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Contact COR Management Services, Ltd., Bob Olman,... === Subject: Re: Random Rational? posting-account=jnanCwwAAADhdDGKIZxmMT8cdQqnhCJe > The probability that a random rational number has an even > denominator is 1/3 (Salamin and Gosper 1972). Not entirely serious.... A rational number cannot have both numerator and denominator even. So the possibilities are E/O, O/E, O/O, so one third of the possibilities have an even numerator. === Subject: Re: Random Rational? > The probability that a random rational number has an even > denominator is 1/3 (Salamin and Gosper 1972). I would say this comes from the fact that a randomly chosen fraction can often be simplified (e.g., 8/6 with even denominator becomes 4/3 without). Now, a fraction has an even demoninator if - the denominator is even and the numerator is odd (p = 1/2*1/2 = 1/4) - the denominator is divisible by four and the denominator is divisible by two but not by four (p = 1/4*1/4 = 1/16) - the denominator is divisible by eight and the denominator is divisible by four but not by eight (p = 1/8*1/8 = 1/64) - etc.etc. Giving a total probability of 1/4 + 1/16 + 1/64 + ... = 1/3 Voila. -- M.vr.gr. Dave (d-dot-langers-at-wxs-dot-nl) === Subject: Re: Random Rational? <325cr3F3gff2hU1@individual.net> posting-account=XHGPzAwAAACBzDwz8l_9KYOmofDRUylj > The probability that a random rational number has an even > denominator is 1/3 (Salamin and Gosper 1972). > I would say this comes from the fact that a randomly chosen fraction can > often be simplified (e.g., 8/6 with even denominator becomes 4/3 without). > Now, a fraction has an even demoninator if > - the denominator is even and the numerator is odd (p = 1/2*1/2 = 1/4) > - the denominator is divisible by four and the denominator is divisible > by two but not by four (p = 1/4*1/4 = 1/16) > - the denominator is divisible by eight and the denominator is divisible > by four but not by eight (p = 1/8*1/8 = 1/64) > - etc.etc. > Giving a total probability of 1/4 + 1/16 + 1/64 + ... = 1/3 > Voila. > -- > M.vr.gr. > Dave > (d-dot-langers-at-wxs-dot-nl) Very nice, but it still rests on the assumption that one can randomly pick a fraction. It seems to be using the assumption/definition that for a subset Q of N, the probability of a random integer being in Q is lim_{n to infty} |Q(x) cap n|/n. This is clearly a property of N, with the usual ordering, not just a property of the set N. One might reasonable argue that a probability should be stable under permutations, and thus we shouldnt consider any particular ordering of N as special. Of course you can easily show orderings that give you contradictory results, unless Q is finite/cofinite. === Subject: Re: Random Rational? > I was reading through MathWorld (in particular > http://mathworld.wolfram.com/RationalNumber.html) and came across the > assertion: > The probability that a random rational number has an even > denominator is 1/3 (Salamin and Gosper 1972). > Apparently this result came from > http://www.inwap.com/pdp10/hbaker/hakmem/number.html#item54, again > without proof. > With a vague memory of basic analysis and measure theory ringing > warning bells, I tried to find any rigorous definition of a random > rational, without success. I noted that the same HAKMEM category made > casual references to random natural numbers and so forth. > Are such concepts justifiable? It seems to me that there cannot exist > a reasonable probability measure on either of these (countable) sets. How about this: for each positive integer N let P_N = N^{-2}|{(m,n): m,n integers, 1<= m,n <= N, m/n has even denominator}| and let P = lim_{N->infinity} P_N (if that limit exists). Is P a good interpretation of the probability that a random rational number has an even denominator? Is P = 1/3? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Random Rational? The probability that a random rational number has an even >> denominator is 1/3 (Salamin and Gosper 1972). >> Are such concepts justifiable? It seems to me that there cannot exist >> a reasonable probability measure on either of these (countable) sets. >How about this: >for each positive integer N let >P_N = N^{-2}|{(m,n): m,n integers, 1<= m,n <= N, m/n has even denominator}| >and let P = lim_{N->infinity} P_N (if that limit exists). >Is P a good interpretation of the probability that a random rational >number has an even denominator? No. Or more precisely, nobody has shown it. The distribution implied by your definition is not a uniform distribution because it is weighted heavily toward low denominators. For example, m/n=1 happens N times in your m*n space, m/n=1/2 happens ßoor(N/2) times, m/n=1/10000 happens ßoor(N/10000) times. We all know that uniform distribution over a countably infinite set is nonsense. Therefore, for the phrase random rational number to mean anything, some other distribution must be specified. >Is P = 1/3? For what its worth, I think the answer is yes. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Division by zero. Go ahead and laugh. >Physics should be a mathematics in its own right. >>Why do you insist on confusing the description with the thing > described. >>The map is NOT the territory. The word is NOT the thing. >>Physics has to be about the physical world, not just a gossimer > pattern >>of abstraction. >Physics shall be built upon axioms which have, as a foundation, the >solid >bedrock of reality to rest upon. >>There is only one axiom: If it does not accurately describe/predict what >>we observe, it is wrong. Physics does not work the same way math does. >> It uses math as a tool, but it is not math. >Thats the way it is. Not the way it needs to be. >I go out on a limb to say this, but I have an imagination, and my >imagination tells me that R3 is an abstraction, and the space in front > of my >own 2 eyes is something else, something real. >>Last I heard, space/time is H^3, where H is a Hausdorf space. Other >>models include R^10 or some other high power of R. R^3 covers neither >>time nor the curvature of space currently used to model how gravity >>affects light. > Hausdorff Space > A topological space in which every pair of distinct points have a pair > of disjoint open neighborhoods. > Wheres the proof that there are points in space ? Open neighborhood > implies things too. One must make assumptions to use this for constructing > models. Points in space would correspond to what youve been calling locations. Open neighborhood implies there is a way to measure the distance between points. > I simply dont buy it. If you dont believe there are locations and distances, you will have a hard time doing anything relevant to physics. > You have to get right down to the profound stupidly simple dirt level truths > of the universe before your eyes. I see no proof that the universe is a > Hausdorff Space. If it is, then I will go away and never show my face again. > If people are using Hausdorff Spaces in their models, I have no objection to > it, but they must confess that this was assumed, and not was not proved with > due rigor. In what sense do you think it is assumed? They made a conjecture, and found that the model agrees with experimental results. If it is a simpler model than other options to get the same accuracy, doesnt it make sense to use it? >>Could you provide an example? Ive never seen a non-rigorous use of >>math in physics. >Yes. They count higher than 1. I dont think that they should do that >neccesarily. They also use calculus while claiming that space is > discrete, >or grainy or something. Honestly, the math is absolutely the most > rigorous >thing which exists in either world. It is the linkage between math and >physical world which is not rigorous. I want to fix this linkage. >It is very crazy. Must understand space/time as if it were R3. Must > think in >reverse somehow. Going mad fast. >>If calculus is providing accurate results within the limits of our >>ability to measure, then whether space is discrete or not becomes >>irrelevent. The linkage between math and physics may be non-rigorous, >>but that is because physics is an experimental science. We dont *know* >>the rules of physics, we discover them and see what models in >>mathematics give us accurate predictions. > Yes, You are right. But let me say this. Prior to this proof, people are > going around calling things identical. I say that this is clearly > impossble, and sloppy usage of language, and cleary non-rigorous thinking. It depends on what is meant by identical. Some use it to mean indistinguishable. > In sci.physics, they call electrons identical. I prove that no two objects > in physical universe are identical, directly contradicting the current > model. Will it lead anywhere ? I do not know. My hope is that it will. Man > started by counting on his fingers. Now he has 100 different methods to > calculate the volume of a sphere using algebra. It took 5,000 years to get > where we are today, but it all started with 10 dirty, smelly fingers. Have you asked what they mean when they say electrons are identical? Surely you cant think that they mean they are the same electron. Note: if you have an axiom in the mathematical sense, you cannot have a proof of it. > ------------------- > Axiom1 > No two objects in physical universe can be identical. > -------------------- > Proof > Consider a dollar bill in your hand. It is one piece of paper and its value > is exactly one dollar. > Suppose that there were 2 identical dollars in the exact same position in > space, and you held it in your hand. > Now a third, 4th, and an Nth dollar bill, all stacked exactly together in > space, occupying the exact same spot. You have an infinite number of dollar > bills in your hand. It looks like a single dollar bill, but it is really > woth $(infinity). > This is clearly ridiculous. You do not have an infinite amount of dollars, > you only have 1. The rest are all trivial. > Reducto ad absurdum, > QED You have infinitely many dollar bills, but they are also infinitely dense. I hope you enjoy being in a black hole. > Now, I think that we know something about the universe that we didnt know > before. And, when the physicists start talking about identical electrons, > they might as well be talking phlogiston. > This is clearly a new approach. I feel as if I were a caveman, and I count > to ten using fingers and toes, knowing that there is a calculus somewhere in > the future of man and I have utterly no comprehension of it, yet I know it > must be there. It sounds like you are warping their obvious meaning. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Unit vector by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IFnwH04857; >What is the length of (a1,a2,a3)? Answer this question and you will >see why the length of (a1,a2,a3)/sqrt(a1^2 + a2^2 + a3^2) is 1. >- MO I know that the length of (a1,a2,a3) is sqrt(a1^2 + a2^2 + a3^2). Damn it,I should see the obvious but I dont PS-I also know that if unit vector is (u1,u2,u3) and you sqrt(u1^2 + u2^2 + u3^3) that you get the length of 1.But why do we always get 1 no matter what the a1,a2,a3 coordinates are? === Subject: Re: Unit vector >What is the length of (a1,a2,a3)? Answer this question and you will >see why the length of (a1,a2,a3)/sqrt(a1^2 + a2^2 + a3^2) is 1. >- MO > I know that the length of (a1,a2,a3) is sqrt(a1^2 + a2^2 + a3^2). > Damn it,I should see the obvious but I dont > PS-I also know that if unit vector is (u1,u2,u3) and you sqrt(u1^2 + > u2^2 + u3^3) that you get the length of 1.But why do we always get 1 > no matter what the a1,a2,a3 coordinates are? Let R= distance of (a1,a2,a3) from the origin, i.e R=sqrt(a1^2 + a2^2 + a3^2) Now what is the length of this vector (a1/R,a2/R,a3/R)? === Subject: Re: Unit vector by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IJuSX27617; >Let R= distance of (a1,a2,a3) from the origin, i.e R=sqrt(a1^2 + a2^2 >+ a3^2) >Now what is the length of this vector (a1/R,a2/R,a3/R)? The length is sqrt(a1/r +a2/r + a3/r) I knew how to calculate the length of a vector from the start but prob made a mistake somewhere cos I didnt get 1 === Subject: Re: Unit vector >Let R= distance of (a1,a2,a3) from the origin, i.e R=sqrt(a1^2 + a2^2 >+ a3^2) >Now what is the length of this vector (a1/R,a2/R,a3/R)? > The length is sqrt(a1/r +a2/r + a3/r) If the length of (a1,a2,a3) is R=sqrt(a1^2 + a2^2 + a3^2), how do you get the length of (a1/R,a2/R,a3/R) to be sqrt(a1/r +a2/r + a3/r) ? > I knew how to calculate the length of a vector from the start but prob > made a mistake somewhere cos I didnt get 1 === Subject: Re: Unit vector >Now what is the length of this vector (a1/R,a2/R,a3/R)? > The length is sqrt(a1/r +a2/r + a3/r) No! === Subject: Re: Unit vector by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JELx604091; >>Now what is the length of this vector (a1/R,a2/R,a3/R)? >> >> The length is sqrt(a1/r +a2/r + a3/r) >No! >>sqrt(a1/r^2 +a2/r^ + a3/r^) >>Im sloppy that s all >Wrong again. The length of (a1/r, a2/r, a3/r) is sqrt(a1^2/r^2 + >a2^2/r^2 + a3^2/r^2). >Read the reply offered by G.E.Ivey on Oct 18. Prof. Ivey walks you >through each and every step very carefully. >- MO Darn it!!!I probably should put parenthesis around fraction to indicate that i meant (a2/r)^2... === Subject: Re: Unit vector Consider Pythagoras. -- Richard. richard_wrigley@tiscali.co.uk I have yet to see any problem, however complicated, which when looked at in the right way, did not become still more complicated Poul Anderson >What is the length of (a1,a2,a3)? Answer this question and you will >see why the length of (a1,a2,a3)/sqrt(a1^2 + a2^2 + a3^2) is 1. >- MO > I know that the length of (a1,a2,a3) is sqrt(a1^2 + a2^2 + a3^2). > Damn it,I should see the obvious but I dont > PS-I also know that if unit vector is (u1,u2,u3) and you sqrt(u1^2 + > u2^2 + u3^3) that you get the length of 1.But why do we always get 1 > no matter what the a1,a2,a3 coordinates are? === Subject: Re: linear programming by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IJuSd27612; arrive at those values? What calculations can be used to find thoses values? >>This problem is to be solved algebraically and geometrically: >> >>You have a property of 175,000 square feet to develop into housing >for >>at least 100 people. Assuming that town-house units need at least >>3,000 square feet and house an average of 2 people per unit, and >>single-family units need 10,000 square feet and house an average of >>4.5 people per unit, how many units of each type can you build? What >>is the optimal distribution of the types of units? >>I have solved the problem algebrically using the following equations: >>3,000x + 10,000y = 175,000 >>2x + 4.5y = 100 >>Im stumped on how to solve it graphically. > To solve it graphically, you graph it: > The graph of 3000x+ 10000y= 175000 is a straight line passing >through (0,17.5) and (58 1/3, 0). > The graph of 2x+ 4.5y= 100 is a straight line passing >through (0,22 2/9) and (50, 0). > The solution is the x,y values of the point where the two lines >cross. === Subject: Re: linear programming by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LEJmi24705; >arrive at those values? What calculations can be used to find thoses >values? A straight line is determined by any two points on it. The particular ones I got were by taking each coordinate to be 0 in succesion. The first equation is 3000x+ 10000y= 175000 . If x= 0 then 10000y= 175000 so y= 175000/10000= 17.5. The line passes through the point (0, 17.5) If y= 0 then 3000x= 175000 so x= 175000/3000= 58 1/3. The line passes through (58 1/3, 0). Mark those two points and draw the line through them. The second equation is 2x + 4.5y = 100. If x= 0, 4.5y= 100 so y= 100/4.5= 22 2/9. The line passes through (0, 22 2/9). If y= 0 then 2x= 100 so x= 100/2= 50. The line passes through (50, 0). Theres nothing special about x= 0 and y= 0 except that 0 is an easy number! I could as easily have taken x= 1 or -100 or anything else, put that number into the equation and solve for y to get a point on the line. >This problem is to be solved algebraically and geometrically: > >You have a property of 175,000 square feet to develop into housing >>for >at least 100 people. Assuming that town-house units need at least >3,000 square feet and house an average of 2 people per unit, and >single-family units need 10,000 square feet and house an average of >4.5 people per unit, how many units of each type can you build? >What >is the optimal distribution of the types of units? >I have solved the problem algebrically using the following >equations: >3,000x + 10,000y = 175,000 >2x + 4.5y = 100 >Im stumped on how to solve it graphically. >> To solve it graphically, you graph it: >> The graph of 3000x+ 10000y= 175000 is a straight line passing >>through (0,17.5) and (58 1/3, 0). >> The graph of 2x+ 4.5y= 100 is a straight line passing >>through (0,22 2/9) and (50, 0). >> The solution is the x,y values of the point where the two lines >>cross. === Subject: Re: linear programming by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9Q158l01861; you >>arrive at those values? What calculations can be used to find thoses >>values? > A straight line is determined by any two points on it. The >particular ones I got were by taking each coordinate to be 0 in >succesion. > The first equation is 3000x+ 10000y= 175000 . If x= 0 then >10000y= 175000 so y= 175000/10000= 17.5. The line passes through the >point (0, 17.5) If y= 0 then 3000x= 175000 so x= 175000/3000= >58 1/3. The line passes through (58 1/3, 0). Mark those two points >and draw the line through them. > The second equation is 2x + 4.5y = 100. If x= 0, 4.5y= 100 so >y= 100/4.5= 22 2/9. The line passes through (0, 22 2/9). > If y= 0 then 2x= 100 so x= 100/2= 50. The line passes through >(50, 0). > Theres nothing special about x= 0 and y= 0 except that 0 is an >easy number! I could as easily have taken x= 1 or -100 or anything >else, put that number into the equation and solve for y to get a point >on the line. >>This problem is to be solved algebraically and geometrically: >> >>You have a property of 175,000 square feet to develop into housing >for >>at least 100 people. Assuming that town-house units need at least >>3,000 square feet and house an average of 2 people per unit, and >>single-family units need 10,000 square feet and house an average of >>4.5 people per unit, how many units of each type can you build? >>What >>is the optimal distribution of the types of units? >>I have solved the problem algebrically using the following >>equations: >>3,000x + 10,000y = 175,000 >>2x + 4.5y = 100 >>Im stumped on how to solve it graphically. > To solve it graphically, you graph it: > The graph of 3000x+ 10000y= 175000 is a straight line passing >through (0,17.5) and (58 1/3, 0). > The graph of 2x+ 4.5y= 100 is a straight line passing >through (0,22 2/9) and (50, 0). > The solution is the x,y values of the point where the two lines >cross. === Subject: Extrema / Diff. f(x) = (e^-x)(sinx) Now, we differentiate to: (e^-x)(cos x) + (e^-x)(sinx) or: (e^-x)(cosx + sinx) Now, we set this to 0. (e^-x)(cosx + sinx) = 0 How do we solve for x ? (Trying to find extrema, but Im having some trouble; it would seem to me that cosx + sinx would never equal out to 0, that I can think of.) I should note that the textbook always phrases these questions in radians, not degrees. (e^-x) cos x = (e^-x) -sin x === Subject: Re: Extrema / Diff. > f(x) = (e^-x)(sinx) > Now, we differentiate to: > (e^-x)(cos x) + (e^-x)(sinx) > or: (e^-x)(cosx + sinx) > Now, we set this to 0. > (e^-x)(cosx + sinx) = 0 > How do we solve for x ? > (Trying to find extrema, but Im having some trouble; it would seem to me > that cosx + sinx would never equal out to 0, that I can think of.) I should > note that the textbook always phrases these questions in radians, not > degrees. > (e^-x) cos x = (e^-x) -sin x When x = -pi/4, for example, cos(x) = sqrt(2)/2 and sin(x) = -sqrt(2)/2 === Subject: Re: Extrema / Diff. alt.math.undergrad: >f(x) = (e^-x)(sinx) >Now, we differentiate to: >(e^-x)(cos x) + (e^-x)(sinx) Nope. The derivative of e^-x is -e^-x, not e^-x. Remember: [e^u] = u * e^u. You forgot to apply the Chain Rule. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Extrema / Diff. > f(x) = (e^-x)(sinx) > Now, we differentiate to: > (e^-x)(cos x) + (e^-x)(sinx) > or: (e^-x)(cosx + sinx) Thats incorrect. f(x) = e^(-x) sin(x) df/dx = e^(-x) d[sin(x)]/dx + sin(x) d[e^(-x)]/dx = e^(-x) cos(x) + sin(x) [-e^(-x)] = e^(-x) (cos(x) - sin(x)) > Now, we set this to 0. > (e^-x)(cosx + sinx) = 0 Actually e^(-x) (cos(x) - sin(x)) = 0 > How do we solve for x ? Noting that e^x != 0 for all x, multiply both sides by e^x, leaving cos(x) - sin(x) = 0 and solve. [This is a special case of the general rule: If ab=0 then a=0 or b=0 ] -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Extrema / Diff. >> f(x) = (e^-x)(sinx) >> Now, we differentiate to: >> (e^-x)(cos x) + (e^-x)(sinx) >> or: (e^-x)(cosx + sinx) > Thats incorrect. > f(x) = e^(-x) sin(x) > df/dx = e^(-x) d[sin(x)]/dx + sin(x) d[e^(-x)]/dx > = e^(-x) cos(x) + sin(x) [-e^(-x)] > = e^(-x) (cos(x) - sin(x)) Why did e^-x become -e^-x when differentiated? >> Now, we set this to 0. >> (e^-x)(cosx + sinx) = 0 > Actually > e^(-x) (cos(x) - sin(x)) = 0 >> How do we solve for x ? > Noting that e^x != 0 for all x, multiply both > sides by e^x, leaving > cos(x) - sin(x) = 0 > and solve. > [This is a special case of the general rule: > If ab=0 then a=0 or b=0 ] Would the appropriate value for x here be 1/4Pi, then? === Subject: Re: Extrema / Diff. >> f(x) = (e^-x)(sinx) >> Now, we differentiate to: >> (e^-x)(cos x) + (e^-x)(sinx) >> or: (e^-x)(cosx + sinx) > Thats incorrect. > f(x) = e^(-x) sin(x) > df/dx = e^(-x) d[sin(x)]/dx + sin(x) d[e^(-x)]/dx > = e^(-x) cos(x) + sin(x) [-e^(-x)] > = e^(-x) (cos(x) - sin(x)) > Why did e^-x become -e^-x when differentiated? f: x ---> e^(-x) is the composite of two functions: g: x ---> -x h: x ---> e^x i.e. f is a function of a function, f(x) = h(g(x)) So when you differentiate, you must use the chain rule... Mike. >> Now, we set this to 0. >> (e^-x)(cosx + sinx) = 0 > Actually > e^(-x) (cos(x) - sin(x)) = 0 >> How do we solve for x ? > Noting that e^x != 0 for all x, multiply both > sides by e^x, leaving > cos(x) - sin(x) = 0 > and solve. > [This is a special case of the general rule: > If ab=0 then a=0 or b=0 ] > Would the appropriate value for x here be 1/4Pi, then? === Subject: Re: Extrema / Diff. > f(x) = (e^-x)(sinx) > Now, we differentiate to: > (e^-x)(cos x) + (e^-x)(sinx) > or: (e^-x)(cosx + sinx) >> Thats incorrect. >> f(x) = e^(-x) sin(x) >> df/dx = e^(-x) d[sin(x)]/dx + sin(x) d[e^(-x)]/dx >> = e^(-x) cos(x) + sin(x) [-e^(-x)] >> = e^(-x) (cos(x) - sin(x)) >> Why did e^-x become -e^-x when differentiated? > f: x ---> e^(-x) > is the composite of two functions: > g: x ---> -x > h: x ---> e^x > i.e. f is a function of a function, f(x) = h(g(x)) > So when you differentiate, you must use the chain rule... > Mike. I thought that the derivative of e^x is always just e^x? Unless the book simply uses x to refer to any real number, as opposed to any real number including variables? === Subject: Re: Extrema / Diff. > I thought that the derivative of e^x is always just e^x? Unless the book > simply uses x to refer to any real number, as opposed to any real number > including variables? James, you dont have e^x. You have e^(-x). d/dx[e^u] = e^u * u Your u is -x. No, Im not going to show you why d/dx[-x] = -1. -- Darrell === Subject: Re: Extrema / Diff. days. My association with the Department is that of an alumnus. [.snip.] >I thought that the derivative of e^x is always just e^x? Yes. But you arent differentiating e^x. You were differentiating e^{-x} ->with respect to x<-. Its like the derivative of sin(-x). The derivative of sin(x) is cos(x), but the derivative of sin(-x) is not just cos(-x); it is cos(-x)*(-x) = -cos(-x). Likewise, if u is any function of x, then the derivative of e^u with respect to x is (e^u)(u) = u*e^u. If u(x)=x, then u = 1, so you dont have to do anything. But if u(x)=-x, then u(x) = -1, so (e^{-x}) = (-1)e^{-x} = -e^{-x}. The Chain Rule. >Unless the book >simply uses x to refer to any real number, as opposed to any real number >including variables? I have no idea what this question is supposed to mean, but Ill bet you dollars to donuts that your book says: d -- e^x = e^x. dx Youll notice the key fact that the exponent is exaclty the same as the variable with respect to which you are differentiating. If you were doing d ---- e^{-x} d(-x) then you might have a case for saying it is equal to e^{-x} (it is); but you arent. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Extrema / Diff. > [.snip.] >>I thought that the derivative of e^x is always just e^x? > Yes. But you arent differentiating e^x. You were differentiating > e^{-x} ->with respect to x<-. > Its like the derivative of sin(-x). The derivative of sin(x) is > cos(x), but the derivative of sin(-x) is not just cos(-x); it is > cos(-x)*(-x) = -cos(-x). > Likewise, if u is any function of x, then the derivative of e^u with > respect to x is (e^u)(u) = u*e^u. > If u(x)=x, then u = 1, so you dont have to do anything. But if > u(x)=-x, then u(x) = -1, so > (e^{-x}) = (-1)e^{-x} = -e^{-x}. > The Chain Rule. >>Unless the book >>simply uses x to refer to any real number, as opposed to any real number >>including variables? > I have no idea what this question is supposed to mean, but Ill bet > you dollars to donuts that your book says: > d > -- e^x = e^x. > dx > Youll notice the key fact that the exponent is exaclty the same as > the variable with respect to which you are differentiating. If you > were doing > d > ---- e^{-x} > d(-x) > then you might have a case for saying it is equal to e^{-x} (it is); > but you arent. === Subject: Re: Extrema / Diff. > f(x) = (e^-x)(sinx) > Now, we differentiate to: > (e^-x)(cos x) + (e^-x)(sinx) > or: (e^-x)(cosx + sinx) First, its best to write e^(-x). Second, theres a mistake above. > Now, we set this to 0. > (e^-x)(cosx + sinx) = 0 > How do we solve for x ? > (Trying to find extrema, but Im having some trouble; it would seem to me > that cosx + sinx would never equal out to 0 No, cosx + sinx can be 0; review basic trig. > that I can think of.) I should > note that the textbook always phrases these questions in radians, not > degrees. === Subject: Re: Extrema / Diff. >> f(x) = (e^-x)(sinx) >> Now, we differentiate to: >> (e^-x)(cos x) + (e^-x)(sinx) >> or: (e^-x)(cosx + sinx) > First, its best to write e^(-x). Second, theres a mistake above. >> Now, we set this to 0. >> (e^-x)(cosx + sinx) = 0 >> How do we solve for x ? >> (Trying to find extrema, but Im having some trouble; it would seem to me >> that cosx + sinx would never equal out to 0 > No, cosx + sinx can be 0; review basic trig. For what value of x? === Subject: Re: Extrema / Diff. days. My association with the Department is that of an alumnus. >> No, cosx + sinx can be 0; review basic trig. >For what value of x? Oh, really. You have to start thinking for yourself at some point, you know... Let cos x = u, sin x = v. Note that u^2 + v^2 = 1. If u=-v, then 1 = u^2 + v^2 = v^2 + v^2 = 2v^2, so v^2 = 1/2, so v = 1/sqrt(2) or v = -1/sqrt(2); in which case we must have u = -1/sqrt(2) or u = 1/sqrt(2). Now an easy review of trigonometry: when is sin(x) = 1/sqrt(2)? At x = pi/4 and at x = 3pi/4. At pi/4, we have cos(x) = 1/sqrt(2), so thats no good. But at x=3pi/4, we get cos(x) = -1/sqrt(2). So thats one value. A similar argument will give you that when x = -pi/4, we have sin(x) = -1/sqrt(2) and cos(x) = 1/sqrt(2), giving you the second point. Those are the only two points with -pi <= x < pi, but there are an infinity of others. Or if you want it geometrically: (cos x, sin x) are the coordinates of a point on the unit circle, at an angle of x radians from the positive is minus the y-coordinate? -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Extrema / Diff. > No, cosx + sinx can be 0; review basic trig. >>For what value of x? > Oh, really. You have to start thinking for yourself at some point, you > know... You know you put up that post *after* I put up the post where I answered it myself, right? === Subject: Re: Extrema / Diff. Good grief, in Usenet no one really knows exactly when any particular post is made. >> No, cosx + sinx can be 0; review basic trig. >For what value of x? >> Oh, really. You have to start thinking for yourself at some point, you >> know... > You know you put up that post *after* I put up the post where I answered > it myself, right? === Subject: Re: Extrema / Diff. days. My association with the Department is that of an alumnus. >> No, cosx + sinx can be 0; review basic trig. >For what value of x? >> Oh, really. You have to start thinking for yourself at some point, you >> know... >You know you put up that post *after* I put up the post where I answered it >myself, right? Point 1) Learn a bit about usenet distribution. Your message had not come up yet ->in my server<-. Point 2) You should have attempted to answer the question BEFORE raising it in the first place. Asking For what value of x? was an act of lazyness on your part, even if you eventually got around to answering the question yourself, given some prompting. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Extrema / Diff. > No, cosx + sinx can be 0; review basic trig. >>For what value of x? > Oh, really. You have to start thinking for yourself at some point, you > know... >>You know you put up that post *after* I put up the post where I answered >>it >>myself, right? > Point 1) Learn a bit about usenet distribution. Your message had not > come up yet ->in my server<-. Just pointing such things out. > Point 2) You should have attempted to answer the question BEFORE > raising it in the first place. Asking For what value of x? was an > act of lazyness on your part, even if you eventually got around to > answering the question yourself, given some prompting. No, I attempt to answer such questions /before/ posting. The answer simply hadnt occurred to me earlier. Do not call me lazy again; I have never done one of those please solve this questions. Every question Ive posted to this group has been worked out to, usually, within a hairs breadth of a solution, and never posted without my having labored over it for a while. I need not mention that I post only a tiny fraction of the problems Im working through; you can extrapolate how much work that is by the volume of my posts. I do not take kindly to condescension for no good reason. === Subject: Re: Extrema / Diff. [.snip.] >>You know you put up that post *after* I put up the post where I answered >>it >>myself, right? > Point 1) Learn a bit about usenet distribution. Your message had not > come up yet ->in my server<-. > Just pointing such things out. And I just pointed out how irrelevant it was for you to point it out in the first place. > Point 2) You should have attempted to answer the question BEFORE > raising it in the first place. Asking For what value of x? was an > act of lazyness on your part, even if you eventually got around to > answering the question yourself, given some prompting. > No, I attempt to answer such questions /before/ posting. The answer simply > hadnt occurred to me earlier. Do not call me lazy again; I have never done > one of those please solve this questions. That is ALL you posted. Was I (or anyone else, for that matter) supposed to read your mind as to what you had managed or not managed to do? ALL you posted was please tell me the answer. That is all that was posted, that is all that I saw. >Every question Ive posted to > this group has been worked out to, usually, within a hairs breadth of a > solution, Please, tell me, how long did you work on For what values of x does sin x + cos x = 0? ? How far did you get before being prompted? What was your stumbling block? Why did you not post your difficulty in answering the question, if you had indeed worked on it, and only posted a plain request for the answer to be provided to you? > and never posted without my having labored over it for a while. I > need not mention that I post only a tiny fraction of the problems Im > working through; you can extrapolate how much work that is by the volume of > my posts. I do not take kindly to condescension for no good reason. I dont take kindly to being lectured about posting times by someone who doesnt know about distribution of usenet; and I do not take kindly to people getting annoyed because someone was unable to read their minds as to what they meant or what they had done, simply because they never went to the trouble of stating it (out of, what, time-saving effort?) If you do work on the problems as you say, thats comendable. But in that case, then you would gain a ->lot<- more from the group if, instead of just posting your questions, you would tell us exactly what you tried, how far you got, and why you could not finish. Then you would get pointers as to why something does or does not work, or whether your idea was heading in the right direction, or how to see that final step, instead of getting solutions. If you have already done some of the work yourself, then let the people know so they dont repeat it. Have some consideration for ->our<- time, just as you obviously hold yours in such high regard. Arturo Magidin, sans .sig === Subject: Re: Extrema / Diff. > No, cosx + sinx can be 0; review basic trig. > For what value of x? Are there points on the unit circle where the x-coordinate is the negative of the y-coordinate? === Subject: Re: Extrema / Diff. >> No, cosx + sinx can be 0; review basic trig. >> For what value of x? > Are there points on the unit circle where the x-coordinate is the negative > of the y-coordinate? The second and fourth quadrants, where the reference angle would be equivalent to Pi/4, since Pi/4 is the only angle for which they have equal values? === Subject: Paper draft, opinions requested For well over a century a rather attractive but mathematically wrong idea has gripped number theory. This paper outlines the rather basic argument which pulls this wrong idea out from the background on to center stage, with far-reaching conclusions that have extraordinary implications for the discipline of number theory itself. ___JSH ------------------------------------------------------------- --------- I. Factorization The following are in a commutative ring. Start with P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where f is nonzero and note that at m=0, P(0) = u^2 f^2(3x + uf), which gives you terms that do not vary as m varies. So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) which shows that at least two of the as have to equal 0 at m=0, while one equals 3. Now let f be coprime to 3 and x so that 3x + uf is coprime to f. Since, at m=0, two of the as must equal 0, its convenient to just arbitrarily select a_1 and a_2 as those two. Then you have uf for the first, uf for the second and 3x + uf for the third as terms that do not vary when m varies. Now then, if m=1, what are the *constant* terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Thats logical because they do not vary with m, so if m=1003909273, what are the constant terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Now divide f^2 from both sides, which gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you note that P(0)/f^2 = u^2(3x + uf), which means that now your constant terms are u for the first, u for the second and 3x + uf for the third. Now then, if m=1, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. If m = 2938479378, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. How can the constant terms of the first two go from uf to u? They must be divided by f. Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided from P(m), it is u; therefore, a_1 x + uf is divided by f, and you have a_1 x/f + u and the constant term of a_2 x + uf is uf, but when f^2 is divided from P(m), it is u; therefore, a_2 x + uf is divided by f, and you have a_2 x/f + u while the constant term of a_3 x + uf is 3x + uf, and after f^2 is divided off, it is 3x + uf, so you have a_3 x + uf so, dividing P(m) by f^2 gives P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). II. Algebraic integers Now take P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside by f^2(a_1 a_2 a_3) and you have P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). Now consider the case that m, f, and u are algebraic integers, then I have the ratios of algebraic integers: uf/a_1, uf/a_2, and uf/a_3, and now let v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_2 where the vs and ws are algebraic integers in each case coprime to each other. Making the substitutions I have P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). And I have from before that P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f so (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f as that is the last coefficient from the last term u^3 f, which proves that (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 then I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) but I still have that P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) without contradiction. III. Galois Theory Now in the ring of algebraic integers consider the possibility that a_1/f is not an algebraic integer to see if that leads to a contradiction. First, if a_1/f is not an algebraic integer and w_1 is, they obviously cannot be equal. But I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) and P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) so far simultaneously true without contradiction, so there must exist z_1, z_2, and z_3 such that w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 and z_1 z_2 z_3 = 1, where algebraically the zs are units, but z_1, z_2 and z_3 are not units in the ring of algebraic integers. That raises a question about units as if z_1, z_2 and z_3 are units algebraically but are not algebraic integers, what does that mean? Consider properties that cover rings like the ring of algebraic integers, and the ring of integers itself, and it can be shown that two are required: 1. No rational in the ring except 1 and -1 is a unit. 2. No non-unit number within the ring is a factor of any two integers that are coprime to each other in the ring of integers. So there are two basic factor properties of rings like the ring of integers and the ring of algebraic integers. Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where properties 1. and 2. hold, which includes the ring of algebraic integers, and I have named that ring, the ring of objects. Some seem intent on challenging the results here with positions that necessarily contradict the algebra in Section 1. At times these people have tried to invoke Galois Theory, claiming it invalidates the argument in Section 1, but that requires a belief that mathematics can be inconsistent. Notice that the conclusions here follow logically and do not extend beyond the mathematics itself. Social issues have no relevance to mathematical truth. People who try to unnecessarily add to what follows mathematically are not mathematicians. James Harris === Subject: Re: Paper draft, opinions requested For well over a century a rather attractive but mathematically wrong idea has gripped number theory. This paper outlines the rather basic argument which pulls this wrong idea out from the background on to center stage, with far-reaching conclusions that have extraordinary implications for the discipline of number theory itself. ___JSH ------------------------------------------------------------- --------- I. Factorization The following are in a commutative ring. Start with P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where f is nonzero and note that at m=0, P(0) = u^2 f^2(3x + uf), which gives you terms that do not vary as m varies. So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) which shows that at least two of the as have to equal 0 at m=0, while one equals 3. Now let f be coprime to 3 and x so that 3x + uf is coprime to f. Since, at m=0, two of the as must equal 0, its convenient to just arbitrarily select a_1 and a_2 as those two. Then you have uf for the first, uf for the second and 3x + uf for the third as terms that do not vary when m varies. Now then, if m=1, what are the *constant* terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Thats logical because they do not vary with m, so if m=1003909273, what are the constant terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Now divide f^2 from both sides, which gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you note that P(0)/f^2 = u^2(3x + uf), which means that now your constant terms are u for the first, u for the second and 3x + uf for the third. Now then, if m=1, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. If m = 2938479378, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. How can the constant terms of the first two go from uf to u? They must be divided by f. Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided from P(m), it is u; therefore, a_1 x + uf is divided by f, and you have a_1 x/f + u and the constant term of a_2 x + uf is uf, but when f^2 is divided from P(m), it is u; therefore, a_2 x + uf is divided by f, and you have a_2 x/f + u while the constant term of a_3 x + uf is 3x + uf, and after f^2 is divided off, it is 3x + uf, so you have a_3 x + uf so, dividing P(m) by f^2 gives P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). II. Algebraic integers Now take P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside by f^2(a_1 a_2 a_3) and you have P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). Now consider the case that m, f, and u are algebraic integers, then I have the ratios of algebraic integers: uf/a_1, uf/a_2, and uf/a_3, and now let v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 where the vs and ws are algebraic integers in each case coprime to each other. Making the substitutions I have P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). And I have from before that P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f so (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f as that is the last coefficient from the last term u^3 f, which proves that (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 then I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) but I still have that P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) without contradiction. III. Galois Theory Now in the ring of algebraic integers consider the possibility that a_1/f is not an algebraic integer to see if that leads to a contradiction. First, if a_1/f is not an algebraic integer and w_1 is, they obviously cannot be equal. But I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) and P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) so far simultaneously true without contradiction, so there must exist z_1, z_2, and z_3 such that w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 and z_1 z_2 z_3 = 1, where algebraically the zs are units, but z_1, z_2 and z_3 are not units in the ring of algebraic integers. That raises a question about units as if z_1, z_2 and z_3 are units algebraically but are not algebraic integers, what does that mean? Consider properties that cover rings like the ring of algebraic integers, and the ring of integers itself, and it can be shown that two are required: 1. No rational in the ring except 1 and -1 is a unit. 2. No non-unit number within the ring is a factor of any two integers that are coprime to each other in the ring of integers. So there are two basic factor properties of rings like the ring of integers and the ring of algebraic integers. Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where properties 1. and 2. hold, which includes the ring of algebraic integers, and I have named that ring, the ring of objects. Some seem intent on challenging the results here with positions that necessarily contradict the algebra in Section 1. At times these people have tried to invoke Galois Theory, claiming it invalidates the argument in Section 1, but that requires a belief that mathematics can be inconsistent. Notice that the conclusions here follow logically and do not extend beyond the mathematics itself. Social issues have no relevance to mathematical truth. People who try to unnecessarily add to what follows mathematically are not mathematicians. James Harris === Subject: Re: Paper draft, opinions requested In my oppinion, you need to read more of the responses you get. > uf/a_1, uf/a_2, and uf/a_3, > and now let > v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_2 This should end with v_3/w_3 = uf/a_3, an error that has been pointed out by numerous people in sci.math. > where the vs and ws are algebraic integers in each case coprime to > each other. There are other problems, but it is clear that you are not reading even the simplest corrections. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Paper draft, opinions requested > For well over a century a rather attractive but mathematically wrong > idea has gripped number theory. This paper outlines the rather basic > argument which pulls this wrong idea out from the background on to > center stage, with far-reaching conclusions that have extraordinary > implications for the discipline of number theory itself. ___JSH > ------------------------------------------------------------- --------- > I. Factorization > The following are in a commutative ring. It might be helpful if you start by providing the definition of a ring. === Subject: Re: Tan to Slope >> What do they get? Do they agree that the point is (1,2)?? If somehow >> you got the right slope but wrong point, that could lead to the >> discrepancy you state. Or maybe the answer book is just wrong. >The point (1,2) is given by them. > I give them m=-89/128, and they say its correct. But when I give them b = >345/128 its marked as incorrect. I tried to enter it in different ways (2 >+(89/128), or 2.69) but its always given back to me as incorrect. So this is online, and you dont actually have their answer. I see. Some of these systems are very picky about format of numbers, and will recognize only certain forms. You might see if they have any instructions. Including a plus sign certainly is unusual. Did you try simply 2 89/128 As to 2.69, that is incorrect. If you are supposed to give it in decimal form with two decimal points (seems unlikely, since the slope was ok as a fraction), it must be 2.70. bob === Subject: Re: Tan to Slope >> The point on the curve where we take the slope is (1,2). >> So, knowing my m = -89/128, finding Ôb should be simple enough. >> y - y1 = m(x-x1) >> y - 2 = -89/128 (x-1) >> y- 2 = -89x/128 + 89/128 >> y = -89x/128 + 345/128 >> And yet, while the slope is correct, the Ôb is incorrect. I dont see >> where >> I could be going wrong in such a short, short process. > What wrong with the constant term? Does not y = -89x/128 + 345/128 go > thru x = 1, y = 2 just like it ought? Well, despite the fact that the site says my slope is correct, Ill repost the entire problem. I dont see why it would be relevant, but that might be why Im not figuring it right: The curve is (x^3)(y^5) + (4y^3) = (x^7) + 63, the point where we find the tangent line is (1,2). Now, using implicit differentiation: (x^3)(5y^4 dy/dx) + (3x^2)(y^5) + (12y^2 dy/dx) = 7x^6 (x^3)(5y^4 dy/dx) + (12y^2 dy/dx) = (7x^6) - (3x^2)(y^5) dy/dx = [(7x^6) - (3x^2)(y^5)] / [(x^3)(5y^4) + (12y^2)] Plugging in (1,2) gets me -89/128, which webworks reports as correct. === Subject: Re: Tan to Slope by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IGFUj07441; > The point on the curve where we take the slope is (1,2). > So, knowing my m = -89/128, finding Ôb should be simple enough. > y - y1 = m(x-x1) > y - 2 = -89/128 (x-1) > y- 2 = -89x/128 + 89/128 > y = -89x/128 + 345/128 > And yet, while the slope is correct, the Ôb is incorrect. I dont see > where > I could be going wrong in such a short, short process. >> What wrong with the constant term? Does not y = -89x/128 + 345/128 go >> thru x = 1, y = 2 just like it ought? >Well, despite the fact that the site says my slope is correct, Ill repost >the entire problem. I dont see why it would be relevant, but that might be >why Im not figuring it right: >The curve is (x^3)(y^5) + (4y^3) = (x^7) + 63, the point where we find the >tangent line is (1,2). >Now, using implicit differentiation: >(x^3)(5y^4 dy/dx) + (3x^2)(y^5) + (12y^2 dy/dx) = 7x^6 >(x^3)(5y^4 dy/dx) + (12y^2 dy/dx) = (7x^6) - (3x^2)(y^5) >dy/dx = [(7x^6) - (3x^2)(y^5)] / [(x^3)(5y^4) + (12y^2)] >Plugging in (1,2) gets me -89/128, which webworks reports as correct. Yes, but that wasnt the point. William Elliot asked WHY you think that the y intercept is not 345/128. That certainly IS the correct y-itercept. === Subject: Re: Tan to Slope >>> The point on the curve where we take the slope is (1,2). >> So, knowing my m = -89/128, finding Ôb should be simple enough. >> y - y1 = m(x-x1) >> y - 2 = -89/128 (x-1) >> y- 2 = -89x/128 + 89/128 >> y = -89x/128 + 345/128 >> And yet, while the slope is correct, the Ôb is incorrect. I dont > see >> where >> I could be going wrong in such a short, short process. > What wrong with the constant term? Does not y = -89x/128 + 345/128 > go > thru x = 1, y = 2 just like it ought? >>Well, despite the fact that the site says my slope is correct, Ill > repost >>the entire problem. I dont see why it would be relevant, but that > might be >>why Im not figuring it right: >>The curve is (x^3)(y^5) + (4y^3) = (x^7) + 63, the point where we > find the >>tangent line is (1,2). >>Now, using implicit differentiation: >>(x^3)(5y^4 dy/dx) + (3x^2)(y^5) + (12y^2 dy/dx) = 7x^6 >>(x^3)(5y^4 dy/dx) + (12y^2 dy/dx) = (7x^6) - (3x^2)(y^5) >>dy/dx = [(7x^6) - (3x^2)(y^5)] / [(x^3)(5y^4) + (12y^2)] >>Plugging in (1,2) gets me -89/128, which webworks reports as correct. > Yes, but that wasnt the point. William Elliot asked WHY you think > that the y intercept is not 345/128. That certainly IS the correct > y-itercept. I answered in another post on this same thread; because webworks keeps telling me that the y-int of 345/128 is incorrect. Though, I should make an addendum: I emailed the admin. There was a mistake on the site. Its since been corrected. === Subject: Tan to Slope that the y intercept is not 345/128. That certainly IS the correct > y-intercept. > I answered in another post on this same thread; because webworks keeps > telling me that the y-int of 345/128 is incorrect. > Though, I should make an addendum: I emailed the admin. There was a mistake > on the site. Its since been corrected. Hurray! Be thankful that the mistake you found wasnt in a prescription drug you took and the reply you got was so few people suffer this mistake were not concerned or that the mistake you found was in a household item which caused you injury or destruction of personal belonging and the extend of the Ôcorrection was to refund your money provided you agree they werent liable for the other losses, which of course couldnt ever have happened, or that you found a mistake a new car by being a design fault accident victim. At a $100.00 or more for text book, rebates should be offered to those with faulty texts. Finding errors in text book answers is a common thread in these newsgroups. === Subject: Re: Tan to Slope > alt.math.undergrad: >>The curve is unimportant here. I figured the slope to be -89/128. I >>checked >>it with webworks, and my slope is correct. >>The point on the curve where we take the slope is (1,2). >>So, knowing my m = -89/128, finding Ôb should be simple enough. > Youre looking for an equation in the form > y = mx + b > You know m, x, and y. Substitute them: > 2 = (-89/128)*1 + b > 2 + 89/128 = b > b = (256+89)/128 = 345/128 > Final equation: > y = (-89/128)x + 345/128 > This is a bit shorter than what you did, but its the same answer. > Either we made different errors with the same effect, or were both > right. > Why do you say that 345/128 is the wrong intercept? This is a webworks assignment. It tells me whether my answer is correct or incorrect the moment I hit the submit button. Thats why I know my slope is correct, but that the intercept of 345/128 is not. Ill try emailing the professor overseeing the site today, to see if the error is theirs. === Subject: Widrow-Hoff? Im having some trouble interpreting a reference to the Widrow-Hoff equation. The reference is on page 9 of http://animatlab.lip6.fr/papers/Gerard_Mey_Sig_EJOR03.ps.gz, but in short, it involves an algorithm in which a variable is increased by: Is = (1 - beta)*Is + beta and decreased by: Is = (a - beta)*Is This confuses me because it doesnt seem to conform to any traditional definition of the Widrow-Hoff delta rule. For instance, what does beta signify? Any help would be greatly appreciated. === Subject: gaussian-Jordian elimination Can someone explain to me the purpose of Gaussian elimination in matrices? Whats the difference in a augmented and inverse matrix? I know the matrix consists of a series of linear equations and an array and is used to solve linear equations faster than normal. Im also interested in financial analysis and statistics and would like to filter out data that is irrelevant. I guess that can be done with a correlation coefficient and some type of oscillator. Bill === Subject: Re: gaussian-Jordian elimination > Can someone explain to me the purpose of Gaussian elimination in > matrices? Whats the difference in a augmented and inverse matrix? I know > the matrix consists of a series of linear equations and an array and is > used > to solve linear equations faster than normal. Im also interested in > financial analysis and statistics and would like to filter out data that > is > irrelevant. I guess that can be done with a correlation coefficient and > some > type of oscillator. Gaussian elimination is a systematic procedure for solving square systems of linear equations. As applied to matrices it can be used to find matrix determinants, their inverses and things like LU decompositions. What GE boils down to is the application of three rules to the row operations of a matrix, or augmented matrix. 1) Row interchange may be performed on any two rows 2) A row may be multiplied throughout by a non zero constant 3) Any row may be replaced by the sum of that row and any other When you say whats the difference between an augmented and inverse matrix, its a little like asking whats the difference between an apple and a banana.... If two matrices when multiplied togerther produce the identity matrix, then they are inverses of eachother. so if we have a matrix A, then its inverse is typically written A^-1 , so: A * A^-1 = A^-1 * A = I An augmented matrix is simply a matrix with another matrix joined to its right hand side. When using GE to solve a system of linear equations such as Ax = b, then typically the augmented matrix A|b is worked upon. Also when finding a matrix inverse then the augmented matrix A|I is used. ÔI being the identity matrix. There are plenty of computer packages around to solve systems of linear equations, and Id guess do the sort of analysis you require. My advice would be to find out a bit about GE, matrix inverses etc., then use a computer package to do whatever you have in mind. > Bill === Subject: Theoretical Probability - Poisson Random Variable Let X be a Poisson random variable with parameter lambda. Show that P{ X = i } increases monotonically and then decreases monotonically as i increases, reaching its maximum when i is the largest integer not exceeding lambda. HINT: Consider P{ X = i }/P{ X = i - 1} Any help would be much appreciated. -Meg === Subject: Theoretical Probability - more Poisson RV Suppose that the number of events that occur in a specified time is a Poisson Random variable with parameter lambda. If each event is counted with probability p, independently of ever other event, show that the number of events that are counted is a Poisson random variable with parameter ( lambda * p ). Also, give an intuitive argument as to why this should be so. As an application, suppose that the number of distinct uranium deposits in a give area is a Poisson random variable with parameter lambda = 10. If in a fixed period of time, each deposit is discovered independently with probability 1/50, find the probability that (a) exactly 1, (b) at least 1, and (c) at most 1 desposit is discovered during that time. Again, any help is much appreciated. Meg === Subject: Proof of Automorphism by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JFt6X11989; I know this should be easy, but my brain is just not seeing this. Can someone please help or point me in the right direction? Suppose that G is a finite Abelian group and G has no element of order 2. Show that the mapping g->g^2 is an automorphism of G. Show, by example, that if G is infinite the mapping need not be an automorphism. === Subject: Re: Proof of Automorphism alt.math.undergrad: > I know this should be easy, but my brain is just not seeing this. Can > someone please help or point me in the right direction? > Suppose that G is a finite Abelian group and G has no element of order > 2. Show that the mapping g->g^2 is an automorphism of G. The mapping is clearly a homomorphism, so it fails to be an automorphism only if it fails to be a bijection. If it isnt 1-1, it has a non-trivial kernel; why is this impossible? > Show, by > example, that if G is infinite the mapping need not be an > automorphism. It still has to be 1-1, but in the infinite case it could fail to be a bijection by failing to map G *onto* G. Whats the most familiar infinite Abelian group you know? (Remember that you have to specify both the set and the operation.) Does it work? Brian === Subject: Re: Proof of Automorphism > I know this should be easy, but my brain is just not seeing this. Can > someone please help or point me in the right direction? > Suppose that G is a finite Abelian group and G has no element of order > 2. Show that the mapping g->g^2 is an automorphism of G. What part of being an automorphism is giving you trouble? > Show, by > example, that if G is infinite the mapping need not be an > automorphism. The (additive) integers has no element of order 2. For additive groups the map is g -> 2*g. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Lagrange multipliers > Its saying that the constrained extrema of the objective Very basic question (I am trying to learn more about the language of Maths.) constrained extrema tranlsates to maximum value of the function given the constraints? Does this word extrema always mean maximum? If yes, they why not just use maximum (sounds too easy? and extrema sounds more nifty/deep/mature?). If no, why? ->HS > function occur when the gradient of the objective function > is a linear combination of the gradients of the constraints. > Consider the one-constraint case -- its saying that > the constrained extrema are located where the gradient > of the objective function is parallel (or anti-parallel) > to the gradient of the constraint function. -- (Remove all underscores,if any, from my email address to get the correct one. Apologies for the inconvenience but this is to reduce === Subject: Re: Lagrange multipliers > Its saying that the constrained extrema of the objective > Very basic question (I am trying to learn more about the language of Maths.) > constrained extrema tranlsates to maximum value of the function given > the constraints? Does this word extrema always mean maximum? If > yes, they why not just use maximum (sounds too easy? and extrema > sounds more nifty/deep/mature?). If no, why? > ->HS An extreme can be either a maximum or a minimum. Since much the same methods are used for finding minima as for finding maxima, it is useful to describe only finding potential extrema, and then noting separately how to distinguish whether they are maxima or minima (or neither). > function occur when the gradient of the objective function > is a linear combination of the gradients of the constraints. > Consider the one-constraint case -- its saying that > the constrained extrema are located where the gradient > of the objective function is parallel (or anti-parallel) > to the gradient of the constraint function. === Subject: Re: Lagrange multipliers >Its saying that the constrained extrema of the objective >>Very basic question (I am trying to learn more about the language of Maths.) >>constrained extrema tranlsates to maximum value of the function given >>the constraints? Does this word extrema always mean maximum? If >>yes, they why not just use maximum (sounds too easy? and extrema >>sounds more nifty/deep/mature?). If no, why? >>->HS > An extreme can be either a maximum or a minimum. Since much the same > methods are used for finding minima as for finding maxima, it is useful > to describe only finding potential extrema, and then noting separately > how to distinguish whether they are maxima or minima (or neither). Usually, there are times when I am really tuned into the maths language and then many things are quite easy. Then there are times when I am not and this is what happens :) -- (Remove all underscores,if any, from my email address to get the correct one. Apologies for the inconvenience but this is to reduce === Subject: Re: Lagrange multipliers >Its saying that the constrained extrema of the objective >>Very basic question (I am trying to learn more about the language of Maths.) >>constrained extrema tranlsates to maximum value of the function given >>the constraints? Does this word extrema always mean maximum? If >>yes, they why not just use maximum (sounds too easy? and extrema >>sounds more nifty/deep/mature?). If no, why? >>->HS > An extreme can be either a maximum or a minimum. Since much the same > methods are used for finding minima as for finding maxima, it is useful > to describe only finding potential extrema, and then noting separately > how to distinguish whether they are maxima or minima (or neither). > Usually, there are times when I am really tuned into the maths language > and then many things are quite easy. Then there are times when I am not > and this is what happens :) Even Homer nods. === Subject: Compromise, and you leave my papers? For those of you who dont know, yes, I have for several years posted on Usenet, often posting a lot, but decided to leave it behind me as it wasnt effective. But when the paper was put online by that journal--its an electronic one--sci.mathers got together to send emails claiming the paper was wrong. They succeeded in convincing the editor Ioannis Argyros, who yanked my paper immediately, and told me about it after, and refused to allow me to answer the sci.mathers claims: Id waited over nine months on that paper. Now those posters are arguing with me claiming my *conclusion* is wrong, and I am tired of this and would like to suggest a compromise. Posters have repeatedly claimed that they disagree with the conclusion of my paper. My compromise is to change that conclusion in light of their objections, and note that the result shown algebraically conßicts with what follows from the definition of the ring of algebraic integers. Id then like to note that it seems to be an area where further study would be advised. My request is that posters here arguing with me, accept that compromise so that I dont have to worry about them mounting any more email campaigns! Then I can leave Usenet and get back to business and my other papers. Arguing here is not effective, but sci.mathers mounting emails campaigns is. Id like to try and solve at least that problem before I go. The problem is that Usenet IS effective in one clear way, as journal editors dont typically get a horde of emails when they publish! And I think it is clearly had an impact on the atmosphere given what editors are telling me now. Id just as soon get you people off my back when it comes to my papers. So sci.mathers mounting emails assaults are a direct threat, as has been proven already. To answer that threat Im willing to make changes to my papers conclusion, which so many of you have fought against for so many months. You should welcome this offer. I am willing to compromise. Are any of you? James Harris === Subject: Re: Compromise, and you leave my papers? > [...] > My compromise is to change that conclusion in light of their > objections, and note that the result shown algebraically conßicts > with what follows from the definition of the ring of algebraic > integers. > [...] > I am willing to compromise. Are any of you? Heres a counter offer: (1) You do not post any more about claiming the prime-counting formula as yours; (2) You do not mention your APF paper again; (3) You do not mention the mathematics community ganging up on you; (4) You will submit only papers that have no major logical ßaw in them, but which can be rewritten without major reconstruction. (I.e., the only errors in your papers are fixable ones.) In return, people on Usenet will not harass you with e-mails, and will not send any sort of communication to the editor(s) of whatever journal(s) you submit your paper to in an attempt to remove your paper. (Note that if you submit a paper which is incorrect, that people have the right to let you and/or the editor know about it.) Im sure that most of Usenet would be willing to hold up their end. Are you willing to hold up yours? -- Christopher Heckman === Subject: Re: Compromise, and you leave my papers? > Then I can leave Usenet and get back to business and my other papers. Most would be upset if you leave this group. Please dont go! (Note that nobody gives a dam if Arhimedes Plutonium, or Kabatoff goes). I suggest the following good-will gesture: subject. === Subject: Re: Compromise, and you leave my papers? > For those of you who dont know, yes, I have for several years posted > on Usenet, often posting a lot, but decided to leave it behind me as > it wasnt effective. > But when the paper was put online by that journal--its an electronic > one--sci.mathers got together to send emails claiming the paper was > wrong. > They succeeded in convincing the editor Ioannis Argyros, who yanked my > paper immediately, and told me about it after, and refused to allow me > to answer the sci.mathers claims: > Id waited over nine months on that paper. How about organizing an email campaign to get your paper unyanked and the corrections placed next to it? You might even be able to get the people who emailed corrections to work with you on this. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Compromise, and you leave my papers? (snip) > I am willing to compromise. Are any of you? So weve seen from James denial, anger, despair, and now bargaining. Hopefully acceptance is not far behind. === Subject: Re: Compromise, and you leave my papers? >For those of you who dont know, yes, I have for several years posted >on Usenet, often posting a lot, but decided to leave it behind me as >it wasnt effective. >But when the paper was put online by that journal--its an electronic >one--sci.mathers got together to send emails claiming the paper was >wrong. >They succeeded in convincing the editor Ioannis Argyros, who yanked my >paper immediately, and told me about it after, and refused to allow me >to answer the sci.mathers claims: >Id waited over nine months on that paper. >Now those posters are arguing with me claiming my *conclusion* is >wrong, _Now_ theyre arguing that its wrong? People have explained why its wrong (to the extent that its comprehensible) from the start - its nobodys fault but your own that you decided to try to publish it anyway. >and I am tired of this and would like to suggest a compromise. >Posters have repeatedly claimed that they disagree with the conclusion >of my paper. >My compromise is to change that conclusion in light of their >objections, and note that the result shown algebraically conßicts >with what follows from the definition of the ring of algebraic >integers. >Id then like to note that it seems to be an area where further study >would be advised. >My request is that posters here arguing with me, accept that >compromise so that I dont have to worry about them mounting any more >email campaigns! >Then I can leave Usenet and get back to business and my other papers. >Arguing here is not effective, but sci.mathers mounting emails >campaigns is. >Id like to try and solve at least that problem before I go. >The problem is that Usenet IS effective in one clear way, as journal >editors dont typically get a horde of emails when they publish! >And I think it is clearly had an impact on the atmosphere given what >editors are telling me now. >Id just as soon get you people off my back when it comes to my >papers. >So sci.mathers mounting emails assaults are a direct threat, as has >been proven already. Uh, right. Youve sent your crap to several journals - in all but one case the papers were simply rejected. Did any of those journals mention sci.math? You really _are_ being paranoid about this. Except for that one case where the journal screwed up, the reason your papers have been rejected has nothing to do with sci.math - its because they dont contain anything thats both correct and significant. Honest. Supposing for the sake of argument that everyone in sci.math agreed to the silly compromise you suggest here. That would _not_ help you get those papers published! Because sci.math has nothing to do with it. Honest. >To answer that threat Im willing to make changes to my papers >conclusion, which so many of you have fought against for so many >months. >You should welcome this offer. >I am willing to compromise. Are any of you? Uh, this is simply hilarious. The notion of compromise doesnt arise here. If you write something correct nobody will complain about it - if the new version is still wrong people will say so. And btw yes, Nora is absolutely right in pointing out that this shows youre a despicable liar - going on for months about how youre right and were all going to jail for saying youre wrong. If you were right this compromise wouldnt come up. >James Harris ************************ David C. Ullrich === Subject: Re: Compromise, and you leave my papers? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >>Now those posters are arguing with me claiming my *conclusion* is >>wrong, > _Now_ theyre arguing that its wrong? Just read Now in the meaning It so happens that. There will still be more than enough left in the post to object to. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Compromise, and you leave my papers? >For those of you who dont know, yes, I have for several years posted >on Usenet, often posting a lot, but decided to leave it behind me as >it wasnt effective. >But when the paper was put online by that journal--its an electronic >one--sci.mathers got together to send emails claiming the paper was >wrong. >They succeeded in convincing the editor Ioannis Argyros, who yanked my >paper immediately, and told me about it after, and refused to allow me >to answer the sci.mathers claims: >Id waited over nine months on that paper. >Now those posters are arguing with me claiming my *conclusion* is >wrong, and I am tired of this and would like to suggest a compromise. >Posters have repeatedly claimed that they disagree with the conclusion >of my paper. >My compromise is to change that conclusion in light of their >objections, and note that the result shown algebraically conßicts >with what follows from the definition of the ring of algebraic >integers. >Id then like to note that it seems to be an area where further study >would be advised. >My request is that posters here arguing with me, accept that >compromise so that I dont have to worry about them mounting any more >email campaigns! This is truly laughable and despicable. For weeks you deny lying about the conclusion of that paper, and call other people liars for daring to disagree. You threaten loss of jobs, even JAIL TERMS, for people who simply say you are wrong. Now you whine, Lets compromise - please dont tell the journal editor that my paper is wrong - please let me just sneak this in - please go easy on me ... There are so many repulsive aspects to what you are proposing: 1. It is the NORMAL thing to do to write a letter to the Editor when you see a mistake. If ANYBODY published your piece-of- there would be letters to the editor - thousands, if it were Wiles - yet you want special treatment. 2. Yes, the conclusion is wrong. That is because the conclusion is based on the main result, which implies that, e.g., a root of a monic irreducible polynomial with integer coefficients is coprime in the algebraic integers to prime factors of the constant term - a demonstrably FALSE result - that even if you change the application at the end involving 65*x^3 + 12*x - 1, the paper is STILL totally wrong. It deserves NO immunity from letters-to-the-editor. No deal, unless you just gut the whole thing. That you could justify. 3. Incredibly enough, the paper STILL contains a dimwit low-level algebraic error in the latter part, where you say (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3(-1 + mf^2)xu^2 + u^3 = 65x^3 - 12x + 1. Astonishing! The paper was written over a year ago. It went through much criticism, many revisions, re-composed in several different word processors, (supposedly) reviewing at one or more journals, etc., all with *no change* in this line. Finally someone HERE noticed it was wrong. AFTER THAT, it was recomposed yet again, and a co-author, A. Beckwith was added. But the error is still there! Did Beckwith actually READ any part of this paper? Why is he a co-author? Did he catch even this low-level error which you yourself have admitted, let alone the important ones? Have you resubmitted this version of the paper to yet ANOTHER journal, hoping it squeaks through by accident ? 4. Suppose we agreed to your compromise. How will we know you keep your end of the bargain, especially in light of 3. just above, and your pattern of repeated lying ??? 5. You have been trying to lie about what the paper originally said. Now you want to change the conclusion. You wouldnt be doing that if the conclusion was actually right. So what you are now doing is admitting the lie. 6. We said you were wrong. You said we were lying. We werent wrong. We werent lying. YOU were wrong and YOU were lying. You should make note of this for future reference. The liar was YOU. The one who screwed up the math, that was YOU. Now, we are STILL saying you are wrong. You will say, no, YOURE wrong, youre lying. Were not. Its the same as before. We werent lying before and were not lying now. We BELIEVE what we say, and we can prove what we claim. Try to remember this ! And by the way, how about an apology? Can we have an apology with that admission? >Then I can leave Usenet and get back to business and my other papers. >Arguing here is not effective, but sci.mathers mounting emails >campaigns is. >Id like to try and solve at least that problem before I go. Just go. >The problem is that Usenet IS effective in one clear way, as journal >editors dont typically get a horde of emails when they publish! If they publish crap like what you sent, they will get a horde of emails. They *should*. >And I think it is clearly had an impact on the atmosphere given what >editors are telling me now. What are they telling you? That it is a piece of crap, possibly? Anybody could come to that conclusion, with no prompting from sci.math. >Id just as soon get you people off my back when it comes to my >papers. Then stop trying to claim they are correct. >So sci.mathers mounting emails assaults are a direct threat, as has >been proven already. They should be. Letters to the editor are how journals in general, not just math journals, are kept honest. >To answer that threat Im willing to make changes to my papers >conclusion, which so many of you have fought against for so many >months. Its not just the conclusion. The whole thing is wrong. Agreeing with you about the conclusion would be tacitly agreeing to the rest of it. That I will never do. >You should welcome this offer. >I am willing to compromise. Are any of you? Im not. Dick Cheney at this point would tell you what to go do with yourself. Go do it, Liar. Nora B. >James Harris === Subject: Re: Compromise, and you leave my papers? > 1. It is the NORMAL thing to do to write a letter to the Editor > when you see a mistake. If ANYBODY published your piece-of- > there would be letters to the editor - thousands, if it were > Wiles - yet you want special treatment. Im not sure that it is the NORMAL thing to do. I suspect that in most cases people contact the author on finding a mistake in a paper. I suspect that most editors would either refer you to the author, or forward your message to hir. Only if the author refused to address the concerns would an editor be likely to bother to get involved. Heck, even Pertii contacted the authors of the erroneous papers that he found. The case at hand is, of course, different. The author had most definitely been contacted before publication, and the people contacting the editor had every reason to believe that contacting the author yet again would accomplish less than nothing. So the usual first step was bypassed. === Subject: Re: Compromise, and you leave my papers? > 1. It is the NORMAL thing to do to write a letter to the Editor > when you see a mistake. If ANYBODY published your piece-of- > there would be letters to the editor - thousands, if it were > Wiles - yet you want special treatment. > Im not sure that it is the NORMAL thing to do. I suspect that in most > cases people contact the author on finding a mistake in a paper. I > suspect that most editors would either refer you to the author, or > forward your message to hir. Only if the author refused to address the > concerns would an editor be likely to bother to get involved. Heck, > even Pertii contacted the authors of the erroneous papers that he found. > The case at hand is, of course, different. The author had most > definitely been contacted before publication, and the people contacting > the editor had every reason to believe that contacting the author yet > again would accomplish less than nothing. So the usual first step was > bypassed. Does anyone else think that this thread (as all the ones before it) is completely pointless? Clearly people at some point started out wanting to help James understand the mistake in his papers. However, by now it is clear that this is a futile attempt, and that he does not appreciate this help. Consequently posters no longer reply out of sympathy for James (and make this clear); however, then what is the point? No one except James could possibly have an interest in prolonged discussions why his results are wrong. If James wants to publish a paper in some third-, fourth- or fifth-rate journal, all he needs to do is submit it to enough of these and keep quiet about it. I am certain that there are many proofs of Fermats last theorem, P=NP, P!=NP etc. hidden away somewhere where no-one will ever find them ... By the way, an entry for Jamess paper does remain on MathSciNet, so the editor wasnt successful in removing all records of the fact that it was once accepted. Finally, I think we should all feel somewhat sorry for James. Clearly he is not well; perhaps if he was willing to seek therapy, he would be able to lead a happier life. Until then, it seems best not to spend much time on the thankless task of refuting his purported proofs. Lasse --- === Subject: Re: Compromise, and you leave my papers? > 1. It is the NORMAL thing to do to write a letter to the Editor > when you see a mistake. If ANYBODY published your piece-of- > there would be letters to the editor - thousands, if it were > Wiles - yet you want special treatment. > Im not sure that it is the NORMAL thing to do. I suspect that in most > cases people contact the author on finding a mistake in a paper. I > suspect that most editors would either refer you to the author, or > forward your message to hir. Only if the author refused to address the > concerns would an editor be likely to bother to get involved. Heck, > even Pertii contacted the authors of the erroneous papers that he found. > The case at hand is, of course, different. The author had most > definitely been contacted before publication, and the people contacting > the editor had every reason to believe that contacting the author yet > again would accomplish less than nothing. So the usual first step was > bypassed. > Does anyone else think that this thread (as all the ones before it) is > completely pointless? Clearly people at some point started out wanting > to help James understand the mistake in his papers. However, by now > it is clear that this is a futile attempt, and that he does not > appreciate this help. Consequently posters no longer reply out of > sympathy for James (and make this clear); however, then what is the > point? No one except James could possibly have an interest in > prolonged discussions why his results are wrong. > If James wants to publish a paper in some third-, fourth- or > fifth-rate journal, all he needs to do is submit it to enough of these > and keep quiet about it. I am certain that there are many proofs of > Fermats last theorem, P=NP, P!=NP etc. hidden away somewhere where > no-one will ever find them ... > By the way, an entry for Jamess paper does remain on MathSciNet, so > the editor wasnt successful in removing all records of the fact that > it was once accepted. > Finally, I think we should all feel somewhat sorry for James. Clearly > he is not well; perhaps if he was willing to seek therapy, he would be > able to lead a happier life. Until then, it seems best not to spend > much time on the thankless task of refuting his purported proofs. > Lasse > --- Oh yeah, all I really need is therapy. LOL. Posters who argue with me cant stop for a simple reason. Figure it out. Ultimately what I do on Usenet is post. Now what that has to do with my personal life, such that people feel they can recommend therapy to me is another matter. Care to discuss your personal life in reply? James Harris === Subject: Re: Compromise, and you leave my papers? big deal, like were the first cohort to dyscover grouptherapy, even online? now, the fact that one of us may be the Lead Investigator in a MK-infra virtual lab would probably also not be new, since its a Ô60s-era porgramme. > Oh yeah, all I really need is therapy. LOL. Posters who argue with > me cant stop for a simple reason. Figure it out. --A HYDROGEN (sic; cracked methane) ECONOMY?... The Three Phases of Exploitation of the Protocols of the Elders of Kyoto (sik): BORE/GUSH/NADIR @ http://www.tarpley.net. Http://www.tarpley.net/bushb.htm (partial contents, below): 17 -- THE ATTEMPTED COUP DETAT, 3/30/81 (87K) 18 -- IRAN-CONTRA (140K) 19 -- THE LEVERAGED BUYOUT MOB (67K) 20 -- THE PHONY WAR ON DRUGS (26K) 21 -- OMAHA (25K) 22 -- GEORGE #9 TAKES THE PRESIDENCY (112K) 23 -- THE END OF HISTORY (168K) 24 -- THE NEW WORLD ORDER (255K) 25 -- THYROID STORM (139K) http://quincy4board.homestead.com/files/curriculum/Cosmo === Subject: Re: Compromise, and you leave my papers? > 1. It is the NORMAL thing to do to write a letter to the Editor > when you see a mistake. If ANYBODY published your piece-of- > there would be letters to the editor - thousands, if it were > Wiles - yet you want special treatment. > Im not sure that it is the NORMAL thing to do. I suspect that in most > cases people contact the author on finding a mistake in a paper. I > suspect that most editors would either refer you to the author, or > forward your message to hir. Only if the author refused to address the > concerns would an editor be likely to bother to get involved. Heck, > even Pertii contacted the authors of the erroneous papers that he found. > The case at hand is, of course, different. The author had most > definitely been contacted before publication, and the people contacting > the editor had every reason to believe that contacting the author yet > again would accomplish less than nothing. So the usual first step was > bypassed. Well, it was not only bypassed but there was a *gang* emailing, as multiple people coordinated to send emails. They didnt get together to pick one person to send an email--they wanted to put pressure. And notice that Nora Baron has turned insulting and clearly is rigid, with an impossible position that I should just toss my work. Thats not compromise. Still think Nora Baron is a woman? Its not hard to create an online persona, that is feminine. Guys hate compromise, and Im sure some of them cringed to see me use the word. A guy would want total victory: no compromise. And remember I have math results that are not even disputed! Like remember that with P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) and P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) the as are the three roots of the cubic a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 and you can check at m=1, f=sqrt(2), that my argument is correct. If you think thats chance, check other solutions that give one of the as rational, as you will find it will always either have f as a factor or be coprime to f. So you have take-no-prisoners and burn everything from Nora Baron and an offer of compromise from me. Think about it. And remember, think about all the effort put out by that person already in replies to my posts!!! So none of that was for anything, but personal satisfaction? Guy behavior. All the way, and if you think Nora Baron is a woman, consider why some guy would want you to think that, and why it is a strategy that works...until firmly tested by an offer as simple as, compromise. James Harris === Subject: Re: Compromise, and you leave my papers? > And remember I have math results that are not even disputed! > Like remember that with > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > and > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > the as are the three roots of the cubic > a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > and you can check at m=1, f=sqrt(2), that my argument is correct. > If you think thats chance, check other solutions that give one of the > as rational, as you will find it will always either have f as a > factor or be coprime to f. You still do not see the difference between irreducible polynomials and reducible polynomials. And you still fail to state in what ring they should be factors. Because what you state is false in the algebraic integers. All three will not be coprime to f in the algebraic integers, if the cubic in a is irreducible. Check the facts when the polynomial in a is irreducible. As it is, the roots of the cubic in a are only governed by the product m.f^2. And there are only finitely many such products that yield a reducible cubic. You have found them all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > And remember I have math results that are not even disputed! > Like remember that with > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > and > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > the as are the three roots of the cubic > a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > and you can check at m=1, f=sqrt(2), that my argument is correct. > > If you think thats chance, check other solutions that give one of the > as rational, as you will find it will always either have f as a > factor or be coprime to f. > You still do not see the difference between irreducible polynomials and > reducible polynomials. And you still fail to state in what ring they > should be factors. Because what you state is false in the algebraic > integers. All three will not be coprime to f in the algebraic integers, > if the cubic in a is irreducible. Check the facts when the polynomial > in a is irreducible. > As it is, the roots of the cubic in a are only governed by the product > m.f^2. And there are only finitely many such products that yield a > reducible cubic. You have found them all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > And remember I have math results that are not even disputed! > Like remember that with > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > and > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > the as are the three roots of the cubic > a^3 - 3(-1 + mf^2)a^2 - f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0 > and you can check at m=1, f=sqrt(2), that my argument is correct. Ah, now I see why I was in error. Your cubic in a is wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Compromise, and you leave my papers? > Still think Nora Baron is a woman? Its not hard to create an > online persona, that is feminine. Guys hate compromise, and Im sure > some of them cringed to see me use the word. A guy would want total > victory: no compromise. Since you brought up the possibility of compromise, what are you? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Compromise, and you leave my papers? >> 1. It is the NORMAL thing to do to write a letter to the Editor >> when you see a mistake. If ANYBODY published your piece-of- >> there would be letters to the editor - thousands, if it were >> Wiles - yet you want special treatment. >> >> Im not sure that it is the NORMAL thing to do. I suspect that in most >> cases people contact the author on finding a mistake in a paper. I >> suspect that most editors would either refer you to the author, or >> forward your message to hir. Only if the author refused to address the >> concerns would an editor be likely to bother to get involved. Heck, >> even Pertii contacted the authors of the erroneous papers that he found. >> The case at hand is, of course, different. The author had most >> definitely been contacted before publication, and the people contacting >> the editor had every reason to believe that contacting the author yet >> again would accomplish less than nothing. So the usual first step was >> bypassed. >Well, it was not only bypassed but there was a *gang* emailing, as >multiple people coordinated to send emails. >They didnt get together to pick one person to send an email--they >wanted to put pressure. >And notice that Nora Baron has turned insulting and clearly is >rigid, with an impossible position that I should just toss my work. >Thats not compromise. Where did this ridiculous notion of compromise in mathematics come from? The paper is nonsense. Also, the reason its getting rejected is does not have anything to do with what happens here on sci.math, honest. If you got the entire group to agree that it was all correct it would _still_ be rejected by any journal, unless the people running the journal were irresponsible or incompetent. Because its nonsense. Not because we say so, because it _is_. Honest. >Still think Nora Baron is a woman? What does that have to do with anything? If we found proof she was a man would that make your paper correct somehow? ************************ David C. Ullrich === Subject: Re: Compromise, and you leave my papers? >For those of you who dont know, yes, I have for several years posted >on Usenet, often posting a lot, but decided to leave it behind me as >it wasnt effective. >But when the paper was put online by that journal--its an electronic >one--sci.mathers got together to send emails claiming the paper was >wrong. >They succeeded in convincing the editor Ioannis Argyros, who yanked my >paper immediately, and told me about it after, and refused to allow me >to answer the sci.mathers claims: >Id waited over nine months on that paper. >Now those posters are arguing with me claiming my *conclusion* is >wrong, and I am tired of this and would like to suggest a compromise. >Posters have repeatedly claimed that they disagree with the conclusion >of my paper. >My compromise is to change that conclusion in light of their >objections, and note that the result shown algebraically conßicts >with what follows from the definition of the ring of algebraic >integers. >Id then like to note that it seems to be an area where further study >would be advised. >My request is that posters here arguing with me, accept that >compromise so that I dont have to worry about them mounting any more >email campaigns! > This is truly laughable and despicable. For weeks you deny lying > about the conclusion of that paper, and call other people liars for > daring to disagree. You threaten loss of jobs, even JAIL TERMS, for > people who simply say you are wrong. > Now you whine, Lets compromise - please dont tell the journal > editor that my paper is wrong - please let me just sneak this in - > please go easy on me ... Now Im willing to consider your objections and you say my behavior is despicable? Im willing to modify the conclusion of *my* paper in light of your objections, and you insult me? You have attacked the conclusion of my paper, and Im agreeing to modify it in light of what you have claimed are your objections. What could be wrong with that deal? What compromise do you offer? What do you want me to do with my paper? James Harris === Subject: Re: Compromise, and you leave my papers? > What compromise do you offer? What do you want me to do with my paper? Fix it: (1) Fix the terminology. As it stands, it purports to be a paper about algebraic integers. However, its reasoning and conclusion are simply false when applied to the algebraic integers. Based on some of your recent posts, it appears that it isnt actually supposed to be about algebraic integers. Apparently, you think the term algebraic integer should apply to some different set of numbers than everyone else. Like it or not, it has already been decided what set of numbers the term algebraic integer refers to, and so if you want to work in some other ring or field, you have to name it something else. (2) Specify precisely what this other ring or field is. So far, all youve given are a set of magical properties it has, and you only give those when you have to give one in order to try to answer an error found in your paper. Get off the grassy gnoll, and leave the magic ring out of your theory. (3) Fix the exposition. You throw in variables all over the place without saying anything about them (are they integers? rational? real? complex? algebraic integers? are there constaints on the them?), leaving it to the reader to try to absorb the paper as a whole, and then go back and try to find some way to answer those questions that will make the paper as a whole make sense. The purpose of a mathematical paper is to communicate mathematics, and your paper fails to do that for the above reasons. Until you fix these problems, it is unpublishable. (Note: this does not imply that fixing the above problems makes it publishable...fixing them merely gets it to the point where it is not a total waste of time for anyone to even try to read it). -- --Tim Smith === Subject: Re: Compromise, and you leave my papers? >For those of you who dont know, yes, I have for several years posted >on Usenet, often posting a lot, but decided to leave it behind me as >it wasnt effective. >But when the paper was put online by that journal--its an electronic >one--sci.mathers got together to send emails claiming the paper was >wrong. They succeeded in convincing the editor Ioannis Argyros, who yanked my >paper immediately, and told me about it after, and refused to allow me >to answer the sci.mathers claims: >Id waited over nine months on that paper. Now those posters are arguing with me claiming my *conclusion* is >wrong, and I am tired of this and would like to suggest a compromise. Posters have repeatedly claimed that they disagree with the conclusion >of my paper. My compromise is to change that conclusion in light of their >objections, and note that the result shown algebraically conßicts >with what follows from the definition of the ring of algebraic >integers. Id then like to note that it seems to be an area where further study >would be advised. My request is that posters here arguing with me, accept that >compromise so that I dont have to worry about them mounting any more >email campaigns! This is truly laughable and despicable. For weeks you deny lying > about the conclusion of that paper, and call other people liars for > daring to disagree. You threaten loss of jobs, even JAIL TERMS, for > people who simply say you are wrong. > Now you whine, Lets compromise - please dont tell the journal > editor that my paper is wrong - please let me just sneak this in - > please go easy on me ... > Now Im willing to consider your objections and you say my behavior is > despicable? > Im willing to modify the conclusion of *my* paper in light of your > objections, and you insult me? > You have attacked the conclusion of my paper, and Im agreeing to > modify it in light of what you have claimed are your objections. > What could be wrong with that deal? > What compromise do you offer? What do you want me to do with my > paper? That question is a gift. You should be a straight man. > James Harris === Subject: Re: Compromise, and you leave my papers? >For those of you who dont know, yes, I have for several years posted >on Usenet, often posting a lot, but decided to leave it behind me as >it wasnt effective. >But when the paper was put online by that journal--its an electronic >one--sci.mathers got together to send emails claiming the paper was >wrong. They succeeded in convincing the editor Ioannis Argyros, who yanked my >paper immediately, and told me about it after, and refused to allow me >to answer the sci.mathers claims: >Id waited over nine months on that paper. Now those posters are arguing with me claiming my *conclusion* is >wrong, and I am tired of this and would like to suggest a compromise. Posters have repeatedly claimed that they disagree with the conclusion >of my paper. My compromise is to change that conclusion in light of their >objections, and note that the result shown algebraically conßicts >with what follows from the definition of the ring of algebraic >integers. Id then like to note that it seems to be an area where further study >would be advised. My request is that posters here arguing with me, accept that >compromise so that I dont have to worry about them mounting any more >email campaigns! This is truly laughable and despicable. For weeks you deny lying > about the conclusion of that paper, and call other people liars for > daring to disagree. You threaten loss of jobs, even JAIL TERMS, for > people who simply say you are wrong. > Now you whine, Lets compromise - please dont tell the journal > editor that my paper is wrong - please let me just sneak this in - > please go easy on me ... > Now Im willing to consider your objections and you say my behavior is > despicable? > Im willing to modify the conclusion of *my* paper in light of your > objections, and you insult me? > You have attacked the conclusion of my paper, and Im agreeing to > modify it in light of what you have claimed are your objections. > What could be wrong with that deal? > What compromise do you offer? What do you want me to do with my > paper? > James Harris You dont seem to have read Nora Barons message to which you replied. She points out that it is not just the conclusion of your paper that is wrong. I think you should change EVERYTHING in the paper that is written you will have to scrap the whole thing and start over. Its not clear to me, at least, whether you can get anything worth publishing out of what youve written. === Subject: Re: Compromise, and you leave my papers? > Now Im willing to consider your objections and you say my behavior is > despicable? > Im willing to modify the conclusion of *my* paper in light of your > objections, and you insult me? > You have attacked the conclusion of my paper, and Im agreeing to > modify it in light of what you have claimed are your objections. > What could be wrong with that deal? > What compromise do you offer? What do you want me to do with my > paper? If you now believe your conclusion is wrong, then its your responsibility to modify or eliminate it, regardless of whether anyone else accepts a compromise or not. Its the honest thing to do. On the other hand, if you still believe its correct, then agreeing to change it would be dishonest. Lying about a correct result is no better than lying about a false one. On top of everything else, you offer to fix it if its wrong, or lie about it if its right, *only* if everyone else agrees to compromise their integrity by pretending not to see other things in your paper that they *know* are wrong! Yes, Id call that despicable, on several levels. Id also call it unbelievably slimy and contemptible if your past actions hadnt taught me to expect just such behavior from you. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Compromise, and you leave my papers? >For those of you who dont know, yes, I have for several years posted >on Usenet, often posting a lot, but decided to leave it behind me as >it wasnt effective. >But when the paper was put online by that journal--its an electronic >one--sci.mathers got together to send emails claiming the paper was >wrong. They succeeded in convincing the editor Ioannis Argyros, who yanked my >paper immediately, and told me about it after, and refused to allow me >to answer the sci.mathers claims: >Id waited over nine months on that paper. Now those posters are arguing with me claiming my *conclusion* is >wrong, and I am tired of this and would like to suggest a compromise. Posters have repeatedly claimed that they disagree with the conclusion >of my paper. My compromise is to change that conclusion in light of their >objections, and note that the result shown algebraically conßicts >with what follows from the definition of the ring of algebraic >integers. Id then like to note that it seems to be an area where further study >would be advised. My request is that posters here arguing with me, accept that >compromise so that I dont have to worry about them mounting any more >email campaigns! This is truly laughable and despicable. For weeks you deny lying > about the conclusion of that paper, and call other people liars for > daring to disagree. You threaten loss of jobs, even JAIL TERMS, for > people who simply say you are wrong. > Now you whine, Lets compromise - please dont tell the journal > editor that my paper is wrong - please let me just sneak this in - > please go easy on me ... > Now Im willing to consider your objections and you say my behavior is > despicable? You have been lying shamelessly for weeks about the conclusion of that paper. Now you admit it in some fashion or other, just to get our agreement not to point out to editors that the core of it is still wrong. No apology of course. Yes, its despicable. Remember this. The math in the paper is wrong. You lied about what it said, not once but many times. Now you admit it. In a few weeks, maybe even days, you will forget what actually happened. You will say that WE were wrong and that WE lied, that WE are the ones who will say or do anything to suppress your work. The fact that we were right, that we DIDNT lie, that it was YOU who lied, is something you are going to forget. It has happened over and over and it will happen again. That too is despicable. > Im willing to modify the conclusion of *my* paper in light of your > objections, and you insult me? I didnt insult you. I called you a liar. > You have attacked the conclusion of my paper, and Im agreeing to > modify it in light of what you have claimed are your objections. Unless you modify the conclusion to say The main result in this paper and its application to the conclusion are all false, I see no reason to compromise. > What could be wrong with that deal? Agreeing with you that your modified conclusion is false would be implicitly agreeing that the rest of it, the main theorem, is correct. It is not. I will never agree to any implication that it is. If it is published somewhere, I will write a letter to the journal pointing out their incompetence. > What compromise do you offer? What do you want me to do with my > paper? Oh, brother. Well, someone else can address what you should do with it. If I were you I would burn it and forget it. Nora B. > James Harris === Subject: Re: Compromise, and you leave my papers? > For those of you who dont know, yes, I blah blah blah blah > blah blah blah blah blah blah blah blah blah blah blah blah > blah blah blah. > But when the paper was blah blah blah, blah blah blah blah. > blah blah blah blah blah blah blah! > Blah blah blah blah blah blah blah blah blah blah blah blah > blah blah blah, and blah blah me about it blah blah, blah blah > blah blah blah blah blah: > See http://www.blah.de/journals/blahblah/blahblah.html > Id waited over nine months blah blah blah. > Now those posters are arguing with me blah blah blah blah blah > blah blah blah blah blah blah blah. > Blah blah blah blah blah blah blah blah blah blah blah > of my blah blah paper. > My compromise is to blah blah blah blah blah blah blah blah > blah blah blah blah blah blah blah blah blah blah blah > blah blah blah blah blah blah blah blah blah blah blah blah blah > blah. > Id then like to note that blah blah blah blah blah blah blah blah > blah blah blah. > My request is that blah blah blah blah blah blah blah > blah blah blah blah blah blah blah blah blah blah blah blah blah > blah blah blah blah blah blah blah! > Then I can leave blah blah blah blah blah blah blah! > Blah blah blah blah blah blah blah blah blah blah blah blah > blah blah blah blah blah blah blah. > Blah blah blah blah blah blah blah blah before I go. > The problem is that Blah blah blah blah blah blah blah blah > Blah blah blah blah blah blah blah blah! > And I think it is clearly Blah blah blah blah blah blah blah blah. > To answer that blah blah blah threat Im willing blah blah > blah blah, which so many of you have blah blah blah. > You should blah blah this blah blah offer. > I am willing to blah blah blah. Are any of you? > James Blah Blah Harris === Subject: Re: Compromise, and you leave my papers? > Now those posters are arguing with me claiming my *conclusion* is > wrong, and I am tired of this and would like to suggest a compromise. Your conclusion IS wrong. > Posters have repeatedly claimed that they disagree with the conclusion > of my paper. > My compromise is to change that conclusion in light of their > objections, and note that the result shown algebraically conßicts > with what follows from the definition of the ring of algebraic > integers. That isnt a compromise. Thats an abandonment of your conclusion. If the result conßicts with the definitions upon which the argument is based, it is ßawed. Period. > Then I can leave Usenet and get back to business and my other papers. No one is stopping you from leaving Usenet. Indeed, most readers would encourage it. > I am willing to compromise. Are any of you? Again, no compromise has been suggested. You offer to change your conclusions -- thats a simple admission that your current conclusions are wrong. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Linear Transformation help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JLj0H14643; I am currently trying to do a course at home on on linear algebra and I cam across these two questions which I am having problems with, if someone could please help it would be greatly appreciated. 1) For each possible pair of s,t gave an example of a linear transformation a:K^r --> K^r with rank s and nullity t. 2) Give an example of a linear transformation a:K^r --> K^r with Ker(a)=img(a), or give a reason why no such a exists. Any help would greatly be appreciated. === Subject: Re: Linear Transformation help! > I am currently trying to do a course at home on on linear algebra and > I cam across these two questions which I am having problems with, if > someone could please help it would be greatly appreciated. > 1) For each possible pair of s,t gave an example of a linear > transformation a:K^r --> K^r with rank s and nullity t. > 2) Give an example of a linear transformation a:K^r --> K^r with > Ker(a)=img(a), or give a reason why no such a exists. > Any help would greatly be appreciated. 2) is trivial. Let K^r = R^2, and let a((x,y)) = (0,x) === Subject: Re: Linear Transformation help! > I am currently trying to do a course at home on on linear algebra and > I cam across these two questions which I am having problems with, if > someone could please help it would be greatly appreciated. > 1) For each possible pair of s,t gave an example of a linear > transformation a:K^r --> K^r with rank s and nullity t. Doesnt rank + nullity = dimension? For a:R^r -> R^r map (a1,a2,.. a_s, a_(s+1),... a_r) to (a1,a2,.. a_s, 0,0,..0). Rank s, nullity t = r-s or have I missed the mark? > 2) Give an example of a linear transformation a:K^r --> K^r with > Ker(a)=img(a), or give a reason why no such a exists. > Any help would greatly be appreciated. > 2) is trivial. Let K^r = R^2, and let a((x,y)) = (0,x) === Subject: PLEASE HELP ME SOLVE THIS RIDDLE!!!!!!!!!!!!!!!!!1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9JMUSj18998; WHAT HAS PLENTY OF SEATS BUT NO PLACE TO SIT??????????? ALSO COWS HAVE ONE AND SO DOES PATTY, MA HAS ONE BUT NOT YOUR DADDY???????????? PLEASE LET ME KNOW ASAP!!!!!!!!!!!!!!!!! === Subject: Re: PLEASE HELP ME SOLVE THIS RIDDLE!!!!!!!!!!!!!!!!!1 >WHAT HAS PLENTY OF SEATS BUT NO PLACE TO SIT??????????? >ALSO >COWS HAVE ONE AND SO DOES PATTY, MA HAS ONE BUT NOT YOUR >DADDY???????????? > PLEASE LET ME KNOW ASAP!!!!!!!!!!!!!!!!! THIS IS A MATH GROUP! IT IS NOT A RIDDLE GROUP! === Subject: Re: PLEASE HELP ME SOLVE THIS RIDDLE!!!!!!!!!!!!!!!!!1 > .... > PLEASE LET ME KNOW ASAP!!!!!!!!!!!!!!!!! Please post riddles in the rec.puzzles news group rather than a mathematical group. Ken Pledger. === Subject: Re: PLEASE HELP ME SOLVE THIS RIDDLE!!!!!!!!!!!!!!!!!1 > WHAT HAS PLENTY OF SEATS BUT NO PLACE TO SIT??????????? A Northern Line tube train during rush hour? > ALSO > COWS HAVE ONE AND SO DOES PATTY, MA HAS ONE BUT NOT YOUR > DADDY???????????? > PLEASE LET ME KNOW ASAP!!!!!!!!!!!!!!!!! === Subject: Re: Analysis, Cesaro Summability and Tauber by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KCIgI20403; TAUBER. IF a(k)>=0 and s(n)=sum( a(k), k=1 to n) and {[s(1)+....+s(n)]/n}->L as n->oo THEN s(n)->L as n->oo. ------------------------------------------------------------- ------ Before the proof I will need the following result. b(n) can be negative or positive. (A) IF b(n)->K as n->oo THEN {[b(1)+....+b(n)]/n}->K as n->oo. Not difficult to prove this result. ------------------------------------------------------------- ------- PROOF by contradiction. TAUBER. NOTE a(k)>=0 so 0<=s(n) and is increasing so lim sup exists. CASE 1: Assume s(n)->K and K is finite real number and 0<=KL. By (A) {[s(1)+....+s(n)]/n}->K as n->oo. Contradiction since KL. CASE 2: Assume s(n)->oo as n->oo. Then given M>0 exists N(M) s.t. n>N(M) implies s(N)>M. Assume n>N(M) then h(n) = [s(1)+....+s(n)]/n = = [s(1)+....+s(N(M))+s(N(M)+1)+...+s(n)]/n [s(1)+....+s(N(M))+M+...+M]/n = = [s(1)+....+s(N(M))]/n + [M+...+M]/n = [Constant]/n + M[(n-N(M))/n] > > C/n + M/2 If n>C then (n-C)/n -> 1 so (n-C)/n > (1/2) large n > M/2 So given (M/2)>0 exists N(M)>0 s.t. h(n) > M/2 so h(n)->oo as n->oo. Contradiction. End of proof. === Subject: Real Analysis HELP!! about integration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KCIi020451; Suppose that the function f : [a,b]-->Reals is bounded and that f : [a,b]-->Reals is continuous except at a finite number of points z1,...,zk in [a,b]. Prove that f : [a,b]-->Reals is integrable. Note: Use only basic notions of integration. === Subject: Re: Real Analysis HELP!! about integration >Suppose that the function f : [a,b]-->Reals is bounded and that f : >[a,b]-->Reals is continuous except at a finite number of points >z1,...,zk in [a,b]. Prove that f : [a,b]-->Reals is integrable. >Note: Use only basic notions of integration. Why dont you show us how far you got with this problem? It may or may not be homework, but it looks like homework. If we just hand you a solution, youll read it but you probably wont be any better equipped to solve the next problem that comes along. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Real Analysis HELP!! about integration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LDuRd22551; >Suppose that the function f : [a,b]-->Reals is bounded and that f : >[a,b]-->Reals is continuous except at a finite number of points >z1,...,zk in [a,b]. Prove that f : [a,b]-->Reals is integrable. >Note: Use only basic notions of integration. >>Why dont you show us how far you got with this problem? It may or >>may not be homework, but it looks like homework. If we just hand you >>a solution, youll read it but you probably wont be any better >>equipped to solve the next problem that comes along. >>-- >>Stan Brown, Oak Road Systems, Tompkins County, New York, USA >> href=http://OakRoadSystems.com>http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are >>gradually coming to understand that nothing we do is ever our >>fault, especially if it is really stupid. --Dave Barry >i dont want answers, i want guidance. dont know where to start, and >by bounded is it m < f(x) < M or |f(x)| < M or are both equivalent >statements... im not worried about the answer so much as i am about >understanding the process of proof. Okay, start by writing down those basic notions of integration that you are told to use. In particular, what could you do if you knew f was continuous everywhere? === Subject: Re: Norms !!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KCIjH20499; > If it is: > //x+y//^2 + //x-y//^2= 2//x//^2+ 2//y//^2, > proof that this norm creates by scalar product(x,y),or to proof that >its possible only for euclyd spaces. Let ABCD be a parallelogram and consider its sides x = AB and y = AD as vectors. The diagonals AC and BD are equal to AC = AB + BC = AB + AD = x + y BD = BA + AD = -AB + AD = y - x Using the cosine theorem for the triangles ABC and ABD: |AC|^2 = |AB|^2 + |AD|^2 - 2|AB|*|AD|*cos(ABC) |BD|^2 = |AB|^2 + |AD|^2 - 2|AB|*|AD|*cos(BAD) Since ABCD is a parallelogram, ABC + BAD = 180 and cos(ABC) = cos(180 - BAD) = -cos(BAD) |AC|^2 = |AB|^2 + |AD|^2 + 2|AB|*|AD|*cos(BAD) Adding the equations for |AC|^2, |BD|^2, we get the parallelogram law: |AC|^2 + |BD|^2 = 2|AB|^2 + 2|AD|^2 or in the norm notation: ||x + y||^2 + ||x - y||^2 = 2||x||^2 + 2||y||^2 The dot product of vectors x = AB and y = AD should be: x.y = ||x||*||y||*cos(x,y) = |AB|*|AD|*cos(BAD) Calculating cos(BAD) form the equation for |AC|^2: cos(BAD) = (|AC|^2 - |AB|^2 - AD|^2)/(2|AB|*|AD|) and substituting into the dot product: x.y = (||AC|^2 - |AB|^2 - AD|^2)/2 = This suggests that given a norm satisfying the parallelogram law, the dot product should be defined as: x.y = (||x + y||^2 - ||x||^2 - ||y||^2)/2 Now, it is necessary to verify, using only the properties of a norm and the parallelogram law, that this expression satisfies all axioms of a dot product, namely: 1) x.x >= 0 2) x.x = 0 iff x = 0 3) x.y = y.x 4) x.(y + z) = x.y + x.z 5) (ax).y = x.(ay) = a(x.y) 1) x.x = (||2x||^2 - 2||x||^2)/2 = (4||x||^2 - 2||x||^2)/2 = = 2||x||^2/2 = ||x||^2 >= 0 2) x.x = 0 iff x = 0, because ||x|| = 0 iff x = 0 3) y.x = (||y + x||^2 - ||y||^2 - ||x||^2)/2 = = (||x + y||^2 - ||x||^2 - ||y||^2)/2 = = x.y 4) x.(y + z) = (||x + y + z||^2 - ||x||^2 - ||y + z||^2)/2 By the parallelogram law: ||x + y + z||^2 + ||x - y - z||^2 = 2||x||^2 + 2||y + z||^2 ||x - z + y||^2 + ||x - z - y||^2 = 2||x - z||^2 + 2||y||^2 ||x - z + y||^2 = ||z - y - x||^2 ||z + y + x||^2 + ||z - y - x||^2 = 2||z||^2 + 2||x + y||^2 ||x + z||^2 + ||x - z)||^2 = 2||x||^2 + 2||z||^2 Substituting these expressions into the first expression for ||x + y + z||^2: ||x + y + z||^2 = = 2||x||^2 + 2||y + z||^2 - ||x - y - z||^2 = = 2||x||^2 + 2||y + z||^2 - + ||x - z + y||^2 - 2||x - z||^2 - 2||y||^2 = = 2||x||^2 + 2||y + z||^2 + + ||z - y - x||^2 - + 2*(||x + z||^2 - 2||x||^2 - 2||z||^2) - 2||y||^2 = = 2||x||^2 + 2||y + z||^2 + + 2||z||^2 + 2||x + y||^2 - ||z + y + x||^2 + + 2||x + z||^2 - 4||x||^2 - 4||z||^2 - 2||y||^2 = = 2||x + y||^2 + 2||y + z||^2 + 2||z + x||^2 - - 2||x||^2 - 2||y||^2 - 2||z||^2 - ||z + y + x||^2 This allows to calculate the ||x + y + z||^2: ||x + y + z||^2 = ||x + y||^2 + ||y + z||^2 + ||z + x||^2 - - ||x||^2 - ||y||^2 - ||z||^2 For the dot product x.(y + z), we get: x.(y + z) = (||x + y + z||^2 - ||x||^2 - ||y + z||^2)/2 = = (||x + y||^2 + ||y + z||^2 + ||z + x||^2 - - ||x||^2 - ||y||^2 - ||z||^2 - - ||x||^2 - ||y + z||^2)/2 = = (||x + y||^2 - ||x||^2 - ||y||^2)/2 + = (||x + z||^2 - ||x||^2 - ||z||^2)/2 = = x.y + x.z 5) (ax).y = (||ax + y||^2 - ||ax||^2 - ||y||^2)/2 By the parallelogram law: 2||ax + y||^2 + 2||x + ay||^2 = = ||(a+1)x + (a+1)y||^2 + ||(a-1)x + (1-a)y||^2 = = (a+1)^2*||x + y||^2 + (a-1)^2*||x - y||^2 Again, by the parallelogram law: ||x - y||^2 = 2||x||^2 + 2||y||^2 - ||x + y||^2 and substituting to the previous equation: 2||ax + y||^2 + 2||x + ay||^2 = = (a+1)^2*||x + y||^2 + + (a-1)^2*(2||x||^2 + 2||y||^2 - ||x + y||^2) = = 4a||x + y||^2 + 2(a-1)^2*(||x||^2 + ||y||^2) Hence, (ax).y + x.(ay) = = (||ax + y||^2 - ||ax||^2 - ||y||^2)/2 + + (||x + ay||^2 - ||x||^2 - ||ay||^2)/2 = = a||x + y||^2 + [(a-1)^2*(||x||^2 + ||y||^2)]/2 - - (a^2 + 1)(||x||^2 + ||y||^2)/2 = = a||x + y||^2 - a||x||^2 - a||y||^2 = = 2a(||x + y||^2 - ||x||^2 - ||y||^2)/2 = = 2a(x.y) All that remains to show is that (ax).y = x.(ay). However, it is getting late and somehow, I cannot figure out the trick. === Subject: Re: Norms !!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KK5C132548; Yes , that right! But it would be little easy: here is the Appoloniys equality: ||z-x||+||z-y||=||x-y||/2 + 2||z-(x+y)/2|| And with it help we can more easy proof that 1):(x+y,z)=(x,z)+(y,z) 2)(ax,y)=a(x,y) I prooved first for a-natural, then for a-racional a=p/q => (ax,y)=(x*p/q,y)=(p*x/q,y)=p(x/q,y)= =p*q/q(x/q,y)=p/q(q*x/q,y)=p/q(x,y) for a-irracional a=lim(pn/qn) and the same operations. But now I have to proof this equality of Appoloniy and to find the fomula when scalar product is a complex! === Subject: Re: Norms !!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LC8Zx12650; |Yes! The Appoloniys equality I proved- its trivial.But with it more things gets easy. For complex variant I am proving in moment, that why I didnt look you solution. === Subject: Re: Norms !!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9IEn1e31889; >> If it is: >> //x+y//^2 + //x-y//^2= 2//x//^2+ 2//y//^2, >> proof that this norm creates by scalar product(x,y),or to proof that >>its possible only for euclyd spaces. >Simple. You can consider the definition of a norm in an n-dimensional >Euclidean space >||v|| = sqrt(...) >then ||v||^2 is a summation of an n number of squares. >Obviously ||x+y||^2 + ||x-y||^2 will give you n pairs of squares in >the form of (a+b)^2 + (a-b)^2, which simplifies to 2*a^2 + 2*b^2 (1) >The RHS should be expanded to contain n pairs of terms that match the >form (1) above. >Hope you can figure out the details on your own. I cant understand, what mean //v//=sqrt(...)?If you mean equality from condition, then axiom of inner product arent filled.Check it. Have to try some else... === Subject: Re: finding the determinant of a matrix with complex elements by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9KELtY32109; SEnd me linear programming question and their solutions. Id be very thankful. >ive not had much experience of work on matrices with complex entries, but >wondered whether the same procedure (or a similar one) applies when finding >ones determinant. >so if you were to use elementary row operations to find the determinant of a >complex matrix, would you proceed in a directly analagous way as to the >determinant finding procedure of a matrix with real entries? === Subject: homeomorfism can somebody solve this problem or give me some hints how to solve it? I have to show that A={(x,y) where abs(x)+abs(y) <=1} and B={(x,y) where x^2+y^2<=1}in R^2 are homeomorphic. (B is a disk of radius 1 and A is a rhombus } in other words I have to find a one to one and onto and continual f and f^-1 such that f:A->B any suggestions? === Subject: Re: homeomorfism in alt.math.undergrad: > can somebody solve this problem or give me some hints how to solve it? > I have to show that A={(x,y) where abs(x)+abs(y) <=1} and B={(x,y) where > x^2+y^2<=1}in R^2 are homeomorphic. > (B is a disk of radius 1 and A is a rhombus } > in other words I have to find a one to one and onto and continual f and f^-1 > such that f:A->B > any suggestions? Think of A as being made up of line segments starting at the origin and work in polar coordinates. The segment from (0, 0) to (1, 0) is fine as it is, but the segment from (0, 0) to (1/2, 1/2), for instance, needs to be stretched by a factor of sqrt(2). The amount of stretching needed is a continuous function of the angle between the segment and the x-axis. Brian === Subject: double integral and volume I am working the following problem: set up a double integral to find the volume of the solid bounded by the graphs of the equations: x^2 + z^2 = 1 y^2 + z^2 = 1 first octant Im not sure how to write the following but heres an attempt at what I got: the first two equations represent quadrants of circles in the planes. If these are swept through the x-y plane you get a circle in the x-y plane as well with the equation x^2 + y^2 = 1. I.e. it looks to me like the equations are describing a chunk of a sphere bounded by the I decided to integrate first by dy and then by dx. So the first integral would be evaluated from y=0 to y=(1-x^2)^(1/2) and I used the equation (1-x^2)^(1/2) as the z-function to integrate. The second integral would be evaluated from x=0 to x = 1. This isnt what the answer book shows. It evaluates the integrals of dy and then dx just like I did. However, it shows the inside integral evaluated from y=0 to y=x. It also shows the double integral being multiplied by 2. They use the same z funtion of (1-x^2)^(1/2). Where is my visualization of all this going wrong? If I take a cube and grind it down so it throws a shadow on the y-z plane of a circle and a shadow on the x-z plane of a circle dont I have a piece of a sphere? I think I may be going wrong with integrating (1-x^2)^(1/2). I think it should be (1-x^2-y^2)^(1/2) but I still cant visualize what the answer book is doing. tim === Subject: Re: double integral and volume > I am working the following problem: > set up a double integral to find the volume of the solid bounded by the > graphs of the equations: > x^2 + z^2 = 1 > y^2 + z^2 = 1 > first octant > Im not sure how to write the following but heres an attempt at what I got: > the first two equations represent quadrants of circles in the x-z and y-z > planes. > If these are swept through the x-y plane you get a circle in the x-y plane > as well with the equation x^2 + y^2 = 1. I.e. it looks to me like the > equations are describing a chunk of a sphere bounded by the > x-y planes. > I decided to integrate first by dy and then by dx. > So the first integral would be evaluated from y=0 to y=(1-x^2)^(1/2) and I > used the equation (1-x^2)^(1/2) as the z-function to integrate. The second > integral would be evaluated from x=0 to x = 1. > This isnt what the answer book shows. It evaluates the integrals of dy and > then dx just like I did. However, it shows the inside integral evaluated > from y=0 to y=x. It also shows the double integral being multiplied by 2. > They use the same z funtion of (1-x^2)^(1/2). > Where is my visualization of all this going wrong? If I take a cube and > grind it down so it throws a shadow on the y-z plane of a circle and a > shadow on the x-z plane of a circle dont I have a piece of a sphere? No. This is true: A sphere casts a circular shadow on the yz and xz planes This it not necessarily true: Anything that casts a circular shadow on the yz and xz planes is a sphere. Im not sure how best to demonstrate this. They say a picture is worth a thousand words, so take a look at and see if it helps. This is an image of an object bounded by two cylinders centred on axes (that produces the curved sides), and constrained to a single octant (that produces the ßat sides on the back). Hopefully you can see how this object will have a quarter-circular shadow when lit from above or from over to the right; but it clearly isnt a portion of a sphere. Once you have realised that, the books method should become clear. > think I may be going wrong with integrating (1-x^2)^(1/2). I think it > should be (1-x^2-y^2)^(1/2) but I still cant visualize what the answer > book is doing. > tim Be wary of trusting your intuition of 3d shapes. For example, shows an image that will cast circular shadows on all three planes. Yet it is not a sphere :) -- Larry Lard Replies to group please === Subject: Re: double integral and volume >> I am working the following problem: >> set up a double integral to find the volume of the solid bounded by the >> graphs of the equations: >> x^2 + z^2 = 1 >> y^2 + z^2 = 1 >> first octant >> Im not sure how to write the following but heres an attempt at what I >> got: >> the first two equations represent quadrants of circles in the x-z and y-z >> planes. >> If these are swept through the x-y plane you get a circle in the x-y >> plane as well with the equation x^2 + y^2 = 1. I.e. it looks to me like >> the equations are describing a chunk of a sphere bounded by the x-z, y-z, >> and x-y planes. >> I decided to integrate first by dy and then by dx. >> So the first integral would be evaluated from y=0 to y=(1-x^2)^(1/2) and >> I used the equation (1-x^2)^(1/2) as the z-function to integrate. The >> second integral would be evaluated from x=0 to x = 1. >> This isnt what the answer book shows. It evaluates the integrals of dy >> and then dx just like I did. However, it shows the inside integral >> evaluated from y=0 to y=x. It also shows the double integral being >> multiplied by 2. They use the same z funtion of (1-x^2)^(1/2). >> Where is my visualization of all this going wrong? If I take a cube and >> grind it down so it throws a shadow on the y-z plane of a circle and a >> shadow on the x-z plane of a circle dont I have a piece of a sphere? > No. > This is true: > A sphere casts a circular shadow on the yz and xz planes > This it not necessarily true: > Anything that casts a circular shadow on the yz and xz planes is a > sphere. > Im not sure how best to demonstrate this. They say a picture is worth > a thousand words, so take a look at > and see if it helps. This > is an image of an object bounded by two cylinders centred on axes > (that produces the curved sides), and constrained to a single octant > (that produces the ßat sides on the back). Hopefully you can see how > this object will have a quarter-circular shadow when lit from above or > from over to the right; but it clearly isnt a portion of a sphere. > Once you have realised that, the books method should become clear. >> I >> think I may be going wrong with integrating (1-x^2)^(1/2). I think it >> should be (1-x^2-y^2)^(1/2) but I still cant visualize what the answer >> book is doing. >> tim > Be wary of trusting your intuition of 3d shapes. For example, > shows an image that will > cast circular shadows on all three planes. Yet it is not a sphere :) Larry, problem could stand having the contraints specified more exactly. In looking at these shapes it is obvious that the shadow on the could be a rectangle (or more specifically a square). Jumping from the constraint statement of first octant, however, to the assumption that you should integrate over a rectangle is still not obvious to me. But at least I understand what they are doing now. I appreciate your help very much. You, and everyone who provides tutoring on here are WINNERS! tim === Subject: Numbering of Octants In Calculus III, we were discussing the octants and the professor mentioned there is no official numbering convention except for the 1st octant in which X, Y, and Z are all positive as opposed to the quadrants which are numbered from 1 to 4, counterclockwise. I cant imagine the math world has allowed such a inconsistency to exist so I ask is there an official numbering system that someone could provide an official reference to it? Christopher === Subject: Re: Numbering of Octants > In Calculus III, we were discussing the octants and the professor > mentioned there is no official numbering convention except for the 1st > octant in which X, Y, and Z are all positive as opposed to the quadrants > which are numbered from 1 to 4, counterclockwise. I cant imagine the > math world has allowed such a inconsistency to exist so I ask is there an > official numbering system that someone could provide an official > reference to it? > Christopher i suppose as long as you clearly define the system you are working in before you start, then it doesnt matter whether you follow one convention or any other - so long as you are consistent === Subject: I need help with a riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LC8Y212561; I have hands....but you will never see me wave === Subject: Re: I need help with a riddle >I have hands....but you will never see me wave This is NOT a riddle group! It IS a mathematics group. === Subject: Re: I need help with a riddle >I have hands....but you will never see me wave > This is NOT a riddle group! > It IS a mathematics group. To be a little more positive, I suggest you send riddles to the rec.puzzles news group. Ken Pledger. === Subject: Re: I need help with a riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LDgFL21559; >I have hands....but you will never see me wave A CLOCK for Gods sake! How much thought did you give this? A have another riddle: What does this have to do with mathematics? === Subject: conversion solution please by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LC8YD12578; hello. thank you for taking the time to read this. Im a carpentry student, and I have come across a problem I cant solve. I need to convert cubic yards to square feet. example; 8.56 cubic yards = x square feet If someone knows of a formula for this, and is willing to e-mail me your time. b. auger === Subject: Re: conversion solution please > I need to >convert cubic yards to square feet. >example; 8.56 cubic yards = x square feet Well, _I_ need to convert lead to gold. I think its a tossup which of us will succeed first. Cubic yards are a measure of volume; square feet are a measure of area. You cant convert between them. You could convert cubic yards to _cubic_ feet if you want; see my http://oakroadsystems.com/math/convert.htm -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: conversion solution please > hello. thank you for taking the time to read this. Im a carpentry > student, and I have come across a problem I cant solve. I need to > convert cubic yards to square feet. > example; 8.56 cubic yards = x square feet > If someone knows of a formula for this, and is willing to e-mail me > your time. > b. auger You cant convert CUBIC yards into SQUARE feet. -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: related rates---moving ladder by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LDmRG21750; >A ladder 25 feet long is leaning against the wall of a house. the base >of the ladder is pulled away from the wall at a rate of 2 feet per >second. >a. how fast is the top moving down the wall when the base of the >badder is 7 feet, 15 feet, and 24 feet from the wall? >b. consider the triangle formed by the side of the house, the ladder, >and the ground. Find the rate at which the area of the triangle is >changing when the base of the ladder ise 7 feet from the wall? >c. find the rate at which the angle between the badder and the wall >of the house is changing when the base of the ladder is 7 feet from >the wall? Have you done anything at all on this? The ladder, leaning against the house at any angle, forms a right triangle so use the Pythagorean theorem: Letting x be the distance from the base is from the wall and y the height of the ladder up the wall, x^2+ y^2= 25^2 ladder is pulled away from the wall at a rate of 2 feet per second. tells you that dx/dt= 2. (a) Differentiate x^2+ y^2= 25^2 with respect to t (CHAIN RULE!) and solve for dy/dt. (b) The area of the triangle is A= (1/2)xy. Differentiate with respect to t to find dA/dt. (c) The angle is theta= arctan(y/x). Differentiate with respect to t to find dtheta/dt. === Subject: Principal Ideals! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LHcDf10501; Hey guys, I just needed a little help with this question, I have no idea where to start so a nudge in the right direction. 1a. Let R=Q[x]. Prove that the ideal (x^2 - 2)R + (x^3 - x^2 - x - 2)R of R is a principal ideal domain. (if you use the fact that R is a Euclidean domain then you must prove that fact) 1b. Let R=Z[x]. prove that the ideal 2R + xR is not a principal ideal. any and all help will be appreciated! === Subject: Re: Principal Ideals! > Hey guys, > I just needed a little help with this question, I have no idea where > to start so a nudge in the right direction. > 1a. Let R=Q[x]. Prove that the ideal (x^2 - 2)R + (x^3 - x^2 - x - 2)R > of R is a principal ideal domain. (if you use the fact that R is a > Euclidean domain then you must prove that fact) Im guessing you meant only principle ideal without the domain. You can use the Euclideal Algorithm without knowing the ring is Euclidean. Just show that what you get, -4, is really gcd(x^2 - 2, x^3 - x^2 - x - 2) and <-4> = = <2, x> show = Z[x] != <2, x>. -- Paul Sperry Columbia, SC (USA) === Subject: Use of Vietes formulas by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LHcDx10513; hi I can somehow understand the use of Vietes formulas to find we can apply this formula to problems not related to parabolas. I know how to calculate the following example with the use of vietes formulas,but I dont know why the vietes formulas even work here,since the problem is not related in any way to parabolas Perimeter of first rectangle is 20m,and its area is 20m^2. Perimeter of the second rectangle is 22/3m,its area is 2m^2. Calculate the lengths of sides for these two rectangles Result:Sizes for first one are 3 and 7,and for the second one 2/3 and 3 Why can we use vietes formulas in the above and similar examples? thank you === Subject: Re: Use of Vietes formulas > Perimeter of first rectangle is 20m,and its area is 20m^2. > Calculate the lengths of sides for these two rectangles > Result:Sizes for first one are 3 and 7 But 3 * 7 = 21 /= 20. Results are 5 +- sqr 5. === Subject: 2sinA versus sin2A by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LIFRW14258; I am getting confused. I assume 2 sinA is not the same as sin2A Right? 2 sin 30.bc = 2 x 0.5 = 1 sin 2A where A = 30 is sin 60.bc or not? === Subject: Re: 2sinA versus sin2A >I am getting confused. I assume 2 sinA is not the same as sin2A >Right? >2 sin 30.bc = 2 x 0.5 = 1 >sin 2A where A = 30 is sin 60.bc which is sqrt(3)/2, about .866 Youre correct. Sine of an angle ought logically to be written with parentheses, which would emphasize the order of operations. But the convention is to write sin 2A where sin(2A) would be more clear and logical, just because sin 2A is shorter. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: 2sinA versus sin2A by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC41F03376; sin 2A where A = 30 is sin 60.bc which is sqrt(3)/2, about .866 Hello Stan >>sin 2A where A = 30 is sin 60.bc >>which is sqrt(3)/2, about .866 Why should sin 60.bc be equivalent to sqrt 3 divided by 2? Got me really confused now.. >>I am getting confused. I assume 2 sinA is not the same as sin2A >>Right? >>2 sin 30.bc = 2 x 0.5 = 1 >>sin 2A where A = 30 is sin 60.bc >which is sqrt(3)/2, about .866 >Youre correct. >Sine of an angle ought logically to be written with parentheses, >which would emphasize the order of operations. But the convention is >to write sin 2A where sin(2A) would be more clear and logical, just >because sin 2A is shorter. >-- >Stan Brown, Oak Road Systems, Tompkins County, New York, USA > http://OakRoadSystems.comgradually coming to understand that nothing we do is ever our >fault, especially if it is really stupid. --Dave Barry === Subject: Re: 2sinA versus sin2A > Why should sin 60.bc be equivalent to sqrt 3 divided by 2? > Got me really confused now.. Hi Jo, Mr. Stan Brown has put an excellent trigonometry tutorial online which I am sure will help clarify this for you. The section relevant to your question is located at: http://www.oakroadsystems.com/twt/special.htm#funcs30 However, the whole tutorial is well worth a read IMHO. Hope this helps. -- Hamish === Subject: Re: 2sinA versus sin2A >Mr. Stan Brown has put an excellent trigonometry tutorial online which I >am sure will help clarify this for you. The section relevant to your >question is located at: >http://www.oakroadsystems.com/twt/special.htm#funcs30 >However, the whole tutorial is well worth a read IMHO. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: 2sinA versus sin2A > Why should sin 60.bc be equivalent to sqrt 3 divided by 2? Draw an equilateral triangle. Then draw the line from one of its vertices to the midpoint of the opposite side. This line divides is into two 30-60-90 triangles. sqrt(3)/2. hth meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: 2sinA versus sin2A by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LIuTm17524; >I am getting confused. I assume 2 sinA is not the same as sin2A >Right? >2 sin 30.bc = 2 x 0.5 = 1 >sin 2A where A = 30 is sin 60.bc >or not? Right. In fact sin(2A) = 2sin(A)cos(A). Similarly, sin(x + y) is not the same as sin(x) + sin(y). The danger of function notation is that it might suggest a distributive law f(x + y) = f(x) + f(y), but this is not how functions behave, except when f(x) = mx, where m is a constant. Your question is similar. Todd Trimble === Subject: Re: 2sinA versus sin2A by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LJRXF20644; >>I am getting confused. I assume 2 sinA is not the same as sin2A >>Right? >>2 sin 30.bc = 2 x 0.5 = 1 >>sin 2A where A = 30 is sin 60.bc >>or not? >Right. In fact sin(2A) = 2sin(A)cos(A). >Similarly, sin(x + y) is not the same as sin(x) + sin(y). The >danger of function notation is that it might suggest a distributive >law f(x + y) = f(x) + f(y), but this is not how functions behave, >except when f(x) = mx, where m is a constant. Your question is >similar. >Todd Trimble I wonder if you could go a bit further. take cos2A if I square it what do I get? cos4A? how can I get rid of the 2 in an equation? Can I divide both sides by 2 and end up with cosA? === Subject: Re: 2sinA versus sin2A >take cos2A >if I square it what do I get? cos4A? Nope. You get cos(2A), which is [cos(2A)]^2. Youre thinking of the laws of logarithms, where log(x^2) = 2*log(x). -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: 2sinA versus sin2A >>take cos2A >>if I square it what do I get? cos4A? > Nope. You get cos(2A), which is [cos(2A)]^2. > Youre thinking of the laws of logarithms, where log(x^2) = > 2*log(x). > -- at all. === Subject: Re: 2sinA versus sin2A >I am getting confused. I assume 2 sinA is not the same as sin2A >Right? >2 sin 30.bc = 2 x 0.5 = 1 >sin 2A where A = 30 is sin 60.bc >or not? >>Right. In fact sin(2A) = 2sin(A)cos(A). >>Similarly, sin(x + y) is not the same as sin(x) + sin(y). The >>danger of function notation is that it might suggest a distributive >>law f(x + y) = f(x) + f(y), but this is not how functions behave, >>except when f(x) = mx, where m is a constant. Your question is >>similar. >>Todd Trimble > I wonder if you could go a bit further. > take cos2A > if I square it what do I get? cos4A? > how can I get rid of the 2 in an equation? > Can I divide both sides by 2 and end up with cosA? cos(2*A) = (cos(A))^2 - (sin(A))^2 or in more general terms: cos(A+B) = [cos(A) * cos(B)] - [sin(A) * sin(B)] -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: 2sinA versus sin2A > I am getting confused. I assume 2 sinA is not the same as sin2A > Right? No sin(2A) = 2*sin(A)*cos(A) > 2 sin 30.bc = 2 x 0.5 = 1 > sin 2A where A = 30 is sin 60.bc > or not? -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Help with inequalities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LJRWh20635; I need major help on inequalities! I am so lost! === Subject: Re: Help with inequalities I believe many of us would be willing to help if you would be kind enough to explain what exactly youre having trouble with so we have a reference point. Christopher === Subject: math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9LKiHC27472; I have 9 triangles in 1 triangle, my teacher says there are a total of 26 triangles in all. How do you get 26? === Subject: Re: math > I have 9 triangles in 1 triangle, my teacher says there are a total of > 26 triangles in all. How do you get 26? Consider triangles of different sizes. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Please help I have two similar problems i need help with: 1) Show the set {(x,y) lie in R^2 : y>x} is open under usual metric. and 2) Show the set {(x,y) lie in R^2 : xy =1, x>0} is closed under usual metric. I have been trying to figure out how to start solving these problems for a while. Im pretty sure I have to use balls to show whether they are open or closed, but I dont know how to do it. I am aware of the many theorems about open balls and closed balls, but the book and my teacher do not provide any examples on how to solve problems like these. I am really confused on what to do. If anybody could give me a push in the right direction, or point me to some good resources, it would be much appretiated. === Subject: Re: Please help > I have two similar problems i need help with: > 1) Show the set {(x,y) lie in R^2 : y>x} is open under usual metric. > and > 2) Show the set {(x,y) lie in R^2 : xy =1, x>0} is closed under usual > metric. > I have been trying to figure out how to start solving these problems > for a while. Im pretty sure I have to use balls to show whether they > are open or closed, but I dont know how to do it. I am aware of the > many theorems about open balls and closed balls, but the book and my > teacher do not provide any examples on how to solve problems like > these. I am really confused on what to do. If anybody could give me a > push in the right direction, or point me to some good resources, it > would be much appretiated. 1) The boundary of A = {(x,y) in R^2 : y>x} is {(x,y) in R^2 : y=x} Show that every point in A is a positive distance from that boundary. 2) Show that its compliment in R^2 is open === Subject: Re: Please help > I have two similar problems i need help with: > 1) Show the set {(x,y) lie in R^2 : y>x} is open under usual metric. > and > 2) Show the set {(x,y) lie in R^2 : xy =1, x>0} is closed under usual > metric. > I have been trying to figure out how to start solving these problems > for a while. Im pretty sure I have to use balls to show whether they > are open or closed, but I dont know how to do it. I am aware of the > many theorems about open balls and closed balls, but the book and my > teacher do not provide any examples on how to solve problems like > these. I am really confused on what to do. If anybody could give me a > push in the right direction, or point me to some good resources, it > would be much appretiated. Hint#1: Draw a picture. If that doesnt do it for you, ask again. -- Paul Sperry Columbia, SC (USA) === Subject: Cartesian equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC41A03382; Find the Cartesian equations for the following pairs of parametric equations: x = cos2A, y = sinA I am not sure on how to go about this one. Thought of squaring both sides of y y^2 = sin^2A sin^2A = 1 - cos^2A y^2 = 1 - cos^2A cos^2A = 1 - y^2 so... x = 1 - y^2 Im not too far as the answer seems to be x = 1 - 2y^2... Help... === Subject: Re: Cartesian equations >Find the Cartesian equations for the following pairs of parametric >equations: x = cos2A, y = sinA >I am not sure on how to go about this one. Thought of squaring both >sides of y >y^2 = sin^2A THANK YOU for showing us what you tried. This makes it enormously easier to give you useful feedback. I think your notation makes a mistake more likely, by confusing sine of 2A with square of sine of A. Try y^2 = (sin A)^2 >sin^2A = 1 - cos^2A >y^2 = 1 - cos^2A >cos^2A = 1 - y^2 (cos A)^2 = 1 - y^2 >x = 1 - y^2 Nope -- this is what I feared. x is not (cos A)^2, its cos(2A). Lets drop back. If x = cos(2A) then x = 1 - 2(sin A)^2 -- look at the double-angle formulas in your textbook. But you know (sin A) = y, and therefore (sin A)^2 = y^2. That gives the answer in your book, x = 1 - 2y^2. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Cartesian equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MEWah17025; >Find the Cartesian equations for the following pairs of parametric >equations: >x = cos2A, y = sinA >I am not sure on how to go about this one. Thought of squaring both >sides of y >y^2 = sin^2A >sin^2A = 1 - cos^2A >y^2 = 1 - cos^2A >cos^2A = 1 - y^2 >so... >x = 1 - y^2 >Im not too far as the answer seems to be x = 1 - 2y^2... >Help... Jo, I have read a couple of your postings. Would you allow me a couple of constructive suggestions? First, learn to use parenthesis to explicitly deliminate your expressions - at least until you learn the rules of operator precedence in math expressions. Second, get a book on trigonometry and crack it open. Third, if you are enrolled in a trig class, stop by your teachers office and ask for help. Bringing him/her an apple, though old fashion and a bit corny, will go a long way in getting lots of generous help. - MO === Subject: Re: Cartesian equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn1V11892; >Find the Cartesian equations for the following pairs of parametric >equations: >x = cos2A, y = sinA >I am not sure on how to go about this one. Thought of squaring both >sides of y >y^2 = sin^2A >sin^2A = 1 - cos^2A >y^2 = 1 - cos^2A >cos^2A = 1 - y^2 >so... >x = 1 - y^2 >Im not too far as the answer seems to be x = 1 - 2y^2... >Help... See my earlier message in the thread sin(2A) = 2sin(A), on doubling formulas. There I mentioned the identity cos(2A) = 1 - 2*sin^{2}(A) which should help you here. I suspect you are being led astray by notation -- were you confusing cos^2A with cos2A ? Consider making liberal use of parentheses so that cos^{2}(A) is clearly contrasted with cos(2A). [Writing cos^{2}(A) to denote (cos(A))^2 is an unfortunate and sometimes confusing convention, but like all time-honored conventions it dies hard.] Todd Trimble === Subject: Re: Cartesian equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MDn1811898; >Find the Cartesian equations for the following pairs of parametric >equations: >x = cos2A, y = sinA >I am not sure on how to go about this one. Thought of squaring both >sides of y >y^2 = sin^2A >sin^2A = 1 - cos^2A >y^2 = 1 - cos^2A >cos^2A = 1 - y^2 It might help to remember that cos 2A= cos^2 A- sin^2 A= (1- sin^2 A)- sin^2 A= 1- 2 sin^2 A. Then cos 2A= x= 1- 2 sin^2 A= 1- 2y^2. x= 1- 2y^2. >so... >x = 1 - y^2 >Im not too far as the answer seems to be x = 1 - 2y^2... >Help... === Subject: radius calculation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MC42r03408; I am trying to calculate a best fit radius for the following problem. I have three points for which I have a length between the points, the angle of the points (3D), and the XYZ position of the points. (I am measuring a pipe along the ground - info recorded is direction, inclination and length, from this I can calculate the XYZ position using minimum curvature calulation). (length, inclination, azimuth, x, x, z.) What I am having problems is a best fit radius calulation over three points as the calulation over two points can give a distorted reading. Can anyone advise. === Subject: Re: radius calculation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MG2xI26350; >I am trying to calculate a best fit radius for the following problem. >I have three points for which I have a length between the points, the >angle of the points (3D), and the XYZ position of the points. (I am >measuring a pipe along the ground - info recorded is direction, >inclination and length, from this I can calculate the XYZ position >using minimum curvature calulation). >(length, inclination, azimuth, x, x, z.) >What I am having problems is a best fit radius calulation over three >points as the calulation over two points can give a distorted reading. >Can anyone advise. Hi Bruce, if X=(x1,x2,x3), Y=(y1,y2,y3), Z=(z1,z2,z3) are the three points in question, R=(r1,r2,r3) is the point in the plane spanned by XY and XZ that has the same distance to X, Y and Z and r is this distance (the radius you search for), you can determine r by solving the 4 equations (1) (x1-r1)**2 + (x2-r2)**2 + (x3-r3)**2 - r**2 = 0 (2) (y1-r1)**2 + (y2-r2)**2 + (y3-r3)**2 - r**2 = 0 (3) (z1-r1)**2 + (z2-r2)**2 + (z3-r3)**2 - r**2 = 0 (4) (r1-x1)*((y2-x2)*(z3-x3) - (y3-x3)*(z2-x2)) - (r2-x2)*((y1-x1)*(z3-x3) - (y3-x3)*(z1-x1)) + (r3-x3)*((y1-x1)*(z2-x2) - (y2-x2)*(z1-x1)) = 0 The first three reßect the fact that R has equal distance to X, Y and Z and the fourth equation says that R lies in the plane spanned by XY and XZ. Best wishes Torsten. === Subject: Re: radius calculation >I am trying to calculate a best fit radius for the following problem. > I have three points for which I have a length between the points, the > angle of the points (3D), and the XYZ position of the points. (I am > measuring a pipe along the ground - info recorded is direction, > inclination and length, from this I can calculate the XYZ position > using minimum curvature calulation). > (length, inclination, azimuth, x, x, z.) > What I am having problems is a best fit radius calulation over three > points as the calulation over two points can give a distorted reading. > Can anyone advise. A general equation for a circle is: (X-A)^2 + (Y-B)^2 = R^2 Plug in each point and solve the 3 resulting equations for A, B, and R -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Number Pattern Help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MFWTi22794; >2 3 7 16 32 ___ ___ ___ _____ >WHAT IS THE PATTERN? x_(n+1) = x_n + n^2 So 2 3 7 16 32 57 93 ... But I have a question: Why the interest in number sequences? Is it just a friendly challenge, a guessing game? Or is there something more to it? - MO === Subject: Lets get Harris published Jimmy, I know that youve been having trouble getting published. I thought that maybe I could help. I checked with some of my buddies in the publishing business and they looked over your material and said that they could get you published! Here are the journals that you should consider sending your material to: Asimovs Science Fiction Entertainment Weekly Humor Times Mad Magazine National Enquirer Pro Wrestling Illustrated Realms of Fantasy Soap Opera Weekly Digest (under the Humor-in-the-US section). Oppie === Subject: Relations & Functions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9MLWfA23662; What is the difference , if any, between a function and a relation. === Subject: Re: Relations & Functions >What is the difference , if any, between a function and a relation. A function is a kind of relation, specifically it is one where no two ys can come from a single x. Which part of the definition in your textbook is unclear? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Relations & Functions > What is the difference , if any, between a function and a relation. Notation: m means is a member of ss means is a subset of Given sets A and B, AxB = {(x,y): x m A and y m B} R is a relation in AxB iff R ss AxB A function, F, from A to B is a relation on AxB such that whenever (x,y) m F and (x,z) m F, then y = z. This is usually described as saying that F is single valued, in the sense that each x m A determines a single y m B such that (x,y) m F. One difference is that the reverse of a relation R in AxB , meaning the set {(y,x) : (x,y) in R}, is always a relation in BxA, but the reverse, in the same sense, of a function need not be a function. An example of a function on the reals, |R, to the reals, for which the reverse relation is not a function is the squaring function, which, as a includes (2,4) and (-2,4). But the reverse relation then has both (4,2) and (4,-2) as members so is not single valued as a function is required to be. === Subject: Re: Relations & Functions > What is the difference , if any, between a function and a relation. Every function is a relation, but not every relation is a function. A relation R (on set AxB) is a function only if for every x (in A) there is at most one y (in B) such that x R y (and then we normally write R(x) = y). For example, the relation R on integers defined by x R y <=> x is divisible by y is not a function because for every x > 1, there is more than y such that x is divisible by y. On the other hand, the relation R on integers given by x R y <=> the number of divisors of x is y is a function because for every x there is precisely one y such that y is the number of divisors of x. hth meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: derivative simplification HELP please? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9NJmt502304; ok I am having a problem simplifying the first derivative of the following poblem... F(x)= x^(2/3)*(3x+10) I get F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) ok I am having a problem simplifying the first derivative of the >following poblem... >F(x)= x^(2/3)*(3x+10) >I get F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) product rule. Nearly right. Actually its F(x) = x^(2/3)*(3) + (2/3)x^(-1/3)*(3x+10) >here is where I get messed up... >The book give me Ill bet its really (1/3)x^(-1/3)*[9x + 2(3x+10)] >I do not understand where the 9x comes from, nor do I understand where >the x^(2/3) goes? The x^(-1/3) is really 1/x^(1/3). Factor it out and you have F(x) = x^(-1/3) * [ x^(1/3)*x^(2/3)*(3) + (2/3)*(3x+10) ] i.e. if you factor out x^(-1/3) from the first term, you have divided by x^(-1/3) or multiplied by its reciprocal x^(1/3). Compare factoring out 1/2 from 3x+(9/2) -- it is (1/2) * (2*3x+9). Or alternatively, multiply out again and notice what happens to the two terms in brackets as you multiply each by x^(-1/3) Simplifying: F(x) = x^(-1/3) * [ x^(1/3)*x^(2/3)*(3) + (2/3)*(3x+10) ] x^(1/3) * x^(2/3) = x F(x) = x^(-1/3) * [ x*(3) + (2/3)*(3x+10) ] F(x) = x^(-1/3) * [ 3x + (2/3)*(3x+10) ] Notice that (2/3) in the second term: you need to factor out 1/3: F(x) = x^(-1/3) * [ 3x + (2/3)*(3x+10) ] F(x) = (1/3)x^(-1/3) * [ 3*3x + 2*(3x+10) ] F(x) = (1/3)x^(-1/3) * [ 9x + 2*(3x+10) ] Thats what your book had, you tell us, but it seems silly not to combine similar terms: F(x) = (1/3)x^(-1/3) * [ 9x + 2*(3x+10) ] F(x) = (1/3)x^(-1/3) * [ 9x + 6x+20 ] F(x) = (1/3)x^(-1/3) * [ 15x+20 ] Final answer: 15x+20 F(x) = -------- 3x^(1/3) >All I can come up with is (3x^(2/3)+ 2(3x+10))/(3x^(1/3)) >please help show me where I am doing this incorrectly... -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: derivative simplification HELP please? > ok I am having a problem simplifying the first derivative of the > following poblem... > F(x)= x^(2/3)*(3x+10) > I get F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) product rule. > here is where I get messed up... > The book give me > 1/3x^(-1/3)*[9x + 2(3x+10)] > I do not understand where the 9x comes from, nor do I understand where > the x^(2/3) goes? > All I can come up with is (3x^(2/3)+ 2(3x+10))/(3x^(1/3)) > please help show me where I am doing this incorrectly... Ok, you have two terms above: first - (1/3)(x^(-1/3))(2)(3x+10) if you factor out (1/3)x(-1/3) from that what is left? second - (3)x(2/3) =(9)(1/3)(x^(-1/3))(x) if you factor out (1/3)x(-1/3) what is left? tim === Subject: Re: derivative simplification HELP please? > ok I am having a problem simplifying the first derivative of the > following poblem... > F(x)= x^(2/3)*(3x+10) > I get F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) product rule. > here is where I get messed up... > The book give me > 1/3x^(-1/3)*[9x + 2(3x+10)] > I do not understand where the 9x comes from, nor do I understand where > the x^(2/3) goes? > All I can come up with is (3x^(2/3)+ 2(3x+10))/(3x^(1/3)) > please help show me where I am doing this incorrectly... They are equal, the book has factorised: F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) factorise out 1/3x^(-1/3) F(x)=1/3x^(-1/3) * F(x) * 1/3x^(1/3) =1/3x^(-1/3) * ( x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) ) *1/3x^(1/3) =1/3x^(-1/3) * ( x^(2/3)*x^(1/3)*(3)*3 + 2/3 *3*x^(-1/3)*x^(1/3)*(3x+10)) =1/3x^(-1/3) * (9x+ 2(3x+10)) Or you can avoid the product rule by multipling out first: F(x)=3x^(5/3)+10x^(2/3) F(x)=15/3 *x^(2/3) + 20/3 *x^(-1/3) which is also equal to the F(x) above Hope this helps === Subject: Re: derivative simplification HELP please? > They are equal, the book has factorised: > F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) > factorise out 1/3x^(-1/3) > F(x)=1/3x^(-1/3) * F(x) * 1/3x^(1/3) Sorry the last term in the above line should be 3x^(1/3) not 1/3 x^(1/3) === Subject: Re: derivative simplification HELP please? in alt.math.undergrad: > ok I am having a problem simplifying the first derivative of the > following poblem... > F(x)= x^(2/3)*(3x+10) > I get F(x) = x^(2/3)*(3) + 2/3x^(-1/3)*(3x+10) product rule. Lets make that 3x^(2/3) + (2/3) x^(-1/3) * (3x + 10) for greater readability. > here is where I get messed up... > The book give me > 1/3x^(-1/3)*[9x + 2(3x+10)] > I do not understand where the 9x comes from, nor do I understand where > the x^(2/3) goes? x^(2/3) = x * x^(-1/3), so 3x^(2/3) = 3x * x^(-1/3), and your expression for F(x) becomes 3x * x^(-1/3) + (2/3) x^(-1/3) * (3x + 10). Now you can factor out x^(-1/3) to get x^(-1/3) [3x + (2/3) * (3x + 10)]. Rewrite the quantity in square brackets as a single term, a fraction with denominator 3: x^(-1/3) * (9x + 2(3x + 10))/3. Now take the factor of 1/3 out front, and you have the books form. [...] Brian === Subject: trig equations HELP.. I am having problems with 2 questions. Solve each equation for solutions over the interval (0-360degrees). Give solutions to the nearest tenth as appropriate. 2tan^2xsinx - tan^2x=0 where x is theta and ^ means squared on this problem i do this so far (dont know if its right) 2tan^2xsinx=tan^2x 2tan^2x=tan^2x/sinx 2tan^2x=sinx/cox^2x 2sin^2x=sinx 2sin^2x - sinx = 0 The answer is: {0, 30, 150, 180} in degrees.. but I dont know how to finish solving the equation. (All the questions before this was factoring, I do not know how to factor this or I might be working with the wrong type of identity or something) PLEASE HELP ME.... Also the other one is: Determine all solutions of each equation in radians (for x) or degrees (for theta) to the nearest tenth as appropiate. 3sin^2x - sinx - 1 = 0 the answer is .9 + 2 n pi, 2.3 + 2npi, 3.6 + 2npi, 5.8 + 2npi Kenny Chunn jkc014@latech.edu KuLL@cox-internet.com Kennyz79@gmail.com === Subject: Re: trig equations > HELP.. > I am having problems with 2 questions. > Solve each equation for solutions over the interval (0-360degrees). Give > solutions to the nearest tenth as appropriate. > 2tan^2xsinx - tan^2x=0 where x is theta and ^ means squared In ASCII that newsgroups use, a less ambiguous notation would be 2*tan(x)^2*sin(x) - tan(x)^2 = 0 The LHS (left hand side) factors to tan(x)^2*(2*sin(x) - 1) so that tan(x)^2 = 0 or 2*sin(x) - 1 = 0 each of which is fairly easy to solve for x in degrees. === Subject: Re: trig equations >2tan^2xsinx - tan^2x What are we supposed to make of this? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: trig equations >HELP.. >I am having problems with 2 questions. >Solve each equation for solutions over the interval (0-360degrees). Give solutions to the nearest tenth as appropriate. >2tan^2xsinx - tan^2x=0 where x is theta and ^ means squared >on this problem i do this so far (dont know if its right) >2tan^2xsinx=tan^2x At this point you can simplify things by noting that one possibility is for tan(x) = 0; and if tan x is not equal to 0, then you can divide through by tan^2x to get 2sinx = 1 sin(x) = 1/2. So the two possibilities are either tan(x) = 0 (find the solutions in the interval), or sin(x) = 1/2 (find the solutions in the interval. But, going as you do: >2tan^2x=tan^2x/sinx Careful! You can only do this if sin(x) is not zero. So now you have to consider the possibility of sin(x) = 0 as well... >2tan^2x=sinx/cox^2x >2sin^2x=sinx >2sin^2x - sinx = 0 Here you can factor out sin(x) to get sin(x)(2sin(x) -1 ) = 0. But the product is zero if and only if one of the two factors is zero, so you have sin(x) = 0 or 2sin(x)-1 = 0. (Which, conveniently enough, includes the case you had excluded before). So solve each of them separately, and then the union of the solutions will be the solutions to your original equation. (By the way, you can get that more easily by first multiplying through by cos^2(x); that will give 2sin^3(x) - sin^2(x) = 0 sin^2(x)(2sin(x) - 1) = 0 so sin(x) = 0 or 2sin(x)-1 = 0... etc.) >The answer is: {0, 30, 150, 180} in degrees.. but I dont know how to finish solving the equation. > (All the questions before this was factoring, I do not know how to >factor this or I might be working with the wrong type of identity or >something) PLEASE HELP ME.... Also the other one is: Determine all >solutions of each equation in radians (for x) or degrees (for theta) >to the nearest tenth as appropiate. 3sin^2x - sinx - 1 = 0 the Set y = sin(x). Substituting into your equation, you get 3y^2 - y - 1 = 0. Solve this for y using some method you know (perhaps the quadratic equation?). That will give you the possible values for y. Say, for arguments sake, that you get y=1/2, y = 2. Then you would need to solve sin(x)=1/2 and sin(x)=2 for x to get the final answers (none of these two are actually solutions, they were just so you see how to get the answers). If you can find all the answers between 0 and 360 (degrees), then adding multiples of 2pi will give you all the solutions. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: trig equations Arturo, Kenny >>HELP.. >>I am having problems with 2 questions. >>Solve each equation for solutions over the interval (0-360degrees). Give solutions to the nearest tenth as appropriate. >>2tan^2xsinx - tan^2x=0 where x is theta and ^ means squared >>on this problem i do this so far (dont know if its right) >>2tan^2xsinx=tan^2x > At this point you can simplify things by noting that one possibility > is for tan(x) = 0; and if tan x is not equal to 0, then you can divide > through by tan^2x to get > 2sinx = 1 > sin(x) = 1/2. > So the two possibilities are either tan(x) = 0 (find the solutions in > the interval), or sin(x) = 1/2 (find the solutions in the interval. > But, going as you do: >>2tan^2x=tan^2x/sinx > Careful! You can only do this if sin(x) is not zero. So now you have > to consider the possibility of sin(x) = 0 as well... >>2tan^2x=sinx/cox^2x >>2sin^2x=sinx >>2sin^2x - sinx = 0 > Here you can factor out sin(x) to get > sin(x)(2sin(x) -1 ) = 0. > But the product is zero if and only if one of the two factors is zero, > so you have > sin(x) = 0 or 2sin(x)-1 = 0. > (Which, conveniently enough, includes the case you had excluded > before). > So solve each of them separately, and then the union of the solutions > will be the solutions to your original equation. > (By the way, you can get that more easily by first multiplying through > by cos^2(x); that will give > 2sin^3(x) - sin^2(x) = 0 > sin^2(x)(2sin(x) - 1) = 0 > so sin(x) = 0 or 2sin(x)-1 = 0... etc.) >>The answer is: {0, 30, 150, 180} in degrees.. but I dont know how to > finish solving the equation. >> (All the questions before this was factoring, I do not know how to >>factor this or I might be working with the wrong type of identity or >>something) PLEASE HELP ME.... > Also the other one is: Determine all >>solutions of each equation in radians (for x) or degrees (for theta) >>to the nearest tenth as appropiate. 3sin^2x - sinx - 1 = 0 the > Set y = sin(x). Substituting into your equation, you get > 3y^2 - y - 1 = 0. > Solve this for y using some method you know (perhaps the quadratic > equation?). That will give you the possible values for y. Say, for > arguments sake, that you get y=1/2, y = 2. Then you would need to > solve sin(x)=1/2 and sin(x)=2 for x to get the final answers (none of > these two are actually solutions, they were just so you see how to get > the answers). > If you can find all the answers between 0 and 360 (degrees), then > adding multiples of 2pi will give you all the solutions. === Subject: Re: Calculus problem help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9NNaH919382; >My teacher recently gave me this problem to do, i was wondering if >someone could help me. >The topic is the net change theorom, the problem is: >Given f(x)= f(x0) + int(f(s)ds) where a=x0 and b=x >where we assume that f is differentiable on the interval (x0, x). >A) supppose that f(x) is a constant function. Then, f(x) = f(x0), >for any choice of x0. using the fundamental theorem of Calculus, >evaluate the simple integral in the above equation and write down the >resulting formula for f(x). ------------------------------------------------------------- -------- x f(x)= f(x0) + S f(s)ds = x0 Substitute f(s) = f(x0): x = f(x)= f(x0) + S f(x0)ds = x0 Pull the constant f(x0) out of the integral: x = f(x0) + f(x0) * S ds = x0 Integrate the function h(s) = 1: x = f(x0) + f(x0) * [s] = x0 Substitute the integration limits into [s]: = f(x0) + f(x0)*(x - x0) = ------------------------------------------------------------- -------- >b) Now, suppose that f(x) is twice differentiable on (x0, x). Then, >no t only does the equation above hold, but we can also apply the net >change theorem to f(x) and obtain: >f(x) = f(x0) + int( f(z)dz) where a=x0 and b=x >Assume that f(x) is a constant function, so that f(x)=f(x0). >Substitute the experssion for f(x0) given in the above equation into >the equation from part A, and evaulate all integrals to obtain a >formula for f(x0). Your formula should include f(x0), f(x0), and >f(x0). ------------------------------------------------------------- -------- x f(x) = f(x0) + S f(s)ds = x0 s Substitute for f(s) = f(x0) + S f(z)dz: x0 x s = f(x0) + S [ f(x0) + S f(z)dz ]ds = x0 x0 Substitute f(z) = f(x0): x s = f(x0) + S [ f(x0) + S f(x0)dz ]ds = x0 x0 Pull the constant f(x0) out of the inner integral over z: x s = f(x0) + S [ f(x0) + f(x0) * S dz ]ds = x0 x0 Integrate the function h(z) = 1: x s = f(x0) + S [ f(x0) + f(x0) * [z] ]ds = x0 x0 Substitute the integration limits into [z]: x = f(x0) + S [ f(x0) + f(x0)*(s - x0) ]ds = x0 Split integral of the sum into the sum of integrals: x x = f(x0) + S f(x0)ds + S f(x0)*(s - x0)ds = x0 x0 Pull the constant f(x0) out of the 2nd integral: x x = f(x0) + f(x0) * S ds + f(x0) * S (s - x0)ds = x0 x0 Integrate the functions h(s) = 1 and g(s) = s - x0: x x = f(x0) + f(x0) * [s] + f(x0) * [1/2*(s - x0)^2] = x0 x0 Substitute the integration limits into [s] and [1/2*(s - x0)^2]: = f(x0) + f(x0)*(x - x0) + f(x0)/2*(x - x0)^2 === Subject: Simple Probablity Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9NNaHj19421; Im going nuts with this, because I cant remember how to figure this simple little probability! Can someone take mercy on me and help? Okay, heres the problem: Event A has a 1 in 3 chance of occuring Event B has a 1 in 3 chance of occuring The chance for *both* Event A and Event B to occur is 1/9 (if I remember correctly) *WHAT* is the chance for *one* of the two events (A or B) to occur? Somehow I dont think its 2/3... === Subject: Re: Simple Probablity Problem > Im going nuts with this, because I cant remember how to figure this > simple little probability! Can someone take mercy on me and help? > Okay, heres the problem: > Event A has a 1 in 3 chance of occuring > Event B has a 1 in 3 chance of occuring > The chance for *both* Event A and Event B to occur is 1/9 (if I > remember correctly) This is only the case provided that the events A and B are independent, but on that assumption, see diagram below. > *WHAT* is the chance for *one* of the two events (A or B) to occur? > Somehow I dont think its 2/3... | 1/3 | 2/3 | ----------------------- --- | | | | | | | | | 2/3 ~A | 2/9 | 4/9 | | | | | | | ----------------------- --- A | | | | 1/9 | 2/9 |1/3 | | | ----------------------- --- B ~B === Subject: Re: Simple Probablity Problem >Im going nuts with this, because I cant remember how to figure this >simple little probability! Can someone take mercy on me and help? >Okay, heres the problem: >Event A has a 1 in 3 chance of occuring >Event B has a 1 in 3 chance of occuring >The chance for *both* Event A and Event B to occur is 1/9 (if I >remember correctly) Maybe it is, maybe it isnt. It depends on whether the events are independent. For instance, if A is I get a ßat tire and B is My engine overheats, then the probability of both is indeed 1/9. But if A is I roll a die and get a 1 or 2 and B is I roll a die and get a 2 or 3 then P(A and B) is 1/6. >*WHAT* is the chance for *one* of the two events (A or B) to occur? >Somehow I dont think its 2/3... You need more information about the relationship of A and B. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Simple Probablity Problem > Event A has a 1 in 3 chance of occuring > Event B has a 1 in 3 chance of occuring Are these events statistically independent? Ill assume they are, else the question cant be answered given the data youve presented. > The chance for *both* Event A and Event B to occur is 1/9 (if I > remember correctly) Yep. > *WHAT* is the chance for *one* of the two events (A or B) to occur? > Somehow I dont think its 2/3... Theres a 1/9 chance of both happening. Theres a (2/3)^2 = 4/9 chance of neither happening. So theres a (9/9 - 4/9 - 1/9) = 4/9 chance of either happening. Or you can figure it as there being a (2/3)(1/3)=2/9 chance of only A and a (2/3)(1/3)=2/9 chance of only B, so a 2/9+2/9=4/9 chance of either happening. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Simple Probablity Problem > Event A has a 1 in 3 chance of occuring > Event B has a 1 in 3 chance of occuring > Are these events statistically independent? > Ill assume they are, else the question cant > be answered given the data youve presented. > The chance for *both* Event A and Event B to occur is 1/9 (if I > remember correctly) > Yep. > *WHAT* is the chance for *one* of the two events (A or B) to occur? > Somehow I dont think its 2/3... > Theres a 1/9 chance of both happening. Theres a (2/3)^2 = 4/9 > chance of neither happening. So theres a (9/9 - 4/9 - 1/9) = 4/9 > chance of either happening. Or you can figure it as there being > a (2/3)(1/3)=2/9 chance of only A and a (2/3)(1/3)=2/9 chance of > only B, so a 2/9+2/9=4/9 chance of either happening. But not both. Whats missing is a quantifier for one. For _exactly_ one, the answer is, as youve said, 4/9. For _at least_ one, P(A / B) = P(A) + P(B) - P(A / B) = 5/9. -- Paul Sperry Columbia, SC (USA) === Subject: Re: 2sinA versus sin2A by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9P1eRf12637; Well, I assume someone has already said this. But sin2x = 2sinxcosx. Therefore, sin2x could equal 2sinx, but does not have to. In fact, unless cosx = 1, sin2x does not equal 2sinx. === Subject: focus of xy=2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9P1eSq12715; Help needed. I am just unable to find out focus of the hyperbola xy=2. can someone please ... thanx === Subject: Re: focus of xy=2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9Q157J01782; Here is one way to find the focal points of the hyperbola with equation x*y=1. Note that this includes the branch where boh coordinates are negative. You can modify this computation easily for any other such case x*y=a. We will of course use the fact that since the hyperbola is symmetric about the line y=x the focal points are on this line, and since it is symmetric about the line y=-x the origin is the midpoint of the segment joining the focal points. So suppose P=(x,1/x) and F=(c,c) and -F=(-c.-c) and let D1=dist(P,F) and D2=dist(P,-F). Then you can easily compute: D1^2= (x+[1/x]-c)^2 -2 +c^2 and D2^2= (x+[1/x]+c)^2 -2 +c^2 and for the point (1,1) where x=1 we calculate that D2=sqrt(2)*(c-1) assuming c>1 while D1=sqrt(2)*(c+1) so that as expected, D2-D1=2*sqrt(2). Now the points F and -F are going to be the focal points of this hyperbola if and only if for ALL points (x,1/x) we have D2-D1=2*sqrt(2). It is easy to see that if -2+c^2=0, that is, c=sqrt(2), then D2-D1=2*c (a constant, and the one we needed). So we have found the foci. ---------------------------------------------- > Help needed. > I am just unable to find out focus of the hyperbola xy=2. > can someone please ... > thanx === Subject: Re: focus of xy=2 > I am just unable to find out focus of the hyperbola xy=2. I do not know myself; but here is an idea that you could try: Overlay a clear plastic sheet upon a sheet of graph paper. Plot the points to draw a smooth curve for xy=2. Now, rotate the plastic sheet anchored at the origin until the curve now is symmetric around either the y-axis, or the this rotated curve based on the standard equation for hyperbola? At this extent, trigonometry may give you what you want. Thats the best I can do for you. Its been too long. G C === Subject: Re: focus of xy=2 > Help needed. > I am just unable to find out focus of the hyperbola xy=2. > can someone please ... > thanx If you are more used to hyperbolas with axes parallel to the coordinate axes, a simple substution here will make things easier. make x = (x- y)/sqrt(2) and y = (x+ y)/sqrt(2) and simplify Find the (x,y) coordinates of the foci. Then substitute back to get the xy-coordinates. The substitution above coresponds to a 45 degree rotation which makes the new axes coincide with the axes of symmetry of the hyperbola. For more general rotations, things are not usually so pretty. === Subject: Math is obsolete today . . . . . . We have computers and calculators who can do the work for us! Kids shouldnt be tortured with Math anymore! === Subject: Re: Math is obsolete today . . . . . . > We have computers and calculators who can do the work for us! > Kids shouldnt be tortured with Math anymore! Yea! Cool! No Math. And while were at it lets just keep re-electing George Bush. === Subject: Re: Math is obsolete today . . . . . . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9QGpU015750; >We have computers and calculators who can do the work for us! >Kids shouldnt be tortured with Math anymore! Are you serious??? Yes some of the basic math can be solved by the computer but you still need to know it, as I am going into computer programming, Math is essential not just that you yourself do all the math but for understanding how to go about writing codes and equations. With out it there wouldnt be computers. And what happens if you dont hav a calculator with you an you dont know how to multiply 12*32 or even something as simple as 6*7? Yes we would be significantly stupid if we omitted math from the curiculum... === Subject: Re: Math is obsolete today . . . . . . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9Q158H01810; >Kids shouldnt be tortured with Math anymore! What would you rather the Kids be, in a manner of speaking, tortured with? Alexander Bogomolny http://www.cut-the-knot.org === Subject: Re: Math is obsolete today . . . . . . > We have computers and calculators who can do the work for us! > Kids shouldnt be tortured with Math anymore! Anyone with that worldview is a prime target for every con artist in the world. === Subject: Re: Math is obsolete today . . . . . . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9PIGls30776; >> We have computers and calculators who can do the work for us! >> Kids shouldnt be tortured with Math anymore! >Anyone with that worldview is a prime target for every con artist in the >world. ÔKids shouldnt be tortured with thinking any longer! Be sure round peebles are lucky!... A computer owner,Alain. === Subject: Re: Math is obsolete today . . . . . . alt.math.undergrad,can.uucp: [...] Please ignore the troll or delete the cross-posting: this is apparently part of an attempt to ßood can.uucp with replies to cross-posted trolling messages. Brian === Subject: Re: Math is obsolete today . . . . . . alt.math.undergrad: >We have computers and calculators who can do the work for us! >Kids shouldnt be tortured with Math anymore! Evidently English is obsolete too, if you can use who with the antecedent computers and calculators. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Seeking on Proof Heres a linear function Fn defined by Fn(x) = {SUM(from(i=0)to(n-1))(-1)^i(X0,X1,...XiXi+1,...,Xn)}+(-1)^n( XnX0,X1,...,Xn-1 ), where x=(X0,X1,...,Xn). How can I prove that Fn-1(Fn(x)) = 0 for n>=2 btw, I have proved the basis step, for n=2, but have no clue on how to expand the formula and get the value of Fn(x) and then apply it to Fn-1. ps. Functions of this type, usually denoted by dn and called differentials, play important role in Topology and (homological) Algebra. The property above is simply denoted by d^2=0 Heres a linear function Fn defined by Fn(x) = {SUM(from(i=0)to(n-1)(-1)^i(X0,X1,...XiXi+1,...,Xn)}+(-1)^n( XnX0,X1,...,Xn-1) , where x=(X0,X1,...,Xn). How can I prove that Fn-1(Fn(x)) = 0 for n>=2 btw, I have proved the basis step, for n=2, but have no clue on how to expand the formula and get the value of Fn(x) and then apply it to Fn-1. === Subject: to find the length of arc by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9PIGki30768; I can find the length of arc, (1)if the distance of line joining the two ends of the arc(1), and (2)the distance of the perpendicular line joining the mid-point of given line(line joining the two ends of the arc) and mid of arc.(2) note:(1),(2).are two required things === Subject: Re: to find the length of arc > I can find the length of arc, (1)if the distance of line joining the > two ends of the arc(1), and (2)the distance of the perpendicular line > joining the mid-point of given line(line joining the two ends of the > arc) and mid of arc.(2) > note:(1),(2).are two required things Given (1) and (2), if youre wanting to find the length of a _circular_ arc, see case 8 at , in which the arc length is s and your (1) and (2) are c and h, resp. David === Subject: Absurdly complicated division of polynomials Okay, in the midst of trying to find the points of inßection, I end up taking the 2nd derivative (and dividing twice along the way) of (x-1) / ((x^2)+3). The final equation set to 0, verified in the book (so I know its right), is: f(x) = [ ((x^2+3)^2)(-2x+2) - (-(x^2)+2x+3)(4x)((x^2) + 3) ] ((x^2) +3)^-4 Setting f to 0... I have been playing around with this for a while, and I have no idea how to get it to 0. I tried simplifying all of this, and got it down to: [2(x^5 -3x^4 -6x^3 -6x^2 -27x +9) / (x^2 +3)^4] = 0 The polynomial is just too damned huge. I dont know where to start. Even looking at the books answers (where x is approximately -2.0642, 0.3054, and 4.7588), I really dont see *how* they solved for it. === Subject: Re: Absurdly complicated division of polynomials >Okay, in the midst of trying to find the points of inßection, I end up >taking the 2nd derivative (and dividing twice along the way) of (x-1) / >((x^2)+3). >The final equation set to 0, verified in the book (so I know its right), >is: >f(x) = [ ((x^2+3)^2)(-2x+2) - (-(x^2)+2x+3)(4x)((x^2) + 3) ] ((x^2) +3)^-4 >Setting f to 0... I have been playing around with this for a while, and I >have no idea how to get it to 0. HINT THE FIRST: An expression a/b is equal to 0 if and only if b is not zero and a=0. Since your denominator is (x^2+3)^4, which is never zero, that means that you just have to st the numerator equal to 0. >I tried simplifying all of this, and got it >down to: >[2(x^5 -3x^4 -6x^3 -6x^2 -27x +9) / (x^2 +3)^4] = 0 Not a good simplifaction. Instead, factor out (x^2) + 3 from the numerator first, to get (x^2+3)*[ (x^2+3)(-2x+2) - 4x(-x^2 + 2x + 3)] = 0 which is (x^2+3)*[ -2x^3 -6x + 2x^2 + 6 + 4x^3 - 8x^2 - 12x] = 0 or (x^2+3)*[ 2x^3 -6 x^2 - 18x + 6 ] = 0 or 2*(x^2+3)[x^3 - 6x^2 + 9x + 3] = 0 Now, for this to be 0, you must have one of the factors equal to 0. x^2+3 is never 0, so it must be that x^3 - 6x^2 + 9x + 3 is 0. At this point, you can try the rational roots (no root is rational), and then use Cardanos Formulas to get the answer. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Ordinary Least Squares -- proofs Hey, Im taking an econometrics theory class. Our book is Introductory Econometrics by Wooldridge and Im not particularly happy with it. The problem I have with it is the author seems to spend more effort on algebraic aspects of proofs (which I can do just fine on my own) and hand wave away / expend little effort explaining key insights. So, could someone either point me towards a book that takes better care of the proofs (preferebly algebraic instead of matrix proofs, because that is what we use in our class) I would be extremely grateful. In particular, proofs for (1) OLS estimators under MLR (2) Variance of the estimators (3) variance of the disturbance term PS: I tried posting this in sci.stat.edu but I didnt get anything after a week. Is there a better place I could post this? Earl Alternatively, here is one of my problems with the text: (apologies for the latex but I tried to write this without and it just turned into a giant mess. Feel free to alternatively grab a pdf from https://mywebspace.wisc.edu/elhathaw/web/math/elhq02.pdf -- no login or pass needed) Latex: usepackage{amssymb} begin{document} MLR of the form: [ y_{i} = beta_{0} + beta_{1}x_{i1} + beta_{2}x_{i2} + cdots + beta_{k}x_{ik} + u_{i} ] then the estimators of $beta_{j}$ for $j in [1,k]$ are equal to: [ hat{beta}_{j} = frac{ sum_{i=1}^{n}{ hat{r}_{ij} y_{i} } }{ sum_{i=1}^{n}{ hat{r}_{ij}^{2} } } ] Where $hat{r}_{ij}$ are the OLS residuals for a regression of $x_{j}$ on $x_{l}, l neq j, l in [1,k]$ So, we have that $x_{ij} = hat{x}_{ij} + hat{r}_{ij}$ The author makes two claims: Since $hat{x}_{ij}$ is just a linear function of the explanatory variables $x_{l}, l neq j, l in [1,k]$, it follows that: [ sum_{i=1}^{n}{ hat{x}_{ij} hat{u}_{i} } = 0 hbox{ since } sum_{i=1}^{n}{ x_{ij} hat{u}_{i} } = 0 forall j in [1,k] hbox{ and } sum_{i=1}^{n}{ hat{u}_{i} } = 0 ] Now, I understand why the sum of residuals $sum_{i=1}^{n}{hat{u}_{i}} = 0$. But could anyone take a stab at explaining the rest of it? end{document} === Subject: Wot is the easiest way to find inverse of a 3x3 martix? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9Q157g01799; Hello pple, Can any1 tell me whats the easiest or shortest way to find the inverse of a 3x3 matrix?..I am a student of Control Engineering and although finding the inverse isnt that hard...its quite lenthy and for a problem we have to find the inverse of about 3 or 4 matrices..and that takes up the bulk of my time...any1 knows any a really short method of finding the inverse of a 3 x 3 matrix..pliz let me know James === Subject: Re: Wot is the easiest way to find inverse of a 3x3 martix? > Hello pple, > Can any1 tell me whats the easiest or shortest way to find > the inverse of a 3x3 matrix?..I am a student of Control Engineering > and although finding the inverse isnt that hard...its quite lenthy > and for a problem we have to find the inverse of about 3 or 4 > matrices..and that takes up the bulk of my time...any1 knows any a > really short method of finding the inverse of a 3 x 3 matrix..pliz let > me know > James The usual method by hand for 3 x 3s is to use cramers rule. Anything beyond 3 x 3 is usually tackled by Gaussian Elimination of the matrix augmented with the identity matrix. But its far easier to use a computer to do it if you dont have to show working. A good online program to do matrix algebra can be found here http://www.bluebit.gr/matrix-calculator/ === Subject: Re: Wot is the easiest way to find inverse of a 3x3 martix? > Hello pple, > Can any1 tell me whats the easiest or shortest way to find > the inverse of a 3x3 matrix?..I am a student of Control Engineering > and although finding the inverse isnt that hard...its quite lenthy > and for a problem we have to find the inverse of about 3 or 4 > matrices..and that takes up the bulk of my time...any1 knows any a > really short method of finding the inverse of a 3 x 3 matrix..pliz let > me know Easiest way: use software or a graphing calculator. If you have to do it by hand, most of the methods will take a lot of time. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Wot is the easiest way to find inverse of a 3x3 martix? > Can any1 tell me whats the easiest or shortest way to find > the inverse of a 3x3 matrix?..I am a student of Control Engineering > and although finding the inverse isnt that hard...its quite lengthy > and for a problem we have to find the inverse of about 3 or 4 > matrices..and that takes up the bulk of my time...any1 knows any a > really short method of finding the inverse of a 3 x 3 matrix..pliz let > me know You can multiply any column by a number and add or subtract that result to any other column without changing the value of the determinate. Do this twice in a judicious way and you get a matrix with a top row of [a_11, 0, 0] which reduces the calculation to a_11 times the determinate of a 2 by 2 matrix. Two divisions, four multiplications, four substractions and then two multiplications, one substraction and one more multiplication. Five substractions, seven multiplications, two divisions in toto. Is that quicker than what youre doing? === Subject: Wot is the easiest way to find inverse of a 3x3 martix? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9QC9jf21179; Hello pple, Can any1 tell me whats the easiest or shortest way to find the inverse of a 3x3 matrix?..I am a student of Control Engineering and although finding the inverse isnt that hard...its quite lenthy and for a problem we have to find the inverse of about 3 or 4 matrices..and that takes up the bulk of my time...any1 knows any a really short method of finding the inverse of a 3 x 3 matrix..pliz let me know James === Subject: Re: Wot is the easiest way to find inverse of a 3x3 martix? >Can any1 tell me whats the easiest or shortest way to find >the inverse of a 3x3 matrix? Please dont post the same question multiple times. Youve already had a helpful response. If you dont understand it or need more details, say so; but posting your question as a fresh query just annoys the people who might help you. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Proof with Row Space by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9QGpW715866; Can anyone start me out here? If a matrix U is in row echelon form, show that the nonzero row vectors of U form a basis for the row space of U. Terri === Subject: Re: Proof with Row Space > Can anyone start me out here? > If a matrix U is in row echelon form, show that the nonzero row > vectors of U form a basis for the row space of U. As an example, suppose u = (1, 2, 3, 0, 6), v = (0, 0, 4, 5, 0) and w = (0, 0, 0, 1, 2) are your non-zero rows. By definition, they, or their transposes, span the row space. Suppose a, b and c are scalars and a*u + b*v + c*w = 0. What can say about a? Then what can you say about b? About c? Now, in general, suppose r_1, r_2, ..., r_t are the non-zero rows from top to bottom. If a_1, ..., a_t are scalars and a_1*r_1 + ... + a_t*r_t = 0. If there is a non-zero a_i, let a_r be the first one. Rememeber what echelon form means and look at the example. You may want to write r_i = (r_i1, r_i2, ..., r_in). -- Paul Sperry Columbia, SC (USA) === Subject: Help on a Combination Word Problem! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9QJ31H29286; The question is... In how many ways can 8 differently colored interlocking beads be put together to form a (closed) necklace? a. 5,155 b. 5,247 c. 4,975 d. 5,115 e. 5,040 I know the answer is E, and it involves using nCr, but I do not know === Subject: Re: Help on a Combination Word Problem! > The question is... > In how many ways can 8 differently colored interlocking beads be put > together to form a (closed) necklace? > a. 5,155 > b. 5,247 > c. 4,975 > d. 5,115 > e. 5,040 > I know the answer is E, and it involves using nCr, but I do not know Since order matters, its nPr. 8P8 = 40320 However, since there is a loop, the starting place doesnt actually matter. Dividing out the 8 repetitions gives you 5040. Note: since you can ßip a necklace over for many types of beads (round), you may have to divide by 2 as well, dropping the answer to 2520. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Help on a Combination Word Problem! >The question is... >In how many ways can 8 differently colored interlocking beads be put >together to form a (closed) necklace? >a. 5,155 >b. 5,247 >c. 4,975 >d. 5,115 >e. 5,040 >I know the answer is E, and it involves using nCr, but I do not know I dont think you do want nCr. Youre using all 8 of the beads, so 8C8 = 1 which is obviously wrong. Rather, you want something with nPr because youre interested in the arrangements. But theres a kicker: since the necklace is a circle, the two linear arrangements ABCDEFGH and BCDEFGHA are really the same arrangement, and in general lots of linear arrangements are the same arrangement when the ends are joined. In other words, you want to use the formula for _circular_ permutations. Theres a good explanation, with helpful pictures and a formula, at http://mathworld.wolfram.com/CircularPermutation.html . -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Help on a Combination Word Problem! alt.math.undergrad: > The question is... > In how many ways can 8 differently colored interlocking beads be put > together to form a (closed) necklace? > a. 5,155 > b. 5,247 > c. 4,975 > d. 5,115 > e. 5,040 > I know the answer is E, and it involves using nCr, but I do not know No, you dont need nCr at all. To make it easier to talk about, lets assume that one of the beads is red. You put the red bead down on the table in front of you and start snapping the others together, working clockwise around the circle. In how many ways can you do this? The first bead can be any of the remaining 7 colors. Once its been snapped into place, there are 6 choices left for the second bead. Once those two are in place, there are 5 choices left for the third bead. And so on. By the time you get to the seventh and last of the non-red beads, you have only one left, so theres only one way to make the Ôchoice. Thus, the total number of different ways is 7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! = 5040. Alternatively, you can think of it this way. You know that if you were arranging the 8 beads in a straight string, there would be 8! possible permutations, right? Now imagine taking one of those straight strings and snapping the ends together to make a circle; you get 8! circles. But some of them are really the same circle. For instance, if we use the integers 1 - 8 to represent the eight colors, the string 2 4 6 8 1 3 5 7 and the string 4 6 8 1 3 5 7 2 produce the same circle. A little thought should show you that each circle corresponds to 8 different strings, one for each of the 8 places where you could break the circle open. Thus, the actual number of distinct circles is not 8!, but 8!/8, or 7!. Brian === Subject: Algorithm writing Help please. Completely lost on how to write algorithms. Tks... Write an algorithm that outputs the minimal value of the difference (s[k] - s[k-3]) and the last (i.e. maximal) value of k realizing such minimal difference for arbitrary sequence of integers s[1],.....,s[n], where 3 where algebraically the zs are units, but z_1, z_2 and z_3 are not > units in the ring of algebraic integers. > But, if z_1, z_2 and z_3 are units algebraically but are not algebraic > integers, what does that mean? Indeed, what does it mean for a number to Ô[be] a unit algebraically ? Simple question of definition, what is it? > It means they are not algebraic integers. Nothing more. > Consider properties that cover rings like the ring of algebraic > integers, and the ring of integers itself, and it can be shown that > two are required: > 1. No rational in the ring except 1 and -1 is a unit. > 2. No non-unit number within the ring is a factor of any two integers > that are coprime to each other in the ring of integers. > So there are two basic factor properties of rings like the ring of > integers and the ring of algebraic integers. > Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where > properties 1. and 2. hold, which includes the ring of algebraic > integers, and I have named that ring, the ring of objects. You have not uniquely identified a ring. -- Larry Lard Replies to group please === Subject: Understanding Galois Theory Some of you may face classes where you will be taught certain things under the category of Galois Theory but you may have to do something hard for any student: challenge or ignore as false. Ive posted a rather basic argument that outlines a simple way to see how the algebra actually behaves, which happens to contradict with what you will probably be taught by your professors. They are following what they were taught, and the error in thinking actually goes back to the late 1800s, which you may believe is impossible. Some of you will check. You may be mathematicians. Those of you who simply reject what youre taught about Galois Theory as to how it relates to factors of roots on your own understanding may potentially be the greatest of all. Some of you, however, will accept what you are taught, despite it being wrong, and you will not check the basic argument Ive posted that proves youre being taught wrong. Or worst of all, you will check it, not understand it or disbelieve it, and accept error. If you are a mathematician or potentially are a mathematician, then you need to be able to spot error, or faced with a subtle error that has lasted for over a hundred years, be able to check when you are *told* it exists. If you are not so capable then I think you need to find another field, but you can wait until the truth is out to understand fully. Oh yeah, Galois Theory actually covers how you can *present* roots of polynomials using simple radicals, and the elementary operators +, -, and /. Nothing more. That is all. James Harris === Subject: Re: Understanding Galois Theory > Some of you may face classes where you will be taught certain things > under the category of Galois Theory but you may have to do something > hard for any student: challenge or ignore as false. Some of you may read Usenet and encounter posts from bizarre crank-troll-kook hybrids. You know what to do. Back to sci.math with you James! -- Larry Lard Replies to group please === Subject: Re: Understanding Galois Theory > Some of you may face classes where you will be taught certain things > under the category of Galois Theory but you may have to do something > hard for any student: challenge or ignore as false. > Ive posted a rather basic argument that outlines a simple way to see > how the algebra actually behaves, which happens to contradict with > what you will probably be taught by your professors. What you have posted is so far removed from what is discussed in Galois Theory that it isnt funny. If you believe Im wrong, please feel free to provide connections between your work and Galois Theory. > Those of you who simply reject what youre taught about Galois Theory > as to how it relates to factors of roots on your own understanding may > potentially be the greatest of all. And what *does* Galois Theory say about roots? > Oh yeah, Galois Theory actually covers how you can *present* roots of > polynomials using simple radicals, and the elementary operators +, -, > and /. > Nothing more. That is all. Actually, Galois Theory covers something very different. One of its applications is in whether the roots of a polynomial can be represented using radicals. Another is whether certain geometric constructions can be done with compass and straight edge. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Understanding Galois Theory days. My association with the Department is that of an alumnus. >Actually, Galois Theory covers something very different. One of its >applications is in whether the roots of a polynomial can be represented >using radicals. Another is whether certain geometric constructions can >be done with compass and straight edge. Is constructibility really a Galois Theory application? After all, Gauss established exactly which regular n-gons are constructible (those for which the odd prime factors of n are all distinct Fermat primes). Constructibility has to do with field extensions (a number is constructible if and only if it is in an extension of Q which can be obtained by successive extensions of degree 2)... I guess Galois Theory might come in through the Fundamental Theorem to tell you something about whether a is constructible by looking at Q(a)... -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Understanding Galois Theory >>Actually, Galois Theory covers something very different. One of its >>applications is in whether the roots of a polynomial can be represented >>using radicals. Another is whether certain geometric constructions can >>be done with compass and straight edge. > Is constructibility really a Galois Theory application? > After all, Gauss established exactly which regular n-gons are > constructible (those for which the odd prime factors of n are all > distinct Fermat primes). > Constructibility has to do with field extensions (a number is > constructible if and only if it is in an extension of Q which can be > obtained by successive extensions of degree 2)... I guess Galois > Theory might come in through the Fundamental Theorem to tell you > something about whether a is constructible by looking at Q(a)... As I recall, it was used to show that you cant square a circle, and that you cant trisect an angle. Its been a few years, but I thought it was in there. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Understanding Galois Theory days. My association with the Department is that of an alumnus. >Actually, Galois Theory covers something very different. One of its >applications is in whether the roots of a polynomial can be represented >using radicals. Another is whether certain geometric constructions can >be done with compass and straight edge. >> Is constructibility really a Galois Theory application? >> After all, Gauss established exactly which regular n-gons are >> constructible (those for which the odd prime factors of n are all >> distinct Fermat primes). >> Constructibility has to do with field extensions (a number is >> constructible if and only if it is in an extension of Q which can be >> obtained by successive extensions of degree 2)... I guess Galois >> Theory might come in through the Fundamental Theorem to tell you >> something about whether a is constructible by looking at Q(a)... >As I recall, it was used to show that you cant square a circle, and >that you cant trisect an angle. Its been a few years, but I thought >it was in there. To show you cant square a circle, you show that pi is not constructible by showing it is transcendental. To show that you cant trisect the angle, you show that cos(20 degrees) is of degree 3 over Q, hence you cannot construct the cosine of 20 degrees, hence you cannot construct an angle of 20 degrees, hence you cannot trisect an angle of 60 degrees (which is constructible). Since the extension is of degree 3, you cannot find cos(20 degrees) in a tower of extensions of degree 2 over Q; but this follows simply from dimension arguments. I dont recall Galois Theory being there, but I guess it could be in some of the smaller results along the way... (Of course, duplicating the cube is easy to show impossible, since cuberoot(2) has degree 3 over Q, and the problem is equivalent to constructing cuberoot(2)). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Understanding Galois Theory >>Actually, Galois Theory covers something very different. One of its >>applications is in whether the roots of a polynomial can be represented >>using radicals. Another is whether certain geometric constructions can >>be done with compass and straight edge. >Is constructibility really a Galois Theory application? >After all, Gauss established exactly which regular n-gons are >constructible (those for which the odd prime factors of n are all >distinct Fermat primes). >Constructibility has to do with field extensions (a number is >constructible if and only if it is in an extension of Q which can be >obtained by successive extensions of degree 2)... I guess Galois >Theory might come in through the Fundamental Theorem to tell you >something about whether a is constructible by looking at Q(a)... >>As I recall, it was used to show that you cant square a circle, and >>that you cant trisect an angle. Its been a few years, but I thought >>it was in there. > To show you cant square a circle, you show that pi is not > constructible by showing it is transcendental. To show that you cant > trisect the angle, you show that cos(20 degrees) is of degree 3 over Q, hence > you cannot construct the cosine of 20 degrees, hence you cannot > construct an angle of 20 degrees, hence you cannot trisect an angle of > 60 degrees (which is constructible). Since the extension is of degree > 3, you cannot find cos(20 degrees) in a tower of extensions of degree > 2 over Q; but this follows simply from dimension arguments. I dont > recall Galois Theory being there, but I guess it could be in some of > the smaller results along the way... > (Of course, duplicating the cube is easy to show impossible, since > cuberoot(2) has degree 3 over Q, and the problem is equivalent to > constructing cuberoot(2)). I looked up some of my old notes: in Artins book, he talks about the impossible geometric constructions while introducing the concepts for Galois Theory. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Understanding Galois Theory days. My association with the Department is that of an alumnus. [.snip.] >I looked up some of my old notes: in Artins book, he talks about the >impossible geometric constructions while introducing the concepts for >Galois Theory. Yes, absolutely. And most books talk about it after the Fundamental Theorem, but I simply do not see the Fundamental Theorem being used in the proofs of impossibility, unless I am missing something. I dont even see automorphisms of fields being used. The concepts are there, because you need to talk about field extensions and subextensions and so on, which are the basic building blocks of Galois Theory. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Question on writing math Loop Greets All, I was wondering if this type of loop is possible in and if so whats the best way of going about it? Heres and example of what a user would type in Example: Variable List of Numbers user enters 10,20,30,.... Example of Calculation formula: 1) (((10+20) /25) /2) =0.6 (1st user number typed in + 2nd user number typed in and calculatiions) 2) (((0.6+30)/25)/2) =.612 (Answer from above + 3rd user number typed in and calculations) Does this type of calculation have a name for it so I can look it up? TIA === Subject: Re: Question on writing math Loop if there are some known C0, C1, C2, .........., Cn ( in ur example the user enter them); Note : C0 = 10, C1 = 20,... ( in ur example) we define, A0 = C0 A1 = (A0 + C1 ) / 50; A2 = (A1 + C2) / 50; A3 = (A2 + C3) / 50; and so on ...... you can verify that for some 0 < K <= n Ak = ( A0 / (50 ^ K) ) + Sum ( i=1 to K ) [ Ci / ( 50 ^ ( k + 1 - i ) ) ] thus you have a sequence where A0 = C0 and Ak = ( A0 / (50 ^ K) ) + Sum ( i=1 to K ) [ Ci / ( 50 ^ ( k + 1 - i ) ) ] for k = 1,2, ..... , n and yes this type of loop is possible, it might be helpful for you to think of C as an array of length n + 1 where Ci is in position i. > Greets All, > I was wondering if this type of loop is possible in and if so whats the > best way of going about it? > Heres and example of what a user would type in > Example: Variable List of Numbers user enters > 10,20,30,.... > Example of Calculation formula: > 1) (((10+20) /25) /2) =0.6 (1st user number typed in + 2nd user number > typed in and calculatiions) > 2) (((0.6+30)/25)/2) =.612 (Answer from above + 3rd user number typed in > and calculations) > Does this type of calculation have a name for it so I can look it up? > TIA === Subject: surface area problem I have a problem completing the integration necessary to find the surface area for this problem. f(x,y) = 9 - x^2 find the surface area above the square region in the x-y plane bounded by y=3, x=3, y=0, x=0. Using the surface area process the partial derivatives is: part fx = -2x and the formula to integrate is (1 + 4x2)^(1/2) The integration boundaries are from 0 to 3 for both dy and dx. If we integrate with respect to dy first, we wind up with a factor of 3 times the integral with respect to dx. It looks simple but I cant seem to get the integral from 0 to 3 of (1 + 4x^2)^(1/2) dx to work out. Ive done the integration by parts with u = (1 + 4x^2)^(1/2) and du = dx but it keeps going in circles. There must be a substitution method or something that Im not seeing. Any hints? tim (Oh, the answer book shows winding up with an answer like [2x(1+4x62)^(1/2)] + ln[2x + (1+4x^2)^(1/2)] or something like that. I dont have it handy at the moment. ) === Subject: Re: surface area problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgTT25964; ... >It looks simple but I cant seem to get the >integral from 0 to 3 of (1 + 4x^2)^(1/2) dx to work out. Ive done >the integration by parts with u = (1 + 4x^2)^(1/2) and du = dx >but it keeps going in circles. >There must be a substitution method or something that Im not >seeing. Any hints? The 1st substitution that will work is x = tan(f)/2 dx = df/(2*cos^2(f)) sqrt(1 + 4x^2) = sqrt(1 + tan^2(f)) = 1/cos(f) 3 3 3 S S sqrt(1 + 4x^2)dxdy = 3 * S sqrt(1 + 4x^2)dx = 0 0 0 b b = 3/2 * S 1/cos^3(f) df = 3/2 * S cos(f)df/cos^4(f) df a a and now z = sin(f), dz = cos(f) df d = 3/2 * S 1/(1 - z^2)^2 dz = c d = 3/8 * S 1/(1-z) + 1/(1+z) + (1-z)^2 + 1/(1+z)^2]dz = etc. c The second substitution that will work is x = sinh(f)/2 dx = df*cosh(f)/2 sqrt(1 + 4x^2) = sqrt(1 + sinh^2(f)) = cosh(f) 3 3 3 S S sqrt(1 + 4x^2)dxdy = 3 * S sqrt(1 + 4x^2)dx = 0 0 0 b b = 3/2 * S cosh^2(f) df = 3/4 * S [1 + cos(2f)] df = etc. a a Comparing the results obtained with 2 different substitutions, you will notice that ln[x + sqrt(1 + x^2)] = sinh^(-1)(x) (inverse function of sinh(x)) === Subject: Re: surface area problem > ... >>It looks simple but I cant seem to get the >>integral from 0 to 3 of (1 + 4x^2)^(1/2) dx to work out. Ive done >>the integration by parts with u = (1 + 4x^2)^(1/2) and du = dx >>but it keeps going in circles. >>There must be a substitution method or something that Im not >>seeing. Any hints? > The 1st substitution that will work is > x = tan(f)/2 > dx = df/(2*cos^2(f)) > sqrt(1 + 4x^2) = sqrt(1 + tan^2(f)) = 1/cos(f) > 3 3 3 > S S sqrt(1 + 4x^2)dxdy = 3 * S sqrt(1 + 4x^2)dx = > 0 0 0 > > b b > = 3/2 * S 1/cos^3(f) df = 3/2 * S cos(f)df/cos^4(f) df > a a > and now z = sin(f), dz = cos(f) df > d > = 3/2 * S 1/(1 - z^2)^2 dz = > c > d > = 3/8 * S 1/(1-z) + 1/(1+z) + (1-z)^2 + 1/(1+z)^2]dz = etc. > c > The second substitution that will work is > x = sinh(f)/2 > dx = df*cosh(f)/2 > sqrt(1 + 4x^2) = sqrt(1 + sinh^2(f)) = cosh(f) > 3 3 3 > S S sqrt(1 + 4x^2)dxdy = 3 * S sqrt(1 + 4x^2)dx = > 0 0 0 > > b b > = 3/2 * S cosh^2(f) df = 3/4 * S [1 + cos(2f)] df = etc. > a a > Comparing the results obtained with 2 different substitutions, you > will notice that > ln[x + sqrt(1 + x^2)] = sinh^(-1)(x) (inverse function of sinh(x)) Vladimir, it!! My old brain has to work through this stuff in pieces. Again, thank you. tim === Subject: Re: surface area problem > I have a problem completing the integration necessary to find the surface > area for this problem. > f(x,y) = 9 - x^2 > find the surface area above the square region in the x-y plane bounded by > y=3, x=3, y=0, x=0. > Using the surface area process the partial derivatives is: part fx = -2x > and the formula to integrate is (1 + 4x2)^(1/2) > The integration boundaries are from 0 to 3 for both dy and dx. > If we integrate with respect to dy first, we wind up with a factor of > 3 times the integral with respect to dx. > It looks simple but I cant seem to get the > integral from 0 to 3 of (1 + 4x^2)^(1/2) dx to work out. Ive done the > integration by parts with u = (1 + 4x^2)^(1/2) and du = dx > but it keeps going in circles. > There must be a substitution method or something that Im not seeing. Any > hints? Review your trig substitutions. === Subject: Re: surface area problem >> I have a problem completing the integration necessary to find the surface >> area for this problem. >> f(x,y) = 9 - x^2 >> find the surface area above the square region in the x-y plane bounded by >> y=3, x=3, y=0, x=0. >> Using the surface area process the partial derivatives is: part fx = -2x >> and the formula to integrate is (1 + 4x2)^(1/2) >> The integration boundaries are from 0 to 3 for both dy and dx. >> If we integrate with respect to dy first, we wind up with a factor of >> 3 times the integral with respect to dx. >> It looks simple but I cant seem to get the >> integral from 0 to 3 of (1 + 4x^2)^(1/2) dx to work out. Ive done the >> integration by parts with u = (1 + 4x^2)^(1/2) and du = dx >> but it keeps going in circles. >> There must be a substitution method or something that Im not seeing. Any >> hints? > Review your trig substitutions. Wade, Its been a while since Ive done trig substitutions. Ill work this through tim === Subject: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgUX26009; Can someone tell me how to solve the integral: S dx/[(x^2)*ln(x)] ??? === Subject: Re: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9q14709; >Can someone tell me how to solve the integral: >S dx/[(x^2)*ln(x)] ??? Substitution y = ln(x) dy = dx/x leads to S 1/[x^2*ln(x)] dx = S e^(-y)/y dy I do not believe this can be expressed by elementary functions. It is possible to express e^(-y) by a power series and then intergate term by term: inf S e^(-y)/y dy = Sum S (-1)^n*y^(n-1)/n! dy = n=0 inf = ln(|y|) + Sum (-y)^n/(n*n!) + C = n=1 inf = ln[|ln(x)|] + Sum (-1)^n*[-ln(x)]^n/(n*n!) + C n=1 Convergence is fast. For an accuracy better than 10^(-6), use about 10 terms for 0.1 < x < 10, 20 terms for 10^(-3) < x < 10^3, 30 terms for 10^(-5) < x < 10^5. === Subject: Re: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UMBOX28844; S 1/[x^2*ln(x)] dx = inf = ln[|ln(x)|] + Sum [-ln(x)]^n/(n*n!) + C n=1 inf lim { ln[|ln(x)|] + Sum [-ln(x)]^n/(n*n!) } = -g ~= -0.57721567 where g is Eulers constant. === Subject: Re: Need help with some integral! > Can someone tell me how to solve the integral: > S dx/[(x^2)*ln(x)] ??? x = e^u, then by parts. === Subject: Re: Need help with some integral! > Can someone tell me how to solve the integral: > S dx/[(x^2)*ln(x)] ??? > x = e^u, then by parts. Details? -- Paul Sperry Columbia, SC (USA) === Subject: Re: Need help with some integral! > Can someone tell me how to solve the integral: > S dx/[(x^2)*ln(x)] ??? x = e^u, then by parts. > Details? integral e^2u u e^u du = integral u.e^3u du = (1/9) integral v.e^v dv thence by parts === Subject: Re: Need help with some integral! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9S14696; >You all ignored the / sign. >The integrand is: 1/[(x^2)*ln(x)] and not just: (x^2)*ln(x). >I still need help... If you are looking for an elementary expression for the indefinite integral, the problem looks hopeless. There might be hope of computing certain definite integrals having the integrand above. Where did this problem come from? Todd Trimble === Subject: Re: Need help with some integral! Can someone tell me how to solve the integral: > S dx/[(x^2)*ln(x)] ??? x = e^u, then by parts. > Details? > integral e^2u u e^u du > = integral u.e^3u du > = (1/9) integral v.e^v dv > thence by parts I think one of us has misread the problem. Note the /. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Need help with some integral! Can someone tell me how to solve the integral: > S dx/[(x^2)*ln(x)] ??? x = e^u, then by parts. > I think one of us has misread the problem. Note the /. integral 1/(x^2 log x) dx = integral e^-2u / u * e^u du = integral 1/(u.e^u) du = integral (e^v / v) dv then by parts from your local atom smasher. ;-) === Subject: Re: Need help with some integral! My immediate guess is parts! >Can someone tell me how to solve the integral: >S dx/[(x^2)*ln(x)] ??? -- Casey === Subject: Re: Linear Algebra problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgW126192; >I am having so much trouble with this problem, if you can help with in >any way possible I will truly appreciate it. This problem deals with >Linear Algebra, most important matrices. >Here is the problem, > 1 7 2 1 8 2 >A = 2 9 -4 B = 3 10 4 > 5 11 6 5 -12 6 >Calculate > > (a) (AB)32 > (b) (BA)21 > (c) the transpose of B > (d) if (A -1)31 = (1/det(A))W , find >det (A) and W Matrix element ij of a matrix produt AB is the dot product of i-th row of A and the j-th column of B (they intersect at the matrix element ij). (AB)_32 = A_row3 . B_co2 = A_31*B_12 + A_32*B_22 + A_33*B_32 = = 5*8 + 11*10 + 6*(-12) = = 40 + 110 - 72 = 78 (BA)_21 = B_row2 . A_co21 = 43 Transposed matrix B^T of a given matrix B is the matrix B tipped over the main diagonal (B_11, B_22, ..., B_nn) 1 3 5 B^T = 8 10 -12 2 4 6 To calculate the determinant of a 3x3 matrix A, add 1. The product of the main diagonal elements with the + sign. 2. The product of the other diagonal elements with the - sign. 3. The products of 2 triangles around the main diagonal with the - sign. 4. The products of 2 triangles around the other diagonal with the + sign. 1. +(1*9*6) = +54 2. -(5*9*2) = -90 3. -(2*7*6) = -84 and -(1*11*(-4)) = +44 4. +(2*2*11) = +44 and +(5*7*(-4)) = -140 det(A) = 54 - 90 - 84 + 44 + 44 - 140 = -172 The last question is not clear - is (A -1)31 the 31 element of the matrix A - 1: 1 7 2 1 0 0 A - 1 = 2 9 -4 - 0 1 0 5 11 6 0 0 1 or the 31 element of the inverse matrix A^(-1) ? The second case is more likely. In the first case (A - 1)_31 = A_31 = 5 and (A - 1)_31 = W/det(A) 5 = W/(-172) W = -860 In the second case, the matrix element ij of the inverse matrix is calculated as the ratio of the minor of the element A_ji and det(A). The minor(A_ji) is the determinant of the matrix obtained by eliminating the j-th row and i-th column (pay attention to the reversed order of indices i, j) from the original matrix. The 31 element of the inverse matrix to the 3x3 matrix A is the ratio minor(A_13)/det(A). The minor(A_13) is obtained by eliminating the 1st row and the 3rd column from the matrix A and calculating the determinant of the resulting 2x2 matrix: minor(A_13) = det | 2 9| = 2*11 - 5*9 = 22 - 45 = -23 | 5 11| A^(-1)_31 = -23/det(A) = -23/(-172) = 23/172 = ~= 0.133721 A^(-1)_31 = W/det(A) 23/172 = W/(-172) W = -23 === Subject: Re: Functions... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgUU26026; >I despise functions. With a passion. Theyre a simple concept but >bafße me so. >Can anyone suggests answers and explain? >f(a) = a if a is odd, -a if a is even. for all integers a (i.e. in Z) >a. show that f o f = ID subscriptZ >b. prove the f is bijective and give its inverse. >If you help I will love you forever :) I take it that f is defined on integers only. Okay, since f is defined differently for odd and even integers, to show that f(f(x))= x, look at two cases: If a is odd, then f(a)= a so f(f(a))= f(a)= f(a). If a is even, then f(a)= -a so f(f(a))= f(-a). Of course, -a is also even so f(-a)= -(-a)= a. In either case f(f(a))= a. The first thing you should do in proving that f is bijective is look up the definition of bijective. A function is bijective if it is both injective and surjective. Hmmm, need to look those up too! A function is injective if f(x)= f(y) if and only if x= y. Suppose f(x)= f(y). If x is odd, in which case f(x)= x so f(y)= x. The only to possible values for y is y or -y so either y= x or -y= x. But since x is odd, in either case y must also be odd and f(y)= y: y= x. If x is even, then f(x)= -x and f(y)= -x. Again, we have either y= x or y= -x. Since x is even, so is y. that means f(y)= -y so -y= -x and again y= x. Yes, f is injective. A function is surjective, if, for every y in Z, there exist an x such that f(x)= y. Okay, given y in Z, either y is odd or y is even. If y is odd, let x= y. Then x is also odd so f(x)= x= y. If y is even, let x= -y. Then x is also even so f(x)= -x= -(-y)= y. In either case, there exist an x such that f(x)= y. f is surjective and therefore is bijective. As far as the inverse is concerned, weve already proved f(f(x))= x. What IS the definition of inverse? === Subject: Re: Functions... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgVA26096; >I despise functions. With a passion. Theyre a simple concept but >bafße me so. >Can anyone suggests answers and explain? >f(a) = a if a is odd, -a if a is even. for all integers a (i.e. in Z) >a. show that f o f = ID subscriptZ >b. prove the f is bijective and give its inverse. >If you help I will love you forever :) if a is odd, then f(f(a)) = f(a) = a if a is even, then f(f(a)) = f(-a) = -(-a) = a in both cases, f(f(a)) = a, i.e., f o f = ID. assume f(a) = f(b), then f(f(a)) = f(f(b)) and a = b. since f o f = ID, f is its own inverse, i.e., f^(-1) = f. === Subject: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) I am a student at a community college. My state has a good community college system that works well with the major university in our state and the surrounding states. However, they have recently adopted a new math curriculum than has me somewhat confused. Since many of our students have been out of high school for quite sometime, most must take remedial or developmental mathematics courses before going on to College Algebra. The typical student who tests into the lowest math class will take the following classes: Pre-Algebra, Basic Algebra, Interim Algebra, and finally College Algebra. The first three classes are remedial or developmental and do not count towards graduation nor to the students overall GPA. However, under the new system, two new classes are being offered. The first of the two is Contemporary Mathematics (I do not see a problem with this course). It can be taken in place of College Algebra. It is described as a class that uses the basics of College Algebra in real world applications. The major universities in the state have agreed to accept this class in place of College Algebra for degrees that only require one course in math. However, Contemporary Mathematics is a stopping point; if the student wishes to take a higher math class (Calculus, Trigonometry, ect.); Contemporary Mathematics cannot be used as a prerequisite. College Algebra must be taken. The other new course in entitled Applied Mathematics. The prerequisite for Applied Mathematics is only Basic Algebra. However, Applied Mathematics is not a sufficient prerequisite for College Algebra. A student can use Applied Mathematics only towards a two-year degree. Applied Mathematics, basically, is a course that is designed for students looking to earn a two-year degree while taking as little math as possible. This is were I see a problem. Applied Mathematics counts towards graduation and on a GPA. Its sister course, Interim Algebra, a much more demanding and rigorous prerequisite for College Algebra, does not count toward graduation or GPA. In addition, when transferred to a four-year institution, Applied Mathematics counts as a General Education elective. When Interim Algebra is transferred, it only counts as a remedial or developmental elective and, as I have noted before, does not count toward graduation or GPA. It is obviously unfair to give students college credits for taking a lesser course. The only reward given to the student that completes Interim Algebra is the opportunity to move on to College Algebra. I see two possible solutions to this problem; either Interim Algebra is counted as an elective (not likely, considering it is the equivalent of Algebra II in high school) or Applied Mathematics (no one seems to know what its high school equivalency is) is no longer counted as an elective when transferred to a four-year institution. I would like to know what others think of this situation. The student body seems to be rather content; the only ones that really care are those getting an As in Interim Algebra and would like to see that A counted toward their GPA. The math professors and instructors seem content with system, however most agree that is somewhat unfair. I believe it is the job of educators to insure that all students are given a fair and challenging education. Moreover, since this new system seems to be unfair and unchallenging, I cannot agree with it. Please let me know what you think. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > I am a student at a community college. My state has a good > community college system that works well with the major university in our > state and the surrounding states. However, they have recently adopted a new > math curriculum than has me somewhat confused. > Since many of our students have been out of high school for > quite sometime, most must take remedial or developmental mathematics courses > before going on to College Algebra. The typical student who tests into the > lowest math class will take the following classes: Pre-Algebra, Basic > Algebra, Interim Algebra, and finally College Algebra. The first three > classes are remedial or developmental and do not count towards graduation > nor to the students overall GPA. However, under the new system, two new > classes are being offered. > The first of the two is Contemporary Mathematics (I do not see a > problem with this course). It can be taken in place of College Algebra. It > is described as a class that uses the basics of College Algebra in real > world applications. The major universities in the state have agreed to > accept this class in place of College Algebra for degrees that only require > one course in math. However, Contemporary Mathematics is a stopping point; > if the student wishes to take a higher math class (Calculus, Trigonometry, > ect.); Contemporary Mathematics cannot be used as a prerequisite. College > Algebra must be taken. > The other new course in entitled Applied Mathematics. The > prerequisite for Applied Mathematics is only Basic Algebra. However, > Applied Mathematics is not a sufficient prerequisite for College Algebra. A > student can use Applied Mathematics only towards a two-year degree. Applied > Mathematics, basically, is a course that is designed for students looking to > earn a two-year degree while taking as little math as possible. > This is were I see a problem. > Applied Mathematics counts towards graduation and on a GPA. Its > sister course, Interim Algebra, a much more demanding and rigorous > prerequisite for College Algebra, does not count toward graduation or GPA. > In addition, when transferred to a four-year institution, Applied > Mathematics counts as a General Education elective. When Interim Algebra is > transferred, it only counts as a remedial or developmental elective and, as > I have noted before, does not count toward graduation or GPA. > It is obviously unfair to give students college credits for > taking a lesser course. The only reward given to the student that completes > Interim Algebra is the opportunity to move on to College Algebra. > I see two possible solutions to this problem; either Interim > Algebra is counted as an elective (not likely, considering it is the > equivalent of Algebra II in high school) or Applied Mathematics (no one > seems to know what its high school equivalency is) is no longer counted as > an elective when transferred to a four-year institution. > I would like to know what others think of this situation. The > student body seems to be rather content; the only ones that really care are > those getting an As in Interim Algebra and would like to see that A counted > toward their GPA. The math professors and instructors seem content with > system, however most agree that is somewhat unfair. I believe it is the job > of educators to insure that all students are given a fair and challenging > education. Moreover, since this new system seems to be unfair and > unchallenging, I cannot agree with it. > Please let me know what you think. If you think of college algebra along with its prerequisites as a unit, with high requirements and reward (access to higher math courses), and the other, terminal course as a unit with low requirements and reward, the paradox disappears. It would be surprising if all the different tracks and courses in an educational system could be fitted into a perfect system with no local incongruities. -- john === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > I am a student at a community college. My state has a good > community college system that works well with the major university in our > state and the surrounding states. However, they have recently adopted a new > math curriculum than has me somewhat confused. > Since many of our students have been out of high school for > quite sometime, most must take remedial or developmental mathematics courses > before going on to College Algebra. The typical student who tests into the > lowest math class will take the following classes: Pre-Algebra, Basic > Algebra, Interim Algebra, and finally College Algebra. The first three > classes are remedial or developmental and do not count towards graduation > nor to the students overall GPA. However, under the new system, two new > classes are being offered. > The first of the two is Contemporary Mathematics (I do not see a > problem with this course). It can be taken in place of College Algebra. It > is described as a class that uses the basics of College Algebra in real > world applications. The major universities in the state have agreed to > accept this class in place of College Algebra for degrees that only require > one course in math. However, Contemporary Mathematics is a stopping point; > if the student wishes to take a higher math class (Calculus, Trigonometry, > ect.); Contemporary Mathematics cannot be used as a prerequisite. College > Algebra must be taken. > The other new course in entitled Applied Mathematics. The > prerequisite for Applied Mathematics is only Basic Algebra. However, > Applied Mathematics is not a sufficient prerequisite for College Algebra. A > student can use Applied Mathematics only towards a two-year degree. Applied > Mathematics, basically, is a course that is designed for students looking to > earn a two-year degree while taking as little math as possible. > This is were I see a problem. > Applied Mathematics counts towards graduation and on a GPA. Its > sister course, Interim Algebra, a much more demanding and rigorous > prerequisite for College Algebra, does not count toward graduation or GPA. > In addition, when transferred to a four-year institution, Applied > Mathematics counts as a General Education elective. When Interim Algebra is > transferred, it only counts as a remedial or developmental elective and, as > I have noted before, does not count toward graduation or GPA. > It is obviously unfair to give students college credits for > taking a lesser course. The only reward given to the student that completes > Interim Algebra is the opportunity to move on to College Algebra. > I see two possible solutions to this problem; either Interim > Algebra is counted as an elective (not likely, considering it is the > equivalent of Algebra II in high school) or Applied Mathematics (no one > seems to know what its high school equivalency is) is no longer counted as > an elective when transferred to a four-year institution. The first point I have is there is a great deal of difference between what is hard and what is worth credit. When I was in school, students had to take either College Algebra or a basic Statistics course. The Statistics course was, in my oppinion, much easier. Then I tutored a lady who couldnt understand what it meant for two events to be independent. If the Applied Math course involves a lot of new concepts, students may have a difficult time simply because they are having to absorb completely foreign material. I teach courses in Intermediate Algebra and Finite Math. When asked which is harder, I respond, It depends on how you think. When asked what the difference is, I ask them, Whats the difference between a rearview mirror and a tailpipe? I dont know what the content of the non-algebra courses you mentioned is, but what is easier will vary from student to student. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) Ease will depend, as you suggest, on background. It is my experience that students [not all, of course] who find prior studies difficult, seem to shine in statistics, it being a brand-new non-dependent [or not so dependent] set of rules and ideas. on the other hand, some who have gone through the prior studies rather mechanically and reasonably successfully, find the new concepts rather difficult to grasp, getting confused with the wording. >The first point I have is there is a great deal of difference between >what is hard and what is worth credit. When I was in school, students >had to take either College Algebra or a basic Statistics course. The >Statistics course was, in my oppinion, much easier. Then I tutored a >lady who couldnt understand what it meant for two events to be >independent. If the Applied Math course involves a lot of new concepts, >students may have a difficult time simply because they are having to >absorb completely foreign material. I teach courses in Intermediate >Algebra and Finite Math. When asked which is harder, I respond, It >depends on how you think. When asked what the difference is, I ask >them, Whats the difference between a rearview mirror and a tailpipe? >I dont know what the content of the non-algebra courses you mentioned >is, but what is easier will vary from student to student. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > Since many of our students have been out of high school for > quite sometime, most must take remedial or developmental mathematics courses > before going on to College Algebra. The typical student who tests into the > lowest math class will take the following classes: Pre-Algebra, Basic > Algebra, Interim Algebra, and finally College Algebra. The first three > classes are remedial or developmental and do not count towards graduation > nor to the students overall GPA. However, under the new system, two new > classes are being offered. Do students at you school not take math courses such as Calculus, Linear Algebra, and Differential Equations. Because you seem to imply that students take no more than college algebra at your college. > I would like to know what others think of this situation. The > student body seems to be rather content; the only ones that really care are > those getting an As in Interim Algebra and would like to see that A counted > toward their GPA. The math professors and instructors seem content with > system, however most agree that is somewhat unfair. I believe it is the job > of educators to insure that all students are given a fair and challenging > education. Moreover, since this new system seems to be unfair and > unchallenging, I cannot agree with it. > Please let me know what you think. While I wouldnt call demanding or rigorous, hell I wouldnt call college algebra or any math below for that matter demanding or rigorous, I find it sad that students are getting credit and units for a such a basic and remedial class. I feel that you dont get close to demanding until you take courses above Pre-Cal or Trig. So in my opinion only math classes above Pre-Cal/Trig should count towards ones graduation. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > Do students at you school not take math courses such as Calculus, > Linear Algebra, and Differential Equations. Because you seem to imply > that students take no more than college algebra at your college. The college I attend is a community college (it only offers two-year programs). Yes Calculus, Linear Algebra, and Differential Equations are offered, however most students transfer to a four-year institution when it comes time to take those classes. >> Since many of our students have been out of high school for >> quite sometime, most must take remedial or developmental mathematics >> courses >> before going on to College Algebra. The typical student who tests into >> the >> lowest math class will take the following classes: Pre-Algebra, Basic >> Algebra, Interim Algebra, and finally College Algebra. The first three >> classes are remedial or developmental and do not count towards graduation >> nor to the students overall GPA. However, under the new system, two new >> classes are being offered. >> I would like to know what others think of this situation. >> The >> student body seems to be rather content; the only ones that really care >> are >> those getting an As in Interim Algebra and would like to see that A >> counted >> toward their GPA. The math professors and instructors seem content with >> system, however most agree that is somewhat unfair. I believe it is the >> job >> of educators to insure that all students are given a fair and challenging >> education. Moreover, since this new system seems to be unfair and >> unchallenging, I cannot agree with it. >> Please let me know what you think. > While I wouldnt call demanding or rigorous, hell I wouldnt call > college algebra or any math below for that matter demanding or > rigorous, I feel that you dont get close to > demanding until you take courses above Pre-Cal or Trig. So in my > opinion only math classes above Pre-Cal/Trig should count towards ones > graduation. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) [Ghastly horror story of demise of college education snipped] > I would like to know what others think of this situation. The > student body seems to be rather content; the only ones that really care are > those getting an As in Interim Algebra and would like to see that A counted > toward their GPA. The math professors and instructors seem content with > system, however most agree that is somewhat unfair. I believe it is the job > of educators to insure that all students are given a fair and challenging > education. Moreover, since this new system seems to be unfair and > unchallenging, I cannot agree with it. > Please let me know what you think. Its an outrage. This trend is why Americans are becomming the dumbest people on Earth. Dont teach mathematics, its too hard for spoiled brats. Instead give social promotion grades so the stups can make quick easy grades mastering stupidy and get employed in a professional job market that going overseas. In otherwords, its an educational theory design defect. But no problem, the victims of that design defect wont be smart enuf to know theyve been injured. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > Its an outrage. This trend is why Americans are becomming the dumbest > people on Earth. Dont teach mathematics, its too hard for spoiled > brats. I understand why you are outraged over this, however I do not particularly agree with it. I am sure that the people who take the Applied Mathematics course will go on to do great things in their chosen profession (remember the Applied Mathematics course only applies to a two-year degree). What I am outraged over is that credit is gave were it is undue and no credit seems to be given where it is due. When transferred to a four-year institution, Applied Mathematics counts as a general education elective. Interim Algebra transfers as a developmental elective, and counts as nothing. > [Ghastly horror story of demise of college education snipped] >> I would like to know what others think of this situation. >> The >> student body seems to be rather content; the only ones that really care >> are >> those getting an As in Interim Algebra and would like to see that A >> counted >> toward their GPA. The math professors and instructors seem content with >> system, however most agree that is somewhat unfair. I believe it is the >> job >> of educators to insure that all students are given a fair and challenging >> education. Moreover, since this new system seems to be unfair and >> unchallenging, I cannot agree with it. >> Please let me know what you think. Instead give social promotion grades so the stups can make quick > easy grades mastering stupidy and get employed in a professional job > market that going overseas. > In otherwords, its an educational theory design defect. But no problem, > the victims of that design defect wont be smart enuf to know theyve been > injured. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) <10oa50fftr812bc@corp.supernews.com Its an outrage. This trend is why Americans are becoming the dumbest > people on Earth. Dont teach mathematics, its too hard for spoiled > brats. > I understand why you are outraged over this, however I do not particularly > agree with it. I am sure that the people who take the Applied Mathematics > course will go on to do great things in their chosen profession (remember > the Applied Mathematics course only applies to a two-year degree). What I > am outraged over is that credit is gave were it is undue and no credit seems > to be given where it is due. When transferred to a four-year institution, > Applied Mathematics counts as a general education elective. Interim Algebra > transfers as a developmental elective, and counts as nothing. Ill concur that applied mathematics may be excusable. Contemporary mathematics however is inexcusable. Besides the watering down of education, the outrage is the points game. With the points game, knowledge is lost sight of, devalued to the paramount of grades which can be earned, whinned for or frauded with equal success; learning is replaced by getting credit and knowing how to teach becomes more important than having something to teach. > [Ghastly horror story of demise of college education snipped] >> I would like to know what others think of this situation. >> The >> student body seems to be rather content; the only ones that really care >> are >> those getting an As in Interim Algebra and would like to see that A >> counted >> toward their GPA. The math professors and instructors seem content with >> system, however most agree that is somewhat unfair. I believe it is the >> job >> of educators to insure that all students are given a fair and challenging >> education. Moreover, since this new system seems to be unfair and >> unchallenging, I cannot agree with it. >> Please let me know what you think. > Instead give social promotion grades so the stups can make quick > easy grades mastering stupidy and get employed in a professional job > market that going overseas. > In otherwords, its an educational theory design defect. But no problem, > the victims of that design defect wont be smart enuf to know theyve been > injured. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > When transferred to a four-year institution, >Applied Mathematics counts as a general education elective. Interim Algebra >transfers as a developmental elective, and counts as nothing. The applicable concept could be that a student should obtain credit appropriate for the degree which he earns. This is why the Applied Mathematics yields credit toward your A.A. degree but is not counted as and not designed for transfer credit to a university. G C === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > The applicable concept could be that a student should obtain credit > appropriate > for the degree which he earns. This is why the Applied Mathematics yields > credit toward your A.A. degree but is not counted as and not designed for > transfer credit to a university. Nevertheless, Applied Mathematics DOES transfer to a university, as a general education elective. It is counted towards graduation as an elective and counts on the students overall GPA. Interim Algebra, the more demanding course, DOES NOT transfer to a university. >> When transferred to a four-year institution, >>Applied Mathematics counts as a general education elective. Interim >>Algebra >>transfers as a developmental elective, and counts as nothing. > G C === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > Interim Algebra, the more demanding >course, DOES NOT transfer to a university. To NO extent? not even for general education credit? Some upper-end administrator will have to make some re-designations on what transfers as what, and why! Also, what is Interim Algebra? Is this ÔIntermediate Algebra? G C === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) > To NO extent? not even for general education credit? Yes, Interim Algebra is the same as Intermediate Algebra and has the same high school equivalency of Algebra II. It counts as developmental elective when transferred, in other words, it does not count towards graduation or to the students GPA, it only allows the student the opportunity (if the student passes Interim Algebra of course) to take College Algebra. Applied Mathematics transfers as general education elective, so it will count toward graduation and on a GPA; however Applied Mathematics is not a sufficient pre-requisite for College Algebra. Therefore, Applied Mathematics is a college-credited course that can be used to get an Associates of Arts degree and will transfer to a four-year institution as an elective. However, the four-year institution will require College Algebra, so the student must either take Interim Algebra, or test directly into College Algebra(Applied Mathematics does not provide sufficient instruction to test into College Algebra). > Some upper-end administrator will have to make some re-designations on > what transfers as what, > and why! I have talked to many of the math instructors at my school. Applied Mathematics is simply a course designed for students who want to get a two-degree in something that does not require much math. The reason the instructors give for Interim Algebra not counting toward graduation is that it is the high school equivalent of Algebra II. They will not say what the high school equivalent Applied Mathematics is. >> Interim Algebra, the more demanding >>course, DOES NOT transfer to a university. > Also, what is Interim Algebra? Is this ÔIntermediate Algebra? > G C === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) alt.math.undergrad: [...] > I have talked to many of the math instructors at my school. Applied > Mathematics is simply a course designed for students who want to get a > two-degree in something that does not require much math. The reason the > instructors give for Interim Algebra not counting toward graduation is that > it is the high school equivalent of Algebra II. They will not say what the > high school equivalent Applied Mathematics is. If its a genuine liberal arts math course and not simply an easy algebra-type course, then it doesnt really have a high school equivalent. Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) What would you say is the more difficult course in your curriculum? Liberal arts Math or Interim Algebra (Intermediate Algebra). There was a course offered at my high school called Applied Mathematics. I am not sure what it covered but I was told it was designed for students looking to get out of high school with a little math as possible. I am told the same thing about the Applied Mathematics course in college; its a course designed for people trying to get their associates degree with as little math as possible. > alt.math.undergrad: > [...] >> I have talked to many of the math instructors at my school. Applied >> Mathematics is simply a course designed for students who want to get a >> two-degree in something that does not require much math. The reason the >> instructors give for Interim Algebra not counting toward graduation is >> that >> it is the high school equivalent of Algebra II. They will not say what >> the >> high school equivalent Applied Mathematics is. > If its a genuine liberal arts math course and not simply an > easy algebra-type course, then it doesnt really have a high > school equivalent. > Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) alt.math.undergrad: > What would you say is the more difficult course in your > curriculum? Liberal arts Math or Interim Algebra > (Intermediate Algebra). It depends on the student. Students with the appropriate background who prefer learning fairly mechanical manipulations and recipes for solving specific types of problems will probably do better in the algebra. Students with some intellectual curiosity who are willing to think about unfamiliar ideas will probably do better in the lib. arts math, especially if -- as is the case with many of our students here -- its been 10, 15, even 30 years since they last encountered any algebra. [...] Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) >I believe it is the job of educators >to insure that all students are given a fair and challenging >education. Moreover, since this new system seems to be unfair and >unchallenging, I cannot agree with it. >Please let me know what you think. (1) The last thing most students seek is challenge; they are instead oriented towards survival or success (measured in terms of grades). In institutions where there is a formal SET process (Student Evaluation of Teaching), the teachers who want to provide a challenging education tend to suppress that urge in the interest of staying employed or getting promoted. (2) Although fair is a common term among students (usually in the context of But thats not FAIR!), its meaning actually highly subjective and personal. For example, almost everyone is far quicker to view an inequality as being unfair when they are disadvantaged by it rather than when they are advantaged by it. There are indeed some commonly-accepted canons of fairness, such as natural justice in the legal arena (due process, etc.) But in the educational arena, a large amount of unfairness is inevitable as long as a main function of the system is to promote social stratification -- the separation of winners from losers in a manner that the losers see as it being their own fault that they are losers (and thus not making a huge stink about it). [added sci.edu] -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) use of the word fair. I am not talking about a classroom full of elementary students shouting thats not fair. I am talking about an entire states college math curriculum. At this level, should not fairness be of the utmost importance? >>I believe it is the job of educators >>to insure that all students are given a fair and challenging >>education. Moreover, since this new system seems to be unfair and >>unchallenging, I cannot agree with it. >>Please let me know what you think. > (1) The last thing most students seek is challenge; they are instead > oriented towards survival or success (measured in terms of grades). > In > institutions where there is a formal SET process (Student Evaluation of > Teaching), the teachers who want to provide a challenging education tend > to suppress that urge in the interest of staying employed or getting > promoted. > (2) Although fair is a common term among students (usually in the > context > of But thats not FAIR!), its meaning actually highly subjective and > personal. For example, almost everyone is far quicker to view an > inequality > as being unfair when they are disadvantaged by it rather than when they > are advantaged by it. > There are indeed some commonly-accepted canons of fairness, such as > natural > justice in the legal arena (due process, etc.) But in the educational > arena, a large amount of unfairness is inevitable as long as a main > function of the system is to promote social stratification -- the > separation > of winners from losers in a manner that the losers see as it being > their > own fault that they are losers (and thus not making a huge stink about > it). > [added sci.edu] > -- > --------------------------- > | BBB b Barbara at LivingHistory stop co stop uk > | B B aa rrr b | > | BBB a a r bbb | Quidquid latine dictum sit, > | B B a a r b b | altum viditur. > | BBB aa a r bbb | > ----------------------------- === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) Youre welcome. >You seem to undermine, if not manipulate, my use of the word fair. Really? I was trying to take your statements at face value, and respond in good faith in light of my experiences of tertiary teaching. >I am not talking about a classroom full of >elementary students shouting thats not fair. Nor am I. My main experience is with community college and university students. >I am talking about an entire states college math curriculum. At this level, >should not fairness be of the utmost importance? Well, the *appearance* of adequate fairness is certainly important for the social stratification system to work effectively. But you dont seem to appreciate that fairness is highly subjective, and that the different players (students, teaching staff, administration, state regulators, voters) have widely differing views of whats fair, both as individuals and when taken together as a class. For just one sort of example (based on ample personal experience), most students view it as unfair when they dont get suffient partial credit for what they know when that knowledge is part of a test answer which is incorrect for some other reason. But as a marker of these anwers, I view it as unfair that they should expect me to excavate through the layers of their confused ramblings in order to extract and assay what might be a nugget of knowledge. The key point-of-view difference here is that the student knows what they thought they were trying to say, but the marker only knows what they actually said. And believe me, there are plenty of aspects of the system you are in that are unfair *to your benefit*, and in a totally fair system (if such a thing were possible) you could easily be *worse* off. >I believe it is the job of educators >to insure that all students are given a fair and challenging >education. Moreover, since this new system seems to be unfair and >unchallenging, I cannot agree with it. >Please let me know what you think. >> (1) The last thing most students seek is challenge; they are instead >> oriented towards survival or success (measured in terms of grades). >> In >> institutions where there is a formal SET process (Student Evaluation of >> Teaching), the teachers who want to provide a challenging education tend >> to suppress that urge in the interest of staying employed or getting >> promoted. >> (2) Although fair is a common term among students (usually in the >> context >> of But thats not FAIR!), its meaning actually highly subjective and >> personal. For example, almost everyone is far quicker to view an >> inequality >> as being unfair when they are disadvantaged by it rather than when they >> are advantaged by it. >> There are indeed some commonly-accepted canons of fairness, such as >> natural >> justice in the legal arena (due process, etc.) But in the educational >> arena, a large amount of unfairness is inevitable as long as a main >> function of the system is to promote social stratification -- the >> separation >> of winners from losers in a manner that the losers see as it being >> their >> own fault that they are losers (and thus not making a huge stink about >> it). >> [added sci.edu] -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) >For just one sort of example (based on ample personal experience), most >students view it as unfair when they dont get suffient partial credit for >what they know when that knowledge is part of a test answer which is >incorrect for some other reason. But as a marker of these anwers, I view it >as unfair that they should expect me to excavate through the layers of >their confused ramblings in order to extract and assay what might be a >nugget of knowledge. The key point-of-view difference here is that the >student knows what they thought they were trying to say, but the marker only >knows what they actually said. If its any consolation, Ive had similar experience at the secondary level. One quick example was a parent complaining bitterly because her offspring had the right answer, but got only 1/10 marks on that question. She could not comprehend that the student had done every single step wrongly, and just by chance came up with the right result. Of course, the rest was a mess as well, but that point was one for her consideration. There are more, lots more. Better are those where you wont give a make-up because you know full well that they couldnt be bothered as a group to study for the first one, which had been dead simple. Life just isnt fair any more. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu: >> For just one sort of example (based on ample personal >> experience), most students view it as unfair when >> they dont get suffient partial credit for what they >> know when that knowledge is part of a test answer which >> is incorrect for some other reason. But as a marker of >> these anwers, I view it as unfair that they should >> expect me to excavate through the layers of their >> confused ramblings in order to extract and assay what >> might be a nugget of knowledge. The key point-of-view >> difference here is that the student knows what they >> thought they were trying to say, but the marker only >> knows what they actually said. I consider it part of my job to try to work out what they know and to reward them appropriately. Of course my notion of whats appropriate may disagree with theirs, but if they clearly know something thats relevant to the problem and recognize that its relevant, theyll get some credit for it. I have on occasion given substantial partial credit for a solution that was wholly wrong -- the method used could not have yielded a correct answer -- because it showed an understanding of the concepts and embodied a fairly thoughtful analysis (albeit incomplete or slightly misguided). > If its any consolation, Ive had similar experience at > the secondary level. One quick example was a parent > complaining bitterly because her offspring had the right > answer, but got only 1/10 marks on that question. She > could not comprehend that the student had done every > single step wrongly, and just by chance came up with the > right result. That seems to me a very different case from what Barb appeared to be describing: its not a matter of finding whats right, but of observing that nothing is: in those circumstances I dont consider the final answer to be right, since I dont separate it from the argument or calculation by which it was obtained. I very rarely have to deal with parents, thank goodness, but my students learn very early that in general I care far less about the final answer (in questions of a generally computational nature) than I do about whether they got it by some legitimate method. [...] Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) >alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu: > For just one sort of example (based on ample personal > experience), most students view it as unfair when > they dont get suffient partial credit for what they > know when that knowledge is part of a test answer which > is incorrect for some other reason. But as a marker of > these anwers, I view it as unfair that they should > expect me to excavate through the layers of their > confused ramblings in order to extract and assay what > might be a nugget of knowledge. The key point-of-view > difference here is that the student knows what they > thought they were trying to say, but the marker only > knows what they actually said. >I consider it part of my job to try to work out what they >know and to reward them appropriately. I should have mentioned that I was referring to a class of 50 or more students. Of course there other environments in which it is very useful to help the students excavate through their incohate knowledge (the limiting case of this being 1-to-1). But with 50 or more students, trying to be an educational archaeologist is a combination of time-consuming, frustrating, thankless, maddening, and pointless. >Of course my notion >of whats appropriate may disagree with theirs, but if they >clearly know something thats relevant to the problem and >recognize that its relevant, theyll get some credit for >it. I have on occasion given substantial partial credit for >a solution that was wholly wrong -- the method used could >not have yielded a correct answer -- because it showed an >understanding of the concepts and embodied a fairly >thoughtful analysis (albeit incomplete or slightly >misguided). I do also find that theres considerable educational value in training students to EXPRESS what they know clearly. This is particularly difficult when students have been habitually rewarded over the years for wafßy answers. I dont expect to be able to hold back that tide, but I do what I reasonably can to foster clear thinking and expression, both for their education and my sanity. >> If its any consolation, Ive had similar experience at >> the secondary level. One quick example was a parent >> complaining bitterly because her offspring had the right >> answer, but got only 1/10 marks on that question. She >> could not comprehend that the student had done every >> single step wrongly, and just by chance came up with the >> right result. One advantage of the tertiary scene is the absence of that sort of thing. >That seems to me a very different case from what Barb >appeared to be describing: its not a matter of finding >whats right, but of observing that nothing is: Well, something MIGHT be if one were to dig deeply enough and give enough benefit-of-the doubt. But what would be the educational value of that? >in those >circumstances I dont consider the final answer to be right, >since I dont separate it from the argument or calculation >by which it was obtained. I very rarely have to deal with >parents, thank goodness, but my students learn very early >that in general I care far less about the final answer (in >questions of a generally computational nature) than I do >about whether they got it by some legitimate method. I am particularly pleased when a student gets an answer different from the one I had indended, but which is still a reasonable interpretation of the question. Not only do I assess students based on what they actually write (rather than on what they plausibly thought they were trying to write), but based on the questions I actually asked (rather than what I thought I had asked). This strikes me as both equitable and educational. >[...] >Brian -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) in alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu: >>alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu : >> For just one sort of example (based on ample personal >> experience), most students view it as unfair when >> they dont get suffient partial credit for what they >> know when that knowledge is part of a test answer which >> is incorrect for some other reason. But as a marker of >> these anwers, I view it as unfair that they should >> expect me to excavate through the layers of their >> confused ramblings in order to extract and assay what >> might be a nugget of knowledge. The key point-of-view >> difference here is that the student knows what they >> thought they were trying to say, but the marker only >> knows what they actually said. >>I consider it part of my job to try to work out what they >>know and to reward them appropriately. > I should have mentioned that I was referring to a class of 50 or more > students. Its bad here, but not usually quite that bad, though I do have one with 50+ this term. Ill still do it on exams, but that means two exams during the semester and one final exam, so its tolerable. And I simply wont collect and mark homework for a class that big. > Of course there other environments in which it is very useful to > help the students excavate through their incohate knowledge (the limiting > case of this being 1-to-1). But with 50 or more students, trying to be an > educational archaeologist is a combination of time-consuming, frustrating, > thankless, maddening, and pointless. Yes, yes, usually, yes, and (for me) no, because I cant live with myself if I dont. [...] > I do also find that theres considerable educational value > in training students to EXPRESS what they know clearly. Youll get no argument from me there! [...] Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) >I consider it part of my job to try to work out what they >know and to reward them appropriately. Of course my notion >of whats appropriate may disagree with theirs, but if they >clearly know something thats relevant to the problem and >recognize that its relevant, theyll get some credit for >it. I have on occasion given substantial partial credit for >a solution that was wholly wrong -- the method used could >not have yielded a correct answer -- because it showed an >understanding of the concepts and embodied a fairly >thoughtful analysis (albeit incomplete or slightly >misguided). I can see your viewpoint, but it must be a rare occasion. Ive done the same, but the solution had to reßect a real understanding of the principles being tested in that particular question. In fact, one I have in mind made an error at the start that made the problem much more difficult, yet he pursued it accurately from then on, showing a keen understanding. So, in essence, he answered a more difficult question and so got the credit; a loss of one mark for the error, then credit for a fine job. We are judges by the nature of the pofession, and sometimes we have to be Solomon. === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) >I believe it is the job of educators >to insure that all students are given a fair and challenging >education. Moreover, since this new system seems to be unfair and >unchallenging, I cannot agree with it. >Please let me know what you think. >> (1) The last thing most students seek is challenge; they are instead >> oriented towards survival or success (measured in terms of grades). >> In >> institutions where there is a formal SET process (Student Evaluation of >> Teaching), the teachers who want to provide a challenging education >> tend >> to suppress that urge in the interest of staying employed or getting >> promoted. >> (2) Although fair is a common term among students (usually in the >> context >> of But thats not FAIR!), its meaning actually highly subjective and >> personal. For example, almost everyone is far quicker to view an >> inequality >> as being unfair when they are disadvantaged by it rather than when they >> are advantaged by it. >> There are indeed some commonly-accepted canons of fairness, such as >> natural >> justice in the legal arena (due process, etc.) But in the educational >> arena, a large amount of unfairness is inevitable as long as a main >> function of the system is to promote social stratification -- the >> separation >> of winners from losers in a manner that the losers see as it being >> their >> own fault that they are losers (and thus not making a huge stink about >> it). >> [added sci.edu] >> -- >> --------------------------- >> | BBB b Barbara at LivingHistory stop co stop uk >> | B B aa rrr b | >> | BBB a a r bbb | Quidquid latine dictum sit, >> | B B a a r b b | altum viditur. >> | BBB aa a r bbb | >> ----------------------------- > use of the word fair. I am not talking about a classroom full of > elementary students shouting thats not fair. I am talking about an > entire states college math curriculum. At this level, should not fairness > be of the utmost importance? Life is not fair. Get used to it. -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) [sci.edu deleted; this was already verging on excessive cross-posting] in alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu: >>I believe it is the job of educators >>to insure that all students are given a fair and challenging >>education. Moreover, since this new system seems to be unfair and >>unchallenging, I cannot agree with it. >>Please let me know what you think. > (1) The last thing most students seek is challenge; > they are instead oriented towards survival or > success (measured in terms of grades). In > institutions where there is a formal SET process (Student > Evaluation of Teaching), the teachers who want to > provide a challenging education tend to suppress that > urge in the interest of staying employed or getting > promoted. Dont confuse the existence of a formal process with the existence of an inßuential process. In my experience it is not uncommon to find that research and at least a minimal ability to get along with ones colleagues count more than everything else put together, followed by service (e.g., committee work); only then does teaching get a look-in, and it actually has a significant effect only at the extremes. That is, extremely good evaluations help a little, and extremely bad ones hurt a bit more, but in the vast middle ground the evaluations dont make much difference. This isnt always the case, of course, but I think that its not unusual. (And its not actually impossible to be demanding and get decent evaluations, especially if you teach some higher-level courses, though it does usually takes a hell of a lot of work.) > (2) Although fair is a common term among students > (usually in the context of But thats not FAIR!), its > meaning actually highly subjective and personal. For > example, almost everyone is far quicker to view an > inequality as being unfair when they are disadvantaged > by it rather than when they are advantaged by it. True, but not particularly responsive to the question. And the question is a reasonable one that deserves to be taken seriously, not dismissed so cavalierly. Unfortunately, without knowing the actual content of the Applied Mathematics course, I cant really give an opinion. If its anything like the courses commonly offered under the name ÔLiberal Arts Math, then the situation is paralleled at many four-year schools: Liberal Arts Math frequently has only a Basic Algebra prerequisite (or rough equivalent) and does count for credit towards the degree, while the equivalent of the Interim Algebra course does not. The justification is typically that the Lib. Arts Math course either teaches some significant mathematical ideas, albeit ones that dont require much algebraic background, or gives the student a better picture of how mathematics fits into the modern world by looking at (the elementary aspects of) some modern applications that are light on algebraic prerequisites. Ive taught both kinds; if theyre done well, the justification has some genuine merit. (An example of a good book intended for the first type of Lib. Arts Math course is _The Heart of Mathematics_, by Edward Berger & Mike Starbird. Examples of decent books intended for the second type are _For All Practical Purposes_, by COMAP, and _Excursions in Modern Mathematics_, by Peter Tannenbaum.) If, on the other hand, that Applied Mathematics course is really just a watered down version of the Interim Algebra course, then the situation is fairly ridiculous (and rather deplorable). > There are indeed some commonly-accepted canons of > fairness, such as natural justice in the legal arena > (due process, etc.) But in the educational arena, a > large amount of unfairness is inevitable as long as a > main function of the system is to promote social > stratification -- the separation of winners from > losers in a manner that the losers see as it being > their own fault that they are losers (and thus not > making a huge stink about it). Fortunately, quite a few of us who teach have no patience with the notion that this is the main function of an educational system. Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) Unfortunately, > without knowing the actual content of the Applied > Mathematics course, I cant really give an opinion. If its > anything like the courses commonly offered under the name > ÔLiberal Arts Math, then the situation is paralleled at > many four-year schools: Liberal Arts Math frequently has > only a Basic Algebra prerequisite (or rough equivalent) and > does count for credit towards the degree, while the > equivalent of the Interim Algebra course does not. > The justification is typically that the Lib. Arts Math > course either teaches some significant mathematical ideas, > albeit ones that dont require much algebraic background, or > gives the student a better picture of how mathematics fits > into the modern world by looking at (the elementary aspects > of) some modern applications that are light on algebraic > prerequisites. Ive taught both kinds; if theyre done > well, the justification has some genuine merit. If, on the other hand, that Applied Mathematics course is > really just a watered down version of the Interim Algebra > course, then the situation is fairly ridiculous (and rather > deplorable). The Applied Mathematics course is most likely the ÔLiberal Arts Math course you are referring. And I do believe it teaches some significant mathematical ideas and gives the student a better picture of how mathematics fits into the modern world. However, I do not know if it is a watered down version of Interim Algebra. I have actually heard it referred to as a beefed up version of Basic Algebra. I understand why it deserves merit, however if that is the case, should not Interim Algebra deserve merit also? The argument around the state is; Interim Algebra is the equivalent of Algebra II in high school; so they do not want to give any type of college credit for Interim Algebra. What the states math professors seem to be quiet about is the high school equivalency of Applied Mathematics. > [sci.edu deleted; this was already verging on excessive > cross-posting] > in > alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu: >I believe it is the job of educators >to insure that all students are given a fair and challenging >education. Moreover, since this new system seems to be unfair and >unchallenging, I cannot agree with it. >Please let me know what you think. >> (1) The last thing most students seek is challenge; >> they are instead oriented towards survival or >> success (measured in terms of grades). In >> institutions where there is a formal SET process (Student >> Evaluation of Teaching), the teachers who want to >> provide a challenging education tend to suppress that >> urge in the interest of staying employed or getting >> promoted. > Dont confuse the existence of a formal process with the > existence of an inßuential process. In my experience it is > not uncommon to find that research and at least a minimal > ability to get along with ones colleagues count more than > everything else put together, followed by service (e.g., > committee work); only then does teaching get a look-in, and > it actually has a significant effect only at the extremes. > That is, extremely good evaluations help a little, and > extremely bad ones hurt a bit more, but in the vast middle > ground the evaluations dont make much difference. This > isnt always the case, of course, but I think that its not > unusual. > (And its not actually impossible to be demanding and get > decent evaluations, especially if you teach some > higher-level courses, though it does usually takes a hell of > a lot of work.) >> (2) Although fair is a common term among students >> (usually in the context of But thats not FAIR!), its >> meaning actually highly subjective and personal. For >> example, almost everyone is far quicker to view an >> inequality as being unfair when they are disadvantaged >> by it rather than when they are advantaged by it. > True, but not particularly responsive to the question. And > the question is a reasonable one that deserves to be taken > seriously, not dismissed so cavalierly. Unfortunately, > without knowing the actual content of the Applied > Mathematics course, I cant really give an opinion. If its > anything like the courses commonly offered under the name > ÔLiberal Arts Math, then the situation is paralleled at > many four-year schools: Liberal Arts Math frequently has > only a Basic Algebra prerequisite (or rough equivalent) and > does count for credit towards the degree, while the > equivalent of the Interim Algebra course does not. > The justification is typically that the Lib. Arts Math > course either teaches some significant mathematical ideas, > albeit ones that dont require much algebraic background, or > gives the student a better picture of how mathematics fits > into the modern world by looking at (the elementary aspects > of) some modern applications that are light on algebraic > prerequisites. Ive taught both kinds; if theyre done > well, the justification has some genuine merit. (An example > of a good book intended for the first type of Lib. Arts Math > course is _The Heart of Mathematics_, by Edward Berger & > Mike Starbird. Examples of decent books intended for the > second type are _For All Practical Purposes_, by COMAP, and > _Excursions in Modern Mathematics_, by Peter Tannenbaum.) > If, on the other hand, that Applied Mathematics course is > really just a watered down version of the Interim Algebra > course, then the situation is fairly ridiculous (and rather > deplorable). >> There are indeed some commonly-accepted canons of >> fairness, such as natural justice in the legal arena >> (due process, etc.) But in the educational arena, a >> large amount of unfairness is inevitable as long as a >> main function of the system is to promote social >> stratification -- the separation of winners from >> losers in a manner that the losers see as it being >> their own fault that they are losers (and thus not >> making a huge stink about it). > Fortunately, quite a few of us who teach have no patience > with the notion that this is the main function of an > educational system. > Brian === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) >[sci.edu deleted; this was already verging on excessive >cross-posting] >alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu: >I believe it is the job of educators >to insure that all students are given a fair and challenging >education. Moreover, since this new system seems to be unfair and >unchallenging, I cannot agree with it. >Please let me know what you think. >> (1) The last thing most students seek is challenge; >> they are instead oriented towards survival or >> success (measured in terms of grades). In >> institutions where there is a formal SET process (Student >> Evaluation of Teaching), the teachers who want to >> provide a challenging education tend to suppress that >> urge in the interest of staying employed or getting >> promoted. >Dont confuse the existence of a formal process with the >existence of an inßuential process. In my experience it is >not uncommon to find that research and at least a minimal >ability to get along with ones colleagues count more than >everything else put together, followed by service (e.g., >committee work); only then does teaching get a look-in, and >it actually has a significant effect only at the extremes. That is generally true at universities (such as Cleveland State University where you are), but is generally not true at community colleges (which is the OPs situation). Note that I used the term teachers, not professors or lecturers. >That is, extremely good evaluations help a little, and >extremely bad ones hurt a bit more, but in the vast middle >ground the evaluations dont make much difference. This >isnt always the case, of course, but I think that its not >unusual. >(And its not actually impossible to be demanding and get >decent evaluations, especially if you teach some >higher-level courses, though it does usually takes a hell of >a lot of work.) Indeed, and the high-contact-hours untenured often-part-time teachers at community colleges seldom have that luxury. >> (2) Although fair is a common term among students >> (usually in the context of But thats not FAIR!), its >> meaning actually highly subjective and personal. For >> example, almost everyone is far quicker to view an >> inequality as being unfair when they are disadvantaged >> by it rather than when they are advantaged by it. >True, but not particularly responsive to the question. unsnipped part, which admittedly is probably not the part that the OP is interested in getting responses to. >And the question is a reasonable one that deserves to be taken >seriously, not dismissed so cavalierly. Cavalierly? Politically Im more of a Roundhead myself . And when I want to be dismissive I usually make comments like [snip of confused rubbish] or somesuch to make that perfectly clear. But Im sure the OP does appreciate your response to the meat of their question. >Unfortunately, without knowing the actual content of the Applied >Mathematics course, I cant really give an opinion. If its >anything like the courses commonly offered under the name >Liberal Arts Math, then the situation is paralleled at >many four-year schools: Liberal Arts Math frequently has >only a Basic Algebra prerequisite (or rough equivalent) and >does count for credit towards the degree, while the >equivalent of the Interim Algebra course does not. >The justification is typically that the Lib. Arts Math >course either teaches some significant mathematical ideas, >albeit ones that dont require much algebraic background, Never having been involved with a Mathematics for poets class, Im curious regarding what would be examples of such ideas? ISTM that the *most* significant and foundational mathematical idea is that of deductive proof in general. Does the class cover that idea, and if so using what topic domains? >gives the student a better picture of how mathematics fits >into the modern world by looking at (the elementary aspects >of) some modern applications that are light on algebraic >prerequisites. Id be interested in what sorts of examples are used there too. >Ive taught both kinds; if theyre done >well, the justification has some genuine merit. (An example >of a good book intended for the first type of Lib. Arts Math >course is _The Heart of Mathematics_, by Edward Berger & >Mike Starbird. Examples of decent books intended for the >second type are _For All Practical Purposes_, by COMAP, and >_Excursions in Modern Mathematics_, by Peter Tannenbaum.) >If, on the other hand, that Applied Mathematics course is >really just a watered down version of the Interim Algebra >course, then the situation is fairly ridiculous (and rather >deplorable). >> There are indeed some commonly-accepted canons of >> fairness, such as natural justice in the legal arena >> (due process, etc.) But in the educational arena, a >> large amount of unfairness is inevitable as long as a >> main function of the system is to promote social >> stratification -- the separation of winners from >> losers in a manner that the losers see as it being >> their own fault that they are losers (and thus not >> making a huge stink about it). >Fortunately, quite a few of us who teach have no patience >with the notion that this is the main function of an >educational system. I carefully said *a* main, not *the* main. And indeed many (most?) instructors do not view their activities in that light. Nonetheless, the institution as a whole certainly works within that overall context. >Brian -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: A Question about Math Curriculum. (Math Instructors and Professors Please Respond) in alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad: [...] >>in >>alt.algebra.help,alt.math,alt.math.recreational, alt.math.undergrad,sci.edu : >>I believe it is the job of educators >>to insure that all students are given a fair and challenging >>education. Moreover, since this new system seems to be unfair and >>unchallenging, I cannot agree with it. >>Please let me know what you think. > (1) The last thing most students seek is challenge; > they are instead oriented towards survival or > success (measured in terms of grades). In > institutions where there is a formal SET process (Student > Evaluation of Teaching), the teachers who want to > provide a challenging education tend to suppress that > urge in the interest of staying employed or getting > promoted. >>Dont confuse the existence of a formal process with the >>existence of an inßuential process. In my experience it is >>not uncommon to find that research and at least a minimal >>ability to get along with ones colleagues count more than >>everything else put together, followed by service (e.g., >>committee work); only then does teaching get a look-in, and >>it actually has a significant effect only at the extremes. > That is generally true at universities (such as Cleveland > State University where you are), but is generally not > true at community colleges (which is the OPs > situation). Note that I used the term teachers, not > professors or lecturers. Ah. It wouldnt occur to me that that was significant, since I routinely list my occupation as Ôteacher and think of myself as such; I took it as a general statement. I agree that the situation at the community colleges is generally quite different. [...] >>The justification is typically that the Lib. Arts Math >>course either teaches some significant mathematical ideas, >>albeit ones that dont require much algebraic background, > Never having been involved with a Mathematics for poets > class, Im curious regarding what would be examples of > such ideas? ISTM that the *most* significant and > foundational mathematical idea is that of deductive proof > in general. Thats a view of how mathematics should be done, not a mathematical idea per se. Doing mathematics really has two parts: one is discovery, and the other is proof. In a course of this type the emphasis is necessarily on discovery, and formal proof has no place. Informal proof, also known as explaining why thus-and-such *has* to be true, does have a place; typically in such courses I do quite a bit of it and ask for a little from my students. > Does the class cover that idea, and if so > using what topic domains? I make it clear that both parts are important, but I emphasize the discovery side, especially in what I ask them to do. Some of the topic areas that Ive used are: * binary numbers and arithmetic, starting from the problem of finding an efficient set of weights for a single-pan balance; * balanced ternary arithmetic, starting from the problem of finding an efficient set of weights for a two-pan balance; * basic properties of (countably) infinite order types, concluding with basic ordinal addition and, with a good group, multiplication; * deterministic two-person games of perfect information (various heap take-away games, including Nim, and some others types); the idea of isomorphic games arises fairly naturally, and there are easy opportunities to drag in a little modular arithmetic. The hardest one is probably the order types. Ive done them in terms of an army of ants with number tags instead of name tags; their normal formation, of course, is 0 1 2 3 ..., marching to the left, but we soon have them performing all sorts of evolutions. The biggest hurdle is the idea that we dont need a physical realization of an abstract order; the problem arises the moment I suggest having Ant 0 fall out and fall back in at the rear of the formation. Theres also an army of fancy ants, with separate numbers on head and thorax; their basic formation is (0,0) (0,1) (0,2) ... (1,0) (1,1) (1,2) ... (2,0) (2,1) (2,2) ... . . . . . . . . . We eventually discover how to parachute the plain ants onto this formation so that each fancy ant gets exactly one rider, and even how to figure out which plain ant lands on ant (p, q). Ive even had a couple of students successfully answer all of the following question on a take-home final exam, and quite a few answer the first two parts, even though the formation and ideas involved have not been discussed at all: Suppose that I try to form up the fancy ants according to the following rule: Ant (p, q) is in front of Ant (m, n) if pn < qm. Thus, Ant (3,5) is in front of Ant (4, 6) because 3 * 6 < 5 * 4. (a) Unfortunately, this rule doesnt define a legitimate formation. Find two ants that according to this rule ought to be standing in the same place in the formation. (That is, neither of the ants is in front of the other according to the rule.) (b) To get around the difficulty mentioned in part (a), well allow ants that ought to be in the same position to stand on one anothers backs. Explain how to find infinitely many ants who will have to stand on the back of Ant (1, 2). (c) Ant (11, 25) is in front of Ant (4, 9); find an ant who is behind Ant (11,25) but in front of Ant (4, 9). >> or gives the student a better picture of how mathematics >> fits into the modern world by looking at (the elementary >> aspects of) some modern applications that are light on >> algebraic prerequisites. > Id be interested in what sorts of examples are used there too. Electoral methods, apportionment, fair division; Euler and Hamilton circuits and paths, minimal spanning trees, precedence graphs and scheduling algorithms; population growth (exponential model, discrete logistic model); square-cube law and growth; symmetry, especially frieze and wallpaper patterns. [...] > But in the educational arena, a large amount of > unfairness is inevitable as long as a main function > of the system is to promote social stratification -- > the separation of winners from losers in a manner > that the losers see as it being their own fault that > they are losers (and thus not making a huge stink about > it). >>Fortunately, quite a few of us who teach have no patience >>with the notion that this is the main function of an >>educational system. > I carefully said *a* main, not *the* main. Hm; I thought that I had used Ôa as well. Apparently my fingers got ahead of my brain. [...] Brian === Subject: Re: proving trig identities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9G14672; >I am having a lot of trouble with this problem: >1+tanx/1+cotx=tanx >I have changed the tanx to sinx/cosx, and cotx to cosx/sinx >then divided them by multiplying the reciprocal. I have tried many >different ways on how to solve this, but I just get stuck. Can you >help? The identity that you write is not really a trig identity. It is easily show that (1+x)/(1+1/x) = x, for all x /= 0 To wit: (1+x)/(1+1/x) = [x(1+x)]/[x(1+1/x)] = [x(1+x)]/[x + 1] = x The only trig in your version of this identity is tanx = 1/cotx - MO === Subject: Re: proving trig identities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgVd26055; >I am having a lot of trouble with this problem: >1+tanx/1+cotx=tanx >I have changed the tanx to sinx/cosx, and cotx to cosx/sinx >then divided them by multiplying the reciprocal. I have tried many >different ways on how to solve this, but I just get stuck. Can you >help? Try rewriting as 1 + tan(x) = (1 + cot(x))*tan(x). Distribute. What do you know about cot(x)*tan(x)? By the way, you need to use parentheses to avoid ambiguity. On the left-hand side it should be (1 + tan(x))/(1 + cot(x)). Todd Trimble === Subject: Re: proving trig identities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgUv26039; >I am having a lot of trouble with this problem: >1+tanx/1+cotx=tanx >I have changed the tanx to sinx/cosx, and cotx to cosx/sinx >then divided them by multiplying the reciprocal. I have tried many >different ways on how to solve this, but I just get stuck. Can you >help? tan(x) = 1/cot(x) [1 + tan(x)]/[1 + cot(x)] = [1 + tan(x)]/[1 + 1/tan(x)] = = [1 + tan(x)]/{[tan(x) + 1]/tan(x)} = 1/[1/tan(x)] = tan(x) === Subject: Re: proving trig identities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9SNgUn26022; >I am having a lot of trouble with this problem: >1+tanx/1+cotx=tanx >I have changed the tanx to sinx/cosx, and cotx to cosx/sinx >then divided them by multiplying the reciprocal. I have tried many >different ways on how to solve this, but I just get stuck. Can you >help? I assume you mean (1+ tan x)/(1+ cot x)= tan x. What you did sounds good: write tan x as sin x/cos x and cot x as cos x/sin x so you have (1+ sin x/cos x)/(1+ cos x/sinx) and multiply both numerator and denominator by (sin x)(cos x). doing that, you get (sin x cos x+ sin^2 x)/(sin x cos x+ cos^2 x) Factor sin x out of the numerator and cos x out of the denominator to get (sin x/cos x){(cos x+ sin x)/(sin x+ cos x)} Since the numerator and denominator of that second fraction are the same, it is equal to 1 and we get sin x/cos x which is, in fact, tan x. === Subject: Inverse Matrices 1.03 for Windows! Inverse Matrices 1.03 for Windows now available! The program provides detailed, step-by-step solution in a tutorial-like format to the following problem: Given a 2x2 matrix, or a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse matrix by using the Gauss-Jordan elimination method. The check of the solution is given. The program is designed for university students and professors. Price: $15. Product homepage: http://www.dekovsoft.com/inverse_matrices.htm http://www.dekovsoft.com/ddekov/ === Subject: Re: Inverse Matrices 1.03 for Windows! First: You are using a freeware version of ClickTeam Install Creator in your trial versions. Do your paid versions use a non-freeware version of ClickTeam Install Creator - since you are asking for money? Second: You do not allow for decimal entry. Third: You do not allow for complex numbers. Fourth: You simply provide a display of all the steps - like you would get in a book. That is there are no animations (Flash or Java) or sounds that would help make it stick in a students mind - i.e. a student who has challenges with learning from a book is still going to have the same challenges with your so called helpful programs! Fifth: You are charging $15 USD? for each little segment - are there not calculator programs (e.g. for the TI-89 or equivalent) that one can download free (with a one-time purchase of a calculator) that will show the steps of a matrix inversion, etc? Plus, at the end of it all, one still has a calculator to show for it. >Inverse Matrices 1.03 for Windows now available! >The program provides detailed, step-by-step solution in a >tutorial-like format to the following problem: Given a 2x2 matrix, or >a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse >matrix by using the Gauss-Jordan elimination method. The check of the >solution is given. The program is designed for university students and >professors. >Price: $15. >Product homepage: >http://www.dekovsoft.com/inverse_matrices.htm >http://www.dekovsoft.com/ddekov/ -- Casey === Subject: Re: Inverse Matrices 1.03 for Windows! Programs description: The program provides detailed, step-by-step solution in a tutorial-like format to the following problem: Given a 2x2 matrix, or a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse matrix by using the Gauss-Jordan elimination method. The check of the solution is given. The program is designed for university students and professors. The above description shows the aim of the program and its scope. Suppose a student wants to see the solution (or the check) in a tutorial-like format. The student cannot find the solution in a textbook. Or suppose the student has some error and cannot see the error in the solution. Or suppose the student wants to see lots of examples to better understand the Gauss-Jordan elimination method. Or suppose a professor wants to give a quiz/homework to students and he/she needs a lots of problems (if the step-by-step solutions are given, the professor can save some time during the check of the quizes). Then the student/professor can use the program. The real question about such a program is as follows: How close is the solution produced by the program to the solutions given in the textbooks? Will the solution produced by the program be better, or worse, than the solutions, given in the textbooks? The authors of textbooks do not follow the strict Gauss-Jordan algorithm. Every author uses a set of deviations. E.g., if the augmented matix is as follows: |3 ... | |1 ... | | ... | in order to make the pivot entry 3 equal to 1, almost all authors prefer to exchange the first and the second row. (Accordingly the strict Gauss-Jordan algorithm we have to divide the first row by 3). Every author uses its own set of deviations. The program uses 11 deviations. The set of deviations used by the program covers almost all sets of deviations used in the textbooks. If we compare the solutions produced by the program with the corresponding solutions given in the textbooks, we can see that in approximately 95% of the cases the solution given by the program coincides with the solution given in the textbooks. (Or it is better - better or worse - from the point of view of the sets of deviations, because if we use the strict Gauss-Jordan algorithm, we can speak only about correct or non-correct solution). Every lecturer uses its own set of deviations. If some university wants the set of the deviations used by the program to coincide with the set of deviations used in the university, a custom build of the program can be produced. Really low price - one can see the offer soon at the web page of the program: http://www.dekovsoft.com/inverse_matrices.htm No doubt, the program will be useful for a large number of students and it will help them to better understand the Gauss-Jordan elimination method and to improve their skills. Dr.Dekov http://www.dekovsoft.com/ddekov/ > First: > You are using a freeware version of ClickTeam Install Creator in your > trial versions. > Do your paid versions use a non-freeware version of ClickTeam Install > Creator - since you are asking for money? > Second: > You do not allow for decimal entry. > Third: > You do not allow for complex numbers. > Fourth: > You simply provide a display of all the steps - like you would get in > a book. > That is there are no animations (Flash or Java) or sounds that would > help make it stick in a students mind - i.e. a student who has > challenges with learning from a book is still going to have the same > challenges with your so called helpful programs! > Fifth: > You are charging $15 USD? for each little segment - are there not > calculator programs (e.g. for the TI-89 or equivalent) that one can > download free (with a one-time purchase of a calculator) that will > show the steps of a matrix inversion, etc? > Plus, at the end of it all, one still has a calculator to show for it. >Inverse Matrices 1.03 for Windows now available! >The program provides detailed, step-by-step solution in a >tutorial-like format to the following problem: Given a 2x2 matrix, or >a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse >matrix by using the Gauss-Jordan elimination method. The check of the >solution is given. The program is designed for university students and >professors. >Price: $15. >Product homepage: >http://www.dekovsoft.com/inverse_matrices.htm >http://www.dekovsoft.com/ddekov/ === Subject: Uniqueness of physical objects in the physical universe. Ive changed some things around, made some changes, polished up the proof a little, and think that there might be something of interest here. Any serious feedback would be most appreciated. In our last episode I attempted to prove that No two objects in the physical universe are identical. Apparently, this may be reducible to a tautology. I am not sure that being a tautology matters much, because we are talking about physical properties of real objects in the universe. If you demonstrate a physical property of an object in the universe, then it dosent matter how you did it - tautology or not. A physical property is a physical property, and it matters not how you arrive at the proof of that property, as long as the demonstration is valid. Tautologies are trivial within the framework or abstract logical systems. If I say that It is a fast photon because it is a fast photon - then at least you know that you have a fast photon. Tautologies are not neccesarily trivial when youre talking about real objects. However, it seems that the original statement might be reducible to a question of uniqueness. So, I have the following statement to work with : ----------------------------------------------- Every object in the physical universe is unique ----------------------------------------------- So, I would like to prove that statement. It is not possible to compare every single physical object in the universe in a physics lab, so it must be proved mathematically. Definitions: Physical object. Any object in the physical universe which exists. This can be a person, place or thing. A region of space/time is an object. A region of empty space is an object. Locations are therefore objects. Events are objects, as per relativity theory. If it exists in the physical universe then it is an object. Unique A physical property of an object in the universe such that if an object is unique, then there is no other object which is identical to that object. There is a physical difference between objects which are distinct, and there are no physical differences between objects which are identical. Most uniqueness proofs require 2 things, first you demonstrate existence, and then you demonstrate uniqueness. However, in this case, I cannot prove that a physical object exists, it must be assumed (possibly via an axiom). So, lets assume that objects really exist in the physical universe, and try something like this- ----------------------------------------------------- Every object in the physical universe is unique Proof Suppose not Let O1 and O2 be distinct objects in the physical universe which are identical. There are 2 possible cases, 1) O1 and O2 are in separate locations 2) O1 and O2 are in in the exact same location Case 1) If O1 and O2 are in two separate locations, then they are not identical, and therefore they are both unique. Case 2) O1 and O2 are in the same location and they are also identical in every possible physical respect. They cannot be distinct, because either O1 or O2 is trivial and one of them does not really exist. If you can have O1 and O2 in the same exact location, doing the same exact thing, then let O3, O4, O(n) be identical to O1 and all in the same exact location. You now have an infinite number of identical physical objects in the exact same spot, which is obviously absurd. A contradiction. QED ----------------------------------------------------- Is this science, or have I finally cracked ? WillieK === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. Are electrons non-identical? Are atoms of the same isotope of the same element non-identical? Are molecules of the same chemical composition of the same isotopes non-identical? At some level of complexity uniqueness becomes almost inevitable on merely statistical grounds, but that is only a statistical conclusion. === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. > Are electrons non-identical? > Are atoms of the same isotope of the same element non-identical? > Are molecules of the same chemical composition of the same isotopes > non-identical? > At some level of complexity uniqueness becomes almost inevitable on > merely statistical grounds, but that is only a statistical conclusion. Two electrons which must be in two different locations. The location of an object which is situated in the universe is a physical property of that object. If you have two electrons in two separate location, then they cannot be identical. If you have two electrons which are both in the same exact location, then one of them must be trivial. Otherwise, you have two electrons in the same exact spot which are identical to the first, you might as well add a third, fourth, and so on, even infinitely many if you wish. Clearly impossible. Two electrons are always unique. Consider two separate regions of space/time which appear to be identical. If the regions are in separate locations then they are fundemantally different by virtue of their locations. They are not in the same location. They cannot be identical. They are unique. All objects in the universe are unique. No two atoms are identical. No two molecules. No two houses. No two people. No two chunks of space/time are identical. No two events. Abstract atoms are identical. Real ones are not. Physicists are always taking things as identical because it makes their models purr, but this is clearly impossible. === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. > Are electrons non-identical? > Are atoms of the same isotope of the same element non-identical? > Are molecules of the same chemical composition of the same isotopes > non-identical? > At some level of complexity uniqueness becomes almost inevitable on > merely statistical grounds, but that is only a statistical conclusion. > Two electrons which must be in two different locations. The location of an > object which is situated in the universe is a physical property of that > object. > If you have two electrons in two separate location, then they cannot be > identical. According to my understanding of quantum mechanics, electrons do not have precise locations. Further, and particularly if their approximate locations are anywhere close to each other, there is a non-zero probability that in any time interval that they have exchanged positions, so that it is impossible to be sure of which electron is which over any time interval. Under such circumstances, what can it mean to say that an electron has a unique identity? === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. Are electrons non-identical? Are atoms of the same isotope of the same element non-identical? Are molecules of the same chemical composition of the same isotopes > non-identical? At some level of complexity uniqueness becomes almost inevitable on > merely statistical grounds, but that is only a statistical conclusion. > Two electrons which must be in two different locations. The location of an > object which is situated in the universe is a physical property of that > object. > If you have two electrons in two separate location, then they cannot be > identical. > According to my understanding of quantum mechanics, electrons do not > have precise locations. > Further, and particularly if their approximate locations are anywhere > close to each other, there is a non-zero probability that in any time > interval that they have exchanged positions, so that it is impossible to > be sure of which electron is which over any time interval. > Under such circumstances, what can it mean to say that an electron has a > unique identity? it must be composed of energy, and that energy has to be somewhere - otherwise there is no electron. It can be treated as a generic physical phenomena like any other. Precise location may or may not be knowable, but must exist. If electrons in close proximity are capable of suddenly exchanging positions, I dont think that this is a problem. You would however have a problem if two objects merged into the same exact location without undergoing some fundamental physical change. If two identical electrons merge to coexist in the same exact location, they must have been transformed into a third object which is very different that either of the original electrons. I am told that bosons can do this and I simply cant agree to it. I think you are referring to the double slit experiment. To summarize, It may well be impossible to tell which electron is which, but they still posses the property of uniqueness. They still have this physical property. Our inability to distinguish one from the other does not make them identical. They are still unique. === Subject: Re: Uniqueness of physical objects in the physical universe. >> In our last episode I attempted to prove that No two objects in the >> physical universe are identical. >> Are electrons non-identical? >> Are atoms of the same isotope of the same element non-identical? >> Are molecules of the same chemical composition of the same isotopes >> non-identical? >> At some level of complexity uniqueness becomes almost inevitable on >> merely statistical grounds, but that is only a statistical conclusion. > Two electrons which must be in two different locations. The location of an > object which is situated in the universe is a physical property of that > object. > If you have two electrons in two separate location, then they cannot be > identical. > If you have two electrons which are both in the same exact location, then > one of them must be trivial. Otherwise, you have two electrons in the same > exact spot which are identical to the first, you might as well add a > third, > fourth, and so on, even infinitely many if you wish. Clearly impossible. > Two electrons are always unique. > Consider two separate regions of space/time which appear to be identical. > If > the regions are in separate locations then they are fundemantally > different > by virtue of their locations. They are not in the same location. They > cannot > be identical. They are unique. All objects in the universe are unique. > No two atoms are identical. > No two molecules. > No two houses. > No two people. > No two chunks of space/time are identical. > No two events. > Abstract atoms are identical. Real ones are not. > Physicists are always taking things as identical because it makes their > models purr, but this is clearly impossible. Have you ever heard about bosons? -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Uniqueness of physical objects in the physical universe. >> In our last episode I attempted to prove that No two objects in the >> physical universe are identical. >> Are electrons non-identical? >> Are atoms of the same isotope of the same element non-identical? >> Are molecules of the same chemical composition of the same isotopes >> non-identical? >> At some level of complexity uniqueness becomes almost inevitable on >> merely statistical grounds, but that is only a statistical conclusion. > Two electrons which must be in two different locations. The location of an > object which is situated in the universe is a physical property of that > object. > If you have two electrons in two separate location, then they cannot be > identical. > If you have two electrons which are both in the same exact location, then > one of them must be trivial. Otherwise, you have two electrons in the same > exact spot which are identical to the first, you might as well add a > third, > fourth, and so on, even infinitely many if you wish. Clearly impossible. > Two electrons are always unique. > Consider two separate regions of space/time which appear to be identical. > If > the regions are in separate locations then they are fundemantally > different > by virtue of their locations. They are not in the same location. They > cannot > be identical. They are unique. All objects in the universe are unique. > No two atoms are identical. > No two molecules. > No two houses. > No two people. > No two chunks of space/time are identical. > No two events. > Abstract atoms are identical. Real ones are not. > Physicists are always taking things as identical because it makes their > models purr, but this is clearly impossible. > Have you ever heard about bosons? I have. Its a problem. Apparently, certain results from QM seem to indicate some strange things which I simply cant accept. If two bosons are in the same location at the same time, doing the exact same thing, then one of them is trivial. Otherwise, the combination of the two creates a third which is identical to either of the first. You might as well add yet another, and another, until you have infinitely many bosons in thye same exact spot. This is clearly absurd. The theory is missing something. I simply cannot accept it. Im not going to make very many friends by saying that, but it is simply not possible. Thinking abstractly, yes - you can imagine cubes or spheres superimposed on top of each other in R3, infinitely many of them if you wish. No problem for a mathematician to do this. But, if we are talking about bricks, you simply cannot jam 2 identical bricks into the same location at the same time without one of them becoming trivial. === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. Are electrons non-identical? Are atoms of the same isotope of the same element non-identical? Are molecules of the same chemical composition of the same isotopes > non-identical? At some level of complexity uniqueness becomes almost inevitable on > merely statistical grounds, but that is only a statistical conclusion. >> Two electrons which must be in two different locations. The location of > an >> object which is situated in the universe is a physical property of that >> object. >> If you have two electrons in two separate location, then they cannot be >> identical. >> If you have two electrons which are both in the same exact location, > then >> one of them must be trivial. Otherwise, you have two electrons in the > same >> exact spot which are identical to the first, you might as well add a >> third, >> fourth, and so on, even infinitely many if you wish. Clearly >> impossible. >> Two electrons are always unique. >> Consider two separate regions of space/time which appear to be > identical. >> If >> the regions are in separate locations then they are fundemantally >> different >> by virtue of their locations. They are not in the same location. They >> cannot >> be identical. They are unique. All objects in the universe are unique. >> No two atoms are identical. >> No two molecules. >> No two houses. >> No two people. >> No two chunks of space/time are identical. >> No two events. >> Abstract atoms are identical. Real ones are not. >> Physicists are always taking things as identical because it makes their >> models purr, but this is clearly impossible. >> Have you ever heard about bosons? > I have. Its a problem. Apparently, certain results from QM seem to > indicate > some strange things which I simply cant accept. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > If two bosons are in the same location at the same time, doing the exact > same thing, then one of them is trivial What exactly do you mean by trivial? . Otherwise, the combination of the > two creates a third which is identical to either of the first. You might > as > well add yet another, and another, until you have infinitely many bosons > in > thye same exact spot. This is clearly absurd. ^^^^^^^^^^^^^^^^ > The theory is missing something. I simply cannot accept it. Im not going > to > make very many friends by saying that, but it is simply not possible. ^^^^^^^^^^^^^^^^^^^^ > Thinking abstractly, yes - you can imagine cubes or spheres superimposed > on > top of each other in R3, infinitely many of them if you wish. No problem > for > a mathematician to do this. But, if we are talking about bricks, you > simply > cannot jam 2 identical bricks into the same location at the same time > without one of them becoming trivial. argument from incredulity doesnt work here. Many quantum results are counterintuitive. The Pauli exclusion principle applies to fermions but not to bosons. Your posts are neither physics or mathematics. -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Uniqueness of physical objects in the physical universe. <41855ec6_3@newsfeed.slurp.net> <4185a6ee_1@newsfeed.slurp.net Many quantum results are counterintuitive. The Pauli exclusion principle > applies to fermions but not to bosons. Your posts are neither physics or > mathematics. Tho two electrons (Fermions) can occupy the same orbital, they will have opposite spins. Bosons are what and how do they evade the Pauli exclusion principle? === Subject: Re: Uniqueness of physical objects in the physical universe. >> Many quantum results are counterintuitive. The Pauli exclusion principle >> applies to fermions but not to bosons. Your posts are neither physics or >> mathematics. > Tho two electrons (Fermions) can occupy the same orbital, they will have > opposite spins. Bosons are what and how do they evade the Pauli exclusion > principle? Google is your friend. -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Uniqueness of physical objects in the physical universe. >> Many quantum results are counterintuitive. The Pauli exclusion principle >> applies to fermions but not to bosons. Your posts are neither physics or >> mathematics. > Tho two electrons (Fermions) can occupy the same orbital, they will have > opposite spins. Bosons are what and how do they evade the Pauli exclusion > principle? > Google is your friend. Many things are counterintuitive. I dont have a problem with that. Its the couter-possible stuff that I wonder about. === Subject: Re: Uniqueness of physical objects in the physical universe. > In our last episode I attempted to prove that No two objects in the > physical universe are identical. Are electrons non-identical? Are atoms of the same isotope of the same element non-identical? Are molecules of the same chemical composition of the same isotopes > non-identical? At some level of complexity uniqueness becomes almost inevitable on > merely statistical grounds, but that is only a statistical conclusion. >> Two electrons which must be in two different locations. The location of > an >> object which is situated in the universe is a physical property of that >> object. >> If you have two electrons in two separate location, then they cannot be >> identical. >> If you have two electrons which are both in the same exact location, > then >> one of them must be trivial. Otherwise, you have two electrons in the > same >> exact spot which are identical to the first, you might as well add a >> third, >> fourth, and so on, even infinitely many if you wish. Clearly >> impossible. >> Two electrons are always unique. >> Consider two separate regions of space/time which appear to be > identical. >> If >> the regions are in separate locations then they are fundemantally >> different >> by virtue of their locations. They are not in the same location. They >> cannot >> be identical. They are unique. All objects in the universe are unique. >> No two atoms are identical. >> No two molecules. >> No two houses. >> No two people. >> No two chunks of space/time are identical. >> No two events. >> Abstract atoms are identical. Real ones are not. >> Physicists are always taking things as identical because it makes their >> models purr, but this is clearly impossible. >> Have you ever heard about bosons? > I have. Its a problem. Apparently, certain results from QM seem to > indicate > some strange things which I simply cant accept. > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > If two bosons are in the same location at the same time, doing the exact > same thing, then one of them is trivial > What exactly do you mean by trivial? Ill describe it like this. I have a dollar bill in my hand. I claim that it is not a single bill, but many bills all compressed together so to speak, they are all identical and they are in the exact same location. It may look like one dollar, but it is really one million dollars. If I try to deposit my stack of millions in the bank its very likely that I will be laughed at. There is just one dollar, and the other 999,999 are trivial - they dont exist. > . Otherwise, the combination of the > two creates a third which is identical to either of the first. You might > as > well add yet another, and another, until you have infinitely many bosons > in > thye same exact spot. This is clearly absurd. > ^^^^^^^^^^^^^^^^ > The theory is missing something. I simply cannot accept it. Im not going > to > make very many friends by saying that, but it is simply not possible. ^^^^^^^^^^^^^^^^^^^^ > Thinking abstractly, yes - you can imagine cubes or spheres superimposed > on > top of each other in R3, infinitely many of them if you wish. No problem > for > a mathematician to do this. But, if we are talking about bricks, you > simply > cannot jam 2 identical bricks into the same location at the same time > without one of them becoming trivial. > argument from incredulity doesnt work here. Many quantum results are > counterintuitive. The Pauli exclusion principle applies to fermions but not > to bosons. Your posts are neither physics or mathematics. Well, Ive heard quantum mechanics described as being mystical, supernatural, and all kinds of things. I think its a natural consequence of not starting from 1st principles. They say that space/time is grainy - but wheres the proof ? We use calculus in physics, but cant even be sure if space is continuous. They generalize physical objects and treat them as abstractions, when clearly physical objects are _not_ abstractions. There are some issues. Uniqueness is an abstract mathematical property, now successfully applied to all physical objects. It is a mixture of math and physics, and it meshes rather well. It is now a physical property of all real objects. === Subject: Re: Uniqueness of physical objects in the physical universe. > >> In our last episode I attempted to prove that No two objects in > the >> physical universe are identical. >> Are electrons non-identical? >> Are atoms of the same isotope of the same element non-identical? >> Are molecules of the same chemical composition of the same isotopes >> non-identical? >> At some level of complexity uniqueness becomes almost inevitable on >> merely statistical grounds, but that is only a statistical > conclusion. Two electrons which must be in two different locations. The location > of >> an > object which is situated in the universe is a physical property of > that > object. If you have two electrons in two separate location, then they cannot > be > identical. If you have two electrons which are both in the same exact location, >> then > one of them must be trivial. Otherwise, you have two electrons in > the >> same > exact spot which are identical to the first, you might as well add a > third, > fourth, and so on, even infinitely many if you wish. Clearly > impossible. Two electrons are always unique. Consider two separate regions of space/time which appear to be >> identical. > If > the regions are in separate locations then they are fundemantally > different > by virtue of their locations. They are not in the same location. > They > cannot > be identical. They are unique. All objects in the universe are > unique. > No two atoms are identical. > No two molecules. > No two houses. > No two people. > No two chunks of space/time are identical. > No two events. Abstract atoms are identical. Real ones are not. Physicists are always taking things as identical because it makes > their > models purr, but this is clearly impossible. Have you ever heard about bosons? >> I have. Its a problem. Apparently, certain results from QM seem to >> indicate >> some strange things which I simply cant accept. >> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >> If two bosons are in the same location at the same time, doing the >> exact >> same thing, then one of them is trivial >> What exactly do you mean by trivial? > Ill describe it like this. > I have a dollar bill in my hand. I claim that it is not a single bill, but > many bills all compressed together so to speak, they are all identical and > they are in the exact same location. It may look like one dollar, but it > is > really one million dollars. > If I try to deposit my stack of millions in the bank its very likely that > will be laughed at. There is just one dollar, and the other 999,999 are > trivial - they dont exist. >> . Otherwise, the combination of the >> two creates a third which is identical to either of the first. You >> might >> as >> well add yet another, and another, until you have infinitely many >> bosons >> in >> thye same exact spot. This is clearly absurd. >> ^^^^^^^^^^^^^^^^ >> The theory is missing something. I simply cannot accept it. Im not > going >> to >> make very many friends by saying that, but it is simply not possible. > ^^^^^^^^^^^^^^^^^^^^ >> Thinking abstractly, yes - you can imagine cubes or spheres >> superimposed >> on >> top of each other in R3, infinitely many of them if you wish. No >> problem >> for >> a mathematician to do this. But, if we are talking about bricks, you >> simply >> cannot jam 2 identical bricks into the same location at the same time >> without one of them becoming trivial. >> argument from incredulity doesnt work here. Many quantum results are >> counterintuitive. The Pauli exclusion principle applies to fermions but > not >> to bosons. Your posts are neither physics or mathematics. > Well, Ive heard quantum mechanics described as being mystical, > supernatural, and all kinds of things. I think its a natural > consequence > of not starting from 1st principles. They say that space/time is grainy - > but wheres the proof ? We use calculus in physics, but cant even be sure > if > space is continuous. They generalize physical objects and treat them as > abstractions, when clearly physical objects are _not_ abstractions. There > are some issues. > Uniqueness is an abstract mathematical property, now successfully applied > to > all physical objects. It is a mixture of math and physics, and it meshes > rather well. It is now a physical property of all real objects. Bye. -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: Uniqueness of physical objects in the physical universe. > Every object in the physical universe is unique (Ax)(E!y)(x = y) for all x, there exists a unique y such that x = y > So, I would like to prove that statement. It is not possible to compare > every single physical object in the universe in a physics lab, so it must be > proved mathematically. Prove it in FOL from the axiom (Ax)(x = x) and the definion of (E!y).P(y). > Definitions: > Physical object. > Any object in the physical universe which exists. This can be a person, > place or thing. A region of space/time is an object. A region of empty space > is an object. Locations are therefore objects. Events are objects, as per > relativity theory. If it exists in the physical universe then it is an > object. I object, said the cute little object, to being objectified! > Unique (E!y).P(y) is defined as (Ex).P(x) & (Ax)(Ay)(P(x)&P(y) -> x = y) > A physical property of an object in the universe such that if an object > is unique, then there is no other object which is identical to that object. > There is a physical difference between objects which are distinct, and there > are no physical differences between objects which are identical. Nobody is like nobody. -- > Every object in the physical universe is unique > Proof > Suppose not > Let O1 and O2 be distinct objects in the physical universe which are > identical. O1 /= O2 & O1 = O2. Thats a contradiction. QED > There are 2 possible cases, > 1) O1 and O2 are in separate locations > 2) O1 and O2 are in in the exact same location > Case 1) > If O1 and O2 are in two separate locations, then they are not identical, > and therefore they are both unique. > Case 2) > O1 and O2 are in the same location and they are also identical in every > possible physical respect. They cannot be distinct, because either O1 or O2 > is trivial and one of them does not really exist. > If you can have O1 and O2 in the same exact location, doing the same > exact thing, then let O3, O4, O(n) be identical to O1 and all in the same > exact location. You now have an infinite number of identical physical > objects in the exact same spot, which is obviously absurd. A contradiction. > Is this science, or have I finally cracked ? No, its philosophy which is never what its cracked up to be. ;-) === Subject: Re: Uniqueness of physical objects in the physical universe. > Every object in the physical universe is unique > (Ax)(E!y)(x = y) > for all x, there exists a unique y such that x = y Actually, For all objects O1 in the physical universe, there is no O2 such that O1 = O2. > So, I would like to prove that statement. It is not possible to compare > every single physical object in the universe in a physics lab, so it must be > proved mathematically. > Prove it in FOL from the axiom (Ax)(x = x) and the definion of (E!y).P(y). Im no logician, > Definitions: > Physical object. > Any object in the physical universe which exists. This can be a person, > place or thing. A region of space/time is an object. A region of empty space > is an object. Locations are therefore objects. Events are objects, as per > relativity theory. If it exists in the physical universe then it is an > object. > I object, said the cute little object, to being objectified! objection sustained ? > Unique > (E!y).P(y) is defined as > (Ex).P(x) & (Ax)(Ay)(P(x)&P(y) -> x = y) > A physical property of an object in the universe such that if an object > is unique, then there is no other object which is identical to that object. > There is a physical difference between objects which are distinct, and there > are no physical differences between objects which are identical. > Nobody is like nobody. > -- > Every object in the physical universe is unique > Proof > Suppose not > Let O1 and O2 be distinct objects in the physical universe which are > identical. > O1 /= O2 & O1 = O2. Thats a contradiction. QED My proving has always been skewed, Cause Im a ridiculous dude, And this proof may need work, I say with a smirk, And Ill try not to come too unglued. > There are 2 possible cases, > 1) O1 and O2 are in separate locations > 2) O1 and O2 are in in the exact same location > Case 1) > If O1 and O2 are in two separate locations, then they are not identical, > and therefore they are both unique. > Case 2) > O1 and O2 are in the same location and they are also identical in every > possible physical respect. They cannot be distinct, because either O1 or O2 > is trivial and one of them does not really exist. > If you can have O1 and O2 in the same exact location, doing the same > exact thing, then let O3, O4, O(n) be identical to O1 and all in the same > exact location. You now have an infinite number of identical physical > objects in the exact same spot, which is obviously absurd. A contradiction. > Is this science, or have I finally cracked ? > No, its philosophy which is never what its cracked up to be. ;-) Ive ßipped over items unique, They tell me Im just a math geek, But physics to warn, A notion is born, And I should get a kiss on the cheek. === Subject: Re: Uniqueness of physical objects in the physical universe. (Ax)(E!y)(x = y) > for all x, there exists a unique y such that x = y > Actually, For all objects O1 in the physical universe, there is no O2 such > that O1 = O2. When O1 exists, theres the object O1 such that O1 = O1. So there! === Subject: Re: Uniqueness of physical objects in the physical universe. Every object in the physical universe is unique > (Ax)(E!y)(x = y) > for all x, there exists a unique y such that x = y > Actually, For all objects O1 in the physical universe, there is no O2 such > that O1 = O2. > When O1 exists, theres the object O1 such that O1 = O1. So there! > Baloney, even an object is an abstraction. .....billlllllly billy billy billy billy ......... Im talking about real objects in the physical universe in general. These are not abstractions. The word object _can_ be deployed as a name for an abstraction, if you are considering abstractions. But Im not. Thats the whole problem. These two worlds are so easily confused, I have a very hard time keep it straight myself. However, the coffee cup on my desk is not an abstraction, it is physical. I say it is unique among all other objects in the whole universe, and I have proved it. > When O1 exists, theres the object O1 such that O1 = O1. So there! Yes - you are exactly correct!! A physical object is exactly identical to itself. This is true, and it is trivial, but it is also very, very useful to know. Being trivial does not make it useless. This fact is as solid as a block of granite. It is a physical property of all physical objects in the real universe. === Subject: Re: Uniqueness of physical objects in the physical universe. <7J6hd.348115$D%.291095@attbi_s51 > Every object in the physical universe is unique > (Ax)(E!y)(x = y) > for all x, there exists a unique y such that x = y Actually, For all objects O1 in the physical universe, there is no > O2 such that O1 = O2. When O1 exists, theres the object O1 such that O1 = O1. So there! > Baloney, even an object is an abstraction. > .....billlllllly billy billy billy billy ......... Butt sir, thats Billy Goat. > Im talking about real objects in the physical universe in general. > These are not abstractions. You are not. You have present not one dirt clod nor pebble to consider. All youve presented is some vague abstractions which you call or claim to be objects, to wit O1 and O2. -- > Actually, For all objects O1 in the physical universe, there is no > O2 such that O1 = O2. > When O1 exists, theres the object O1 such that O1 = O1. So there! > Yes - you are exactly correct!! Thus Ive refuted your statement, demostrating that the object O1 is the exact object O2 that, contrary to your assertion, does need exist. === Subject: Re: Uniqueness of physical objects in the physical universe. Every object in the physical universe is unique > (Ax)(E!y)(x = y) > for all x, there exists a unique y such that x = y Actually, For all objects O1 in the physical universe, there is no > O2 such that O1 = O2. When O1 exists, theres the object O1 such that O1 = O1. So there! > Baloney, even an object is an abstraction. > .....billlllllly billy billy billy billy ......... > Butt sir, thats Billy Goat. heh heh heh ---- : ) > Im talking about real objects in the physical universe in general. > These are not abstractions. > You are not. You have present not one dirt clod nor pebble to consider. > All youve presented is some vague abstractions which you call or claim > to be objects, to wit O1 and O2. > -- > Actually, For all objects O1 in the physical universe, there is no > O2 such that O1 = O2. When O1 exists, theres the object O1 such that O1 = O1. So there! > Yes - you are exactly correct!! > Thus Ive refuted your statement, demostrating that the object O1 is the > exact object O2 that, contrary to your assertion, does need exist. (picture the wicked witch of the west) Immmmm Meltingggg !! === Subject: Re: Uniqueness of physical objects in the physical universe. Baloney, even an object is an abstraction. .....billlllllly billy billy billy billy ......... > Butt sir, thats Billy Goat. > heh heh heh ---- : ) > Im talking about real objects in the physical universe in general. > These are not abstractions. > You are not. You have present not one dirt clod nor pebble to consider. > All youve presented is some vague abstractions which you call or claim > to be objects, to wit O1 and O2. > Actually, For all objects O1 in the physical universe, there is no > O2 such that O1 = O2. > When O1 exists, theres the object O1 such that O1 = O1. So there! Yes - you are exactly correct!! > Thus Ive refuted your statement, demonstrating that the object O1 is the > exact object O2 that, contrary to your assertion, does need exist. > (picture the wicked witch of the west) > Immmmm Meltingggg !! An ice cube in a soda drink? Leftys at bat, the bases are loaded and its a ßy ball into left field ... -- The Left Fielder === Subject: Re: Uniqueness of physical objects in the physical universe. Baloney, even an object is an abstraction. .....billlllllly billy billy billy billy ......... > Butt sir, thats Billy Goat. > heh heh heh ---- : ) > Im talking about real objects in the physical universe in general. > These are not abstractions. > You are not. You have present not one dirt clod nor pebble to consider. > All youve presented is some vague abstractions which you call or claim > to be objects, to wit O1 and O2. Actually, For all objects O1 in the physical universe, there is no > O2 such that O1 = O2. > When O1 exists, theres the object O1 such that O1 = O1. So there! Yes - you are exactly correct!! > Thus Ive refuted your statement, demonstrating that the object O1 is the > exact object O2 that, contrary to your assertion, does need exist. > (picture the wicked witch of the west) > Immmmm Meltingggg !! > An ice cube in a soda drink? A unique ice cube, in a unique soda. > Leftys at bat, the bases are loaded and its a ßy ball into left > field ... > -- The Left Fielder Probably more like a swing and a miss with my luck, Ill rack my brain on this for a while until I wear myself out and then move on. I doubt anyone really takes any of it too seriously. === Subject: Re: Uniqueness of physical objects in the physical universe. field ... > -- The Left Fielder > Probably more like a swing and a miss with my luck, Ill rack my brain on > this for a while until I wear myself out and then move on. I doubt anyone > really takes any of it too seriously. Philosophers do, theyve discussed this other abstrusely futile matters for ages. -- Split wood, not hairs. Philosophys a joke. If it isnt, youre taking life to seriously. === Subject: Re: Uniqueness of physical objects in the physical universe. > Leftys at bat, the bases are loaded and its a ßy ball into left > field ... > -- The Left Fielder > Probably more like a swing and a miss with my luck, Ill rack my brain on > this for a while until I wear myself out and then move on. I doubt anyone > really takes any of it too seriously. > Philosophers do, theyve discussed this other abstrusely futile matters for > ages. I know. I dont think that you can ever get anywhere with first order logic, because its like trying to construct a physical object with algebra. You cant build an abstraction out of lumber and nails, and you cant construct reality from abstractions. I think that this is why they failed. I also believe that I have failed somehow, but I just dont know how yet. I never did believe that it was possible to make a statement about the universe, but this one seems very difficult to refute. All physical objects are unique. <<< How can you trash this based on the justification I gave ? Im stumped. I would like to show that it is false, but I dont know how. I think that Im guilty of abusing math and probably just pissing people off. Then again, this may be the millionth time that someone proposed this nonsense and its just the first time that I came across it. Too bad there isnt a collection of hairbrained ideas out there somewhere, so that when you get one of these silly ideas you could easily refer to a book and determine that you are working with nonsense. > -- Split wood, not hairs. > Philosophys a joke. If it isnt, youre taking life to seriously. Good quote. The best philosophers didnt care about science - Martin Buber hit the nail on the head. Theres one other thing Id like to add here - which is probably yet more nonsense, but Im certainly no philosopher and so I dont know if this has ever been stated, wierd idea - here goes. ----------------------------------------- The universe is not infinite. Justification: Our understanding of time, and our ability to measure it is based on cycles in nature. There are 365 Earth rotations per revolution. There are 60 minutes per every hour. There are 12 Moon revolutions per Earth revolution. The minute hand rotates 45 times per eruption of the Old Faithful geyser. Etc etc. Now, as things become larger and larger, like galaxies, the cycles become slower and slower. The Suns orbit period about the center of the galaxy (the galactic year) is reckoned to be about 230 million years. As things become even larger, the cycles become slower, and slower, to the point where it takes so much time that its motion may be taken as very near zero. At this point, time is practically unmeasurable, and therefore unobservable. If the universe is composed of space/time, and at some point time itself is unmeasurable or unobservable relative to you, then you have defined the edge of the universe, and it is finite relative to you. ----------------------------------------- I washed a lot of hogs to come up with that one - probably just more philosophical garbage, but maybe the converse is true also - about things being so small that they cease to exist. I dont know. Maybe existence is relative depending on scale, or size of an object. Could account for some quantum wierdness. I really dont know. Just a thought to taunt you with : ) === Subject: Uniqueness of physical objects in the physical universe. brain on this for a while until I wear myself out and then move on. > I doubt anyone really takes any of it too seriously. Philosophers do, theyve discussed this other abstrusely futile > matters for ages. > I know. I dont think that you can ever get anywhere with first order logic, > because its like trying to construct a physical object with algebra. You > cant build an abstraction out of lumber and nails, and you cant construct > reality from abstractions. I think that this is why they failed. > I also believe that I have failed somehow, but I just dont know how yet. Youre making an abstract theory about objects out of words instead of sticks and stones. > Philosophys a joke. If it isnt, youre taking life to seriously. > Good quote. In no time at all, time had begun, for its time had come. > The universe is not infinite. So what? Its bigger than both of us. > If the universe is composed of space/time, and at some point time itself is > unmeasurable or unobservable relative to you, then you have defined the > edge of the universe, and it is finite relative to you. Is the universe the space time stage for the universe? Is the universe all the stuff in the universe? Is the universe all the happenings in the universe? Is the universe the biggest whopping abstraction of all? === Subject: Re: Uniqueness of physical objects in the physical universe. > Probably more like a swing and a miss with my luck, Ill rack my > brain on this for a while until I wear myself out and then move on. > I doubt anyone really takes any of it too seriously. Philosophers do, theyve discussed this other abstrusely futile > matters for ages. > I know. I dont think that you can ever get anywhere with first order logic, > because its like trying to construct a physical object with algebra. You > cant build an abstraction out of lumber and nails, and you cant construct > reality from abstractions. I think that this is why they failed. > I also believe that I have failed somehow, but I just dont know how yet. > Youre making an abstract theory about objects out of words instead of > sticks and stones. > Philosophys a joke. If it isnt, youre taking life to seriously. > Good quote. > In no time at all, time had begun, for its time had come. > The universe is not infinite. > So what? Its bigger than both of us. > If the universe is composed of space/time, and at some point time itself is > unmeasurable or unobservable relative to you, then you have defined the > edge of the universe, and it is finite relative to you. > Is the universe the space time stage for the universe? > Is the universe all the stuff in the universe? > Is the universe all the happenings in the universe? > Is the universe the biggest whopping abstraction of all? nooo.....noooooo.......noooooo....certainly not, heavens no !!! philosophical mish-mash. Dude - I just explained exactly why the universe seems to be finite in size, why it has an edge. It is very simple !! One more time. Please just read it. I think its all very clear. ----------------------------------------- The universe is not infinite. Justification: Our understanding of time, and our ability to measure it is based on cycles in nature. There are 365 Earth rotations per revolution. There are 60 minutes per every hour. There are 12 Moon revolutions per Earth revolution. The minute hand rotates 45 times per eruption of the Old Faithful geyser. Etc etc etc. This is how we measure and observe time. Now, as things become larger and larger, like galaxies, the cycles become slower and slower. The Suns orbital period about the center of the galaxy (the galactic year) is reckoned to be about 230 million years. As things become even larger the cycles become slower and slower, to the point where it takes so much time that its motion may be taken as very near zero. At this point, time is practically unmeasurable, and therefore unobservable. If time is unmeasurable & unobservable relative to you, then you have defined the edge of the universe, and it is finite relative to you. Its a matter of scale. Its really very, very simple. ----------------------------------------- === Subject: Re: Uniqueness of physical objects in the physical universe. Dude - I just explained exactly why the universe seems to be finite in size, > why it has an edge. It is very simple !! Of course it seems to be finite. Thats because the egocenter of the observable universe doesnt bother to look further than the edge. > The universe is not infinite. > Our understanding of time, and our ability to measure it is based on cycles > in nature. There are 365 Earth rotations per revolution. There are 60 > minutes per every hour. There are 12 Moon revolutions per Earth revolution. > The minute hand rotates 45 times per eruption of the Old Faithful geyser. > Etc etc etc. This is how we measure and observe time. > Now, as things become larger and larger, like galaxies, the cycles become > slower and slower. The Suns orbital period about the center of the galaxy > (the galactic year) is reckoned to be about 230 million years. > As things become even larger the cycles become slower and slower, to the > point where it takes so much time that its motion may be taken as very near > zero. At this point, time is practically unmeasurable, and therefore > unobservable. > If time is unmeasurable & unobservable relative to you, then you have > defined the > edge of the universe, and it is finite relative to you. > Its a matter of scale. Its really very, very simple. As the universe isnt repeating itself, its taking infinitely long. === Subject: Re: Uniqueness of physical objects in the physical universe. > In no time at all, time had begun, for its time had come. > Dude - I just explained exactly why the universe seems to be finite in size, > why it has an edge. It is very simple !! > Of course it seems to be finite. Thats because the egocenter of the > observable universe doesnt bother to look further than the edge. > The universe is not infinite. > Our understanding of time, and our ability to measure it is based on cycles > in nature. There are 365 Earth rotations per revolution. There are 60 > minutes per every hour. There are 12 Moon revolutions per Earth revolution. > The minute hand rotates 45 times per eruption of the Old Faithful geyser. > Etc etc etc. This is how we measure and observe time. > Now, as things become larger and larger, like galaxies, the cycles become > slower and slower. The Suns orbital period about the center of the galaxy > (the galactic year) is reckoned to be about 230 million years. > As things become even larger the cycles become slower and slower, to the > point where it takes so much time that its motion may be taken as very near > zero. At this point, time is practically unmeasurable, and therefore > unobservable. > If time is unmeasurable & unobservable relative to you, then you have > defined the > edge of the universe, and it is finite relative to you. > Its a matter of scale. Its really very, very simple. > As the universe isnt repeating itself, its taking infinitely long. All Im saying is that it has an outer boudary - relative to an observer. Beyond this approximate distance, nothing exists relative to an observer. You dont need any fancy math to derive this result. It is extremely simple to comprehend, and it becomes absolutely obvious why the universe is finite. However, if you traveled to the edge, you would find yourself at the center of a similar looking place. And if two observers became separated by a distance which was great enough, they would cease to exist relative to one another. It is very straightforward and I would say even stupidly simple, but kind of wierd. === Subject: Uniqueness of physical objects in the physical universe. <6Tzid.570150$8_6.407381@attbi_s04 Its a matter of scale. Its really very, very simple. As the universe isnt repeating itself, its taking infinitely long. > All Im saying is that it has an outer boudary - relative to an observer. > Beyond this approximate distance, nothing exists relative to an observer. Read up about inßation during the early stages of the Big Bangup Job. > However, if you traveled to the edge, you would find yourself at the center > of a similar looking place. And if two observers became separated by a > distance which was great enough, they would cease to exist relative to one > another. It is very straightforward and I would say even stupidly simple, > but kind of wierd. Get to the edge? You cant even get to Mars. However, such real tangibles aside have you noticed how the horizon edge of the observable universe keeps moving ahead of you while youre moving toward it and were you to look behind in your soggy saga, you would find the horizon edge creeping up on you. Eventually when you run out of time, it will catch you and for all of your travels, you will have gone from the center of the observable to the center of the observable universe, ie nowheres. -- If everybodys vote was counted, the neotheofascist would have been counted out. === Subject: Re: Uniqueness of physical objects in the physical universe. > Its a matter of scale. Its really very, very simple. As the universe isnt repeating itself, its taking infinitely long. > All Im saying is that it has an outer boudary - relative to an observer. > Beyond this approximate distance, nothing exists relative to an observer. > Read up about inßation during the early stages of the Big Bangup Job. > However, if you traveled to the edge, you would find yourself at the center > of a similar looking place. And if two observers became separated by a > distance which was great enough, they would cease to exist relative to one > another. It is very straightforward and I would say even stupidly simple, > but kind of wierd. > Get to the edge? You cant even get to Mars. However, such real > tangibles aside have you noticed how the horizon edge of the observable > universe keeps moving ahead of you while youre moving toward it and were > you to look behind in your soggy saga, you would find the horizon edge > creeping up on you. Eventually when you run out of time, it will catch > you and for all of your travels, you will have gone from the center of the > observable to the center of the observable universe, ie nowheres. Exactly. Existence is relative. The justification for the boundary itself is not really a bad one, could be wrong, but seems pretty credible. Can be derived with no math whatsoever. === Subject: Re: Uniqueness of physical objects in the physical universe. > Its a matter of scale. Its really very, very simple. As the universe isnt repeating itself, its taking infinitely long. All Im saying is that it has an outer boudary - relative to an > observer. > Beyond this approximate distance, nothing exists relative to an > observer. Read up about inßation during the early stages of the Big Bangup Job. > However, if you traveled to the edge, you would find yourself at the > center > of a similar looking place. And if two observers became separated by a > distance which was great enough, they would cease to exist relative to > one > another. It is very straightforward and I would say even stupidly > simple, > but kind of wierd. Get to the edge? You cant even get to Mars. However, such real > tangibles aside have you noticed how the horizon edge of the observable > universe keeps moving ahead of you while youre moving toward it and were > you to look behind in your soggy saga, you would find the horizon edge > creeping up on you. Eventually when you run out of time, it will catch > you and for all of your travels, you will have gone from the center of the > observable to the center of the observable universe, ie nowheres. > Exactly. > Existence is relative. > The justification for the boundary itself is not really a bad one, could be > wrong, but seems pretty credible. Can be derived with no math whatsoever. Just to recap, lets say you wanted to build a clock so that you could observe the passage of time. You want to make observations of time passing. You could use our solar system, as has been done for thousands of years. The Earth goes around the sun marking off the years as if it were one of the hands of a clock. You could also use a galaxy, but you would have to wait a long, long time for it to make even a single rotation. You could, in principle, base your clock on the rotation of a galaxy. It should actually work quite well. However, if your clock becomes too big, like the outer limits of the universe, it is simply so vast that motion is very near zero relative to man, and so it simply cannot be used as a clock because its motion is so close to zero. It is motionless (relative to man). And so your clock wont work if you base it on rotations of outer boundaries of universe. Therefore, time becomes unobservable, and spacetime no longer exists relative to an observer here on Earth, because spacetime really has no meaning if time is unobservable. This is why the universe is finite, but open, and existence is relative. Now, On a scale of 0 to 10, 0 being not a crackpot at all, and 10 being completely wacky crackpot material, what do you think ? 8 ? === Subject: Re: Uniqueness of physical objects in the physical universe. <8Ahjd.55759$HA.16732@attbi_s01> === Subject: Uniqueness of physical objects in the physical universe. If you push this button time goes backwards goes time button this push you if. > Just to recap, lets say you wanted to build a clock so that you > could observe the passage of time. You want to make observations of > time passing. Objective or surjective time? > You could use our solar system, as has been done for thousands of > years. The Earth goes around the sun marking off the years as if it > were one of the hands of a clock. Wouldnt work, too slow, Id be late for work. > You could also use a galaxy, but you would have to wait a long, long > time for it to make even a single rotation. You could, in principle, > base your clock on the rotation of a galaxy. It should actually work > quite well. Naw totally inadequate and too clumbersome. More practical is the vibration of a cesium atom. > However, if your clock becomes too big, like the outer limits of the > universe, it is simply so vast that motion is very near zero > relative to man, and so it simply cannot be used as a clock because > its motion is so close to zero. It is motionless (relative to man). > And so your clock wont work if you base it on rotations of outer > boundaries of universe. Youre getting too big for your britches, think small. Whats the smallest interval of time that can be measured and if the interals of time got shorter and shorter, would time vanish? > Therefore, time becomes unobservable, and spacetime no longer exists > relative to an observer here on Earth, because spacetime really > has no meaning if time is unobservable. Nope, last I looked its still there. > This is why the universe is finite, but open, and existence is > relative. Ya, I inherited existence from my relatives. > On a scale of 0 to 10, 0 being not a crackpot at all, and 10 > being completely wacky crackpot material, what do you think ? Hey, if youre going to be a crackpot, lets at least get a laugh out of it. ;-) In no time at all, time had begun, for its time had come. ---- === Subject: Re: Uniqueness of physical objects in the physical universe. === > Subject: Uniqueness of physical objects in the physical universe. > If you push this button time goes > backwards > goes time button this push you if. > Just to recap, lets say you wanted to build a clock so that you > could observe the passage of time. You want to make observations of > time passing. > Objective or surjective time? > You could use our solar system, as has been done for thousands of > years. The Earth goes around the sun marking off the years as if it > were one of the hands of a clock. > Wouldnt work, too slow, Id be late for work. > You could also use a galaxy, but you would have to wait a long, long > time for it to make even a single rotation. You could, in principle, > base your clock on the rotation of a galaxy. It should actually work > quite well. > Naw totally inadequate and too clumbersome. > More practical is the vibration of a cesium atom. > However, if your clock becomes too big, like the outer limits of the > universe, it is simply so vast that motion is very near zero > relative to man, and so it simply cannot be used as a clock because > its motion is so close to zero. It is motionless (relative to man). > And so your clock wont work if you base it on rotations of outer > boundaries of universe. > Youre getting too big for your britches, think small. Whats the > smallest interval of time that can be measured and if the interals of time > got shorter and shorter, would time vanish? > Therefore, time becomes unobservable, and spacetime no longer exists > relative to an observer here on Earth, because spacetime really > has no meaning if time is unobservable. > Nope, last I looked its still there. > This is why the universe is finite, but open, and existence is > relative. > Ya, I inherited existence from my relatives. > On a scale of 0 to 10, 0 being not a crackpot at all, and 10 > being completely wacky crackpot material, what do you think ? > Hey, if youre going to be a crackpot, > lets at least get a laugh out of it. ;-) > In no time at all, time had begun, for its time had come. working on the James Harris equation of squaring the circle via revolutionary new cubic equation prime number generator such that disingenuous algebraic violinists have been harrassing him with tinfoil hats and torturing his mind intentionally with circular logic mind maze. If you dont believe me, just look at this http://www.timecube.com/ My tin foil hat is shaped like that of Paul Revere. This is because my ideas are so revolutionary. === Subject: Re: Uniqueness of physical objects in the physical universe. <8Ahjd.55759$HA.16732@attbi_s01 Hey, if youre going to be a crackpot, > lets at least get a laugh out of it. ;-) > In no time at all, time had begun, for its time had come. > If you dont believe me, just look at this > http://www.timecube.com Oh, thats your problem, having dirty thoughts. -- === Subject: Re: Uniqueness of physical objects in the physical universe. > In no time at all, time had begun, for its time had come. Dude - I just explained exactly why the universe seems to be finite in > size, > why it has an edge. It is very simple !! Of course it seems to be finite. Thats because the egocenter of the > observable universe doesnt bother to look further than the edge. > The universe is not infinite. Our understanding of time, and our ability to measure it is based on > cycles > in nature. There are 365 Earth rotations per revolution. There are 60 > minutes per every hour. There are 12 Moon revolutions per Earth > revolution. > The minute hand rotates 45 times per eruption of the Old Faithful > geyser. > Etc etc etc. This is how we measure and observe time. Now, as things become larger and larger, like galaxies, the cycles > become > slower and slower. The Suns orbital period about the center of the > galaxy > (the galactic year) is reckoned to be about 230 million years. As things become even larger the cycles become slower and slower, to the > point where it takes so much time that its motion may be taken as very > near > zero. At this point, time is practically unmeasurable, and therefore > unobservable. If time is unmeasurable & unobservable relative to you, then you have > defined the > edge of the universe, and it is finite relative to you. Its a matter of scale. Its really very, very simple. As the universe isnt repeating itself, its taking infinitely long. > All Im saying is that it has an outer boudary - relative to an observer. The best available telescopes have not yet come to any such boundary for observers on earth. Each improvement in telescopes has merely made the observable universe larger. So what evidence is there that any such boundary exists? The only suggestion towards that end that I can see is that if the universe had a beginning, one could not see farther than light could have traveled since that beginning, but even in that case, the big bang theory might allow us to see all there has ever been. > Beyond this approximate distance, nothing exists relative to an observer. > You dont need any fancy math to derive this result. It is extremely simple > to comprehend, and it becomes absolutely obvious why the universe is finite. > However, if you traveled to the edge, you would find yourself at the center > of a similar looking place. And if two observers became separated by a > distance which was great enough, they would cease to exist relative to one > another. It is very straightforward and I would say even stupidly simple, > but kind of wierd. === Subject: Re: Uniqueness of physical objects in the physical universe. > The best available telescopes have not yet come to any such boundary > for observers on earth. Each improvement in telescopes has merely made > the observable universe larger. So what evidence is there that any such > boundary exists? The only suggestion towards that end that I can see is > that if the universe had a beginning, one could not see farther than > light could have traveled since that beginning, but even in that case, > the big bang theory might allow us to see all there has ever been. Well, two things come to mind. First, quantum wierdenss. I actually believe that something can be so small that it ceases to exist relative to an observer. This might explain alot of things. Second, the way that we observe time. The only way to observe time passing is by studying cyclical behaviour in nature. Our calendars, atomic clock, wristwatch, etc. All of these are based on observable cycles in nature. If you cant observe cyclical behaviour, then you cannot observe time, and space/time simply stops because it has become unobservable. This happens when something is so large that its motion is very near zero, relative to us. This is why the universe is finite. But its finiteness is only relative depending on who is observing it. You dont need any math to understand it - which is the amazing thing, just division I suppose. I do _not_ think that it is caused by division by zero. It is caused by something being so large that its natural cyclical frequency is very nearly zero, and its motion simply stops, and time stands still, and this negates existence. I think that eventually astronomers will discover many things which are just as wierd as quantum wierdness. Its really the same thing from two different perspectives I think. The whole thing is very simple. Consider that existence is relative, as described above. Out near the edge of the universe there is a light source. Maybe a galaxy which just a little bit beyond the edge of the universe relative to you. I think that it is possible that the light from that galaxy might be able to cross over the boundary and travel right into our universe, even though the galaxy itself does not exist relative to us. Strange, but possible, that you could actually see something which does not even exist relative to you. So, the big telescopes might actually be seeing into another universe without even realizing it. Seems rather paradoxical. === Subject: Re: Uniqueness of physical objects in the physical universe. > Probably more like a swing and a miss with my luck, Ill rack my > brain on this for a while until I wear myself out and then move on. > I doubt anyone really takes any of it too seriously. Philosophers do, theyve discussed this other abstrusely futile > matters for ages. > I know. I dont think that you can ever get anywhere with first order > logic, > because its like trying to construct a physical object with algebra. > You > cant build an abstraction out of lumber and nails, and you cant > construct > reality from abstractions. I think that this is why they failed. I also believe that I have failed somehow, but I just dont know how yet. > Youre making an abstract theory about objects out of words instead of > sticks and stones. > Philosophys a joke. If it isnt, youre taking life to seriously. > Good quote. In no time at all, time had begun, for its time had come. > The universe is not infinite. So what? Its bigger than both of us. > If the universe is composed of space/time, and at some point time itself > is > unmeasurable or unobservable relative to you, then you have defined > the > edge of the universe, and it is finite relative to you. Is the universe the space time stage for the universe? > Is the universe all the stuff in the universe? > Is the universe all the happenings in the universe? > Is the universe the biggest whopping abstraction of all? > nooo.....noooooo.......noooooo....certainly not, heavens no !!! > philosophical mish-mash. > Dude - I just explained exactly why the universe seems to be finite in size, > why it has an edge. It is very simple !! > One more time. Please just read it. I think its all very clear. > ----------------------------------------- > The universe is not infinite. > Justification: > Our understanding of time, and our ability to measure it is based on cycles > in nature. There are 365 Earth rotations per revolution. There are 60 > minutes per every hour. There are 12 Moon revolutions per Earth revolution. > The minute hand rotates 45 times per eruption of the Old Faithful geyser. > Etc etc etc. This is how we measure and observe time. > Now, as things become larger and larger, like galaxies, the cycles become > slower and slower. The Suns orbital period about the center of the galaxy > (the galactic year) is reckoned to be about 230 million years. > As things become even larger the cycles become slower and slower, to the > point where it takes so much time that its motion may be taken as very near > zero. At this point, time is practically unmeasurable, and therefore > unobservable. > If time is unmeasurable & unobservable relative to you, then you have > defined the > edge of the universe, and it is finite relative to you. Or perhaps only the running down of the clock. > Its a matter of scale. Its really very, very simple. > ----------------------------------------- === Subject: Re: Uniqueness of physical objects in the physical universe. Consider two objects in the physical universe, O1 and O2. I would like to prove that if O1 and O2 are oriented in fixed positions with respect to one another, that this relationship is invariant with respect to how they are oriented with respect to the rest of the universe. Maybe it matters - maybe it dosent - and maybe I need to put it down before I snap, again. === Subject: Re: Uniqueness of physical objects in the physical universe. > Prove it in FOL from the axiom (Ax)(x = x) and the definion of (E!y).P(y). If I prove it using FOL, then I have proved an abstraction. You have no physical objects in a FOL. Physical objects exist only in the universe, not abstract space. That is the disconnect. You need a fresh start. I dont think it can be done purely with FOL. === Subject: Re: Uniqueness of physical objects in the physical universe. (E!y).P(y). > If I prove it using FOL, then I have proved an abstraction. Baloney, even an object is an abstraction. === Subject: question of this system of linear congruences I have following difficult(to me) problem. Would you please give me some detailed solution or explanation. _________________________________________________________ Find solution of following system of linear congruences. x=11 (mod 36) x=7 (mod 40) x=32 (mod 75) === Subject: Re: question of this system of linear congruences > I have following difficult(to me) problem. > Would you please give me some detailed solution or explanation. > x=11 (mod 36) > x=7 (mod 40) > x=32 (mod 75) Solve the equations in pairs x = 11 + 36k = 7 +40l => 40l -36k = 4 => 10l -9k =1 => 10( 1 + 9s)- 9( 1 +10s) =1 => x = 11 + 36(1 + 10s) => x = 47 +360s X = 47 + 360s x = 32 + 75m x= 47 + 360s = 32 + 75m => 15 = 75m -360s => 1 = 5m -24s => 5(5 +24t) - 24(1 +5t) = 1 => x = 47 + 360(1 + 5t) => x = 407 + 1800t Which has form x = u + LCM(36,30,75) The eqns of the form Ax + By =1 could be solved by inspection. If the equations hadnt had obvious solutions, you oould have used Euclids algorithm or Eulerts method 15x + 17 y = 1 15x + (15 +2)y =1 15(x+y) +2y = 1 (1 + 7*2)(x+y) + 2y = 1 (x+y) + 2(7x +8y) = 1 x +y = 1 7x +8y =0 y = -7, x =8 15( 8 + 17t) + 17( -7 - 15t)=1 === Subject: Re: question of this system of linear congruences > I have following difficult(to me) problem. > Would you please give me some detailed solution or explanation. > _________________________________________________________ > Find solution of following system of linear congruences. > x=11 (mod 36) > x=7 (mod 40) > x=32 (mod 75) There may be better ways but . . . Look at the first equation. It says x = 11 + 36*r. Now, look at the last equation. It says x = 32 + 75*s. Put the two together and simplify a little to get 36*r - 75*s = 21. Divide by 3 to get 25*s + 7 = 12*r. Since s cant be even write s = 2*n + 1, substitute and simplify to get 32 + 50*n = 12*r. Since 32 and 12 are divisible by 4 but 50 isnt, n = 2*m. Divide by 4 to get 8 + 25*m = 3*s. Note 8 + 25 = 3*11. Let m = 1 backtrack to get s = 5 and x = 407. Check - a _must_: 407 - 11 = 11*36; 407 - 7 = 10*40 407 - 32 = 5*75 -- Paul Sperry Columbia, SC (USA) === Subject: Math is a hard discipline You can look over my work for yourself, and see that the problem Ive found in algebraic number theory is real. So why wont top math professors acknowledge my results? How is it that a paper of mine was so easily censored? (Note: I didnt withraw it, the editors yanked it when some sci.mathers sent some disparaging emails!) The simple answer has to do with your youth and their age. Know how when you were a little kid you could dream of being an astronaut, or of winning a gold medal at the Olympics? When you were a kid, your options were wide. Now for most of you, though still in many ways kids, those dreams are reasonably dead. The older you get, the fewer are your options. You as undergraduates have a bigger potential future than any math professor who is much older, who by now probably knows about where theyll end up in the pack. Now consider people supposedly at the top of the heap like Taylor, Wiles or Ribet. My work undermines all of their work, and throws them down into the pack with others. So its far worse for people who feel accomplished in algebraic number theory, as they not only have to look what they can no longer accomplish, but be burdened by what they thought they had accomplished, but didnt. Many of you will think its easier to go along with the old folks whos futures are lost, as what else can you do? Well, you are losing your own future as well, if you go into algebraic number theory, as you can learn all you want about elliptic curves and Galois Theory, and all these difficult subjects that older people will keep teaching you because they have nowhere to go. Youll expend all of that energy, and maybe get out your own work, or even get you own notoriety TODAY but the clock will be ticking. It could happen in a few months, or a few years or a few decades. You could be smiling holding your grandkids, with your pride in a history of notable accomplishment when the news comes out, and you haven nothing. Its math. Your fantasies, hope and dreams dont change it. Eventually the truth will come out, and what if its a hundred years from now, before youre a joke? Is it worth it to you to have congratulations from old men who have nothing except the ability to use their social power to hide the truth for them to smile on you? Andrew Wiles didnt prove Fermats Last Theorem, most of the supposed accomplishments of algebraic number theory over the last hundred years plus are nonesuch. Math is a hard discipline. Little mistakes grow, when not addressed. Over a hundred years ago some people focused on roots of monic polynomials with integer coefficients, and others built a lot of work on this very simple idea that has weird problems. Your professors will keep trying to ignore my work, as what future do they have? Does Andrew Wiles want to go home to his wife and kids and tell them hes no longer some great hero to the math world? That he accomplished nothing? Thats whats brutal about mathematics! When youre wrong, you can have spent years, and lots of effort, and come out at the end with nothing. Theyre old men. Youre young. Why shouldnt they sacrifice your future so that they can stay comfortable? Isnt that the way it has always been? James Harris === Subject: Re: Math is a hard discipline ... his usual garbage. Idiot. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Math is a hard discipline Much too hard and much too disciplined for James S Harris. === Subject: Re: Math is a hard discipline alt.math.undergrad: >Much too hard and much too disciplined for James S Harris. Galdor of the Havens, who sat near by, overheard him. ÔYou speak for me also, he cried. -- /The Lord of the Rings, Bk II Chapter 2 -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Math is a hard discipline > (Note: I didnt withraw it, the editors yanked it when some sci.mathers > sent some disparaging emails!) No, what they sent were simple counterexamples that showed that the result of the paper was false. Youve went on to claim that the counterexamples dont apply, because your paper isnt actually about what it said it was about. However, that also is fatal to the paper, because you do not *define* what you are talking about in the paper. ... > My work undermines all of their work, and throws them down into the pack > with others. So far, you work has fallen into three categories: 1. Results that are wrong. 2. Results that only you can understand, because you havent the foggiest notion how to write mathematics. 3. Results that are true, but routine or trivial. If there is anything worthwhile lurking in category #2, then the only way you will make any headway at getting anyone to notice is by LEARNING HOW TO WRITE MATHEMATICS. -- --Tim Smith === Subject: Refuting argument, algebraic number theory The following post steps through the math without much commentary, until the end. The math proof is very basic with mostly simple algebraic manipulations, and the bedrock concepts--like constants are constant--are basic as well. It is enough for those who rely on what can be mathematically proven to be correct. But those unlike them, the truth is never enough. ___JSH ------------------------------------------------------------- --------- I. Factorization What follows is in a commutative ring. Let P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where f is nonzero and note that at m=0, P(0) = u^2 f^2(3x + uf), which, of course, does not vary as m varies. So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) which requires that at least two of the as must equal 0 at m=0, while one equals 3. Now let f be coprime to 3 and x so that 3x + uf is coprime to f. Since, at m=0, two of the as must equal 0, its convenient to just arbitrarily select a_1 and a_2 as those two. Then you have uf for the first, uf for the second and 3x + uf for the third as terms that do not vary when m varies. Im going to emphasize that point because its important. Now then, if m=1, what are the *constant* terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Thats logical because they do not vary with m, so if m=1003909273, what are the constant terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Now divide f^2 from both sides, which gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you note that P(0)/f^2 = u^2(3x + uf), which means that now your constant terms are u for the first, u for the second and 3x + uf for the third. Now then, if m=1, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. If m = 2938479378, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. How can the constant terms of the first two go from uf to u? They must be divided by f. Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided from P(m), it is u; therefore, a_1 x + uf is divided by f, and you have a_1 x/f + u and the constant term of a_2 x + uf is uf, but when f^2 is divided from P(m), it is u; therefore, a_2 x + uf is divided by f, and you have a_2 x/f + u while the constant term of a_3 x + uf is 3x + uf, and after f^2 is divided off, it is 3x + uf, so you have a_3 x + uf so, dividing P(m) by f^2 gives P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). II. Algebraic integers Now take P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside by f^2(a_1 a_2 a_3) and you have P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). Now consider the case that m, f, and u are algebraic integers, then I have the ratios of algebraic integers: uf/a_1, uf/a_2, and uf/a_3, and now let v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 where the vs and ws are algebraic integers in each case coprime to each other. Making the substitutions I have P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). And I have from before that P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f so (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f as that is the last coefficient from the last term u^3 f, which proves that (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 then I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) but I still have that P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) without contradiction. III. Galois Theory Now in the ring of algebraic integers consider the possibility that a_1/f is not an algebraic integer to see if that leads to a contradiction. First, if a_1/f is not an algebraic integer and w_1 is, they obviously cannot be equal. But I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) and P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) so far simultaneously true without contradiction, so there must exist z_1, z_2, and z_3 such that w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 and z_1 z_2 z_3 = 1, where algebraically the zs are units, but z_1, z_2 and z_3 are not units in the ring of algebraic integers. But, if z_1, z_2 and z_3 are units algebraically but are not algebraic integers, what does that mean? It means they are not algebraic integers. Nothing more. Consider properties that cover rings like the ring of algebraic integers, and the ring of integers itself, and it can be shown that two are required: 1. No rational in the ring except 1 and -1 is a unit. 2. No non-unit number within the ring is a factor of any two integers that are coprime to each other in the ring of integers. So there are two basic factor properties of rings like the ring of integers and the ring of algebraic integers. Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where properties 1. and 2. hold, which includes the ring of algebraic integers, and I have named that ring, the ring of objects. Since they may be in such a ring, there is no contradiction with previous results. However, at least some believe that Galois Theory shows how factors can distribute within roots, and these results contradict with that belief. Unfortunately much of modern math in the area of algebraic number theory relies on the false belief, so that major works thought to be true, are in fact false. But given the simplicity of the argument presented here, the conclusions follow without room for doubt for the mathematician. Given what Ive seen up to this time, it seems unlikely that the math community will just suddenly give in to what is true, but some of you still may be saved by knowing it. Those not mathematicians, of course, may just ignore the truth. But if any of you are, you will not. James Harris === Subject: Re: Refuting argument, algebraic number theory > The following post steps through the math without much commentary, > until the end. The math proof is very basic with mostly simple > algebraic manipulations, and the bedrock concepts--like constants are > constant--are basic as well. > It is enough for those who rely on what can be mathematically proven > to be correct. > But those unlike them, the truth is never enough. ___JSH > ------------------------------------------------------------- --------- > I. Factorization > What follows is in a commutative ring. > Let > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) > with the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where f is nonzero and note that at > m=0, P(0) = u^2 f^2(3x + uf), > which, of course, does not vary as m varies. > So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) Nonsense! The expression on the left of the Ô= sign does NOT equal the expression on the right. You previously defined the left side to be P(m) (actually one possible factorization of P(m): (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) is P(m) while the right side is NOT P(m), it is P(0): u^2 f^2 (3x + uf) is P(0) Now if you had correctly written a_1, a_2, a_3 as a_1(m), a_2(m), and a_3(m) you would not have fallen into this trap. KeithK > which requires that at least two of the as must equal 0 at m=0, while > one equals 3. > Now let f be coprime to 3 and x so that 3x + uf is coprime to f. > Since, at m=0, two of the as must equal 0, its convenient to just > arbitrarily select a_1 and a_2 as those two. > Then you have uf for the first, uf for the second and 3x + uf for the > third as terms that do not vary when m varies. > Im going to emphasize that point because its important. > Now then, if m=1, what are the *constant* terms? > They are uf, for the first, uf for the second, and 3x + uf for the > third. > Thats logical because they do not vary with m, so if m=1003909273, > what are the constant terms? > They are uf, for the first, uf for the second, and 3x + uf for the > third. > Now divide f^2 from both sides, which gives > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f > P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and you note that P(0)/f^2 = u^2(3x + uf), which means that now your > constant terms are u for the first, u for the second and 3x + uf for > the third. > Now then, if m=1, what are the constant terms now? > They are u for the first, u for the second, and 3x + uf for the third. > If m = 2938479378, what are the constant terms now? > They are u for the first, u for the second, and 3x + uf for the third. > How can the constant terms of the first two go from uf to u? > They must be divided by f. > Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided > from P(m), it is u; therefore, a_1 x + uf is divided by f, and you > have > a_1 x/f + u > and the constant term of a_2 x + uf is uf, but when f^2 is divided > from P(m), it is u; therefore, a_2 x + uf is divided by f, and you > have > a_2 x/f + u > while the constant term of a_3 x + uf is 3x + uf, and after f^2 is > divided off, it is 3x + uf, so you have > a_3 x + uf > so, dividing P(m) by f^2 gives > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). > II. Algebraic integers > Now take > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) > and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside > by f^2(a_1 a_2 a_3) and you have > P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) > and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is > P(m)/f^2 = > (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). > Now consider the case that m, f, and u are algebraic integers, then I > have the ratios of algebraic integers: > uf/a_1, uf/a_2, and uf/a_3, > and now let > v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 > where the vs and ws are algebraic integers in each case coprime to > each other. > Making the substitutions I have > P(m)/f^2 = > (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). > And I have from before that > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f > so > (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f > as that is the last coefficient from the last term u^3 f, which proves > that > (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let > (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 > then I have > P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) > but I still have that > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) > without contradiction. > III. Galois Theory > Now in the ring of algebraic integers consider the possibility that > a_1/f is not an algebraic integer to see if that leads to a > contradiction. > First, if a_1/f is not an algebraic integer and w_1 is, they obviously > cannot be equal. > But I have > P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) > and > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) > so far simultaneously true without contradiction, so there must exist > z_1, z_2, and z_3 such that > w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 > and z_1 z_2 z_3 = 1, > where algebraically the zs are units, but z_1, z_2 and z_3 are not > units in the ring of algebraic integers. > But, if z_1, z_2 and z_3 are units algebraically but are not algebraic > integers, what does that mean? > It means they are not algebraic integers. Nothing more. > Consider properties that cover rings like the ring of algebraic > integers, and the ring of integers itself, and it can be shown that > two are required: > 1. No rational in the ring except 1 and -1 is a unit. > 2. No non-unit number within the ring is a factor of any two integers > that are coprime to each other in the ring of integers. > So there are two basic factor properties of rings like the ring of > integers and the ring of algebraic integers. > Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where > properties 1. and 2. hold, which includes the ring of algebraic > integers, and I have named that ring, the ring of objects. > Since they may be in such a ring, there is no contradiction with > previous results. > However, at least some believe that Galois Theory shows how factors > can distribute within roots, and these results contradict with that > belief. > Unfortunately much of modern math in the area of algebraic number > theory relies on the false belief, so that major works thought to be > true, are in fact false. > But given the simplicity of the argument presented here, the > conclusions follow without room for doubt for the mathematician. > Given what Ive seen up to this time, it seems unlikely that the math > community will just suddenly give in to what is true, but some of you > still may be saved by knowing it. > Those not mathematicians, of course, may just ignore the truth. > But if any of you are, you will not. > James Harris === Subject: Re: Refuting argument, algebraic number theory > The following post steps through the math without much commentary, > until the end. The math proof is very basic with mostly simple > algebraic manipulations, and the bedrock concepts--like constants are > constant--are basic as well. > It is enough for those who rely on what can be mathematically proven > to be correct. > But those unlike them, the truth is never enough. ___JSH > ------------------------------------------------------------- --------- > I. Factorization > What follows is in a commutative ring. Which one? > Let > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) Ok, either its in the ring Z[x][m][f][u] or (a strange polynomial ring where you havent defined f,u,x)[m]. > with the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) In what ring is this factorization being done? > where f is nonzero and note that at > m=0, P(0) = u^2 f^2(3x + uf), > which, of course, does not vary as m varies. What do you mean it doesnt vary as m varies? P(0) is just P(0). > So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) > which requires that at least two of the as must equal 0 at m=0, while > one equals 3. Note: this tells us nothing about what happens when m<>0. Do you mean for the as to be functions of m? > Now let f be coprime to 3 and x so that 3x + uf is coprime to f. It would be easier to tell if f is coprime to 3 and x if f and x were defined. > Since, at m=0, two of the as must equal 0, its convenient to just > arbitrarily select a_1 and a_2 as those two. > Then you have uf for the first, uf for the second and 3x + uf for the > third as terms that do not vary when m varies. What do you mean by this? They are independent of m. Also, 3x+uf appears to be a binomial. > Im going to emphasize that point because its important. > Now then, if m=1, what are the *constant* terms? Whichever terms dont have variables in them. > They are uf, for the first, uf for the second, and 3x + uf for the > third. In that case, u,f,x must not be variables. What are they? > Thats logical because they do not vary with m, so if m=1003909273, > what are the constant terms? > They are uf, for the first, uf for the second, and 3x + uf for the > third. > Now divide f^2 from both sides, which gives > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f > P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and you note that P(0)/f^2 = u^2(3x + uf), which means that now your > constant terms are u for the first, u for the second and 3x + uf for > the third. > Now then, if m=1, what are the constant terms now? uf, uf, 3x+uf. You pointed them out above. > They are u for the first, u for the second, and 3x + uf for the third. Perhaps you need to specify what you are looking for constant terms in. > If m = 2938479378, what are the constant terms now? > They are u for the first, u for the second, and 3x + uf for the third. > How can the constant terms of the first two go from uf to u? > They must be divided by f. > Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided > from P(m), it is u; therefore, a_1 x + uf is divided by f, and you > have > a_1 x/f + u > and the constant term of a_2 x + uf is uf, but when f^2 is divided > from P(m), it is u; therefore, a_2 x + uf is divided by f, and you > have > a_2 x/f + u > while the constant term of a_3 x + uf is 3x + uf, and after f^2 is > divided off, it is 3x + uf, so you have > a_3 x + uf > so, dividing P(m) by f^2 gives > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). In what ring? > II. Algebraic integers > Now take > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) > and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside > by f^2(a_1 a_2 a_3) and you have > P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) > and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is > P(m)/f^2 = > (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). > Now consider the case that m, f, and u are algebraic integers, then I > have the ratios of algebraic integers: > uf/a_1, uf/a_2, and uf/a_3, > and now let > v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 > where the vs and ws are algebraic integers in each case coprime to > each other. > Making the substitutions I have > P(m)/f^2 = > (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). > And I have from before that > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f > so > (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f > as that is the last coefficient from the last term u^3 f, which proves > that > (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let > (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 > then I have > P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) > but I still have that > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) > without contradiction. > III. Galois Theory > Now in the ring of algebraic integers consider the possibility that > a_1/f is not an algebraic integer to see if that leads to a > contradiction. So if a_1 = 1, and f=2, 1/2 is an algebraic integer? > First, if a_1/f is not an algebraic integer and w_1 is, they obviously > cannot be equal. > But I have > P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) > and > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) > so far simultaneously true without contradiction, so there must exist > z_1, z_2, and z_3 such that > w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 > and z_1 z_2 z_3 = 1, > where algebraically the zs are units, but z_1, z_2 and z_3 are not > units in the ring of algebraic integers. > But, if z_1, z_2 and z_3 are units algebraically but are not algebraic > integers, what does that mean? > It means they are not algebraic integers. Nothing more. > Consider properties that cover rings like the ring of algebraic > integers, and the ring of integers itself, and it can be shown that > two are required: > 1. No rational in the ring except 1 and -1 is a unit. > 2. No non-unit number within the ring is a factor of any two integers > that are coprime to each other in the ring of integers. Is pi or 1/pi in your ring? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Refuting argument, algebraic number theory The following post steps through the math without much commentary, until the end. The math proof is very basic with mostly simple algebraic manipulations, and the bedrock concepts--like constants are constant--are basic as well. It is enough for those who rely on what can be mathematically proven to be correct. But for too many others, the truth is never enough. ___JSH ------------------------------------------------------------- --------- I. Factorization What follows is in a commutative ring. Let P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) with the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where f is nonzero and note that at m=0, P(0) = u^2 f^2(3x + uf), which, of course, does not vary as m varies. So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)? (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf) which requires that at least two of the as must equal 0 at m=0, while one equals 3. Now let f be coprime to 3 and x so that 3x + uf is coprime to f. Since, at m=0, two of the as must equal 0, its convenient to just arbitrarily select a_1 and a_2 as those two. Then you have uf for the first, uf for the second and 3x + uf for the third as terms that do not vary when m varies. Im going to emphasize that point because its important. Now then, if m=1, what are the *constant* terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Thats logical because they do not vary with m, so if m=1003909273, what are the constant terms? They are uf, for the first, uf for the second, and 3x + uf for the third. Now divide f^2 from both sides, which gives P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you note that P(0)/f^2 = u^2(3x + uf), which means that now your constant terms are u for the first, u for the second and 3x + uf for the third. Now then, if m=1, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. If m = 2938479378, what are the constant terms now? They are u for the first, u for the second, and 3x + uf for the third. How can the constant terms of the first two go from uf to u? They must be divided by f. Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided from P(m), it is u; therefore, a_1 x + uf is divided by f, and you have a_1 x/f + u and the constant term of a_2 x + uf is uf, but when f^2 is divided from P(m), it is u; therefore, a_2 x + uf is divided by f, and you have a_2 x/f + u while the constant term of a_3 x + uf is 3x + uf, and after f^2 is divided off, it is 3x + uf, so you have a_3 x + uf so, dividing P(m) by f^2 gives P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf). II. Algebraic integers Now take P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside by f^2(a_1 a_2 a_3) and you have P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3) and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3). Now consider the case that m, f, and u are algebraic integers, then I have the ratios of algebraic integers: uf/a_1, uf/a_2, and uf/a_3, and now let v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_3 where the vs and ws are algebraic integers in each case coprime to each other. Making the substitutions I have P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3). And I have from before that P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f so (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f as that is the last coefficient from the last term u^3 f, which proves that (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3 then I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) but I still have that P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) without contradiction. III. Galois Theory Now in the ring of algebraic integers consider the possibility that a_1/f is not an algebraic integer to see if that leads to a contradiction. First, if a_1/f is not an algebraic integer and w_1 is, they obviously cannot be equal. But I have P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3) and P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf) so far simultaneously true without contradiction, so there must exist z_1, z_2, and z_3 such that w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3 and z_1 z_2 z_3 = 1, where algebraically the zs are units, but z_1, z_2 and z_3 are not units in the ring of algebraic integers. But, if z_1, z_2 and z_3 are units algebraically but are not algebraic integers, what does that mean? It means they are not algebraic integers. Nothing more. Consider properties that cover rings like the ring of algebraic integers, and the ring of integers itself, and it can be shown that two are required: 1. No rational in the ring except 1 and -1 is a unit. 2. No non-unit number within the ring is a factor of any two integers that are coprime to each other in the ring of integers. So there are two basic factor properties of rings like the ring of integers and the ring of algebraic integers. Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where properties 1. and 2. hold, which includes the ring of algebraic integers, and I have named that ring, the ring of objects. Since they may be in such a ring, there is no contradiction with previous results. However, at least some believe that Galois Theory shows how factors can distribute within roots, and these results contradict with that belief. Unfortunately much of modern math in the area of algebraic number theory relies on the false belief, so that major works thought to be true, are in fact false. But given the simplicity of the argument presented here, the conclusions follow without room for doubt for the mathematician. Given what Ive seen up to this time, it seems unlikely that the math community will just suddenly give in to what is true, but some of you still may be saved by knowing it. Those not mathematicians, of course, may just ignore the truth. But if any of you are, you will not. James Harris === Subject: Re: Refuting argument, algebraic number theory > But for too many others, the truth is never enough. ___JSH > But those unlike them, the truth is never enough. ___JSH That is the only difference. Perhaps you are unclear on the concept of fixing mistakes? When people point out problems in your posts, you should fix the areas with the problems, not simply change the phrasing of the preamble. I dont recall anyone finding a problem with but those unlike them, the truth is never enough. -- --Tim Smith === Subject: Re: Refuting argument, algebraic number theory >The following post steps through the math Oh crap. Time to update my kill file again. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Refuting argument, algebraic number theory > I. Factorization > What follows is in a commutative ring. > Let > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) Is m an element of the ring, so that you are talking about P(m) as an element of your commutative ring, or are you talking about P(m) as an element in the ring of polynomials that is associated with your commutative ring? Are f, m, and x arbitrary elements of the ring? See...right here, in the first freaking section of your argument, you go off without saying what the you are doing. Thats not how to write mathematics. -- --Tim Smith === Subject: Re: Refuting argument, algebraic number theory > The following post steps through the math without much commentary, > until the end. The math proof is very basic with mostly simple > algebraic manipulations, and the bedrock concepts--like constants are > constant--are basic as well. > It is enough for those who rely on what can be mathematically proven > to be correct. > But those unlike them, the truth is never enough. ___JSH > ------------------------------------------------------------- --------- > I. Factorization > What follows is in a commutative ring. Im not sure what that means. Can you please define ring. And by the way, do you mean that the coefficents are in some particular commutative ring? Or that the roots of your polynomial lie in some ring? === Subject: Implicit Diff. Im probably going to have a couple of questions today, but heres the one currently getting me: A person walks along a straight sidewalk at a constant speed of 8 feet/second. A spotlight is trained on the person from across the street, which is 20 feet wide. How fast is the spotlight turning (in radians/second) when the distance between the person and the spotlight is 52 feet? Ive done this one a few times over, and below is the version that I feel probably came closest to getting it right - I felt confident with the math: A = 20. dA = 0. C = 52. dC = 7.3846. B = 38. dB = 8. c^2 = a ^ 2 + b^ 2 2704 = 400 + b^2 48 = b c^2 = a^2 + b^2 2CdC = 2AdA + 2BdB 104dC = 16B dC = .1538461538B dC = 7.384615385 My diagram, as best I can recreate in ASCII, looks like this: Man | Side | Side C. 52 feet. B | ----- Spotlight Side A, 20 ft. the angle of the spotlight). Going for implicit diff, I get: (CdB - BdC) / C^2 = cosTheta dTheta 52(8) - 48(7.3846) / 2704 = cosTheta dTheta .0227585799 = cosThetadTheta cosTheta = A/C = 20/52 = .3846153846 .0227585799 = .3846153846 dTheta .0591723077 = dTheta And thats the right answer. You know, typing it out carefully helped me see the mistake Id been making. itll be of help to someone else doing a similar problem, so Im letting it go through. === Subject: Re: Implicit Diff. > ... > My diagram, as best I can recreate in ASCII, looks like this: > Man > | > Side | Side C. 52 feet. > B | > ----- Spotlight > Side A, 20 ft. You should do that sort of thing in a constant width font. === Subject: This one, I actually cant solve. The circumference of a sphere was measured to be 83 cm with a possible error of 0.7 cm. Use differentials to estimate the maximum error in the calculated surface area. Estimate the relative error in the calculated surface area. Okay, so, I /did/ figure out that the maximum error in the calculated surface area is 36.98744. dA / dR approx. deltaA/deltaR 8PiR = deltaA/deltaR 8PiR = deltaA / .111408 [ It took me a while to figure out that I should divide the .7 by 2Pi, if I was going to be dividing the circum by 2 Pi to get the radius] 8Pi(13.2098)(.111408) = deltaA 36.98744 = deltaA. However, Im making no headway solving for the relative error. The relative error, IIRC, is f(x)/f(x). So far, Ive tried: 2/r (as 8PiR / 4PiR^2 reduces to 2/R), Ive tried 13.2098 / (13.2098 +.111408), 13.2098^2 (13.2098 +.111408)^2, 8Pi(13.2098) / 4Pi(13.2098+.111408)^2. The most straightforward one - 2/R - is the one that I presumed would actually be correct, since it fits the definition of f/f exactly, but none of the above have yielded the correct answer. Any pointers in the right direction would be appreciated. === Subject: Re: This one, I actually cant solve. alt.math.undergrad: >The circumference of a sphere was measured to be 83 cm with a possible error >of 0.7 cm. ...Estimate the relative error in the calculated surface area. >... Im making no headway solving for the relative error. The relative >error, IIRC, is f(x)/f(x). Correct. Well, actually not quite. You want df(x)/f(x) or f(x)dx/f(x) to be precise -- the differential of the function divided by the value of the function. And its usually easier to find the relative error if you stay in letters rather than numbers until the very end. C = 2*pi*r S = 4*pi*r^2 = (1/pi)C^2 dS = [2C/pi] dC dS/S = (2C/pi) * dC / [(1/pi)C^2] = 2C*dC/C^2 = 2*dC/C = 2*0.7/83 = about 1.69% >So far, Ive tried: 2/r (as 8PiR / 4PiR^2 reduces to 2/R), Ive tried >13.2098 / (13.2098 +.111408), 13.2098^2 (13.2098 +.111408)^2, 8Pi(13.2098) / >4Pi(13.2098+.111408)^2. I havent looked thoroughly into your algebra, but Ill bet theres a mistake in converting between radius and circumference. Remember that what is measured is not radius but circumference, so your relative error is relative to the (surface area as function of the) circumference and not the radius. (Thats true in general. In this case it shouldnt matter, since C is proportional to r.) >The most straightforward one - 2/R - is the one that I presumed would >actually be correct, since it fits the definition of f/f exactly, but >none of the above have yielded the correct answer. Any pointers in the right >direction would be appreciated. Hmm ... lets see if I can do this in terms of r: S = 4*pi*r^2; dS = 8*pi*r*dr; dS/S = 2*dr/r But now you need dr/r in terms of dC/C. Since C is a linear function of r, theyre the same, 0.7/83. 2 times that gives the figure I mentioned above. Does your textbook disagree? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: This one, I actually cant solve. days. My association with the Department is that of an alumnus. >The circumference of a sphere was measured to be 83 cm with a possible error >of 0.7 cm. Use differentials to estimate the maximum error in the calculated >surface area. >Estimate the relative error in the calculated surface area. >Okay, so, I /did/ figure out that the maximum error in the calculated >surface area is 36.98744. >dA / dR approx. deltaA/deltaR >8PiR = deltaA/deltaR >8PiR = deltaA / .111408 [ It took me a while to figure out that I should >divide the .7 by 2Pi, if I was going to be dividing the circum by 2 Pi to >get the radius] >8Pi(13.2098)(.111408) = deltaA >36.98744 = deltaA. Im going to rework the problem, because Im having trouble figuring out what you did. Notice that I explain what Im doing to make it easier for others to figure out what I was thinking as I was doing it. The area in terms of the radius is A(r) = 4pi*r^2; but we want the area have that f(r)^2 = 4pi^2*r^2, so A(f) = f(r)^2/pi. That is: if we square the circumference and divide by 4pi, we get the area. So we have A(c) = c^2/pi as a function that gives the area A in terms of the circumference c. Now: the error in A is Delta(A), which is approximately equal to the differencia dA. The differential is dA = A(c)*dc, where dc is the differencial in c. Here, A(c) = 2c/pi, and -0.7 <= dc <= 0.7. So A(83)*(-0.7) <= dA <= A(83)*0.7. -166.2*0.7/pi <= dA <= 166.2*0.7/pi -116.2/pi <= dA <= 116.2/pi So the total error is bounded by +/- (116.2)/pi which is approximately +/- 36.9876. (the discrepancy between your solution and mine is that you are approximating twice: once when you approximate .7/2pi by .111408 (a bit too small), and then again when you approximate at the end). >However, Im making no headway solving for the relative error. The relative >error, IIRC, is f(x)/f(x). The relative error is approximately equal to the absolute error divided by the quantity calculated. That is: you have found that the absolute error is +/- 36.9876 square centimeters. If you use the value of 83 centimeters that you found to begin with to calculate the area, you obtain A(83) = (83)^2/pi = 6889/pi. So we have (-116.2/pi)/(6889/pi) <= dA/A <= (116.2/pi)/(6889/pi) -116.3/6889 <= dA/A <= 116.2/6889 Since 116.3/6889 is approximately 0.01687, this means that the relative error is bounded approximately by -0.01687 <= dA/A <= 0.01687 or a percentage error of about plus or minus 1.687%. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: This one, I actually cant solve. > The circumference of a sphere was measured to be 83 cm with a possible > error of 0.7 cm. Use differentials to estimate the maximum error in the > calculated surface area. Nm. I came back to it after doing some other problems, and I finally managed to get it. === Subject: Units, and algebraic integers Consider x^2 + 2^{sqrt(2)}x + 1 = 0. It doesnt have algebraic integer roots because 2^{sqrt(2)} is transcendental, so the roots cant be roots of a monic polynomial with integer coefficients. So they are not algebraic integers. But despite not being algebraic integer, they are properly units, i.e. factors of 1. That example bugs me because it highlights how weird things are in algebraic number theory that the units which are the roots of that equation are just ignored. Either theyre tossed into the grab bag of complex numbers, or the other of transcendentals. Gauss started the ball rolling considering numbers of the form a+bi, where Ôa and Ôb are integers, now called gaussian integers in his honor. with integer coefficients, but then stopped. Why? Some of you may turn to complex numbers but Gauss new of complex numbers and transcendentals when he started studying gaussian integers. So why did he freaking bother? Why did he screw around with those numbers? Why didnt progress continue such that the roots of x^2 + 2^{sqrt(2)}x + 1 = 0 would properly be included in some integer-type category? My guess is that the ring of algebraic integers--because of the problems Ive highlighted--were like a pleasant drug. It seemed easy to work in them, and big results seemed within reach, except they were wrong. People settled for what was easy because it appeared to work. They didnt really progress the field, but claimed they did. By focusing on roots of monic polynomials with integer coefficients, you can think youre proving a lot, when you make assumptions about numbers not roots of monic polynomials with integer coefficients, but what about the roots of x^2 + 2^{sqrt(2)}x + 1 = 0? My work shows the limitations with the arbitrary focus on roots of monic polynomials with integer coefficients. It is such a bad focus that the problems can be shown with basic algebra. Its so easy to show that Ive posted it yet again. The proof almost looks like some algebra homework problem from high school. Its that easy. Ignoring it just makes you a fool, if youre young. But it has been around for so long that a lot of older people who have no future starting over have good reason to block my work and act like there is no problem. So they break their own rules. I put a paper through formal peer review--at least Southwest Journal of Pure and Applied Mathematics claims they do formal peer review--and some sci.mathers manage to get it withdrawn immediately with a few emails. Think thats normal? Do you really think math editors care a lot about what sci.mathers say? Older people like Wiles or Ribet no longer have a future. Theyre beyond learning new things, getting over the hearbreak of error, and starting over. But many of you are just learning. You have the future ahead of you, and the possibility of real brilliance in mathematics. The older men in the club you wish to join clearly see themselves as having nothing to gain from the truth. Like yeah, sure, Andrew Wiles wants to face the world, or even Ribet or Taylor, if the truth is fully known. Right. Who would? Would you? Why should they tell you the truth? Whats in it for them? James Harris === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. > So they are not algebraic integers. > But despite not being algebraic integer, they are properly units, i.e. > factors of 1. In what ring? > That example bugs me because it highlights how weird things are in > algebraic number theory that the units which are the roots of that > equation are just ignored. No, they arent. You just have to carefully define *which ring* roots exist in, and then determine whether they are units *in that ring*. > Either theyre tossed into the grab bag of complex numbers, or the > other of transcendentals. Not necessarily. They could be used as part of the basis for a particular ring. > Gauss started the ball rolling considering numbers of the form a+bi, > where Ôa and Ôb are integers, now called gaussian integers in his > honor. > with integer coefficients, but then stopped. > Why? Why do you think they stopped? > Some of you may turn to complex numbers but Gauss new of complex > numbers and transcendentals when he started studying gaussian > integers. > So why did he freaking bother? Why did he screw around with those > numbers? > Why didnt progress continue such that the roots of > x^2 + 2^{sqrt(2)}x + 1 = 0 > would properly be included in some integer-type category? It quite easily can be. > My guess is that the ring of algebraic integers--because of the > problems Ive highlighted--were like a pleasant drug. It seemed easy > to work in them, and big results seemed within reach, except they were > wrong. > People settled for what was easy because it appeared to work. They > didnt really progress the field, but claimed they did. > By focusing on roots of monic polynomials with integer coefficients, > you can think youre proving a lot, when you make assumptions about > numbers not roots of monic polynomials with integer coefficients, but > what about the roots of > x^2 + 2^{sqrt(2)}x + 1 = 0? If youre talking about algebraic integers, the above roots are simply not a consideration. If you want to examine such roots, feel free to do so. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Units, and algebraic integers > Older people like Wiles or Ribet no longer have a future. Theyre > beyond learning new things, getting over the hearbreak of error, and > starting over. Pffff! Unlike James Harris, Wiles did get over the heartbreak of error. He experienced that heartbreak, and had to start over on a key part of his initial attempt at FLT, after his first public claim of his proof. This error jeopardized his lifes work. What did he do when the error was pointed out? Did he react like the paranoid baby James Harris? Did he call his critics liars? Did he pathetically lie by claiming that he was really just working in some other meaningless magic ring where the error could be disposed of in fantasy land? Of course not, since, unlike James Harris, Wiles is a real mathematician. Instead, he went back and worked on it for another year, collaborating with Taylor, finally getting it all correct. === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. > So they are not algebraic integers. > But despite not being algebraic integer, they are properly units, i.e. > factors of 1. As has been repeatedly pointed out to you in the past, and as you at one point seemed to have taken on board, the concept of Ôfactor only has meaning if you specify the set of numbers you are discursing correct this if your vague hand-waving has given me the wrong impression - that you are using Ôx is properly [a] unit to mean Ôthere exists a real y such that xy = 1 By this definition, ALL non-zero reals are Ôproperly units; so to say some number is Ôproperly [a] unit is to say virtually NOTHING AT ALL. Is this what you want? -- Larry Lard Replies to group please === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. > So they are not algebraic integers. True and not algenraic numbers either. > But despite not being algebraic integer, they are properly units, i.e. > factors of 1. Why would you expect x,y in xy =1 to be algebraic integers - let alone rational integers ? (pi)*(1/pi) = 1 so what ? These numbers are called integers because they are generalisations of the rational integers. Z is contained in the algebraic integrs so if, for example, x^3 + y^3 = z^3 is impossible over some integer ring it is impossible over Z. Whether or not a particular non-integer can be inverted in some ring or field is irrelevant to number theory. > That example bugs me because it highlights how weird things are in > algebraic number theory that the units which are the roots of that > equation are just ignored. Why would it bug you ? One has no relevance to the other. > Gauss started the ball rolling considering numbers of the form a+bi, > where Ôa and Ôb are integers, now called gaussian integers in his > honor. In fact, it was Euler who started the ball rolling by realising that numbers of the form (x*sqrt(2) +y*sqrt(5)) could, under certain circumstances, be manipulated like rational integers. If A,B are relatively prime integers (not units) and AB = C^3 then you can deduce that A and B are both cubes He applied this approach to 2x^2 - 5 = z^3 but the method did not produce the solution x = 4. Euler admitted his method failed but was unable to say why. This shows that just trusting the maths is not a viable strategy and leads to fatal contradictions. === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. > So they are not algebraic integers. > But despite not being algebraic integer, they are properly units, i.e. > factors of 1. > That example bugs me because it highlights how weird things are in > algebraic number theory that the units which are the roots of that > equation are just ignored. Since the roots of an equations with transcendental coefficients need not be algebraic numbers at all, they need not be a part of, or have any effect on, algebraic number theory. Since JSH has not yet sorted out algebraic number theory successfully, he should do that before venturing further. === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. How do you know 2^sqrt(2) is transcendental? Yes, I know that the Gelfond-Schneider theorem says so, but have you looked into the proof of that theorem, to make sure it does not rely ultimately on Galois theory or some other area that you claim is wrong? The proof of the Gelfond-Schneider theorem is available in Gelfonds book Transcendental and Algebraic Numbers, if you want to check it out. There is an old Dover edition that is cheap ($10 at http://tinyurl.com/59buj), and a new Dover Phoenix edition that is $37.50 (http://store.yahoo.com/doverpublications/0486495264.html). -- --Tim Smith === Subject: Re: Units, and algebraic integers Consider x^2 + 2^{sqrt(2)}x + 1 = 0. It doesnt have algebraic integer roots because 2^{sqrt(2)} is transcendental, so the roots cant be roots of a monic polynomial with integer coefficients. So they are not algebraic integers. But despite not being algebraic integers, they are properly units, i.e. factors of 1. Yet notice how theyre categorized. Either theyre tossed into the grab bag of complex numbers, or the other of transcendentals. Think about it. Gauss started the ball rolling considering numbers of the form a+bi, where Ôa and Ôb are integers, now called gaussian integers in his honor. He knew of both complex numbers and transcendentals, so why bother? with integer coefficients, but then progress stopped so that I can give you this example over a hundred years later: x^2 + 2^{sqrt(2)}x + 1 = 0 Why? Why didnt progress continue such that the roots of x^2 + 2^{sqrt(2)}x + 1 = 0 would properly be included in some integer-type category? My guess is that the ring of algebraic integers--because of the problems Ive highlighted--was like a pleasant drug. It seemed easy to work in the ring, and get big results. Trouble is, theyre not right. People settled for what was easy because it appeared to work. They didnt really progress the field, but claimed they did. There was no real progress in this area since Gauss. Thats over a hundred years of stagnation with false claims of results. By focusing on roots of monic polynomials with integer coefficients, you can think youre proving a lot, when you make assumptions about numbers not roots of monic polynomials with integer coefficients, but what about the roots of x^2 + 2^{sqrt(2)}x + 1 = 0? My work shows the limitations with the arbitrary focus on roots of monic polynomials with integer coefficients. It is such a bad focus that the problems can be shown with basic algebra. So people argue with me, not because Im wrong--after all, you can check the math for yourself--but because Im right, and they dont like whats mathematically correct because of the social implications. But whats mathematically correct is so easy to show that Ive posted it yet again. The proof almost looks like some algebra homework problem from high school. Its that easy. Ignoring it just makes you a fool, if youre young, but might seem smart if youre older, and either a grad student with a lot invested or even more so if youre an actual professor. Besides, its a problems thats freaking over a hundred years old! Maybe professors feel they arent responsible! And starting over is so hard. Can you imagine? Having to unlearn and then re-learn, as the mathematical foundations are set aright? Easier to block and act like there is no problem. So they break their own rules. I put a paper through formal peer review--Southwest Journal of Pure and Applied Mathematics claims they do formal peer review--and some sci.mathers managed to get it withdrawn immediately with a few emails. Think thats normal? Do you really think math editors typically care a lot about what sci.mathers say? Older people like Wiles or Ribet probably dont have a future. Theyre beyond learning new things, getting over the hearbreak of error, and starting over. Its hard to start over, to get over the emotions, and accept truths that dont sit well, even in mathematics. But you are different. Many of you are just learning. You have the future ahead of you, and the possibility of real brilliance in mathematics. The older men in the club you wish to join clearly see themselves as having nothing to gain from the truth. Like yeah, sure, Andrew Wiles wants to face the world, or even Ribet or Taylor, if the truth is fully known. Right. Who would? Would you? Why should they tell you the truth? Whats in it for them? James Harris === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. > So they are not algebraic integers. > But despite not being algebraic integers, they are properly units, > i.e. factors of 1. So theyre units in some larger ring? So what? Do you mean to suggest that every number thats a unit in some ring must be considered while doing work with algebraic numbers? Do you think that has any real value? > Yet notice how theyre categorized. Yeah, theyre categorized, theyre second-class numbers. > Either theyre tossed into the grab bag of complex numbers, or the > other of transcendentals. So what? Theorems about algebraic numbers make NO statements about the transcendental numbers. > Think about it. Gauss started the ball rolling considering numbers of > the form a+bi, where Ôa and Ôb are integers, now called gaussian > integers in his honor. > He knew of both complex numbers and transcendentals, so why bother? > with integer coefficients, but then progress stopped so that I can > give you this example over a hundred years later: > x^2 + 2^{sqrt(2)}x + 1 = 0 > Why? > Why didnt progress continue such that the roots of > x^2 + 2^{sqrt(2)}x + 1 = 0 > would properly be included in some integer-type category? They are integers in the appropriate field. Take K = Q(2^sqrt(2)), the smallest field containing the rational numbers and 2^sqrt(2). This field contains Z[2^sqrt(2)], and K is its field of fractions. The roots of the above quadratic are integers in the splitting field of the quadratic over K, that is, they are roots of a monic polynomial with coefficients in Z[2^sqrt(2)]. > My guess is that the ring of algebraic integers--because of the > problems Ive highlighted--was like a pleasant drug. > It seemed easy to work in the ring, and get big results. > Trouble is, theyre not right. Why not produce a result of algebraic number theory that you claim is false, and show that it is false? All your rhetoric and claims are of no use if you cannot display a specific error. The above complaint about the ring of algebraic integers not containing any transcendentals is silly. The algebraic numbers, and the corresponding ring of integers, were defined for the purpose of understanding the nature of solutions of polynomial equations over the rationals, and for that purpose, they are certainly successful. > People settled for what was easy because it appeared to work. They > didnt really progress the field, but claimed they did. Sez you. > There was no real progress in this area since Gauss. Thats over a > hundred years of stagnation with false claims of results. Again with your non-mathematics? Why not display a specific result that you can show to be false? Every one that you have attempted to produce over the past seven years has been shown to be ßawed. Flawed. Just like your arguments. > By focusing on roots of monic polynomials with integer coefficients, > you can think youre proving a lot, when you make assumptions about > numbers not roots of monic polynomials with integer coefficients, but > what about the roots of > x^2 + 2^{sqrt(2)}x + 1 = 0? In the study of algebraic numbers, and in the study of algebraic integers, NO assumptions are made about the transcendentals. Your example shows no ßaw with the algebraic numbers, just as the fact that an automobile is not a tasty desert topping doesnt yield a valid complaint about automobiles. > My work shows the limitations with the arbitrary focus on roots of > monic polynomials with integer coefficients. It is such a bad focus > that the problems can be shown with basic algebra. > So people argue with me, not because Im wrong--after all, you can > check the math for yourself--but because Im right, and they dont > like whats mathematically correct because of the social implications. No, people argue with you because you are wrong. Well, not just that, but also because you are so proud of your ignorance that it amounts to an insult to the concept of human thought. Why would people care about arguing against a correct argument? Your remark about so-called social implications is just your wishful thinking. > But whats mathematically correct is so easy to show that Ive posted > it yet again. You keep posting the same monologue. Its just that: a monologue, devoid of any original mathematical content, and totally lacking in any real insight into what youre doing. You confuse algebraic manipulation with mathematical insight, and your claims about the implications of your work are just exercises in grandiose thinking. > The proof almost looks like some algebra homework problem from high > school. > Its that easy. And just as penetrating as your typical high school algebra homework problem. > Ignoring it just makes you a fool, if youre young, but might seem > smart if youre older, and either a grad student with a lot invested > or even more so if youre an actual professor. Of course. No one can look at your work and fail to see its brilliance! Instead, any disagreement with your work must be due to some sort of corruption on the part of the Old Guard. No matter that you entered this fray with the expressed belief that the world of mathematics is corrupt. No matter that you have denied every single mistake youve had pointed out to you, called those who have pointed those mistakes liars, threatened every dire consequence you could imagine, contacted the FBI, whined, pleaded, and cajoled to get your critics to let up and just accept your work. And now what? Youre hanging out in this newsgroup for undergrads, like a pervert hanging out in the elementary school playground. Candy, anyone? > Besides, its a problems thats freaking over a hundred years old! > Maybe professors feel they arent responsible! But you havent displayed any specific problem. Its all down to your garbled argument that clearly produces ßawed results, results that are in direct contradiction to ordinary arithmetic, and so (the argument) is itself deeply ßawed. You just dont get it. You could have taken an algebra course . Ive read your non- sensical algebra is algebra quips, and theyre irrelevant: you do not understand the underpinnings of algebra, and until you actually put in the time and effort, your work will betray the abysmal ignorance that exists in your mind. > And starting over is so hard. Can you imagine? Having to unlearn and > then re-learn, as the mathematical foundations are set aright? Just like you have asserted on countless occasions: accept my argument, or all of mathematics is doomed. It doesnt matter that each time you claim something specific enough to make some form of computation, the claims youve made have been shown to be in error. You have NEVER produced an error due to mathematics, or algebraic number theory. NEVER. Instead, each so-called error has been shown to be an error of yours. > Easier to block and act like there is no problem. > So they break their own rules. I put a paper through formal peer > review--Southwest Journal of Pure and Applied Mathematics claims they > do formal peer review--and some sci.mathers managed to get it > withdrawn immediately with a few emails. Your paper contained a conclusion that I showed to be incorrect, namely that a pair of numbers, a_i and 5, are coprime in the ring of algebraic integers. I found a common factor in that ring, and showed that this common factor is not a unit in that ring. Thus, the two numbers are not coprime as algebraic integers. As far as I understand, you have admitted as much. You have admitted that your conclusion is incorrect, yet you continue this foolish argument. > Think thats normal? Do you really think math editors typically care > a lot about what sci.mathers say? However, its not the opinion of the readers of sci.math that matter, but the presentation of ßawed arguments as correct. > Older people like Wiles or Ribet probably dont have a future. > Theyre beyond learning new things, getting over the hearbreak of > error, and starting over. Yeah, poor them. Really. Id hate to be in their shoes. > Its hard to start over, to get over the emotions, and accept truths > that dont sit well, even in mathematics. You sound like youre talking about your own predicament: no results, no technique, just rehashed high school algebra with unsupported claims that it says anything new. But wait... heres the candy from that nice man in the trenchcoat: > But you are different. > Many of you are just learning. You have the future ahead of you, and > the possibility of real brilliance in mathematics. But we have to look out for the grownups, dont we: > The older men in the club you wish to join clearly see themselves as > having nothing to gain from the truth. > Like yeah, sure, Andrew Wiles wants to face the world, or even Ribet > or Taylor, if the truth is fully known. Right. Who would? Corrupt, right? The whole establishment is corrupt! Look at them! Every bit as corrupt as you came into this game believing ahead of time. DONT TRUST ANYONE OVER 30! > Would you? Why should they tell you the truth? Whats in it for > them? Innuendo. No facts. No argument, other than the same old assertions. Youre inviting the kids to take a little ride in your car. You have some nice candy, and they can play with your pet doggy. Just a little ride in that car of yours. Candy. Yummm. Itll be our little secret, wont it? You do like candy, dont you? > James Harris Dale. === Subject: Re: Units, and algebraic integers > Consider > x^2 + 2^{sqrt(2)}x + 1 = 0. > It doesnt have algebraic integer roots because 2^{sqrt(2)} is > transcendental, so the roots cant be roots of a monic polynomial with > integer coefficients. Consider x^3 + pi x^2 - pi x - 1 = 0 Note that pi is transcendental, yet this has 1 as a root. Last time I checked, 1 was an algebraic integer. Now, are are right that the roots of x^2 + 2^sqrt(2) x + 1 are not algebraic integers, but your because... clause doesnt actually state the reason for this. > So they are not algebraic integers. > But despite not being algebraic integers, they are properly units, > i.e. factors of 1. sqrt(2) > 1, so 2^sqrt(2) > 2, so (2^sqrt(2))^2 > 4, so the roots of your polynomial are both non-zero real numbers. Since every non-zero real number is a factor of 1, you have not pointed out anything non-trivial. ... > Why didnt progress continue such that the roots of > x^2 + 2^{sqrt(2)}x + 1 = 0 > would properly be included in some integer-type category? Define integer-type category. > So they break their own rules. I put a paper through formal peer > review--Southwest Journal of Pure and Applied Mathematics claims they > do formal peer review--and some sci.mathers managed to get it > withdrawn immediately with a few emails. Jesus Christ, James...get the over it already. Your paper was shot down with simple counterexamples. The journal could have handled it better, but you could have avoided the whole problem by fixing your paper when the problems were pointed out, instead of ignoring them. When the problems were pointed out, did you write to the journal and say There are fatal errors in my paper. Here are the corrections and give them the fixes? Nope. Instead, you submit the paper elsewhere, without fixing it. (Clue for you: thats why no one else is interested in publishing it). ... > Older people like Wiles or Ribet probably dont have a future. > Theyre beyond learning new things, getting over the hearbreak of > error, and starting over. Wiles had errors in his first public FLT proof. He fixed them. Thats the step that always eludes you--fixing errors. -- --Tim Smith