mm-189 === Subject: A little help is neededI am not quite sure how to solve this problem. Can someone give me alittle help, It would greatly be appreciated. The question is as follows:Determine whether f: Q x Q -> Q given by f(a/b,c/d)=(a+c)/(b+d)is a well-defined function.Again thank you very much for your help. === Subject: Re: A little help is neededIn sci.math, problem. Can someone give me a> little help, It would greatly be appreciated. The question is as follows: Determine whether f: Q x Q -> Q given by f(a/b,c/d)=(a+c)/(b+d)> is a well-de'ned function. Again thank you very much for your help.There is an issue regarding how one de'nes negative fractions, butotherwise it looks 'ne. The main issue is incompleteness of thede'nition; to be full about it, one should probably dosomething like:f: Q x Q -> Qf(r1, r2) = (a+c)/(b+d) where r1 = a/b, r2 = c/d, a,b,c,d integer, b > 0, d > 0, gcd(a,b)=1, gcd(c,d)=1 [*]Obviously this would be very hard to characterize in termsof the standard arithmetic operations on Q, but a function'sa function, and since every rational r can be uniquely writtenas the quotient of two integers a/b, b > 0, gcd(a,b) = 1,this de'nition works.Just try not to integrate it over the unit square... :-)[*] probably should also state that 0 = 0/1 is the unique representation of 0 as well.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: A little help is needed Adjunct Assistant Professor at the University of Montana.>f: Q x Q -> Q>f(r1, r2) = (a+c)/(b+d) where r1 = a/b, r2 = c/d,> a,b,c,d integer, b > 0, d > 0,> gcd(a,b)=1, gcd(c,d)=1 [*] [.snip.]>[*] probably should also state that 0 = 0/1 is the unique> representation of 0 as well.That follows from the de'nition: gcd(0,b) = |b| for all integersb. For gcd(x,y) is the nonnegative integer d that satis'es: (1) d|x and d|y;and (2) if c is an integer such that c|x and x|y, then x|d.Clearly, |b| divides b and divides 0. And if x is any integer thatdivides both b and 0, then it divides |b|. Thus, gcd(0,b)=|b|. So ifa=0, for gcd(a,b) to equal 1, with b>0, you must have b=1.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === Re: A little help is neededLet S and T be sets. A function f:S->T is given by associating to eachelement s in S a (unique) element f(s) in T. Thus to check that afunction is well-de'ned we must check both(i) For every s in S, f(s) is in T.(ii) If s_1 and s_2 are in S with s_1=s_2 then f(s_1)=f(s_2).We note that equality in Q is determined by cross-multiplying;That is, a/b = c/d iff ad = bc. So for the example f:QxQ->Q given byf(a/b,c/d) = (a+c)/(b+d) we must show (i) and (ii).(i) Assume (a/b,c/d) is in QxQ. That is a,b,c,d in Z and neither b nord is zero. Does (a+c)/(b+d) qualify for membership in Q? Well a+c andb+d are clearly integers, since a,b,c, and d are. Moroever, sinceneither b nor d is zero, b+d does not equal zero either. So(a+c)/(b+d) is in Q.(ii) Assume (a/b,c/d) and (e/f,g/h) are in QxQ and (a/b,c/d) =(e/f,g/h)That is a/b = e/f and c/d = g/h, or af=be and ch=dg.Now, f(a/b,c/d) = (a+c)/(b+d) and f(e/f,g/h) = (e+g)/(f+h).We must determine whether (a+c)/(b+d) = (e+g)/(f+h)Or (a+c)(f+h) = (e+g)(b+d)Or af + ah + cf + ch = be + de + bg + dg.This does not seem to hold in general since we only have af=be andch=dg.In fact, we can exhibit a counterexample:(1/2,1/3) = (2/4,3/9)But f(1/2,1/3) = 2/5 whereas f(2/4,3/9) = 5/13.Since 26 > 25 these are not equal. So f holds (i) but not (ii), and so f is not well-de'ned.This is why one might decide to de'ne g:QxQ->Q by g(a/b,c/d) = (ad+bc)/(bd). (Since b and d are nonzero bd is not zero).Which is well-de'ned.Hope this helps,Sam> I am not quite sure how to solve this problem. Can someone give me a> little help, It would greatly be appreciated. The question is as follows:> > Determine whether f: Q x Q -> Q given by> > f(a/b,c/d)=(a+c)/(b+d)> is a well-de'ned function.> > Again thank you very much for your help. === Subject: Re: A little help is needed Adjunct Assistant Professor at the University of Montana.>Let S and T be sets. A function f:S->T is given by associating to each>element s in S a (unique) element f(s) in T. Thus to check that a>function is well-de'ned we must check both>(i) For every s in S, f(s) is in T.>(ii) If s_1 and s_2 are in S with s_1=s_2 then f(s_1)=f(s_2).We note that equality in Q is determined by cross-multiplying;>That is, a/b = c/d iff ad = bc. So for the example f:QxQ->Q given by>f(a/b,c/d) = (a+c)/(b+d) we must show (i) and (ii).>(i) Assume (a/b,c/d) is in QxQ. That is a,b,c,d in Z and neither b nor>d is zero. Does (a+c)/(b+d) qualify for membership in Q? Well a+c and>b+d are clearly integers, since a,b,c, and d are. Moroever, since>neither b nor d is zero, b+d does not equal zero either.The sum of nonzero integers is always nonzero? So much for additive inverses!-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: A little help is needed> I am not quite sure how to solve this problem. Can someone give me a> little help, It would greatly be appreciated. The question is as follows:> > Determine whether f: Q x Q -> Q given by> > f(a/b,c/d)=(a+c)/(b+d)> is a well-de'ned function.> > Again thank you very much for your help.Just a hint. Well-de'ned means that f(q_1, q_2) is independent ofthe particular realizations of q_1 and q_2 as quotients of integers. For instance if q_1 = 1/2, it could also equal 2/4, 3/6, 4/8, etc. Ifyou plug in all of these different values into your formula, you mustget the same thing if you are to call f a well-de'ned === Subject: Re: A little help is needed Adjunct Assistant Professor at the University of Montana.>I am not quite sure how to solve this problem. Can someone give me a>little help, It would greatly be appreciated. The question is as follows:Determine whether f: Q x Q -> Q given by f(a/b,c/d)=(a+c)/(b+d)>is a well-de'ned function.Again thank you very much for your help.Whether it is well-de'ned can mean two different things, both ofwhich are relevant here:(1) Whether given any input, the output belongs to the speci'ed range (in this case, wether given any two rationals, the output is a rational); and/or(2) Whether given the same input, written in two different ways, will yield the same output. This because (a) there are many different ways to write a rational as a/b; and (b) the function is de'nedHere, you are being extremely imprecise in describing the elements ofQ; for all I know, a, b, c, d may be rationals themselves! Though,presumably, you mean the to be integers, b and d nonzero.If that's all you ask, then (1) is equivalent to asking wether underthose conditions, b+d will be nonzero. Alas, not if you are at libertyto choose a, b, c, d arbitrarily. For example, you might take a=1,b=-1, c =d =1, in which case the rules for your function would ask youto evaluate (1+1)/(-1+1). So in order for us to meet the requirement of (1), we must somehowrestrict how the rationals are being written. One obvious way to avoidthe pitfall above is to specify that b and d must be ->positive<-rather than just nonzero. In that case, the number (a+c)/(b+d) is arational number for all integer values of a, b, c, and d, with b and dpositive.So now we turn to (2). Your de'nition depends on how we represent therationals r=a/b and s=c/d. If we choose some other representation of rand/or of s, say r=p/q, s=u/v, then we would need(a+c)/(b+d) = f( (a/b), (c/d) ) = f(r,s) = f( (p/q),(u/v) ) = (p+u)/(q+v).So we would need to check whether, if a, b, c, d, p, q, u, v areintegers, b, d, q, v positive, such that: (i) aq = bp; and (ii) cv = du,whether this implies that (a+c)(q+v) = (p+u)(b+d)(so that (a+c)/(b+d) = (p+u)/(q+v)).Alas, no. Say a = b = 1, p = q = 2, c=u=1, d=v=2. Thenf ( 1/1, 1/2) = 2/3butf (2/2, 1/2) = 3/4and 2/3 is not equal to 3/4. So your function is not well de'ned even if we ask that b and d bothbe nonzero.One easy way to make it well-de'ned then is to specify a ->unique<-way to write the rational inputs a/b and c/d, so that the value of fwill not depend on how we write them: there will only be one validway to write them, and that's the way we write them in order toevaluate the function. Of course, that means that to do an actualevaluation you 'rst must check to see if your rationals satisfy theseconditions.The easiest way to set this unique way is to say:f( (a/b), (c/d) ) = (a+c)/(b+d) where a, b, c, d are integers, b, d are positive integers, and gcd(a,b)=gcd(c,d)=1 (that is, written in lowest terms with positive denominators).Since every rational number x has a unique expression as a quotient oftwo coprime integers, with the divisor positive, this will guaranteethat the value of f is uniquely determined. And since b and d arespeci'ed to be positive, the value is a rational number (thedenominator is not zero).-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === Re: A little help is needed> I am not quite sure how to solve this problem. Can someone give me a> little help, It would greatly be appreciated. The question is as follows:> > Determine whether f: Q x Q -> Q given by> > f(a/b,c/d)=(a+c)/(b+d)> is a well-de'ned function.> > Again thank you very much for your help.f(2/3,4/5) = 6/8 = 3/4f(4/6,4/5) = 8/11 === Subject: Re: A little help is needed> I am not quite sure how to solve this problem. Can someone give me a> little help, It would greatly be appreciated. The question is as follows: Determine whether f: Q x Q -> Q given by f(a/b,c/d)=(a+c)/(b+d)> is a well-de'ned function. Again thank you very much for your help.It would be well-de'ned if you described some unique way of expressing anarbitrary rational number in the form a/b. For example, you might say thatand a and b are integers, b is positive and that a and b have no commonfactor greater than one.-- Clive Toothhttp://www.clivetooth.dk === Subject: Re: Point-set Topology Hints Wanted . Adjunct Assistant Professor at the University of Montana.>> >>I am trying to show that Cl(A)' is a subset of A.'>> >>A' is the set of all limit point of A .>> limit pts A = derived A = A' = { x | x in cl Ax }>> So what you actually want is that every punctured neighborhood>> intersect A.>> When space is T1, A' is closed.>> Now for all A>> A' subset cl A>> so>> A subset cl A' = A'>> Is A subset A' even when space is not T1 ?>> Let X = {x,y}, with the indiscrete topology T = {empty, X}. Let A>> ={x}. Then x is not in A', but y is. So A' = {y}. And by a symmetric>> argument, A'' = {x}. So A'' is disjoint from A'.>Yes. Perhaps for an encore you have counter example for T0 space?As a matter of fact, yes, I 'gured one out yesterday while trying tolearn to cross-country ski...Let X be the naturals union a single point, called in'nity.The topology on X consists of the empty set, and all subsets U of Xthat satisfy the following two conditions: (1) in'nity is in U; and (2) There exists k in N such that for all n>k, n is in U.This is a topology: the empty and total sets are there. An arbitraryunion of sets satisfying (1) and (2) satis'es (1) and (2); and theintersection of two sets satisfying (1) and (2) also satis'es (1) and(2).It is also a T0 space: given any two natural numbers r and s, the setsU_r = X-{r}; U_s = X-{s} are open sets containing, respectively,r and not s; and s and not r. And given a natural number r andin'nity, the set U_r = X-{r} is an open set containing in'nity butnor r. There is no open set containing r and not in'nity, so thetopology is T0 but no T1.Now let A = {in'nity}. Then A' is all of N. And A'' is all of X, soA'' is not contained in A'.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Point-set Topology Hints Wanted .Well , great now you have jumped ahead Cl(A)' is a subset of A.'>> >>A' is the set of all limit point of A .>> limit pts A = derived A = A' = { x | x in cl Ax }>> So what you actually want is that every punctured neighborhood>> intersect A.>> When space is T1, A' is closed.>> Now for all A>> A' subset cl A>> so>> A subset cl A' = A'>> Is A subset A' even when space is not T1 ?>> Let X = {x,y}, with the indiscrete topology T = {empty, X}. Let A>> ={x}. Then x is not in A', but y is. So A' = {y}. And by a symmetric>> argument, A'' = {x}. So A'' is disjoint from A'.>Yes. Perhaps for an encore you have counter example for T0 space? As a matter of fact, yes, I 'gured one out yesterday while trying to> learn to cross-country ski... Let X be the naturals union a single point, called in'nity. The topology on X consists of the empty set, and all subsets U of X> that satisfy the following two conditions: (1) in'nity is in U; and> (2) There exists k in N such that for all n>k, n is in U. This is a topology: the empty and total sets are there. An arbitrary> union of sets satisfying (1) and (2) satis'es (1) and (2); and the> intersection of two sets satisfying (1) and (2) also satis'es (1) and> (2). It is also a T0 space: given any two natural numbers r and s, the sets> U_r = X-{r}; U_s = X-{s} are open sets containing, respectively,> r and not s; and s and not r. And given a natural number r and> in'nity, the set U_r = X-{r} is an open set containing in'nity but> nor r. There is no open set containing r and not in'nity, so the> topology is T0 but no T1. Now let A = {in'nity}. Then A' is all of N. And A'' is all of X, so> A'' is not contained in A'. -- > => It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> = Arturo Magidin> magidin@math.berkeley.edu> === Subject: Re: appeal to check argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Juxt17918;Fair point, Toni. Well here it is then - I posted it last night already. Ta, Mark G. >Does anyone have a couple of minutes to point out the errors if I >>post a two-screen-length argument about properties odd perfect >>numbers need to have?What makes you think if people haven't got time to do that, then they>have time to respond to you telling this is the case?Oh, and if you post something to sci.math, someone will point out the>errors. Don't worry about that.-- >I'm not interested in mathematics that might have anything>to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Silly question for someone with a big calculator. Theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Juwl17910;My Guess is as Good as Yours Theorem 2.1:Choose any alpha, beta in Z (whole numbers) such that 2|alpha| > |beta| > 1 in Z (whole numbers).De'ne x_m = coef'cient of x in (alpha x + alpha y + beta e)^m y_m = coef'cient of y in (alpha x + alpha y + beta e)^m e_m = coef'cient of y in (alpha x + alpha y + beta e)^mThen x_m = y_m for all m in naturals and the sequence(2x_m/y_m) converges to some number z in reals such that 1,4142 < z < 1,4200.Note: Surely, it should work as well when alpha, beta in reals (insteadof Z) full'll the above condition. But since I am only workingon the ring at the moment, that is how I thought I`d formulateit.//The rules of any input on this one.C. Dement === Subject: Correction, reply to lemma by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv2L18055;like a nice name for a woman) could check her result by noting that the sum of the coef'cients of x and ydivided by the coef'cient of e is always a constant. Thisis incorrect. However, perhaps more interestingly, it appearsto approach a constant:For ^2, below, 12*2/17 = 1,411764...For ^3, 408*2/577 = 1,414211...For ^4, 470832*2/665857 = 1,414213...>For example (2 x + 2 y + (-3) e) ^ 2 => = 4x*x + 4 x*y - 6x +4 y*x + 4y*y - 6y - 6x - 6y + 9 => = 17e - 12x - 12y> > Proceeding, I get > > (2 x + 2 y + (-3) e) ^ 4 = -408x - 408y + 577e> > (2 x + 2 y + (-3) e) ^ 8 = -470832x -470832y +665857e === Subject: Re: Silly question for someone with a big calculator. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv2518042;> For any given m in naturals which is a power a 2, we would like >> to 'nd the coef'cient (i.e., a whole number by de'nition) >> of the unit e in (2 x + 2 y + (-3) e) ^ m raised to the power>> of m, assuming that the associative/distibutive laws hold. The>> task is to 'nd for what m this coef'cient is not a prime number.>> >> The rules of the game are: x*x = y*y = e*e = e and >> x*y = -y*x>> >> >> For example (2 x + 2 y + (-3) e) ^ 2 = >> = 4x*x + 4 x*y - 6x +4 y*x + 4y*y - 6y - 6x - 6y + 9 =>> = 17e - 12x - 12y>> >> Proceeding, I get >> >> (2 x + 2 y + (-3) e) ^ 4 = -408x - 408y + 577e>> >> (2 x + 2 y + (-3) e) ^ 8 = -470832x -470832y +665857e>> >> which I quickly checked to be prime from >> http://www.numbertheory.org/php/prime_generator.html:Well if I've done it right, for 2^4 the coef'cient of e is you. However, you can check for yourself if it is correct or not: notice the sum of the coef'centsof x and y divided by the coef'cient of e is always a constant:Namely 1,414211... (! - I just saw this, myself. Of course, there will probably be some trivial reason for this.). C.Dement === Subject: Re: Silly believe you. However, you can check for yourself > if it is correct or not: notice the sum of the coef'cents> of x and y divided by the coef'cient of e is always a constant:> Namely 1,414211... (! - I just saw this, myself. Of course, there > will probably be some trivial reason for this.).De'ne always a constant:4/3 = 1.333333...24/17 = 1.411764...816/577 = 1.414211...941664/665857 = 1.414213...If we let k represent that ratio for one of the exponents, the ratio forthe next in this list is (4k)/(k^2+2); it can be shown that the ratioswill converge to sqrt(2).1 = 3^2-2^2-2^2 = 17^2-12^2-12^2 = 577^2-408^2-408^2 = ...-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: strange identi'cation topology class, andit was shown, with a rectangle with its bottom left corner on theorigin, four units in length and 2 units in height, how to create acylinder, Moebius strip, torus, and Klein bottle. For example,identifying (0, y) with (4, y) was a cylinder and (0, y) with (4, 1 -y) was a Moebius strip, while a Klein bottle was made by identifying(0, y) with (4, y) and then (x, 1) with (1-x, 0). So the question wasbrought up about what would the object be if we did both of thefollowing things:identify (0, y) with (4, 1-y)identify (x, 1) with (1-x, 4)Our professor had no idea how to visualize this or if it wasrecognized topological entity. So does anyone know what this is, whatit looks like, if it has a name, === Subject: Re: strange identi'cation spaces>We are talking about identi'cation spaces in my topology class,[...]>So the question was>brought up about what would the object be if we did both of the>following things:identify (0, y) with (4, 1-y)>identify (x, 1) with (1-x, 4)If I understand the corrections in your other posts, you mean youwant to identify the edges of a rectangle as follows:*----->>------*| || || |^ v| || || |*-----<<------*This is often presented in other ways; for example, it's equivalentto identifying antipodal points on the boundary of a disc, or toidentifying antipodal points on a sphere. They're all the ProjectivePlane RP^2. It doesn't embed in R^3, which may be why you can'tvisualize it easily. On the other hand,>Our professor had no idea how to visualize this or if it was>recognized topological entity. Please tell me your professor is teaching topology as part of an exchangeprogram with, I don't know, Control Theory or Mathematical Logic orsomething? This is sort of basic!dave === Subject: Re: strange identi'cation spaces>>Our professor had no idea how to visualize this or if it was>>recognized topological entity. Please tell me your professor is teaching topology as part of an exchange>program with, I don't know, Control Theory or Mathematical Logic or>something? This is sort of basic!His teacher is (if I did a good job of cyberstalking) a dynamicalsystems type. My theory is that this is an instance of the SocraticMethod (as they called it in Athens when the perfeseer trolledthe stoonts). Lee Rudolph === Subject: Re: strange identi'cation spacesCorrection: The rectangle is *one* unit in length. === Subject: Re: strange identi'cation spacesargh! I meant one unit in *height*. sorry for the multiple posts.So to reiterate: the rectangle is one unit in height, and four units in length. === Subject: laplace transforms and multiple intergrationDoes anyone know of any sites that you can enter the laplace transformand get the answer, and again the same === Subject: Calculus helpThis seems like it should be easy, but it is not doing it for me>the lim as x approaches 0 for the function (sin x) / xIf x= -0.1, the answer is supposed to be .9983.The answer book says Ensure you use radian mode.Why?? Why does this for any guidance! === Subject: Re: Calculus help> This seems like it should be easy, but it is not doing it for me> the lim as x approaches 0 for the function (sin x) / x If x= -0.1, the answer is supposed to be .9983. The answer book says Ensure you use radian mode. Why?? Why does this problem require guidance!If the angle is measured in degrees, the limit is pi/180, approx 0.01745.If degrees were used in calculus formulas, all of the trig derivatives, forexample, would have that ugly number as a constant in them. === Subject: Re: Calculus helpWhy?? Why does this problem require radian mode vice degree mode???Stumped>If you will look in your calculus book at the argument where it showsthe limit of sin(x)/x is 1, you will most likely 'nd a geometricargument. That argument will usually involve the formula for the areaof a sector of a circle A = (1/2)r^2*theta, probably with r = 1.Where did that formula come from? It comes from the proportion of thearea of the sector to the area of the circle:AreaSector/AreaCircle = SectorCentralAngle/CircleCentralAngleAreaSector/Pi*r^2 = theta / 2*Pi (using radian measure here!!)Hence AreaSector = (1/2)r^2 theta.So this formula is incorrect if you don't use radian measure, and sowould be the sin(x)/x limit.The same applies to any result using the s = r*theta formula for arclength.--Lynn === Subject: Re: Calculus help> This seems like it should be easy, but it is not doing it for me > the lim as x approaches 0 for the function (sin x) / x> > If x= -0.1, the answer is supposed to be .9983.> > The answer book says Ensure you use radian mode.> > Why?? Why does this problem require guidance!> > Because x_radians is not the same as x_degrees in measuring an angle, and sin(x_radians) is not the same as sin(x_degrees) in numerical value, any more than x_feet is the same distance as x_meters.Usually, in evaluating trigonometric functions, values of the function are based on the underlying angles, and the choice between radian and degree measure is ony a matter of convenience.However, this difference cannot be ignored in evaluating expressions like sin(x)/x, in which x appears both inside and outside of the sine function.For use here, and generally in calculus, angular measure should always be in radians. === Subject: Re: Calculus help> This seems like it should be easy, but it is not doing it for me > the lim as x approaches 0 for the function (sin x) / x> > If x= -0.1, the answer is supposed to be .9983.> > The answer book says Ensure you use radian mode.> > Why?? Why does this problem require radian mode vice degree mode???Let's see ... because degrees are not the same as radians? Consider for example sin(30 degrees)/(30 degrees), which numerically is (1/2)/30 = 1/60. Replacing degrees with radians we get sin(pi/6)/(pi/6) = (1/2)/(3.14../6), which is pretty close to 1. If you look at the geometry of the situation, you'll see why radians are the natural candidate to insure the above limit is 1. === Subject: <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> HI,> > Can someone tell me the results of the following integration?> > int_{-infty+j*epsilon}^{infty+j*epsilon}frac{1}{v(v-j*a)(v+j*b )}dv> > where v is complex variable,a,b,epsilon are real to Maple...> with(student):> assume(epsilon>0);> assume(b>0);additionally(b>epsilon);> assume(a>0);additionally(a>epsilon);> assume(t,real);> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)),> t=-in'nity .. in'nity); / I I 2 I Pi |- ------------- - --------------| a (I a - I b) b (-I a + I b)/ with assumptions on a and b> simplify(%); 2 I Pi ------ a b with assumptions on a and b> But if a, Tao Cui > HI,>> >> Can someone tell me the results of the following integration?>> >> int_{-infty+j*epsilon}^{infty+j*epsilon}frac{1}{v(v-j*a)(v+j*b )}dv>> >> where v is complex variable,a,b,epsilon are real According to Maple...> with(student):>> assume(epsilon>0);>> assume(b>0);additionally(b>epsilon);Since the pole at -I*b is below the real axis and the path t + I*epsilonlies above the real axis, the relative sizes of b and epsilon do notaffect the integral. That is, we don't need to assume b>epsilon.>> assume(a>0);additionally(a>epsilon);>> assume(t,real);>> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)),>> t=-in'nity .. in'nity); / I I > 2 I Pi |- ------------- - --------------|> a (I a - I b) b (-I a + I b)/ with assumptions on a and b> simplify(%); 2 I Pi> ------> a bThe residues at the poles of 1/(v(v-ja)(v+jb)) are 1 - ------ at ja a(a+b) 1 -- at 0 ab 1 - ------ at -jb b(a+b)With the assumption that a>epsilon, the path given for the integral canbe viewed as a limiting case of a contour circling the pole at ja oncecounterclockwise, and so the integral should be 2 pi j - ------ a(a+b)The value given by Maple is for a contour circling 0 counterclockwise.I don't off hand see anything wrong with the formula you gave to Maple,so, assuming there was not a copy and paste error, it would seem thatMaple is in error.> with assumptions on a and b> But if atake out the trash before replying === Subject: Re: a in'nite integration <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> > The value given by Maple is for a contour circling 0 counterclockwise.> I don't off hand see anything wrong with the formula you gave to Maple,> so, assuming there was not a copy and paste error, it would seem that> Maple is in error.> I tried to reproduce the error today, but couldn't...> with(student):> assume(epsilon>0);> assume(b>0);additionally(b>epsilon);> assume(a>0);additionally(a>epsilon);> assume(t,real);> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)),> t=-in'nity .. in'nity); 2 Pi ------------- a (I a + I b) with assumptions on a and b> simplify(%); 2 I Pi - --------- a (a + b) with assumptions on a and b> assume(epsilon>0);> assume(b>0);additionally(b assume(a>0);additionally(a>epsilon);> assume(t,real);> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)),> t=-in'nity .. in'nity); 2 Pi ------------- a (I a + I b) with assumptions on a and b> simplify(%); 2 I Pi - --------- a (a + b) with assumptions on a and b> assume(epsilon>0);> assume(b>0);additionally(b>epsilon);> assume(a>0);additionally(a assume(t,real);> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)),> t=-in'nity .. in'nity); 0> assume(epsilon>0);> assume(b>0);additionally(b assume(a>0);additionally(a assume(t,real);> int(1/((t+I*epsilon)*(t+I*epsilon-I*a)*(t+I*epsilon+I*b)),> t=-in'nity .. in'nity); 0These agree with yours. So maybe it was a copy and paste error...Or sunspots... Or little green men...-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Combinatoric:4 groups of 4Look up references on 'nite geometries or combinatorial designs, andaf'ne planes in particular.> This is a planning problem for scheduling players in teams of 4.> Not sure if it is feasible.> > 16 players, 5 days of play, 4 teams each day.> Each player teamed with each other player only once over the 5 days....> > Any suggestions or pointers to places of study are much appreciated. === Subject: Re: Combinatoric:4 groups of 4> This is a planning problem for scheduling players in teams of 4.> Not sure if it is feasible.> > 16 players, 5 days of play, 4 teams each day.> Each player teamed with each other player only once over the 5 days.> Look up references on 'nite geometries or combinatorial designs, and> af'ne planes in particular.Or look for references on bridge rotations (or whist rotations), as this sort of thing has been studied in depth by tournament organizers over the years. Maybe even post the question to a bridge group.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Recursive ExpansionIs there a way to explicitly compute an arbitrary element of a list that isgiven by the expansionR_0 = nullR_{n+1} = 0,R_n,R_n,R_n,1where we are putting three copies of the previous list between 0 and 1. Iwant to know how to randomly access an element in that list without usingrecursion. It is also necessary that no requirements external storage spacegrow larger than n. === Subject: Re: given by the expansionR_0 = null>R_{n+1} = 0,R_n,R_n,R_n,1where we are putting three copies of the previous list between 0 and 1. I>want to know how to randomly access an element in that list without using>recursion. It is also necessary that no requirements external storage space>grow larger than n.First of all, it is easily checked by recursion that |R_n| = 3^n - 1.We seek R_n[k]. R_n[0] = 0 and R_n[3^n-2] = 1. If k is neither 0 nor3^n-2, then R_n[k] = R_{n-1}[k-1 mod 3^{n-1}-1]. Continue until eitherk = 0 or k = 3^n-2. This will be true at least by the time n = 1.For example,R_3[10] = R_2[1] = R_1[0] = 0R_10[32768] = R_9[13085] = R_8[6524] = R_7[2151] = R_6[694] = R_5[209]= R_4[48] = R_3[21] = R_2[4] = R_1[1] = 1R_6[940] = R_5[213] = R_4[52] = R_3[25] = 1I suspect that there might be a neater method involving the base 3representation of the index.Rob Johnson take out the trash before replying === Subject: Re: Recursive Expansion>Is there a way to explicitly compute an arbitrary element of a list that is>given by the expansion>R_0 = null>R_{n+1} = 0,R_n,R_n,R_n,1>where we are putting three copies of the previous list between 0 and 1. I>want to know how to randomly access an element in that list without using>recursion. It is also necessary that no requirements external storage space>grow larger than n.R_n has length 3^n-1. If we start numbering at 0, element number j of R_n is f(j,n) = { 0 if j = 0 { 1 if j = 3^n-2 { f((j-1) mod (3^(n-1)-1), n-1) otherwiseThis can be written using iteration instead of recursion.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Recursive Expansion> Is there a way to explicitly compute an arbitrary element of a list that> is given by the expansion> > R_0 = null> R_{n+1} = 0,R_n,R_n,R_n,1> > where we are putting three copies of the previous list between 0 and 1. I> want to know how to randomly access an element in that list without using> recursion. It is also necessary that no requirements external storage> space grow larger than n.This gets very big very quickly:R0R1 0,,,,1R2 0,0,,,,1,0,,,,1,0,,,,1,1R3 0,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0 ,,,,1,0,,,,1,1,1R40,0,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,, 1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,1,0,0,0,,,,1,0,,,,1,0,,,, 1,1,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,1,0,0,0, ,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,,1,0,,,,1,1,0,0,,,,1,0,,,, 1,0,,,,1,1,1,1Are you sure about those commas?-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, === etc: http://users.powernet.co.uk/etonSubject: Re: Recursive > > Is there a way to explicitly compute an arbitrary element of a list that is> given by the expansion> > R_0 = null> R_{n+1} = 0,R_n,R_n,R_n,1> > where we are putting three copies of the previous list between 0 and 1. I> want to know how to randomly access an element in that list without using> recursion. It is also necessary that no requirements external storage space> grow larger than n.Generate the 'rst few items in the list and look for a pattern. Determine how to generate the pattern, given n. You may need to keeptrack of a level counter as you go through the pattern, equivalent to arecursion level. You may also need to keep track of a modest amount o'nformation in an array.Thad === Subject: Re: Tautological spaces> I did a post recently where I said the base tautological space that> mathematicians operate in is x = 0(mod x), and I realized later that's> wrong as it's 1 = 0(mod 1). That's the base tautological space where by tautological space I mean> a region of truth. In mathematics it's then a region of precision of your thought processes approximates the precision withwhich a demented goat could carve a small stellated dodecahedron out of Brieby using a broken plastic fork Scotch taped to one of its horns.-- Clive Toothhttp://www.clivetooth.dk === Subject: Re: Tautological spacesX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html> I did a post recently where I said the base tautological space that>> mathematicians operate in is x = 0(mod x), and I realized later that's>> wrong as it's 1 = 0(mod 1).>> That's the base tautological space where by tautological space I mean>> a region of truth.>> In for clarifying that, James.The precision of your thought processes approximates the precision with>which a demented goat could carve a small stellated dodecahedron out of Brie>by using a broken plastic fork Scotch taped to one of its horns.I think that's an accurate description of the precision of his writing. I for one don't want to speculate on the thoughtprocesses that lead him to write the things he does -the idea scares me.************************David C. Ullrich === Subject: Re: Tautological spaces Discussion, post recently where I said the base tautological space that>> mathematicians operate in is x = 0(mod x), and I realized later that's>> wrong as it's 1 = 0(mod 1).>> That's the base tautological space where by tautological space I mean>> a region of truth.>> In mathematics it's then a region of precision of your thought processes approximates the precision with> which a demented goat could carve a small stellated dodecahedron out of Brie> by using a broken plastic fork Scotch taped to one of its horns.Not fair! He's since clari'ed this notion. Region of mathematicaltruth was only loose talk. A tautological space is *really* alogical domain.-- ...you are around so that I have something else to do when I'm not'guring something important out. I was especially intrigued on thisiteration by cursing, which I think I'll continue at some later dateas it's so amusing. --- James S. Harris === Subject: Re: Dense Subset of Sobolev Space?> > Hi Everybody, I'm interested in 'nding a dense set in the Sobolev>space of functions on the interval [-1,1], such that>f' is in L2[-1,1], with the inner product (f,g)=int( f'(x)*g'(x) + f(x)*g(x), x = -1..1). I think -- but am not sure -- that the set of ('nite)>linear combinations of x and exp(i*n*x) (where n runs over all>integers) is dense. > > I think that the exp(i*n*x) are dense, but that has to do> with nonharmonic Fourier series and the fact that> 1 < pi; I think, but am not sure, that you simply left out> a pi somewhere. I'm going to assume you meant> L2[-Pi, Pi] instead of L2[-1,1]; you could instead talk> about [-1,1] and put some Pi's into the exponentials.> > You also need to note that since you're clearly including> an inner product! You need to use the complex> conjugate of g and g'.> > With those modi'cations it's clear that the span of> the exponentials is dense in the space; let's call> the space H.That seems false to me. Suppose P_n are trig polys converging to x in this space. Then P_n' -> 1 in L2 = L2[-Pi, Pi]. But the trig poly P_n' has no constant term, so the distance from P_n' to 1 is at least 1 by orthogonality. (In fact no space of 2Pi-periodic functions in H can be dense in H; they all miss x.)Suppose we do this: For f in L2 let I(f)(x) = integral_[0,x] f(t) dt. If D is a dense subspace of L2, then {I(f) : f in D} + {constants} is dense in H. Proof: If f is in H, then f = I(f') + constant. So if f_n in D converge to f' in L2, then I(fn) -> I(f') = f + c in L2, so that I(fn) - c -> f in L2 and (I(fn) - c)' = fn -> f' in L2 as desired. This shows that {trig polys} + span{x} is dense in H; same for ordinary polynomials, ie, span {1, x, x^2, ...}. === Subject: Re: Dense Subset of Sobolev Space?X-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html> >> Hi Everybody,>> I'm interested in 'nding a dense set in the Sobolev>>space of functions on the interval [-1,1], such that>>f' is in L2[-1,1], with the inner product>> (f,g)=int( f'(x)*g'(x) + f(x)*g(x), x = -1..1).>> I think -- but am not sure -- that the set of ('nite)>>linear combinations of x and exp(i*n*x) (where n runs over all>>integers) is dense. >> >> I think that the exp(i*n*x) are dense, but that has to do>> with nonharmonic Fourier series and the fact that>> 1 < pi; I think, but am not sure, that you simply left out>> a pi somewhere. I'm going to assume you meant>> L2[-Pi, Pi] instead of L2[-1,1]; you could instead talk>> about [-1,1] and put some Pi's into the exponentials.>> >> You also need to note that since you're clearly including>> an inner product! You need to use the complex>> conjugate of g and g'.>> >> With those modi'cations it's clear that the span of>> the exponentials is dense in the space; let's call>> the space H.That seems false to me. Suppose P_n are trig polys converging to x in this >space. Then P_n' -> 1 in L2 = L2[-Pi, Pi]. But the trig poly P_n' has no >constant term, so the distance from P_n' to 1 is at least 1 by >orthogonality. (In fact no space of 2Pi-periodic functions in H can be >dense in H; they all miss x.)Well of course you're right, which raises the question of how Icould have got this wrong... Ah - I was actually thinkingabout the Sobolev space on the circle, not the interval; thefunction x (or rather the 2-Pi periodic function which equalsx on the interval) is not in that space. What I said is ok forthe Sobolev space on the circle.>Suppose we do this: For f in L2 let I(f)(x) = integral_[0,x] f(t) dt. If D >is a dense subspace of L2, then {I(f) : f in D} + {constants} is dense in >H. Proof: If f is in H, then f = I(f') + constant. So if f_n in D converge >to f' in L2, then I(fn) -> I(f') = f + c in L2, so that I(fn) - c -> f in >L2 and (I(fn) - c)' = fn -> f' in L2 as desired. This shows that {trig >polys} + span{x} is dense in H; same for ordinary polynomials, ie, span {1, >x, x^2, ...}.Or we could 'x what I said: Instead ofNow it follows that the trigonometric polynomialsare dense in H: Given f in H, you can 'nd atrig poly Q such that ||f' - Q||_2 < epsilon/c, andthen the above shows that there is a trig polyP such that ||f||_H < epsilon.(oops, there was also a typo) we sayNow it follows that the trigonometric polynomialsplus the span of x is dense in H: Given f in H, by subtracting a multiple of x you canassume wlog that f(-Pi) = f(PI). Now int(f') = 0, so you can 'nd a trig poly Q with no constant termsuch that ||f' - Q||_2 < epsilon/c, and then the above shows that there is a trig poly P such that ||f - P||_H < epsilon.So the exponentials plus any for clarifying that.************************David C. Ullrich === Subject: Re: discrete mathematics fat sets>Given a set S {1,2,3,...,n}A subset P is said to be a fat set if every element in it is >= the>cardinality of the subset P.The problem is to 'nd the number of such fat sets. Also to come up>with a recurrsion logic.> >Hint: How can you recursively generate the fat subsets of {1, 2, 3, ..., n+1} from those of {1, 2, 3, ..., n}?Letting c_n be the number of fat subsets of {1, 2, 3, ..., n}, I get the recursionc_(n+1) = 2 c_n - sum(k=1..§oor((n+1)/2), C(n-k, k-1)),where C(i,j) is the binomail coef'cient, i choose j. This checks for 0 <= n <= 5. Interestingly, empirically, it seems that c_n = F_(n+2), the (n+2)nd Fibonacci number (with F1 = F2 = 1). I haven't veri'ed this from the recursion, nor can I come up with an explanation why this is so.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: discrete mathematics fat sets* Stephen J. Herschkorn> >Given a set S {1,2,3,...,n}A subset P is said to be a fat set if every element in it is >= the>cardinality of the subset P.> Interestingly, empirically, it seems that c_n => F_(n+2), the (n+2)nd Fibonacci number (with F1 = F2 = 1). I> haven't veri'ed this from the recursion, nor can I come up with an> explanation why this is so.Here is an attempt:For F_{n+1}, the set is S_{n+1}={1,2,...,n+1} we have that all thesets in S_n that are fat are still fat in S_{n+1}, there are F_n suchsets.So we need to include all the sets where n+1 is a member. Look at setS_{n-1} and list all the fat sets: fat(n-1)={, [1], [2], ..., [n-1], [2,3], ...} We know that each such set U in fat(n-1) is such that any memaber xof U is such that x>=|U|. Make the familiy fatnew(n-1) that is equalto fat(n-1) with the element n+1 added. I.e: fatnew(n-1)={[n+1], [1,n+1], [2,n+1], ..., [n-1,n+1], [2,3,n+1], ...} We of course expect that some of the members of fatnew(n-1) are nolonger fat. However, add 1 to all member not equal to n+1: fatnew'(n-1)={[n+1], [2,n+1], [3,n+1], ..., [n,n+1], [3,4,n+1], ...} We claim that the collection fatnew'(n-1) contains only fat sets.Why? Well, any set U in fatnew is such that x in U => x>=|U|-1.After every element is increased, we can conclude that x in U => x-1 >= |U|-1, or x>=|U|. (The new element n+1 does notdestroy the property of any set since no set is greater that n.)The collection fatnew'(n-1) has in fact only sets from S_{n+1} andfatnew'(n-1) has no common elements with fat(n). We have now twocollections of fat sets from S_{n+1}.Are there any more fat sets with the element n+1. No, but onlyhandwaving can show it. :-)Therefore, F_{n+1} = |fat(n)| + |fatnew'(n-1)| = F_n + F_{n-1}QED-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@i'.uio.no http://www.i'.uio.no/~jonhaug/, Phone: +47 22 95 21 52 === Subject: Re: discrete mathematics fat sets>Given a set S {1,2,3,...,n}A subset P is said to be a fat set if every element in it is >= the>cardinality of the subset P.The problem is to 'nd the number of such fat sets. Also to come up>with a recurrsion logic.Hints: 1) S is a fat subset of {1,...,n-1} iff (S+1) union {n+1} is a fat subsetof {1,...,n+1}, where S+1 = {s+1: s in S}. 2) Does Fibonacci ring a bell?Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: discrete mathematics fat sets> > Given a set S {1,2,3,...,n} A subset P is said to be a fat set if every element in it is >= the> cardinality of the subset P. The problem is to 'nd the number of such fat sets.> > In'nitely Countable> nulset> {n}, n >= 1> {n,m}, min n,m >= 2> ...> {n1,.. nj }, min (n1,.. nj) >= j> ...Back to the question... GIven n, how many fat subsets of S ={1,2,3,...,n} exist? === Subject: Sequence A001511If anyone had looked up the sequence 1, 2, 1, 3, ... , 5 in the On-line Encyclopedia of Integer Sequences, they would have found itlisted as sequence A001511. === Subject: Re: Sequence A001511>If anyone had looked up the sequence 1, 2, 1, 3, ... , 5 in the >On-line Encyclopedia of Integer Sequences, they would have found it>listed as sequence A001511.So what?!?Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === Subject: Re: Sequence A001511>If anyone had looked up the sequence 1, 2, 1, 3, ... , 5 in the>On-line Encyclopedia of Integer Sequences, they would have found it>listed as sequence A001511. So what?!? I believe there was originally some context to his comment (ïWhat is the> next term in this sequence' or some such) but this has been === snipped and> the Subject: header changed. That context, alas, has now been lost, as apparently neither you nor I> have the time or inclination to work out which particular thread the post> belongs to. -- > P.A.C. Smiththread: SeriesWhat are the next ten characters in the following series?1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 === Subject: Proof for Median Minimizing sum[] abs(x_i - M)I can't for the life of me 'nd a proof of this on the net and I don'thave any books to look it up in. Could someone please point me in theright direction?The median M of a collection of numbers minimizes the following:sum[i=1 to N] abs(x_i - M) === Subject: Re: Proof for Median Minimizing sum[] abs(x_i - M)>I can't for the life of me 'nd a proof of this on the net and I don't>have any books to look it up in. Could someone please point me in the>right direction?The median M of a collection of numbers minimizes the following:sum[i=1 to N] abs(x_i - M)>The sum is piecewise linear and convex in M, so it is minimized at M = x_i for some i. Show that if i < n-i, then the sum is greater when M = x_(i+1) than when M = x_i.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Fat subsets of ordinals>Given a set S {1,2,3,...,n}A subset P is said to be a fat set if every element in it is >= the>cardinality of the subset P.The problem is to 'nd the number of such fat sets. Also to come up>with a recurrsion logic.> >Now let's consider fat subsets of ordinals. That is, taking the ordinal de'nition of cardinals and letting |.| denote cardinality, say a set A of ordinals is fat iff a >= |A| for all a in A. We work in ZFC.For an ordinal a, de'neS_a = {A subset of a: A is fat};c_a = |S_a|, andd_a = |P(a) S_a|.(P(.) denotes power set.)Seems to me that, given an in'nite cardinal k,k <= a < k2 (ordinal multiplication) implies c_a = sup {2^m: ma partial difference equation is key. No one else in recorded human>history managed such a feat, but mathematicians refuse to give me the>time of day. The bastards!!!>Hell yeah!......what? I am little and innocent and emotionally charged speech thrills me =)>Because it's a partial difference equation, it leads to a partial>difference equation, Because it's a partial difference equation, it leads to a partial difference equation. teeheehee. I suppose the second equation should be function. Then again, it could be bzioung and the statement would be just as valid :)>which could be key in research to prove or>disprove the Riemann Hypothesis.Oh Lord I just had an inspiration! Using your partial difference equation i have managed to prove that the Riemann Hypothesis is probably either true or false!But mathematicians today belong to a democracy, and they don't *like*>me, so they don't like my research.Try this: go anonymous or switch names, rename your functions etc etc and then post your research again. Chances are, while they won't know it is you and will not immediately recognize the research as yours(no comments please :P) , they still won't like it.What boring people.I can be quite entertaining. === Subject: Modular ArithmeticI was wondering if anyone knows a way to reduce 2^100 mod 5. I knowthe answer is 1 (2^100 mod 5 = 1), but I had to use Maple to 'nd what2^100 alot. === Subject: Re: Modular ArithmeticIn sci.math, Shawn Windle< wondering if anyone knows a way to reduce 2^100 mod 5. I know> the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to 'nd what> 2^100 was. Does anyone know an easier way to 'nd the is to note that:2 = 2 (mod 5)2^2 = 4 (mod 5)2^3 = 8 = 3 (mod 5)2^4 = 16 = 1 (mod 5)Ah ha...if 2^4 = 1 (mod 5) then so is 2^8, 2^12, etc.which means 2^100 = 1 (mod 5) since 100 = 25 x 4.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Modular Arithmetic> > 100 > How to reduce 2 mod 5 4 25HINT (1 == 2 ) (mod 5)-Bill Dubuque === fearloathing2001@yahoo.com (Shawn Windle)>I was wondering if anyone knows a way to reduce 2^100 mod 5. I know>the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to 'nd what>2^100 a>lot.As shown in other subthreads, there are quick ways using the fact thata^{p-1} = 1 mod p when p is a prime. Here is a method that is of someuse when the modulus is not necessarily a prime and is still quite fast.100 in base 2 is 1100100.Each time you add a bit to the exponent, you square the number if theadded bit is a one, then you multiply by the base as well:2^(0 base 2) = 1 mod 52^(1 base 2) = 1^2*2 = 2 mod 52^(11 base 2) = 2^2*2 = 3 mod 52^(110 base 2) = 3^2 = 4 mod 52^(1100 base 2) = 4^2 = 1 mod 52^(11001 base 2) = 1^2*2 = 2 mod 52^(110010 base 2) = 2^2 = 4 mod 52^(1100100 base 2) = 4^2 = 1 mod 5So 2^100 = 1 mod 5.Rob Johnson take out the trash before replying === Subject: Re: Modular ArithmeticEn el Windle escribi.97:> I was wondering if anyone knows a way to reduce 2^100 mod 5. I know> the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to 'nd what> 2^100 was. Does anyone know an easier way to 'nd the >2^100 = (2^2)^50 Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Modular Arithmetic Adjunct Assistant Professor at the University of Montana.>I was wondering if anyone knows a way to reduce 2^100 mod 5. I know>the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to 'nd what>2^100 was. Does anyone the residues modulo 5:2^1 = 2 (mod 5)2^2 = 4 (mod 5)2^3 = 3 (mod 5)2^4 = 1 (mod 5)Therefore, 2^{4k} = (2^{4})^k = 1^k = 1 (mod 5).Since 100 = 4*25, you get 2^{100} = (2^{4})^{25} = 1^{25} = 1 (mod 5).-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === Proofs - Please help quickly!I'm really having trouble with 4 trigonometric functions. Here they are:1) Find an exact expression for: sin(pi/12)2) Solve to 4 decimal places (0_<7123b25e.0402081559.219a24e3@ trigonometric functions. Here they are: 1) Find an exact expression for: sin(pi/12)180/pi deg/rad * pi/12 rad = 180/12 deg = 15degcos^2 15deg - sin^2 15deg = cos 30deg = sqrt(3)/21 - 2*sin^2 15deg = sqrt(3)/22*sin^2 15deg = 1 - sqrt(3)/2sin^2 15deg = 1 - sqrt(3)/2sin 15deg = sqrt(1 - sqrt(3)/2) 2) Solve to 4 decimal places (0_ sin2x + cosx = 0sin2x = 2 sinx cosx0 = 2 sinx cosx + cosx = cosx * (1 + 2 sinx)0 = 1 + 2 sinx or 0 = cosxcosx = 0 at pi/2, 3pi/21 + 2 sinx = 0 => sinx = -1/2sinx = -1/2 at 7 pi/6, 11 pi/6 3) Solve sin^2 x - 2sinx - 1 = 0 and 'nd the general solution.Standard quadratic formula.sin x = (2 +/- sqrt(4 + 4))/2 = 1 + sqrt(2) or 1 - sqrt(2)1 - sqrt(2) yields the approximate solution x = -0.42707858...1 + sqrt(2) yields no real solution, but the complex solutionx = Pi/2 + 1.528570919480 * i is available. (This accordingto GP/Pari.)It is not clear how the problem generalizes beyond the obviousaddition of 2 * N * Pi for any integer N, or the speci'cationof an arbitrary A, B, and C in the equationA * sin^2 x + B * sin x + C = 0, yieldingx = arcsin( (-B +/- sqrt(B^2 - 4*A*C))/(2*A) ) + 2*N*Pias one might expect. 4) Prove the following (this is a real toughy): tanx/secx = secxtanx = sinx/cosxsecx = 1/cosxtanx/secx = sinx/cosx * cosx = sinxThe problem appears misspeci'ed. If you wish to solve this equation,as opposed to proving it for general x, one can simply note that ifcosx = secx, then cos^2x = cosx * secx = 1, cos x = 1 or -1,and therefore x=0 or pi. Please help! Soon!Well, it's late Sunday here; by the time you get this the homework isprobably past due... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Proofs - Please help quickly!En el escribi.97:> I'm really having trouble with 4 trigonometric functions. Here they> are: 1) Find an exact expression for: sin(pi/12)Use sin(a/2) = sqrt((1 - sin(a))/2 or sin(a - b) = sin(a)cos(b) +cos(a)sin(b) with suitable a (and b)> 2) Solve to 4 decimal places (0_ sin2x + cosx = 0Use sin(2x) = 2sin(x)cos(x), then factorize> 3) Solve sin^2 x - 2sinx - 1 = 0 and 'nd the general solution.Let sin(x) = t, and solve quadratic.> 4) Prove the following (this is a real toughy): tanx/secx = secxYou mean tan(x)/sec(x) = sin(x). Express tan(x) and sec(x) in terms ofsin(x)-- Best (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Proofs - Please help quickly!>I'm really having trouble with 4 trigonometric functions. Here they are:1) Find an exact expression for: sin(pi/12)HINT: pi/12 = (1/2)(pi/6), and sin(pi/6) = 1/2. Use the half angleformulas.>2) Solve to 4 decimal places (0_sin2x + cosx = 0HINT: Use the doulbe angle formulas: sin(2x) = 2sin(x)cos(x); so0 = sin(2x)+cos x = 2sinx(x)cos(x) +cos(x) = cosx(2sin(x) + 1).This is zero if and only if either cos(x)=0, or 2sin(x)+1 = 0.>3) Solve sin^2 x - 2sinx - 1 = 0 and 'nd the general solution.HINT: substitute y = sin(x).>4) Prove the following (this is a real toughy): tanx/secx = secxHINT: tan(x) = sin(x)/cos(x) sec(x) = 1/cos(x) (a/b)/(c/d) = (ad)/(bc)By the way: you'll have a hard time proving it. I suspect you aremeant to show that tan(x)/sec(x) = sin(x).-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: The Lost Proof of FermatIf space is *quantized* yet also continuous, then it too, has theindivisible units, then a measurement of space means that Fermat'slast theorem holds, for it.According to the Pythagorean theorem:x^2 + y^2 = z^2All possible integer solutions are then rerpresented as:[a^2 - b^2]^2 + [2ab]^2 = [a^2 + b^2]^2a^4 -2(ab)^2 + b^4 + 4(ab)^2 = a^4 + 2(ab)^2 + b^4 = [a^2 + b^2]^2all odd numbers can be represented as:[a^2 - b^2] or Z^p - Y^pif Y is an even natural n and Z is odd, same for a and b .Fermat's last theorem, for integers a,b,Z,Y,p:[a^2 - b^2]^p + Y^p = Z^p[a^2 - b^2]^p = Z^p - Y^pa^2 - b^2 = [Z^p - Y^p]^[1/p]When Z^p - Y^p is a prime number, it cannot have an integer root. a^2 - b^2 is not an integer, for [Z^p - Y^p]^[1/p] , for a,b,Z,Y,p,unless p = 2. === Subject: Re: The Lost Proof of Fermat> If space is *quantized* yet also continuous, then it too, has the> indivisible units, then a measurement of space means that Fermat's> last theorem holds, for it.Blimey! So physics has some application afeter all!-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: interval for student living in Washington Dc. In school weare learning about different ways to compound interest, interval andcontinuously. I had to look deeper, however, to say what is theinterval for continuously compounding interest?. This is easilydetermined by setting the 2 well know equations equal to each other:(P(1+(R/x))^(x*T))=(P*e^(R*T))However, I have not been able to solve this equation for X. Can anyonehelp me? === Subject: Re: interval for continuously equations now... === Subject: Re: interval for continuously Washington Dc. In school we> are learning about different ways to compound interest, interval and> continuously. I had to look deeper, however, to say what is the> interval for continuously compounding interest?. This is easily> determined by setting the 2 well know equations equal to each other:> (P(1+(R/x))^(x*T))=(P*e^(R*T))> However, I have not been able JacobConsider interest compounded every year, quarter, month, week, day, hour,minute, second, etc. Continuous compounding is the limit of this sequenceas the interval shrinks to zero. To make your equation correct, you musttake the limit of the left-hand side as x goes to in'nity. As an example,let r=0.06, say, and see how money you have after one year with differentfrequencies of compounding.Gordon EverstineGaithersburg, MD === Subject: Re: interval for continuously compounding interest?> I had to look deeper, however, to say what is the> interval for continuously compounding interest?. This is easily> determined by setting the 2 well know equations equal to each other:> (P(1+(R/x))^(x*T))=(P*e^(R*T))> However, I have not been able to solve this equation for X.There is no interval for continuously compounding interest. That's why we call it continuous. If there were such an interval, then x would be just an ordinary number and you would use the left hand side above; e^(R*T) would not enter the picture. === Subject: Re: JSH: Don't talk to me> > Jim, my son, what's troubling you? > > Let me know.> > I'm waiting.> > GodEloi, Eloi, lama sabachthani? === Subject: Re: JSH: Don't talk to me> > So it's ok for me to continue calling you a disloyal American who is a disgrace> to> the armed services of which you claim to be a veteran?> Hah, that's our President. === Subject: Re: JSH: Don't sha1:a1Dl4o/SuSC8P0yM67C8BG25m9c=>> >> So it's ok for me to continue calling you a disloyal American who>> is a disgrace to the armed services of which you claim to be a>> veteran?>> Hah, that's our President.James Harris is our president?President Gar'eld was the author of an original proof of thePythagorean theorem[1]. I suppose that if James S Harris is thepseudonym of George W. Bush, the *real* prover of Fermat's lasttheorem and other mathematical marvels, then this is a continuation ofa great tradition (if Gar'eld by himself counts as a tradition, thatis).Footnotes: [1] At least, that's what I understood from my copy of ProofsWithout Words, which presents the proof.-- Jesse F. HughesIf anything is true in general about Usenet, it's that people can goon and on about just about anything. -- James Harris speaks thetruth. === Subject: Re: JSH: Don't talk to me I don't necessarily mind mocking responses. Exactly. It isn't mockery or abuse that gets to you, it is simple and> clear proof that you are wrong. That's the unforgiveable offence, isn't it? Gib Hell no!!! That's a relief!!! I'm not a mathematician. I'm some guy> who decided he'd go looking for something that might have been missed> in the great rush of math society to build upon itself. And I found it. Like don't try the bull of saying I don't admit when I'm wrong,> when time after time over a period of years I have.Yeah, after you've been smacked with the 2x4-of-truth enough times to leave a dent.> And besides there's my prime counting function which any person out> there with the balls to go do a Google search on can see is unique in> that it uses a partial difference equation, and it doesn't take long> to 'nd out that no one else in recorded history managed to 'nd such> a gem. I think the problem is that today's mathematicians are pencil> pushers--and not in a good way--who do NOT have guts. Freaking cowards are running as fast as they can. So I'm in the process of chasing them down.You look more like a clown stumbling around in a McDonald's play room.> James Harris--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: JSH: Don't talk to me by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv2s18029;> Back in the 70's when it came out that Nixon had a secret list of >> journalists he disliked, it became a badge of honor to be on the list. >> In fact, many journalists NOT on the list were embarassed to have been >> omitted.I don't 'nd your analogy very convincing. Posts from the cult of JSH >detractors are often sad documents that should prove embarrassing to their >authors when and if their heads are ever removed from their posteriors. And >I don't think many sci.math posters are embarrassed at being ommited from >JSH's list, do you? It is true that there have been many nasty retorts to JSH's posts but oddly enough he has never seemed to be disturbed by those (except perhaps for reply as nastily). The three people that JSH lists here are people he speci'cally dislikes, apparently, because they are respond honestly and intelligently- attacking his statements rather than him. That appears to be what JSH can't stand. === Subject: Re: JSH: Don't talk to me>I don't 'nd your analogy very convincing. Posts from the cult of JSH >detractors are often sad documents that should prove embarrassing to their >authors when and if their heads are ever removed from their posteriors. And >I don't think many sci.math posters are embarrassed at being ommited from >JSH's list, do you?> > It is true that there have been many nasty retorts to JSH's posts but oddly > enough he has never seemed to be disturbed by those (except perhaps for > reply as nastily). The three people that JSH lists here are people he > speci'cally dislikes, apparently, because they are respond honestly and > intelligently- attacking his statements rather than him. That appears to > be what JSH can't stand.The idea that these posters always respond intelligently - attacking his statements rather than him - cannot be taken seriously. === Subject: Re: JSH: Don't talk to meX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>I don't 'nd your analogy very convincing. Posts from the cult of JSH >>detractors are often sad documents that should prove embarrassing to their >>authors when and if their heads are ever removed from their posteriors. And >>I don't think many sci.math posters are embarrassed at being ommited from >>JSH's list, do you?>> >> It is true that there have been many nasty retorts to JSH's posts but oddly >> enough he has never seemed to be disturbed by those (except perhaps for >> reply as nastily). The three people that JSH lists here are people he >> speci'cally dislikes, apparently, because they are respond honestly and >> intelligently- attacking his statements rather than him. That appears to >> be what JSH can't stand.The idea that these posters always respond intelligently - attacking his >statements rather than him - cannot be taken going to point this outwhen I read the above.But even in my case there's a germ of truth in what he said: thetruly rabid strings of obscenities that James has delivered to mehave been in reply to things I've said about the math. Bestexample:>>You see, the conventional thinking is that you can divide 7 from both>>sides and>>still be in the ring of algebraic integers, because algebraic integer>>are in'nitely decomposable, >> >> What??? Who thinks that?>> >> The only person I know who thinks this thing you call the>> conventional thinking is _you_.You stupid head!!! What the is wrong with you Ullrich?No matter how many ing times I tell you to off, you keep>replying to me!!!What the is your problem you head?You Ullrich are a stupid piece of dumb who refuses to get the>message when someone does NOT want to talk to you, you stupid ing>ty asshole.You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass>stupid self somewhere to GET A ING CLUE and QUIT ING REPLYING>TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!! OFF!!!!Can't you get it through your stupid head? OFF!!!!!!!!!!!!!!!!!(Yes, ïThe only person I know who thinks this thing you call theconventional thinking is _you_' might be regarded as a statementabout him if one looked at the words and ignored the meaning,but it's really a statement about the falsity of what he said, nota personal comment - he said everyone thinks something, andthe only person who thinks that is you is a simple denialof that statement of his.)************************David C. Ullrich === Subject: Re: JSH: Research question answered[snip incomprehensible pseudo-math about tautological space ]The only tautological space you have established is between your ears -- unfortunately, it is *not*tautological, but contradictory. Any given JSH identity lead to a contradiction.> James HarrisJSH Motto: When making love to a beautiful woman, I enhance the pleasure by fantasizing I am masturbating.--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: Concentric-Circles Game (to help teach basic number theory)I made this simple game up today, in part so as to help teach kidsbasic number-theory concepts (such a congruences and divisibility),but it might be far more useful as useless entertainment...if eventhat...Start with m concentric circles drawn on paper,where n is from 5 to 12 (or higher perhaps).The rings between the circles are subdivided so that the inner circleis subdivided into 2 equal-area sections, the next ring out issubdivided into 3 equal sections, the next ring into 4, etc..until theouter ring is subdivided into (m+1) equal sections.(And the sections are aligned so that one boundry of each is along asingle line.)Players start by each placing a game-piece (small enough to 't in the smallest section) into any unoccupied section on the board.A spinner or dice are used to pick a random integer n, where each nshould be between 1 and somewhere near m. On each move, a player can place a new game-piece (leaving earlierpieces) n rings inward (if possible) from their last game-piece'sposition, or n rings outwards (if possible).(Outward and inward are along a chain of sections which touch. Donot pass center circle or outer circle.)OR the player can place a game-piece clockwise or counterclockwise (inthe same ring)a *multiple* of n spaces from their last-placed piece.But in any case, a player can't change directions during his/her move.(And the player's {counter}clockwise moves may indeed circle around asingle ring as many times as needed.)A player can jump over occupied spaces when measuring the number ofspaces from their last piece, but they must ONLY place a piece on anunoccupied section.And the winner is the last player able to place a piece on the === Subject: Re: Concentric-Circles Game (to help teach basic number theory)> Start with m concentric circles drawn on paper,> where n is from 5 to 12 (or higher must've de'n'ly been somethingup when I saw the m and n, and now, after a fairbit of googling I sayWelcome Brother M or NIf you add four you get Von Neuman's number(I Know* zere are two of der n) === Subject: Re: :: towards a constructive education :: (news server friendly)You ask me to cut many parts out, so I have. I did read them, and still'nd that I don't quite understand this game of question a topic followedby anger at me responding, somehow implying that I am bringing up the topicswhen they are obviously yours. But I won't reply to them anymore.I didn't 'nd much about Heyting algebras in your response. You did mention[...]: Now let's get to what is wrong with your post. You've already worn: down my rage with your stamina, (I never expected you to reply, I: presumed it would be: below you) and it's Sunday, so my tone is softened down a bit. You: have had more negative responses than positive, and you really haven't: started any constructive discussion anywhere.If you look to my replies to Sergio Roa Ovalle and John Wilkins (evensomewhat Brian Scott) in the group 2 newgroups, you will 'nd the start of abibliography and detailed references to expand particular topics.: Some have openly: declared that they didn't even bother to read the whole lot of: nonsense. I actually printed it out and: read it through a few times. It really angered me a lot. Why?: Because of the hocus pocus with the W at the beginning.: Because of the endless list of assertions for which you neither: provide proofs, examples, or clues as to why they should be of: interdisciplinary interest.I'm also being told that I am mentioning topics beyond the level to beexpected by newsgroups. I am being told a lot of contradictory things aboutwhat people can comprehend and what they can't. I am detailing sources inthe leafs of this thread that request them, and although they arenecessarily only skeletal, they do form the foundations for the talk I amlooking for.: Because of your explanation in terms of insecurity, (You choose not to: interpret insecurity as something negative. But that is your: subjective choice. I think most people would prefer not to be: insecure), which is arrogant and belittling even if your personal: opinion is another.Contrast this with:: Because you keep rubbing under ones nose, how naturally it all comes: to you, as opposed to everybody else.: Because I couldn't get rid of the feeling that you are biassed against: boolean logic for private reasons.: The last one, combined with your failure to provide compelling reasons: for considering your point of view, and your galathaea: prankster,: fablist, magician, liar give me the impression of someone showing of: her stylish chic new theory just to impress others, like a teeny with: a newly discovered worldview. There are moral implications in what: your attitude re§ects, and I don't have the impression that I like: them.You have to see that this at least tells me you have, at least somewhat, aproblem with non-Boolean logics. My signature is a tease, yes, but it hasits explanation in my choice of the name galathaea for my usenet andelsewhere electronic persona, and I've written about this on codeguru.There is something about the moral implications of all of these non-Booleanlogics which you don't like.And that's all I really meant by that hunch of mine. It unsettles you insome way. You were angry that I might propose this, yet you use it inattack of my programme. So now I'm back believing it with a little higherordering in my belief hierarchy.: As for your intended topic, you mentioned the halting problem as an: example where one needs polyvalent logic. When I studied computer: science, there never: was a mention of the need for polivalent logic in this context. Maybe: I missed something. As I see it, for a given algorithm, and a given: input, you know it's halting behaviour or you don't. That is pretty: boolean to me. So I would like you to elaborate more on this example.: Then I would like you to establish: a relationship based on heyting structures and another area of: knowledge from among the ngs you posted to, and to explain why this: relationship is important: to both areas.Actually, attempting a Boolean algebra (and deriving a contradiction) is keyto the proof of the undecidability of the halting problem. A goodintroduction to the topic and proof is found in the cute Wikipedia, found athttp://en.wikipedia.org/wiki/Halting_problem. I explain in the other leafsI mention above what the lambda calculus has to do with Heyting algebrasincluding the Curry-Howard isomorphism.: My advice to you, if you want to be more successful in your quest to: gather a multidisciplinary group of people willing to discuss your: topic:: Try to establish a path of such examples, relationships and: explanations that goes through all of the disciplines you addressed.: Expose this path as: succinctly as you can, without compromising understandability.: Leave out all the hocus pocus stuff.: Leave out as much of your personal opinion as you can.: And consider dropping (this is again very personal, I'm afraid) the: prankster, fablist, magician, liar it is no great recommendation of: yourself. Maybe it works wonders among intelectuals of the liberal: arts, but I think it's a real turnoff for the more scienti'c oriented: mind.See, I thought building a path from vision research through conceptualstructures and cognitive work, moving on to external models of reality wouldwork to bring them all together, just as you suggest. Yet I found that manydescribed this as too long, and because they were unaware that the visionresearch was legitimate, they blew it off as hocus-pocus. Some stoppedreading, others stopped believing. I tried to build all of the connectionsin a summary fashion and have still been chastised for length. Others aretelling me to include more references.relationships among those that already are familiar with the logicalresearch, whatever 'eld they concentrate in. But I probably did need toadd some other path for those who may not be aware of the research and seemy comments as hand-waving.My signature will not change. All of it is true.-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar === Subject: Re: :: towards a constructive education :: (news server friendly)> You ask me to cut many parts out, so I have. I did read them, and still> 'nd that I don't quite understand this game of question a topic followed> by anger at me responding, somehow implying that I am bringing up the topics> when they are obviously yours. But I won't reply to them anymore.I asked you wether we could cut out 2 (not many) parts, provided wecould agree on a settlement. If you still feel the need to discussthem, just do so.Don't go implying I am censoring the conversation, as if you wouldcomply. : your attitude re§ects, and I don't have the impression that I like> : them.> > You have to see that this at least tells me you have, at least somewhat, a> problem with non-Boolean logics. No. That's like saying someone has a problem with evolution theory,just because he dislikes social darwinists.> My signature is a tease, yes, but it has> its explanation in my choice of the name galathaea for my usenet and> elsewhere electronic persona, and I've written about this on codeguru.> There is something about the moral implications of all of these non-Boolean> logics which you don't like.> > And that's all I really meant by that hunch of mine. It unsettles you in> some way. You were angry that I might propose this, yet you use it in> attack of my programme. So now I'm back believing it with a little higher> ordering in my belief hierarchy.I said I didn't like the moral implications of what your attitudere§ects, not the non-boolean logics. Come on, it is clearly writtenon the paragraph you mention. I don't thing the kind of formal systemsyou talk about have any moral properties or consequences. That youvalue something more if it proves to be provocative, 'ts into mypicture of you. But what is the value of provocation in itself? Lookat the response to your post. Most of the reaction you provoked isnegative. What if you had concentrated in provoking positive response,instead of just provoking?And it's not your programme I attacked, it was your arrogant bigotattitude. I would sincerely very much like your programme to get alongand produce some interesting results. > Actually, attempting a Boolean algebra (and deriving a contradiction) is key> to the proof of the undecidability of the halting problem. A good> introduction to the topic and proof is found in the cute Wikipedia, found at> http://en.wikipedia.org/wiki/Halting_problem. I read their cute (oops aren't we patronizing again?) proof sketch.I also read another proof which uses register machines. I don't seethe point you are trying to make. The proofs work by deriving acontradiction to the asumption that the halting problem is decidable.Everything happens within boolean logic. If I am missing something,please point it out to me and do not refer me to other posts whichonly contain vague assertions and more references to external sources. I think the facts speak for themselves, unless you discount themajority of negative responses as irrelevant. > My signature will not change. All of it is true.Yeah, as if anyone will believe you liar === Subject: Re: :: towards a constructive education :: (news server friendly) And stop grinning, until you put up a decent argument. It's just sad.What is it with imbeciles like you? decent argument ???Look Daddy, I got on the debating team!!!ad nauseum:-)mitch === Subject: Re: :: towards a constructive education :: (news server friendly)Hi Mitchie, still not back from kiddie league? I'm still waiting foran answer more worthy of yourself.> And stop grinning, until you put up a decent argument. It's just sad.> > What is it with imbeciles like you? decent argument ??? That must be an idiom with some funny connotations or the phrase is ofa subtlety that is lost upon me, since I'm not a native englishspeaker and also an imbecile. Come on Mitchie don't be a prick and letme in on the fun! === Subject: Re: :: towards a constructive education :: (news server friendly)> Hi Mitchie, still not back from kiddie league? I'm still waiting for> an answer more worthy of yourself.>Mostly, it isn't worth the effort.I spent the last year dealing with someone exhibiting the same crude behavior when I wastrying to talk about logic and set theory. Why don't you give me some sense of what youknow about Boolean-valued forcing, partition characteristics of large cardinals, or evendescriptive set theory?I do not have Galathaea's wide background. So, it is unlikely I would be able to mount adecent argument against someone as esteemed as yourself.My suspicion, however, is that you do not even understand the numbers they keep score within my league. But, then maybe you do.:-)mitch === Subject: [arccos((sqrt(5)-1)/2 )] / pi is irrationalHelloHow can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspectthis has something to with the golden ratio, and maybe with Fibonaccinumbers or continued === Subject: Re: <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> Hello> > How can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]?The real part of a root of unity cannot bean algebraic integer unless it is 0, 1, or -1.If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part(sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u)is not a root of unity, that is: u is irrational.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued fractions? But I could't start.There are several possible methods.What do you know of either Chebyshev polynomials or cyclotomic polynomials? Or what can you say about the polynomial1+2 t-2 t^2+2 t^3+t^4?Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued fractions? But I could't contain a misprint? G C === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued expression? Does the expression contain a misprint? > > G COH sorry! It's to prove [arccos((sqrt(5)-1)/2 )] / pi is irrational.Amanda === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect> this has something to with the golden ratio, and maybe with> Fibonacci numbers or continued the expression? Does the expression contain a misprint?>> G C OH sorry! It's to prove [arccos((sqrt(5)-1)/2 )] / pi is irrational.> AmandaHint : if a= arccos((sqrt(5)-1)/2), we have cos(a)=2cos 2*Pi/5 === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued expression? Does the expression contain a misprint? I was thinking the same thing, but then I looked at the subject title. === Subject: Mathematical Models and Methods in Applied Sciences - Vol 14 No 1Mathematical Models and Methods in Applied SciencesView table-of-contents and abstracts athttp://www.worldscinet.com/m3as.htmlContents:An Image Segmentation Variational Model With Free Discontinuities AndContour CurvatureGiovanni Bellettini And Riccardo MarchHomogenization Of Boundary Value Problems And Spectral Problems ForNeutron Transport In Locally Periodic MediaMustapha Mokhtar-KharroubiThe Main Inequality Of 3d Vector AnalysisGiles Auchmuty3d-Shell Elements And Their Underlying Mathematical ModelD. Chapelle, A. Ferent and K. J. BatheFlux Spikes In Viscous FlowsBai-He Tan And Zi-Niu WuFor more information, go to http://www.worldscinet.com/m3as.html === Subject: Re: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i192Qsb15702;>:=> Sorry, I forgot that e^x and arctan(x) are algebraic ...>:>:?? In what sense are e^x & arctan x algebraic? In usual mathematical >:parlance, these are called transcendental functions. >:I suspect irony; however, there is a classi'cation of transcendental>functions which puts e^x and arctan(x) in one compartment. Each satis'es>a differential equation (here even of 'rst order), where the differential>expression equated to zero is a non-trivial algebraic function (can be>made a polynomial) in the variables and the derivative: (d/dx) y - y = 0, satis'ed by y=e^x and others, (1+x^2) * (d/dx) y = 0 , satis'ed by y=arctan(x).I have seen a name algebraic transcendental function for these. It>sounds like an oxymoron, in the company of regular singular points. One>can make a hierarchy of algebraic transcendental functions by assigning>two integers to such an f: (1) the minimal order p of an algebraic differential equation satis'ed> by f,> (2) within the class of algebraic differential equations of order p, the> minimal degree of the polynomial in the variables which is satis'ed> by f.(Finer subdivisions are possible) . Examples of higher order algebraic trancendental functions are Bessel>functions.> Exercise: Many know the second order linear differential equation>satis'ed by sin(x): y''+y=0. Show that sin(x) is actually an algebraic>transcendental function of order 1.I think it was Charles Hermite who proved that the Gamma function,>although its reciprocal extends to an entire analytic function, is not>algebraic transcendental. I saw the proof in an old-fashioned Advanced>Calculus book.>hey do you guys have a copy of the proof of e is transcendental? ..id really appreciate if somebody could help me. ive been searching and found nothing. === Subject: Re: classes of transcendental numbers ?>:=> Sorry, I forgot that e^x and arctan(x) are algebraic ...>:>:?? In what sense are e^x & arctan x algebraic? In usual mathematical >:parlance, these are called transcendental functions. >:I suspect irony; however, there is a classi'cation of transcendental>functions which puts e^x and arctan(x) in one compartment. Each satis'es>a differential equation (here even of 'rst order), where the differential>expression equated to zero is a non-trivial algebraic function (can be>made a polynomial) in the variables and the derivative: (d/dx) y - y = 0, satis'ed by y=e^x and others, (1+x^2) * (d/dx) y = 0 , satis'ed by y=arctan(x).I have seen a name algebraic transcendental function for these. It>sounds like an oxymoron, in the company of regular singular points. One>can make a hierarchy of algebraic transcendental functions by assigning>two integers to such an f: (1) the minimal order p of an algebraic differential equation satis'ed> by f,> (2) within the class of algebraic differential equations of order p, the> minimal degree of the polynomial in the variables which is satis'ed> by f.(Finer subdivisions are possible) . Examples of higher order algebraic trancendental functions are Bessel>functions.> Exercise: Many know the second order linear differential equation>satis'ed by sin(x): y''+y=0. Show that sin(x) is actually an algebraic>transcendental function of order 1.I think it was Charles Hermite who proved that the Gamma function,>although its reciprocal extends to an entire analytic function, is not>algebraic transcendental. I saw the proof in an old-fashioned Advanced>Calculus book.> hey do you guys have a copy of the proof of e is transcendental? ..id really appreciate if somebody could help me. ive been searching and found nothing.Or you might look into Irrational Numbers, by Ivan Niven, a veryreadable book with a lot of good stuff about irrationality andtranscendance - in particular the transcendance of e and pi areconsidered. If you have available the complete papers of DavidHilbert, he came up with his own === Subject: e is transcendental (was: classes of transcendental numbers <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> hey do you guys have a copy of the proof of e is transcendental? ..id really> appreciate if somebody could help me. ive been searching and found nothing.Spivack, CALCULUS, 2nd edition, Chapter 21.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: classes of transcendental numbers ?>hey do you guys have a copy of the proof of e is transcendental? ..id>really appreciate if somebody could help me. ive been searching and>found nothing.You might look in Herstein, Topics in Algebra, section 5.2.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: classes of transcendental numbers ?> >>hey do you guys have a copy of the proof of e is transcendental? ..id>>really appreciate if somebody could help me. ive been searching and>>found nothing.> > You might look in Herstein, Topics in Algebra, section 5.2.Or type transcendence of e into Google. :-)-- --Tim Smith === Subject: Re: classes of transcendental numbers ?>hey do you guys have a copy of the proof of e is transcendental? ..id>>really appreciate if somebody could help me. ive been searching and>>found nothing.You might look in Herstein, Topics in Algebra, section 5.2.Or Stewart and Tall, Galois Theory have easily readable proofs oftranscendality of both e and pi. (That's the best feature of that book!) Derek Holt. === Subject: Re: grateful for comments: argument> If an odd perfect number exists and has three prime> factors a,b,c then we can partition the total list of> factors, both prime and composite, into those dividing> by a [call this ïa.sum'], remaining factors dividing> by b [b.sum], remaining factors dividing by c [c.sum],> and 1.> > For perfect number status, what we call ïcomplements'> of each sum must divide by that prime:> for example 1 + a.sum + b.sum = c.comp must divide by c. > In general x must divide x.comp for each of x = a,b,c.> > By examining each of > a.comp = 1 + b.sum + c.sum,> b.comp = 1 + a.sum + c.sum,> c.comp = 1 + a.sum + b.sum > we 'nd it is never the case that all three divide by > they primes they should, and hence there are no odd > perfect numbers. Further there are no even perfect > numbers dividing by more than two primes. What if n = 3^2 x 13^2 x 61? The divisors of n not divisible by 3 are (1 + 13 + 169) x (1 + 61) = 183 x 62, which is divisible by 3. The divisors of n not divisible by 13 are (1 + 3 + 9) x (1 + 61) = 13 x 62, which is divisible by 13. The divisors of n not divisible by 61 are (1 + 3 + 9) x (1 + 13 + 169) = 13 x 183 = 13 x 3 x 61, which is divisible by 61.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Limit Of Sum Over Some RationalsFirst, a simple looking limit, where I have revealed the (somewhatunexpected) closed-form at the post's bottom. --- t 1 r limit ----- / ---- , m -> oo m^2 --- e^r rwhich is, in linear-mode:limit{m-> oo} (1/m^2) sum{r} r^t /e^rAnd the r-sum is over all distinct positive rationals rwhich have denominators, in their simplest forms, which are each <= pi*m.--Okay, the lesson for today:Start with the 'nite double-sum identity:sum{k=1 to n} sum{j=1 to m, GCD(k,j)=1} f(k,j) =sum{i=1 to min(m,n)} mu(i) (sum{1<=k<=m/i} sum{1<=j<=n/i} f(ki,ji))- De'ning f(k,j) asa(j/m) b(k/m) c(k/j),and taking limits, after dividing both sides by m^2, we get:For positive q and s, --- 1 limit ----- / a(den(r)/m) b(num(r)/m) c(r) m -> oo m^2 --- r= distinct rationals, 1<=den(r)<=q*m 1<=num(r)<=s*m 6 /q /s= ---- | | pi^2 | | a(x) b(y) c(y/x) dy dx /0 /0(limit{m->oo} (1/m^2) sum{r= distinct rationals,1<=den(r)<=q*m,1<=num(r)<=s*m} a(den(r)/m) b(num(r)/m) c(r) =(6/pi^2)* integral{0 to q}integral{0 to s} a(x) b(y) c(y/x) dy dx) where a(), b(),c() are such that the integral exists, is de'ned, andthe identity above is correct...);)(And num(r) is numerator of reduced r, den(r) is denomiator ofreduced r.)-- So, we might have, if I am right so far,limit{m-> oo} (1/m^(t+u+2)) sum{r} (den(r))^u (num(r))^t /e^r= t! q^(2+t+u) (6/pi^2)/(2+t+u),where the r-sum is over all positive distinct rationals r,where den(r) <= m*q, q = 'xed positive real.--So, in answer to the original question:limit{m-> oo} (1/m^2) sum{r} r^t /e^r,where the r-sum is over all distinct positive rationals rwhich have denominators, in their simplest forms, which are each <= pi*m,is3 * t!. (?)(I 'nd this more interesting than otherwise because of the fact that3 is the integer closest to both pi Quet === Subject: Order/Chaos QuestionI would like to know of a number distributiongenerator that can be adjusted to like a rheostatto produce numbers between order or chaos.The problem is similar to mutating a perfectsine wave into less and less order, eventuallyturning it into random noise.I would like a randomness index built into itthat goes from 0 to 1, where zero is chaos and 1 isorder. All fractions in between can also be input.I would like it to generate integers within a speci'edrange from 0 to r, where r is the ceiling integer.I am only interested in the simplest case of numbersalong a one dimensional line segment. The distributionof integers is along the line segment.There should be another integer n that designateshow many points are plotted on the line segment,with the option of the points being either incombination (no point generated more than once) oroverlapping (the same point can be generated morethan once).++++Is it possible for noise from a known source, to betuned into resonance with noise from a referencesignal? === Subject: Re: Series> What are the next ten characters in the following series?> > 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5> > Lookup ID Number A001511 in the On-Line Encyclopedia of Integer SequencesIt should be noted that this sequence is such thata(2^n -k) = a(k),for all k where 1<= k <= 2^n -1.In other words, each sequence of the 'rst 2^n -1 terms, for n = Quet === Subject: Re: Series> > > What are the next ten characters in the following series?> 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5> > I think you mean ïsequence.'> 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 6> > Probably ïsymbols' vice ïcharacters' as well.> > Either way, you original use of series is incorrect.> > The 32nd ïsymbol' can either be ï6' or ï5'> > It is 6 if you want to make it at least remotely interesting. But if you > meant that one can come up with a rule so that the 32nd symbol can be > either ï5' or ï6', then you might as well say that the 32nd symbol can be > any other real number as well.> I was thinking of toggling bits. If there are only 5 bits totoggle, the 32nd number is 5. More then 've bits, then the 32ndnumber === $$f_w91tn_g > > while c=N-1 is not so representable.> > I can only prove this trivial case.> Suppose c=N-1 is representable in the form that c=ax+by> with x,y>=0 while N-1=(a-1)(b-1)-1=ab-a-b.> Then ab-a-b=ax+by and hence the equality> ab=(x+1)a+(y+1)b ...(1)> holds.> Since gcd(a,b)=1 and therefore x+1 is divisible by b> while y+1 is divisible by a, which implies that> (x+1)a+(y+1)b>=2ab, in contradiction to (1).> > > You can 'nd integers m, n >= 0 so that am - bn = 1 (since gcd(a,b) = 1). Then, for any integer c, the possible representatiuons of c are all of the form c = a(mc - kb) - b(nc - ka) for some integer k. So the problem reduces to 'nding some choice for k that forces the inequalities mc - kb >= 0, nc - ka <= 0 to be satis'ed. That's equivalent to being able to choose k satisfying nc/a <= k <= mc/b . Note that (mc/b) - (nc/a) = ((am - bn)c)/(ab) = c/(ab) -- thus, the interval [nc/a, mc/b] contains at least one integer as soon as c >= ab (since, then, that interval has length >= 1). This proves the result for c >= ab. How about the remaining values for c ?? If N <= c < ab, consider c + a + b which is > ab (by the way in which N is de'ned). If c + a + b is not a multiple of either a or b, then c + a + b = ra + sb with r, s > 0 and it's immediate that c = (r-1)a + (s-1)b which is a representation of the correct sort. If c + a + b = sa, then c = (s-1)a - b; since s > b, you can write s-1 = qb+r with q >= 1, 0 <= r< b and then verify that c = ra + (q-1)b, taking care of this case. The argument when c + a + b turns out to be a multiple of b is (pretty much) the same. [][/quote:3bb60765ba]My second post has *previouly* exhibited a proof> > For any integer c>=N, it can be written as > c=ab-a-b+m while m is an > integer greater than 1. > > Since gcd(a,b)=1, there exists some integers s,t such that as-bt=1.> Multipy m on both sides and we have > ams-bmt=m ...(2) > > According to euclidean division, there exists some integers q,r suchthat > mt=aq+r, 0<=r > Let x*=ms-bq,y*=r=mt-aq and we have > ax*-by*=m ...(4) > by (2). > > Finally Let x=x*-1,y=a-1-y*. > Since y*=r<=a-1 by (3), y=a-1-y*>=0. > Thus ax*=by*+m>0, which implies x*>0 and it follows that > x=x*-1>=0. > > Thus ax+by=a(x*-1)+b(a-1-y*) > =ax*-by*+ab-a-b > =ab-a-b+m=c ( by (4) ), > indicating that x,y are the coef'cients required.> for the whole problem.> > Let N=(a-1)(b-1), where a,b are positive integers> and gcd(a,b)=1> Show that every integer c>=N is representable in the form > c=ax+by with x,y>=0, while c=N-1 is not so Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === szs107@psu.edu (Stas Sheynkop)>I found the following problem in an old number theory text by William>J. LeVeque. Neither I nor anybody I know can solve this problem. Let N=(a-1)(b-1), where a,b are positive integers>and gcd(a,b)=1>Show that every integer c>=N is representable in the form >c=ax+by with x,y>=0, while c=N-1 is not so representable.Since gcd(a,b)=1, we have some (u,v) so that au+bv=1. The set of allsolutions to ax+by=c is { (cu+bk,cv-ak) : k in Z }. Thus, we have anon-negative solution (x,y), i.e. one in which both x and y are non-negative, precisely when there is an integer k so that k >= -cu/b (sothat cu+bk >= 0) and k <= cv/a (so that cv-ak >= 0). Thus, ax+by=chas a non-negative iff there is an integer in [-cu/b,cv/a].Since au+bv=1, [-cu/b,cv/a] has length c/(ab). Suppose there is nointeger in this interval. This means that it must be contained in someinterval (j,j+1). Since cx and cy are integers, we must have -cu/b-j >= 1/b [1a]and j+1-cv/a >= 1/a [1b]Adding [1a] and [1b] and multiplying by ab gives ab-cau-cbv >= a+b [2]Since au+bv=1, [2] becomes c <= ab-a-b [3]Therefore, if c >= ab-a-b+1 = (a-1)(b-1), then there is a non-negativesolution (x,y) to ax+by=c.Suppose c = (a-1)(b-1)-1 = ab-a-b. Then, since au = 1-bv, -cu/b = -(ab-a-b)u/b = -u(a-1) - v + 1/b [4a]and since bv = 1-au cv/a = (ab-a-b)v/a = v(b-1) + u - 1/a [4b]Combining [4a] and [4b] and au+bv=1, we get [-cu/b,cv/a] = [-au+u-v + 1/b,-au+u-v + 1-1/a] [5]Thus, the interval in [5] contains no integers since it is contained inthe interval (j,j+1) where j = -au+u-v.Therefore, if c = ab-a-b = (a-1)(b-1)-1, there are no non-negativesolutions (x,y) to ax+by=c.Rob Johnson take out the trash before replying === Subject: Re: Number Theory Problem> > while c=N-1 is not so representable.> > I can only prove this trivial case. > Suppose c=N-1 is representable in the form that c=ax+by > with x,y>=0 while N-1=(a-1)(b-1)-1=ab-a-b.> Then ab-a-b=ax+by and hence the equality> ab=(x+1)a+(y+1)b ...(1) > holds.> Since gcd(a,b)=1 and therefore x+1 is divisible by b > while y+1 is divisible by a, which implies that > (x+1)a+(y+1)b>=2ab, in contradiction to (1).> > You can 'nd integers m, n >= 0 so that am - bn = 1 (since gcd(a,b) = 1). Then, for any integer c, the possible representatiuons of c are all of the form c = a(mc - kb) - b(nc - ka) for some integer k. So the problem reduces to 'nding some choice for k that forces the inequalities mc - kb >= 0, nc - ka <= 0 to be satis'ed. That's equivalent to being able to choose k satisfying nc/a <= k <= mc/b . Note that (mc/b) - (nc/a) = ((am - bn)c)/(ab) = c/(ab) -- thus, the interval [nc/a, mc/b] contains at least one integer as soon as c >= ab (since, then, that interval has length >= 1). This proves the result for c >= ab. How about the remaining values for c ?? If N <= c < ab, consider c + a + b which is > ab (by the way in which N is de'ned). If c + a + b is not a multiple of either a or b, then c + a + b = ra + sb with r, s > 0 and it's immediate that c = (r-1)a + (s-1)b which is a representation of the correct sort. If c + a + b = sa, then c = (s-1)a - b; since s > b, you can write s-1 = qb+r with q >= 1, 0 <= r < b and then verify that c = ra + (q-1)b, taking care of this case. The argument when c + a + b turns out to be a multiple of b is (pretty much) the === same. []Subject: Re: Number Theory [quote:0f5f20a1f4]> while c=N-1 is not so representable.> I can only prove this trivial case. Suppose c=N-1 is representable in the form that c=ax+by with x,y>=0 while N-1=(a-1)(b-1)-1=ab-a-b.Then ab-a-b=ax+by and hence the equality ab=(x+1)a+(y+1)b ...(1) holds.Since gcd(a,b)=1 and therefore x+1 is divisible by b while y+1 is divisible by a, which implies that (x+1)a+(y+1)b>=2ab, in contradiction to (1).[/quote:0f5f20a1f4]Now I can complete the whole proof.For any integer c>=N, it can be written as c=ab-a-b+m while m is an integer greater than 1.Since gcd(a,b)=1, there exists some integers s,t such that as-bt=1.Multipy m on both sides and we have ams-bmt=m ...(2)According to euclidean division, there exists some integers q,r suchthat mt=aq+r, 0<=r=0.Thus ax*=by*+m>0, which implies x*>0 and it follows thatx=x*-1>=0.Thus ax+by=a(x*-1)+b(a-1-y*) =ax*-by*+ab-a-b =ab-a-b+m=c ( by (4) ),indicating that x,y are the coef'cients Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: greek numerals by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1931MG18556;Really sounds very interesting - what is acrophonics?The Greek alphabet itself is usually regarded as an adaptation of the earlier Phoenician alphabet, but you probably knew that already. My website http://www.otherlanguages.org has some hobby stuff about languages and language resources if you're curious. Please 'll me in on acrophonics. I can't 'nd it in the Mark>Mark im sorry for being vague-im exploring the early Greek system of>using acrophonics and then the alphanumerical usage-in general im>looking at exploring how the greeks acquired their systems-how they>were modi'ed and how they were superseded>neil>> .>> Perhaps you have a speci'c thing in mind by the ïGreek system', >> not just ancient Greek mathematics in general, Neil? >> >> Sounds intriguing. Can you say more? >> >> Mark>> > Hi i require information wrt the above-im writting an essay on how the> greek system came into being-and what impact it had on mathematics-can> anyone direct me to good sites etc>> http:// +mathematics+notation&btnG=Google+SearchI made this simple game up today, in part so as to help teach kids>basic number-theory concepts (such a congruences and divisibility),>but it might be far more useful as useless entertainment...if even>that...Start with m concentric circles drawn on paper,>where n is from 5 to 12 (or higher perhaps).The rings between the circles are subdivided so that the inner circle>is subdivided into 2 equal-area sections, the next ring out is>subdivided into 3 equal sections, the next ring into 4, etc..until the>outer ring is subdivided into (m+1) equal sections.>(And the sections are aligned so that one boundry of each is along a>single line.)Players start by each placing a game-piece (small enough to 't in the> smallest section) into any unoccupied section on the board.A spinner or dice are used to pick a random integer n, where each n>should be between 1 and somewhere near m.> >On each move, a player can place a new game-piece (leaving earlier>pieces) n rings inward (if possible) from their last game-piece's>position, or n rings outwards (if possible).>(Outward and inward are along a chain of sections which touch. Do>not pass center circle or outer circle.)OR the player can place a game-piece clockwise or counterclockwise (in>the same ring)>a *multiple* of n spaces from their last-placed piece.But in any case, a player can't change directions during his/her move.>(And the player's {counter}clockwise moves may indeed circle around a>single ring as many times as needed.)A player can jump over occupied spaces when measuring the number of>spaces from their last piece, but they must ONLY place a piece on an>unoccupied section.And the winner is the last player able to place a === Subject: Re: 1 + 1/3 + 1/5 + ... + 1/(2*N - 1)> I'm interested in the value of the sum> > 1 + 1/3 + 1/5 + ... + 1/(2*N - 1)> > I can bound this sum above and below using integrals, but I wonder> if there are better ways to approximate its value for large N.> > information:http://www.research.att.com/projects/OEIS?Anum= A025550http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A004041As well as the other replies' representation involving logarithm, Icould add that your sum is also:sum{k=1 to 2n-1} H(k) (-1)^(k+1),where H(k) = 1 +1/2 +...+ 1/k, as is noted in some other replies inthis thread.And your sum has the in'nite-sum representation:sum{k=1 to oo} (1/(k2) -1/(k+2n) +1/(2k-2n)) = n *sum{k=1 to oo} (n +k*3/2)/(k(k+n*2)(k+n))(But I would not trust my last sum, since I am making lots of errorslately.)And with the in'nite sum, wecan attribute a value to1 + 1/3 + 1/5 +...+1/(2n-1)for === Subject: Re: NaturalNumbers are the P-adics and why Kissing density jumps in KPP Re: applying RiemannHypothesis modi'cation to Kepler Packing Problem> The below is an old post talking about the Kepler === PROOF OF THE KEPLER PACKING> PROBLEM> <1993Aug19.021638.256@rp.CSIRO.AU Radiophysics/Australia> Telescope> National Facility> example in 9> dimensions, where the densest known> sphere packing is> that produced by> the Lamda_9 lattice (packing density> 0.14577, kissing> number 272),> whilst the greatest known kissing number> is that> achieved by the> non-lattice packing P_{9a} (packing> density 0.12885,> max kissing number> 306). For more details, an excellent book> > Okay, well, if the NaturalNumbers are the P-adics, and if RH implies> that there are no straightlines at in'nity because the P-adics> compose the 1/2 Realline.> > Then what an application of RH would do to the Kepler Packing Problem> is to 'rst ask the question of does a p-adic dimensional space make> much sense. Is there a ...99999 dimensional space in 10-adics? Is> there a ....11111 dimensional space in 2-adics?> > Then further, a RH application of p-adics to the KPP of kissing points> versus densest-nonkissing plan of attack to prove would then ask the> Very Important Question:> > Question: does the above quoting suggest that the divergence of 9th> dimension becomes even more divergent when in the 10th dimension. Then> the 11th dimension, how much of a divergence if any from the previous> dimensions.> > You see, if NaturalNumbers are really the P-adics, then in KPP there> should be a linear increase in divergence as we go higher in> dimensions with the density of packing.> > For example: the writer above noted that in 9th dimension the kissing> diverges from regular KPP, then the kissing should also diverge in> 10th dimension, and also in 11th dimension and so forth. But, if the> KPP does not diverge in 10th dimension from that of kissing in 10th> dimension Suggests or Implies that the P-adics are involved.> > If Straightlines exist out to in'nity and if NaturalNumbers are the> FiniteIntegers then the KPP should not be a pockmarked gapping of> kissing points divergence as we increase in dimensions.> > On the other hand, if NaturalNumbers are the P-adics and that all> straightlines curve as they approach in'nity (i.e. straightlines do> not exist), then the divergence of the KPP from that of kissing points> versus densest pack would not be a smooth linear relationship as we> increase in dimensions, and instead have gaps where in say dimension> 22 the kissing points is the densest pack and where dimension 23 the> kissing points are not the densest.> > Demonstration: If we take oranges to pack and we had a square box> (Euclidean Space) and a similar volumed sphere and asked to pack those> oranges in which container could we get the densest packing? The cube> or the sphere? So that in the KPP, applying the RH would suggest that> the divergence of kissing is because of the fundamental reason that> NaturalNumbers are really the P-adics.> > Because if space is Euclidean and that straightlines remain straight> out to in'nity and that NaturalNumbers are FiniteIntegers then as you> increase in dimensions from say 9 to 10 to 11 to 12 etc etc, that the> divergence from packing should also be a Smooth and linear> progression. But it is not. It is gap ridden and swinging back and> forth between kissing as the densest and kissing not the densest.> I know I shouldn't feed the trolls, but...The p-adics are a different completion of the rationals than thereals. The p-adics are not in the real numbers. They are adifferent class of objects.'cid ïooh === Subject: Re: NaturalNumbers are the P-adics and why Kissing density jumps in KPP Re: applying RiemannHypothesis modi'cation to Kepler Packing Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18NYNb02455;Oooh I see. While I couldn't understand the speci'cs of your post, I understand it belongs in the same research 'eld with tautological spaces and the maximum natural number theory. === Subject: Re: NaturalNumbers are the P-adics and why Kissing density jumps in KPP Re: applying RiemannHypothesis modi'cation to Kepler Packing Problem> Oooh I see. While I couldn't understand the speci'cs of your post, I understand it belongs in the same research 'eld with tautological spaces and the maximum natural number theory.Let me try again because my 'rst post was garbled in some paragraphs.In the early 1990s I offered a proof of the Kepler Packing Problembased on Kissing points. That the densest packing would have thelargest number of kissing points. And so I thought KPP was done andvanquished but when I posted that proof to the Internet circa 1993,objections were raised in that kissing-points diverged from fcc as thedensest packing in higher dimensions such as 9th dimension. Ifgeometry consisted only of 3rd dimension, I had proved KPP becausekissing-points was equal to fcc in 3rd dimension. So, the onlyobjection to my proof is that in higher dimension it seems as thoughkissing-points is no longer fcc.But I have a rejoinder to the higher dimensional objection. Myrejoinder is that the NaturalNumbers are the P-adics. And the P-adicsdo not form a straightline Euclidean space. So the 9th dimensionalobjection is no longer a straightline space but is a curved space. Andso where the kissing points in 9th dimensions yields a 0.12885 densityyet the fcc yields a 0.14577 creates this divergence in 9th dimension.But if you allow that the NaturalNumbers are really P-adics and thatthis 9th dimension is a curved space and not a straightline Euclideanspace then the kissingpoint density equals the fcc density in 9thdimension.I am claiming that if you apply P-adics to the dimensions such as 9thdimension then the kissingpoints become equal to the fcc packing justas they were equal in 3rd dimension.I claim the proof of KPP is based all upon Kissing Points and whatneeds altering is the fact that straightlines become curved thefurther you go out because the NaturalNumbers are the P-adics and soby the time one enters 9th dimensions it is not a 9th dimensionalEuclidean cube to pack with balls but rather a sphere like objectsince the space is curved.In other words, the disparity between kissing points and fcc packingwith higher dimensions is a means of gauging or measuring how muchspace has curved starting at 2 dimensions then 3rd dimension on up to9th dimension and on higher.Kissing Points proves Kepler Packing and the reason there exists adivergence in higher dimensions is because they are curved and nolonger Euclidean. KPP is a 'ne example of where we can measure therate of curvature of Euclidean straight lines. We thus must see thatstarting at 9, a supposedly straightline has already begun to curveand is no longer straight.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Subject: Re: need help!!!> Mark im sorry for being vague-im exploring the early Greek system of> using acrophonics and then the alphanumerical usage-in general im> looking at exploring how the greeks acquired their systems-how they> were modi'ed and how they were superseded> neil> > > .> Perhaps you have a speci'c thing in mind by the ïGreek system', > not just ancient Greek mathematics in general, Neil? > > Sounds intriguing. Can you say more? > > Mark> > Hi i require information wrt the above-im writting an essay on how the>> greek system came into being-and what impact it had on mathematics-can>> anyone direct me to good sites etcthttp:// +mathematics+notation&btnG=Google+Search hi every body can plot this function please say me with> www,hupo19@yahoo.com it is y=arcsin^-1(3/cosx),thank youdear Mark and Neil can u plot this function y=arcsin^-1(3/cosx) === Subject: Re: need help!!!> hi every body can plot this function please say me with> www,hupo19@yahoo.com it is y=arcsin^-1(3/cosx),thank you> > dear Mark and Neil can u plot this function y=arcsin^-1(3/cosx)The arcsine function is only real for arguments no larger that 1 in absolute value, but 3/cos(x) is never smaller than 3 in absolute value.So unless arcsin^-1 means something like inverse of arcsine, there are no real points on the graph. === Subject: Research Scientist in Knowledge Representation & Reasoning Researcher Level B: Knowledge Representation and Reasoning National Information and Communication Technologies Australia ------------------------------------------------------------- National ICT Australia Limited (NICTA) invites applications from highlymotivated researchers for appointment as Researchers (Level B). Thepositions are to work within the Knowledge Representation and ReasoningProgram (KRR) in traditional areas of knowledge representation for a'xed term period of two to three years. Research conducted in the KRRprogram includes, but is not limited to, the following:- Nonmonotonic reasoning- Belief revision and merging- Negotiation and games- Ontologies- Cognitive robotics and planning- Multi-agent systems- Logic programmingPreference will be given to applicants with a strong background in logic.For more information about NICTA, go to http://www.nicta.com.au.The KRR program is described in http://nicta.com.au/programs/krr/kr.html.Essential criteria: A PhD in an appropriate area with a strong publicationrecord. Evidence of impact in the relevant areas. Willingness to supervisegraduate students, and to participate in graduate teaching.Desirable criteria: Strong communication skills. Interest in graduateteaching. Record of liaison with industry. Ability to work in a team.Experience in writing research proposals.Level B salary packages are in the range of $70,000 - $100,000 (inclusiveof employer contribution to superannuation). Funds are available forresearch support and infrastructure, and travel support.Applications should be sent to jobs@nicta.com.au and should include a fullCV, list of publications, names and contact details of three referees,and a short (at most 1 page) statement of how the applicant's researchinterests, achievements and future goals align with NICTA's KRR program.Applicants should look at the general conditions (which apply to Level Bresearchers in all NICTA programs) in http://www.nicta.com.au to ensurethat they are in agreement with, and can ful'll them.NICTA reserves the right to appoint by invitation or not to make anyappointment.For further information, contact Professor Norman Foo(norman@cse.unsw.edu.au) or Dr Maurice Pagnucco (morri@cse.unsw.edu.au). === Subject: Re: 1/0 now allowed> Its de'ned as 1.6367348238383838> > Dont ask why, but must be used or your calculations will be wrong.The phrase way too much free time just popped into my head.-- http://hertzlinger.blogspot.com === Subject: Re: 1/0 now allowed>Its de'ned as 1.6367348238383838Dont ask why, but must be used or your calculations will be wrong.sight!-Staunch Giga Troll Defender === Subject: Re: 1/0 now allowed by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv1d17995;I de'ne you to be a nutcase. === Subject: Re: Plotting a 7 vertice graph in which detailed help, have 'gured it out.Diana David,> I came up with three sub-graphs, which I have uploaded the images ofwith> Mathematica.> The 'rst of the four images is just the complete graph for K_7, anddoesn't> relate.> Do I have the right idea?> http://home.earthlink.net/~diana53/mathematica/1-1-17.html I'm not sure why you have the complete graph on 7 vertices, since its> not 4-regular. The rest of the graphs look like different drawings of the> same graph (the complement of the 7 cycle), as Professor Eppstein> suggested. His other suggestion is the complement of a 4 cycle and 3> cycle; i.e. you will have a graph whose vertices can be partitioned into> two sets, one set with 3 independent vertices and another set with 4> vertices and two disjoin edges, and then form all edges between these two> sets. J> === Subject: Re: Plotting a 7 vertice graph in which every vertex has degree 4 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1931NF18564;.Or to be really crass, 1. draw a regular seven-sided polygon in pencil, then -2. join two edges from one corner to the two corners furthest away with pencil where they are not already marked in pen - 3. go over the 2 or 1 new pencil lines from [1.] in pen if they look right - 4. repeat [2.] with the corner to the immediate right [unless you are left-handed, in which case to the immediate left] unless it is already joined to two corners by pen lines -5. go over the outside seven edges in pen. Best wishes, Mark .> David,>> I came up with three sub-graphs, which I have uploaded the images of with>> Mathematica.>> The 'rst of the four images is just the complete graph for K_7, and doesn't>> relate.>> Do I have the right idea?>> http://home.earthlink.net/~diana53/mathematica/1-1-17. html I'm not sure why you have the complete graph on 7 vertices, since its >not 4-regular. The rest of the graphs look like different drawings of the >same graph (the complement of the 7 cycle), as Professor Eppstein >suggested. His other suggestion is the complement of a 4 cycle and 3 >cycle; i.e. you will have a graph whose vertices can be partitioned into >two sets, one set with 3 independent vertices and another set with 4 >vertices and two disjoin edges, and then form all edges between these two >sets.J === Subject: Re: Plotting a 7 vertice graph in which detailed help, have 'gured it out.Diana> . Or to be really crass, 1. draw a regular seven-sided polygon in pencil, then - 2. join two edges from one corner to the two corners furthest> away with pencil where they are not already marked in pen - 3. go over the 2 or 1 new pencil lines from [1.] in pen if> they look right - 4. repeat [2.] with the corner to the immediate right [unless> you are left-handed, in which case to the immediate left] unless it> is already joined to two corners by pen lines - 5. go over the outside seven edges in pen.> Best wishes, Mark .>> David,>> I came up with three sub-graphs, which I have uploaded the images ofwith>> Mathematica.>> The 'rst of the four images is just the complete graph for K_7, anddoesn't>> relate.>> Do I have the right idea?>> http://home.earthlink.net/~diana53/mathematica/1-1-17. html I'm not sure why you have the complete graph on 7 vertices, since its>not 4-regular. The rest of the graphs look like different drawings of the>same graph (the complement of the 7 cycle), as Professor Eppstein>suggested. His other suggestion is the complement of a 4 cycle and 3>cycle; i.e. you will have a graph whose vertices can be partitioned into>two sets, one set with 3 independent vertices and another set with 4>vertices and two disjoin edges, and then form all edges between these two>sets.J> === Subject: wronskianFor some problems we need to evaluate the wronskian to show whether a coupleof solutions are a fundamental set.Currently I'm doing this by actually evaluating the determinant|f g ||f ï g ï|and it does 'ne. However, a theorem I have says that W(t) = ce^P(t), whereP(t) = Integral( -p(t)*dt ), and p(t) comes from the homogeneous secondorder linear diferential equation:y'' +p(t)y' +q(t)y = 0This formula is much faster than doing the determinant, and so I would liketo use it. But, I do not know how to get the constant c. I assume that cis the constant that pops out of the integral, but without some initialconditions I don't know how to get it. === Subject: Re: wronskian>For some problems we need to evaluate the wronskian to show whether a couple>of solutions are a fundamental set.>Currently I'm doing this by actually evaluating the determinant>|f g |>|f ï g ï|I 'nd it baf§ing that the standard DE textbooks waste time doing thisfor solutions of second order DE's. f and g are linearly independentunless one is a constant multiple of the other. In any case where itisn't blatantly obvious whether that is the case, I suspect it also won't be obvious whether the Wronskian simpli'es to 0. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: wronskian|For some problems we need to evaluate the wronskian to show whether a couple|of solutions are a fundamental set.||Currently I'm doing this by actually evaluating the determinant|||f g |||f ï g ï|||and it does 'ne. However, a theorem I have says that W(t) = ce^P(t), where|P(t) = Integral( -p(t)*dt ), and p(t) comes from the homogeneous second|order linear diferential equation:||y'' +p(t)y' +q(t)y = 0||This formula is much faster than doing the determinant, and so I would like|to use it. But, I do not know how to get the constant c. I assume that c|is the constant that pops out of the integral, but without some initial|conditions I don't know how to get it.In general, determining the constant of integration amounts to computingW for one value of t, and === using that as an initial value.Keith RamsaySubject: Re: For some problems we need to evaluate the wronskian to show whether acouple> of solutions are a fundamental set.> > Currently I'm doing this by actually evaluating the determinant> > |f g |> |f ï g ï|> > and it does 'ne. However, a theorem I have says that W(t) =ce^P(t), where> P(t) = Integral( -p(t)*dt ), and p(t) comes from the homogeneoussecond> order linear diferential equation:> > y'' +p(t)y' +q(t)y = 0> > This formula is much faster than doing the determinant, and so Iwould like> to use it. But, I do not know how to get the constant c. I assumethat c> is the constant that pops out of the integral, but without someinitial> conditions I don't know how to get it.c=W(t_0) - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Alternate representations of convex decreasing functions g(x)that can be represented as:g(x) = int f(x,y) dy.Speci'cally, is there a principled way of 'nding an alternateparameterization y2 such that,g(x) = int f_2(x,y2) dy2 = max_y2 f(x,y2).I have found speci'c cases where this can be done but I don't knowany general principles for 'nding the requisite transformation orfeasibility conditions. Any info Macricknu === an LP as follows...max vstv-0.5x-0.5y<=0x=1y=1bounds0<=v<=1integersvendv,x and y can take on the value 0 or 1 only.v gets the value 1 if and only if both x and y are 1 respectively, inall other cases, v=0.How can we convert this problem into a minimiztion problem (min v) sothat v again gets the value 1 if and only if x and y are 1respectively ? For all other values of x and y, v must get the value0.Can you please let me know the formulation to get === Subject: Re: lattices, rings of algebraic integers|>Is it always the case that Z[alpha] is a lattice in C if and only if|>Z[beta] is a lattice?|>|>David Bernier|| No. Try the three cube roots of 2.This counterexample doesn't work because Z[u] where u is oneof the complex roots isn't a lattice either. It contains the element1+u+u^2 which lies inside the unit circle, as well as all of its powerswhich tend to 0.Keith Ramsay === Subject: In'nite Dimensional In'nitesimalAny smooth connected 1 dimensional manifold is diffeomorphic either tothe circle, or to some interval of real numbers.Take a line segment of length 1. It is one dimensional.A-----------BFind the midpoint of the line segment and rotate it into 2 dimensionsA|||------BEach leg is 1/2Rotate into 3 dimensions, and each leg is 1/3A|||------C||Brotate into N dimensions and each leg is 1/N Continue this process as a limitN---->ooBy the above process, an in'nite dimensional universe is a point. === >Subject: Re: Rationals are Uncountable If q is rational and the original sequencing, x_n, contains every> rational in (-1,1), then any translation of (-1,1) by a rational,> x -> x + q, with |q| < 1, produces a new x'_n which still> contains every rational in (-1+q,1+q), including 0. 0 is provably not in x'_n.I have simpli'ed this proof.Let X be every rational in (-1,1)and x_n a sequence that contains every member of X.Construct sequences a_n and b_n as described by Cantor.De'ne a new sequence:x'_n = x_n - a_i for some iConside the sequnce a_n.Let a_i = x_m and a_j = x_n.If i rational in (-1,1), then any translation of (-1,1) by a rational,> x -> x + q, with |q| < 1, produces a new x'_n which still> contains every rational in (-1+q,1+q), including 0. 0 is provably not in x'_n.> > I have simpli'ed this proof.Then youo have proven that 0 is greater than 1.> Let X be every rational in (-1,1)Your statement is, at best, ambiguous.Do you mean that X is to be the SET of rationals in (-1,1)? > and x_n a sequence that contains every member of X.> Construct sequences a_n and b_n as described by Cantor.Then there is an irrational, c, that is the common limit of the a_n and teh b_n.> > De'ne a new sequence:> x'_n = x_n - a_i for some i> > Conside the sequnce a_n.> Let a_i = x_m and a_j = x_n.> If i If a_i = x_k then x'_k = 0.> Therefore the position of 0 in x'_n> depends of which i in a_i we chose from a_n.> > We know the LUB of a_n is an irrational, r.> The closer a_i is to r, the later 0 appears in x'_n.> There is no member of a_n such that x'_n> contains every rational in X' = (-1-r, 1-r).> In particular, x_n can not be translated> to cover the interval (-1-r, 1-r) in such> a way that x'_n contains 0.(1) x'_n need not cover (-1-r,1-r), it covers (-1-a_i, 1 - a_i) for some i.(2) Since |a_i| < 1 and a_i is rational, 0 MUST be a member of (-1-a_i, 1 - a_i)> > If x_n contained every rational in (-1,1)> there would be a rational number, a_i, such> that x'_n contains the rationals in (-1-r,1-r).WRONG.> Another sequence is required to represent> these rationals in (-1-r,1-r).> x_n does not contain every rational in (-1,1).WRONG.Let X be the set of rationals in (-1,1), and, for any real r, let X+r = {x+r: x in X}, and X-a = {x-a: x in X}.Then, for any a in X, 0 is a member of X+a and X-a. If a_n is any increasing sequence in X and b_n a decreasing sequence in X such that a_n < b_n for all n, then there is at least one real number r such that a_n < r < b_n for all n.And for every such r, -1 < a_n < r < b_n < 1 so that -1-r < 0 < 1-r, contrary to your delusions. === Subject: Re: Rationals are Uncountable x_n does not contain every rational in (-1,1). But what happens when you start with a sequence which does contain> all rationals in (-1,1) ?>Please present one.Then demonstrate how to translate that sequenceto contain every rational in (-1-r,1-r) wherer is irrational and |r| < 1.Or do it the other way.Give a sequence that contains every rational in(-1-r,1-r) and show how to translate it to includeevery rational in (-1,1).Russell- 2 many 2 count === Subject: Re: Rationals are Uncountable x_n does not contain every rational in (-1,1). But what happens when you start with a sequence which does contain> all rationals in (-1,1) ? > Please present one.> Then demonstrate how to translate that sequence> to contain every rational in (-1-r,1-r) where> r is irrational and |r| < 1.No one claims that for irrational r. One needs only claim that -1-r < 0 < 1-r, which follows immediately from |r| < 1 > > Or do it the other way.> Give a sequence that contains every rational in> (-1-r,1-r) and show how to translate it to include> every rational in (-1,1).Not needed to disprove your claim. One only needs -1-r < 0 < 1-r, which follows immediately from |r| < 1. Though it is qite possible that you, Russell, cannot follow that simple argument without help.> > > Russell> - 2 many 2 count> > === Subject: Re: Rationals are Uncountable> Not needed to disprove your claim. One only needs -1-r < 0 < 1-r, which> follows immediately from |r| < 1. Though it is qite possible that you,> Russell, cannot follow that simple argument without help.I agree that if X' is the set of all rationals in (-1-r,1-r) withr irrational and |r|<1 then 0 is in X'.What I am proving is that if x_n is a sequence thatcontains every rational in X=(-1,1) then there is noway to convert x_n to x'_n such that x'_n contains 0.Russell- 2 many 2 count === Subject: Re: Rationals are Uncountable> I agree that if X' is the set of all rationals in (-1-r,1-r) with> r irrational and |r|<1 then 0 is in X'.> What I am proving is that if x_n is a sequence that> contains every rational in X=(-1,1) then there is no> way to convert x_n to x'_n such that x'_n contains 0.Assume without loss of generality that r is positive.If x_n > r, x'_n = x_n - 2if 0 <= x_n < r, x'_n = x_nIf x_n < 0 and 1/x_n is an integer, x'_n = x_n/(1 + x_n)If x_n < 0 and 1/x_n is not an integer, x'_n = x_n -- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Rationals are Uncountable> >> x_n does not contain every rational in (-1,1).>> But what happens when you start with a sequence which does contain>> all rationals in (-1,1) ?> > Please present one.Take any sequence a_n enumerating all rationals.Then de'ne b_n as follows:b_n = a_N where N = min{k: |{b_1,..., b_k} intersect (-1,1) | = n }.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Rationals are Uncountable> > >> x_n does not contain every rational in (-1,1).>> But what happens when you start with a sequence which does contain>> all rationals in (-1,1) ?> > Please present one.> > Take any sequence a_n enumerating all rationals.I haven't been following this thread, but that last sentence amuses mebecause the title of this thread is exactly the statement that there is no such sequence a_n enumerating all rationals. > > Then de'ne b_n as follows:> b_n = a_N where N = min{k: |{b_1,..., b_k} intersect (-1,1) | = n }. === Subject: Re: Rationals are Uncountable>> >> >> x_n does not contain every rational in (-1,1).>> But what happens when you start with a sequence which does contain> all rationals in (-1,1) ?> >> Please present one.>> >> Take any sequence a_n enumerating all rationals.> > I haven't been following this thread, but that last sentence amuses me> because the title of this thread is exactly the statement that there is no such > sequence a_n enumerating all rationals. > >> >> Then de'ne b_n as follows:>> b_n = a_N where N = min{k: |{b_1,..., b_k} intersect (-1,1) | = n }.for th rationals in (-1,1) it's very easy0,1/2,-1/2,2/3,1/3,-1/3,-2/3...note this counts every non-zero rational in'nitely many times, so itreally is over killif you still don't believe it then, just doing the positive ones.for x_n let m be the largest integer such that t(m) := m(m+1)/2 <= n, then set x_nto be {n-t(m)}/m is that explicit for you? Get the negative ones, and theinterlace them. The algorithm gives all the multiples of 1/2 then 1/3 then1/4 and so one. I may count 1 a lot as well, I can't quite decide, butpass to a subsequence as necessary.m === Subject: Re: Rationals are Uncountable <8cqdnfL04qic_Lrd4p2dnA@comcast.com> haven't been following this thread, but that last sentence amuses> me because the title of this thread is exactly the statement that> there is no such sequence a_n enumerating all rationals.You're right, you haven't been following this thread.Evidently, Russell is happy to allow that the countability of therationals is a theorem of set theory. He's not after such smallpotatoes as correcting a century year old error. Rather, he's afterthe inconsistency of the whole ball of wax.-- Jesse F. Hughes[blah blah blah] which you are trying to dispute, but I've found thatyou've used ***de'nitions*** rather than mathematical logic, often inyour posts. -- James Harris on the insuf'ciency of math. de'nitions. === Subject: Re: Math Too Advanced For Mainstream EconomistsI think this thread would amusing to those who 'nd JSH threadsamusing, but with a twist. I was delighted to 'nd in a dictionary the word MUMPSIMUS, which means stubborn persistence in an error that has been exposed. -- Joan Robinson > ...the original poster was unable to 'nd a single piece of> empirical support for the labor market model presented...The above, of course, is untrue, and it is hard to see how poorMark Witte cannot know it is untrue, if he had any clue aboutwhat he is talking about.> A> demand curve is a relationship between price and quantity demanded. > If something else changes, such as prices of related goods in demand> (here the discount rate would be an example), this shifts the curve.Poor Mark Witte is confused. A discount rate is not a price.> To be charitable to the original poster, his confusion probably stems> from his notion that a change by the 'rm in question in the number of> workers it hires would change the discount rate for the economy.The above is a strawperson. No such claim is to be found at the aboveURL.Notice no claim is made about the discount rate for the economybeing changed by a change in how many workers 'rm hires. In fact,an economy-wide market for 'nancial capital is not described atmy URL.Poor Mark Witte just doesn't understand accounting. If the bestreturn a 'rm can obtain is 10%, then that's what the 'rm givesup in investing elsewhere. If costs are different for the bestinvestment (e.g., because the level of wages is different), then the'rm gives up some other rate of return than 10%.To help poor Mark Witte out, here's some explanations of a conceptimportant for understanding Step 2 of the algorithm given at myURL: My URL is more a tutorial to results well-established in the literature.No claim is made to novelty there. For that matter, no claim is madethere that the relationship shown is a demand curve for labor. But thatclaim is made in the literature.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question 't perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Math Too Advanced For Mainstream EconomistsRobert Vienneau empirical support for the labor market model presented... The above, of course, is untrue, and it is hard to see how poor> Mark Witte cannot know it is untrue, if he had any clue about> what he is talking about.Of course, it would be standard practice to produce here a single suchexample. If one could. === Subject: How far can one go with self-study nowadays?Just thinking about all the neat things available online, compared to when Iwas in school 20ish years ago, I'm wondering how far one can go nowadays inmathematics with self-study.20 years ago, it would have been hard. If you lived near a university thathad a graduate program in math, you would have been able to at least 'gureout what books to read, but if you didn't live near a university, even'guring out what books to read might have been reasonably hard.And if you had needed help, 'nding someone to ask would have been hard.It would have been doable, I think, but it would have been very hard.Compare that to nowadays. You can 'nd textbooks easily online, and 'ndcustomer reviews of them, to get an idea of which ones are good. If you getstuck on something, there is a fair chance that a Google search will 'nda solution, and if not, there is usenet--a post to sci.math will probablyget an answer.I think one should be able to reasonably get to at least a fair way intograduate level. When you get to the point that you need to regularly readjournals, that could be a problem--those things are expensive, and without auniversity library you might be limited. Is there a way around this, or isthat the 'rst big barrier?-- --Tim Smith === Subject: Re: How far can one go with self-study nowadays?>Just thinking about all the neat things available online, compared to when I>was in school 20ish years ago, I'm wondering how far one can go nowadays in>mathematics with self-study.[...]>Compare that to nowadays. You can 'nd textbooks easily online, [21st century wonders and impediments omitted]>is that the 'rst big barrier?If learning mathematics were centered on textbooks, we could hand studentsa big fat encyclopedia when they come through the door and then hand thema degree after a few years.The 'rst big barrier will really be the lack of what makes universitiesthe best place to learn mathematics: people. Most students need someone who will set expectations and timetables; they need someone to review theirwork and offer improvements and conversations; they need someone whocan open up the texts, providing examples and motivation, answeringquestions and so on. Of course, there are students who can manage allthis on their own. I'd have to say that's pretty rare.By the way, while you can 'nd textbooks on line, they're not nearly ascommon as printed texts, and they don't usually compare well withthe best texts in any given area. If you're going to save all that tuition money, you could afford to buy some good books :-)dave === Subject: Re: How far can one go with self-study nowadays?> By the way, while you can 'nd textbooks on line, they're not nearly as> common as printed texts, and they don't usually compare well with> the best texts in any given area. If you're going to save all that > tuition money, you could afford to buy some good books :-)I was probably unclear--by 'nding textbooks online, I meant 'nding placesonline to buy textbooks. E.g., Amazon. I realize it was always possible tobuy textbooks by mail-order, but I don't think that compares well to buyingonline, because places like Amazon have customer reviews that can be veryhelpful. (And Amazon has those handy customers who shopped for this alsoshopped for... and customers who bought this also bought..., which can bevery useful).-- --Tim Smith === Subject: Re: How far can one go with self-study nowadays?>> By the way, while you can 'nd textbooks on line, they're not nearly as>> common as printed texts, and they don't usually compare well with>> the best texts in any given area. If you're going to save all that >> tuition money, you could afford to buy some good books :-)I was probably unclear--by 'nding textbooks online, I meant 'nding places>online to buy textbooks. E.g., Amazon. I realize it was always possible to>buy textbooks by mail-order, but I don't think that compares well to buying>online, because places like Amazon have customer reviews that can be very>helpful. (And Amazon has those handy customers who shopped for this also>shopped for... and customers who bought this also bought..., which can be>very useful).Don't overlook the fact that most new editions of books are nobetter than the previous editions. For example on ebay you can buy acalculus book that is an edition or two old for just a few dollarscompared to well over $100 for the version in your college bookstore.--Lynn === Subject: Re: How far can one go with self-study nowadays?> Just thinking about all the neat things available online, compared to whenI> was in school 20ish years ago, I'm wondering how far one can go nowadaysin> mathematics with self-study. 20 years ago, it would have been hard. If you lived near a universitythat> had a graduate program in math, you would have been able to at least'gure> out what books to read, but if you didn't live near a university, even> 'guring out what books to read might have been reasonably hard. And if you had needed help, 'nding someone to ask would have been hard. It would have been doable, I think, but it would have been very hard. Compare that to nowadays. You can 'nd textbooks easily online, and 'nd> customer reviews of them, to get an idea of which ones are good. If youget> stuck on something, there is a fair chance that a Google search will 'nd> a solution, and if not, there is usenet--a post to sci.math will probably> get an answer. I think one should be able to reasonably get to at least a fair way into> graduate level. When you get to the point that you need to regularly read> journals, that could be a problem--those things are expensive, and withouta> university library you might be limited. Is there a way around this, oris> that the 'rst big barrier?Well, certainly anything is possible, at least in principle. In my opinionthere are two requirements:1) talent2) hard workGiven ample amounts of both of the above, then there is really no limit tothe levels achievable. Note that this is really independent of onlineservices.You correctly state the presence of a nearby university as an aid, but Iconsider that a merely practical question. I've often had to commute 15miles to reach a suf'ciently equipped library. If that distance had been150 miles, I might not have visited the library quite as frequently, but Ido believe I would have reached the same level as I have today nevertheless.Having a friend to study with is quite important. A sparring partner is veryvaluable. I've been blessed with several very skilled such friends and Iacknowledge at least a part of my skills to their untiring sparring.-Michael. === Subject: Looking for primes of a particular form.I am trying to 'nd primes numbers of the following forms: p = 6*(10^N)-1, q = 6*(10^N)+1All I can 'nd so far are: N=1, p= 5, q= 7 N=2, p= 59, q= 61 N=3, p=599, q=601sieves possible values of N between 4 and 10000. It usesmodular arithmetic as a substitute for trial division by each prime. That program left me with a manageable listof candidate values of N. (By the way, at my level of skill, this is the most advanced modular arithmetic problemI've ever successfully programmed a solution for.)But how do I further test to see which of these candidates,(e.g., values of N such that p or q have no small primefactors) are really prime?I thought I heard once of a fast primality test that requiresone to know the prime factors of n-1. Well in my case Ihave the luxury of knowing that the only prime factors of q-1 are 2,3, and 5. But I do not know how implement thetest in my computer code.Is there any theoretical proof, that I may have missed,that tells me whether or not the primes I am looking foreven exist? Also, a separate question is whether any twinprimes exist in these forms for N>3? (E.g. both p and qare prime for the same value of N.) Is there be aproof of the (non)existence of such twin primes?Summary of my questions:1. I want to know how to program the fast primalitytest that relies on knowing the factors of n-1.2. I want to know if any theoretic considerationscan be brought forth regarding number of primesthat even exist for the two forms I gave. Samefor twin primes.P.S. What primality tests are known that are optimizedfor numbers of the form a*(b^n) +/- 1? === Subject: Re: Looking for primes of a particular form.I am trying to 'nd primes numbers of the following forms: p = 6*(10^N)-1, q = 6*(10^N)+1p is prime for n=1,2,4,5,7,10,13,22,23,28,34,40,61,73,361,...q is prime for n=1,2,8,9,15,20,26,38,45,65,112,244,...> uses modular arithmetic as a substitute for trial division OK, but there are much faster ways to decide whether a number isprime or not. See e.g. http://www.math-atlas.org/index/11Y05.htmldave === Subject: Re: Looking for primes of a particular form.See also http://www.research.att.com/projects/OEIS?Anum=A056716 http://www.research.att.com/projects/OEIS?Anum=A056805--Don Reble djr@nk.ca === Subject: Re: Looking for primes of a particular form.>I am trying to 'nd primes numbers of the following forms: p = 6*(10^N)-1, q = 6*(10^N)+1>All I can 'nd so far are: N=1, p= 5, q= 7> N=2, p= 59, q= 61> N=3, p=599, q=601Typo. My values of N are off by one. Shouldn't affect the questions I was trying to ask though.>sieves possible values of N between 4 and 10000. It uses>modular arithmetic as a substitute for trial division by >each prime. That program left me with a manageable list>of candidate values of N. (By the way, at my level of >skill, this is the most advanced modular arithmetic problem>I've ever successfully programmed a solution for.)But how do I further test to see which of these candidates,>(e.g., values of N such that p or q have no small prime>factors) are really prime?I thought I heard once of a fast primality test that requires>one to know the prime factors of n-1. Well in my case I>have the luxury of knowing that the only prime factors >of q-1 are 2,3, and 5. But I do not know how implement the>test in my computer code.Is there any theoretical proof, that I may have missed,>that tells me whether or not the primes I am looking for>even exist? Also, a separate question is whether any twin>primes exist in these forms for N>3? (E.g. both p and q>are prime for the same value of N.) Is there be a>proof of the (non)existence of such twin primes?Summary of my questions:1. I want to know how to program the fast primality>test that relies on knowing the factors of n-1.2. I want to know if any theoretic considerations>can be brought forth regarding number of primes>that even exist for the two forms I gave. Same>for twin primes.P.S. What primality tests are known that are optimized>for numbers of the form a*(b^n) +/- 1? === Subject: Probability QuestionCan anyone give me some help with this question,The probability of winning exactly 10 on the National Lottery is0.01765. If I buy 100 tickets, what is the probability that I do notwin a 10 prize? The probability of winning the jackpot on theNational Lottery is 7.15x10^-8. How many tickets do I need to buy tohave at least a 10% chance of winning the jackpot? === Subject: Re: Probability (JohnDeCarlos)>Can anyone give me some help with this question,>The probability of winning exactly 10 on the National Lottery is>0.01765. If I buy 100 tickets, what is the probability that I do not>win a 10 prize? The probability of winning the jackpot on the>National Lottery is 7.15x10^-8. How many tickets do I need to buy to>have at least a 10% chance of winning the jackpot?Assuming that you are asking about a single lottery and that there area suf'ciently large number of tickets that can win a 10 prize, thenbuying 100 _different_ tickets should leave you with (1-.01765)^100 =16.85% chance of not winning the 10 prize.the odds you give, there are 14 million different lottery ticketspossible. Again assuming that you are asking only about a singlelottery, you would need to buy 1.4 million _different_ tickets to havea 10% chance of winning the jackpot.Rob Johnson take out the trash before replying === Subject: Re: Probability QuestionRob Johnson wibbled:{OrigPoster was talking abut the UK national lottery - pick 6 to match 6 from 49}> Only one set of six numbers can win the jackpot. It's relaively common in the UK for the jackpot to be split between holders of identical tickets: there is no all tickets different restriction in the total sale distribution.> the odds you give, there are 14 million different lottery ticketsRoughly. Whatever 49C6 is - just under 14million.> possible. Again assuming that you are asking only about a single> lottery, you would need to buy 1.4 million _different_ tickets to have> a 10% chance of winning the jackpot.It'll be N, where the Nth root of .9 is 1-p where p is the probability he gave. He might get away with 1.38 million-- Now hunting 25 === Subject: Re: Probability QuestionJohnDeCarlos wibbled:> Can anyone give me some help with this question,> The probability of winning exactly 10 on the National Lottery is> 0.01765. If I buy 100 tickets, what is the probability that I do not> win a 10 prize? What limitations are you putting on number duplication?The smallest known set of tickets that can guarantee a 3 number match 163. It's not proven that this is the minimum.http://lottery.merseyworld.com/Wheel/Past that, I don't really care. Ask me about the soccer pools instead. === Subject: Re: Probability Question> Can anyone give me some help with this question,> The probability of winning exactly 10 on the National Lottery is> 0.01765. If I buy 100 tickets, what is the probability that I do not> win a 10 prize? Depends on whether you buy 100 tickets for the same lotteryor 1 ticket for each of 100 lotteries.> The probability of winning the jackpot on the> National Lottery is 7.15x10^-8. How many tickets do I need to buy to> have at least a 10% chance of winning the jackpot?Same comments apply.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: coordinates problem?I am stuck in a coordinates problem in my research. In an ellipsoidalcoordinates, is it true that the semi-pricipal axes must strictly satisfya>b>c? In my case, I am trying to use a equation sovled in an ellipsoidalcoordinates for a spheroidal object (two semi-pricial axes are identical).Given an abitratry point (x,y,z) in the space and a=b, one can only 'nd twosolutions (xi and eta) for the corresponding ellipsoidal coordinates.x^2/(a^2+xi)+y^2/(a^2+xi)+z^2/(c^2+xi)=1x^2/(a^2+ eta)+y^2/(a^2+eta)+z^2/(c^2+eta)=1x^2/(a^2+zeta)+y^2/(a^2+zeta )+z^2/(c^2+zeta)=1Does it mean that I cannot treat a spheroidal coordinates as a special caseof an ellipsoidal === Subject: Re: grateful for comments: argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id arguing with that!It's late here - I'll unpick my confusion tomorrow. Mark> If an odd perfect number exists and has three prime>> factors a,b,c then we can partition the total list of>> factors, both prime and composite, into those dividing>> by a [call this ïa.sum'], remaining factors dividing>> by b [b.sum], remaining factors dividing by c [c.sum],>> and 1.>> >> For perfect number status, what we call ïcomplements'>> of each sum must divide by that prime:>> for example 1 + a.sum + b.sum = c.comp must divide by c. >> In general x must divide x.comp for each of x = a,b,c.>> >> By examining each of >> a.comp = 1 + b.sum + c.sum,>> b.comp = 1 + a.sum + c.sum,>> c.comp = 1 + a.sum + b.sum >> we 'nd it is never the case that all three divide by >> they primes they should, and hence there are no odd >> perfect numbers. Further there are no even perfect >> numbers dividing by more than two primes. What if n = 3^2 x 13^2 x 61? The divisors of n not divisible by 3 are >(1 + 13 + 169) x (1 + 61) = 183 x 62, which is divisible by 3. The divisors of n not divisible by 13 are >(1 + 3 + 9) x (1 + 61) = 13 x 62, which is divisible by 13. The divisors of n not divisible by 61 are >(1 + 3 + 9) x (1 + 13 + 169) = 13 x 183 = 13 x 3 x 61, >which is divisible by 61.-- >Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19Bhuu22728;>HelloHow can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued fractions? But I could't whether arcos[IRRATIONAL-G.R.] is IRRATIONAL and if so the ratio:IRRATIONAL devided by IRRATIONAL(Pi) = Stefanideshttp://www.stefanides.gr === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>HelloHow can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>this has something to with the golden ratio, and maybe with Fibonacci>numbers or continued fractions? But I could't whether arcos[IRRATIONAL-G.R.] is IRRATIONAL and if so the ratio:> > IRRATIONAL devided by IRRATIONAL(Pi) = IRRATIONAL.> haven't worked out the details as yet, but a possible answer mightlie in the factthat sin(pi/10) = (sqrt(5)-1)/4 = cos(2*pi)/5)But> IRRATIONAL divided by IRRATIONAL(Pi) = IRRATIONALis simoly not true! Try pi/pi .But is an algebraic number divided by a transcendental number alwaysirrational?Note that we also havesin(pi/6)= (sqrt(4) -0)/4andsin(pi/12)= (sqrt(6) -sqrt(2) )/4Challenge from an old trig book:Show that if d > 12 and sin(pi/d)= (sqrt(n) - sqrt(n-4) )/4 then d istrranscendental.Hint: Gelfond's === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> But is an algebraic number divided by a transcendental number always> irrational?It's always transcendental, because the algebraic numbers are closedunder the usual operations of arithmetic.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>IRRATIONAL devided by IRRATIONAL(Pi) = IRRATIONAL.Hey, this is fun! So Sqrt(8) / Sqrt(2) = 2 is irrational?-- I'm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.math === Subject: Background about integer part function by John Stewart by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19BhtO22692;Hi gud day......i am having a hard time searching about the subject and also about sum of two distinct odd numbers.....could u help me..please === Subject: Re: Background about integer part function by John StewartX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>Hi gud day......>i am having a hard time searching about the subject and also about sum of two distinct odd numbers.....could u help me..pleasePossibly someone could help you if you said what the _question_ is.(Hint: people will be more likely to help if they don't think you're agud and u...)************************David C. Ullrich === Subject: Computational complexity of Init (Pre'x) languages by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19Bhvl22736;Let L be any recursively enumerable language and let Init(L) be the language of all pre'xes of L, that is:Init(L) = {x | xy = z, z is in L, y is in S*}where S* is the set of all words made from the letters of the alphabet S of L. What is the computational complexity of the family {Init(L) | L is recursively enumerable}?Is this family in NP or in DSPACE(n)? Is there any literature which deals with this matter? === Subject: Re: Silly question for someone with a big calculator. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, believe you. However, you can check for yourself >> if it is correct or not: notice the sum of the coef'cents>> of x and y divided by the coef'cient of e is always a constant:>> Namely 1,414211... (! - I just saw this, myself. Of course, there >> will probably be some trivial reason for this.).De'ne always a constant:... a mistaken presumption.>4/3 = 1.333333...>24/17 = 1.411764...>816/577 = 1.414211...>941664/665857 = 1.414213...If we let k represent that ratio for one of the exponents, the ratio for>the next in this list is (4k)/(k^2+2); it can be shown that the ratios>will converge to sqrt(2).Sorry, I think I must be overlooking something (I do a lot). Itcertainly *looks* to me as if the above sequence will converge tosqrt(2); however, I am neither sure what your k is in (4k)/(k^2+2)nor how one can show that that the sequence of ratios converges, of which no closed form exists- unless I`m overlooking something, of course. For the general (alpha x + alpha y + beta e), the ratio Iget is (4alpha*beta)/(2alpha^2 + beta^2) but, as you stated,this only gives the next term of the sequence i.e., following termsmust be calculated from anew. Moreover, it seems possible toplug in any nonzero values for alpha and beta intoz = (alpha x + alpha y + beta e) such that the sequence of ratiosappears to approach sqrt(2). Note: if we further generalize the case,letting the (nonzero) coef'cients of x and y differ, it appearsthat the sequence of ratios will approach u( coef'cient of x/ coef'cient of y) where u is a function from Fto the reals and u(1) = sqrt(2). This function u may be of interest.I thought your following line was neat:>1 = 3^2-2^2-2^2 = 17^2-12^2-12^2 = 577^2-408^2-408^2 = ...By chance, I noticed 1 = 665857^2 - 470832^2 - 470832^2, but didn`tthink to go back and check it for the earlier Johnson>panoptes@iquest.net>http:// members.iquest.net/~panoptes/4/3 = 1.333333...>24/17 = 1.411764...>816/577 = 1.414211...>941664/665857 = 1.414213...If we let k represent that ratio for one of the exponents, the ratio for>the next in this list is (4k)/(k^2+2); it can be shown that the ratios>will converge to sqrt(2).> > Sorry, I think I must be overlooking something (I do a lot). It> certainly *looks* to me as if the above sequence will converge to> sqrt(2); however, I am neither sure what your k is in (4k)/(k^2+2)> nor how one can show that that the sequence of ratios converges, > of which no closed form exists- unless I`m overlooking something, > of course.If k = 4/3, (4k)/(k^2+2) = 24/17.If k = 24/17, (4k)/(k^2+2) = 816/577.etc.If we let delta = 2-k^2, the delta for the following k will be2*delta^2/(k^2+2)^2 . For example, for k=4/3, we have delta = 2/9. Fork=24/17, we have delta = 2/289 = 2*(2/9)^2/((4/3)^2+2)^2 . Sincerepeating this often enough will get you a delta as close to 0 as youwant, k will get as close to sqrt(2) as you want.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: I need a site about number theory .I want to learn about number theory , If you know a good website aboutthat please let me === Subject: Re: I need a site about number theory .> I want to learn about number theory , If you know a Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: James> > [...] Yes, it would sound better, for example, with have the 'nal word.> But wait! Compare it to how the original ends:> > But still in matters vegetable, animal, and mineral,> He is the very model of a modern Major-General.> > Reading this, it is clear that there is a caesura between vegetable> and animal, which corresponds to pausing after so, in the parody.There's no need to ïmiss a beat' after vegetable if you put thestress on ïveg' and ïtab' and plod through the word using foursyllables: ïVEG-er-TAB-ul'. That's how I've always heard it === Subject: Re: ïPirates Of Penzance' tune ïModern Major General'>(author>unknown):>> [snip most of it]>>I always have the last word; so, with utmost 'nality,>That's all from me, the model of a Newsgroup Personality.>> >> I think the last line is missing a two-syllable word near the end. Apart>> from that the whole thing scanned beautifully.> Eh? The second last line seems to me to trip along better if last is> replaced by 'nal, but the last line looks just 'ne; > > Sorry, I meant the second last line. Replacing last with 'nal is a> start, but you still need another syllable. Adding and after the> semi-colon completes the 'x. (The rest of the lyric scans just as well as> Gilbert's, who was an unparallelled master of scansion, no joke about it.)Well, the second last line scans OK for me with stresses onal, have, 'n, word, with, most, nal.Perhaps you are stressing ut and shortening the vowel of ' to aschwa?-- Chris HenrichThose readers who are unacquainted with the mathematical technicalities will'nd that they can manage quite well by ignoring them. -- John Nash === Subject: Re: James> >> >>To be sung to the ïPirates Of Penzance' tune ïModern Major General'>>(author>>unknown):> [snip most of it]>>I always have the last word; so, with utmost 'nality,>>That's all from me, the model of a Newsgroup Personality.> > I think the last line is missing a two-syllable word near the end. Apart> from that the whole thing scanned beautifully.>> Eh? The second last line seems to me to trip along better if last is>> replaced by 'nal, but the last line looks just 'ne; >> >> Sorry, I meant the second last line. Replacing last with 'nal is a>> start, but you still need another syllable. Adding and after the>> semi-colon completes the 'x. (The rest of the lyric scans just as well as>> Gilbert's, who was an unparallelled master of scansion, no joke about it.)> > Well, the second last line scans OK for me with stresses on> al, have, 'n, word, with, most, nal.> Perhaps you are stressing ut and shortening the vowel of ' to a> schwa?you know what, I found it so impossible to read utmost with the stresson most, that I thought there was another syllable missing. You'reright, changing last to 'nal makes it the right number of syllables. The mis-stress on utmost is more distracting than anything else in this parody though, and I think it'd be better if we kept last, but changed utmost to uttermost, which isn't really a word, but its rhythm is perfect... === Subject: Re: James>>I always have the last word; so, with utmost 'nality,>That's all from me, the model of a Newsgroup Personality.>> >> I think the last line is missing a two-syllable word near the end. Apart>> from that the whole thing scanned beautifully.> > Yes, it would sound better, for example, with have the 'nal word.> But wait! Compare it to how the original ends:> > But still in matters vegetable, animal, and mineral,> He is the very model of a modern Major-General.> > Reading this, it is clear that there is a caesura between vegetable> and animal,> > [You correctly guessed that I meant *second*-last line]> But actually I've always heard the original second-last line performed > with 4 very clear syllables to the word ve-ge-ta-ble, so that the line > has the same constant rhythm as all the other lines.Ah! I hadn't thought of this (having never heard it performed). So wemay indeed say that the parody is §awed there.> This is called a patter song by the way, because it just patters along> at a constant rhythm with no let-up. Gilbert and Sullivan included one in> most of their shows.Always nice to learn a new bit of jargon. What a shame that the parodyfalls short of pure and pattering iambic octameter by a mere syllable.If this piece is indeed an anonymous bit of netlore, then I'd be in favorof promulgating a version with 'nal word instead of last word.-Jim Ferry === Subject: limit of sequence in metric spaceLet E be a metric space with norm || . ||, and consider a sequence {x_n} n>=1, x_n in E. Write lim(n to oo) x_n is just a formal expression at thispoint, saylim(n to oo) x_n = x_1 + sum(1 to oo) ( x_{k+1}-x_k ),sincex_n = x_1 + sum(1 to n-1) (x_{k+1}-x_k).If I can show that || lim(n to oo) x_n || < oo, does this suf'ce forclaiming lim (n to oo) x_n exists and is in E ? Or do I 'rst have to showlim(n to oo) x_n exists or converges by some other means BEFORE I talkabout taking norms.For example, consider the space l_1 whose elements are in'nite sequencesx=(x_1,x_2,...) such that ||x||=sum(1 to oo) |x_i| < oo. Suppose x^(n) is aCauchy sequence: Given eps, there exists N s.t. for n,m >= N,||x^(m)-x^(n)|| Let E be a metric space with norm || . ||,Why not just say E is a normed linear space?> and consider a sequence {x_n} n>=1, x_n in E. Write lim(n to oo) x_n is just a formal expression at this> point,> > lim(n to oo) x_n = x_1 + sum(1 to oo) ( x_{k+1}-x_k ),> > since> > x_n = x_1 + sum(1 to n-1) (x_{k+1}-x_k).> > If I can show that || lim(n to oo) x_n || < oo, does this suf'ce for> claiming lim (n to oo) x_n exists and is in E ?Who knows? lim(n to oo) x_n now has two meanings, the usual one and now this idea of yours, which appears meaningless.> Or do I 'rst have to show> lim(n to oo) x_n exists or converges by some other means BEFORE I talk> about taking norms.||x|| is de'ned iff x is in E; it's as simple as that. === Subject: Re: limit of sequence in metric space>Let E be a metric space with norm || . ||, and consider a sequence {x_n} n>>=1, x_n in E. Write lim(n to oo) x_n is just a formal expression at this>point, sayIs this a general metric space or a vector space? Talking about x_1 +x_2 doesn't make sense unless it's a vector space so I'll assume thatit is.>lim(n to oo) x_n = x_1 + sum(1 to oo) ( x_{k+1}-x_k ),sincex_n = x_1 + sum(1 to n-1) (x_{k+1}-x_k).If I can show that || lim(n to oo) x_n || < oo, does this suf'ce for>claiming lim (n to oo) x_n exists and is in E ?No. Consider the sequence in Q:x_n = (1 + 1/n)^nwhere lim(n to oo) x_n not in Q>Or do I 'rst have to show>lim(n to oo) x_n exists or converges by some other means BEFORE I talk>about taking norms.The standard de'nition of converging sequence in a vector space isthat for all e > 0 there exists n_e > 0 s.t.n > n_e => ||x_n - x|| < e for some x in EThis is what you must show to prove a sequence converges. The tricknow is to not just 'nd n_e but to 'nd x as well.>For example, consider the space l_1 whose elements are in'nite sequences>x=(x_1,x_2,...) such that ||x||=sum(1 to oo) |x_i| < oo. Suppose x^(n) is a>Cauchy sequence: Given eps, there exists N s.t. for n,m >= N,>||x^(m)-x^(n)||||x^(n(k+1))-x^(n(k))|| < 1/2^k. Writex^(n(k))=x^(n(1)) + sum(1 to k-1) ( x^(n(j+1)) - x^(n(j)) ).Then formally write lim(k to oo) x^(n(k)) as above. Then||lim(k to oo) x^(n(k))|| <= ||x^(n(1))|| + sum(1 to oo) ||x^(n(j+1)) ->x^(n(j))|| < ||x^(n(1))|| + 1 < oo.I think this just shows that all sequences in x^(n) convergeabsolutely - but of course they do since they were picked from l1.You're supposed to show there exists a sequence x in l1 s.t.||x^k - x|| = Sum(i:1->oo, |(x^k)_i - (x)_i|) < e.In other words, the sequences converge towards exactly one sequence inl1, in the sense of the || ||_1 -norm. This follows by observing thei:th element of each of the sequences x^(n), and noting they formCauchy sequences in R so they must converge. The sequence formed bythe limits of these Cauchy sequences is also the limit sequence ofx^(n).-- I'm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.math === Subject: Re: Silly question for someone with a big calculator. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id check for yourself > if it is correct or not: notice the sum of the coef'cents> of x and y divided by the coef'cient of e is always a constant:> Namely 1,414211... (! - I just saw this, myself. Of course, there > will probably be some trivial reason for this.).>>De'ne always a constant:... a mistaken presumption.>4/3 = 1.333333...>>24/17 = 1.411764...>>816/577 = 1.414211...>>941664/665857 = 1.414213...>>If we let k represent that ratio for one of the exponents, the ratio for>>the next in this list is (4k)/(k^2+2); it can be shown that the ratios>>will converge to sqrt(2).Sorry, I think I must be overlooking something (I do a lot). It>certainly *looks* to me as if the above sequence will converge to>sqrt(2); however, I am neither sure what your k is in (4k)/(k^2+2)>nor how one can show that that the sequence of ratios converges, >of which no closed form exists- unless I`m overlooking something, >of course. For the general (alpha x + alpha y + beta e), the ratio I>get is (4alpha*beta)/(2alpha^2 + beta^2) but, as you stated,>this only gives the next term of the sequence i.e., following terms>must be calculated from anew. Moreover, it seems possible to>plug in any nonzero values for alpha and beta into>z = (alpha x + alpha y + beta e) such that the sequence of ratios>appears to approach sqrt(2). Note: if we further generalize the case,>letting the (nonzero) coef'cients of x and y differ, it appears>that the sequence of ratios will approach >u( coef'cient of x/ coef'cient of y) where u is a function from FCorrection, u is a function from reals {0} to reals.>to the reals and u(1) = sqrt(2). This function u may be of interest.I thought your following line was neat:>>1 = 3^2-2^2-2^2 = 17^2-12^2-12^2 = 577^2-408^2-408^2 = ...By chance, I noticed 1 = 665857^2 - 470832^2 - 470832^2, but didn`t>think to go back and check it for the earlier Johnson>>panoptes@iquest.net>>http:// members.iquest.net/~panoptes/>039 53 36 N / 086 11 55 W === Subject: Re: Energy by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19DGRM29906;Energy is the mysterious ability of matter to expand;Matter may be deformed by the application of some formof forces which require spending some form of energy,thusdoing work.Energy is the capacity of doing work[work and energyare exchangeable],and matter may in generalexpand when energy is involved by heating it[applyingheating energy] ,or conversely ,contract it when energy has been spent to cool it.So Energy is not the mysterious ability of the matter.Probably Your phraseing was Stefanideshttp://www.stefanides.gr === Subject: Re: Looking for primes of a particular form. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19DGRK29893;>I am trying to 'nd primes numbers of the following forms:>> p = 6*(10^N)-1, q = 6*(10^N)+1>>All I can 'nd so far are:>> N=1, p= 5, q= 7>> N=2, p= 59, q= 61>> N=3, p=599, q=601May I (gently) suggest that you need to do some background reading?Based upon your comments and questions it is clear that you lackthe background (and language) to discuss this topic. Do you knowwhat a primitive root is? 'nite 'eld? cyclic group? sylow sub-group?Until you have obtain basic background knowledge it would be hardto explain what you need to know. However:If you know the factors of p-1 all you need do to prove it primeis 'nd a primitive root.If you know the factors of p+1 all you need do to prove it prime is'nd a generator of the twisted multiplicative subgoup of GF(p^2).i.e. the order of GF(p^2) = p^2-1 = (p-1)(p+1). There is a subgroup of order p+1. Find a generator of this subgroup.See: (e.g.)H.Riesel Prime Numbers and Computer Methods for FactorizationYou will also need the ability/code to do multi-precision arithmetic. === Subject: Re: Looking for primes of a particular form.> You will also need the ability/code to do multi-precision arithmetic.May I suggest GMP.-Michael. === I wonder if somebody would like to help me with a question aboutthe spectum of a radarsignal.A radar sends out a pulsed sinosoidial waveform.The carrier cos(wt-kz) which is a cosine that propagates in thez-directionis modulated by a pulsetrain which is a typical square-wave wich is asum of cosine with its odd harmonics cos(wt)+(1/3)*cos(wt)+ ...If I do the amplitude modulation (multiplication) of the pulse trainand the carrier. I get a discrete spectrum...But somebody told me that the following rule exists:A continous signal has a discrete spectrum and a discrete signal has acontinous spectrum.The radarsignal is a discrete signal. The emitter is on and off, onand off,...it should according to the rule of thumb have a continous spectrum.What have I missed here ?Greatful for any === Subject: Re: Fourier analysis of with a question about> the spectum of a radarsignal.> A radar sends out a pulsed sinosoidial waveform.> The carrier cos(wt-kz) which is a cosine that propagates in the> z-direction> is modulated by a pulsetrain which is a typical square-wave wich is a> sum of cosine with its odd harmonics cos(wt)+(1/3)*cos(wt)+ ...> > If I do the amplitude modulation (multiplication) of the pulse train> and the carrier. I get a discrete spectrum...> > But somebody told me that the following rule exists:> A continous signal has a discrete spectrum and a discrete signal has a> continous spectrum.> > The radarsignal is a discrete signal. The emitter is on and off, on> and off,...> it should according to the rule of thumb have a continous spectrum.> > What have I missed here ?You missed that the discrete spectrum requires an in'ntelylong signal. As soon as the signal is shorter, basically thebefore in'nitely thin frequency lines widen.A frequency can only be represented as accurate as many cyclesyou have witnessed. A 1000 cycles of a sine make the linewidthin the order of 1/1000 th or so.Rene-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com& commercial newsgroups - http://www.talkto.net === Subject: Re: Fourier analysis of a with a question about> the spectum of a radarsignal.> A radar sends out a pulsed sinosoidial waveform.> The carrier cos(wt-kz) which is a cosine that propagates in the> z-direction> is modulated by a pulsetrain which is a typical square-wave wich is a> sum of cosine with its odd harmonics cos(wt)+(1/3)*cos(wt)+ ...> > If I do the amplitude modulation (multiplication) of the pulse train> and the carrier. I get a discrete spectrum...> > But somebody told me that the following rule exists:> A continous signal has a discrete spectrum and a discrete signal has a> continous spectrum.A periodic signal that exists for all time has a discretespectrum. All others are smeared out to some extent.However, it is sometimes useful to approximate a signalthat looks periodic enough in some time window as if itwere a true periodic signal for all time. Since the prfand the carrier are unrelated, your radar signal will notbe truly periodic, but it may be close enough to model itthat way.-- local optimization seldom leads to global optimizationmy e-mail address is: AT mmm DOT com === wonder if somebody would like to help me with a question about> the spectum of a radarsignal.> A radar sends out a pulsed sinosoidial waveform.> The carrier cos(wt-kz) which is a cosine that propagates in the> z-direction> is modulated by a pulsetrain which is a typical square-wave wich is a> sum of cosine with its odd harmonics cos(wt)+(1/3)*cos(wt)+ ...> If I do the amplitude modulation (multiplication) of the pulse train> and the carrier. I get a discrete spectrum...> But somebody told me that the following rule exists:> A continous signal has a discrete spectrum and a discrete signal has a> continous spectrum.> The radarsignal is a discrete signal. The emitter is on and off, on> and off,...> it should according to the rule of thumb have a continous spectrum.> What have I LasseA couple of realities:The prf (pulse repetition frequency) of your typical pulse radar has norelation to the carrier frequency. The carrier is going to be Gigahertzand the prf is going to be Kilohertz and the two are generated by independant oscillators.The pulse duty cycle is typically very small, not a square wave.Some representative numbers for a 100 km search radar would be a prf inthe order of a Kilohertz and a pulse width in the order of 5 microseconds,which would give a duty cycle in the order of .5% (unless I screwed thedecimal somewhere along the line).Say a RF carrier of 10 Ghz; run those numbers through a fourier analysisand see what you === Subject: Re: Looking for primes of a particular form. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19FH8x07853;>> You will also need the ability/code to do multi-precision arithmetic.May I suggest GMP.-Michael.>For Unix? Absolutely yes!!!For Windoze? The con'guration make'le does not work. VC++does not recognize the assembler syntax. Building a GMP libraryunder Windoze is a near impossible task. I've given up. I don'thave the time..... === Subject: Re: Looking for primes of a particular form.>> You will also need the ability/code to do multi-precision arithmetic.May I suggest GMP.-Michael.> For Unix? Absolutely yes!!! For Windoze? The con'guration make'le does not work. VC++> does not recognize the assembler syntax. Building a GMP library> under Windoze is a near impossible task. I've given up. I don't> have the time.....I have a version of GMP for Windows on my site at:http://fp.gladman.plus.com/computing/gmp4win.htmI have converted the Intel processor assembler 'les into Intel syntax sothat they work with the open source NASM assembler. Brian Gladman === Subject: Re: Looking for primes of a particular form.> >You will also need the ability/code to do multi-precision arithmetic.>>May I suggest GMP.>>-Michael.> > > For Unix? Absolutely yes!!!> > For Windoze? The con'guration make'le does not work. VC++> does not recognize the assembler syntax. Building a GMP library> under Windoze is a near impossible task. I've given up. I don't> have the time.....> Bob, You probably should install cygwin, it'll install all the unix utilities you know and love, and comes with a precompiled libgmp IIRC. === Subject: Counterexample found! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19FUNX09181;Try the matrix3 5 7 9 11 13 1529 27 25 23 21 19 17But 25 belongs to a covering with only one element (includingitself).The revision will be available soon.Hisanobu Shinya === Subject: Re: Mathematical mathematical screensaver? You might like SphereEversion. It works within the context of thexscreensaver package (www.jwz.org) and produces very impressive pictureson the screen. === Subject: Re: JSH: Tautological spaces> >> I did a post recently where I said the base tautological space that>> mathematicians operate in is x = 0(mod x), and I realized later that's>> wrong as it's 1 = 0(mod 1).> > Tautological space. That is very funny. I have noted this before. JSH is > a humorist.Nevertheless, it is a bit disappointing.When I saw the title, I wondered if he had perhaps extended hisactivities === towards topology.But it is just the same old song.MarcSubject: $$f_w91tn_g > > In a tautological space like x+y+z = 0(mod x+y+z), everything has> x+y+z as a factor, but also you have 3 distinct elements x, y andz,> which give form to the space without regard to their values.> > I've spent a lot of time working the tautological space> > x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2)> > which you'll notice has 4 elements.> > James HarrisIf f(x) is a homeomorphism from T onto S and for every point p in T,f(U(p)) = U(f(p)), and the monad is invariant under standardtopological transformations, with the caveat that the de'nition alsocomprizes a type of dynamic situation sematics, where concepts, suchas proper set, ordinal and cardinal are relativised to context,taking care of paradox at all levels via symmetry, and the top,would naturally not exist, of course, since there is nothing outsidethe universe. So it becomes an in'nite chain or composition of evermore inclusive situated sets expressing an interesting informational- topological dynamic. Belief = B Knowing = K K(B) > B(K) [T|F = T ] ---> K ---> B Symmetry forms the basis of logic, thus symmetry forms the basis oftruth. To know is to believe but to believe, is not necessarily, to know. Outside of Total Existence there is nothing. This is an irrefutablefact. Or we could say that there is no outside of Total Existence.Therefore the largest possible set does not exist, where does notexist is equivalent to nothing. Nothing contains everything. Before the beginning there is nothing. After the end, there isnothing. Therefore Alpha = Omega.If space is *quantized* yet also continuous, then it too, has theindivisible units, then a measurement of space means that Fermat'slast theorem holds, for it. According to the Pythagorean theorem: x^2 + y^2 = z^2 All possible integer solutions are then rerpresented as: [a^2 - b^2]^2 + [2ab]^2 = [a^2 + b^2]^2 a^4 -2(ab)^2 + b^4 + 4(ab)^2 = a^4 + 2(ab)^2 + b^4 = [a^2 + b^2]^2 all odd numbers can be represented as: [a^2 - b^2] or Z^p - Y^p if Y is an even natural n and Z is odd, same for a and b . Fermat's last theorem, for integers a,b,Z,Y,p: [a^2 - b^2]^p + Y^p = Z^p [a^2 - b^2]^p = Z^p - Y^p a^2 - b^2 = [Z^p - Y^p]^[1/p] When Z^p - Y^p is a prime number, it cannot have an integer root. a^2 - b^2 is not an integer, via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption === $$f_w91tn_g > > In a tautological space like x+y+z = 0(mod x+y+z), everything has> x+y+z as a factor, but also you have 3 distinct elements x, y andz,> which give form to the space without regard to their values.> > I've spent a lot of time working the tautological space> > x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2)> > which you'll notice has 4 elements.> > James HarrisIf f(x) is a homeomorphism from T onto S and for every point p in T,f(U(p)) = U(f(p)), and the monad is invariant under standardtopological transformations, with the caveat that the de'nition alsocomprizes a type of dynamic situation sematics, where concepts, suchas proper set, ordinal and cardinal are relativised to context,taking care of paradox at all levels via symmetry, and the top,would naturally not exist, of course, since there is nothing outsidethe universe. So it becomes an in'nite chain or composition of evermore inclusive situated sets expressing an interesting informational- topological dynamic. Belief = B Knowing = K K(B) > B(K) [T|F = T ] ---> K ---> B Symmetry forms the basis of logic, thus symmetry forms the basis oftruth. To know is to believe but to believe, is not necessarily, to know. Outside of Total Existence there is nothing. This is an irrefutablefact. Or we could say that there is no outside of Total Existence.Therefore the largest possible set does not exist, where does notexist is equivalent to nothing. Nothing contains everything. Before the beginning there is nothing. After the end, there isnothing. Therefore Alpha = Omega.If space is *quantized* yet also continuous, then it too, has theindivisible units, then a measurement of space means that Fermat'slast theorem holds, for it. According to the Pythagorean theorem: x^2 + y^2 = z^2 All possible integer solutions are then rerpresented as: [a^2 - b^2]^2 + [2ab]^2 = [a^2 + b^2]^2 a^4 -2(ab)^2 + b^4 + 4(ab)^2 = a^4 + 2(ab)^2 + b^4 = [a^2 + b^2]^2 all odd numbers can be represented as: [a^2 - b^2] or Z^p - Y^p if Y is an even natural n and Z is odd, same for a and b . Fermat's last theorem, for integers a,b,Z,Y,p: [a^2 - b^2]^p + Y^p = Z^p [a^2 - b^2]^p = Z^p - Y^p a^2 - b^2 = [Z^p - Y^p]^[1/p] When Z^p - Y^p is a prime number, it cannot have an integer root. a^2 - b^2 is not an integer, via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: JSH: Tautological spaces> > I did a post recently where I said the base tautological space that> mathematicians operate in is x = 0(mod x), and I realized later that's> wrong as it's 1 = 0(mod 1).> > Tautological space. That is very funny. I have noted this before. JSH is > a humorist.It basically means logical domain.Mathematicians re§exively work in a simple logical domain1 = 1which seems so intuitive, so obvious that most don't seem to realizethat it IS a logical domain, where there's a dependency on atautology.Unless of course you're talking about fallacious reasoning which leadsto1 = 2as contradiction proves falsity.What happened to me is I ended up with a problem that pushed me intorecognition of higher logical domains, which are *inclusive* of thebase explicit domain.I call them tautological spaces.If you ever wonder where I get cubic examples of non-polynomialfactorization, the answer is that they're from the tautological spacex^2 + y^2 + vz^2 = x^2 + y^2 + vz^2, which workably, like if you're going to do some mathematics in thatspace, isx^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2),which seems to be an idea that leaves most people gasping andretreating to the notion that tautologies are meaningless for provinganything.But tautological space link mathematics and physics in a way that youcan't begin to grasp without some hint of their power and validity.Like the tautological space I use is a 4 dimensional one, as it hasthe 4 elements x, y, z, and v.Whether you realize it or not, I think it might actually limit ourphysical world in a way that I don't have the mental energy andfocus--or maybe the ability--to deal with right now.James Harris === Subject: Re: JSH: Tautological spacesX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>> >> I did a post recently where I said the base tautological space that>> mathematicians operate in is x = 0(mod x), and I realized later that's>> wrong as it's 1 = 0(mod 1).>> >> Tautological space. That is very funny. I have noted this before. JSH is >> a humorist.It basically means logical domain.Mathematicians re§exively work in a simple logical domain1 = 1Uh, 1 = 1 is not a logical domain, it's an equation.>which seems so intuitive, so obvious that most don't seem to realize>that it IS a logical domain, where there's a dependency on a>tautology.A dependency on a tautology is not a dependence on anything.Really is kind of shocking, the fact that mathematicians nneverrealised that their work depends on the fact that 1 = 1.>Unless of course you're talking about fallacious reasoning which leads>to1 = 2as contradiction proves falsity.What happened to me is I ended up with a problem that pushed me into>recognition of higher logical domains, which are *inclusive* of the>base explicit domain.Um, if this means anything, the base explicit domain consists of allthe things that can actually be _proved_. If so then it is indeed clear that your results are outside the base explicit domain, butthat's not something to be proud of.>I call them tautological spaces.Making up silly terms does not make your reasoning correct.Sorry.>If you ever wonder where I get cubic examples of non-polynomial>factorization, the answer is that they're from the tautological spacex^2 + y^2 + vz^2 = x^2 + y^2 + vz^2, which workably, like if you're going to do some mathematics in that>space, isx^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2),which seems to be an idea that leaves most people gasping and>retreating to the notion that tautologies are meaningless for proving>anything.But tautological space link mathematics and physics in a way that you>can't begin to grasp without some hint of their power and validity.Like the tautological space I use is a 4 dimensional one, as it has>the 4 elements x, y, z, and v.Whether you realize it or not, I think it might actually limit our>physical world in a way that I don't have the mental energy and>focus--or maybe the ability--to deal with right now.>James Harris************************David C. Ullrich === Subject: Re: JSH: Tautological spaces> > I did a post recently where I said the base tautological space that> mathematicians operate in is x = 0(mod x), and I realized later that's> wrong as it's 1 = 0(mod 1).> > Tautological space. That is very funny. I have noted this before. JSH is > a humorist.> > It basically means logical domain.> > Mathematicians re§exively work in a simple logical domain> > 1 = 1> > which seems so intuitive, so obvious that most don't seem to realize> that it IS a logical domain, where there's a dependency on a> tautology.Where mathematicians actually work has proven itself out of reach of JSH's intuition and intellect for lo these 8 years.> > Unless of course you're talking about fallacious reasoning No. That is what JSH is always dealing with. Actual mathematicians avoid it as much as possible.> which leads to 1 = 2A common consequence of JSH's chains of reasonning.> as contradiction proves falsity.> > What happened to me is I ended up with a problem that pushed me into> recognition of higher logical domains, which are *inclusive* of the> base explicit domain.So far, nothing has been able to drag you, screaming and kicking, into any _logical_ domain at any level. One expects that JSH's perceptions here will again prove to be delusions.> > I call them tautological spaces.We they should be called I am too plite to mention here.> > If you ever wonder where I get cubic examples of non-polynomial> factorization, the answer is that they're from the tautological space> > x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2,I have often wondered why, but never where. > > which workably, like if you're going to do some mathematics in that> space, is> > x^2 + y^2 + vz^2 = 0(mod x^2 + y^2 + vz^2),> > which seems to be an idea that leaves most people gasping and> retreating to the notion that tautologies are meaningless for proving> anything.Your tautological spaces have so far proved meaningless for proving anything, and those who have proved things, like those who have provided the many proofs of your many errors, have not relied of your tautological spaces to do so.> > But tautological space link mathematics and physics in a way that you> can't begin to grasp without some hint of their power and validity.> > Like the tautological space I use is a 4 dimensional one, as it has> the 4 elements x, y, z, and v.> > Whether you realize it or not, I think it might actually limit our> physical world in a way that I don't have the mental energy and> focus--or maybe the ability--to deal with right now.What you think has so far always been proved nonsense. And it has been so long this way, over 8 years now I believe, that no one expects anything else from you, except some amusement from the NG's private clown. === Subject: Re: JSH: Tautological spaces by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv1w18002;My hunch tells me that to construct M you will need elements from tautological space theory. *giggle* === Subject: How to determine whether a point is contained inside an object/polygon?Hello all,I am trying to determine whether a given point is contained inside atetrahedron, and I believe I am following the correct method in orderto do so but need it double-checked just in case!The tetrahedron is speci'ed by the points:(10,20,5),(5,10,5),(10,10,5) and (10,10,10).What I did was obtain the inner surface normal for each face and thendetermine whether the point lies inside or outside the face. Idiscovered that it lies outside three faces but inside one face. Iassume for it to lie inside the tetrahedron it must lie inside all thefaces, which it doesn't in this case, so the clari'cation. === Subject: Re: How to determine whether a point is contained inside an object/polygon?> Hello all,> I am trying to determine whether a given point is contained inside a> tetrahedron, and I believe I am following the correct method in order> to do so but need it double-checked just in case!> The tetrahedron is speci'ed by the points:> (10,20,5),(5,10,5),(10,10,5) and (10,10,10).> What I did was obtain the inner surface normal for each face and then> determine whether the point lies inside or outside the face. I> discovered that it lies outside three faces but inside one face. I> assume for it to lie inside the tetrahedron it must lie inside all the> faces, which it doesn't in this case, so clari'cation.Try computing the barycentric coordinates.The example there talks about barycentric coordinates in the plane, butthe same idea works in R^n, given n+1 points that do not all lie in thesame (n-1)-dimensional hyperplane. I will assume n = 3 in what follows.Given the vertices at points p, q, r, s in R^3, and given a point x,there is a unique representation of x in in the form x = a * p + b * q + c * r + d * s (1)where a, b, c, d are real numbers such that a + b + c + d = 1. (2) The condition (1) gives you three equations in four unknowns, and thenormalizing condition (2) provides the fourth equation.If the resulting coordinates a, b, c, d are all positive, then x liesinside the tetrahedron. If any coordinate is negative, then x liesoutside the tetrahedron. If one or more coordinates is zero and the restare positive, then x lies on a face of the tetrahedron or coincides witha vertex. -- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: My Childhood and In'nityWhen I was a young three and a half year old lad, I used to writenumbers on paper and line them across the room. My goal was to get tothe largest number possible which I now call M. Then my mother toldme that the largest number is in'nity. I asked her what things arein'nite. She could not answer me. Then I asked her what do you meangoes on for ever? How can something go on for ever? She could notanswer me and told me to ask my teacher, as that is her job.So the next day I went to nursery school and was playing on the swingand I bumped my nose and started crying. The nursery school teachersaw me crying and tried to make me feel better. I decided at thatpoint that it would be a good opportunity to ask her what in'nity is.So I did. But she didn't understand me because I was still crying.For years, I blocked this out of my memory, this disturbing incident.And I went to college and eventually got a PhD from the University ofSan Moritz, a non-accredited but well-respected university, writing mythesis on the role of life experience in scienti'c thought. Then Idifferential equations and started to shake. What I was doing wasmeaningless! What my teachers told me was a lie! Partial differentialequations presuppose the existence of in'nity! And how do I know thatthere is an in'nity? No one knows in fact. Then I remembered myselfas a child lining up numbers trying to 'nd M, the largest number inthe universe. I decided that I would convert my partial differentialequation paper about difference equations instead. Then my lifechanged. I decided that I would pick up the pieces of where I left offas a child and renew my search for M. I am currently using my computerto 'nd this number, as I have concluded that this is not a job for ahuman being.I am looking for collaborators for this big project. Anyone who wantsto participate in this revolution, literally, in mathematics andscience should please contact me. I can be found at my brother-in-lawCraig Feinstein's email address. Just do not make it known to him thatI am using it, as he thinks it is not healthy for me to continue thisproject, and that I should just do what my doctor says.Dr. Ben Zona === Subject: Re: My Childhood and In'nity> Just do not make it known to him that> I am using it, as he thinks it is not healthy for me to continue this> project, and that I should just do what my doctor says. Dr. Ben ZonaHoax.l8r, Mike N. Christoff === Subject: Re: My Childhood and In'nity> > When I was a young three and a half year old lad, I used to write> numbers on paper and line them across the room. My goal was to get to> the largest number possible which I now call adult. How big is2M, git?-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: Re: My Childhood and In'nityOriginator: dmcanzi@remulak.uwaterloo.ca (David Canzi)>> >> When I was a young three and a half year old lad, I used to write>> numbers on paper and line them across the room. My goal was to get to>> ingorant as a child. You are stupid as an adult. How big is>2M, git?2M=M, obviously. The computer program to search for M is trivial:Set N=0. Loop, incrementing N, until N+1=N. When the loop exits, N=M.It's easy and requires no expensive equipment. Somebody ought to look.Lead, follow, or get out of the way, Al.-- David Canzi === Subject: Re: My Childhood and In'nity>> When I was a young three and a half year old lad, I used to write>> numbers on paper and line them across the room. My goal was to get to>> the largest number possible are stupid as an adult. How big is>2M, git? 2M=M, obviously. The computer program to search for M is trivial:> Set N=0. Loop, incrementing N, until N+1=N. When the loop exits, N=M.> It's easy and requires no expensive equipment. Somebody ought to look.Set up a computation-sharing program. I will rent you the unused cycleds onmy machine at $1 an hour. I'll bill weekly, monthly, or whatever isconvenient.As a favor, I have already started the count on my machine. No M yet. Yourbill so far is about $52. === Subject: Re: My Childhood and In'nity>> When I was a young three and a half year old lad, I used to write>> numbers on paper and line them across the room. My goal was to get to>> the largest number possible which I now an adult. How big is>2M, git? 2M=M, obviously. The computer program to search for M is trivial:> Set N=0. Loop, incrementing N, until N+1=N. When the loop exits, N=M.> It's easy and requires no expensive equipment. Somebody ought to look. Lead, follow, or get out of the way, Al.Assume that the largest number is an integer. Assumefurther that it's a positive integer. This means youcan use an unsigned integer format for the searchvariable. It also means that you can use the followingshortcut method: Initialize the variable to zero, thenrather than repeatedly adding one until you reach thehighest value, subtract one instead. This is the shortway ïround via negative in'nity! Why waste cycles? === Subject: Re: up the pieces of where I left off> as a child and renew my search for M. I am currently using my computer> to 'nd this number, as I have concluded that this is not a job for a> human being.Just my words! Lets make a big Internet project GISM Great Internet Searchfor M. Everyone donates their CPU time and there's a prize to the owner ofthe computer that actually reaches M.With laughs,-Michael. === Subject: Re: My Childhood and In'nity> For years, I blocked this out of my memory, this disturbing incident.> And I went to college and eventually got a PhD from the University of> San Moritz, a non-accredited but well-respected university, writing my> thesis on the role of life experience in scienti'c thought.http://www.dailyfreepress.com/media/paper87/DFPArchive /frontpage/1015982.htmlhttp://www.degree.net/news.htmhttp:// www.caribvoice.org/Editor's%20Chronicles/diplomamill.html> I can be found at my brother-in-law> Craig Feinstein's email address. Just do not make it known to him that> I am using it, as he thinks it is not healthy for me to continue this> project, and that I should just do what my doctor says.What does he/she say? : keep taking the pills?Anyway, if you don't want Mr Feinstein to know, I suggest not postingto public forums like sci.math.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Rings by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19GEJA12680;A Gaussian Ring is an integral domain verifying(i) the intersection of two principal ideals is a principal idealEquivalent conditions are(ii) each couple of elements has a GCD(iii) each couple of elements has a LCM(iv) any couple of elements is elements is contained in a minimum principal ideal(v) any ideal generated by two elements is contained in a minimum principal ideal.I. Consider G=Z[X][[X1,X2,...]].a) Is G a Gaussian Ring ?b) Is G a Factor Ring ?c) Does G verify the Bezout condition ?II. Any (other?) example of a Gaussian non Factor non Bezout Ring ? === Subject: Re: Rings> A Gaussian Ring is an integral domain verifying> (i) the intersection of two principal ideals is a principal ideal> Equivalent conditions are> (ii) each couple of elements has a GCD> (iii) each couple of elements has a LCM> (iv) any couple of elements is elements is contained in a minimum principal ideal> (v) any ideal generated by two elements is contained in a minimum principal ideal.> I. Consider G=Z[X][[X1,X2,...]].> a) Is G a Gaussian Ring ?yes, because every factor ring verify (ii)> b) Is G a Factor Ring ?YesIt's a well known problem : if A is a factor ring, so is A[X]> c) Does G verify the Bezout condition ?NoGCD(X1,X2)=1But PX1+QX2 = 1 is false. Assuming PX1+QX2 =1, you substitute 0 to X1 ans X2 you 'nd 0=1.> II. Any (other?) example of a Gaussian non Factor non Bezout Ring ? Now you should 'nd an example of a gaussian ring that is not a factor ring. === Subject: Re: JSH: Pattern argument I've put out the Decker quadratic (source information at bottom),>> but replies from various people indicate there's still room for>> confusion, so I'll give a pattern argument.>> >> The original quadratic is >> >> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >> >> where his a's are roots of >> >> a^2 - (x - 1)a + 7(x^2 + x).>> >> You can modify it easily enough to get>> >> b^2 - (x - 1)b + 17(x^2 + x), where you then have>> >> (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 25x) + 17(5)(x-1) + 17^2 >> >> which is>> >> (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12) >> >> or you can go down to>> >> c^2 - (x - 1)c + 2(x^2 + x), which gives you>> >> (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 25x) - 2(5)(x-1) + 2^2 >> >> which is>> >> (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3).>> >> So I have>> >> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >> >> (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12) and>> >> (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3).>> >> The more astute of you should see a pattern. Clearly the constant>> factor on the right, which shows up as a coef'cient on the left must>> equal 2 mod 5 for this particular setup.>> >> But why?>> >> It turns out that I'm exploiting a symmetry with *integer*>> coef'cients in the factorization.>> >> Imagine that in each case there's a base equation>> >> (5y_1(x) + 1)(5y_2(x) + 2)>> >> where you can linearly tranform the second factor (5y_2(x) + 2) rather>> easily leaving you free to multiply the 'rst factor by various>> integers as long as they equal 2 mod 5.>> >> Now if you beliee that *any* of the explanations given by various>> posters out there can relate that to reducibility over Q or Galois>> Theory, then you're not good with a basic pattern.>> >> The problem is that y_1(x) and y_2(x) can't both be algebraic integer>> functions.>> >> Now if you *still* are having a problem understanding, ask yourself,>> how can anyone force a constant factor like 7, or 2, or 17 to only>> work with *one* factorization?>> >> Like consider again>> >> b^2 - (x - 1)b + 17(x^2 + x), where you have>> >> (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12).>> >> Why does that 17 lead to *one* factorization on the left? How is it>> possible for it to be limited?>> >> The answer is that there's only one way to distribute it and have>> *integer* coef'cients in the factorization on the left.>> >> Mathematicians here need to be at least a little curious, or you'll>> never 'gure it out!!!>> >> Think logically, and force people disagreeing with me to answer basic>> questions, like how do they explain the pattern:>> >> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >> >> (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12) and>> >> (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3)?>> >> >> >> The *pattern* is self-evident. You are of course treating Of course it is you stupid . That's the ING POINT OF MY> POST!!! off Nora Baron. Hey, imagine if you were the only person who replied to my posts, and> I'd always just tell you to off! Then you could make your OWN ING THREADS and point out that I> would just tell you to OFF!!! It could be a partnership of some kind. I tell you to off, and> you just keep replying anyway, over, and over again, like the stupid> who likes to give another person yet another opportunity to tell> them to OFF!!!!!!!!!Curious response, given this quote from a couple of minutes prior tothis post.,----[ Dik Winter and Rick Decker now? So far it looks like only| Nora Baron at least made an attempt to address the math, while a LOT| of other posters felt a need to pile up posts on different subjects.`------ But remember, as long as one human being follows the rules ofmathematics, then mathematics as a human discipline survives.Right now I'm that one human being, so mathematics survives. -- James S. Harris === Subject: Re: JSH: Pattern argumentI've been watching the one and only James Harris Show for some timenow, but I'm still at a loss what James is actually trying to prove(if anything). Is this a crucial part of some purported proof of FLTor what?Just curious.> So I've put out the Decker quadratic (source information at bottom),> but replies from various people indicate there's still room for> confusion, so I'll give a pattern argument.[ remaining part of message recycled ] === Subject: Re: JSH: Pattern argument>Subject: Re: JSH: Pattern argumentI went on a trip to the nearby beach to 'nd rocks that have approximately>your IQ. It wasn't easy but I think I have found a suf'ciently stupid one.> But does it have a BS in Physics and is it a veteran? --> Mensanator> Ace of ClubsBetter yet, did James serve his country during the Korean War? I heard hehelped the Marines escape from the Chosin Reservoir and served from 1950-1953.He also belongs to the Allied Seniors Recreational Vehicle Association - soplease, afford him some respect. Please don't toss him a bone either; a mandoesn't do that to a Korean War veteran.-ET === Subject: Re: JSH: Pattern argument <40270A45.47104F0B@osu.edu> sha1:BEbN7FrtJHAwudk80z5m+J/NT4U=> Better yet, did James serve his country during the Korean War? I> heard he helped the Marines escape from the Chosin Reservoir and> served from 1950-1953. He also belongs to the Allied Seniors> Recreational Vehicle Association - so please, afford him some> respect. Please don't toss him a bone either; a man doesn't do that> to a Korean War veteran.James says that he is around 35 years old. -- Jesse F. HughesLeaving things always seems to 'x me, Running seems to ease my worried mind. -- Bad Livers, Honey, I've Found a Brand New Way === Subject: Re: JSH: Pattern argument> > Better yet, did James serve his country during the Korean War? I> heard he helped the Marines escape from the Chosin Reservoir and> served from 1950-1953. He also belongs to the Allied Seniors> Recreational Vehicle Association - so please, afford him some> respect. Please don't toss him a bone either; a man doesn't do that> to a Korean War veteran.> > James says that he is around 35 years old.My post was a presentation of a mathematical pattern, but I've noticedhow certain posters *immediately* replied to push it in a differentdirection.Where are Dik Winter and Rick Decker now? So far it looks like onlyNora Baron at least made an attempt to address the math, while a LOTof other posters felt a need to pile up posts on different subjects.That's what I'm talking about people!It'd be one thing if I could make a post like my original one and getone or two replies or even none!But instead I see evidence of an effort to hide the post by peopleputting up a continual stream of irrelevant replies.Sometimes in such a situation I'd start another thread, and they'djust follow, while others would accuse me of ignoring refutations!Whether you realize it or not many people on the sci.math newsgroupare behaving in a way that de'es rational explanation.James Harris === Subject: Re: JSH: Pattern argument> But instead I see evidence of an effort to hide the post by people> putting up a continual stream of irrelevant replies.You mean like all those new threads you make to avoid answering your mathematical critics? === Subject: Re: JSH: Pattern argument> My post was a presentation of a mathematical pattern, but I've noticed> how certain posters *immediately* replied to push it in a different> direction.Just following your lead, old sod.How many postings presenting only mathematical ideas have you replied to by trying to push the discussion in da different direction? An overwhelming majority of your posts in response to purely mathematical analyses of your mathematical writings are of this nature.What goes around, comes around. That you do not like such equal treatment is irrelevant. === Subject: Re: JSH: Pattern argument <40270A45.47104F0B@osu.edu> <87y8rcwx1r.fsf@phiwumbda.org> Better yet, did James serve his country during the Korean War? I>> heard he helped the Marines escape from the Chosin Reservoir and>> served from 1950-1953. He also belongs to the Allied Seniors>> Recreational Vehicle Association - so please, afford him some>> respect. Please don't toss him a bone either; a man doesn't do that>> to a Korean War veteran.>> >> James says that he is around 35 years old. My post was a presentation of a mathematical pattern, but I've noticed> how certain posters *immediately* replied to push it in a different> direction. Where are Dik Winter and Rick Decker now? So far it looks like only> Nora Baron at least made an attempt to address the math, while a LOT> of other posters felt a need to pile up posts on different subjects.Moron. I corrected someone's misstatements about you. I didn't rushto pile up posts on your moronic pattern argument one way or theother. > That's what I'm talking about people! It'd be one thing if I could make a post like my original one and get> one or two replies or even none! But instead I see evidence of an effort to hide the post by people> putting up a continual stream of irrelevant replies.I've no idea who Eddie Tomayko is, but he has no evident history ofreplying to your posts ever before. Must be part of the conspiracy.I'll bet it's Nora Baron with a new pseudo-pseudonym. Even more'endishly, I'll bet that this new pseudonym is her real name....In any case, only an utter moron with no concept of how realnewsreaders work would believe that followups to posts hide thepost. Real newsreaders (not Google's website) typically put theoriginal post on the top of the thread where it's easily seen,regardless of how many replies there are[1]. Furthermore, on Google, new posts ensure that a thread is presentedearly in the menu, so that readers are *more* likely to see the threadand so probably more likely to read the original post. If you want tobury a post on Google, then you ought to start many threads withdifferent subject lines, so that the thread with the post gets pushedto a subsequent page.Of course, you've been told this before. I don't expect you to gain asudden and unexpected ability to avoid making an ass of yourself. Butit is remarkable that you accuse me (the biggest collector of yournuggets of, well, wisdom or whatever) of trying to suppress your postwhen I correct misstatements about you.Footnotes: [1] Typically, newsservers have a 'nite capacity, so if you makeenough posts in a single newsgroup on any topic whatsoever, the servermay be forced to expunge some older posts. In this crude way, onecould try to suppress posts, but it wouldn't be very effective. Mostnews servers have suf'cient capacity that it would take at leasthundreds of posts for this to work.-- Jesse F. HughesIt's easy folks. Just talk about my approach to your favoritemathematician. If they can't be interested in it, they'vedemonstrated a lack of mathematical skill. -- James Harris === Subject: Re: JSH: Pattern argument> My post was a presentation of a mathematical pattern, but I've noticed> how certain posters *immediately* replied to push it in a different> direction. Where are Dik Winter and Rick Decker now? So far it looks like only> Nora Baron at least made an attempt to address the math, while a LOT> of other posters felt a need to pile up posts on different subjects.She was not the only one who addressed the math, not that there was any comprehensible math toaddress. All you did was put made-up labels on some alleged patterns. Here's a pattern:1. JSH posts a claim of an irrefutable proof,2. Several posters 'nd errors and point them out,3. JSH ignores refutations/counter-examples and launches personal attacks,4. Repeat from 1. until truth sinks in,5. JSH acknowledges ïminor' error and claims proof is now complete,6. Goto 1.How's that for a pattern? (No tautological space required, just observation.)> That's what I'm talking about people!Talking? Babbling is more like it. Maybe dissembling.> It'd be one thing if I could make a post like my original one and get> one or two replies or even none! But instead I see evidence of an effort to hide the post by people> putting up a continual stream of irrelevant replies.How can anyone's post ïhide' yours? Are you feeling paranoid?Anyway, you/ve had relevant replies. If you're so concerned about relevance, why not respond tothem?> Sometimes in such a situation I'd start another thread, and they'd> just follow, while others would accuse me of ignoring refutations!You *do* ignore refutations. The accusation has been con'rmed by your own behavior.> Whether you realize it or not many people on the sci.math newsgroup> are behaving in a way that de'es rational explanation.There's no need to bother yourself with explaining anyone's behavior. You are not a psychologist.Just stick to the math and respond to the posters who refute your ridiculous claims.> James I know you are, but what am I? Harris--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: JSH: Pattern argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i18Jv1Y17985;No it doesn't have a BS in physics but I don't think it will be hard for it to get one(after all, JSH got one). As for being a veteran, *packs the rock and mails it to a military base in Iraq*. Be patient for umm two days and you'll get yourself a veteran === rock.(Yes I am a drop of water in the Paci'st Ocean)Subject: Re: wanted: more mathematical false proofs > > >>I was recently introduced this cute little oddity, and I would like to>>know if anyone has more of these false proofs.>>x+x+x+x+...+x+x = x^2>>|___x times___|>>so d/dx(x+x+x+x+...+x)=1+1+1+1...+1=x>> |__x times__| |__x times__|>>while d/dx(x^2) = 2x, so x = 2x> > > I kind of like the following:> > Int (1/x) dx = Int (1/x) (1 dx) = (1/x) x - Int (-1/x^2) x dx = 1 + Int> (1/x) dx> so 0 = 1.... but when you see that integration is up to an arbitray constant only, int(1/x)= 1+ int(1/x) ==> ln|x|+C1= 1+ ln|x|+C2 so C1-C2=1, so the constants of integration are not arbitrary, but via Newsfeeds.Com, Uncensored Usenet News =-----http://www.newsfeeds.com - The #1 Newsgroup Service in the World!-----== Over 100,000 Newsgroups - 19 Different === Servers! =-----Subject: Re: wanted: more mathematical false proofs > > >>I was recently introduced this cute little oddity, and I would like to >>know if anyone has more of these false proofs.> > > For a binary relation R, the symmetric and transitive laws imply the> re§exive one, for xRy => yRx by the symmetric law and so xRy and yRx> => xRx by the trasitive law.> Newsfeeds.Com, Uncensored Usenet News =-----http://www.newsfeeds.com - The #1 Newsgroup Service in the World!-----== Over 100,000 Newsgroups - 19 Different === Subject: Re: Question for logarithm experts> I have heard the following remarks about this problem: it's an> unfair question. Its an equation that's not an equation. It is a> single equation with two variables.> >>The second is right, it's not an equation. There is no value of X that can>>make this true.> > Nonsense. An equation with no solutions is still an equation.Sorry, it's an equation with no solution.-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people abuse@teranews.com === Subject: Re: Question for logarithm experts> I am looking for a step-by-step method (proof) on the solution of> this particular equation - (Solve for x) and equations like it:> > 7^X = 4*X.>>Using a graphing calculator it's easy to show that there is *no* solution>>for this equation. I plugged both into mine, and then zoomed into where>>they're at their closest, and it's very obvious that there's no solution.> > Oh, so your graphing calculator thinks for you, does it? No. But, it provides a visual proof that there's no solution.> I happen to like> the function f(x) = 40 + log(x) - x/100 . Does your calculator tell> you anything about the solutions to, say f(x) = 1 ?> > Calculators are 'ne tools, but they're not a replacement for thinking!You're right. I never said it was a tool to replace thinking. I said it wasa tool to reinforce that, when you can't 'nd a solution, that there reallyis no solution. It's a backup when you start thinking, maybe I didsomething wrong to show whether you did or didn't in fact do somethingwrong...-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: Question for logarithm experts by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19Bhve22741;>> I have heard the following remarks about this problem: it's an> unfair question. Its an equation that's not an equation. It is a> single equation with two variables.>The second is right, it's not an equation. There is no value of X that can>>make this true.Nonsense. An equation with no solutions is still an equation.Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~ israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2Since the given form 7^X=4X ,may be logarithmisized to base seven: X=Log_7[4*X]=Log_7[4] + Log_7[X] , or to base e: X=[ln4/ln7]+[lnX/ln7] X=O.712414374 + 0.513898342[lnX]Trying to solve it by approximate solutions and starting with X=1We get: 1=O.712414374 +0 , or 1=O.712414374 This cannot happenFor X=0 ,we get: 0=O.712414374 -[in'nity]or 0=-[in'nity]Here,yes , we have a case of controversies,and not a case of simply lacking solutionI, believe that the designer of this form was consciousof the problem he was Stefanides === Subject: construct a circle of unit area?how to construct a circle with area = exactly one unit?any applications for it outside of probability? === Subject: Re: construct a circle of unit area?> > how to construct a circle with area = exactly one unit?> > any applications for it outside of probability? 1) De'ne an arbitrary circle to have unit area. 2) Pi is irrational and transcendental. There is no way toconstruct a unit area circle by Greek rules (straightedge andcompass, linear and/or quadratic functions). 3) Construct a unit square. Convert to a unit circle using aquadratrix curve. There are lots! Alas, no quadratrix curve can beconstructed by Greek rules.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: Re: construct a circle of unit area?> > how to construct a circle with area = exactly one unit?> Circle: http://mathworld.wolfram.com/Circle.html set equation (5) A = 1 and solve for r === Subject: Re: construct a circle of unit area? Adjunct Assistant Professor at the University of Montana.>how to construct a circle with area = exactly one unit?Depends on (a) you mean by construct; and (b) what you mean byunit.For example, by construct mathematicians usually mean using onlycompass and straightedge, which is equivalent to saying that giventwo points, you can draw the line between them, given a point and adistance, you can draw a circle with center at the given point andradius equal to the given distance; and that you can 'ndintersections of constructed lines and circles. One way to construct a circle of exactly one unit in area would thenbe to construct any circle whatsoever, and then declare its area to bethe equal to one unit of whatever you are measuring. But that would bedeemed as cheating by many.So, presumably, you mean: given a speci'c distance to be used as aunit, to construct (perhaps in the sense above) a circle whose areais equal to the area of the square whose sides are one unitlong. (Such as square can be constructed using only compass andstraightedge).If so, the answer is no: a circle of area exatly one unit squarewould have to have a radius equal to 1/sqrt(pi). Which means that inorder to construct such a circle, you must be able to construct twopoints whose distance is exactly 1/sqrt(pi). If you could do that,then you would be able to construct (through well known methods), twopoints whose distance is exactly the reciprocal of that, sqrt(pi); andfrom that, again through well known methods, you would be able toconstruct two points whose distance is the square of sqrt(pi), whichis pi.However, pi is a transcendental number, and it is known that the onlydistances that are constructible (in the sense above) are thosereal algebraic numbers alpha which can be obtained from rationalnumbers by taking a sequence of square roots; say, things like 3 +sqrt(2+sqrt(4-sqrt(2+3sqrt(7)))). So it is impossible to construct twopoints whose distance is pi (this is also why it is impossible tosquare the circle: that means, given a circle, construct a linesegment whose length squared has the same area as the given circle;the circle of radius 1 one would require you to construct sqrt(pi)). Therefore, it is impossible to construct a circle of area exactly 1unit, in the sense above (i.e., given the unit 'rst, and allowing theuse of only a compass and straightedge).-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: any linear function is continuous, right?> >>For example, f(x)=Ax, where x is a vector, A is a matrix, is this a>>continous function? is this a 1 to 1 function?> > > Yes it is always a continuous function. First prove that there is an> M, dependent on A, such that |Ax| < M|x| where I leave you the choice> of norms for the occasion. Proof of continuity is then easy. Of> course it is not necessarily a 1-1 function. What if A = 0? How> about is a square matrix containing all 1's or whatever. Good grief!> space containing x is of 'nite> dimension, that is correct. However, there are plenty of linear maps> on in'nite-dimensional vector spaces which are not continuous. For> instance, on the space of in'nitely differentiable real-valued> functions on the unit interval [0,1], with the so-called sup norm:> > ||f|| = sup{ |f(x)| : 0 <= x <= 1}> > the differentiation map f |---> f' is not continuous. That is, for any> f, there is a sequence of elements {g_n : n = 1, 2, ... } approaching> f, for which ||f' - (g_n)'|| does not approach 0. For instance, just> take g_n = f + sin(nx)/n, to 'nd the differences ||f - g_n|| ~ 1/n, but> || f' - (g_n)'|| ~ 1 for all n. Alternatively, one could make the norms> || f - g_n || approach zero, but force|| f' - (g_n)' || ==> in'nity, by> suitable selection of the frequencies of the sinusoids.> > In short, not only does the differentiation operator fail to be> continuous at *some* point f of this space, it actually fails to be> continuous at *all* points of the space.> > I'm sure all the analysts can respond much more authoritatively than I> have, so I'll let this go now. I just couldn't let the expression> linear ==> continuous pass without mention.> > DaleHold on a minute, Dale! I never said that linear implied continuous. I said that multiplying by a matrix implied continuous. I was tryingto answer the OP at the OP's level. I am entirely aware that thereare linear operators in in'nite dimensional spaces that are notcontinuous. I am not really familiar with all the intimate details ofrepresenting linear operators with matrices when dealing in in'nitedimensional spaces. It looks to me, looking very super'cially, thatwe would need to have at most a countable dimension, and the matrixmaybe (multiplying with the matrix on the left) as if every row wouldhave to have only 'nitely many non-zero entries. It also looks to meas if it is entirely possible that such a linear operator could beunbounded (or even discontinuous). However, I am by nature more of analgebraist, so this is driving me crazy. I need an === Subject: Re: Repunits prime factors: interesting?Just say: I will know if the found result is Ribenboim's book The book of prime number records and I> found a chapter about repunits (numbers that are formed only by 1's in> base 10: 1, 11, 111, ....) and its primality. Essentially the Book says> that there's a little few known about it. And in general, it's not known> when a repunit is prime or is composite, or if the number of prime> repunits is 'nite or in'nite.> > I had interest in this class of numbers and I wanted to investigate more> about these. With some lucky and investigated another thing, I discovered> (at least) a surprised result (at least for me). It points out that the> number of factors of the repunits is, in mean, very bigger. Sorry, but I> don't check it computationally.> > I offer it with some comments to you. Please, comment any part that you> think that it's good (or bad) or anything you want. I hope that anyone> think that it's useful. Please, if reply me, not by mail if it's possible.> > 1. Notation: I denote as R(n) the n-th repunit, that is a number formed by> n 1's. [R(1)=1, R(2)=11, R(3)=111, ...]> > Now two simply lemmas that are obvious true, and that I only prove these> for formality. Clearly, these results are more general, but we only are> interested in interval [0,1]> > 2. Lemma 1: Any rational number of [0,1] belongs to any of the following> classes:> a) Periodic numbers (mixed or pure) (in base 10) b) Numbers with 'nit> number of digits (in base 10)> > Dem: We need a lemma:> Lemma 1a: Let be a a number in [0,1] with 'nite number of decimal> digits or a periodic number (mixed or pure). Then, for all b>=1, a/b is a> number with 'nite number of decimal digits or a periodic number (mixed> or pure). Dem: Induction on N(b)=number of prime factors of b=n.> > n=0: Then b=1, and a/b=a and all it's ok. n-->n+1: N(b)=n+1, so> b=p_1....p_np_{n+1}.> a/b = a/(p_1...p_np_{n+1}) = [a/(p_1...p_n)]/p_{n+1}. The> numerator is periodic number or a number with 'nit number of> digits (induction hipotesis) and then (by the case n=1) all the> fraction is too.> It 'nishes the dem.> > Prove of the lemma properly:> We prove that a/b is of the two previous classes by induction on N(b).> N(b)=0. Then b=1. So, because a/b is in [0,1], a=0 or a=1 that is> clearly a number with 'nite number of digits. N(b)=n-->N(b)=n+1: So> b=p_1...p_np_{n+1}. So a/b = a/(p_1...p_np_{n+1})=[a/p_1...p_n]/p_{n+1}.> a/p_1...p_n is a periodic number or a number with 'nit number of digits> by induction hipotesis. And so [a/p_1...p_n]/p_{n+1} is a number with> 'nit number of digits or a periodic number applying lemma (1a).> > 3. Lemma 2: Any rational numbers of [0,1] has the form:> 1a) a/10^n for some n>=0, a>=1.> 1b) a/(9R(n)10^r), for some a>=1, r>=0, n>=1> > Dem: Any rational number of [0,1] is a periodic number or a number with> 'nit number of digits. We prove that the 'rst class has the second> form, and the second class has the 'rst form.> > - If we have x rational number in [0,1] that have only a 'nite number of> digits, then> > x=0.a_1.....a_n, where a_i are the digits of x (in base 10).> > So x = (a_1...a_n)/10^n. And saying a = a_1...a_n we have that x => a/10^n, as we want.> > - If we have x rational number in [0,1] that it's a periodic number, then> x has the form> > x=0.b_1...b_ra_1....a_na_1....a_na1_....a_n....> > where a_1...a_n is the period (n>=1) and b_1...b_r is the non-peridic> part (r>=0. We can have r=0 in the case that x is pure periodic number).> > So 10^(r+n)x = b_1....b_ra_1....a_n.a_1....a_na_1...a_n....> -10^rx = - b_1....b_r.a_1....a_na_1...a_n....> > If we add these, we have:> > x = (b_1....b_ra_1....a_n - b_1...b_r)/(10^r(10^n-1))> = (b_1....b_ra_1....a_n - b_1...b_r)/(10^r9R(n)) => a/(9R(n)10^r), where a = b_1....b_ra_1....a_n - b_1...b_r> > as we want prove.> > 4. Theorem: For every prime p>=7, there exist n such that p divides R(n)> [it not implies that R(n) could not be prime. In that case (R(n) prime), p> were equal to R(n)]> > Dem: Let be the fraction 1/p, p prime distinct of 2, 3 and 5 (p>=7).> This number is (clearly) a rational number.> So, by lemma 2, 1/p is has the form (1a) or (1b). Clearly it could not> have the form (1a): If 1/p=a/10^n for some n>=0, then 10^n = pa. So p> divides 10^n, that it's impossible because p is not equal to 2,5. So it> has the form (1b). So there exist some n>=1, r>=0, a>=1 such that> > a/(9R(n)10^r) = 1/p> > So ap = 9R(n)10^r. So p divides 9R(n)10^r. But, because p is> not equal to 2,3,5, p does not divide 9 and 10^r. So, because p is prime,> p divides R(n).> > So we proved that if p>=7, then there are some n such that p divides> R(n), as we want.> > 5. Notes:> a) I believe that it's interesting investigate the sequences of numbers> (a_n) with the similar propierty of that result, because a general study> of this topic could provide us a information about repunits.> > Fixed m>0, we could de'ne that (a_n) is m-anything iff for all p >=m,> there exist n>=0 such that p divides a_n. Our case is m=7.> > I don't know if anyone investigated/discovered anything about it.> > b) If we count the repunits and the primes, obviously there are much more> primes than repunits:> Let Ro(x)=Card({n repunit <= x}), and Pi(x) is the counting prime> function, then it's easy to prove that Ro(x) = [log(9x+1)], where [x] is> the integer part of x, that it's clear more more smaller function than> Pi(x).> > So it's reasonably to think that the composite repunits have more> prime factors in its descomposition. But I don't know if it's true and> how prove it.> > c) We can generalizate this result, that have a great corollary. Now,> it's.> > > 6. Lemma: For all b>=1 such that 2,3,5 do not divide b, 1/b has the form> (1b)> > Dem: Suppose that 1/b has the form (1a). Then there were a, n>=0 such> that> > a/10^n = 1/b> > So ab = 10^n. In particular, b divides 10^n, which is impossible> because 2, 5 do not divide b.> > 7. Theorem: For all b>=1 such that 2,3,5 do not divide b, then there> exist n>=0 such that b divides R(n)> > Dem: Considering 1/b with b such that 2,3,5 do not divide b.> Then 1/b has the form (1b) by lemma.> So there is n>=0 such that> > 1/b = a/(9R(n)10^r)> > So 9R(n)10^r = ab. In particular, b divides 9R(n)10^r. But b> does not divide 9 (because 3 does not divide b) nop 10^r (because 2,3 do> not divide b). So b divides R(n). Then we prove that for such b there are> n such that b divides R(n), as we want.> > 8. Corollary: Let be w(x) the number of distinct prime factors or x. Then> there are repunits with arbitrary values of w(x), that is, for all m>2,> there is some n>=0 such that w(R(n))=m [for m=1 is true too: R(2)=11 that> is prime]> > Dem: Chossing b=p_1...p_m and applying the previous result.> > > The question is: Are there in'nite number of repunits with this> propierty?. That is, for all m>=1, if de'ne W(m)={r repunit such that> w(r)=m}, have we got that card(W(m))=in'nity? (Now we know that> card(W(m))>=1). The case m=1 is if prime repunits are in'nit.> > > > Well, thank you very much for === Subject: Re: Repunits prime factors: a result.Xan [point 7] in somehow interesting?> Just say: I will know if the found result is original at least.> 7. Theorem: For all b>=1 such that 2,3,5 do not divide b, then there>> exist n>=0 such that b divides R(n)Yes, it is known, and the exception of 3 is unnecessary. Actually, if gcd(b,10) = 1 exist n such taht b divides R(n).Dem.:gcd(9b, 10) = 1 ==> 10^(phi(9b)) = 1 (mod 9b) (Th. Euler-Fermat) ==>Exist k such that 10^(phi(9b)) - 1 = k*(9b) >k*b = R(phi(9b))If gcd(b, 3) = 1, is enough with n = phi(b).By example, it is a problem inDivulgaciones Matem.87ticas (Universidad del Zulia, Venezuela), Vol. 7, Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Counterexample found! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19Hjm823865;>Try the matrix3 5 7 9 11 13 1529 27 25 23 21 19 17But 25 belongs to a covering with only one element (including>itself).The revision will be available soon.Hisanobu Shinya Now, the revised version is available athttp://www.geocities.com/erdosfan/solution.htmlHisanobu Shinya === Subject: Need Matlab Software for Fractional Brownin Motion Analysis and SynthesisI need some Matlab code for identi'cation, analysis and synthesis of fractional Brownian motion (fBM) or any combination thereof. I triedsome code from a website, but it kept giving different Hurst exponentseven with the same input everytime I ran it.Please send or post === Subject: Re: Goldberg dual> Does anyone have a name for the dual of a Goldberg polyhedron?A follow up question. Is there a name for an asymmetric Goldberg-likepolyhedron? What is the name of the dual of the Goldberg-likepolyhedron?My guess is this. For 3(N-2) edge elements whose lengths are variable,and where N is is an integer greater that 2, ther is a maximumdiameter polyhedron that can be contructed and a minimum diameterpolyhedron that can be constructed. The maximim diameter polyhedronthat can be built with 3(N-2) edges is a Goldberg Polyhedron likestructure. The minimum diameter sphere will be an omnitriangulatedicosahedral arrangement of the edge elements.Additionally, even smaller spheres can be built by doubling up edges,tripling up edges, etc.Dick === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19JS3200963;>IRRATIONAL devided by IRRATIONAL(Pi) = IRRATIONAL.Hey, this is fun! So Sqrt(8) / Sqrt(2) = 2 is irrational?Toni,I,thank You.What I stated underneath includes the words WEATHER ......ANDpossibly ,AND StefanidesThis boils down to me ,to prove whether arcos[IRRATIONAL-G.R.] is IRRATIONAL and if so the ratio:IRRATIONAL devided by IRRATIONAL(Pi) = Stefanideshttp://www.stefanides.gr === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19Ja2g01712;>>Hello>>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>>this has something to with the golden ratio, and maybe with Fibonacci>>numbers or continued fractions? But I could't prove whether arcos[IRRATIONAL-G.R.] is IRRATIONAL and if so the ratio:>> >> IRRATIONAL devided by IRRATIONAL(Pi) = href=http://www.stefanides.gr>http://www.stefanides.grlie in the fact>that sin(pi/10) = (sqrt(5)-1)/4 = cos(2*pi)/5)But>> IRRATIONAL divided by IRRATIONAL(Pi) = IRRATIONAL>is simoly not true! Try pi/pi .Ray I,thank You.What I stated includes the words WEATHER ......AND.Possibly ,AND Stefanides>But is an algebraic number divided by a transcendental number always>irrational?Note that we also have>sin(pi/6)= (sqrt(4) -0)/4>and>sin(pi/12)= (sqrt(6) -sqrt(2) )/4Challenge from an old trig book:Show that if d > 12 and sin(pi/d)= (sqrt(n) - sqrt(n-4) )/4 then d Steiner === Subject: Proving Hall's Theorem using Tutte'sCan someone tell me how to use tutte's theorem to derive hall'stheorem(marriage theorem). === Subject: Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i19JgVj02273;> > how to construct a circle with area = exactly one unit?if you have a segment of length pi, then the sitehttp://164.8.13.169/Enciklopedija/math/math/g/g144.htmmay be helpful.if you are asking for a way of constructing pi with straight edge and compass, such has been proven impossible.> any applications for it outside of probability?are there applications for this in probability?:) === Subject: Re: there is no such thing as in'nity> I've thought really hard about this one and came to the conclusion> that there is no scienti'c evidence of in'nity existing. The highest> number that anyone has ever measured to according to Isaac Asimov in> his book Science and Human Thought is only about 5.0 x 10^48. No one> has ever gotten past that number. Doesn't this sound weird?Brace yourself!! Here I go.... (5.0 x 10^48) + 1 There, I just went past that number!! > > What's to say that eventually there is a number where it is impossible> to count higher than? If someone were to 'nd this number and prove> that it is in fact the highest number, then that person would> undoubtably be rich and famous.> > I am currently running a computer program that will eventually 'nd> this magic number (I hope and pray) that I call M for short. It> counts and counts and counts and my theory is that it will eventually> stop at M. I am looking for collaborators in this experiment so that I> can use their computer time. The program in FORTRAN is simple:> > 00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 end> > It has currently reached about 2.0 x 10^18. Just as Einstein proved> that there is no aether, I am convinced that I will prove that there> is no in'nity and then write a book or two.> > Dr. Ben ZonaHmmm... Once you're done running your program run it again for M + 1 ;-) === Subject: Re: there is no such thing as in'nityNever understood why nobody else said that. M * 2, now that would reallymake the world end.> I've thought really hard about this one and came to the conclusion> that there is no scienti'c evidence of in'nity existing. The highest> number that anyone has ever measured to according to Isaac Asimov in> his book Science and Human Thought is only about 5.0 x 10^48. No one> has ever gotten past that number. Doesn't this sound weird? Brace yourself!! Here I go.... (5.0 x 10^48) + 1 There, I just went past that number!!> What's to say that eventually there is a number where it is impossible> to count higher than? If someone were to 'nd this number and prove> that it is in fact the highest number, then that person would> undoubtably be rich and famous. I am currently running a computer program that will eventually 'nd> this magic number (I hope and pray) that I call M for short. It> counts and counts and counts and my theory is that it will eventually> stop at M. I am looking for collaborators in this experiment so that I> can use their computer time. The program in FORTRAN is simple: 00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 end It has currently reached about 2.0 x 10^18. Just as Einstein proved> that there is no aether, I am convinced that I will prove that there> is no in'nity and then write a book or two. Dr. Ben Zona Hmmm... Once you're done running your program run it again for M + 1 ;-) === Subject: Re: there is no such thing as hard about this one and came to the conclusion> that there is no scienti'c evidence of in'nity existing. The highest> number that anyone has ever measured to according to Isaac Asimov> in his book Science and Human Thought is only about 5.0 x 10^48.> No one has ever gotten past that number. Doesn't this sound weird? [Sam]> In'nity: http://mathworld.wolfram.com/In'nity.html> Special Relativity> http://scienceworld.wolfram.com/physics/SpecialRelativity.html > Physcal reality is observed -- Equations 7, 8 as v --> c [hanson]> Well, Sam, so Physcal reality is observed since 1905, ...wow!> Or did you mean Fiscal reality is observed? Well, that one> is always observed but rarely to never really taken seriously.> As for the third possibility that Physical reality is observed,> and observable, for that one we should indeed be grateful,> and I think that Albert did not have anything to do with it.> You probably meant to say Physical in'nity is observed.....> ....mmmm...dunno about this one, and if so, then only perhaps,> maybe, and including a lot of ahh, oehhh and hmm's. IMO, nobody ever is going to measure/observe anything in'nite,> big or small. That is so by de'nition. But, by making (more or less)> logical deductions and conclusions from what we can observe, we> can reasonably assume and conjecture that realms/domains beyond> our measuring and our observational capabilities do indeed exist.> Actually, one can make a case that is was exactly for this reason> that the notion of in'nity was created/invented.> Moreover, if we deny such existence then the question immediately> arises, ......but what's beyond the measured/observed limit.>==>(*I*) > Another neat thing about in'nity is that it is not numerically 'xed,> but instead goes with the §ow.......and hence allows a smooth> transition from physics to meta physics and then into philosophy....>[Sam]> In'nity is a concept.... there are an in'nite number of point in> a line segment (mathematically)... as there are an in'nite number> of points in space between two points (physically). > I've got no problem with that.>[hanson]Cool, Sam. Glad to hear that. So, then in order to have an in'nitenumber of your points, by necessity, they must be in'nitely small. See ==> (*I*), and continuing with it we could ask justi'ably wheredoes the size/amount/extent become so small that it safely can be neglected and thrown out. Practical, real world consideration begin tobe the Determinant of the neglect-limit. -- Right, Sam, we have noproblem with that. But, in the larger picture, outside, away from the direct look into the microscope, or with a higher resolution machinethe picture may be far, far different, .....for these thrown out of sight, negletable and neglected vanishing tails or curled up dimensions may turn out to be the existential cornerstones and foundations upon which all our other constructs do rest...............metaphorically a bit like thesituation in pre Koch times when we didn't know, not recognized and neglected existence of the then in'nitely small cells, that they either can do us in or, even more fundamentally, do make us up. ==> (*I*). So, in'nity as a concept appears to be a very many sided blade to cut into the unknown. Yet, it is precisely these in'nities, big and small,that all our knowledge does arise from....and that's why I think theyare so bitchen ........in'nities-unknowns-mysteries.........Yeah!.....Let[Ca pitalOTilde]s go after these ers! .............ahahahha......ahahahahanson === Subject: Re: there is no such thing as in'nityThat is a low blow. Many greater people than you and me have namesthat mean different things in different languages. Senator Kerry, whomay even be president, has a name Kerry that means semen in Hebrew.And Ko' Annan means we are monkeys in Aramaic (literally monkeys,we are). Now, are you going to say that the U.N. is just a bunch ofmonkeys when they were in fact he ones who said that Saddam Husseindidn't have WMD's and GW Bush thought they did? Who was the realmonkey here? I am proud of my name. So if you are insulting me, youshould be consistent and insult these great people too.Dr. Ben Zona === Subject: Re: there is no such thing as in'nity> >I've thought really hard about this one and came to the conclusion>that there is no scienti'c evidence of in'nity existing. The highest>number that anyone has ever measured to according to Isaac Asimov in>his book Science and Human Thought is only about 5.0 x 10^48. No one>has ever gotten past that number. Doesn't this sound weird?> > You are trying to relate in'nity to a quantity. In'nity is not a> quantity. Nor is it the absence of quantity. If it could be so> related, then it would be a number that is not a number, and hence> would have no identity. But in'nity IS NOT A NUMBER.> > You should take a look at Cantor's theory of sets. In it, Cantor does> treat in'nities as numbers, cardinals and ordinals. Try a Google> search on it. Most people already have. But Cantor never claimed that in'nity was a number. Only recursive mathema-rejects have ever done that.> > One of the brilliant things Cantor did was to de'ne an in'nite set> as a set whose elements can be put into 1-1 correspondence with a> proper subset of itself (obviously something one cannot do with a> 'nite set). For instance, you can put the set of positive integers> into 1-1 correspondence with the set of positive even integers by the> correspondence> > n -> 2n But Cantor didn't do that. Ancient Greeks did that.> It seems to go against common sense to say that these two sets have> the same cardinalities. Then again, we don't have any common> experience working with in'nities.> > Patrick === Subject: Re: there is no such thing as in'nity> > I've thought really hard about this one and came to the conclusion> that there is no scienti'c evidence of in'nity existing. The highest> number that anyone has ever measured to according to Isaac Asimov in> his book Science and Human Thought is only about 5.0 x 10^48. No one> has ever gotten past that number. Doesn't this sound weird?> > What's to say that eventually there is a number where it is impossible> to count higher than? If someone were to 'nd this number and prove> that it is in fact the highest number, then that person would> undoubtably be rich and famous.> > I am currently running a computer program that will eventually 'nd> this magic number (I hope and pray) that I call M for short. It> counts and counts and counts and my theory is that it will eventually> stop at M. I am looking for collaborators in this experiment so that I> can use their computer time. The program in FORTRAN is simple:> > 00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 end> > It has currently reached about 2.0 x 10^18. Just as Einstein proved> that there is no aether, I am convinced that I will prove that there> is no in'nity and then write a book or two.> > Dr. Ben Zona> > > Excellent. Absolutely §awless. A real gem. lexicon, though.)Hebrew, I would have thought.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: there is no such thing as in'nity> You should take a look at Cantor's theory of sets. In it, Cantor does> treat in'nities as numbers, cardinals and ordinals. Try a Google> search on it.Cantor was a rusha. He confused generations of mathematicians bytelling them there are lots of in'nities. It is impossible to get aBS in math without admitting that his theories are correct.The diagonal argument was a sham because it presupposes the existenceof in'nity without proving its existence. His theorem is proof thatin'nity does not in fact exist, because if it did, then there wouldbe only one type, not aleph null, aleph one etc., because in'nity byde'nition is as far as one can go. He even de'led the Hebrewalphabet by putting subscripts next to aleph when the gematria ofaleph is not in'nity or even M but one.His theory of in'nities was in fact a precursor to the communistrevolution. Marx believed in thesis and antithesis, that the premise ofdialectical materialism is true.Had Cantor never been alive, there would be world peace and for surethe number M would have been determined.> > One of the brilliant things Cantor did was to de'ne an in'nite set> as a set whose elements can be put into 1-1 correspondence with a> proper subset of itself (obviously something one cannot do with a> 'nite set). For instance, you can put the set of positive integers> into 1-1 correspondence with the set of positive even integers by the> correspondence> > n -> 2n> > It seems to go against common sense to say that these two sets have> the same cardinalities. Then again, we don't have any common> experience working with in'nities.That is even more evidence for the ridiculousness of there being anin'nity. Any math that completely contradicts the senses is wrongjust as any scienti'c theory that contradicts observation is wrong.You just cannot have an in'nite set.Dr. Ben Zona === Subject: Re: there is no such thing as in'nity> You should take a look at Cantor's theory of sets. In it, Cantor does> treat in'nities as numbers, cardinals and ordinals. Try a Google> search on it.> > Cantor was a rusha. He confused generations of mathematicians by> telling them there are lots of in'nities. It is impossible to get a> BS in math without admitting that his theories are correct.> > The diagonal argument was a sham because it presupposes the existence> of in'nity without proving its existence. His theorem is proof that> in'nity does not in fact exist, because if it did, then there would> be only one type, not aleph null, aleph one etc., because in'nity by> de'nition is as far as one can go. He even de'led the Hebrew> alphabet by putting subscripts next to aleph when the gematria of> aleph is not in'nity or even M but one.> > His theory of in'nities was in fact a precursor to the communist> revolution. Marx believed in thesis and antithesis, both the premise of> dialectical materialism is true.> Had Cantor never been alive, there would be world peace and for sure> the number M would have been determined.> > > One of the brilliant things Cantor did was to de'ne an in'nite set> as a set whose elements can be put into 1-1 correspondence with a> proper subset of itself (obviously something one cannot do with a> 'nite set). For instance, you can put the set of positive integers> into 1-1 correspondence with the set of positive even integers by the> correspondence> > n -> 2n> > It seems to go against common sense to say that these two sets have> the same cardinalities. Then again, we don't have any common> experience working with in'nities.> > That is even more evidence for the ridiculousness of there being an> in'nity. Any math that completely contradicts the senses is wrong> just as any scienti'c theory that contradicts observation is wrong.> You just cannot have an in'nite set. You can put, put math goobers still seem to think that Information Theory is science. When it's really structuralist recursive gibberish invented by the Chinese about 4000 years ago. Or as we need to remind the French every day anymore: The Russians are coming, the Russians are coming, so hurry and up get some kind of Austrian contract with some non-Swiss scientists who actually know something about rockets. Or else Paris is going to start looking a lot like downtown Baghdad.> > Dr. Ben Zona === Subject: Re: there is no such thing as in'nity> You should take a look at Cantor's theory of sets. In it, Cantor does> treat in'nities as numbers, cardinals and ordinals. Try a Google> search on it.> > Cantor was a rusha. He confused generations of mathematicians by> telling them there are lots of in'nities. What's so confusing about them?> It is impossible to get a> BS in math without admitting that his theories are correct.> > The diagonal argument was a sham because it presupposes the existence> of in'nity without proving its existence.I already told you that one assumes the existence of an in'nite set,such as the set of all integers. Do you accept the set of allintegers? If one assumes the existence of that set, one does not needto then prove it. Do you accept the continuum of the real line ofnumbers?> His theorem is proof that> in'nity does not in fact exist, because if it did, then there would> be only one type, not aleph null, aleph one etc., because in'nity by> de'nition is as far as one can go.I don't know whose de'nition of in'nity you're using, but it sureisn't a de'nition that mathematicians use. In math, in'nity is aboutthe cardinalities and ordinalities of abstract sets, not aboutdistances in physical space.> He even de'led the Hebrew> alphabet by putting subscripts next to aleph when the gematria of> aleph is not in'nity or even M but one.I'm not up on Hebrew mysticism.> > His theory of in'nities was in fact a precursor to the communist> revolution. Marx believed in thesis and antithesis, both in'nite> quantities that each other out, assuming that the premise of> dialectical materialism is true.I thought Marx was on about economics and the fair dispersal ofresources in a society.> Had Cantor never been alive, there would be world peace and for sure> the number M would have been determined.There have been and still are many threats to world peace, most ofthese threats have nothing to do with Marxism.De'ne this number M.> > > One of the brilliant things Cantor did was to de'ne an in'nite set> as a set whose elements can be put into 1-1 correspondence with a> proper subset of itself (obviously something one cannot do with a> 'nite set). For instance, you can put the set of positive integers> into 1-1 correspondence with the set of positive even integers by the> correspondence> > n -> 2n> > It seems to go against common sense to say that these two sets have> the same cardinalities. Then again, we don't have any common> experience working with in'nities.> > That is even more evidence for the ridiculousness of there being an> in'nity. Any math that completely contradicts the senses is wrong> just as any scienti'c theory that contradicts observation is wrong.> You just cannot have an in'nite set.> > Dr. Ben ZonaMathematics is founded on self consistency, not on sensibility.Science uses calculus, and calculus uses in'nity.Patrick === Subject: Re: there is no such thing as in'nity>> >> His theory of in'nities was in fact a precursor to the communist>> revolution. Marx believed in thesis and antithesis, both in'nite>> quantities dialectical materialism is true.> > I thought Marx was on about economics and the fair dispersal of> resources in a society.He was also keen on Hegelian dialectics.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: there is no such thing as in'nity> Cantor was a rusha. He confused generations of mathematicians by> telling them there are lots of in'nities. It is impossible to get a> BS in math without admitting that his theories are correct.> > The diagonal argument was a sham because it presupposes the existence> of in'nity without proving its existence. His theorem is proof that> in'nity does not in fact exist, because if it did, then there would> be only one type, not aleph null, aleph one etc., because in'nity by> de'nition is as far as one can go. He even de'led the Hebrew> alphabet by putting subscripts next to aleph when the gematria of> aleph is not in'nity or even M but one.> Okay, let's assume there is no in'nity, which is your whole point, no? And now you have this magic number M.People have estimated that there are about 6x10^79 atoms in the entire universe (http://www.sunspot.noao.edu/sunspot/pr/answerbook/ universe.html#q70). I suppose this number 6x10^79 is your magic M. If you tried to write down all the numbers 1 through 6x10^79, you would run out of space to write them. (I won't even go into counting them, since even counting one number every picosecond means this would take ~2x10^61 years)Now, you might be wondering why I am requiring you to write all these umbers down (and why I think your computer program might be cheating)I can write 1 6x10^90, but that does not mean this number exists. It might just be too big of a number, much larger than your M. So, unless you have proof that all numbers up to your number M exists, proving M exists is all for naught.Okay, maybe that number is too big. Afterall, people guess the universe is only 13.7 billion years old (http://wiener.math.csi.cuny.edu/UsingR/Data/age.universe.html ), or roughly 4.3x10^42 yocto-seconds (10^-24). Maybe this should be M, since the age all of existence is less than this number.Personally, I prefer to think of 42 as in'nity. Until you convince me otherwise, I will just continue believing this way. 43? Doesn't even exist. - Tim Timothy M. BrauchGraduate StudentDepartment of MathematicsWake Forest University === Subject: Re: there is no such thing as in'nity> > Cantor was a rusha. He confused generations of mathematicians by> telling them there are lots of in'nities. It is impossible to get a> BS in math without admitting that his theories are correct.> > The diagonal argument was a sham because it presupposes the existence> of in'nity without proving its existence. His theorem is proof that> in'nity does not in fact exist, because if it did, then there would> be only one type, not aleph null, aleph one etc., because in'nity by> de'nition is as far as one can go. He even de'led the Hebrew> alphabet by putting subscripts next to aleph when the gematria of> aleph is not in'nity or even M but one.> > > Okay, let's assume there is no in'nity, which is your whole point, no? > And now you have this magic number M.> > People have estimated that there are about 6x10^79 atoms in the entire > universe > (http://www.sunspot.noao.edu/sunspot/pr/answerbook/ universe.html#q70). > > I suppose this number 6x10^79 is your magic M. If you tried to write > down all the numbers 1 through 6x10^79, you would run out of space to > write them. (I won't even go into counting them, since even counting > one number every picosecond means this would take ~2x10^61 years)> > Now, you might be wondering why I am requiring you to write all these > umbers down (and why I think your computer program might be cheating)> > I can write 1 6x10^90, but that does not mean this number exists. It > might just be too big of a number, much larger than your M. So, unless > you have proof that all numbers up to your number M exists, proving M > exists is all for naught.> > Okay, maybe that number is too big. Afterall, people guess the universe > is only 13.7 billion years old > (http://wiener.math.csi.cuny.edu/UsingR/Data/age.universe.html ), or > roughly 4.3x10^42 yocto-seconds (10^-24). Maybe this should be M, since > the age all of existence is less than this number.> > Personally, I prefer to think of 42 as in'nity. Until you convince me > otherwise, I will just continue believing this way. 43? Doesn't even > exist. Mathematicians has always that though. And since Plank's Constant has less to do with math than idiots with names like Feynmann, that's obviously why what you is say you is true, but it's only true in North Carolina and Calthech. So it's also only true on Tuesdays.> > - Tim> > > Timothy M. Brauch> Graduate Student> Department of Mathematics> Wake Forest University === Subject: Re: there is no such thing as in'nity>> It seems to go against common sense to say that these two sets have>> the same cardinalities. Then again, we don't have any common>> experience working with in'nities.That is even more evidence for the ridiculousness of there being an>in'nity. Any math that completely contradicts the senses is wrong>just as any scienti'c theory that contradicts observation is wrong.>You just cannot have an in'nite set.Agreed that contradictory mathematics is fallacious. But ultimatelyyou come down to saying that you just can't have those damn in'nitesets. Where is your proof?Answer: You can't prove it. It's false. Your Fortran program is anexercise in futility because you want it to eventually REFUTE its ownde'nition! You want it to stop counting at some point, even thoughthe existence of any number n implies the existence of the number n+1.THAT is a contradiction. That is why your program will never stopcounting.Yes, we DO live in a FINITE universe. Otherwise we could have noidentity. But if what you're trying futilely to prove is true, thenthere is no such thing as calculus :-)Xevious === Subject: Re: there is no such thing as in'nity> Yes, we DO live in a FINITE universe.Therefore, there is no such thing as in'nity! Otherwise we could have no> identity.?? But if what you're trying futilely to prove is true, then> there is no such thing as calculus :-)I believe in Calculus but not the way it is taught in school. Itshould be taught where dx=1/M, the good old fashioned way within'nitesimals:If f(x)=x^2, f'(x)=((x+h)^2-x^2)/h=(2xh+h^2)/h=2x+h approximatelyequals 2x,where h=1/M. See! I just revolutionized modern thinking andmathematics into conforming to common sense and human decency!We live in a 'nite universe and there is no room for in'nity here.We must blot out in'nity from our world as it is causing thebreakdown of morals and standards in our fragile society - in anin'nite universe, anything is possible!Dr. Ben Zona === Subject: Re: there is no such thing as in'nity>> Yes, we DO live in a FINITE universe.> Therefore, there is no such thing as in'nity!> Otherwise we could have no>> identity.> ??> But if what you're trying futilely to prove is true, then>> there is no such thing as calculus :-)> I believe in Calculus but not the way it is taught in school. It> should be taught where dx=1/M, the good old fashioned way with> in'nitesimals:Are you claiming M is in'nite?If so, then you have contradicted yourself, since you also claim in'nity doesnot exist.If M is not in'nite, then 1/M is not an in'nitesimal.So which is it?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: there is no such thing as of you people.> > > Clue #1: Blatant Simpsons reference. > How so? Which part of which episode?> Ben Zona, PhD> > Clue #2: YiddishUm... how is it Yiddish?(...Starblade Riven Darksquall...) === Subject: Re: there is no such thing as in'nity> I've thought really hard about this one and came to the conclusion> that there is no scienti'c evidence of in'nity existing. The highest> number that anyone has ever measured to according to Isaac Asimov in> his book Science and Human Thought is only about 5.0 x 10^48. No one> has ever gotten past that number. Doesn't this sound weird?> > What's to say that eventually there is a number where it is impossible> to count higher than? If someone were to 'nd this number and prove> that it is in fact the highest number, then that person would> undoubtably be rich and famous.> > I am currently running a computer program that will eventually 'nd> this magic number (I hope and pray) that I call M for short. It> counts and counts and counts and my theory is that it will eventually> stop at M. I am looking for collaborators in this experiment so that I> can use their computer time. The program in FORTRAN is simple:> > 00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 end> > It has currently reached about 2.0 x 10^18. Just as Einstein proved> that there is no aether, I am convinced that I will prove that there> is no in'nity and then write a book or two.> > Dr. Ben ZonaNumbers don't exist outside of the human mind. They are abstractionswhich give us useful information about the outside world, but thenumber itself does not exist.(...Starblade Riven Darksquall...) === Subject: Re: there is no such thing as in'nity> I've thought really hard about this one and came to the conclusion> that there is no scienti'c evidence of in'nity existing. The highest> number that anyone has ever measured to according to Isaac Asimov in> his book Science and Human Thought is only about 5.0 x 10^48. No one> has ever gotten past that number. Doesn't this sound weird?> > What's to say that eventually there is a number where it is impossible> to count higher than? If someone were to 'nd this number and prove> that it is in fact the highest number, then that person would> undoubtably be rich and famous.> > I am currently running a computer program that will eventually 'nd> this magic number (I hope and pray) that I call M for short. It> counts and counts and counts and my theory is that it will eventually> stop at M. I am looking for collaborators in this experiment so that I> can use their computer time. The program in FORTRAN is simple:> > 00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 end> > It has currently reached about 2.0 x 10^18. Just as Einstein proved> that there is no aether, I am convinced that I will prove that there> is no in'nity and then write a book or two.> > Dr. Ben Zona....Dr of what??? === Subject: Re: there is no such thing as in'nity> I didn't know retarded people posted to newsgroups. In fact, I didn't> know that retarded people could get phds. What mail-order diploma factory> did you get your degree from again?> > I am not retarded and have an above average IQ, thank you. I got my> PhD from Univerisity of San Moritz, a gullible.--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: there is no such thing as in'nityX-Enigmail-Version: 0.83.2.0X-Enigmail-Supports: pgp-inline, pgp-mime> >I didn't know retarded people posted to newsgroups. In fact, I didn't>know that retarded people could get phds. What mail-order diploma factory>did you get your degree from again?>>I am not retarded and have an above average IQ, thank you. I got my>>PhD from Univerisity of San Retarded and gullible.-------http://www.dailyfreepress.com/media/paper87/ DFPArchive/frontpage/1015982.htmlThe diplomas, standard size at 101/2 by 151/2 inches, bestow all the merits of a degree from the University of San Moritz, Puerto Rico, a non-accredited school.The company offered a Free Press reporter a University of San Moritz transcript-- complete with credits transferred from BU-- for another $200. University Degree Program asked for no proof that the reporter, who initially called using an alias, took the BU classes or that the classes even exist.-------No, he thinks everyone else is gullible.-- Suppose we've chosen the wrong god. Every time we go to church we're just making him madder and madder--Homer === SimpsonSubject: Re: there is no such thing as in'nity > >> >> I didn't know retarded people posted to newsgroups. In fact, I didn't>> know that retarded people could get phds. What mail-order diploma>> factory did you get your degree from again?> > I am not retarded and have an above average IQ, thank you. I got my> PhD from Univerisity of San Moritz, a non-acredited but well-respected> university in England. > > ? > > What part of England is San Moritz in?The University of San Moritz does not exist. It is a bogus name used by an Internet diploma mill. === Subject: Re: there is no such thing as in'nity>> > > I didn't know retarded people posted to newsgroups. In fact, I didn't> know that retarded people could get phds. What mail-order diploma> factory did you get your degree from again?>> >> I am not retarded and have an above average IQ, thank you. I got my>> PhD from Univerisity of San Moritz, a non-acredited but well-respected>> university in England.>> >> ?>> >> What part of England is San Moritz in?> > The University of San Moritz does not exist. It is a bogus name used by> an Internet diploma mill.Drat! I had heard San Moritz had excellent winter sports, and I washoping that I might be able to sample these without going abroad :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: there is no such thing as in'nityIn sci.math, N.LENN @ WKX.KM.EU:>>can use their computer time. The program in FORTRAN is simple:>>00001 n=1>>00002 1 n=n+1>>00003 print(3,4)n>>00004 if(n.eq.M) then print(3,4)M>>00005 else go to 1>>00006 end if>>00007 end What bastard version of FORTRAN is this?> Last time I used unit 3, it was a card READER on an ICL1906A.> Where's your format de'nition for label 4?> Where's the STOP ?> Why are the sequence numbers in columns 1 to 5 instead of 72 to 80?> Why is GO TO spelt as two words?> Why is ENDIF spelt as two words?> Why doesn't the compiler bork when it spots the use of an unde'ned > variable in the IF? Jokes are supposed to be internally consistent, you know.>Looks more like a variant of BASIC to me...and probably nota working one, as I've not seen a BASIC with an ELSE statementon a separate line. The (3,4) could be a cursor reference, butthe only BASIC I remember doing that is a TRS-80, which I'm notthat familiar with personally, and I think it used an ï@' anyway.If it is FORTRAN, your objections for the most part are valid,and there's the issue that statements must begin in Column 7.BTW, older FORTRANS were happy with GOTO, GO TO, G O T O, ifI'm not mistaken. I'm not sure they'd be happy with the IF statementas structured; if it's straight FORTRAN it wouldn't know what THENis, and if it's WATFOR or equivalent the print() should be on adifferent line.The example is probably better written anyway. In this caseI'll use C, since it's what I generally use:void doit(int m){ int n = 1; do { n++; } while(n < m); printf(%dn, m);}Should give one an idea of how one might structure this, erm, problem.The encapsulation of this thing in a working main() I'll leave tothe interested reader. For what all this is worth. :-)(Am I missing the joke? :-) )-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: there is no such thing as in'nity> In sci.math, N.LENN @ WKX.KM.EU>:>can use their computer time. The program in FORTRAN is simple:>>00001 n=1>00002 1 n=n+1>00003 print(3,4)n>00004 if(n.eq.M) then print(3,4)M>00005 else go to 1>00006 end if>00007 end>> What bastard version of FORTRAN is this?>> Last time I used unit 3, it was a card READER on an ICL1906A.>> Where's your format de'nition for label 4?>> Where's the STOP ?>> Why are the sequence numbers in columns 1 to 5 instead of 72 to 80?>> Why is GO TO spelt as two words?>> Why is ENDIF spelt as two words?>> Why doesn't the compiler bork when it spots the use of an unde'ned >> variable in the IF?>> Jokes are supposed to be internally consistent, you know.> Looks more like a variant of BASIC to me...and probably not> a working one, as I've not seen a BASIC with an ELSE statement> on a separate line. The (3,4) could be a cursor reference, but> the only BASIC I remember doing that is a TRS-80, which I'm not> that familiar with personally, and I think it used an ï@' anyway.I think it's intended to be Fortran, which almost works if you assume thenumbers in columns 1-5 are an artifact of the text editor and are notactually seen by the compiler.> If it is FORTRAN, your objections for the most part are valid,> and there's the issue that statements must begin in Column 7.That's true if you are using 'xed format, but it could be free format,in which case it doesn't matter where the statements begin.> BTW, older FORTRANS were happy with GOTO, GO TO, G O T O, if> I'm not mistaken. I'm not sure they'd be happy with the IF statement> as structured; if it's straight FORTRAN it wouldn't know what THEN> is, and if it's WATFOR or equivalent the print() should be on a> different line.In 'xed-format Fortran, blanks are not signi'cant, and therefore thecompiler sees no difference at all between GOTO, GO TO, and G O T O. In free-format source, the forms GOTO and GO TO areexplicitly allowed, and similarly with ENDIF or END IF. However, theFortran IF-THEN-ELSE would look like IF (n .eq. M) THEN print (3,4) M ELSE GO TO 1 END IFor perhaps the whole thing would be nested in a DO WHILE or a DO withEXIT that eliminates the need for a GOTO.It's legal to have a single-line IF statement, but such a statement wouldhave no THEN and no ELSE.> (Am I missing the joke? :-) )I think the joke is that the original program could be shortened to: print *, huge(1) endwhich is a standard-conforming way to print the largest value of typedefault integer that can be represented in a particular Fortranimplementation.However, just because it's the largest default integer doesn't meanit's the largest integer, or even that it's the largest integer that canbe represented in that implementation. Fortran implementations areallowed to offer other integer types besides the default type, and thereis a standard function call, SELECTED_INT_KIND(n), which tells whetherthere is an INTEGER kind that can represent integers in the range[-10^n,10^n]. The function returns a positive value if a suitable kindis available, and -1 if not.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: there is no such thing as in'nity> That is a problem, but there also is no de'nition of M.M is the largest number. The computer will stop when it hits M. Idon't need to de'ne it as the computer will know it when it 'nds it.That's why I am using a computer for the calculations as they are muchmore powerful than human beings.Dr. Ben Zona === Subject: Re: there is no such thing as in'nity> That is a problem, but there also is no de'nition of M.> > M is the largest number. The computer will stop when it hits M. I> don't need to de'ne it as the computer will know it when it 'nds it.> That's why I am using a computer for the calculations as they are much> more powerful than human beings.> > Dr. Ben ZonaAnd does this zoned out doctor claim that all computers will stop at the same M? === Subject: Re: there is no such thing as in'nity> And does this zoned out doctor claim that all computers will stop at the > same M?That is an excellent point. That is why I am looking for collaboratorswho will repeat the experiment on different computers. My theory isthis:Even if different computers were to output different answers for M, wecould always take the average of these numbers and calculate thevariance too. Then we would at least have a good statistical estimateof the range of M.Dr. Ben Zona === Subject: New jokes on mathNew jokes on math added on www.bymath.com!!! === Subject: Riemann Hypothesis and P vs NPA few months ago I 'nished reading The Music of the Primes byMarcus du Sautoy. Chapter 10 (titled Cracking Number and Codes) talksabout cryptography with special emphasis on RSA. Somwhere in thechapter the author mentions P versus NP since factorising numbers is adif'cult problem (personally, I don't know if it is NP-hard orIf the Riemann Hypothesis is true, then there is a fast way todiscover the primes used to build the RSA codes on which the securityof e-business currently relies.My question is: If the RH is proved to be true does it automaticallyfollow that P=NP?Comments please. === Subject: Re: Riemann Hypothesis and P vs NP> A few months ago I 'nished reading The Music of the Primes by> Marcus du Sautoy. Chapter 10 (titled Cracking Number and Codes) talks> about cryptography with special emphasis on RSA. Somwhere in the> chapter the author mentions P versus NP since factorising numbers is a> dif'cult problem (personally, I don't know if it is NP-hard or> > If the Riemann Hypothesis is true, then there is a fast way to> discover the primes used to build the RSA codes on which the security> of e-business currently relies.> > My question is: If the RH is proved to be true does it automatically> follow that P=NP?There seems (to me) to be a §aw in the author's argument.Either the Riemann Hypothesis is true or it is not true.If one were to assume that it were true, from which it automaticallyfollows that there is a fast way to discover the primes (from theauthor's argument).For any given number that needs to be factored, this fast way could beused, and then the results quickly veri'ed. If the results areincorrect (or if it is not fast), and the author's assumption istrue, then the RH is false.So, you can either crack every code fairly quickly or you can disprovethe RH. Either one sounds like a huge advance.Perhaps the author is intimating that the proof of the RH shouldreveal a fast way to discover the primes, but I don't see how thisfollows.-------------------------------------------------- ------------------- | Good and evil both increase at compoundBen Hocking, Grad Student | interest. That is why the littlehocking@cs.virginia.edu | decisions you and I make every day are of | such in'nite importance. - C. S. Lewis--------------------------------------------------------- ------------ === Subject: Re: Riemann Hypothesis and P vs NP Adjunct Assistant Professor at the University of Montana.>> A few months ago I 'nished reading The Music of the Primes by>> Marcus du Sautoy. Chapter 10 (titled Cracking Number and Codes) talks>> about cryptography with special emphasis on RSA. Somwhere in the>> chapter the author mentions P versus NP since factorising numbers is a>> dif'cult problem (personally, I don't know if it is NP-hard or>> >> If the Riemann Hypothesis is true, then there is a fast way to>> discover the primes used to build the RSA codes on which the security>> of e-business currently relies.>> >> My question is: If the RH is proved to be true does it automatically>> follow that P=NP?There seems (to me) to be a §aw in the author's argument.Either the Riemann Hypothesis is true or it is not true.>If one were to assume that it were true, from which it automatically>follows that there is a fast way to discover the primes (from the>author's argument).The primes in an RSA modulus, which is the product of two primes ofabout the same size...>For any given number that needs to be factored, this fast way could be>used,Not necessarily: the method in question may be specialized for numberswhich are the product of two primes of about the same size. (Justlike, for example, Lenstra's elliptic curve factorization isparticularly good when the number has a prime factor which is smallcompared to the other prime factors). So this method may not be goodfor any given number.If memory serves, RH is usually invoked in monte-carlo algorithms;this because RH implies a certain regularity of the primes, so thatcertain numbers, randomly chosen, will have a well-behaved probabilitydistribution for something-or-other. So in fact we may be dealing withprobabilistic arguments to begin with...> and then the results quickly veri'ed. If the results are>incorrect (or if it is not fast), and the author's assumption is>true, then the RH is false.So, you can either crack every code fairly quickly or you can disprove>the RH. Either one sounds like a huge advance.Perhaps the author is intimating that the proof of the RH should>reveal a fast way to discover the primes, but I don't see how this>follows.I think the author is saying that a proof of the RH would imply thatcertain probabilistic and/or specialized methods are fast for dealingwith the product of two primes of about the same size. It does notcomplexity of a given (often probabilistic) algorithm was givendependent on RH.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: help! h(x) = cos^4x sinx [0, pi]h(x) = cos^4x sinx [0, pi] 'nd the average volue. === Subject: Re: help! h(x) = cos^4x sinx [0, pi]216.196.97.132:> h(x) = cos^4x sinx [0, pi] 'nd the average value. === Subject: Re: help! h(x) = cos^4x sinx [0, pi]En el escribi.97:> h(x) = cos^4x sinx [0, pi] 'nd the average value.Int(h(x), x, 0, pi)/piThe integral is inmediate ... To check, the result is aprox. 0.12732...-- Best (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: cos^4x sinx [0, pi] 'nd the average volue.What's volue? === Subject: Four Color TheoremRobinson, Sanders, Seymour, & Thomas in their proof of the FCT, stateIt has been known since 1913 that every minimal counterexample to theFour Color Theorem is an internally 6-connected triangulation.Kainen, 1986. pp 59,If G is minimal [ie a minimal counterexample to the FCT], then G is5-connected.If G is minimal [ie a minimal counterexample to the FCT], then G ismaximalplanar [ie a triangulation].Is there a con§ict between the two descriptions of an MCE or does internally 6-conncted mean the same as 5-connected? === Subject: Re: JSH: Pattern argument, revisited So I can have> > > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x), or> > > > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 2) and> > where the b's are roots of b^2 - (x - 2)b + 7(x^2 + x), or> No. x - 2 should be x - 3 and 7 should be 17 in the line above. > > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x + 2)> > where the c's are roots of c^2 - xc + 7(x^2 + x).> No. 7 should be 2 in the line above. > > The more astute of you should see a pattern. Clearly the constant> factor on the right, which shows up as a coef'cient on the left must> equal 2 mod 5 for this particular setup.> > But why?> > Clearly there is an *underlying* equation being multiplied in each> case, which I'll call> > (5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2> No. You yourself see the dif'culties with this approach below. > What's happening is that I'm multiplying that underlying equation by> integers that are 2 mod 5, which allows for *one* factorization if> you're to have integer coef'cients.> > Don't believe me, then try to multiply by 13 instead of 7, like 13> equals 3 mod 5, so it will not work.>Right. No more than your underlying equation will work whenyou multiply by 1. This equals 1 mod 5, so in your own words,it will not work. > > The problem is that it's clear that y_1(x) and y_2(x) can't both be> algebraic integer functions.>Right again. Of course there's no reason why they === Subject: existence of bounded linear functionalLet X normed linear space, X' it's dual. a_1,...,a_n some scalars,x_1,...,x_n some elements of X. How do you show that| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||for some M for each n and for scalars b_1,...,b_n implies that there existsx' in X' satisfying x'(x_k)=a_k and ||x'|| = M.i don't know how to show this without explicitly constructing the boundedlinear functional on X, and of course i'm stuck on constructing it (hencethe problem in the 'rst place)..if X is 'nite dimensional, then each x in X can be written x = b_1 x_1 +... + b_n x_n on some basis {x_k}..then de'ne x'(x)=b_1 x'(x_1) + ... + b_nx'(x_n) and de'ning x'(x_k)=a_k. This is bounded by the given constraint.But what to do if X is in'nite-dimensional??? === Subject: Re: Elevator problem> >For the record, for the case n = 2 and k = 3, I get the following >fractions of discrete time that there is at least one elevator on the >indicated §oor:>1: 2/3>2: 11/30>3: 7/15> > Here are the simulation results for the same case:> > 1: 0.64999> 2: 0.35906> 3: 0.51598I've written a program to compute the exact values. For n = 2 and k = 3,here are the probabilities that an elevator is at a §oor for threedifferent cases: (a) when the closest elevators above and below areequidistant, an elevator is chosen at random (among all closest elevators);(b) ties are broken by choosing the higher elevator; and (c) by choosingthe lower elevator.2 elevator, 3 §oors: (a) random (b) from higher (c) from lower1: 137/211 = 0.649289 20/30 = 0.666667 57/91 = 0.6263742: 77/211 = 0.364929 11/30 = 0.366667 33/91 = 0.3626373: 108/211 = 0.511848 14/30 = 0.466667 52/91 = 0.571429wait: 251/422 = 0.594787 3/5 = 0.6 107/182 = 0.587912Note that protocol (c) has the minimal expected wait time in this case.> About the original case (n = 2, k = 9). Here are the corrected> probabilities over time:> > 1: 0.63577> 2: 0.09947> 3: 0.10187> 4: 0.10184> 5: 0.10392> 6: 0.1114> 7: 0.12414> 8: 0.16067> 9: 0.24452Here are the exact values for n = 2, k = 9 (protocol (a)):1: 2504513355865523/3946109628685459 = 0.6346792: 396055289014916/3946109628685459 = 0.1003663: 393399705607620/3946109628685459 = 0.0996934: 397267600228380/3946109628685459 = 0.1006735: 410780121680700/3946109628685459 = 0.1040976: 440361567884340/3946109628685459 = 0.1115947: 500649271482690/3946109628685459 = 0.1268728: 631167142952430/3946109628685459 = 0.1599479: 974908585506240/3946109628685459 = 0.247056The expected wait time (i.e., the average number of §oors an elevatorhas to travel to pick up its passenger) is 31248735453510871 / 15784438514741836 = 1.97972.For protocols (b) and (c), the wait times are 1.97946 and 1.98037,respectively, so in this case (b) is the best protocol of the three.Here is a comparison of the §oor probabilities and expected waittimes in a nine-story building which is considering various numbersof elevators to install (using protocol (a)): n = 1 n = 2 n = 3 n = 41: 0.5 0.634679 0.704977 0.7496212: 0.0625 0.100366 0.136403 0.1711703: 0.0625 0.099693 0.136679 0.1729504: 0.0625 0.100673 0.139223 0.1763895: 0.0625 0.104097 0.147724 0.1895926: 0.0625 0.111594 0.161834 0.2082127: 0.0625 0.126872 0.189268 0.2440698: 0.0625 0.159947 0.238806 0.3014199: 0.0625 0.247056 0.355433 0.430385wait: 2.90625 1.97972 1.54956 1.29191The exact values become increasingly cumbersome as n increases. E.g.,for n = 4, the values are ratios of (co-prime) 400-digit numbers.-Jim Ferry === Subject: Probability QuestionCan anyone give me some help with this question,The probability of winning exactly 10 dollars on the National Lottery is0.01765. If I buy 100 tickets, what is the probability that I do notwin a 10 dollars prize? The probability of winning the jackpot on theNational Lottery is 7.15x10^-8. How many tickets do I need to buy tohave at least a 10% chance of winning the jackpot? === Subject: JSH: End of argument? Decker example revisitedIt turns out that I can prove my case using the pattern argument quitesimply.First assume(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2with y_1(x) and y_2(x) algebraic integer functions, that is, when x isan algebraic integer they result in algebraic integers.Now consider z_2(x), where z_2(x) = y_2(x) - 1, so it's also analgebraic integer function.Then y_2(x) = z_2(x) + 1, and substituting gives(5y_1(x) + 1)(5z_2(x) + 5 + 2) = 25x^2 + 30x + 2, which is(5y_1(x) + 1)(5z_2(x) + 7) = 25x^2 + 30x + 2.Now multiply both sides by 7, multiplying through the 'rst factor onthe left by 7 to get(5(7)y_1(x) + 7)(5z_2(x) + 7) = 7(25x^2 + 30x + 2).Now let a_1(x) = 7y_1(x), and a_2(x) = z_2(x), substituting gives(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)which is the Decker example (see below for reference information).Here solving for the a's requires noticing that you can re-group onthe right side to get(5a_1(x) + 7)(5a_2(x) + 7) = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2(from [1] in Decker's post, reference at bottom).Now 'nishing up is easy as the a's are roots ofa^2 - (x - 1)a + 7(x^2 + x).But there's where poster's have responded attacking my arguments(often with personal attacks and insults), as the implication is thatonly *one* of the roots should have 7 as a factor!You may now think that y_1(x) is forced to be in a ring where 7 is aunit.But notice, I *arbitrarily* worked things out for 7 to be thepolynomial multiple!Look back now at(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2as it's actually just as easy to multiply by 2, and get(5(2)y_1(x) + 2)(5y_2(x) + 2) = 2(25x^2 + 30x + 2)and now let b_1(x) = 2y_1(x), and b_2(x) = y_2(x), and I have(5b_1(x) + 2)(5b_2(x) + 2) = 2(25x^2 + 30x + 2).And the b's are roots ofb^2 - x b + 2(x^2 + x).Now then, why so much energy from posters trying to attack myargument?Well if you take that result, with the assumption that only one of theroots has 2 as a factor, then your solution for y_1(x)--rememberb_1(x) = 2y_1(x)--proves that for an in'nity of integer x's y_1(x)cannot be an algebraic integer, as it is the root of a non-monicprimitive irreducible over Q.Now I've given various proofs of that fact, but posters haveenergetically worked to confuse the issue, and from what I've seenthey've been very successful.The key here (hopefully to my success in convincing you) is to see the*second* factorization(5b_1(x) + 2)(5b_2(x) + 2) = 2(25x^2 + 30x + 2).where the b's are roots ofb^2 - x b + 2(x^2 + x), knowing how it is derived from(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2and not accept the position of posters like Dik Winter that wouldrequire that y_1(x) be in a ring with 7 a unit in one case, and 2 aunit in the other!The correct answer is that I have found a way to solve for(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2which limits y_1(x) in a particular way, while allowing y_2(x) to bean algebraic integer function.It proves that mathematicians need a bigger category as the ring ofalgebraic integers is too small to include ALL the values of y_1(x),even when x is only integers.Notice that even if you limit x to being integers y_1(x) will havevalues outside of the ring of algebraic integers, not necessarily in aring where any other integer than -1 or 1 is a unit.If mathematicians accept the truth here then they have to change mathtextbooks.Notice how simple the argument is. It's your choice though, if you do not believe in mathematics, in theimportance of its healthiness and correctness, then you can just walkaway now.James HarrisDecker Quadratic Source Information---------------------Recently Rick Decker, a professor at Hamilton College, apparentlytrying to refute my research came up with a quadratic example, which Ilike because it's a quadratic, and easier to manipulate than thecubics I've used before.If you wish to see his original post here are some headers which alsoshow that he posts from === Hamilton College:Subject: Re: Mathematical consistency, courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).