mm-2779 === Subject: Integral with poly. degree 2 If a,b , (a 1. A simple pendulum consists of a mass m swinging at the end of a > massless string of length L. The time T for one complete swing of the > pendulum is known as the period, and if the pendulum swings through only > a small angel, the period is given by T=2(pi)sqrt(L/g) where g is the > acceleration due to gravity. > (a) Show that a small change dL, in the length produces a chane in the > period dt satisfying dT/T = dL/2L What's dT/dL ? > (b)Suppose that the pendulum looses 15 seconds per hour. How should > the length of the pendulum be adjusted? -15 sec / 1 hr = dL/2L > (c)The preceeding formula for the period can be used to measure the > acceleration due to gravity. Assume that the error in measuring the > length L is negligible, and express the error dg in the acceleration of > gravity in terms of the error dT in measuring the period. What's dT/dg ? === Subject: Re: Calculus Help!!! What have you done so far? -- Casey === Subject: Re: Nearing breakthrough...need magic If n = p1^a1 * p2^a2 * ... * pk^ak, where the pi are distinct primes and ai are positive integers, then phi(n)=n*(1-1/p1)*(1-1/p2)*...*(1-1/pk) === Subject: Continuum Hypothesis and Lebesgue measure Suppose you have a subset S of R^n, such that the cardinality of S is strictly less than the cardinality of R^n. Does it follow that the measure of S is 0? If we suppose the Continuum Hypothesis is true, then it does follow, because S must be countable. What if we suppose that the Continuum Hypothesis is false? === Subject: Re: Problem in solving equation with e and log method as well. It's really confusing as the log rules are applied to so many abstract variables, but I finally got it. Below are the steps, hope to be of help to anyone here. Subtitute the equilibrium Q variable into P. q = ln(K1/K3) / (K2 + K4) p = K1e^(-K2q) Method 1 -------------------------- P = K1e^(-K2(ln(K1/K3)/K2+K4))) *ln both sides ln P = ln K1 - K2 (ln(K1/K3)/K2+K4))) ln P (K2+K4) = ln K1 (K2+K4) - K2 (ln(K1/K3)) ln P (K2+K4) = K2lnK1 + K4lnK1 - K2lnK1 + K2lnK3 ln P (K2+K4) = K4lnK1 + K2lnK3 ln P = (K4lnK1 + K2lnK3) / (K2+K4) ln P = (lnK1^K4 + lnK3^K2) / (K2+K4) *exponent both sides e^ln P = e ^ ((ln (K1)^K4 + ln (K3)^K2) / (K2 + K4)) *note: e^ln x = x p = e ^ (ln ((K1^K4)( K3^K2)) ^ ( 1/(K2+K4)) p = K1 ^ (K4/(K2+K4)) K3 ^ (K2/(K2+K4)) -----------------------------------------End Method 1 Method 2 ------------------------- P = K1e^(-K2(ln(K1/K3)/K2+K4))) * Using log rules, bring all the variable left of ln, to become power * x ln y = ln y^x P = K1e ^ (ln(K1/K3) ^ (-K2(K2+K4)^ -1)) * e ^ ln x = x P = K1 (K1/K3) ^ (-K2(K2+K4)^ -1)) * multiply the powers down P = K1 ^ (K4/(K2+K4)) K3 ^ (K2/(K2+K4)) -------------------------- end Method 2 That's all. http://mrbluex.blogspot.com === Subject: Complex roots question How i can use the root of (z-1)^n + (z+1)^n = 0 (which i found to be z = -i*cot(pi*r/2*n) to solve the equation x^3 + 15x^2 + 15x + 1 = 0? Does anyone have any ideas what the question wants me to do? === Subject: Re: Complex roots question >How i can use the root of (z-1)^n + (z+1)^n = 0 (which i found to be z >= -i*cot(pi*r/2*n) to solve the equation x^3 + 15x^2 + 15x + 1 = 0? >Does anyone have any ideas what the question wants me to do? The question seems a little pointless to me since the standard precalculus method works just fine here: The equation has the obvious solution x=-1, and hence has x+1 as a factor. Thus, by long division, you can easily find the other factor and solve for the remaining roots using the quadratic formula. Perhaps the method proposed in the problem provides some extra insight about the solutions? quasi === Subject: Re: Complex roots question >How i can use the root of (z-1)^n + (z+1)^n = 0 (which i found to be z >= -i*cot(pi*r/2*n) to solve the equation x^3 + 15x^2 + 15x + 1 = 0? >Does anyone have any ideas what the question wants me to do? What is the first row of Pascal's triangle that has 15s in it? If you use that power for n and multiply out (z-1)^n + (z+1)^n, what do you get? === Subject: Re: Complex roots question > How i can use the root of (z-1)^n + (z+1)^n = 0 (which i found to be z > = -i*cot(pi*r/2*n) to solve the equation x^3 + 15x^2 + 15x + 1 = 0? > Does anyone have any ideas what the question wants me to do? Well, you might notice that (z-1)^6 + (z+1)^6 = z^6 + 15z^4 + 15z^2 + 1 after expanding the two terms using the Binomial Theorem and observing that half the terms cancel each other. Having noticed that, there's an obvious substitution to apply to the equation that you actually want to solve ... === Subject: Re: Complex roots question >> How i can use the root of (z-1)^n + (z+1)^n = 0 (which i found to be >> z = -i*cot(pi*r/2*n) to solve the equation x^3 + 15x^2 + 15x + 1 = 0? >> Does anyone have any ideas what the question wants me to do? > Well, you might notice that > (z-1)^6 + (z+1)^6 = z^6 + 15z^4 + 15z^2 + 1 > after expanding the two terms using the Binomial Theorem and > observing that half the terms cancel each other. Having noticed > that, there's an obvious substitution to apply to the equation that > you actually want to solve ... I.e. it's the even part of f(z) = (z+1)^6, i.e. (f(z)+f(-z))/2 (you forgot to divide f(z)+f(-z) by 2). More generally, instead of such bisections, one can obtain n-multisections of f(z) via certain linear combinations of f evaluated at n'th roots of unity, e.g. see my prior posts --Bill Dubuque === Subject: Re: Complex roots question Excellent! And here I've been looking at cubics and quadratics. How did you figure out to use ^6? === Subject: Re: Complex roots question > Excellent! And here I've been looking at cubics and quadratics. How did > you figure out to use ^6? This won't be helpful, but ... By recognizing that the cancellation was going to be important and knowing the first few rows of Pascal's Triangle (so that '15' was a big hint) ... === Subject: Re: Complex roots question === Subject: Re: Complex roots question Excellent! And here I've been looking at cubics and quadratics. How did you figure out to use ^6? === Subject: Re: Five Women Out To Dinner Darell and Vincent, Bryan >Need help! >Elizabeth, Gladys, Barbara, Virginia and Gwendolyn each ordered a >different main course. >Steak, ham, chicken, pork, and lamb were the only courses on the menu. >Each woman also ordered mashed potatoes, salad, and a different >vegetable. >The five vegetables were peas, carrots, corn, squash, and spinach. >The woman who had steak and corn sat on the left of Elizabeth and to the >right of Gwendolyn. >The woman who had chicken doesn't play cards, but the woman who had ham, >the woman who had squash, and Barbara played bridge yesterday with >Gwendolyn. >The woman who had squash did not have it with pork; Virginia did not >order pork either. >The woman who had ham was recently married to Elizabeths brother; the >woman who had squash helped Elizabeth wrap her present. >Gladys and Gwendolyns vegetables begin with the same letter as do the >vegetables of Elizabeth and Barbara. >What course and vegetable did each woman have? >Help please. No time for hints; need solution. Explanation optional >> Slow down...this is not an emergency homework answer session. You'll >> take or leave whatever we give you. >> Make a chart, with names up/down and both the entrees and the veggies >> across. Carefully using the information given, places a N or Y in the >> appropriate box. For ex., for [t]he woman who had steak and corn sat on >> the left of Elizabeth and to the right of Gwendolyn this tells you that >> neither Eliz. nor Gwen had steak or corn, so place N on Eliz' and Gwen's >> rows under both steak and corn. Proceed in similar fashion with all the >> information. You may have to re-read the information several times, but >> eventually you get enough N's to determine where the Y's go. (BTW, yes, >> I did solve the problem.) > ... and as soon as you have four Ns in any row or column, you know where > to put a Y. Wherever you have a Y, you know the intersecting categories > are equivalent, so you can copy the pattern of Ns and Ys from the row to > the column and from the column to the row, filling in blank cells that > way. You can find examples of these puzzles in magazines titled Logic > Puzzles. > And after you know how to solve the puzzle via pencil & paper (or their > electronic equivalents), you might want to revisit the same problem using > the Excel spreadsheet program or other software with similar features. > Set it up as an optimization problem and have the Solver function > produce a solution. (Sorry, because of the types of constraints involved, > you probably won't be able to set it up as a set of linear constraints. > And it won't tell you if the solution is unique, as it looks for local > extrema, not global ones.) > -- Vincent Johns Darell and Vincent, > Bryan OK, now what I'd be interested in knowing, if you have time to let us know, - Were you able to solve it? - If not, where did you get stuck? (post parl solution if convenient) Usually these things can be solved in a step-by-step fashion. -- Vincent Johns Please feel free to quote anything I say here. === Subject: Re: Five Women Out To Dinner >> Darell and Vincent, >> Bryan > OK, now what I'd be interested in knowing, if you have time to let us > know, > - Were you able to solve it? > - If not, where did you get stuck? (post parl solution if convenient) > Usually these things can be solved in a step-by-step fashion. > -- Vincent Johns >OK, now what I'd be interested in knowing, if you have time to let us >>know, >> - Were you able to solve it? >> - If not, where did you get stuck? (post parl solution if convenient) >>Usually these things can be solved in a step-by-step fashion. >> -- Vincent Johns Please feel free to quote anything I say here. > My daughter is still working on it, so no credit for Friday's homework. > She'll work on it some more orrow. Will let you know if she gets it. OK. Remember that a lot of the fun (or, from a parent's viewpoint, the educational benefit) of something like this is the process of finding a solution, or a solution method. My first attempts at puzzles like these were kind of clumsy, and I didn't always solve them. I later developed kind of a system, and solved them more easily, but after it became kind of mechanical I found they weren't as entertaining as they'd been at first. If your daughter gets stuck, where it's not obvious what to do next, post that here and you'll probably get some clues. This has happened to me at times -- it's looked as if there's no obvious next step, and I've had to start over (and found that I'd made a mistake). And if you wish, I can tell you how I'd set up Excel to solve it. (But an Excel solution is almost certainly not what her teacher is looking for.) BTW, if you would like to play a computerized version of this kind of puzzle, you might want to download the shareware program Sherlock from http://www.kaser.com/sherwin.html . My wife and I have had a lot of fun with this game. -- Vincent Johns Please feel free to quote anything I say here. === Subject: Re: Five Women Out To Dinner > And after you know how to solve the puzzle via pencil & paper (or their > electronic equivalents), you might want to revisit the same problem using > the Excel spreadsheet program or other software with similar features. > Set it up as an optimization problem and have the Solver function > produce a solution. (Sorry, because of the types of constraints involved, > you probably won't be able to set it up as a set of linear constraints. > And it won't tell you if the solution is unique, as it looks for local > extrema, not global ones.) Funny you mentioned Excel, since I actually used it to solve the problem--but I used it merely as electronic scratch paper, no formulas at all just manually inputting Y and N. -- Darrell === Subject: Re: Five Women Out To Dinner >>And after you know how to solve the puzzle via pencil & paper (or their >>electronic equivalents), you might want to revisit the same problem using >>the Excel spreadsheet program or other software with similar features. >>Set it up as an optimization problem and have the Solver function >>produce a solution. [...] > Funny you mentioned Excel, since I actually used it to solve the > problem--but I used it merely as electronic scratch paper, no formulas at > all just manually inputting Y and N. So do I, sometimes -- it's good at formatting such things -- but for the optimization I described, 1 and 0 work better. -- Vincent Johns Please feel free to quote anything I say here. === Subject: sloppy integration method leads to confusion I was always brought up on what i think is proper integration ie. if you have dy/dx = 3, then you integrate both sides wrt x to get integral sign (dy/dx) . dx = integral 3 dx and then the top dx on the left cancels with the bot bx to give integral sign dy = integral 3 dx leaving y = 3x + c. This is instead of thinking from the first line just to multiply across the dx to give dy = 3dx and then put an integral sign in both sides. However, with some work i have been given they seem to be using a less precise way of doing things and I dont really have time at the moment to learn the proper way (although I may do later). One point of the example sheet says d(Htan(theta)) = w dx next line is H.d(dy/dx) = wdx. (tan theta does = dy/dx in the example) next line is d2y / dx2 = w/H (obviously having divided through by dx) However on a different problem on which Im working I get d[Htan(theta)] = wdx + dx^2 w/L Im quite sure this is right. but using the dodgy integration i get d[dy/dx] = wdx + dx^2 w/L d[dy/dx]/dx = w + dx w/L after having divided through by dx so this leaves d2y/dx2 = w + dx w/L i need to integrate this to get dy/dx and then y. but if you integrate the rhs wrt x you get integral sign w dx + integral sign dx w/L dx and its this right hand integration I cant do ie what is the answer to integral sign dx dx. I know the answer to integral sign dx is x + const. but I cant do it with this extra dx in the expression. I tried thinking to integrate twice but it doesnt lead to the right answer. It also doesnt help because they use dx as a length instead of curly dx and then taking these to the limit to get proper dxes in the integration sign. Any help much appreciated. === Subject: Re: sloppy integration method leads to confusion See reply in uk.education.maths. Please cross-post instead of posting separately to different newsgroups. > I was always brought up on what i think is proper integration ie. > if you have dy/dx = 3, then you integrate both sides wrt x to get integral > sign (dy/dx) . dx = integral 3 dx and then the top dx on the left cancels > with the bot bx to give integral sign dy = integral 3 dx > leaving y = 3x + c. This is instead of thinking from the first line just > to multiply across the dx to give dy = 3dx and then put an integral sign > in both sides. > However, with some work i have been given they seem to be using a less > precise way of doing things and I dont really have time at the moment to > learn the proper way (although I may do later). > One point of the example sheet says d(Htan(theta)) = w dx > next line is H.d(dy/dx) = wdx. (tan theta does = dy/dx in the example) > next line is d2y / dx2 = w/H (obviously having divided through by > dx) > However on a different problem on which Im working I get d[Htan(theta)] = > wdx + dx^2 w/L > Im quite sure this is right. > but using the dodgy integration i get > d[dy/dx] = wdx + dx^2 w/L > d[dy/dx]/dx = w + dx w/L after having divided through by dx > so this leaves d2y/dx2 = w + dx w/L > i need to integrate this to get dy/dx and then y. but if you integrate > the rhs wrt x you get > integral sign w dx + integral sign dx w/L dx and its this right hand > integration I cant do > ie what is the answer to integral sign dx dx. > I know the answer to integral sign dx is x + const. but I cant do it with > this extra dx in the expression. I tried thinking to integrate twice but > it doesnt lead to the right answer. > It also doesnt help because they use dx as a length instead of curly dx > and then taking these to the limit to get proper dxes in the integration > sign. > Any help much appreciated. === Subject: Re: sloppy integration method leads to confusion === Subject: sloppy integration method leads to confusion > However, with some work i have been given they seem to be using a > less precise way of doing things and I dont really have time at the > moment to learn the proper way (although I may do later). That's physics. It could be worse such as equations of motion written by biologists or even much worse such as statistics written by psychologists. > One point of the example sheet says d(Htan(theta)) = w dx h and w are constants and t (theta) and x are independent variables. d(h.tan t) / dx = w (1) d(tan t)/dx = w/h (2) > next line is H.d(dy/dx) = wdx. (tan theta does = dy/dx in the > example dy/dx = tan t (3) > next line is d2y / dx2 = w/H (obviously having divided through > by dx) That's absurd. Do you mean d^2 y / (dx)^2 ? Substituting (3) into (2) d^2 y / (dx)^2 = w/h > However on a different problem on which Im working I get > d[Htan(theta)] = wdx + dx^2 w/L (a) dx^2 = (dx)^2 /= d(x^2) ? > Im quite sure this is right. > but using the dodgy integration i get > d[dy/dx] = wdx + dx^2 w/L (b) You naughty magician, you made h disappear. That however isn't dodgy integration, it's substituting (3) into (a) The dodgy integration is (a). Doesn't it leap out at you, that w dx + w/L (dx)^2 is the sum of a single integral plus a double integral? > d[dy/dx]/dx = w + dx w/L after having divided through by dx > so this leaves d2y/dx2 = w + dx w/L Again that notation is incoherence reeked forth of the babbles of pdf or word processor formatting. > i need to integrate this to get dy/dx and then y. but if you > integrate the rhs wrt x you get > integral sign w dx + integral sign dx w/L dx and > its this right hand integration I cant do > ie what is the answer to integral sign dx dx. It's a double integral. > I know the answer to integral sign dx is x + const. but I cant do it > with this extra dx in the expression. I tried thinking to integrate > twice but it doesnt lead to the right answer. Go back to (a), there's a conceptual problem there. > It also doesnt help because they use dx as a length instead of curly > dx and then taking these to the limit to get proper dxes in the > integration sign. > Any help much appreciated. Sure why not take it to the limit, the (dx)^2 was 2nd order approximation which upon first order approximation dy for dy/dx, leaves the vanishing dx. ---- === Subject: point again Let ABC be a triangle with incentre I and inradius r. If P is an arbitrary point inside ABC prove that at least one of AP, BP and CP is not smaller than 2r. Johnek === Subject: More complex goodness! The question is as follows: ---- Find all the roots, real and complex, of the equation z^3 - 1 = 0. If w is one of the complex roots, prove that 1 + w + w^2 = 0. Find the sums of the following series: S1 = 1 + x^3/3! + x^6/6! + ... S2 = x + x^4/4! + x^7/7! + ... S3 = x^2/2! + x^5/5! + x^8/8! ... Hint: note that S1 + S2 + S3 = e^x and calculate e^(wx) and e^((w^2)x) --- I managed to do the whole first bit of the question (finding the roots and doing the proof). What I need help with is the series bit - I have no idea how to use the hint. I tried doing something like the following, which gives you a series with 1 + w + w^2 as coefficients, (this sum is of course part of the question), but I don't see how this helps me: e^(wx) = (S1 + S2 + S3)^w = 1 + wx + w^2*x^2/2! + ... Does anybody have any ideas? === Subject: Re: More complex goodness! > Find all the roots, real and complex, of the equation z^3 - 1 = 0. If w > is one of the complex roots, prove that 1 + w + w^2 = 0. Find the sums > of the following series: > S1 = 1 + x^3/3! + x^6/6! + ... > S2 = x + x^4/4! + x^7/7! + ... > S3 = x^2/2! + x^5/5! + x^8/8! ... > Hint: note that S1 + S2 + S3 = e^x and calculate e^(wx) and e^((w^2)x) > --- > I managed to do the whole first bit of the question (finding the roots > and doing the proof). What I need help with is the series bit - I have > no idea how to use the hint. > I tried doing something like the following, which gives you a series > with 1 + w + w^2 as coefficients, (this sum is of course part of the > question), but I don't see how this helps me: > e^(wx) = (S1 + S2 + S3)^w = 1 + wx + w^2*x^2/2! + ... e^{wx} = 1 + wx + w^2x^2/2 + x^3/3! + wx^4/4! + w^2x^5/5! + x^6/6! + ... e^{w^2x} = 1 + w^2x + wx^2/2 + x^3/3! + w^2x^4/4! + wx^5/5! + x^6/6! + ... So e^{wx} = S1 + w*S2 + w^2*S3 and e^{w^2x} = S1 + w^2*S2 + w*S3. Now, all you have to do is to solve a linear system with three equations and three unknowns in order to determine S1, S2, and S3. === Subject: Re: More complex goodness! <3t48ieFqpvq4U1@individual.net> === Subject: A Note on Goedel's Theorem If Goedel's Theorem is valid, then it is included in _P_, and is therefore fallacious. -- Let no one post here who does not know the _Laws of Form_. === Subject: Re: A Note on Goedel's Theorem > If Goedel's Theorem is valid, then it is included in _P_, and is therefore > fallacious. > -- > Let no one post here who does not know the _Laws of Form_. Let no one post here who requires people not to post here. BOfL === Subject: Re: A Note on Goedel's Theorem >> Let no one post here who does not know the _Laws of Form_. > Let no one post here who requires people not to post here. There's two different philosophy jokes combined in that .sig . . . . -- Let no one post here who does not know the _Laws of Form_. === Subject: Re: A Note on Goedel's Theorem > Let no one post here who does not know the _Laws of Form_. >> Let no one post here who requires people not to post here. > There's two different philosophy jokes combined in that .sig . . . . > -- > Let no one post here who does not know the _Laws of Form_. Giving due consideration to the nature of this ng, please define joke...:-) BOfL === Subject: Re: A Note on Goedel's Theorem >> Let no one post here who does not know the _Laws of Form_. , === Subject: Re: A Note on Goedel's Theorem > Let no one post here who does not know the _Laws of Form_. > , ,, -- tinmimus99@hotmail.com smeeter 11 or maybe 12 mp 10 mhm 29x13 This calls for a very special blend of psychology and extreme violence. < Vyvyan === Subject: Re: A Note on Goedel's Theorem On Sun, 6 Nov 2005 19:21:22 -0500, in alt.alien.vampire.flonk.flonk.flonk, <1rqdnX7HmPNgAPPeRVn-tQ@comcast.com>, Turtoni humped my leg thusly: > Let no one post here who does not know the _Laws of Form_. > , ^ | id just like to know what that one character meant. -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem >> Let no one post here who does not know the _Laws of Form_. , > ^ > | > id just like to know what that one character meant. Distinction. > -- > dave hillstrom > Vote Dave Hillstrom for Whining Whinger in AUK October 2005 > the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem On Sun, 6 Nov 2005 19:39:38 -0500, in alt.alien.vampire.flonk.flonk.flonk, <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni humped my leg thusly: > Let no one post here who does not know the _Laws of Form_. > , >> ^ >> | >> id just like to know what that one character meant. >Distinction. go on, elaborate. -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem > On Sun, 6 Nov 2005 19:39:38 -0500, in alt.alien.vampire.flonk.flonk.flonk, > <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni > Let no one post here who does not know the _Laws of Form_. >> , > ^ > | > id just like to know what that one character meant. >>Distinction. > go on, elaborate. _Laws of Form_ treats binary analysis-- _not_ binary math, unless you drop the carry-- which distinguishes only two states and a single operation, transition between those states. Technically such system should be referred to as the F2 group, which is more primitive than and fundamental to the L group (which involves maintaining the binary distinction between true and false). There can also be the F3 group-- trinary distinctions-- which Spencer vaporizing in a cloud of mysticism, while someone supposedly has generalized it all to the FN group (and I suppose there could be the F- aleph groups, but I don't want to think about them). Good book. Read Appendix 2 first, and then chapters 1-10, first backwards and then forwards. Best thing since _Principia Mathematica_, fundamental- analysis- wise. Questions? -- tinmimus99@hotmail.com smeeter 11 or maybe 12 mp 10 mhm 29x13 This calls for a very special blend of psychology and extreme violence. < Vyvyan === Subject: Re: A Note on Goedel's Theorem On Sun, 6 Nov 2005 22:18:16 -0500, in alt.alien.vampire.flonk.flonk.flonk, <1n4v8c1rvn9f8.f7ghm6aqn$.dlg@40tude.net>, mimus humped my leg thusly: >> On Sun, 6 Nov 2005 19:39:38 -0500, in alt.alien.vampire.flonk.flonk.flonk, >> <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni humped my leg thusly: > Let no one post here who does not know the _Laws of Form_. > , >> ^ >> | >> id just like to know what that one character meant. >Distinction. >> go on, elaborate. >_Laws of Form_ treats binary analysis-- _not_ binary math, unless you drop >the carry-- which distinguishes only two states and a single operation, >transition between those states. >Technically such system should be referred to as the F2 group, which is >more primitive than and fundamental to the L group (which involves >maintaining the binary distinction between true and false). >There can also be the F3 group-- trinary distinctions-- which Spencer >vaporizing in a cloud of mysticism, while someone supposedly has >generalized it all to the FN group (and I suppose there could be the F- >aleph groups, but I don't want to think about them). >Good book. Read Appendix 2 first, and then chapters 1-10, first backwards >and then forwards. >Best thing since _Principia Mathematica_, fundamental- analysis- wise. >Questions? penis? -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem > On Sun, 6 Nov 2005 22:18:16 -0500, in alt.alien.vampire.flonk.flonk.flonk, > <1n4v8c1rvn9f8.f7ghm6aqn$.dlg@40tude.net>, mimus > humped my leg thusly: > On Sun, 6 Nov 2005 19:39:38 -0500, in alt.alien.vampire.flonk.flonk.flonk, > <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni >> Let no one post here who does not know the _Laws of Form_. >> , > ^ > | > id just like to know what that one character meant. >>Distinction. > > go on, elaborate. >>_Laws of Form_ treats binary analysis-- _not_ binary math, unless you drop >>the carry-- which distinguishes only two states and a single operation, >>transition between those states. >>Technically such system should be referred to as the F2 group, which is >>more primitive than and fundamental to the L group (which involves >>maintaining the binary distinction between true and false). >>There can also be the F3 group-- trinary distinctions-- which Spencer >>vaporizing in a cloud of mysticism, while someone supposedly has >>generalized it all to the FN group (and I suppose there could be the F- >>aleph groups, but I don't want to think about them). >>Good book. Read Appendix 2 first, and then chapters 1-10, first backwards >>and then forwards. >>Best thing since _Principia Mathematica_, fundamental- analysis- wise. >>Questions? > penis? Nope, this is even lower- level than that. (I was under the influence of about one- quarter of an ounce of ethanol at the time I winged the previous post, so please replace the unless you drop the carry by unless you stick to one bit, the clause being unnecessary anyway, and the L group is of course the axioms and so on of ordinary -- tinmimus99@hotmail.com smeeter 11 or maybe 12 mp 10 mhm 29x13 This calls for a very special blend of psychology and extreme violence. < Vyvyan === Subject: Re: A Note on Goedel's Theorem > On Sun, 6 Nov 2005 22:18:16 -0500, in > alt.alien.vampire.flonk.flonk.flonk, > <1n4v8c1rvn9f8.f7ghm6aqn$.dlg@40tude.net>, mimus > humped my leg thusly: > On Sun, 6 Nov 2005 19:39:38 -0500, in > alt.alien.vampire.flonk.flonk.flonk, > <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni > > humped my leg thusly: > Let no one post here who does not know the _Laws of Form_. >> , > ^ > | > id just like to know what that one character meant. >>Distinction. >> go on, elaborate. >>_Laws of Form_ treats binary analysis-- _not_ binary math, unless you > drop >>the carry-- which distinguishes only two states and a single operation, >>transition between those states. >>Technically such system should be referred to as the F2 group, which is >>more primitive than and fundamental to the L group (which involves >>maintaining the binary distinction between true and false). >>There can also be the F3 group-- trinary distinctions-- which Spencer >>vaporizing in a cloud of mysticism, while someone supposedly has >>generalized it all to the FN group (and I suppose there could be the F- >>aleph groups, but I don't want to think about them). >>Good book. Read Appendix 2 first, and then chapters 1-10, first > backwards >>and then forwards. >>Best thing since _Principia Mathematica_, fundamental- analysis- wise. >>Questions? > penis? >> Nope, this is even lower- level than that. >> (I was under the influence of about one- quarter of an ounce of ethanol at >> the time I winged the previous post, so please replace the unless you > drop >> the carry by unless you stick to one bit, the clause being unnecessary >> anyway, and the L group is of course the axioms and so on of ordinary > Ya know, all the words in this post make sense when taken one at a time, but > when they're put together in the above sequence, it may as well be in a > foreign language. And that's one of the reasons I don't like math(s). > Speek Anglisch!! > Smee Nobody ever gets my joke about reading _LoF_ backwards. Anyway, this isn't math. Well, not arithmetic, anyway. Much cruder and simpler and more fundamental. -- tinmimus99@hotmail.com smeeter 11 or maybe 12 mp 10 mhm 29x13 This calls for a very special blend of psychology and extreme violence. < Vyvyan === Subject: Re: A Note on Goedel's Theorem > On Sun, 6 Nov 2005 19:39:38 -0500, in alt.alien.vampire.flonk.flonk.flonk, > <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni > Let no one post here who does not know the _Laws of Form_. >> , > ^ > | > id just like to know what that one character meant. >>Distinction. > go on, elaborate. You said that was all you wanted to know. > -- > dave hillstrom > Vote Dave Hillstrom for Whining Whinger in AUK October 2005 > the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem On Sun, 6 Nov 2005 21:35:02 -0500, in alt.alien.vampire.flonk.flonk.flonk, , turtoni humped my leg thusly: >> On Sun, 6 Nov 2005 19:39:38 -0500, in alt.alien.vampire.flonk.flonk.flonk, >> <8vWdnQXSb7LbP_PeRVn-tA@comcast.com>, Turtoni humped my leg thusly: > Let no one post here who does not know the _Laws of Form_. > , >> ^ >> | >> id just like to know what that one character meant. >Distinction. >> go on, elaborate. >You said that was all you wanted to know. i lied -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem , Brian Fletcher humped my leg thusly: >> Let no one post here who does not know the _Laws of Form_. > Let no one post here who requires people not to post here. >> There's two different philosophy jokes combined in that .sig . . . . >> -- >> Let no one post here who does not know the _Laws of Form_. >Giving due consideration to the nature of this ng, please define >joke...:-) which newsgroup? -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem <%Tbbf.9660$Hj2.54@news-server.bigpond.net.au> What does BOfL mean? === Subject: Re: A Note on Goedel's Theorem On 6 Nov 2005 07:49:53 -0800, in alt.alien.vampire.flonk.flonk.flonk, Starbles@Earthlink.net humped my leg thusly: >What does BOfL mean? attribute. also, as a hidden and integral part of this, it implies the same thing three times or more, probably because they dont know what they are doing. i would suggest you try a real newsreader, not OE, and a real news provider, neither of which need cost you a dime. -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem > On 6 Nov 2005 07:49:53 -0800, in alt.alien.vampire.flonk.flonk.flonk, > Starbles@Earthlink.net humped my leg thusly: >>What does BOfL mean? > attribute. also, as a hidden and integral part of this, it implies > the same thing three times or more, probably because they dont know what > they are doing. i would suggest you try a real newsreader, not OE, and a > real news provider, neither of which need cost you a dime. > -- > dave hillstrom > Vote Dave Hillstrom for Whining Whinger in AUK October 2005 > the belgians are STILL thieves. Three times I stand to say Brian Of Life === Subject: Re: A Note on Goedel's Theorem , Brian Fletcher humped my leg thusly: >> On 6 Nov 2005 07:49:53 -0800, in alt.alien.vampire.flonk.flonk.flonk, >> Starbles@Earthlink.net humped my leg thusly: >What does BOfL mean? >> attribute. also, as a hidden and integral part of this, it implies >> the same thing three times or more, probably because they dont know what >> they are doing. i would suggest you try a real newsreader, not OE, and a >> real news provider, neither of which need cost you a dime. >> -- >> dave hillstrom >> Vote Dave Hillstrom for Whining Whinger in AUK October 2005 >> the belgians are STILL thieves. >Three times I stand to say >Brian Of Life i still say its cooler otherwise. and you want to be cool and popular, right? beyond any reason? wouldnt that be nice? -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem What does BOfL mean? === Subject: Re: A Note on Goedel's Theorem <%Tbbf.9660$Hj2.54@news-server.bigpond.net.au> What does BOfL mean? === Subject: Re: A Note on Goedel's Theorem > What does BOfL mean? Brian Of Life..... (thank for asking;-) === Subject: Re: A Note on Goedel's Theorem >> What does BOfL mean? Bastard operator from life. > Brian Of Life..... http://en.wikipedia.org/wiki/BOFH === Subject: Re: A Note on Goedel's Theorem >> What does BOfL mean? > Bastard operator of life. even. === Subject: Re: A Note on Goedel's Theorem , Brian Fletcher humped my leg thusly: >> What does BOfL mean? >Brian Of Life..... >(thank for asking;-) i thought it meant bastard operator from luna. -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: Re: A Note on Goedel's Theorem alt.alien.vampire.flonk.flonk.flonk, <1ijbemvnhz58 $.1jmhc1mekidhi.dlg@40tude.net> : > If Goedel's Theorem is valid, then it is included in _P_, and is therefore > fallacious. Kurt G.9adel ? Well, the man was quite right... some questions are not decideable. -- ./ === Subject: Re: A Note on Goedel's Theorem > If Goedel's Theorem is valid, then it is included in _P_, and is therefore > fallacious. If the alien vampires flonk into philosophy and mathematics, they will eat Goedel and raise large families. === Subject: Re: A Note on Goedel's Theorem On 05 Nov 2005 20:50:44 +0100, in alt.alien.vampire.flonk.flonk.flonk, , Torkel Franzen humped my leg thusly: >> If Goedel's Theorem is valid, then it is included in _P_, and is therefore >> fallacious. > If the alien vampires flonk into philosophy and mathematics, they >will eat Goedel and raise large families. mmmmmmmmmmmmmmm goedel steaks. maybe a teriyaki marinade... -- dave hillstrom Vote Dave Hillstrom for Whining Whinger in AUK October 2005 the belgians are STILL thieves. === Subject: traingle In an acuteangled triangle ABC, altitudes intersect in point H. Straight line cuts point H and interesects AC and BC in points D and E. Perpendicular to this line intersects point H, and AB in point F. Proove that: DH/HE=AF/FB === Subject: Prime Number Discoverers Does anyone know of any source of who found which prime number (and preferrably when also)? Googling only turns up lists of prime numbers, or possibly who found the largest one at this or that time, but I would like to have a more conclusive list. /Kjell === Subject: Re: Prime Number Discoverers > Does anyone know of any source of who found which prime number (and > preferrably when also)? Googling only turns up lists of prime numbers, > or possibly who found the largest one at this or that time, but I would > like to have a more conclusive list. Here's a list of discoverers of Mersenne Primes: http://mathworld.wolfram.com/MersennePrime.html I don't know who discovered that 2 was prime, though ... === Subject: Re: Prime Number Discoverers Great link! And fun comment. /Kjell > Does anyone know of any source of who found which prime number (and > preferrably when also)? Googling only turns up lists of prime numbers, > or possibly who found the largest one at this or that time, but I would > like to have a more conclusive list. > Here's a list of discoverers of Mersenne Primes: > http://mathworld.wolfram.com/MersennePrime.html > I don't know who discovered that 2 was prime, though ... > === Subject: Ineq. with symmetric means Assume that A,B,C,D , F are functions (0,infty)-->(-infty,infty) which are continuous and strictly monotonic on (0,infty) . For n fixed in {2,3,...} denote w=1/n and M_{F}(X):=F^{-1}( w*sum_{k=1 to k=n}F(x_k)) where X:= (x_1,x_2,...,x_n ) in (0,infty)^n . Questions: 1) Find all pairs (A,B) , (C,D) such that (*) M_{A}(X)-M_{B}(X)ge M_{C}(X)-M_{D}(X) , for all X in (0,infty)^n . 2) Try to prove/or disprove/ that (*) is verified when (A(x),B(x))= (x^2,x ) , ( C(x),D(x))=(x,ln(x)) . === Subject: Calculus: Approach Path for a PLane An approach path for an aircraft landing satisfies the following conditions: (i)The cruising altitude is h when descent starts at a horizontal distance l from touchdown at the origin. (ii) The pilot must miantain a constant horizontal speed v throughout descent. (iii) The absolute value of th vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity). a. FInd a cubic polynomial P(x) = ax^3 + bx^2 + cx + d that satisfies condition (i) by imposing suitable conditions on P(x) at the start of descent and at touchdown. b. use conditions (ii) and (iii) to show that 6hv^2 / l^2 is less than or equal to k. c. Suppose that an airline decides not to allow vertical acceleration of a plane exceed k = 860 mi/hr^2. If the cruising altitude of a plane is 35,000 feet and the speed is 300 mi/hr how far away from the airport should the pilot start descent? === Subject: Calculus: Approach Path for a Plane An approach path for an aircraft landing satisfies the following conditions: (i)The cruising altitude is h when descent starts at a horizontal distance l from touchdown at the origin. (ii) The pilot must miantain a constant horizontal speed v throughout descent. (iii) The absolute value of th vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity). a. FInd a cubic polynomial P(x) = ax^3 + bx^2 + cx + d that satisfies condition (i) by imposing suitable conditions on P(x) at the start of descent and at touchdown. b. use conditions (ii) and (iii) to show that 6hv^2 / l^2 is less than or equal to k. c. Suppose that an airline decides not to allow vertical acceleration of a plane exceed k = 860 mi/hr^2. If the cruising altitude of a plane is 35,000 feet and the speed is 300 mi/hr how far away from the airport should the pilot start descent? === Subject: Re: Calculus: Approach Path for a Plane > (ii) The pilot must maintain a constant horizontal speed v throughout > descent. Just a thought. A pilot will not be able to maintain a horizontal speed of say 450 kts from cruise altitude to touchdown, especially below 10,000 ft. -- a > An approach path for an aircraft landing satisfies the following > conditions: > (i)The cruising altitude is h when descent starts at a horizontal > distance l from touchdown at the origin. > (ii) The pilot must miantain a constant horizontal speed v throughout > descent. > (iii) The absolute value of th vertical acceleration should not exceed > a constant k (which is much less than the acceleration due to gravity). > a. FInd a cubic polynomial P(x) = ax^3 + bx^2 + cx + d that satisfies > condition (i) by imposing suitable conditions on P(x) at the start of > descent and at touchdown. > b. use conditions (ii) and (iii) to show that 6hv^2 / l^2 is less than > or equal to k. > c. Suppose that an airline decides not to allow vertical acceleration > of a plane exceed k = 860 mi/hr^2. If the cruising altitude of a plane > is 35,000 feet and the speed is 300 mi/hr how far away from the airport > should the pilot start descent? === Subject: Re: Integral ineq.with poly.degree =< 3 > Denote bye M_3(+) the set of all (real) > polynomials P of degree =< 3 such that P > 0 on (0,1). > 1) It's true that there exists a constant K , > K in (0.84494 , 0.84495) , such that > Integral_{[0,1]}x*P(x) dx /Integral_{[0,1]}P(x) dx =< K > for all P in M_3(+) ?? Yes > 2) If exists, which is the exact value of K ? Try (14+sqrt(6))/20 Old Ear; === Subject: Re: Integral ineq.with poly.degree =< 3 > Denote bye M_3(+) the set of all (real) > polynomials P of degree =< 3 such that P > 0 on (0,1). > 1) It's true that there exists a constant K , > K in (0.84494 , 0.84495) , such that > Integral_{[0,1]}x*P(x) dx /Integral_{[0,1]}P(x) dx =< K > for all P in M_3(+) ?? > Yes > 2) If exists, which is the exact value of K ? > Try (14+sqrt(6))/20 Oop! - Wrong max. Sorry > Old Earl === Subject: Re: Integral ineq.with poly.degree =< 3 I think I got it right this time! k=[3*(5*sqrt(6*(7-2*sqrt(6)))-2*(17-2*sqrt(6))]/[2*(10*sqrt(6*(7-2*sqrt(6))) -3*(23-3*sqrt(6))] k~.844948974278 Old Earl === Subject: Re: Rolle's Theorem Are you saying that a lot of things in Math are common sense, yet when taken to its conclusions they can be very powerful. Sort of like the Identity Property of Multiplication is very simple, yet it is the very property that allows us to change franctions with different denominators into fractions with a common denominator, thus allowing us to do addition and subtraction with fractions. Kees > That's the trick about math: some things which sound like common sense > aren't. The reason that it seems obvious to you is that you have some > intuitive sense of what derivative means. Why couldn't a function > increase for a while, with positive derivative, then decrease again, > with a negative derivative? Why must the derivative vanish? You might > say, If the derivative is positive first, then negative, it must be > 0. Why is this true? If a function is differenble, it does not > follow that the derivative is continuous. === Subject: Re: Rolle's Theorem In connection with Rolle Theorem: there is a result called Universal Chord Theorem or Rolle Theorem for continuous functions. This is: Theorem(Paul Levy-[4]) Suppose that f is in C[a,b] with f(a)=f(b).Then for every h of the form h=(b-a)/n ,n in {1,2,...},the equation f(x+h)-f(x)=0 , x in I:= [a, b-h] , has at least a solution. Proof. Suppose that g(x):=f(x+h)-f(x) has no solution in I. Because g is continuous we have g(x) > 0 or g(x) < 0 on I. Assume g(x) > 0 for all x in I. Because the points x_k:= a+k(b-a)/n ,k in {0,1,...,n-1}, are in I we must have g(x_k) > 0 that is (1) f(x_{k+1}) - f(x_k) > 0 for k in {0,1,...,n-1}. By summing (1) we obtain f(x_n) - f(x_0) = f(b)-f(a) > 0 which is a contradiction. Some references: [1] R. P. Boas , A primer of real functions , The Carus Mathematical Monographs, MAA, vol.13, 1960. [2] H. Hopf , Uber die Sehnen ebener Kontinuen und die Schliefen geschlossener Wege , Comment.Math. Helvetici 9 (1937), 303-319. [3] R. J. Levit , The finite difference extension of Rolle` s theorem, Amer. Math. Monthly 70 (1963) , 26-30 . [4] P. Levy, Sur une generalisation du theoreme de Rolle, C.R. Acad. Sci. Paris 198 (1934) , 424-425. Try to find a geometric interpretation of this theorem. [Mounting-climbing ???] === Subject: Re: Rolle's Theorem In connection with Rolle Theorem: there is a result called Universal Chord Theorem or Rolle Theorem for continuous functions. This is: Theorem(Paul Levy-[4]) Suppose that f is in C[a,b] with f(a)=f(b).Then for every h of the form h=(b-a)/n ,n in {1,2,...},the equation f(x+h)-f(x)=0 , x in I:= [a, b-h] , has at least a solution. Proof. Suppose that g(x):=f(x+h)-f(x) has no solution in I. Because g is continuous we have g(x) > 0 or g(x) < 0 on I. Assume g(x) > 0 for all x in I. Because the points x_k:= a+k(b-a)/n ,k in {0,1,...,n-1}, are in I we must have g(x_k) > 0 that is (1) f(x_{k+1}) - f(x_k) > 0 for k in {0,1,...,n-1}. By summing (1) we obtain f(x_n) - f(x_0) = f(b)-f(a) > 0 which is a contradiction. Some references: [1] R. P. Boas , A primer of real functions , The Carus Mathematical Monographs, MAA, vol.13, 1960. [2] H. Hopf , Uber die Sehnen ebener Kontinuen und die Schliefen geschlossener Wege , Comment.Math. Helvetici 9 (1937), 303-319. [3] R. J. Levit , The finite difference extension of Rolle` s theorem, Amer. Math. Monthly 70 (1963) , 26-30 . [4] P. Levy, Sur une generalisation du theoreme de Rolle, C.R. Acad. Sci. Paris 198 (1934) , 424-425. Try to find a geometric interpreattion of this theorem. [Mounting-climbing ???] === Subject: Re: Rolle's Theorem > That's the trick about math: some things which sound like common sense > aren't. The reason that it seems obvious to you is that you have some > intuitive sense of what derivative means. Why couldn't a function > increase for a while, with positive derivative, then decrease again, > with a negative derivative? Why must the derivative vanish? You might > say, If the derivative is positive first, then negative, it must be > 0. Why is this true? If a function is differenble, it does not > follow that the derivative is continuous. I thought that a condition for being differenble is that it is also continuous. Are you saying that a function could be non continuous over a certain part of the domain, yet in the same part is differenble? How is that possible? Kees === Subject: Re: Rolle's Theorem On Fri, 4 Nov 2005 21:36:34 -0500, Kees Boer >> That's the trick about math: some things which sound like common sense >> aren't. The reason that it seems obvious to you is that you have some >> intuitive sense of what derivative means. Why couldn't a function >> increase for a while, with positive derivative, then decrease again, >> with a negative derivative? Why must the derivative vanish? You might >> say, If the derivative is positive first, then negative, it must be >> 0. Why is this true? If a function is differenble, it does not >> follow that the derivative is continuous. >I thought that a condition for being differenble is that it is also >continuous. >Are you saying that a function could be non continuous over a certain part >of the domain, yet in the same part is differenble? No, that isn't what he is saying. You are correct that at points where a function is differenble, the function is continuous. But that doesn't imply that its derivative is continuous. --Lynn === Subject: Re: Rolle's Theorem > On Fri, 4 Nov 2005 21:36:34 -0500, Kees Boer > That's the trick about math: some things which sound like common sense > aren't. The reason that it seems obvious to you is that you have some > intuitive sense of what derivative means. Why couldn't a function > increase for a while, with positive derivative, then decrease again, > with a negative derivative? Why must the derivative vanish? You might > say, If the derivative is positive first, then negative, it must be > 0. Why is this true? If a function is differenble, it does not > follow that the derivative is continuous. >>I thought that a condition for being differenble is that it is also >>continuous. >>Are you saying that a function could be non continuous over a certain part >>of the domain, yet in the same part is differenble? > No, that isn't what he is saying. You are correct that at points where > a function is differenble, the function is continuous. But that > doesn't imply that its derivative is continuous. > --Lynn Kees === Subject: Re: Rolle's Theorem >Are you saying that a function could be non continuous over a certain >part >of the domain, yet in the same part is differenble? >> No, that isn't what he is saying. You are correct that at points where >> a function is differenble, the function is continuous. But that >> doesn't imply that its derivative is continuous. On a related note, continuity does not imply differenbility. A differenble (hence continuous) function not twice differenble: see http://web01.shu.edu/projects/reals/cont/fp_c1.html A continuous function that is not differenble: f(x)=|x| at x=0. -- Darrell === Subject: Re: Rolle's Theorem On Sat, 5 Nov 2005 01:48:01 -0600, Darrell see http://web01.shu.edu/projects/reals/cont/fp_c1.html Not only is it not twice differenble, its derivative is not even continuous which is in fact an example for which Kees asked. --Lynn === Subject: Order-preserving auorphisms Here is a question. Suppose F is an ordered field. Do we know that the only order-preserving auorphism on F is trivial? === Subject: Re: OT: Cutting glass, specify a math model Forgot to mention that we have double glazing here in Scandinavia. Should probably multiply the cost of wasted glass with at least a factor of 2 ;-p Some of the factories get the glass cut off-site, so making a template is not easy. Also, there is not one but several cutting machines involved. The bummer we (s/w developers) made here was to specify the cutting parameters for one type of machine. Adding support for more machines, different formats, is probably not a wise move. We will go for a general self invented rich format and let the operator figure out what parameters they need in order to get the glass cut to our specifications. At least I will be able to sleep better at night doing it this way. I was hoping someone had been up this road before and made a general descriptive format. RFC 3.14 is on its way.