Summary: The Lorentz transforms themselves are proof t' and x' cannot possibly be just coordinates. Examination of their derivation verifies their identity as ==== Opponents should first actually find out what the content is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selfless, *.answers moderators evidences ignorance and atrocious netiquette. Archive-name: physics-faq/criticism/lorentz-intervals (SR) Lorentz t', x' = Intervals (c) Eleaticus/Oren C. Webster Thnktank@concentric.net ------------------------------ of the use of 'just coordinates' in any scientific formula. Defenders of the Special Relativity faith are especially fond of telling opponents of their space-time fairy tales that they do not know the difference between coordinates and magnitudes. That may often be so, but the fault lies ultimately with SR dogma. The Lorentz-Einstein transformations cannot possibly be 'just coordinates', which is the interpre- tation required to support the many sideshow carnival acts with which the SR faithful bedazzle the public, and establish their moral and intellectual superiority. If I get in my car and drive steadily for a few hours at 50 kilometers per hour, is 50t the distance I travel? Of course not. The last time my hours-counting 'just coord- inates' clock was set to zero was when Zeno first reported one of his paradoxes to Parmenides. That was a long time ago, so my t is not useful for such purposes unless you also use my clock to established the starting time, perhaps t0, and use the formula 50(t-t0) to calculate the distance. In any case, my t is even then not 'just a coordinate' because it always represents particular elapsed times that can be used in the (t-t0) form to calculate perfectly good time intervals (elapsed times). Alternatively, I could (re)set my clock to zero at the start of some meaningful time interval, in which case my t shows a scientifically perfect current and/or end time. In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function of an elapsed time interval (not 'just a coordinate') and a time interval (-vx/cc; the interval amount the t' clock is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] If it takes me 50 minutes to cross the Interstate highway, was x/50 my velocity crossing it? Of course not. The origin of all my axes is at the very spot where Zeno first presented his first paradox to Parmenides. That makes my x equal a couple of thousands of miles, plus, and is not useful for such purposes unless you establish the starting x value, perhaps x0, and use the formula (x-x0)/50 to calculate my velocity. In any case, even then my x is not 'just a coordinate' because it always repesents particular distance intervals that can always be used in the (x-x0) form for any and every scientific purose. Alternatively, I could move my x-axis origin to the starting (zero) point of some meaningful distance, in which case my x shows a scientifically perfect current and/or end distance. In which case, the Lorentz-Einstein x'=(x-vt)/g is a function of a current/ending distance interval (not 'just a coordinate') and a distance interval (-vt; the interval amount the x' axis is being screwed up at time t) and thus cannot be 'just a coordinate' since neither of the independent variables is such a 'just' thing. {Their meaning is shown below, step-by-step.] ------------------------------ 1. Introduction with the obvious debunking of the use of 'just coordinates' in any scientific formula. 3. The Lorentz-Einstein transforms. 4. The 'just coordinates' argument. 5. Single-system, little-purpose ambiguity. 6. Relating two coordinate measures/systems. 7. Distances and moving coordinate axes. 8. Time intervals. 9. Einstein's (1905) derivations. 10. A word about intervals. 11. Intervals versus the Twins Paradox. 12. Summary ------------------------------ Special Relativity's space-time circus is based on the 'transformation' equations by which it is believed one can relate a nominally 'stationary' system's space and time coordinates to those of an inertially (not accelerating) moving other observer. That moving observer's own physical body and coordinate system might have been identical in size to those of the stationary observer before the traveller began moving, but are 'seen' as very different by the stationary observer when the relative velocity of the two is great enough, a high percentage of the velocity of light. Concerning ourselves - as is customary - with just the spatial coordinate axis that lies parallel to the direction of motion, and with time, Einstein arrived at these formulas that relate the moving system measures or coordinates (x' and t') to the stationary system coordinates (x and t): x' = (x - vt)/sqrt(1-vv/cc) (Eq 1x) t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) The v is for the two systems' relative velocity as seen by the stationary observer, and is positive if the dir- ection is toward higher values of x. By concensus, the moving system x'-axis higher values also lie in that direction, and all axes parallel the other system's corresponding axis. We used vv to mean the square of v but might use v^2 for that purpose below. Similarly for c. Because it is believed that no physical object can reach or exceed c, the square-root term in both denominators is presumed always less than one, which means that the formulas say both x' and t' will tend to be greater than x and t, respectively. However, SRians call the x' result 'contraction' - which means shortening - and the t' result 'dilation' - which means increasing. ------------------------------ The 'just coordinates' argument is so patently ridiculous that even opponents have a hard time accepting just how simple and obvious its debunking can be, as shown in this section. However, further sections take a more arithmet- ical approach that you'll maybe find more professorial. The 'just coordinates' argument is that t is mot a duration, not a time interval; it's just an arbitrary clock reading. But what if the moving system observer comes speeding by while you make your annual 'spring forward' or 'fall back' change? The formula says that the moving system clock's 'just coordinate' reading can be calculated from yours: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Imagine the moving system oberver's confusion if his clock changes its reading while he's looking at it! If his clock doesn't change when yours does, the formula is wrong; if it is truly a 'just coordinates' formula. And then what happens if you realize you were a day early and put your clock back to what it had said previously? And suppose you are in NYC and your twin in LA and both are watching the moving observer. You'll both be using the same v because you are at rest wrt (with respect to) each other. You're on Eastern Standard Time and your twin is on Pacific Standard Time maybe. You have three hours more on your clock than does your twin. On which 'just coordinate' clock will the Lorentz transforms base the 'just coordinate' time the moving system clock says? The formula applies to both of your t-times: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Sure, the idea that you can change someone else's clock with no connection of any kind is really ridiculous, but Eqs 1x and 1t aren't MY equations. Are they yours? And we aren't the ones to say x, t, x', and t' are just coordinates. If the t' formula is actually either an elapsed time formula, or the basis of a t'/t ratio, then there is no implication that one clock's reading has anything to do with the other's. It can only be rates of clock ticking, or how one time INTERVAL compares to the other that the formula is about. ------------------------------ Since we're going to be comparing measurements on two coordinate systems in the next section, let's go to our supply cabinet and get our yard-stick (which we use to measure things in inches) and our meter-stick (which we use to measure things in centimeters). Here, I'm getting mine. Oh! Oh! There's an ant on mine, and he ... she ... sure is hanging on, right at the 3.5 inch mark of the yard- stick. Let's see if I can wave the stick around enough that she'll let go. Nope. However, before I gave up I waved the stick and the ant 'all over the place. Always, however, the ant was at the 3.5 mark on the yard-stick, and always 3.5 away from the end of the stick, however far and wide I have transported her. Neither of those 3.5 facts means very much. Of the two, the distance aspect meant almost nothing. So the distance was 3.5 from the end. So what? That length, distance, was not in use. And only maybe the ant might have been concerned with just what location, 'just coordinate', on the stick she was at. Just so with x and t. So, is the 3.5 reading just a coordinate? Or a distance/length? It's ambiguous in and of itself, and really makes no difference what you say until you try to make use of the number. Hey, my address is 5047 Newton Street. If you are looking for me and you're at 4120 Newton, it is helpful information, because it tells you which direction to go. Is that 'just coordinate'? Where it really becomes useful, perhaps, is in telling you how far away I am. That's not just a coordinate value, that's a distance, length, interval. However, it is subtracting 4120 from 5047 that tells you which direction and how far. It is only because both 5047 and 4120 are distances from the same point - ANY same point - that the result means anything. My x - my yardstick reading - is always a distance or length; it is impossible to be otherwise with an honest, competently designed yardstick. Whether or not its reading is of good use in some particular scientific formula depends on whether I put the zero end of the yardstick at some useful place. As in the introduction, we should either put it at the starting location/end, or use two readings from it: (x-x0). ------------------------------ Taking care to not damage our brave little ant, I place my yard-stick onto the table, zero end to the left, 36 end to the right. Now I place the 'just coordinate' meter-stick on the table in the same orientation, in a random location, and find that the ant's coordinate on the meter-stick is 51. The formula relating centimeters to inches is cm=i*2.54 but we want a formula similar to x'=(x-vt)/sqrt(1-vv/cc). That would be c=i/.03937 approximately, but let's use x' for the meter-stick reading, and x for the inch reading: x'=x/.3937. 3.5/.3937 = 8.89 Wait a minute. It's not just science but definition that says c=i/.3937=8.89, so something is wrong. 8.89 is not 51. We already knew that 51 cm was just an arbitrary coordinate. Arbitrary not because that point isn't 51 cm from the zero end of the meter-stick, but because the zero point was in an arbitrary position. Let's put the meter-stick in a position where it's zero point is at the yard-stick zero point. What is the centimeter coordinate now? Hey. 8.89, just like the formula says. The only way for a 'transform' like x'=x/g to work, whatever g might be, is for both coordinate systems to have their zero points aligned, in which case saying the two measures are not intervals is pure idiocy. Noe that with both zero points at the same position both x' and x are great measures for scientific purposes, in any and every case where we were smart enough to put those zero points at a useful location. There is one extension of x'=x/g that will let us use the meter-stick in arbitrary position. When the cm reading was 51, the zero point of the yard-stick read (51-8.89=) 42.11 cm. If we call that point x.z' we get x' = x.z' + x/.3937. = 42.11 + 3.5/.3937 = 42.11 + 8.89 = 51. Obviously, in this formula x/.3937 is the distance from the x' coordinate of the location where x=0. An interval. Just as obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate, does not mean that x' is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z') in the scientific formula. (x'-x.z') equals the useful, Ratio Scale value x/.3937. So, we have discovered a basic fact: a transformation formula like x'=x/g works only if the two zero points of the coordinate systems coincide. That makes it non- sense to say the two coodinates are only coordinates and not intervals. Both must be values that represent distances from their respective zero points unless you take the proper steps to adjust for the discrepancy. Make sure you understand that although the inclusion of x.z' made it possible to correctly calculate x', the result is nonsense when it comes to use of x' for general length/distance purposes; it is x'-x.z' that is a useful number in such cases. It could be that we're measuring a sheet of paper with one end at x=0 and the other at x=3.5; x'=51 is nonsense as a centimeter measure of the paper. But, you say, the Lorentz transform contain a -vt term. ------------------------------ We discovered x'=x.z' + x/g as the correct formula for relating one coordinate to another system's. But the Lorentz transform contains another term, -vt/sqrt(1-vv/cc). What is it? Let's start with our x'=51 cm, x=3.5, x.z'=42.11 example. Every minute, let's move the meter-stick one inch to our right. At minute 0, the cm reading was 51 cm. At minute 1, the cm reading is now 50 cm. At minute 2, the cm reading is now 49 cm. In this instance, v=1 inch/minute. And t was 0, 1, 2. What has happened is that we have made our x.z' a lie, and increasingly so. -vt/.3937 is the change in x.z'. x' = (x.z - vt/.3937) + x/.3937. Obviously, vt/.3937 is not a coordinate; even most SRians wouldn't imagine it was. It is an interval, the distance over which the moving system has moved since t=0. And, of course, x/.3937 is the distance of our brave little ant from the point where x=0 and the centimeter reading is x.z'-vt/.3937. Yes, every minute the meter- stick moves to the right and the meter-stick coordinate of the spot where x=0 gets less and less - and eventually negative. Make sure you understand that every minute the x' coordinate, because of -vt/g, becomes a better measure of, say, the 3.5 paper we might be measuring with the yard-stick, given that 51 was too big a number and -vt is negative. That is, until the two origins coincide at x'=x=0, and then it gets worse and worse. With -vt positive (because v<0) the situation is different. With 51 and -vt positive, x' just gets worse and worse over time. Quite obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary x' coordinate even when the x' axis is moving, does not mean that x'is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (x'-x.z'+vt/.3937) in the scientific formula. (x'-x.z'+vt/.3937) equals the useful, Ratio Scale value x/.3937. ------------------------------ Instead of using our sticks, let's get out two clocks. Mind you, we're not going to deal with different clock rates here, just establish the same basics as for distance. Your clock says 9:00 Eastern Standard Time (EST) and we note that t=540 minutes when we put down the clock. Blindly, let's turn the setting knob of your twin's Pacific Standard Time clock and put it down before us. According to what we see, EST's 540 minutes (9:00) corre- sponds to PST's 14:30; t'=870. We know the formula relating PST to EST is t' (pacific) = t (eastern) - 180 (minutes). Thus, it is not correct that the second clock can have an arbitrary setting, because 870 <> 540-180. We know that the two clocks are related by t' = t/1 since both are using the same second, hour, etc units. But 870 (14:30 in minutes) is not 540/1-180, so once again we know something is wrong. However, t'=t.z' + t/1 works. EST midnight equals PST 0.0 (midnite) - 180, so t.z' = -180, and t' = -180 + 540/1 = 360. Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes. We see thus that like distance measures/coordinates, time axis origins (zero points) must either be 'lined up' or adjusted for. So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving system elapsed time interval since the time axes were both at a common zero. There is no t.z' adjustment: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Make sure you understand that in the clock case, if the EST is showing a good number for elapsed time since the travelling observer passed NYC, then the PST clock is silliness. t.z' must be zero or must be taken out of time lapse calculations for the PST clock to be used intelligently, just as was true for x.z'. What is lacking as yet for Lorentz t' is the -vx/cc term that corresponds to the x' formula -vt term. Break it up into two parts: v/c and x/c. v/c is a scaling factor that changes velocity from whatever kind of unit you are using over to fractions of c. x/c is distance divided by velocity, which is time. x/c is thus the time interval since the two time axes had a common zero point - which they have to have in the Lorentz transforms which do not have the t.z' term we learned to use above. Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving system clock has been changed - since the common/ adjusted time - over and beyond the elapsed time interval represented by x/sqrt(1-vv/cc). We have discovered that the only way for t' to be t/g is for t' and t to have a common zero point, just as for x' and x. It would be otherwise if the t' formula contained an adjustment t.z' under some name or other, but the necessity to include such a term correlates 100% with t' numbers that aren't directly usable. As for x and x', our knowledge of how to setup a proper formula relating t and t' is of no use unless we use the knowledge in scientific formulas; (t'-t.z'+xv/gcc) gives us the only directly useful value: t/g. ------------------------------ When we return to Einstein's derivations of the transform formulas with a well-focused eye, we find he was a wee bit confused - or at least self-contradictory. When he set up his (at first unknown) tau=moving system time formulas, he created three particular instances of tau. Tau.0 is the time at which light is emitted at the moving origin toward a mirror to the right that is moving at rest wrt that moving origin and at a constant distance from that origin. He lets the stationary time slot have the value t, a constant, the stationary system starting time. Tau.1 is the time at which the light is reflected. He lets the stationary time be t+x'/(c-v); t is still a constant and x'/(c-v) is the time interval since t. Tau.2 is the time at which the light gets (back) to the moving origin. The stationary time value is put as t + x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v) + x'/(c+v) is the time interval since t. On the thesis that the moving observer sees the time to the mirror as the same as the time back to the origin, he sets .5[ tau.0 + tau.2 ] = tau.1. Tau.0 completely drops out of the analysis and leaves no trace, and has no effect. Further, the t you see in tau.0, tau.1, and tau.2 also completely drops out with no trace and no effect, leaving us with exactly what you'd get if you had explicilty said t' is an interval and so is t. What doesn't drop out in the stationary time values is x'/(c-v) and x'/(c+v), the time interval it takes for light to get to the fleeing mirror, and the time interval it takes for light to get back to the approaching origin. Thus, his resultant t' formula is strictly based on time intervals in the stationary system. Time intervals since some starting time, yes, but time intervals. There is absolutely nothing in the derived formulas that depends on arbitrary coordinates like the constant t in the stationary time arguments. Let's look at the x dimension; it is x'=x-vt [as x increases by vt, the effect over time is x'=(x+vt)-vt)], which Einstein explicitly sets up as a constant stationary distance. He uses that x' not just in the time interval parts of the stationary time arguments, but also in the x (distance) stationary system argument for the tau at the time light is reflected. x' can't be the stationary system coordinate of the mirror at that time. That value is x'+vt. x' is explicitly an interval, distance. Thus, the whole tau derivation of the t' formula is fully and explicitly based on x' - a spatial length/distance/interval - and the two time interals x'/(c-v) and x'/(c+v). While we're at it, if the starting t is not zero, his x'=x-vt formula is complete nonsense also. Given that there was some L that was the mirror x-location and length when the light is emitted, if t was already, say, 500, then x'=L-vt could have been a very negative length. ------------------------------ There are intervals, and there are intervals. If we put our yard stick zero point at one end of a piece of paper and read off the coordinate at the other end of the paper, we have a good measure of the paper's length, a Ratio Scale measure. [Absolute temperature scales are ratio scale.] If instead we put the one end of the paper at the one inch mark (or the zero end of the stick one inch 'into' the length of the paper) we get measures that are one inch off the true, ratio scale length. The two messed up measures are still intervals, but they are Interval Scale measures. [Household temperature scales are interval scale, which is why your physics and chemistry professors won't let you use them without first converting to the ratio scale absolute temperatures.) t'=t/g and x'=x/g represent ratio scale measures, given that t and x were ratio scalae to start with. t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale measures, even given a good ratio scale t and a good ratio scale x. x'=x.z'+x/g and x'=x/g-vt/g are both interval scale measures, even given a good ratio scale x and a good ratio scale t. Look for the (SR) Lorentz t', x' = degraded measures document soon at a newsgroup near you. ------------------------------ t'=(t-vx/cc)/g shows t' being greater than t. The reason Special Relativity will not allow the use of its basic time equation in determining what SR has to say about the twins' ages, is that t' and x' are supposedly just coordinates, and they say you have to take the coordinate pairs (t',x') and (x,t) into consideration in both the time and place the twins' separation started and the time and place the twins reunited. Since t' and x' are actually both intervals, not just coordinates, the 'excuse' is spurious, and is so even without use of the obvious (x_b-x_a) and (t_b-t_a) usages. However, SR is right to be embarrassed by their transformation formulas. Look for the (SR) Lorentz t', x' = degraded measures document at a newsgroup near you. ------------------------------ A. t'=t/g and x'=x/g can be almost 'just coordinates' in the sense that the values obtained may not be of much use except in the most primal and useless way: how long and how far since/from the time/ place they were zero. Even here, however, the zero points within each of the two scale pairs (t',t) and (x'.x) must have been lined up. If the zero points have been intelligently selected (such as at the starting point and time of a trip) they can be rationally used 'as is' in any valid sci- entific equation. B. Even the interval scale t'=t.z' - xv/gcc + t/g and x'=x.z' - vt/g + x/g are not 'just coordinates'. They can be used to good effect by establishing the relevant starting times/points and using (t'-t.z'+xv/gcc) and (x'-x.z'+vt/g), as the situation may require. C. When you see vx/gcc or vt/g in use in any guise with non-zero values, you know the resultant t' or x' is a degraded, interval scale value. E-X: Anytime you do not see what amounts to t.z' and xv/gcc in the time case, or x.z' and vt/g in the distance case, you know that the t' and/or x' in use are intervals. Period. Y: Either set your clock to zero at the start of the relevant time interval, or use (t-t0), with both being readings on the same clock. Either move your x-axis origin to the starting end or point, or use (x-x0), with both being readings on the same axis. Z: In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see that t' and x' satisfy the mathematical tests for/of interval scales when -vt and -vx/cc are not zero; thus, they must be intervals. When -vt and -vx/cc are zero, t' and x' satisfy the much better mathematical definition of ratio scales, and are thus not just mere intervals, but (rescaled) good ones. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ==== Let f(x) be a function of one real variable and let f*(x) be its inverse. Suppose f(x)=f*(x) for all x. Is anything known about the general solution set of this equation? The only solution I can think of is f(x)=g*(a/g(x)). Can you determine any other solutions? Eugene Shubert http://www.everythingimportant.org ==== > Let f(x) be a function of one real variable and let f*(x) be its > inverse. > > Suppose f(x)=f*(x) for all x. Is anything known about the general > solution set of this equation? > > The only solution I can think of is f(x)=g*(a/g(x)). Can you > determine any other solutions? 0. What exactly are a and g above? What exactly is a/g(X) when g(X) is 0 (as it must be for some X, in order for g to have an inverse)? 1. f(x)=f(-1)(x) -> f(f(x))=x. Unless you have other conditions on f, beyond merely being a function from R to R, that's about all you can say. 2. E.g., suppose f is defined for 0<=x, f(0)=0, and f is monotone decreasing. Then f can be extended to x<0 in such a way that f(f) is the identity function. ==== >Let f(x) be a function of one real variable and let f*(x) be its >inverse. So I presume f is supposed to be 1-1 and onto. >Suppose f(x)=f*(x) for all x. Is anything known about the general >solution set of this equation? This is equivalent to f(f(x)) = x. >The only solution I can think of is f(x)=g*(a/g(x)). Can you >determine any other solutions? Let A and B be disjoint sets of the same cardinality, and h: A -> B 1-1 onto. For a in A define f(a) = h(a) and f(h(a)) = a, and for x not in A union B take f(x) = x. If you want f to be continuous, it must have exactly one fixed point p, A = (-infinity,p) and B = (p,infinity), and h is a decreasing function from A to B. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > > 2. E.g., suppose f is defined for 0<=x, f(0)=0, and f is monotone > decreasing. Then f can be extended to x<0 in such a way that > f(f) is the identity function. but from f(f)=id it does not follow that f is invertible because f may bounded so that its inverse does not exist. But the algebraic definition of invertibility is that left and right inverse exist, which is obviously the case here. So do we have two different definitions of invertibility here? ==== >> >> 2. E.g., suppose f is defined for 0<=x, f(0)=0, and f is monotone >> decreasing. Then f can be extended to x<0 in such a way that >> f(f) is the identity function. > but from f(f)=id it does not follow that f is invertible because f may > bounded so that its inverse does not exist. But the algebraic > definition of invertibility is that left and right inverse exist, > which is obviously the case here. So do we have two different > definitions of invertibility here? f is invertible if there is a g with gf = id = fg. So if ff =id, f /is/ invertible (ff=id -> f not bounded). What is true (and maybe what you meant?) was that f monotone decreasing only implies that it's injective. For my construction to be correct, I also need f to be surjective; this is assured if f is continous and not bounded below. ==== > The only solution I can think of is f(x)=g*(a/g(x)). Can you > determine any other solutions? 1. The term you seek is involution. 2. A bit more general (in terms of a function that actually can be written down explicitely) would be a Moebius transform f(x)=(ax+b)/(cx+d), the condition to make this an involution being left as an amusing homework :-) -- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the address For spamming, remove anything else ==== > f is invertible if there is a g with gf = id = fg. So if ff =id, > f /is/ invertible (ff=id -> f not bounded). What is true (and > maybe what you meant?) was that f monotone decreasing only implies > that it's injective. For my construction to be correct, I also need > f to be surjective; this is assured if f is continous and not bounded > below. Right, of course, fg=id implies surjectivity of f. ==== > > Let f(x) be a function of one real variable and let f*(x) be its > inverse. > > Suppose f(x)=f*(x) for all x. Is anything known about the general > solution set of this equation? > > The only solution I can think of is f(x)=g*(a/g(x)). Can you > determine any other solutions? > > 0. What exactly are a and g above? What exactly is a/g(X) > when g(X) is 0 (as it must be for some X, in order for g to > have an inverse)? a is a constant. If g is an invertible function with inverse g* then the function f(x)=g*(a/g(x)) has the property that f(f(x)=x. For solutions to f=f* my primary interest is in differentiable functions on the interval (b, infinity). I was also thinking that f(x) could equal f*(x) for all x. I now suppose that solutions for that case are greatly limited. What's going on? I see that f(x)=-x+a and f(x)=x is such that f(f(x)=x for all x but I can't think of any other solution. Please help. Eugene Shubert http://www.everythingimportant.org ==== I believe that the function f(x)=arcsinh(1/sinh(x)) is plenty explicit. Eugene Shubert http://www.everythingimportant.org boundary=----=_NextPart_000_0044_01C35F5C.03259350 ==== --------------------------------------------------------------------- my question is wheather the equality == is established or not. (where, A,B,C is a vector and <,> means inner product, x means outer product) I look for this in my linear algebra book, but there isn't about this. If this is true, then please let me know why this is true. The reason for wondering this, is following statement in my differential geometry book. I read : k g(geodesic curvature)= = = thanks in advance... ==== > > my question is wheather the equality == is > established or not. (where, A,B,C is a vector and <,> means inner > product, x means outer product) If outer product means the vector product in R^3, then this is almost true. What is true is that = = but actually = - . -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > this is originally informatic problem, but it's a math challenge too. > I cant figure out what are the secret numbers they want me to search > for such that the conversation can be possible ( see the problem). Problem C > Secret Numbers Input File: C.DAT > Program Source File: C.PAS or C.C or C.CPP Two natural numbers a and b are chosen (1 multiple of a and b (a*b), and person S is told the sum of a and b > (a+b). The discussion between M and S goes like this: M: I do not know the numbers a and b. > S: I do not know them either, but I knew you would not know them. > M: Now I know the numbers! > S: Now I know them, too! Input: The input is given in a text file. The input file contains > pairs of natural numbers x and y (2<=x input is guaranteed to be correct. Output: For each pair x, y, find all pairs of a and b, such that > x<=a in a single line, and finish that line with no more pairs. if there > are a and b found in the given range, or write simply no pairs. if > there are not. Separate the numbers of a pair with a comma, terminate > each pair with a semi-colon, and separate different pairs with a blank > after the semi-colon, as shown in the example below. Example: > input > 2 10 <-no pairs > 2 20 <-4,13; no more pairs this input is giving two cases: > 1) x=2 and y=10 and it says no pairs for that that range > 2) x=2 and y=20 and it returns the single possible pair a=4 and b=13, > no more pairs in the range I tried all kinds of tricks from crytography to understand what kind > of numbers I am seeking for, but I failed. > Does any1 see the maths behind those numbers? You do not need crytography (whatever that is) tricks for this problem. What you do need, in the problem definition, is an upper bound on a and b. Anyway. Suppose a=3 and b=4. M is told 12 and S is told 7. M deduces a=2, b=6 or a=3, b=4 So M would say I do not know the numbers a and b. S deduces a=2, b=5 a=3, b=4 So S would say, I do not know them either BUT, if a=2 and b=5, M would have been told 10, and would have been able to deduce a and b. SO S cannot say but I knew you would not know them. S can only say that if all pairs that add up to the sum give an ambiguous product to M. There is no magic formula for this puzzle. Hope this helps. You may like to look at the thread http://tinylink.com/?WZNSsZjEv2 -- Clive Tooth http://www.clivetooth.dk ==== What is the most complex three dimensional combination of rotations you can get before it can be described by a simpler combination of rotations? IE, how many different kinds of rotations and rotational accelerations and torques, etc. can you all put together gyroscopically before you have to describe it in terms of something simpler? A better way of putting it is this: What is the most complex rotation function or class of rotation functions you can make? That is, when you have the most complex rotation function or class of rotation functions, and additional rotation function or class of rotation functions you add to it does not increase its complexity. To envision this question, imagine I have a gyroscope with dozens of different circles, or chambers, and I set them in motion, one by one, and describe the complex rotation by a number of mathematical function. Exactly how many times can I do this before I don't have to describe it by any more numbers of mathematical functions? Or is there a limit at all? And now the physics part of the question: Is there an angular energy and angular momentum four vector such that we get an invariant quantity? Or do we count any energy and momentum gotten from rotations and spinning as just regular energy and momentum? Meaning that things that spin, even if they're massless, can not go the speed of light if they have any energy in their spinning, meaning they also cannot have momentum? Or if they do have momentum, they must have energy, and cannot go the speed of light? I'm just wondering this, because it seems to me that this means that light, which is a number of photons, is NOT the fastest thing in the universe then, since it has angular momentum, but no net momentum in the axises orthogonal, or normal, to its motion, yet it still would have energy, making it have nonzero mass, and therefore not going at the universal top speed. Or is the spin something that doesn't follow these rules, IE, something that is entirely nonphysical? (...Starblade Riven Darksquall...) ==== In message <4aa861fb.0308110143.3b189ff5@posting.google.com>, Starblade >What is the most complex three dimensional combination of rotations >you can get before it can be described by a simpler combination of >rotations? IE, how many different kinds of rotations and rotational >accelerations and torques, etc. can you all put together >gyroscopically before you have to describe it in terms of something >simpler? Addition of infinitesimal rotations commutes. Angular rates of change are limits involving infinitesimal rotations. So any number of angular momenta, torques etc. can be summed according to the normal rules of vector arithmetic. A better way of putting it is this: What is the most complex rotation >function or class of rotation functions you can make? A (pseudo) vector. >That is, when >you have the most complex rotation function or class of rotation >functions, and additional rotation function or class of rotation >functions you add to it does not increase its complexity. To envision this question, imagine I have a gyroscope with dozens of >different circles, or chambers, and I set them in motion, one by one, >and describe the complex rotation by a number of mathematical >function. Exactly how many times can I do this before I don't have to >describe it by any more numbers of mathematical functions? Or is there >a limit at all? No limit. Its angular momentum is a single vector quantity. And now the physics part of the question: Is there an angular energy >and angular momentum four vector such that we get an invariant >quantity? Or do we count any energy and momentum gotten from rotations >and spinning as just regular energy and momentum? Meaning that things >that spin, even if they're massless, can not go the speed of light if >they have any energy in their spinning, Why not? >meaning they also cannot have >momentum? Or if they do have momentum, they must have energy, and >cannot go the speed of light? Why not? I'm just wondering this, because it seems to me that this means that >light, which is a number of photons, is NOT the fastest thing in the >universe then, since it has angular momentum, but no net momentum in >the axises orthogonal, or normal, to its motion, yet it still would >have energy, making it have nonzero mass, Why would having energy give it nonzero mass? Here's a hint: E = mc^2 is not generally true. >and therefore not going at >the universal top speed. Or is the spin something that doesn't follow these rules, IE, >something that is entirely nonphysical? (...Starblade Riven Darksquall...) Reading too many fantasy novels is bad for you. -- Richard Herring ==== [snip question asked several ways] > A better way of putting it is this: What is the most complex rotation > function or class of rotation functions you can make? The 3-D rotation group. IIRC, it's SO(3), special, orthogonal, 3 by 3 matrices. > To envision this question, imagine I have a gyroscope with dozens of > different circles, or chambers, and I set them in motion, one by one, > and describe the complex rotation by a number of mathematical > function. Exactly how many times can I do this before I don't have to > describe it by any more numbers of mathematical functions? Or is there > a limit at all? I'm sorry, but you faded there. What does it mean for a gyroscope to have circles? What does it mean for a gyroscope to have chambers? Are you talking about adding different gyrscopes with different rotating wheels? [vague, hazy question about angular momentum and relativity snipped] Yes there is a relativistic formulation of angular momentum. No, it does not mean that light can't travel at c. Socks ==== Virgil refers to a proof of Cantor of the uncountability of the reals. I believe he refers to the nested sequences have real endpoints argument. Virgil claims to not know the meaning of the words antidiagonal or antisemitridiagonal, or rather, to have very little idea. The diagonal of the matrix is the sequence with the 'ith element of the sequence being the element in the i'th row and ccolumn of the matrix. The tridiagonal is the set of three sequences consisting of one that is having as its i'th element the element in the i+1'th column and i'th row, the second being the diagonal, and the third having the i'th element being the element in the i'th column and i+1'th row of the matrix. The semitridiagonal is the set of the first and second or second and third sequences of the tridiagonal. An antidiagonal is a sequence differing at each place from the diagonal. An antisemitridiagonal is a set of two sequences each differing in each place from the two sequences of the semitridiagonal. Virgil, the antisemitridiagonal argument is about the 01 and 10 subsequences. Virgil claims on behalf of the status quo that an infinite set can not map to another infinite set that is not its powerset or a superset of its powerset, in that he claims that the rationals are not allowed to map to the powerset of the integers. I say powerset of integers is not powerset of rationals and powerset of integers is not superset of powerset of rationals, as well as powerset of integers is proper subset of powerset of rationals. The powerset proof has multiple lines of attack against it from that an infinite set is not precluded from mapping to the powerset of one of its proper subsets, the powerset argument fails to show otherwise, except that the mapping would invalidate it. The powerset argument only applies to a give infinite set and its very own powerset, or that of a set isomorphic to it in the way the evens are isomorphic to the odds, in the naturals. I just say there is no immediate injection from a set to its powerset. I think there are much better ways to gauge the relative sizes or densities of infinite sets than cardinality, for example the value of the asymptotic density of the subset of naturals in the naturals, and the various algebraic properties of the sequences that can be said to comprise the set. I say there are more reals than rationals because there are infinitely many proper supersets of the rationals that are proper subsets of the reals. I say the same for N and Z and E and N. I have a different context and reason for claiming that not all the elements of Z+ are elements of N, that being about how Z+ is a proper subset of N. There's been some discussion on the list recently about the probability that a random real form a uniform distribution over the reals is a rational. I think that the rationals and irrationals alternate on the real number line, and that the probability of a real being a rational is one half. One half is certainly positive. The asymptotic density of the evens in the naturals is one half, the probability of a random element from a uniform distribution over the naturals being an even number is that value. Virgil, you're pretty funny. It's nice of you to drop defense of the diagonal argument. Virgil, or anyone who cares, please put forward Cantor's other proof. If it's the one I think it is, it applies to rationals as well as reals, nested sequences have rational endpoints. Ross ==== > Virgil refers to a proof of Cantor of the uncountability of the reals. > I believe he refers to the nested sequences have real endpoints > argument. > Cantor's first proof of the uncountability of the reals was indeed based on strictly nested sequences of compact real intervals which do, of necessity have real endpoints, but it is the intersection of all of the intervals of such a sequence which is of interest, since as Cantor constructed it (1) it cannot be empty and (2) none of its points are in the listing from which it was constructed, so that the listing must be incomplete. > > Virgil claims on behalf of the status quo that an infinite set can not > map to another infinite set that is not its powerset or a superset of > its powerset, Not at all, what I claim, and Contor proved, is that no set can be mapped onto any set of cardinal greater thanor equal to that of its power set. What Ross claims I said would prevent some sets from being mapped onto themselves. And while I do make mistakes, I did not make that one. in that he claims that the rationals are not allowed to > map to the powerset of the integers. Take any bijection from the rationals to the integers, say f:Q -> I, then g(x) = {f(x)} maps from the rationals to the power set of the integers, g Q -> P(I), but not surjectively. I say powerset of integers is > not powerset of rationals and powerset of integers is not superset > of powerset of rationals, as well as powerset of integers is proper > subset of powerset of rationals. Am I supposed to deny any of this? What is the case which Ross does not mention, however, is that those powersets are of equal cardinality. > The powerset proof has multiple > lines of attack against it from that an infinite set is not precluded > from mapping to the powerset of one of its proper subsets, the > powerset argument fails to show otherwise, except that the mapping > would invalidate it. The powerset argument only applies to a give > infinite set and its very own powerset, or that of a set isomorphic to > it in the way the evens are isomorphic to the odds, in the naturals. I cannot parse that last sentece in any way that allows it to make sense. can anyone else? > I just say there is no immediate injection from a set to its powerset. I diagree. given set A and its powerset P(A), then f: x -> {x} is immediately and trivially an injection from A to P(A), unles Ross means by immediate injection something which he has not explained here. > > I think there are much better ways to gauge the relative sizes or > densities of infinite sets than cardinality, for example the value of > the asymptotic density of the subset of naturals in the naturals, and > the various algebraic properties of the sequences that can be said to > comprise the set. Ross has been here before. There areother ways of sizing sets, but Cantor's has, in the long run, been the most fruitful. > > I say there are more reals than rationals because there are infinitely > many proper supersets of the rationals that are proper subsets of the > reals. I say the same for N and Z and E and N. As long as Ross gives a clear definition of what he means by more and as long as that definition is a partial order on the class of sets, he has something that will work, at least for him. But if Ross wants everyone else to give up what works for them in favorof his own ordering, it will have to be a lot more useful than Cantor's. > > I have a different context and reason for claiming that not all the > elements of Z+ are elements of N, that being about how Z+ is a proper > subset of N. If, by Z+, Ross means the positive integers, then for Z+ to be a subset of N we also need N a subset of Z, and not all constructions of number systems require this. On the other hand, for every such construction there is a natural mapping from N to Z which preserves additions and multiplications and under which, Z+ is a subset of the image of N. > > There's been some discussion on the list recently about the > probability that a random real form a uniform distribution over the > reals is a rational. I think that the rationals and irrationals > alternate on the real number line, and that the probability of a real > being a rational is one half. In order for Ross to support this allegation of alternation, Ross needs to give a method for finding the next number after any given number in the reals, or at least show that such a method is possible I await with interest Ross' futile attempts to do so. One half is certainly positive. The > asymptotic density of the evens in the naturals is one half, the > probability of a > random element from a uniform distribution over the naturals being > an even number is that value. There is no probabilistic uniform distribution over the naturals possible. > > Virgil, you're pretty funny. It's nice of you to drop defense of the > diagonal argument. The diagonal argument is quite valid, but unnecessary, since other proofs exist. In order to invalidate a theorem, one must either (1) find a counterexample or (2) invalidate EVERY proof. Ross has (1) found no counterexamples R's uncountbility. Ross has (2) not invalidated any proof of R's uncountability. Therefore, R remains uncoountable. > Virgil, or anyone who cares, please put forward Cantor's other > proof. If it's the one I think it is, it applies to rationals as well > as reals, nested sequences have rational endpoints. Try Google. They are very good at finding things. ==== So how many of you have designed your own numerals? Sixteen should cover the most used bases (including the cancerous one based on the first and third prime). Surely you don't use the terribly designed Arabic ones? I have my own set brewing myself, just wanted to check in on all of yours. ==== Let f(n), g(n) and h(n) be three positive functions. Prove or disprove: min{f(n), g(n), h(n)} = Omega(f(n) + g(n) + h(n)) There's a dispute that this can be proven whereas I think it can't be. For three positive functions, I believe it can be graphically shown that the minimum of either of these functions will not (and cannot) have an asymptotic lower bound of these functions altogether. I believe the following would be correct: min{f(n), g(n), h(n)} = O(f(n) + g(n) + h(n)) which states that the minimun of the three positive functions will have an asymptotic upper bound of a function I(n) such that: I(n) = f(n) + g(n) + h(n) Does anyone have a different view? -Andre ==== >Let f(n), g(n) and h(n) be three positive functions. Prove or disprove: >min{f(n), g(n), h(n)} = Omega(f(n) + g(n) + h(n)) Of course not. What if f(n) = g(n) = 1 and h(n) = n? >There's a dispute that this can be proven whereas I think it can't be. >For three positive functions, I believe it can be graphically shown that >the minimum of either of these functions will not (and cannot) have an >asymptotic lower bound of these functions altogether. No. It _can be_ true that min{f(n), g(n), h(n)} = Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)). > I believe the >following would be correct: >min{f(n), g(n), h(n)} = O(f(n) + g(n) + h(n)) Yes, of course, and the constant is 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > >>Let f(n), g(n) and h(n) be three positive functions. Prove or disprove: > > >>min{f(n), g(n), h(n)} = Omega(f(n) + g(n) + h(n)) > > > Of course not. What if f(n) = g(n) = 1 and h(n) = n? Yes, then the conjecture would be false: min(f(n), g(n), h(n)) would be = 2 2 not greater than C1(2+n), where C1 is some constant. >>There's a dispute that this can be proven whereas I think it can't be. >>For three positive functions, I believe it can be graphically shown that >>the minimum of either of these functions will not (and cannot) have an >>asymptotic lower bound of these functions altogether. > > > No. It _can be_ true that min{f(n), g(n), h(n)} = Omega(f(n)+g(n)+h(n)), > namely if f(n) = Theta(g(n)) = Theta(h(n)). It still will not be true because the minimum of f(n) = Theta(g(n)) = Theta(h(n)) will be f(n) as it'll always be bounded by g(n) and h(n). In that case, f(n) will still not be larger than Omega(f(n)+g(n)+h(n)). For example, if f(n) = g(n) = 1 and h(n) = n: 1 != Omega(2+n) -Andre >>I believe the >>following would be correct: > > >>min{f(n), g(n), h(n)} = O(f(n) + g(n) + h(n)) > > > Yes, of course, and the constant is 1. > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 ==== >> >Let f(n), g(n) and h(n) be three positive functions. Prove or disprove: >min{f(n), g(n), h(n)} = Omega(f(n) + g(n) + h(n)) >There's a dispute that this can be proven whereas I think it can't be. >For three positive functions, I believe it can be graphically shown that >the minimum of either of these functions will not (and cannot) have an >asymptotic lower bound of these functions altogether. >> No. It _can be_ true that min{f(n), g(n), h(n)} = Omega(f(n)+g(n)+h(n)), >> namely if f(n) = Theta(g(n)) = Theta(h(n)). >It still will not be true because the minimum of f(n) = Theta(g(n)) = >Theta(h(n)) will be f(n) as it'll always be bounded by g(n) and h(n). In >that case, f(n) will still not be larger than Omega(f(n)+g(n)+h(n)). min{f(n), g(n), h(n)} could be any of the three, or sometimes one and sometimes another. Let's consider f(n). If f(n) = Theta(g(n)) = Theta(h(n)), there are constants C and D such that f(n) >= C g(n) and f(n) >= D h(n). So f(n) + g(n) + h(n) <= (1 + 1/C + 1/D) f(n), i.e. f(n) >= 1/(1 + 1/C + 1/D)(f(n) + g(n) + h(n)) which says f(n) = Omega(f(n) + g(n) + h(n)). Similarly, g(n) and h(n) are Omega(f(n) + g(n) + h(n)), and so is min(f(n), g(n), h(n)). >For example, if f(n) = g(n) = 1 and h(n) = n: Then f(n) is not Theta(h(n)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > min{f(n), g(n), h(n)} could be any of the three, or sometimes one and > sometimes another. > Let's consider f(n). > If f(n) = Theta(g(n)) = Theta(h(n)), there are constants C and D such that > f(n) >= C g(n) and f(n) >= D h(n). So True, but then the following also holds for this: f(n) <= E g(n) and f(n) <= F g(n), where E and F are constants. Shouldn't these inequalities be held true? -Andre > f(n) + g(n) + h(n) <= (1 + 1/C + 1/D) f(n), i.e. > f(n) >= 1/(1 + 1/C + 1/D)(f(n) + g(n) + h(n)) > which says f(n) = Omega(f(n) + g(n) + h(n)). > Similarly, g(n) and h(n) are Omega(f(n) + g(n) + h(n)), and so is > min(f(n), g(n), h(n)). > > >>For example, if f(n) = g(n) = 1 and h(n) = n: > > > Then f(n) is not Theta(h(n)). > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 > > > ==== what I actually meant was that if what you said was the case, then the following would also hold: f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants => f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) and this would then mean that: min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + g(n) + h(n)) The reason being that when you choose f(n) = Theta(g(n)) = Theta(h(n)), you're saying that f(n) is asymptotically tightly bounded (upper and lower) by g(n) as well as h(n). However when we say Omega(f(n) + g(n) + h(n)), it is my understanding that this may or may not be true for Big-Oh(f(n) + g(n) + h(n)). Please correct me if I'm wrong... can we really prove a conjecture like that using Theta and then only proving for the lower bounds (i.e. Omega) and leave out the upper bounds? -Andre > > > >> min{f(n), g(n), h(n)} could be any of the three, or sometimes one and >> sometimes another. Let's consider f(n). >> If f(n) = Theta(g(n)) = Theta(h(n)), there are constants C and D such >> that >> f(n) >= C g(n) and f(n) >= D h(n). So > > > True, but then the following also holds for this: f(n) <= E g(n) and > f(n) <= F g(n), where E and F are constants. Shouldn't these > inequalities be held true? > > > -Andre > >> f(n) + g(n) + h(n) <= (1 + 1/C + 1/D) f(n), i.e. >> f(n) >= 1/(1 + 1/C + 1/D)(f(n) + g(n) + h(n)) >> which says f(n) = Omega(f(n) + g(n) + h(n)). >> Similarly, g(n) and h(n) are Omega(f(n) + g(n) + h(n)), and so is >> min(f(n), g(n), h(n)). > For example, if f(n) = g(n) = 1 and h(n) = n: >> Then f(n) is not Theta(h(n)). >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada V6T 1Z2 > ==== >> min{f(n), g(n), h(n)} could be any of the three, or sometimes one and >> sometimes another. >> Let's consider f(n). >> If f(n) = Theta(g(n)) = Theta(h(n)), there are constants C and D such that >> f(n) >= C g(n) and f(n) >= D h(n). So True, but then the following also holds for this: f(n) <= E g(n) and >f(n) <= F g(n), where E and F are constants. Shouldn't these >inequalities be held true? Indeed. f(n) = Theta(g(n)) means C g(n) <= f(n) <= E g(n) for some constants C and E, and sufficiently large n. So? Why does that bother you? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >what I actually meant was that if what you said was the case, then the >following would also hold: >f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants >=f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. >f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) >and this would then mean that: >min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) >and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + g(n) + >h(n)) There is no contradiction between these. If it's Theta, then it's also Omega. Let's back up a bit. You said >For three positive functions, I believe it can be graphically shown that >the minimum of either of these functions will not (and cannot) have an >asymptotic lower bound of these functions altogether. To which I replied > No. It _can be_ true that min{f(n), g(n), h(n)} = Omega(f(n)+g(n)+h(n)), > namely if f(n) = Theta(g(n)) = Theta(h(n)). So my statement was that in this case, where f(n) = Theta(g(n)) and f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n)) Now, as you noticed, more than that is true in this case: min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n)) But that's fine: saying it's Theta(...) means that it's both O(...) and Omega(...). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== g(n), and h(n) can be any positive functions.. by that definition f(n) = Theta(g(n)) = Theta (h(n)) proves the conjecture true :) cheers, -Andre > >>what I actually meant was that if what you said was the case, then the >>following would also hold: > > >>f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants > > >>=>f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. >>f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) > > >>and this would then mean that: > > >>min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) > > >>and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + g(n) + >>h(n)) > > > There is no contradiction between these. If it's Theta, then it's also > Omega. > > Let's back up a bit. You said > > >>For three positive functions, I believe it can be graphically shown that >>the minimum of either of these functions will not (and cannot) have an >>asymptotic lower bound of these functions altogether. > > > To which I replied > > >>No. It _can be_ true that min{f(n), g(n), h(n)} = Omega(f(n)+g(n)+h(n)), >>namely if f(n) = Theta(g(n)) = Theta(h(n)). > > > So my statement was that in this case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)), > min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n)) > > Now, as you noticed, more than that is true in this case: > > min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n)) > > But that's fine: saying it's Theta(...) means that it's both O(...) and > Omega(...). > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 > ==== A quickie: Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n)) cheers, Anderson > g(n), and h(n) can be any positive functions.. by that definition f(n) = > Theta(g(n)) = Theta (h(n)) proves the conjecture true :) > > cheers, > > -Andre > > > what I actually meant was that if what you said was the case, then > the following would also hold: > f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants > = f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. > f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) > and this would then mean that: > min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) > and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + > g(n) + h(n)) >> There is no contradiction between these. If it's Theta, then it's also >> Omega. >> Let's back up a bit. You said > For three positive functions, I believe it can be graphically shown > that the minimum of either of these functions will not (and cannot) > have an asymptotic lower bound of these functions altogether. >> To which I replied > No. It _can be_ true that min{f(n), g(n), h(n)} = > Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)). >> So my statement was that in this case, where f(n) = Theta(g(n)) and >> f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n)) >> Now, as you noticed, more than that is true in this case: >> min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n)) >> But that's fine: saying it's Theta(...) means that it's both O(...) >> and Omega(...). >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada V6T 1Z2 > ==== > A quickie: > > Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n)) This can be clearly disproved: suppose f(n) = 1, g(n) = n then 1 = O(g(n)) can be true but not the other way round, i.e.: n != Theta(1) for any large value of n. This was quite obvious, don't you think? -Andre > > cheers, > > Anderson > > >> f(n), g(n), and h(n) can be any positive functions.. by that >> definition f(n) = Theta(g(n)) = Theta (h(n)) proves the conjecture >> true :) >> cheers, >> -Andre > >> what I actually meant was that if what you said was the case, then >> the following would also hold: >> f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants >> => f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. >> f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) >> and this would then mean that: >> min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) >> and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + >> g(n) + h(n)) > There is no contradiction between these. If it's Theta, then it's also > Omega. Let's back up a bit. You said >> For three positive functions, I believe it can be graphically shown >> that the minimum of either of these functions will not (and cannot) >> have an asymptotic lower bound of these functions altogether. > To which I replied >> No. It _can be_ true that min{f(n), g(n), h(n)} = >> Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)). > So my statement was that in this case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n)) Now, as you noticed, more than that is true in this case: min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n)) But that's fine: saying it's Theta(...) means that it's both O(...) > and Omega(...). Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. > Strictly speaking you've proven that > w_1 w_2 b_3 = 3 and > w_1 w_2 w_3 = f I already had that w_1 w_2 w_3 = f, and given that f is coprime to 3, b_1 + b_2 + b_3 = 3, and knowing that b_1 = b_2 = 0, b_3 = 3 Where did this come from? When I do the factorization I get: w_2w_3b_1 + w_1w_3b_2 + w_1w_2b_3 = 3 And you know that 2 of the terms are zero. So say b_1=b_2=0 So w_1w_2b_3 = 3 Show how you get your factorization b_1 + b_2 + b_3 ? Step by step, not making prior assumptions about the w's which means w_1 w_2 = 1. Remember you have the coefficients from P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) to use with > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3). > Essentially objections to how f^2 divides off now come down to > claiming that the w's are functions of m, but consider that w_1 w_2 = > 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. Strictly speaking w_1 w_2 w_3 = 3 in this case. Yup. And that is without regard to m, right? Or do you disagree? Agree. Without regard to m. You've just chosen to make different substitutions for w_1 w_2 in the two > separate examples. Which is fine, it neither proves nor disproves anything. > I'll stand by the above statement until you can show your factorization. > Nope. All I did was switch from f being coprime to 3 to f=3. That shows that the w's are constant with regard to m. Remember, w_1 w_2 = 1, at m=0, iff f is coprime to 3, but w_1 w_2 = > 3^{2/3}, when f=3, at m=0, and every other m, as they don't care about > m, so they can't be functions of m. It's just not possible. > Does anyone think they are functions of m? That is, the w's are now all constant with regard to m and have the > same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, > f=sqrt(2), but that's a special case as generally they are not, which > shows a problem with the ring of algebraic integers. I didn't ever suggest that the w's varied. I was hoping to see you prove how > you could verify that a_3 was coprime to f at m== 0. Maybe next time? Well, if the w's don't vary, then I can just check, as I do, with f > coprime to 3, which gives me the factorization P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) as the w's don't vary with m, so I can just calculate them, right? Now then, if you disagree with the values for the w's that I use > there, then you must still be holding on to the notion that they are > dependent on m. Nope. You can plug in any 2 values you want into two of the w's and then calculate the 3rd. Independent of m. Oh, I guess that's a 3rd problem I have then. I disagree with your assertion > that there's a problem with the ring of algebraic integers. Perhaps when we see > that proof that a_3 must be coprime to f, we can consider this assertion. Well a_3 = b_3, and you can just look at the factorization above, with > the information that P(0)/f^2 is coprime to f, to see that b_3 must > be. Now if you're still trying to doubt that result, then you're still > holding on to the notion of a variable dependency from the w's on m. You probably gave the answer to why you're not following the math by > showing you believe there's no problem with the ring of algebraic > integers. Perhaps you just need to show me the math of how you get to b_1 + b_2 + b_3 > I've found the Ring of Objects which includes the ring of algebraic > integers, and does not have this problem, as the b's are all included > in it. The Ring of Objects is the set of all numbers where 1 is the only > member that is both a unit, i.e. factor of 1, and an integer, where no > non-unit member is a factor of any two integers that are coprime. As I noted yesterday, there's a mistake with that definition as I > forgot about -1, so it should be -1 and 1 are the only members that > are bot ha unit and an integer. And no, I didn't catch the mistake > myself as a poster pointed it out yesterday. That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. Yup. > James Harris Phil Nicholson ==== And now I have to make a correction but luckily, it's not a big deal, but maybe it'll clear out any remaining confusion. More below... > Well, the argument which settles things is *luckily* short and rather > direct, so I'll give it here. Some may think it's exactly what > they've seen before, as I've been posting it a lot of places, but I've > seen need to put in minor corrections. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. > Strictly speaking you've proven that > w_1 w_2 b_3 = 3 and > w_1 w_2 w_3 = f > > I already had that w_1 w_2 w_3 = f, and given that f is coprime to 3, > > b_1 + b_2 + b_3 = 3, and knowing that b_1 = b_2 = 0, b_3 = 3 > > which means w_1 w_2 = 1. Yup, and I should have remembered that further down. > Remember you have the coefficients from > > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) > > to use with > > > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3). > > > Essentially objections to how f^2 divides off now come down to > claiming that the w's are functions of m, but consider that w_1 w_2 = > 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. > > Strictly speaking w_1 w_2 w_3 = 3 in this case. > > Yup. And that is without regard to m, right? Or do you disagree? Well I messed up, as w_1 w_2 is coprime to 3, and does NOT equal 3^{2/3}, which I should have known from P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) as all that happens is that when f=3, that f can now divide out, and it has to go through b_3. > You've just chosen to make different substitutions for w_1 w_2 in the two > separate examples. Which is fine, it neither proves nor disproves anything. > > Nope. All I did was switch from f being coprime to 3 to f=3. > > That shows that the w's are constant with regard to m. > > Remember, w_1 w_2 = 1, at m=0, iff f is coprime to 3, but w_1 w_2 = > 3^{2/3}, when f=3, at m=0, and every other m, as they don't care about > m, so they can't be functions of m. It's just not possible. Well I got the numbers wrong but they are constant with regard to m. The easiest way is just to look though at P(m)/3^3 = (m^3 3^3 - 3m^2(3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (b_1 x + u)(b_2 x + u)(b_3/3 x + u). You might think that maybe the 3 goes some other way, but look at P(m)/3^3 = (m^3 3^3 - 3m^2(3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) as you can see that because 3 divides off from *each* factor, there's only one way it can go, and it must be constant with regard to m. It's interesting how it can *seem* like maybe f can go a lot of ways when f doesn't equal 3, if you wish to try and force an m dependency. > That is, the w's are now all constant with regard to m and have the > same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, > f=sqrt(2), but that's a special case as generally they are not, which > shows a problem with the ring of algebraic integers. > > I didn't ever suggest that the w's varied. I was hoping to see you prove how > you could verify that a_3 was coprime to f at m== 0. Maybe next time? > > Well, if the w's don't vary, then I can just check, as I do, with f > coprime to 3, which gives me the factorization > > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) > > as the w's don't vary with m, so I can just calculate them, right? Yup. > Now then, if you disagree with the values for the w's that I use > there, then you must still be holding on to the notion that they are > dependent on m. That's going back to the poster Phil Nicholson who has been so adamnant that m=0 is a special case, but that's really just wishful thinking on his part, as the math says otherwise. Here's Phil's comments that follow. > Oh, I guess that's a 3rd problem I have then. I disagree with your assertion > that there's a problem with the ring of algebraic integers. Perhaps when we see > that proof that a_3 must be coprime to f, we can consider this assertion. > > Well a_3 = b_3, and you can just look at the factorization above, with > the information that P(0)/f^2 is coprime to f, to see that b_3 must > be. Yup. > Now if you're still trying to doubt that result, then you're still > holding on to the notion of a variable dependency from the w's on m. > > You probably gave the answer to why you're not following the math by > showing you believe there's no problem with the ring of algebraic > integers. > I've found the Ring of Objects which includes the ring of algebraic > integers, and does not have this problem, as the b's are all included > in it. The Ring of Objects is the set of all numbers where 1 is the only > member that is both a unit, i.e. factor of 1, and an integer, where no > non-unit member is a factor of any two integers that are coprime. > > As I noted yesterday, there's a mistake with that definition as I > forgot about -1, so it should be -1 and 1 are the only members that > are bot ha unit and an integer. And no, I didn't catch the mistake > myself as a poster pointed it out yesterday. > > That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. > > Yup. Cool beans. James Harris ==== And now I have to make a correction but luckily, it's not a big deal, > but maybe it'll clear out any remaining confusion. More below... Let's see... Well, the argument which settles things is *luckily* short and rather > direct, so I'll give it here. Some may think it's exactly what > they've seen before, as I've been posting it a lot of places, but I've > seen need to put in minor corrections. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. > Strictly speaking you've proven that > w_1 w_2 b_3 = 3 and > w_1 w_2 w_3 = f I already had that w_1 w_2 w_3 = f, and given that f is coprime to 3, b_1 + b_2 + b_3 = 3, and knowing that b_1 = b_2 = 0, b_3 = 3 The correct factorization is b_1w_2w_3 + b_2w_1w_3 + b_3 w_1 w_2 = 3 Yours is correct if w_1=w_2=w_3 = 1 But w_1w_2w_3 = f. Hence a contradiction. which means w_1 w_2 = 1. Yup, and I should have remembered that further down. Remember you have the coefficients from P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) to use with > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3). > Essentially objections to how f^2 divides off now come down to > claiming that the w's are functions of m, but consider that w_1 w_2 = > 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. Strictly speaking w_1 w_2 w_3 = 3 in this case. Yup. And that is without regard to m, right? Or do you disagree? Well I messed up, as w_1 w_2 is coprime to 3, and does NOT equal > 3^{2/3}, which I should have known from P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) as all that happens is that when f=3, that f can now divide out, and > it has to go through b_3. > You've just chosen to make different substitutions for w_1 w_2 in the two > separate examples. Which is fine, it neither proves nor disproves anything. Nope. All I did was switch from f being coprime to 3 to f=3. That shows that the w's are constant with regard to m. Remember, w_1 w_2 = 1, at m=0, iff f is coprime to 3, but w_1 w_2 = > 3^{2/3}, when f=3, at m=0, and every other m, as they don't care about > m, so they can't be functions of m. It's just not possible. Well I got the numbers wrong but they are constant with regard to m. The easiest way is just to look though at P(m)/3^3 = (m^3 3^3 - 3m^2(3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (b_1 x + u)(b_2 x + u)(b_3/3 x + u). > You might think that maybe the 3 goes some other way, but look at P(m)/3^3 = (m^3 3^3 - 3m^2(3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) as you can see that because 3 divides off from *each* factor, there's > only one way it can go, and it must be constant with regard to m. It's interesting how it can *seem* like maybe f can go a lot of ways > when f doesn't equal 3, if you wish to try and force an m dependency. That is, the w's are now all constant with regard to m and have the > same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, > f=sqrt(2), but that's a special case as generally they are not, which > shows a problem with the ring of algebraic integers. I didn't ever suggest that the w's varied. I was hoping to see you prove how > you could verify that a_3 was coprime to f at m== 0. Maybe next time? Well, if the w's don't vary, then I can just check, as I do, with f > coprime to 3, which gives me the factorization P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) as the w's don't vary with m, so I can just calculate them, right? Yup. Now then, if you disagree with the values for the w's that I use > there, then you must still be holding on to the notion that they are > dependent on m. That's going back to the poster Phil Nicholson who has been so > adamnant that m=0 is a special case, but that's really just wishful > thinking on his part, as the math says otherwise. Here's Phil's comments that follow. Oh, I guess that's a 3rd problem I have then. I disagree with your assertion > that there's a problem with the ring of algebraic integers. Perhaps when we see > that proof that a_3 must be coprime to f, we can consider this assertion. Well a_3 = b_3, and you can just look at the factorization above, with > the information that P(0)/f^2 is coprime to f, to see that b_3 must > be. Yup. Yup. It's well proven, many times that a_3 = 3 when m = 0. So a_3 is coprime to f when f is coprime to 3, when m = 0. You just need to extend this proof to show it's behaviour when m<>0. You have further proven I think that a_3 is not coprime to f, when f=3 for any m. So you've only got that last bit to go. Prove that a_3 is (or is not) coprime to f when f coprime to 3, and m<>0. Phil Nicholson. Now if you're still trying to doubt that result, then you're still > holding on to the notion of a variable dependency from the w's on m. You probably gave the answer to why you're not following the math by > showing you believe there's no problem with the ring of algebraic > integers. > I've found the Ring of Objects which includes the ring of algebraic > integers, and does not have this problem, as the b's are all included > in it. The Ring of Objects is the set of all numbers where 1 is the only > member that is both a unit, i.e. factor of 1, and an integer, where no > non-unit member is a factor of any two integers that are coprime. As I noted yesterday, there's a mistake with that definition as I > forgot about -1, so it should be -1 and 1 are the only members that > are bot ha unit and an integer. And no, I didn't catch the mistake > myself as a poster pointed it out yesterday. That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. Yup. Cool beans. > James Harris ==== And now I have to make a correction but luckily, it's not a big deal, > but maybe it'll clear out any remaining confusion. More below... > Let's see... > so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. > Strictly speaking you've proven that > w_1 w_2 b_3 = 3 and > w_1 w_2 w_3 = f I already had that w_1 w_2 w_3 = f, and given that f is coprime to 3, b_1 + b_2 + b_3 = 3, and knowing that b_1 = b_2 = 0, b_3 = 3 > > The correct factorization is > b_1w_2w_3 + b_2w_1w_3 + b_3 w_1 w_2 = 3 > Yours is correct if w_1=w_2=w_3 = 1 > But w_1w_2w_3 = f. > Hence a contradiction. That is correct. What I gave was wrong. The correct one gives b_3 w_1 w_2 = 3, at m=0. Well a_3 = b_3, and you can just look at the factorization above, with > the information that P(0)/f^2 is coprime to f, to see that b_3 must > be. Yup. > > Yup. > It's well proven, many times that a_3 = 3 when m = 0. So a_3 is coprime to f > when f is coprime to 3, when m = 0. > > You just need to extend this proof to show it's behaviour when m<>0. > You have further proven I think that a_3 is not coprime to f, when f=3 for any > m. > So you've only got that last bit to go. Prove that a_3 is (or is not) coprime to > f when f coprime to 3, and m<>0. > > Phil Nicholson. Ok, so you accept that f^2 divides off as a *constant* with respect to m, when f=3. However, you're now apparently still considering the possibility that it divides off as a function of m, if f does not equal 3. Is that correct? And now it seems you think I need to prove that there isn't this really weird math function that can switch from not being a function of m to being a function of m dependent on whether or not it is coprime to 3. Is that a correct assessment? James Harris ==== > Ok, so you accept that f^2 divides off as a *constant* with respect to > m, when f=3. However, you're now apparently still considering the > possibility that it divides off as a function of m, if f does not > equal 3. Is that correct? And now it seems you think I need to prove that there isn't this > really weird math function that can switch from not being a function > of m to being a function of m dependent on whether or not it is > coprime to 3. Is that a correct assessment? No not at all a correct assessment. Let v = -1 + mf^2 The a's you refer to are the 3 solutions to. a^3 + 3va^2 - (v^3 + 1) = 0 Correct? The second term is always a multiple of 3, regardless of m. When f=3 the third term is a multiple of 3, regardless of m. Therefore for the equation to exist, a is also a multiple of 3 when f=3. When m=0, we have v=-1 a^3 -3a^2 = 0 so one of the a's is equal to 3, independent of f. You claim that one of the solutions to the equation, call it a_3, is coprime to f for all other values of m and f. (At least when m and f are relatively coprime) If you have a proof for this, you're invited to present it here. The cubic is not that weird. Oh and a hint, you can prove a_3 is coprime to f if a_1/f and a_2/f are algebraic integers. So your proof will need to address the situation when a_2/f and a_1/f are not algebraic integers. Until you do that your work has a clear gap in it. So your conclusions have no value until you fill this gap. > James Harris ==== > Ok, so you accept that f^2 divides off as a *constant* with respect to > m, when f=3. However, you're now apparently still considering the > possibility that it divides off as a function of m, if f does not > equal 3. Is that correct? And now it seems you think I need to prove that there isn't this > really weird math function that can switch from not being a function > of m to being a function of m dependent on whether or not it is > coprime to 3. Is that a correct assessment? > > No not at all a correct assessment. Actually it has to be, which I'll show in a bit. > Let v = -1 + mf^2 > > The a's you refer to are the 3 solutions to. > > a^3 + 3va^2 - (v^3 + 1) = 0 > Correct? Yup. > The second term is always a multiple of 3, regardless of m. > When f=3 the third term is a multiple of 3, regardless of m. > Therefore for the equation to exist, a is also a multiple of 3 when f=3. Yup. > When m=0, we have v=-1 > a^3 -3a^2 = 0 so one of the a's is equal to 3, independent of f. Yup. > You claim that one of the solutions to the equation, call it a_3, is coprime to > f > for all other values of m and f. (At least when m and f are relatively coprime) > > If you have a proof for this, you're invited to present it here. What you didn't show is the full system which defines the a's. The expressions that show that system are P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). And as you've pointed out, > a^3 + 3va^2 - (v^3 + 1) = 0 is also true as an equation within the system. You *have* to consider the entire system, rather than try to pull pieces out of it, and claim a single piece is the entire thing. Now then, P(m) is a consideration of what I call the uber-polynomial f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) where the focus is on it as a polynomial with respect to m. That's just one of many possible looks. Now the expression is complicated enough that it pays to use m=0 to get a look at the constant term with respect to m, so I set m=0, and get P(0) = u^2 f^2(3x + uf) which tells me that with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) the factor f^2 divides off through only *two* of the a's, which is a fact that no one can dispute. What you now need though to try and disagree with the conclusion that they divide off that way for all m is the belief that they divide off differently *dependent* on m. However, looking at the constant term of the polynomial P(m), remember the uber-polynomial is considered to be a polynomial with respect to m, which is just one of many looks, does NOT constrain m, which is an important point. Consider P(x) = x^2 + 2x + 1, and P(0) = 1. Does that constrain x? Why would you think that my use of m=0, with P(m), would actually put a constraint on the system? I'm merely using m=0 to pull out the constant term with respect to m. Why do I want to look at it? Because now I can check P(m)/f^2, by again looking at *its* constant term. When I do so, I find that the constant term is P(0)/f^2 = u^2(3x + uf) so I know that f^2 divided off from it, leaving it now coprime to f, as long as x and 3 are coprime to f. But P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) and the *only* way to get P(0)/f^2 is for f to divide out through only two of the a's, so that I have P(m)/f^2 = (a_1/f x + u)(a_2/f x + u)(a_3 x + uf) as from before I know that two of the a's equal 0, when m=0, so I can see that the constant term P(0)/f^2 = u^2(3x + uf) corresponds to that factorization. Dispute with that fact require the belief that f^2 divides off with a dependency on m. Readers might find that hard to understand, but remember the *constant* term of any polynomial is that part which doesn't have the key variable. So the constant term for P(x) = x^2 + 2x + 1, doesn't have x in it. Since I can check the constant term for P(m), by setting m=0, for posters to dispute with the conclusion gathered, they must be claiming that f^2 divides off in some way with the variable m in the factors of f^2 that divide off. Remember my use of m=0 at P(m) gives me the *constant* term, just like with P(x) = x^2 + 2x + 1, x=0, gives me P(0) = 1, which is distinctive in that it does not have a dependency on x. So the refusal by Phil Nicholson to admit that he's asserting a dependency on m, is clearly against mathematical logic. But consider this exchange from earlier in the post, with my comments starting: > And now it seems you think I need to prove that there isn't this > really weird math function that can switch from not being a function > of m to being a function of m dependent on whether or not it is > coprime to 3. Is that a correct assessment? > > No not at all a correct assessment. Here the poster is fighting the *mathematical* requirements of his own assertions!!! > The cubic is not that weird. Oh and a hint, you can prove a_3 is coprime to f > if a_1/f and a_2/f are algebraic integers. So your proof will need to address > the situation when a_2/f and a_1/f are not algebraic integers. You continue to try and reduce the expressions P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) to a simpler system. Here you've tried to reduce everything down to the expression > a^3 + 3va^2 - (v^3 + 1) = 0 presumably because it's something that you understand readily. However, the system cannot be so reduced. > Until you do that your work has a clear gap in it. So your conclusions have no > value until you fill this gap. Repeatedly you make claims without being able to base them on mathematical logic. Repeatedly you try to ignore equations or claim they're more limited than they are. Why won't you at least admit the m dependency your position would require on how f^2 divides off? Repeatedly you've claimed you're not asserting such a dependency, which goes against the math. James Harris ==== > >>Lots deleted this time. >> >>I wouldn't mind doing a sanity check, then restate my problem if you don't >> mind. >Fine. >>We have >>P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = m^3 f^6 x^3 - 3m^2 f^4 x^3 + 3mf^2 x^3 +3xu^2 f^2 -3mf^4 xu^2 + u^3 f^3 This is a polynomial with four variables. The coefficients are 1, -3, 3, 3, -3, 1. If you wish to view any of m,f,x,u as a number, then it changes the terms of the polynomial and the coefficients. It is *still* a polynomial. Please stop talking about uber-polynomials as they simply don't exist. > What the math sees is an uber-polynomial, not a polynomial, as a > polynomial has constant coefficients, while with the uber-polynomial > you have all those variables. Perhaps I shouldn't respond to this, but it's just one example of how James doesn't understand the terminology, or the importance of clear definitions. As defined, P(m) is a function from polynomials to polynomials, but that's unimportant. What's been bothering me for a while is that James doesn't appear to know what a polynomial *is*. -- Will Twentyman ==== > >Your assumption that the field of algebraic numbers is required, which >you've repeated several times, but have been unable to prove. >As long as we're mindful that the b's are not always algebraic integers. > > > I have said so myself, and in fact that is why there's a problem with > the ring of algebraic integers. > > It may seem esoteric to readers on sci.physics and sci.skeptic, but > mathematics requires zero errors, and what I've managed to show with > some fascinatingly basic algebra is an error created by that > definition of algebraic integers as roots of monic polynomials with > integer coefficients. > > That definition leaves gaps by not including certain numbers that > should be included which leads to fascinating contradiction like that > ***in the ring of algebraic integers*** you can have abc = 5, where a, > b and c are coprime to 5. Can you provide even *one* example of such an a, b, c? Not something abstract, but actual numbers? > > That coprime just means they don't share non-unit factors, i.e. not > factors of 1, with 5, but they multiply together to give 5, and a, b > and c are each algebraic integers. > > Mathematicians missed this little thing for over a hundred years, but > I can prove there's a problem in the ring with a short argument using > basic algebra, which comes at the end of this post. > > -- Will Twentyman ==== > Barbier's theorem is that all curves of constant width of width w > have the same perimeter, pi * w. > > http://www.cut-the-knot.org/ctk/Barbier.shtml > > gives a proof without calculus. Is there a simple proof that uses calculus? > > http://mathworld.wolfram.com/CurveofConstantWidth.html As the width rotates around through 180 degrees = pi radians, the locus of the two ends of the width is exactly the curve of constant width. So consider what amount of perimeter is traced out by the two ends as the width rotates through an infinitesimal angle d theta. Consider the two widths, one at angle theta, and the other at angle theta + d theta. These two widths meet at some point along the widths, lets say the point of intersection divides one of the widths into pieces of length a and w-a. Not the piece of curve traced as the width goes from theta to theta + d theta is perpendicular, the lengths of these pieces of curve is a a d theta and (w-a) d theta respectively. So adding all these pieces together for all theta, we get perimeter = int_0^pi a d theta + int_0^pi (w-a) a theta which is pi*w. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen ==== let p(x)=x^2 + ax + b in R[x] if factor ring R[x]/(p(x)) is field , find basis of field R[x]/(p(x)) on R ------------------------------------ i think that basis is {1,x} but in solution paper, deg p(x) =2 a exist such that p(a)=0 thus {1,a} is basis ------------------------------- what do you think about it ? ==== >let p(x)=x^2 + ax + b in R[x] if factor ring R[x]/(p(x)) is field , find basis of field R[x]/(p(x)) on R ------------------------------------ i think that basis is {1,x} but in solution paper, deg p(x) =2 a exist such that p(a)=0 thus {1,a} is basis ------------------------------- >what do you think about it ? _If_ basis of field R[x]/(p(x)) on R means a basis for R[x]/(p(x)), regarded as a vector space over R then I think you're right and the solution paper is obviously wrong because {1, a} is not independent. There are at least three possibilities: (i) I'm wrong (ii) the solution paper is wrong (iii) my assumption that basis of field R[x]/(p(x)) on R means a basis for R[x]/(p(x)), regarded as a vector space over R is wrong, the phrase actually means something else here. ************************ David C. Ullrich ==== Let me revise that a little bit: >let p(x)=x^2 + ax + b in R[x] First we should note that there are too many a's in the notation - let's change p to p(x)=x^2 + Ax + B. >>if factor ring R[x]/(p(x)) is field , >>find basis of field R[x]/(p(x)) on R >>------------------------------------ >>i think that basis is {1,x} >>but in solution paper, >>deg p(x) =2 >>a exist such that p(a)=0 >>thus {1,a} is basis >>------------------------------- >>what do you think about it ? _If_ basis of field R[x]/(p(x)) on R means >a basis for R[x]/(p(x)), regarded as a vector >space over R then I think you're right and >the solution paper is obviously wrong because >{1, a} is not independent. There are at least three possibilities: (i) I'm wrong (ii) the solution paper is wrong (iii) my assumption that basis of field R[x]/(p(x)) on R >means a basis for R[x]/(p(x)), regarded as a vector >space over R is wrong, the phrase actually means >something else here. It's true that {1, x} is a basis (or rather {1 +

, x +

} is a basis, where

is the ideal generated by p.) But when I read the statement a exist such that p(a)=0 thus {1,a} is basis I was assuming that a was an element of R, which is obviously impossible. I suspect you didn't tell us exactly what the solution paper says: If a is an element of some _extension_ of R and p(a) = 0 then {1, a} is a basis as well. >************************ David C. Ullrich ************************ David C. Ullrich ==== oh...sorry. solution paper is ------------------------------ deg p(x) =2 for a sucht that p(a)=0 {1,a} is R[x]/(p(x)). ------------------------------ what do you think about it?? sorry, i have infirm english~ ==== oh...sorry. solution paper is ------------------------------ deg p(x) =2 for a sucht that p(a)=0 {1,a} is basis of R[x]/(p(x)). ------------------------------ what do you think about it?? sorry, i have infirm english~ ==== > let p(x)=x^2 + ax + b in R[x] > > > but in solution paper, > > deg p(x) =2 > > a exist such that p(a)=0 > > thus {1,a} is basis I think the author is using the letter a in two different senses (as a coefficient and as a zero of the polynomial p). This is very dodgy practice. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== sorry, i am fool p(x) = x^2 + Ax +B ==== > > Can anyone provide me with the citation of Cantor's paper where he >> 1/2 >> 1/3, 2/3 >> 1/4, 2/4, 3/4 >> 1/5, 2/5, 3/5, 4/5 >> 1/6, 2/6 ... >> ... So the number of rationals between 0 and 1 is the square of the number >of integers between 0 and infinity? Don't be silly; it's not the square, it's the triangular number. Shift second column up one, third column up two, etc. > So there, you've a square. Alright, it's kind of a triangle. But aren't there some redundant numbers? Namely, 2/4 = 1/2. Doesn't this kind of skew the results? If these were points in the x-y plane, then if you can draw a straight line between two equivilant numbers, then any point also on that same number is also equivilant. How do you elimiate equivilant numbers in such a manner? Also, a square number divided by two and a triangular number are essentially the same for large enough numbers, at least proportionately. That is to say, if you take a triangular number and divided it by a square number, then at large enough numbers you'll get closer and closer to 1/2 until you 'get to infinity'... or supposedly get to infinity, but it's not possible so I put it in quotes. (...Starblade Riven Darksquall...) ==== >>So the number of rationals between 0 and 1 is the square of the number >>of integers between 0 and infinity? >> Don't be silly; it's not the square, it's the triangular number. >> Shift second column up one, third column up two, etc. >> So there, you've a square. > Alright, it's kind of a triangle. But aren't there some redundant > numbers? Namely, 2/4 = 1/2. Doesn't this kind of skew the results? If > these were points in the x-y plane, then if you can draw a straight > line between two equivilant numbers, then any point also on that same > number is also equivilant. How do you elimiate equivilant numbers in > such a manner? It should be clear that every rational number appears at least once (in fact, infinitely many times) in the triangle. This is all that's needed, because of the Cantor-Bernstein theorem. If you have an injection from Q into N (which the triangular map provides; just take the smallest available n for a given q), and you also have an injection from N into Q (the inclusion map), then the C-B theorem guarantees that there is a bijection between Q and N. > Also, a square number divided by two and a triangular number are > essentially the same for large enough numbers, at least > proportionately. That is to say, if you take a triangular number and > divided it by a square number, then at large enough numbers you'll get > closer and closer to 1/2 until you 'get to infinity'... or supposedly > get to infinity, but it's not possible so I put it in quotes. The remark about triangular numbers was a joke. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== How do you define Computational Mathematics and what is it consisted of(sciences) ? This might sound a stupid question but I am dead serious. I am interested in Math for computers in general but that is so HUGE. I mean, it's all about Algebra, algorithms, etc. etc. Can anyone help me by leading me to the correct direction? ==== >> My strategy is quite straightforward, > > You've just assumed what you need to prove. I find that to be the most straightforward method of proof. Infinitely efficient. Much faster than dividing both sides of an equation by 0, for example. Phil ==== [snip] Earth to James... Please post one, count 'em: *one*, number which belongs in the ring of algebraic integers but which somehow got 'left out'. You have repeatedly claimed that the ring of algebraic integers is incomplete because it fails to include numbers which belong in it. With all due contempt I believe you have some obligation to produce at least one of them -- a number, mind you, not an ambiguous expression consisting of ill-defined symbols. Name one such number, if you can. If not, stop claiming that such numbers exist. (In your breathlessly anticipated reply, please do not include terminology like: amazing, fascinating, and odd. These expose your mental state, not the underlying mathematics of the subject at hand.) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== >> >> [.snip.] >> >>>theory: >>>> THEOREM: If r is a root of a non-monic polynomial >>> with integer coefficients, ***irreducible*** over >>> the rationals, the r cannot be an algebraic integer. >>>>and conclude that c1 cannot be an algebraic integer. >>> >>> You must also include the hypothesis that the polynomial is >>> primitive. Since nonzero constants are units in Q[x], they are not >>> considered nontrivial factors, so the hypothesis must be explicitly >>> included. >>That is correct. And it is true that c1 cannot be an algebraic >>integer. >>That has not been under debate. >> >> Which is why I said I was nitpicking: pointing out a minor error >> that is well understood. It is true that c1 cannot be an algebraic integer. Which was not what I addressed. I included the stuff before it to provide context, and then pointed out that a hypothesis was missing from the Theorem quoted. What exactly is your problem with that? Are you asserting that the Theorem was correct as written? Are you asserting that the theorem is wrong as corrected according to the mistake I pointed out? Or are you simply committing the fallacy of Argumentum ad logicam? http://www.infidels.org/news/atheism/logic.html#logicam [.snip.] >> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) >>any a's exist, within the ring of algebraic integers, such that >>sqrt(5) is a factor of them in that ring. >> >> This is rather confused. You have a not at the beginning, a whether >> or not after that, and a qualifier any for the a's. It's pretty >> close to nonsense. What you are really saying, presumably, is: >> >> Given a1, a2, a3 algebraic integers such that >> >> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) >> >> [as a polynomial identity], then none of a_1,a_2,a_3 are multiples >> (in the ring of algebraic integers) of sqrt(5). It is true that neither a_1, a_2, nor a_3 has sqrt(5) as a factor >***in the ring of algebraic integers***. > >> This is also true, and has been established. Yup as I've stated. As I've ->PROVEN<- (and was proven by Bengt, and Dale, and Dik, and Nora, and David, and many others), and as you DENIED, mightily, for months on end. But thanks for finally coming around, even if now you are trying to claim credit for a result you do not understand and cannot prove. >>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >>factors in common with 5 in the ring of algebraic integers. >> >> And that's false. I am pretty sure that Dale produced explicit common >> factors; but in any case, your claim here is certainly false, since >> their product is not coprime to 65. I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit >factors in common with 5 in the ring of algebraic integers. This is paraphrased from the post http://groups.google.com/groups?selm=3F148963.6000800%40farir.com Take the polynomial 65x^3 - 12x + 1, and factor it as 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1) with a1, a2, a3 (necessarily) algebraic integers (in fact, minus the roots of x^3 - 12x^2 + 65). Let z be any root of that polynomial. That is, z will be either -a1, -a2, or -a3. It is trivial that any common factor between z and 5 will be a common factor of the corresponding ai. Take the following three polynomials q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 Then any product of q(z), r(z), s(z) and integers will be an algebraic integer, necessarily. We have that q(z)*r(z) = 64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325 Note that (64 x + 128)*(x^3 - 12 x^2 + 65) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320 so (65z + 128)(z^3 - 12z^2 + 65) = 64z^4 - 650z^3 - 1536z^2 + 4160z+8320. But z^3-12z^2+65 = 0. So 64z^4 - 650z^3 - 1536z^2 +4160z+8320 = 0. Therefore, q(z)*r(z) = 64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325 = (64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8320) + 5 = 5. Likewise, take r(z)*s(z) = 32 z^4 - 312 z^3 - 864 z^2 + 2081 z + 4680 This time, note that (32 x + 72)*(x^3 - 12 x^2 + 65) = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680 So (32z+72)*(z^3-12z^2+65) = 32z^4 - 312z^3 - 864z^2 + 2080z + 4680. But since z^3-12z^2+65 is equal to 0, it follows that 32z^4 - 312z^3 - 864z^2 + 2080z + 4680=0. Therefore, r(z)*s(z) = 32z^4 - 312z^3 - 864z^2 + 2081z + 4680 = (32z^4 - 312z^3 - 864z^2 + 2080z + 4680) + z = z. Therefore, r(z), which is an algebraic integer, is a common factor of 5 and of z. It only remains to show that it is not a unit. The claim is that r(z) is a root of f(x) = x^3 - 969 x^2 + 315 x +5. If this is so, then r(z) is not a unit in the ring of all algebraic integers, since this is a monic irreducible polynomial with integer coefficients whose constant term is neither 1 nor -1. As Dale noted, we have f(r(z)) = (r(z))^3 - 969 (r(z))^2 + 315 (r(z)) + 5 = (8 z^2 - 4 z - 45)^3 - 969 (8 z^2 - 4 z - 45)^2 + 315 (8 z^2 - 4 z - 45) + 5 = 512 z^6 - 768 z^5 - 70272 z^4 + 70592 z^3 + 731136 z^2 - 374400 z - 2067520. Letting w(z) = 512 z^3 + 5376 z^2 - 5760 z - 31808 we have that p(z)*w(z) = 512 z^6 - 768 z^5 - 70272 z^4 + 70592 z^3 + 731136 z^2 - 374400 z - 2067520 and therefore, f(r(z)) = p(z)*w(z). But we know that p(z)=0, so f(r(z))=0. This proves that r(z) is not a unit, and yet is a common factor of z and 5. Letting z be -a1, -a2, and -a3, in turn, you obtain common factors of EACH of a1, a2, a3 with 5 in the ring of all algebraic integers, which are not units. >> Lemma. Let R be the ring of all algebraic integers, and let a, b, c be >> any elements of R. If a and b are coprime to c, then a*b is coprime to >> c. >> >> Proof. We use the characterization of coprime valid for commutative >> rings with 1: a and b are coprime in R if and only if there exist x >> and y in R such that ax+by = 1. By that definition That property is EQUIVALENT to being coprime in the algebraic integers, which is EQUIVALENT to having no common factors other than units. > only *one* of the a's is coprime to 5, but none of >them has a factor in common with 5 either. Nonsense. By that definition, NONE of the a's are coprime to 5, and ALL have common factors with 5 which are not units. See the explicit calculations above. >The ring of algebraic integers is really screwed up. Something is, but it is not the ring of algebraic integers. >For those who don't understand, consider that in the ring of evens, >which does not have 1, you can't use that definition of coprime that >Arturo Magidin gives, though it is, interestingly enough, true that in >fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is >not in the ring. The ring 2Z is not a ring with unit, so the characterization given above does not apply. As it happens, 2 and 6 are coprime in the ring in question, but NOT for the incoherent reason you give. They are coprime because the ideal generated by 2 is maximal, contains the ideal generated by 6, and is not prime. >However, rather than use dueling definitions or argue about >definitions I can simply switch to saying that 2 does not share >non-unit factors in the ring of evens with 6. But it does not matter: above I have reproduced EXPLICIT common factors of each of a1, a2, and a3 with 5 in the ring of algebraic integers, and none of them are units. Your claim is as dead as it always was. [.snip.] >> So, assume you were correct and neither a_1, a_2, nor a_3 have ANY >> non-unit factors in common with 5 in the ring of algebraic >> integers. Then, by the lemma, neither does a1*a_2; and applying the >> lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which >> clearly has 5 as a nonunit common factor with 5. This contradicts the >> assumption that none of a_1, a_2, a_3 have common non-unit factors >> with 5 in the ring of algebraic integers. Therefore, your assertion is >> false. Well by your definition of coprime It's not my definition. It's a property which is equivalent to the STANDARD definition (which is NOT has no common factors other than units, as you have been told numerous times), and which in turn is equivalent, IN THE RING OF ALL ALGEBRAIC INTEGERS, to the property that they have no common factors. > NONE of the a's have a factor in >common with 5, in the ring of algebraic integers, and you cannot prove >that any of them do. Weird. Because above I exhibited explicit factors. >Now if you don't want to call that coprime fine. It doesn't change >the situation. No, it does not. You're wrong. [.snip.] You continue to argue that a singular case implies a general case, and to argue that a numeric identity implies a polynomial one. Both are colossal errors. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== Take your favorite quadratic extension of Q. Let p be a rational prime > which splits in the ring of integers of K, (p)=PQ, with P<>Q. One can > then invert elements which are in P but not in Q to obtain a ring > larger than the ring of integers of K, which still satisfies the > condition that the only rational integers which are units are 1 and -1. That's the point of the example I posted. > One can do a symmetric argument and do the same with elements in Q > but not in P. Then one can consider the subrings of the ring of all > algebraic numbers which include the algebraic integers, and EITHER the > elements inverted from P-Q, OR the elements inverted from Q-P, but > noth both. These rings will each, separately satisfy your conditions, > and be incomparable (neither one is contained in the other). > > But the problem is that by choosing the elements carefully, one > obtains elements with the property that any ring which contains BOTH > those inverses will necessarily contain 1/p. For example, choose a > quadratic extension which is a PID, and let (p)=(a)(b), with (a)<>(b) > primes. Then take R[1/a], and R[1/b], where R is the ring of all > algebraic integers. Each of them satisfies the property you want, but > any ring which contains BOTH 1/a and 1/b will necessarily contain 1/p, > which should not be allowed in The Object Ring. That's rather obvious. To wit: any ring that contains a non-integral algebraic number w along with all its conjugates necessarily contains a non-integral rational number, since the ring then contains all the coefficients of the minimal polynomial of w, at least one of which is a non-integral rational number. Thus once one allows adjunctions of non-integral algebraic numbers there are restrictions on what later extensions are possible if such extensions are required to not contain any non-integral rational number. Contrast this with integral extensions, where there are no restrictions whatsoever on just which algebraic integers may be adjoined -- integral extensions never change a nonunit in the base ring into a unit in the extension. That is one of the key properties of integral extensions . It is of course an obvious consequence of lying-over for an integral extension R < T since: c nonunit => cR < P for a max ideal P, which is lain over by prime Q in T which necessarily contains the extension cT, so cT < 1, so c stays nonunit in T (one can also easily prove this directly by massaging the the minimal poly of c). So for algebraic integers the property of being unit or nonunit is absolute -- independent of the ambient field containing the algebraic integer. Any enlargement of the algebraic integers will fail to satisfy these crucial properties (among others). In fact one can easily show that integral extensions may be characterized in terms of the lying-over (LO) property and the incomparable (INC) property, e.g. consider the following THEOREM For commutative rings R < T the following are equivalent (1) R < T is an integral extension (2) A < B has INC and LO for all A,B with R < A < B < T (3) A < A[u] has INC and LO for all A,B with R < A < T, all u in T -Bill Dubuque ==== >> Take your favorite quadratic extension of Q. Let p be a rational prime >> which splits in the ring of integers of K, (p)=PQ, with P<>Q. One can >> then invert elements which are in P but not in Q to obtain a ring >> larger than the ring of integers of K, which still satisfies the >> condition that the only rational integers which are units are 1 and -1. That's the point of the example I posted. Yes; I believe I mentioned that the example was suggested by the post quoted (yours); at least the previous version, which contained my incorrect followup to the example, did. Sorry if I wasn't clearer in giving credit where it is due. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== function f:R -> R f(x) = (x^2)*[1/x] if x /= 0 f(x) = 0 if x=0 show, continuous on x=0 ([a] gauss notation, ex [2.4]=2) ------------------------------------ if 00 => k -> infinity thus, limf(x) = lim(x^2)[1/x] if x->0+ =lim (q^2)k / (q^2)(k^2)+2qkr+(r^2) = 0 if k->infinity similarly if -1>x>0 ~~~~lim f(x) =0 if x->0- thus left limit = right limit = function value therefore, continuous on 1 ------------------------------------------------- this solved course is possible ?? impossible?? i think that proved method was rational approach. i doubt that this satisfy all approach?? my question is that this proved course is correct ? or incorrect? thank to sir. ==== > function f:R -> R > > f(x) = (x^2)*[1/x] if x /= 0 > > f(x) = 0 if x=0 > > show, continuous on x=0 > > ([a] gauss notation, ex [2.4]=2) > > ------------------------------------ > > if 0 > if x=q/p ( p<=q) [...] > > this solved course is possible ?? impossible?? > > i think that proved method was rational approach. > > i doubt that this satisfy all approach?? > > my question is that this proved course is correct ? or incorrect? You are close to a solution in the case where f: Q->Q, where Q is the set of all rational numbers. However, since f:R->R and R is larger than Q, it turns out that the proof doesn't work for the case where the domain is R. I would suggest that you plot the graph of [1/x] for 0 function f:R -> R > f(x) = (x^2)*[1/x] if x /= 0 > f(x) = 0 if x=0 show, continuous on x=0 > ([a] gauss notation, ex [2.4]=2) > 1/x - 1 < [1/x] <= 1/x > if 0 if x=q/p ( p<=q) > Why is x rational? > natural number k,r exist such that p=qk+r (0<= r 0 => k -> infinity thus, limf(x) = lim(x^2)[1/x] if x->0+ =lim (q^2)k / (q^2)(k^2)+2qkr+(r^2) = 0 if k->infinity similarly > if -1>x>0 ~~~~lim f(x) =0 if x->0- > I've doubt that x < 0 is as easy. > thus left limit = right limit = function value > therefore, continuous on 1 > I'm not convinced. ==== The following exercise is in a book on one complex variable (by Krantz), in the section on evaluating real integrals by integrating over a contour: Calculate the integral from 0 to infinity of 1/(x^3 + x + 1). I don't know which contour to use. Any suggestions? Mike ==== >The following exercise is in a book on one complex variable (by >Krantz), in the section on evaluating real integrals by integrating >over a contour: Calculate the integral from 0 to infinity of 1/(x^3 + x + 1). I don't know which contour to use. Any suggestions? This is one where you need to do that thing where you stick in a log(z), integrate around that keyhole-shaped thing and watch the log(z) drop out in the end. >Mike ************************ David C. Ullrich X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 10:34 PM, Leon said: >Both of these theorems would be referred to as the T theorem as if >there was a unique theorem. But these are not the same sentences. They are logically equivalent. You will find cases where the same word is defined differently, or where the same name is used for different theorems. But this isn't one of them. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org ==== >>But I suspect the conjecture may imply Bertrand's postulate, i.e. if >>A_{1,1} >= 2, A_{1,j} = 1 otherwise, A_{i,j+1} = |A_{i,j} - A_{i+1,j}|, >>and A_{i,1} is strictly increasing, then A_{i+1,1} < 2 A_{i,1} >2 3 5 7 9 15 33 >1 2 2 2 6 18 >1 0 0 4 12 >1 0 4 8 >1 4 4 >1 0 ^^^ That 1 would be a 3. (Maybe I didn't say it explicitly enough, but the equation A_{i,j+1} = |A_{i,j} - A_{i+1,j}| is supposed to include the case i=1). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > > Take the difference sequence of the primes, and iteratively take the > difference sequence of the previous sequence, but make sure you take > all differences in absolute value. > > 2 3 5 7 11 13 15 17 19 . . . . > 1 2 2 4 2 2 2 2 . . . . > 1 0 2 2 0 0 0 . . . . > 1 2 0 2 0 0 . . . . > 1 2 2 2 0 . . . . > > The claim is that the leftmost diagonal is always 1 (except the initial 2). > This seems to be some sort of delicate statement > > Not all that delicate! As we shall see. It's almost worthless. > > about the distribution of primes. Has anyone seen this claim? > > Yes, in some popularization or other; and I didn't think much of it then, > and don't now. It struck me as having been made by someone with little > real number sense. (But in that, it is similar to Goldbach's.) > > I don't want to rain on anyone's parade, (well actually I don't really > mind doing that as you probably know), but this is a parade that should > have been rained out before it left the marshalling yard. > > > The conjecture will apply to almost any sequence of numbers all of > the same parity, except for a first one of different parity, that doesn't > grow too fast. And fast is loose - basically, size doubling is the limit. > > In a sense, the algorithm and initial conditions have been chosen quite > cunningly; the absolute value of the differences ensures that any local > variations (as in the primes) get smoothed out fairly quickly, and the > parity conditions ensure that there will always be a one at the very > left, provided there isn't any sudden fast growth or decay anywhere. > > As examples, here is a sequence beginning with primes 2,3,5 and continuing > to grow as fast as possible consistent with the conjecture. > > 2 3 5 9 17 33 65 129 ... essentially just doublings. > 1 2 4 8 16 32 64 ... > 1 2 4 8 16 32 ... not really very exciting, is it? > > And if some of the terms are lowered, it has no effect so long as > the subsequent terms are lowered accordingly (within VERY easy limits!), > as the absolute differences will smooth out local variations, as I say. > > As a second example, we go to the other extreme from primes. > (Contrary to what someone was hoping about composite numbers.) > > Here is a sequence of very round numbers according to a criterion > which I have devised but will keep hidden for now. > > 1 2 4 8 12 20 24 36 48 60 90 120 180 200 240 300 ... > 1 2 4 4 8 4 12 12 12 30 30 60 20 40 60 ... > 1 2 0 4 4 8 0 0 18 0 30 40 20 20 ... > 1 2 4 0 4 8 0 18 18 30 10 20 0 ... > 1 2 4 4 4 8 18 0 12 20 10 20 ... > 1 2 0 0 4 10 18 12 8 10 10 ... > 1 2 0 2 6 8 6 4 2 0 > 1 2 2 4 2 2 2 2 2 > 1 0 2 2 0 0 0 0 > 1 2 2 2 0 0 2 > 1 2 0 2 0 2 > > As you can see, they still satisfy the conjecture. > (OK I admit I had to work a little to make it so, but not a lot.) > > > Finally, I was going to make an example with the primes, but each one > plus-or-minus one, more or less at random. But it was far too easy. > > So here is an example that shows the slow growth of the primes isn't > necessary either; each term is 1.5 times the previous one, rounded > to the nearest odd number. It grows geometrically fast, though not > close to the doubling limit. > > 2 3 5 7 11 17 25 37 55 83 125 187 ... > 1 2 2 4 6 8 12 18 28 42 62 ... > 1 0 2 2 2 4 6 10 14 20 ... > 1 2 0 0 2 2 4 4 6 ... > > Again, you can plus-or-minus two to all the terms to introduce more > local randomness, but provided you don't damage the very low ones > too badly, it won't make any difference. > > > Well, ICBW, but AFAICS the conjecture is a crock. > > ----------------------------------------------------------------------------- - > Bill Taylor W.Taylor@math.canterbury.ac.nz > ----------------------------------------------------------------------------- - > If we knew what we were doing it wouldn't be called research > ----------------------------------------------------------------------------- - Bill, As I am sure you have gathered, I have very little number sense also, I will be the first to admit it. Now for the sake of me possibly learning something here I propose this one question. which uses all the primes just twice) when you force an error (reverse two adjoining primes) down about 75 terms into the sequence the 4 distinct left diagonal deltas are consistent with their patterns until about 105 terms (rows). At this point the integrity of any patterns are lost. This is understandable, but as the 4 delta diagonals continue to about 130 rows the integrity of these patterns return intact and will probably continue until ---->oo baring any other errors in the sequence! Why? Yes, I agree there is a parity issue with the even prime, but with my sequence I believe something else is happening. Can you duplicate this with another sequence that has 4 distinct left diagonal delta patterns ---->oo? Sorry, that was two questions. ;-) If you can create another sequence, with 4 distinct left diagonal delta patterns and causing an error 75 or more terms into the sequence and somewhat duplicating what happens in my sequence where the error causes a breakup of the 4 patterns for 25 of more rows and then returning to the original 4 patterns. Then I will say your argument has merit. A non-response to this post I will consider as no one can create this new sequence with the above criteria. I believe this to be an interesting challenge. Dan ==== here, there are quite a lot and I'll read them tonight. But I just thought I'd post a follow-on to my earlier remarks. I should have done this years ago when I first became underwhelmed by the conjecture as applied to primes, but I didn't, so I'm grateful to Dan for bringing up this topic, so I can now do a little more and then pass on the baton to someone else. So I'm going to look at the set of admissable sequences under the given criterion, with a view to seeing if we can find out how remarkable it IS to find that the primes are among them; and you know I suspect - not very. Actually, I'm not going to look at a sequence, but rather a sequence of of finite sequences.... yuk.... ANYWAY. Let's look at how sequences can begin, and maintain the delta-1's property. For Sloane-like standardisation, I take only sequences of even numbers that begin with a 1, so that all its deltas might be of the same sort. So any sequence must begin 1. There is just 1 of these. :) It can continue 1 0 or 1 2 - 2 initial subsequences of length 2. These may continue 1 0 0 1 0 2 1 2 0 1 2 2 1 2 4 - so 5 3-sequences. These may continue... 1000 1002 1020 1022 1024 1026 1200 1202 1204 1220 1222 1224 1240 1242 1244 1246* 1248 . There are 17 of these 4-sequences. I did these by hand so there may be mistakes, but I'm pretty confident of those up to here. I also did a quick look ahead at how many 5-sequences there were, which is not indeed very difficult when you have the previous lot written down in front of you; so computering them will be fairly easy too. Anyway, those 17 4-sequences can be followed by, respectively, 2 5 3 4 5 8 3 4 7 4 3 6 5 4 5 5 9 different continuations, giving a total of 82 admissable 5-sequences. [*]BTW - there is one particularly interesting case:- 1246 is the ONLY 4-sequence that cannot be continued with a 0! This sort of thing will become much commoner later, OC. (For example, one finds in going from length 5 to 6, there appear cases where the next admissable entries form a non-contiguous range of even numbers!) I stopped there, and leave the next many cases to someone with more efficient programming skills than me! The sequence of counts itself is an intriguing sequence... 1 2 5 17 82 ... and has a smell of factorial growth about it! I wonder if it already appears in Sloane? But just how common this makes admissable sequences among those of the same length and beginning with 1 but otherwise even, is still open to definition. I suppose we would have to limit it to sequences whose every term was bounded by the appropriate power of 2, as given by the maximum sequence 1 2 4 8... These universes of finite sequences are indeed factorial in number! And the admissable sequences as a proportion of their universe does not (as one might thus expect) plummet at all fast - they are surprisingly common, as it happens. Here are the counts:- size of # of length universe admissables 1 1 1 100% 2 2 2 100% 3 6 5 83% 4 24 17 71% 5 120 82 65% 6 720 ??? 7 5040 ? ...and I eagerly await a great mountain of figures! 8 . . ==================================== Another snippet. There is OC the delta way to reduce an (n+1)sequence to an n-sequence, which can be reversed in many different ways. (It's a many-one function.) But there is also a very gross way in which an n-sequence can be extended to an (n+1)-sequence, namely just by adjoining another entry, as I've been doing above. (The reverse, truncation, is also a many-one function OC.) So one can thus make up an intriguing directed graph, or rather two such, a trivial one obtained by adjoining admissable entries, and a more tricky one obtained by differencing a sequence to a one-shorter one. One can thus also superimpose these two to get a two-sorted digraph of all admissable finite sequences, with 2 sorts of arcs, by adjoining & delta-ing respectively. The bi-digraph so obtained has some very intriguing near-patterns, but not quite perfect enough to be general rules. It would also be a fun study though. However, none of the above has any very direct relevance to the INFINITE admissable sequences (such as primes?)! :-) Well then; there y'go. ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- Education is a process of telling a carefully chosen sequence of lies in which the amount of deliberate deception gradually tends towards zero. - John Baez ---------------------------------------------------------------------------- -- ==== |>>I am iterating a certain function, and I keep all the resultant values |>>in an array: e:=array(0..499); |>>I need a local proc(e,n) which takes as input the array of entries, its |>>actual size (actual size may be less than 500) and heuristically tries |>>to determine if the entries are close to a k-cycle. On second thought, why not just do it this way? > findperiod:= proc(e, n, eps) local i,k; for k from 1 to n-1 do for i from 0 to n-k-1 while abs(e[i] - e[i+k]) < eps do od: if i = n-k then RETURN(k) fi; od; 0 end; This should work with complex data, and with all versions of Maple. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== [snip] > |>>I need a local proc(e,n) which takes as input the array of entries, its > |>>actual size (actual size may be less than 500) and heuristically tries > |>>to determine if the entries are close to a k-cycle. > > On second thought, why not just do it this way? > > findperiod:= proc(e, n, eps) > local i,k; > for k from 1 to n-1 do > for i from 0 to n-k-1 while abs(e[i] - e[i+k]) < eps do od: > if i = n-k then RETURN(k) fi; > od; > 0 > end; > > This should work with complex data, and with all versions of Maple. Hmm, I am getting some strange results with your findperiod. Let's look at the code. First, the code I am using for the iteration (f(z)=c^z, and I am iterating using z=1): # c: complex base # z: first complex exponent # n: iteration count # m: cycle count # o: show cyclic component count > iterf:=proc(c,z,n,m,o,eps) local i,j,k,l,e; i:=evalf(z); e:=array(0..499); for k from 0 to n-1 do i:=evalf(c^i); e[k]:=i; if abs(evalf(Re(i)))>1e5 then printf(`real component out of bounds`); break; fi; if abs(evalf(Im(i)))>1e5 then printf(`imaginary component out of bounds`); break; fi; if abs(evalf(Re(i)))<1e-5 then j:=evalf(Im(i)); #if Re close to zero, zap Re i:=evalf(j*I); e[k]:=i; #update array entry fi; if abs(evalf(Im(i)))<1e-5 then j:=evalf(Re(i)); #if Im close to zero, zap Im i:=evalf(j); e[k]:=i; #update array entry fi; if (k+o) mod m = 0 then print(i); fi; od; findperiod(e,n,eps); #call Robert's code after iteration is done. end: >findperiod:= proc(e, n, eps) local i,k,f; for k from 1 to n-1 do for i from 0 to n-k-1 while abs(e[i] - e[i+k]) < eps do od: if i = n-k then RETURN(k); fi; od; 0; end: >eps:=1e-2; eps := .1 > iterf(3-3*I,1,10,1,0,eps); #works correctly! 3. - 3. I 6.642369647 - 2.875670132 I -1540.034993 - 80.14662298 I 0 1. 3. - 3. I 6.642369647 - 2.875670132 I -1540.034993 - 80.14662298 I 0 1. 5 But: > t:=(n,k)->exp(2*k*Pi*I/n); >eps:=1e-1; > iterf(t(3,2),1,33,1,0,eps); [...snip results] 32 While: > iterf(t(3,2),1,30,1,0,eps); > iterf(t(3,2),1,31,1,0,eps); > iterf(t(3,2),1,32,1,0,eps); All return 0 (cannot find a cycle) Now, 3-3*I is a true 5-cycle, whereas the 2nd, 3rd root of unity t(3,2) appears to be a pseudo 17-cycle, which coalsesces into a 1-cycle. Your findCycle returns 32, for eps=0.1 and n > 32 and 0 for eps < 0.1 or n < 32. Is k=32 correct in this case? It appears it should rather be k=17, at least before the values eventually coalesce into the 1-cycle. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. ==== >[snip] >While: >> iterf(t(3,2),1,30,1,0,eps); >> iterf(t(3,2),1,31,1,0,eps); >> iterf(t(3,2),1,32,1,0,eps); All return 0 (cannot find a cycle) I'm not sure what's going wrong, but it looks to me like for example iterf(t(3,2),1,30,1,0,eps) doesn't assign values to e[k] for k >= 30. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > > I am iterating a certain function, and I keep all the resultant values > in an array: e:=array(0..499); > > I need a local proc(e,n) which takes as input the array of entries, its > actual size (actual size may be less than 500) and heuristically tries > to determine if the entries are close to a k-cycle. > > I tried to figure out a quick and dirty solution to this, but after a > while it became obvious that the problem is rather difficult. > > For starters one needs to make several assumptions. One needs to have an > eps, which determines how close the entries need to be to be considered > a match for the cycle and I need the proc to tell the user in case > there is no match. > > There are other difficulties also: Cases where there are accidental > matches, but where the rest of the entries do not comprise a cycle. > > It looks to me that in the worst case, the algorithm seems to be O(n^2), > since I need to check all entries against all previous ones for matches. > > I'd appreciate if anyone can help a bit with this. > > -- > Ioannis > http://users.forthnet.gr/ath/jgal/ > ___________________________________________ > Eventually, _everything_ is understandable. You have options here ... restart; with(networks): MUTE:=(proc() NULL end): findperiod:= proc(e, n, eps) local i,k,store,j; j:=1; for k from 1 to n-1 do for i from k+1 to n do if abs(e[i] - e[k]) < eps then store[j]:=[i,k]; j:=j+1; break;fi od od; store; end: #Remove the break in findperiod if you want all the loops RANDPOINT:=subs(R1=rand(360),R2=rand(1024),unapply(('proc(alpha,r) evalf(r*(cos(alpha)+I*sin(alpha))) end')('R1()'/180*Pi,'R2'()/1024*m+delta),delta,m)): print(`Setting up random point set :`); assign('Px'=mul(x-RANDPOINT(1,1),i=1..8)),assign('Pxf'=unapply(Px,x)),print( `Morfing poly set up`): assign('fx'=RootOf(Px+x0,x)), assign('fx0f'=(evalf@RootOf)(Px,x)), assign('pf'= unapply( subs( subs( [(proc(P,x,x0,p,xn) local i;seq((D@@i)(p)(xn)=unapply(diff(P,x$i),x)(x0),i=1..degree(P,x)) end)(collect(Pxf(x),x),x,fx0f,p,fx0)], [seq(convert(diff(g(x0),x0$i),D)=subs(RootOf(p(_Z)+x0)=fx0,collect((diff)(Ro otOf(p(x)+x0,x),x0$i),D)),i=1..8)]), x0=0,g(0)=fx0f, convert(series(g(x),x=x0,9),polynom) ), x)),print(`Morfing function`=eval(pf)); assign('PS'=[seq((evalf@pf)(RANDPOINT(0,1)),i=1..100)]),plots[complexplot](P S,style=point,title=`Point set`); print(`Random point set ready`); assign('sol'=findperiod(PS,nops(PS),0.001)),print(`Loops located`); print(`Loop count`=nops(map(op,convert(sol,set)))),assign('CS=CS'),MUTE(new('CS'),addver tex(map(op,convert(sol,set)),'CS'),map(proc(L) addedge({op}(L),'CS'); end,((proc(L) local i; [seq([L[i-1],L[i]],i=2..nops(L)),(op@map)(op,[entries](sol))] end)@sort@[op]@map)(op,convert(sol,set)))),print(plots[display](draw(CS),col or=red)); Chris ==== Now, 3-3*I is a true 5-cycle, whereas the 2nd, 3rd root of unity t(3,2) > appears to be a pseudo 17-cycle, which coalsesces into a 1-cycle. It looks rather like a 16-cycle. That explains number 32 - two cycles. In Maple 9, one can use FFT for finding the cycle length. Here is the procedure, C:=proc(V,h) local VR,VC,a,b,c; VR,VC:=map(Re,V),map(Im,V): a:=DiscreteTransforms:-FourierTransform(VR); b:=DiscreteTransforms:-FourierTransform(VC); # print(plots[listplot](map(abs,convert(a,list)))); # print(plots[listplot](map(abs,convert(b,list)))); c:=proc(L) local k; for k from 2 to op([2,2],L)-1 do if abs(L[k])>abs(L[k-1]) and abs(L[k])>=abs(L[k+1]) and abs(L[k])>h then return round(op([2,2],L)/(k-1)) fi od; 1 end; ilcm(c(a),c(b)) end: V should be a complex Array and h is the minimal height of peaks. I'll use h=1.5 in most of the examples below. Example: V:=Array(1..500,x->x mod 20 + (x mod 7)*I,datatype=complex[8]): C(V,1.5); 140 Uncommenting prints in the procedure C, one can see the spectrograms of the FFT of real and imaginary parts of V. Now, I'll use the following iteration procedure, it:=proc(c,n) local V,i; V:=Array(1..n,datatype=complex[8]); V[1]:=evalf[20](c); for i from 2 to n do V[i]:=evalf[20](c^V[i-1]) od; V end: For your first example, V:=it(3-3*I,10): C(V,1.5); 5 Correct! Now, the example with t(3,2), t:=(n,k)->exp(2*k*Pi*I/n); V:=it(t(3,2),500): C(V,1.5); 16 Finally, Robert Israel's example, V:=it(15.384972737405246873 + 0.42943156081631657205*I,100): C(V,1000); 4 Alec Mihailovs http://webpages.shepherd.edu/amihailo/ ==== > I am iterating a certain function, and I keep all the resultant values > in an array: e:=array(0..499); I need a local proc(e,n) which takes as input the array of entries, its > actual size (actual size may be less than 500) and heuristically tries > to determine if the entries are close to a k-cycle. Is k fixed? Then k should be a parameter also. Or do you want to check if there are cycles for any k? > I tried to figure out a quick and dirty solution to this, but after a > while it became obvious that the problem is rather difficult. For starters one needs to make several assumptions. One needs to have an > eps, which determines how close the entries need to be to be considered > a match for the cycle and I need the proc to tell the user in case > there is no match. There are other difficulties also: Cases where there are accidental > matches, but where the rest of the entries do not comprise a cycle. It looks to me that in the worst case, the algorithm seems to be O(n^2), > since I need to check all entries against all previous ones for matches. Just sort the list. Then matches will be next to each other. That's O(n*log(n)). If the list is complex numbers, sorting them by magnitude should be sufficient if it is unlikely that list contains numbers whose magnitudes differ by less than epsilon when the numbers themselves are futher than epsilon apart. > I'd appreciate if anyone can help a bit with this. Is the list tail-recursive? That is, does each entry depend only on the previous entry: x[n+1]:= f(x[n]). For example, a sequence generated by Newton's method would be like this. Then any single repeat is enough to make a cycle. The average case complexity may be much better than the worse case. If there are in fact no entries are even close to being a cycle, this can be quickly determined. Here's a first pass: CycleCheck:= proc(e,n,eps) local L,i; L:= sort( [seq]([e[i],i], i= 0..n-1) ,(a,b)-> abs(a[1]) < abs(b[1]) ); for i to n-1 do if abs(L[i][1]-L[i+1][1]) < eps then print(`Probable match at positions`); return L[i][2], L[i+1][2] fi od; print(`No match`) end proc; ==== > Just sort the list. Then matches will be next to each other. That's > O(n*log(n)). If the list is complex numbers, sorting them by magnitude > should be sufficient if it is unlikely that list contains numbers whose > magnitudes differ by less than epsilon when the numbers themselves are > futher than epsilon apart. My answer was written days ago, before seeing the other answers. For some reason, there was a big delay in posting it. The subsequent answers supercede my answer. ==== The Primary Universal Base Unit Values 1) mass… (hc/G)^1/2 = 5.4563026(39) x 10^-8 kg 2) length… (hG/c^3)^1/2 = 4.0507625(38) x 10^-35 m 3) time… (hG/c^5)^1/2 = 1.3511889(41) x 10^-43 s 4) current… e/(hG/c^5)^1/2 = 1.1857530(90) x 10^24 A 5) temperature… b/(hG/c^3)^1/2 = 3.5518626(82) x 10^32 K 6) substance… M/(hc/G)^1/2 = 1.6605387(31) x 10^-27 kmol 7) intensity… (hG/c^5)^1/2/sr = 1.9720204(06) x 10^-45 cd 8) Einstein solid angle = 6.8517999(97) x 10^1 sr 9) Planck plane angle = 1.3703599(97) x 10^2 rad The Primary Universal Physical Constants 001) irradiance constant i. = 4.5211591(52) x 10-122 s^3/kg 002) radiance constant i.= 3.0978078(26) x 10-120 s^3-sr/kg 003) radiant volume constant = 1.3554094(15) x 10^-113 m-s^2/kg 004) measurement volume = 6.6467654(65) x 10^-104 m^3 005) graviton volume constant = 1.2181812(31) x 10^-96 m^3/kg 006) luminous efficacy = 3.7229937(53) x 10^-96 cd-sr-s^3/kg-m^2 007) electric current volume = 1.3838190(49) x 10^-93 m^2/A 008) luminous energy = 1.8257115(55) x 10^-86 cd-sr-s 009) electric charge volume = 4.1485851(42) x 10^-85 m^3/A-s 010) molar volume = 4.0027765(33) x 10^-77 m^3/kmol 011) moment of inertia = 8.9530708(38) x 10^-77 kg-m^2 012) graviton fluidity = 1.0031235(26) x 10^-70 m-s/kg 013) measurement area = 1.6408677(14) x 10^-69 m^2 014) electric moment = 6.4900363(91) x 10^-54 A-s-m 015) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 016) magnetic moment = 1.9456639(62) x 10^-45 A-m^2 017) luminous intensity = 1.9720204(06) x 10^-45 cd 018) graviton frequency i. = 1.3511889(41) x 10^-43 s 019) luminous flux = 1.3511889(41) x 10^-43 cd-sr 020) graviton moment = 2.2102186(33) x 10^-42 kg-m 021) inductance constant = 3.4877980(18) x 10^-39 kg-m^2/A^2-s^2 022) absorption-emission = 2.4763819(58) x 10^-36 s/kg 023) graviton wavelength = 4.0507625(38) x 10^-35 m 024) Planck constant = 6.6260687(65) x 10^-34 kg-m^2/s 025) relative expansion = 2.8154241(58) x 10^-33 /K 026) electric resistivity = 1.0456155(41) x 10^-30 kg-m^3/A^2-s^3 027) unified substance = 1.6605387(31) x 10^-27 kmol 028) kinematic viscosity = 1.2143880(58) x 10^-26 m^2/s 029) inverse electric current = 8.4334589(42) x 10^-25 /A 030) Boltzmann constant= 1.3806502(93) x 10^-23 kg-m^2/s^2-K 031) thermal resistance = 9.7866124(96) x 10^-21 s^3-K/kg-m^2 032) graviton molality = 3.0433405(93) x 10^-20 kmol/kg 033) elementary charge = 1.6021764(62) x 10^-19 A-s 034) primary radiation = 5.9552136(16) x 10^-17 kg-m^4/s^3 035) specific heat = 2.5303770(42) x 10^-16 m^2/s^2-K 036) magnetic flux q. = 2.0678336(42) x 10^-15 kg-m^2-sr/A-s^2-rad 037) magnetic flux = 4.1356672(77) x 10^-15 kg-m^2/A-s^2 038) electric permittivity = 1.2922425(96) x 10^-13 A^2-s^4/kg-m^3 039) magnetic exposure = 2.9363775(58) x 10^-12 A-s/kg 040) electric constant = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 041) magnetic pole strength = 4.8032041(96) x 10^-11 A-m 042) Newton constant = 6.6723641(43) x 10^-11 m^3/kg-s^2 043) density of states = 2.0392015(07) x 10^-10 s^2/kg-m^2 044) S-B primary constant = 1.3897143(30) x 10^-9 kg/s^3-K^4 045) radiant distribution constant = 3.335640952... x 10^-9 s/m 046) graviton mass constant = 5.4563026(39) x 10^-8 kg 047) molar Planck constant = 3.9903126(87) x 10^-7 kg-m^2/s-kmol 048) magnetic constant = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr 049) electric conductance = 3.8740458(43) x 10^-5 A^2-s^3/kg-m^2 050) conductance q. = 7.7480916(72) x 10^-5 A^2-s^3-rad/kg-m^2-sr 051) magnetic permeability = 8.6102258(15) x 10^-5 kg-m/A^2-s^2 052) fine-structure constant = 7.2973525(36) x 10^-3 /rad 053) second radiation constant = 1.4387752(29) x 10^-2 m-K 054) dielectric constant = 1.4594705(05) x 10^-2 /sr 055) gravitational momentum = 1.6357583(80) x 10^1 kg-m/s 056) relative permeability = 6.8517999(97) x 10^1 sr 057) inverse fine-structure = 1.3703599(97) x 10^2 rad 058) impedance of vacuum = 3.767303134… x 10^2 kg-m^2/A^2-s^3-sr 059) molar gas constant = 8.3144720(88) x 10^3 kg-m^2/s^2-kmol-K 060) spin angle constant = 9.3894326(23) x 10^3 sr-rad 061) i. conductance q. = 1.2906403(83) x 10^4 kg-m^2-sr/A^2-s^3-rad 062) von Klitzing constant = 2.5812807(61) x 10^4 kg-m^2/A^2-s^3 063) inverse gravitational mass = 1.8327429(14) x 10^7 /kg 064) Faraday constant = 9.6485341(30) x 10^7 A-s/kmol 065) speed of light in vacuum = 2.99792458 x 10^8 m/s 066) graviton energy constant = 4.9038802(52) x 10^9 kg-m^2/s^2 067) Josephson primary = 2.4179894(88) x 10^14 A-s^2/kg-m^2 068) Josephson q. = 4.8359789(67) x 10^14 A-s^2-rad/kg-m^2-sr 069) electric displacement = 3.9552465(66) x 10^15 A-s/m 070) absorbed dose = 8.987551787… x 10^16 m^2/s^2 071) luminous density = 2.7467669(26) x 10^17 cd-sr-s/m^3 072) gravity displacement = 4.4930470(15) x 10^18 kg-s/m^2 073) molar mass constant = 3.2858629(17) x 10^19 kg/kmol 074) magnetic potential = 1.0209601(87) x 10^20 kg-m/A-s^2 075) thermal conductance = 1.0218040(21) x 10^20 kg-m^2/s^3-K 076) electric current constant = 1.1857530(90) x 10^24 A 077) luminance constant = 1.2018155(94) x 10^24 cd/m^2 078) luminous flux density = 8.2346000(82) x 10^25 cd-sr/m^2 079) Avogadro constant = 6.0221419(79) x 10^26 /kmol 080) gravitational field = 1.3469816(08) x 10^27 kg/m 081) electric potential = 3.0607616(38) x 10^28 kg-m^2/A-s^3 082) electric conductivity = 9.5637446(19) x 10^29 A^2-s^3/kg-m^3 083) Celcius temperature = 3.5518626(82) x 10^32 K 084) graviton wave number = 2.4686709(99) x 10^34 /m 085) mass flow rate constant = 4.0381492(72) x 10^35 kg/s 086) molar energy = 2.9531863(13) x 10^36 kg-m^2/s^2-kmol 087) surface concentration = 1.0119881(80) x 10^42 kmol/m^2 088) graviton frequency = 7.4008894(66) x 10^42 /s 089) superforce constant = 1.2106066(96) x 10^44 kg-m/s^2 090) luminous intensity i. = 5.0709414(41) x 10^44 /cd 091) angular velocity = 1.0141882(87) x 10^45 rad/s 092) electric flux density = 9.7642024(91) x 10^49 A-s/m^2 093) radiant intensity = 5.2968673(52) x 10^50 kg-m^2/s^3-sr 094) graviton field strength = 2.2187308(44) x 10^51 m/s^2 095) superpower constant = 3.6293075(70) x 10^52 kg-m^2/s^3 096) magnetic flux density = 2.5204148(03) x 10^54 kg/A-s^2 097) thermal conductivity = 2.5224979(53) x 10^54 kg-m/s^3-K 098) magnetic field strength = 2.9272342(65) x 10^58 A/m 099) absorbed dose rate = 6.6515877(33) x 10^59 m^2/s^3 100) graviton surface density = 3.3252544(33) x 10^61 kg/m^2 101) electric field strength = 7.5560134(90) x 10^62 kg-m/A-s^3 102) measurement area i. = 6.0943364(99) x 10^68 /m^2 103) dynamic viscosity = 9.9688619(97) x 10^69 kg/m-s 104) molar concentration = 2.4982658(70) x 10^76 kmol/m^3 105) surface tension constant= 2.9885896(41) x 10^78 kg/s^2 106) electric charge density = 2.4104603(51) x 10^84 A-s/m^3 107) angular acceleration = 7.5058954(07) x 10^87 rad/s^2 108) thermal transfer = 6.2272175(40) x 10^88 kg/s^3-K 109) electric current density = 7.2263783(36) x 10^92 A/m^2 110) luminous efficacy i. = 2.6860104(16) x 10^95 kg-m^2/cd-sr-s^3 111) graviton density constant = 8.2089591(81) x 10^95 kg/m^3 112) measurement density = 1.5044911(77) x 10^103 /m^3 113) radiant density constant = 7.3778445(74) x 10^112 kg/m-s^2 114) radiance constant = 3.2280892(03) x 10^119 kg/s^3-sr 115) irradiance constant = 2.2118221(59) x 10^121 kg/s^3 The Primary Universal Gravitation Equations E=(hc^5/G)^1/2 4.9038802 x 10^9 kg-m^2/s^2 = [(6.6260687 x 10^-34 kg-m^2/s) (2.4216061 x 10^42 m^5/s^5)/(6.6723641 x 10^-11 m^3/kg-s^2)]^1/2 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 [rsu 3.9 x 10^-8] E=c^5/Gv 4.9038802 x 10^9 kg-m^2/s^2 = (2.4216061 x 10^42 m^5/s^5)/ (6.6723641 x 10^-11 m^3/kg-s^2)(7.4008894 x 10^42 /s) 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 [rsu 3.9 x 10^-8] E=hc^3/Gm 4.9038802 x 10^9 kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s) (2.6944002 x 10^25 m^3/s^3)/(6.6723641 x 10^-11 m^3/kg-s^2) (5.4563026 x 10^-8 kg) 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 [rsu 3.9 x 10^-8] E=h(Gd)^1/2 4.9038802 x 10^9 kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s) [(6.6723641 x 10^-11 m^3/kg-s^2)(8.2089591 x 10^95 kg/m^3)]^1/2 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 [rsu 3.9 x 10^-8] G=c^3/mv 6.6723641 x 10^-11 m^3/kg-s^2 = (2.6944002 x 10^25 m^3/s^3)/ (5.4563026 x 10^-8 kg)(7.4008894 x 10^42 /s) 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 [rsu 3.9 x 10^-8] G =wc^4/E 6.6723641 x 10^-11 m^3/kg-s^2 = (4.0507625 x 10^-35 m) (8.0776087 x 10^33 m^4/s^4)/(4.9038802 x 10^9 kg-m^2/s^2) 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 [rsu 3.9 x 10^-8] G=c^4/F 6.6723641 x 10^-11 m^3/kg-s^2 = (8.0776087 x 10^33 m^4/s^4)/ (1.2106066 x 10^44 kg-m/s^2) 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 [rsu 3.9 x 10^-8] G=c^5/P 6.6723641 x 10^-11 m^3/kg-s^2 = (2.4216061 x 10^42 m^5/s^5)/ (3.6293075 x 10^52 kg-m^2/s^3) 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 [rsu 3.9 x 10^-8] E is graviton energy constant, h is Planck constant, c is speed of light in vacuum, G is Newton constant, v is graviton frequency, m is graviton mass constant, d is graviton density constant, w is graviton wavelength, F is superforce constant, P is superpower constant, e is elementary charge, b is second radiation constant, M is molar mass constant, i. is inverse, q. is quantum http://physics.nist.gov/cuu/Constants/ [UPDATED 1998 CODATA-NIST VALUES] ==== So what is your point and what has this to do with mathematics? This is a math group. Go talk to the physickers. >The Primary Universal Base Unit Values 1) mass… (hc/G)^1/2 = 5.4563026(39) x 10^-8 kg >2) length… (hG/c^3)^1/2 = 4.0507625(38) x 10^-35 m >3) time… (hG/c^5)^1/2 = 1.3511889(41) x 10^-43 s >4) current… e/(hG/c^5)^1/2 = 1.1857530(90) x 10^24 A >5) temperature… b/(hG/c^3)^1/2 = 3.5518626(82) x 10^32 K >6) substance… M/(hc/G)^1/2 = 1.6605387(31) x 10^-27 kmol >7) intensity… (hG/c^5)^1/2/sr = 1.9720204(06) x 10^-45 cd >8) Einstein solid angle = 6.8517999(97) x 10^1 sr >9) Planck plane angle = 1.3703599(97) x 10^2 rad The Primary Universal Physical Constants 001) irradiance constant i. = 4.5211591(52) x 10-122 s^3/kg >002) radiance constant i.= 3.0978078(26) x 10-120 s^3-sr/kg >003) radiant volume constant = 1.3554094(15) x 10^-113 m-s^2/kg >004) measurement volume = 6.6467654(65) x 10^-104 m^3 >005) graviton volume constant = 1.2181812(31) x 10^-96 m^3/kg >006) luminous efficacy = 3.7229937(53) x 10^-96 cd-sr-s^3/kg-m^2 >007) electric current volume = 1.3838190(49) x 10^-93 m^2/A >008) luminous energy = 1.8257115(55) x 10^-86 cd-sr-s >009) electric charge volume = 4.1485851(42) x 10^-85 m^3/A-s >010) molar volume = 4.0027765(33) x 10^-77 m^3/kmol >011) moment of inertia = 8.9530708(38) x 10^-77 kg-m^2 >012) graviton fluidity = 1.0031235(26) x 10^-70 m-s/kg >013) measurement area = 1.6408677(14) x 10^-69 m^2 >014) electric moment = 6.4900363(91) x 10^-54 A-s-m >015) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 >016) magnetic moment = 1.9456639(62) x 10^-45 A-m^2 >017) luminous intensity = 1.972020religion. > FOR THERE IS A NICE PLACE IN HELL FOR IDIOTS LIKE YOU! Tsk tsk, do you really talk with your mom with that mouth? Like I said.. don't talk about Hell, which according to you, was given (haha) by the Indo-Aryas (double haha)... maybe what you call heaven might turn out to be Hell? - Just a thought.. -Andre > > > > > > > >>-Andre