degrees Length OA = OB = OC since A, B and C all lie on a circle centred at O. Similarly OA = BA = AC. Therefore OBA and OCA are equilateral triangles with common side OA, so the angle BAC is 120 degrees. 29. > Assume that p is a polynomial function on the set of real numbers. If > p(0) = p(2) = 3 and p'(0) = p'(2) = -1, then integral(0 to 2) > {xp''(x)dx =? > ans. -2 Use integration by parts to write the integral as [x p'(x)]_0^2 - integral_0^2 (p'(x))dx = [x p'(x)]_0^2 - [p(x)]_0^2 = 2 p'(2) - p(2) + p(0) = -2. 32. > When 20 children in a classroom line up for lunch, Pat insists on > being somewhere ahead of Lynn. If Pat's demand is to be satisfied, in > how many ways can the children line up? > ans. 20!/2 I know the answer is not 20! There are 20! ways of lining up the children in total (20 ways to choose which child is first in line, then 19 ways to choose which child is second, and so on). For each ordering in which Pat is ahead of Lynn there is a corresponding different ordering in which Lynn is ahead of Pat, found by switching the two; this switch is reversible and so is a bijection between the orderings in which Pat is ahead of Lynn and those in which Lynn is ahead of Pat. Therefore there are an equal number of orderings of either type, and since every ordering is one type or the other it follows that half of the total number of orderings has Pat ahead of Lynn. 36. > For each real number x, let mu(x) be the mean of the numbers 4, 5, 7, > 9 and x; and let n(x) be the median of these five numbers. For how > many values of x is mu(x) = to n(x)? > ans. 3 mu(x) = (25 + x)/5 There are three (not necessarily distinct) possibilities to consider: if x <= 5 then n(x) = 5, if 5 <= x <= 7 then n(x) = x, and if 7 <= x then n(x) = 7. For each of these possibilities we must solve the resulting equation mu(x) = n(x) and check whether the solutions satisfy the corresponding bounds on x (that is, if for example we solve mu(x) = 5 and find x > 5, then this is not really a solution since n(x) would not be 5 in this case), and whether they are distinct. In the first case, 5 + x/5 = 5, we have x = 0; in the second, 5 + x/5 = x so x = 25/4; in the third, 5 + x/5 = 7 so x = 10. These are all genuine solutions and are distinct. 37. > Sum(k=1 to infinity) k^2/k! > ans. 2e I used the ratio test and came up with zero. Any thoughts? First note that, for k >= 1, k^2/k! = k/(k-1)!. By a change of variable we may write sum(k=1 to oo) k/(k-1)! = sum(k=0 to oo) (k+1)/ k! = sum(k=0 to oo) 1/k! + sum(k=0 to oo) k/k!. The first term is e. For the second term note that for the case k=0 the summand is zero, so we can write this as sum(k=1 to oo) k/k! = sum(k=1 to oo) 1/(k-1)! = sum(k=0 to oo) 1/k!, where the last equality follows from another change of variable. So this term is also e. 50. > How many continuous real-valued functions f are there with domain > [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]? > ans. four at some point x > 0 that f(x) = +x, then by the intermediate value property it follows that f(x) = +x for all positive x, and similarly for f(x) = -x or x < 0. So the origin is the only point where f(x) can change direction. Therefore the possibilities are that f = x, f = - x, f = |x| or f = -|x|. 63. > At how many points in the xy-plane do the graphs of y = x^12 and y = > 2^x intersect? > ans. Three I tried to solve this and get to the point, 12 log x = x log 2. I think I have read once that equations of the type above have no solution in terms of standard functions, so trying to find actual points where they cross is unlikely to be fruitful. Not sure how one would answer this question with anything approaching rigour I'm afraid, but intuitively one can see that the answer will either be one or three: for x <= 0, x^12 is monotonically decreasing and 2^x monotonically increasing, and 0^12 < 2^0 so they intersect once to the left of the y-axis. It is easy to see that the functions must cross at least once for positive x since, e.g. x^12 > 2^x for x = 2. Exponential functions always overtake polynomial ones for sufficiently large values of x, i.e. there will be an x above which 2^x is always greater than x^12, meaning that the functions must cross again. If you want something less handwavy I would guess that there is an argument involving monotonicity of derivatives. === Subject: Re: Made Up Probability Question Interestingly, it turns out that your solution is a special case of Robert Israel's solution, in which his p = 1/2. === Subject: Re: Made Up Probability Question This solution is a generalization of the solution given by matt271829-news@yahoo.co.uk below. Setting p = 1/2 gives that solution, with m = a (n-1) & k = b (n-1). === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - my precondition is that key is uniform and c is fixed. > under that condition M is uniform. Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong Assume the cyphertext is fixed. 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key 2. The distribution of the key is > determined by the distribution of the plaintext. 3. The distribution of the plaintext is unknown. 4. The distribution of the key is unknown. 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. - William Hughes- - - - you are sophistic,it like a blind man,he does not see any thing, if he is told something existing, he just say it is contradiction to what he feels, and deny it. whatmore, your unknown is weak. === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - > my precondition is that key is uniform and c is fixed. > under that condition M is uniform. Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong Assume the cyphertext is fixed. 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key 2. The distribution of the key is > determined by the distribution of the plaintext. 3. The distribution of the plaintext is unknown. 4. The distribution of the key is unknown. 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. - William Hughes- - - - Please either agree that the two conditions are contradictory or point out the first step that you disagree with. - William Hughes === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - > my precondition is that key is uniform and c is fixed. > under that condition M is uniform. > Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. > - William Hughes- - > - - two conditions are contradictory or point out > a step that you disagree with > Please either agree that the two conditions > are contradictory or point out the first step > that you disagree with. - William Hughes- - - - from your sophism , i found you are just Allcorrect tricks have been exhausted. === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - > my precondition is that key is uniform and c is fixed. > under that condition M is uniform. > Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. > - William Hughes- - > - - two conditions are contradictory or point out > a step that you disagree with > Please either agree that the two conditions > are contradictory or point out the first step > that you disagree with. - William Hughes- - - - what you say is not proof, just sophism. can you prove your unknown is contradictory with knoun. your unknow is just you get with out full analysis. I never see such a proof like yours. === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - > my precondition is that key is uniform and c is fixed. > under that condition M is uniform. > Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. > - William Hughes- - > - - two conditions are contradictory or point out > a step that you disagree with > Please either agree that the two conditions > are contradictory or point out the first step > that you disagree with. - William Hughes- - - - what you say is not proof, just sophism. My putative proof is structured as an assumption (Assume the cyphertext is fixed.) followed by 5 steps, each of which follows from the assumption and the previous steps. Which is the first step that you do not think follows from the assumption and previous steps? - William Hughes === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - > my precondition is that key is uniform and c is fixed. > under that condition M is uniform. > Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. > - William Hughes- - > - - > two conditions are contradictory or point out > a step that you disagree with > Please either agree that the two conditions > are contradictory or point out the first step > that you disagree with. > - William Hughes- - > - - what you say is not proof, just sophism. My putative proof is structured as an assumption > (Assume the cyphertext is fixed.) followed by > 5 steps, each of which follows from > the assumption and the previous steps. Which > is the first step that you do not think follows > from the assumption and previous steps? - William Hughes- - - - you are confusing , form step 2. 1. If the cyphertext is fixed then there is a one to one correpondence between the plaintext and the key 2. The distribution of the key is determined by the distribution of the plaintext. -----------wrong There are determined each other. 3. The distribution of the plaintext is unknown. wrong. There are determined each other.so If the distribution of key is known,the distribution of palintext is known.for other condtion, the distribution of key is uniform 4. The distribution of the key is unknown. wrong uniform. 5. It is a contradiction to say that the key distribution is uniform and the cyphertext is fixed wrong. === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - > my precondition is that key is uniform and c is fixed. > under that condition M is uniform. > Stating contradictory conditions in the opposite order does not > help. If c is fixed then the key is not uniform. > If the key is uniform, then c is not fixed. > Please indicate the first step you think is wrong > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the key distribution > is uniform and the cyphertext is fixed. > - William Hughes- - > - - > two conditions are contradictory or point out > a step that you disagree with > Please either agree that the two conditions > are contradictory or point out the first step > that you disagree with. > - William Hughes- - > - - > what you say is not proof, just sophism. My putative proof is structured as an assumption > (Assume the cyphertext is fixed.) followed by > 5 steps, each of which follows from > the assumption and the previous steps. Which > is the first step that you do not think follows > from the assumption and previous steps? - William Hughes- - - - you are confusing , form step 2. 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key 2. The distribution of the key is > determined by the distribution of the plaintext. > -----------wrong There are determined each other. Indeed it is more accurate to say that either distribution determines the other. In particular, if one is uniform, the other must be. 3. The distribution of the plaintext is unknown. > wrong. No. You have been very clear that you are considering the distribution of the plaintext to be unknown, and that it might not be uniform. > There are determined each other.so If the distribution of key is > known, the distribution of palintext is known.for other condtion, the > distribution of key is uniform > So assuming that the key is uniformly distributed is equivalent to assuming that the plaintext is uniformly distributed. Either you abandon your assumption that the distribution of the plaintext might not be uniform or you have a contradiction. - William Hughes === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. your unknown is too weak to slove problems. Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. Assume the cyphertext is fixed. 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key 2. The distribution of the key is > determined by the distribution of the plaintext. 3. The distribution of the plaintext is unknown. 4. The distribution of the key is unknown. 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. - William Hughes- - - - as you have known for you said: You do not have one precondition you have two: cyphertext fixed and key distribution uniform. you disavow === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. Assume the cyphertext is fixed. 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key 2. The distribution of the key is > determined by the distribution of the plaintext. 3. The distribution of the plaintext is unknown. 4. The distribution of the key is unknown. 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. - William Hughes- - - - as you have known for you said: You do not have one precondition you > have two: cyphertext fixed > and key distribution uniform. you disavow No. My claim was and is that you have two preconditions not one and that these two conditions are contradictory. The above is a proof that the conditions are contradictory. Either agree that the conditions are contradictory or point out the first step that you disagree with. - William Hughes === Subject: Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > - William Hughes > The key distribution is uniform only when the cyphertext is > not fixed. > -----you are wrong for wrong understanding. > No. It is very easy to see that for a fixed cyphertext > the key distribution depends on the plaintext distribution. > Since the plaintext distribution is unknown, the key > distribution is unknown. > ---my precondtions regardless the prior probabilityo of M. > Then compromise the probabilities of incomplete condition. > your unknown is too weak to slove problems. > Here is a proof that the assumptions > that the cyphertext is fixed and the > key distribution is uniform are contradictory. > Please indicate the first > step you think is wrong. > Assume the cyphertext is fixed. > 1. If the cyphertext is fixed then there > is a one to one correpondence between the > plaintext and the key > 2. The distribution of the key is > determined by the distribution of the plaintext. > 3. The distribution of the plaintext is unknown. > 4. The distribution of the key is unknown. > 5. It is a contradiction to say that the cyphertext > is fixed and the key distribution is uniform. > - William Hughes- - > - - as you have known for you said: You do not have one precondition you > have two: cyphertext fixed > and key distribution uniform. you disavow No. My claim was and is that you have > two preconditions not one and that these > two conditions are contradictory. > The above is a proof that the conditions > are contradictory. Either agree that the conditions > are contradictory or point out the first step > that you disagree with. - William Hughes- - - - No. My claim was and is that you have two preconditions not one and that these two conditions are contradictory. The above is a proof that the conditions are contradictory. Either agree that the conditions are contradictory or point out the first step that you disagree with. close your eyes and do not analyze the problem completely and use the conditions fully, so the unknown is contradictory with known. the distribution of key is known. using this sophism way , you can find a lot of problems seem to be contradictory. === Subject: solutions manual can someone please send me the solutions for Elementary Principles of Chemical Processes, 3rd Ed., by Felder, Rousseau to em4n0n@gmail.com === Hi all, I met a probability question, could someone help me out? Three random variable A, B and C if A and B are dependent, B and C are dependent, is it possible that A and C are independet? i.e., If P(B|A)<>P(B), P(C|B)<>P(C), is it possible that P(C|A)=P(C)? Prove it if it's wrong or give an example if true. === >Hi all, I met a probability question, could someone help me out? Three random variable A, B and C >if A and B are dependent, B and C are dependent, is it possible that A >and C are independet? >i.e., >If P(B|A)<>P(B), P(C|B)<>P(C), is it possible that P(C|A)=P(C)? Prove >it if it's wrong or give an example if true. Flip 2 coins. Let A = Heads on first coin. Let C = Heads on the second coin. Let B = At least one head. Clearly A,C are indpendent. But events B,A are dependent, as are events B,C. quasi === Subject: Intersection of a half-ray and a triangle both lying in same plane Hi - I've got a triangle. Somewhere *inside* this triangle there is a point in the same plane as the triangle.) This half-ray will intersect one of the three edges of the triangle. What's the best method of computing this intersection? - Olumide === Subject: Re: Primitive polynomials over GF(2^m) of the smaller field. If you are flexible in this choice, then you can > do the following. If you have access to a primitive polynomial of degree > 32, you can easily raise that primitive element g to power 2^16+1=65537, > and you have a primitve element h of the field GF(2^16). Then compute > the minimal polynomial of h, if you want to do arithmetic in that field. > The only conjugate of g (over GF(2^16)) is g'=g^65536. Thus a prim?tive > polynomial of GF(2^32) over GF(2^16) is (x+g)(x+g')=x^2+ (g+g')x + h. Of course, you still probably want to compute the element g+g' as a > polynomial in h in order to actually use this polynomial! Sorry, Jyrki. I could't fully understand your meaning. Let me redescribe your words. If we want to find a primitive polynomial of degree m over GF(2^n), we first find a primitive element g from a primitive polynomial of degree (mn) over GF(2). Here, in our assumption, m=2, n=16. There is only one conjugate g' of g over GF(2^16) needed since the degree m is 2. So the primitive polynomial over GF(2^16) is (x+g)(x+g')=x^2+ (g+g')x + gg' However, g and g' are the primitive elements of GF(2^32). If we want to use the primitive polynomial over GF(2^16), we must transfer the coefficients from GF(2^32) to GF(2^16). That is what I can not figure out. === Subject: Re: Primitive polynomials over GF(2^m) > A primitive polynomial is an irreducible polynomial of degree m with > the added constraint that the smallest integer n for which P(x) > divides X^n + 1 is n = 2^m - 1. ..(1) > ... > I don't agree! The usual definition is that a primitive polynomial is > one whose roots are generators of the multiplicative group of the > extension field, and the definition given above is equivalent to that. > Except it should be x^n - 1, I guess ... Same thing in characteristic 2, but I agree it would be preferable to > write x^n-1. Sorry, didn't read carefully enough. Of course the result holds in any characteristic (with p^m - 1). Incidentally, I never realized that primitive polynomial had another meaning: a polynomial in Z[x] with coprime coefficients. I hope that meaning is obsolete. === Subject: Re: Math GRE subject test practice questions v2 III Another very common type > If x is a real number and P is a polynomial, then lim h->0 [P(x + 3h) > + P(x - 3h) - 2P(x)]/h^2 = a) 0 > b) 6P'(x) > c) 3P''(x) > ans d) 9P''(x) > e) infinity I saw another problem where they asked the same except the limit was > lim h->0 [P(x+h) - P(x-h)]/h and I intuitively answered correctly, > that the answer was 2P'(x). What is the procedure for this type of > question? Firstly I would define j = 3h so that the limit is now written lim j- >0 9[P(x + j) + P(x - j) - 2 P(x)]/j^2. Then note that the quantity in whose limit we are interested can be written 9{[(P(x + j) - P(x))/j - (P(x) - P(x-j)/j)]/j} so intuitively (P(x + j) - P(x))/j is close to the derivative of P at x for small j, and (P(x) - P(x-j)/j) is the same thing with values of x shifted by -j, so the whole expression behaves like the difference of two derivatives evaluated at x values differing by j, divided by j, i.e. the second derivative. This is far from being rigorous but hopefully is convincing enough to make one confident that d) is the right answer. Apply L'Hospital Rule twice. Remember that P(x) is a polynomial function, hence it is differentiable infinitely. === Subject: #513 more Qwest today ; New Book Re: Optimal Strategy for Playing the StockMarket: VonNeumann Game theory and Crossover Technique Portfolio of PAF as of 1NOV 07 T 11,515 Q 900 SGP 9,050 BMY 3,500 total share-wealth-units last reported which was 10SEP07 was 24,265 total share-wealth-units today 24,365** (** where Q shares count as 1/3 share-wealth-units) realestate land 3APR03 of 3 lots $19,000. science-art of pictures,porcelain etc starting JAN03 approx $12,000 realestate land 30JUL03 another lot $11,500. realestate land Sept05 another lot $75,000. Today I had some extra dividend cash to spend and buy more shares. And because of the system of Optimal Strategy that I am forced to buy the cheapest of the stocks above. And because Qwest missed its quarterly earnings by one penny and its stock sank over a dollar earlier this week, that Qwest is the best bargain today. I bought 300 more shares of Qwest today at $7.20 per share. If I had waited until the last hours of the market I could have picked those up at 7.07. But I have ceased on quibbling over pennies. One thing that has caught my attention recently is the upturn in BMY stock. It is a toss up of whether SGP or BMY will be the first to merge with another drug company. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: mypetispace The best website i found in a while http://www.mypetispace.com === Subject: shortest path in bipartite directed graph Are there any specialized algorithms for shortest paths in a bipartite graph? Specifically, I have a bipartite graph with two directed edges between each pair of nodes, with d(x,y) = -d(y,x), and I want to compute all-pairs shortest paths. I was going to use floyd-warshall but thought there may be a better algorithm out there. === Subject: Re: Canal Variation > The first 5 moves are the symmetrical Giuoco Pianissimo. > Then 6 B-KN5 O-O? 7 N-Q5 B-K3? - black should have done > P-KR3, driving back the bishop. Other 7th moves are no > better, in particular B-KN5 is met by white's P-KR3 later, > moving the bishop back and giving the same situation. > Now the mate goes 8 NxN ch PxN 9 B-KR6 KR-K 10 N-KR4 > anything 11 Q-KB3 anything 12 Q-KN3 ch K-KR (forced) Be careful with those anything moves. 10.... N-Q5?! 11.Q-KB3? NxQ ch 10.... K-R1 11.Q-KB3? N-Q5 12.Q-KN3? R-KN1 and Black wins Obviously my analysis isn't meant to be complete and cover every possible move, just the useful ones. 10. ... N-Q5 is met by simply taking one more move to bring around the queen (Q-Q2-K3-KN3). 10. ... K-KR on the other hand does seem good; if then 11. N-KB5 black must take it, as it guards the mating square (11. ... BxB 12. Q-KN4 R-KN 13. B-KN7 ch RxB (forced) 14. QxR mate). anyone's studied this specific pattern, perhaps given it a name? Andrew Usher === Subject: Re: Teach/Tour in China: Summer 2008 We are pleased to announce our 2008 Summer Teaching Program in China. > Once again, we will be hosting our program in cooperation with > Guangdong University of Foreign Studies in Guangzhou, China. The > program structure remains the same, 18 days of teaching followed by a > private cultural tour. In addition to the 1-day excursion in > Guangzhou, this year's cultural tour includes Shenzhen, Hong Kong, and > Macau. For further details, please follow any of these links: > 2008 Program Flyer http://www.ipie.us/2008flyer.pdf > 2008 Program Handbook http://www.ipie.us/2008handbook.pdf > 2008 Application http://www.ipie.us/2008application.pdf have been incorporated into this year's program and documentation. cultural mission. John McBride and Chen Yan > International Partnerships in Educationhttp://www.ipie.us China travel service http://www.itourschina.com Offers China tours,hotels to Beijing,Shanghai,Tibet,Xian,Guangzhou,Shenzhen,Guilin and other Chinese cities in China.Itourschina.com show you the best discovery tour to travel China! === Subject: Re: Topology with base and ... <5or1eqFo7nohU1@mid.individual.net> all that's needed is /B = S. > For B to be base for some topology for S, what's needed is > /B = S; for all U,V in B, > x in U / V ==> some W in B with x in W subset U / V For B to be a base for the topology of S, what's needed is > /B = S; for all U in B, U open > for all open U nhood x, some V in B with x in V subset U Exercise: f:X -> Y, B subbase, base, topology for Y > ==> { f^-1(U) | U in B } subbase, base, topology resp. > of a topology for X. Oh, exercise... > Strange. is this true ? > There doest not exist the assumption that f is quotient map. > Yes, f can be any map. X was not given a topology, only Y. Thus from Y, a topology for X can be generated, as described in the exercise, that makes f continuous. This is the smallest or coarest topology that will make f continuous and is call the initial topology coinduced by f. In the event f is surjective, the initial topology makes f a quotient map. For any finer topology for X, f is not a quotient map, just a continuous function. === Subject: Re: Topology with base and ... > For B to be a subbase for some topology for S, > all that's needed is /B = S. > For B to be base for some topology for S, what's needed is > /B = S; for all U,V in B, > x in U / V ==> some W in B with x in W subset U / V > For B to be a base for the topology of S, what's needed is > /B = S; for all U in B, U open > for all open U nhood x, some V in B with x in V subset U > Exercise: f:X -> Y, B subbase, base, topology for Y > ==> { f^-1(U) | U in B } subbase, base, topology resp. > of a topology for X. > Oh, exercise... > Strange. is this true ? > There doest not exist the assumption that f is quotient map. > Yes, f can be any map. X was not given a topology, only Y. Thus from Y, a topology > for X can be generated, as described in the exercise, that makes > f continuous. This is the smallest or coarest topology that will > make f continuous and is call the initial topology coinduced by f. In the event f is surjective, the initial topology makes f a > quotient map. For any finer topology for X, f is not a quotient > map, just a continuous function. Yes, a topology... I can understand your explanation. === Subject: Re: Topology with base and ... <5or1eqFo7nohU1@mid.individual.net> <6nlWi.3580$Pl6.824@weber.videotron.net> If B_1 and B_2 are in a subbase for a topology, that does not mean that > (B_1) / (B_2) is necessarily in the subbase. > Indeed. Subbase sets for R^N are of the form B_k,U = { f:N -> R | f(k) in U } for k in N and open nonnul U. B_0,U / B_1,U is not a subbase set. > I want to know the counter-examplce about base. > The usual base for R^2 is open balls. R(r,1) / B(s,1) is not an open ball when r /= s, d(r,s) < 1 === Subject: Science is the base for invention Cc: seesan8@yahoo.co.in science will revolutionise through the world http://www.geocities.com/naturewol/ http://indianfriendfinder.com/go/g906261-pmem === Subject: science and sex are interelated http://www.geocities.com/naturewol/ http://indianfriendfinder.com/go/g906261-pmem === Subject: Your Brain: Leftie or Rightie? (was: science and sex are interelated) > http://www.geocities.com/naturewol/ http://indianfriendfinder.com/go/g906261-pmem Interesting you should say that subject header. A left-brain vs. right-brain test (or so it assumes) http://www.news.com.au/couriermail/story/0,23739,22556678-23272,00.html?from =mostpop How do you see the dancer swinging? Mine came out remotely different than anything described there. I first saw her swinging counter clockwise, then after the image loaded fully, clock-wise. Then it switched back and forth chaotically. Then she was going back and forth with the swinging leg in the front, every now and then swinging around a full loop one way or the other. Then going back and forth with the swinging leg in the back, doing something similar. Then she reverted to a 2-dimensional image simply oscillating back and forth in 2-D. Then she started moving in whatever way I directed she should (just like any grown woman ought to) and I made her do all sorts of things pleasing to watch. Now, according to the site, clockwise and counter-clockwise are supposed to correlated to different dominance in the hemispheres. Now ... what do YOU see in the image in that link above? === Subject: Re: Does This Series Look Familiar? Am 01.11.2007 18:31 schrieb Narcoleptic Insomniac: > My recent musings have lead me to studying the following > general series: f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! Does anyone know if these series go by any particular > name? Kyle Czarnecki http://www.mathbin.net/15482 I was interested in this subject some time ago, too and found it much interesting - but left it at an early stage due to lack of connection to other fields and very limited experience, when I started looking at it. Would you mind to enter an email exchange about this? I don't have much to contribute, but may be there is something worth to follow. Gottfried -- --- Gottfried Helms, Kassel === Subject: Re: Does This Series Look Familiar? My recent musings have lead me to studying the following general series: f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! Does anyone know if these series go by any particular name? Yes, it's simply a multisection of the exponentional series. It can be expressed as a linear combination of exp(z w^k) where w is a primitive n'th root of 1. You should be able to find a discussion of multisections in many books on combinatorics (e.g. Comtet iirc). --Bill Dubuque === Subject: Re: Does This Series Look Familiar? My recent musings have lead me to studying the following general series: f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! Does anyone know if these series go by any particular name? Yes, it's simply a multisection of the exponentional series. > It can be expressed as a linear combination of exp(z w^k) > where w is a primitive n'th root of 1. You should be able > to find a discussion of multisections in many books on > combinatorics (e.g. Comtet iirc). well this one is actually -m g (w x) m n 2n because of the alternating signs so you do need a 2n-th root of unity for expression but yes it is an evaluation of the multisected exponential -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Does This Series Look Familiar? My recent musings have lead me to studying the following general series: > f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! > Does anyone know if these series go by any particular name? Yes, it's simply a multisection of the exponentional series. > It can be expressed as a linear combination of exp(z w^k) > where w is a primitive n'th root of 1. You should be able > to find a discussion of multisections in many books on > combinatorics (e.g. Comtet iirc). well this one is actually > -m > g (w x) > m n 2n because of the alternating signs > so you do need a 2n-th root of unity for expression > but yes it is an evaluation > of the multisected exponential oops posting too quick the form has a leading term too but that is the general approach for the cosine-like forms > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: Does This Series Look Familiar? On Nov 1, 10:31 am, Narcoleptic Insomniac > My recent musings have lead me to studying the following > general series: f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! Does anyone know if these series go by any particular > name? yes! this is my generalised trigonometric function! among others... i suspect they were known to gauss but i have yet to see a full exposition outside my own some properties of the n=4 case were shown by ramanujan but he apparently never derived they were a part of the general product and sum properties i give in the posts above... if you have any particular questions about these i have developed many formulae i can share -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Does This Series Look Familiar? > On Nov 1, 10:31 am, Narcoleptic Insomniac My recent musings have lead me to studying the following > general series: f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! Does anyone know if these series go by any particular > name? yes! this is my generalised trigonometric function! > among others... i suspect they were known to gauss > but i have yet to see a full exposition outside my own some properties of the n=4 case were shown by ramanujan > but he apparently never derived they were a part > of the general product and sum properties > i give in the posts above... if you have any particular questions about these > i have developed many formulae i can share also please note i have been looking for references now since the summer/fall of 2001 and have not found much outside a few specific forms so if you find any information i would appreciate them for the paper i have put together on this -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Does This Series Look Familiar? > My recent musings have lead me to studying the following > general series: f_{k, m}(z) := sum_{n = 0}^{oo} (-1)^n z^{nk + m} / (nk + m)! Does anyone know if these series go by any particular > name? I seriously doubt it. You've got infinitely many functions there. f_{1,0}(z)=exp(-z) f_{2,0}(z)=cos(z) f_{2,1}(z)=sin(z) If k=1 and m>1, f_{1,m}(z) can probably be expressed in closed form in terms of the exponential function exp(z) minus a finite number of terms of exp(-z). For other specific values of k and m, deriving a closed form by hand looks doable. Maple gives a closed form for f_{k,m} in terms of the generalized hypergeometric function, which looks reasonable, since the exponents follow an arithmetic progression and ratios of successive coefficients follows the pattern shown in the ref, below: http://mathworld.wolfram.com/GeneralizedHypergeometricFunction.html > Kyle Czarnecki http://www.mathbin.net/15482 -- I.N. Galidakis === Subject: No good talking about it....so I thought I would put this thread into action!! :D No good talking about it....so I thought I would put this thread into action!! :D To all the beautiful girls who have taken the journey from the 'Presentation' at the Cohen's lifestyle clinic to 'Re-Feed' ---this forum is for YOU!! Tell us how things have gone for you since finishing the re-feed program - we're all busting to know!! :D http://www.chags.cn/weightloss/topic/life-after-cohens_3325.html === Subject: Re: The Metric System > Interested in hearing comments about the ramifications, past, present, > future, on US [sic] failure to fully convert. There is no failure to convert. The US is on the metric system. The inch is a metric unit legally defined as 2.54 centimeters. === Subject: Re: The Metric System Nntp-Posting-Host: hera.cwi.nl ... > That's what I mean you worked the problem and found a solution. The mixture > hasn't caused the UK to grind to a halt? As it happens I agree we should > switch over properly. Road signs and cars first perhaps. Sun readers won't > like it though :-) Oh, well, Ireland managed it, Australia managed, so it should not be too difficult ;-). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The Metric System Nntp-Posting-Host: hera.cwi.nl ... > If the metric convention had been set to km/L, maybe the switch would > have been easier. It seems to me that people used to think how far > they will travel on one gallon of fuel tend to more easily relate to how > far they can travel on 1 liter. That is not really a convention (but perhaps it is in the UK). I see and use regularly measures like 1:13 (or 1 litre for 13 kilometres). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The Metric System CWatters skrev i en meddelelse > Hasn't been a problem for the UK. We do science in Metric/SI units then > drive home in miles per hour and discuss fuel consumption in miles per > gallon - even if knowbody knows the price per gallon because fuel it's > priced and sold per litre. > I always wondered about that. Do they display MPG ratings on new > cars as well, or km/l? > BTW, I believe the conversion factor is > 1 mpg = 0.43 km/l > 1 km/l = 2.35 mpg New cars in the UK are still in mpg (or at least mine is). The Belgian car > we had until last year was in L/100KM. I dare say it might be possible to hack the computer in it to do either. To my knowledge most European cars will give you km/l which will make it easy for you to estimate how far you can go when your car is filled up if you know the size of your tank. The L/100KM is of very little practical use for the average driver. === Subject: Re: The Metric System > New cars in the UK are still in mpg (or at least mine is). The Belgian car > we had until last year was in L/100KM. > I dare say it might be possible to hack the computer in it to do either. > To my knowledge most European cars will give you km/l which will make it > easy for you to estimate how far you can go when your car is filled up if > you know the size of your tank. The L/100KM is of very little practical use > for the average driver. But they don't! I've been doing a bit of personal research (on the back of the discovery when driving in Europe the last 2 years that the km/h markings on my German-built car are virtually unreadible) to find out just how many UK-sold cars are metric friendly and the answer is, sadly, very few. EU and British law requires that km/h is available (the latter obviously requires mph too) but that's it. It doesn't require the markings to actually be usable and doesn't require a metric odometer or trip computer. I've recently seen a Seat Altea with a trip computer where the LCD display had mpg and miles physically manufactured into it in the positions where the manual showed l/100km and km. I'm told by a number of people that this is common practice for a number of motor manufacturers and people have actually been told the only way to switch is to replace the dash! I'm betting that there will be a few Irish car owners who can bear me out on this! (I've taken the liberty of cross-posting this to misc.metric where it really belongs).