Subject: dimension of self-similar objects can anybody explain to me the meaning of dimension of self-similar objects (like fractals), how it¹s derived, and how this is used? for instance, why is the dimension of the cantor set log2/log3? Frank === Subject: Re: dimension of self-similar objects >can anybody explain to me the meaning of dimension of self-similar >objects (like fractals), how it¹s derived, and how this is used? for >instance, why is the dimension of the cantor set log2/log3? >Frank If a self-similar Žgure is composed of copies all of a single size, and which are disjoint sets, then the similarity dimension is log(number of copies)/log(ratio of linear dimensions) I.e. a line is composed of n line segments, and is n times the length of each segment, and has dimension log(n)/log(n) = 1; A square is composed of n^2 smaller squares, and its sides are n times the length of the sides of each smaller square, and had dimension log(n^2)/log(n) = 2. The Cantor Set is composed of two smaller copies of itself, and it 3 times larger than the copies, and therefore has dimension log(2)/log(3). I presume that there is a generalisation of this for when the copies are not all of the same size, for when the copies are non-disjoint, or for self-afŽne objects, but I can¹t state it. -- Stewart Robert Hinsley http://www.meden.demon.co.uk/Fractals/reptiles.html http://www.meden.demon.co.uk/Fractals/tiling.html === Subject: Re: dimension of self-similar objects >>can anybody explain to me the meaning of dimension of self-similar >>objects (like fractals), how it¹s derived, and how this is used? for >>instance, why is the dimension of the cantor set log2/log3? >If a self-similar Žgure is composed of copies all of a single size, and >which are disjoint sets, then the similarity dimension is > log(number of copies)/log(ratio of linear dimensions) >I.e. a line is composed of n line segments, and is n times the length of >each segment, and has dimension log(n)/log(n) = 1; >A square is composed of n^2 smaller squares, and its sides are n times >the length of the sides of each smaller square, and had dimension >log(n^2)/log(n) = 2. This is the right idea. Try to cover your Žgure with disks of radius r and ask how the number of disks needed changs with r. As the previous poster noted, when you try to cover a square with these disks, you¹ll need something like 1/r^2 of them, just by thinking about area. That exponent 2 is there because the square is two-dimensional. Do the same thing with a smooth curve: just by thinking about length, you can see you¹ll need something like 1/r^1 disks of radius r just to cover a curve (of length 1). I threw in an extraneous exponent of 1 there just to remind you that you are working with a 1-dimensional thing when you speak of a curve. One more example: you need about 1/r^3 balls of radius r to cover some lumpy blob in 3-space; so again we see the dimension of the blob (3) showing up as the exponent. So here¹s a deŽnition of dimension (of a compact subset X of R^n). The dimension of X is the exponent d in this sentence: When we cover X by balls of radius r, we need about 1/r^d of them. More precisely, let N(r) be the minimal number of balls of radius r whose union contains all of X ; then dim(X) = - log( N(r) )/log( r ) . Exercise: compute the dimension of the Cantor set using this deŽnition. It¹s not an integer! dave === Subject: Re: dimension of self-similar objects >>>can anybody explain to me the meaning of dimension of self-similar >>>objects (like fractals), how it¹s derived, and how this is used? for >>>instance, why is the dimension of the cantor set log2/log3? >>If a self-similar Žgure is composed of copies all of a single size, and >>which are disjoint sets, then the similarity dimension is >> log(number of copies)/log(ratio of linear dimensions) >>I.e. a line is composed of n line segments, and is n times the length of >>each segment, and has dimension log(n)/log(n) = 1; >>A square is composed of n^2 smaller squares, and its sides are n times >>the length of the sides of each smaller square, and had dimension >>log(n^2)/log(n) = 2. >This is the right idea. Try to cover your Žgure with disks of radius r >and ask how the number of disks needed changs with r. As the previous >poster noted, when you try to cover a square with these disks, you¹ll >need something like 1/r^2 of them, just by thinking about area. >That exponent 2 is there because the square is two-dimensional. >Do the same thing with a smooth curve: just by thinking about length, >you can see you¹ll need something like 1/r^1 disks of radius r just >to cover a curve (of length 1). I threw in an extraneous exponent of 1 >there just to remind you that you are working with a 1-dimensional >thing when you speak of a curve. One more example: you need about >1/r^3 balls of radius r to cover some lumpy blob in 3-space; so >again we see the dimension of the blob (3) showing up as the exponent. >So here¹s a deŽnition of dimension (of a compact subset X of R^n). >The dimension of X is the exponent d in this sentence: > When we cover X by balls of radius r, we need about 1/r^d of them. >More precisely, let N(r) be the minimal number of balls of radius r >whose union contains all of X ; then dim(X) = - log( N(r) )/log( r ) . Almost. The dimension can¹t be quite - log( N(r) )/log( r ) because that depends on r. The dimension (or one version of the dimension) is the _limit_ of - log( N(r) )/log( r ) as r -> 0, if this limit exists (in general it¹s the lim inf of - log( N(r) )/log( r ), unless I¹m off by a minus one in which case it¹s the lim sup.) >Exercise: compute the dimension of the Cantor set using this deŽnition. >It¹s not an integer! >dave ************************ David C. Ullrich === Subject: Re: billion Eric Nielsen a .8ecrit : >>>Actually, that was the French. >>I don¹t suppose we could persuade the current US administration to abandon >>billion = 10^9 on the basis that is is a nasty French innovation? > > > The French must have really pissed the US off...40 years of Cold War, and we > still called it Russian dressing. > > Attention ya des francais qui lisent (meme si ils pigent pas tous lol) ! There is French who are reading that (even if they dosn¹t understand all) ! === Subject: Re: Einstein 1905 > I found a difŽculty in reading the famous Einstein¹s paper On the > electrodynamics.... He wants to study the effect of the Lorentz > transformations on the Maxwell¹s equations. Here are the transformations > (b=beta and V=speed of light) > t¹ = b(t-vx/V^2) > x¹ = b(x-vt) > y¹ = y > z¹ = z. > The Žrst equation is (d means partial derivative, E = (X,Y,Z) and > B=(L,M,N)) > 1/V dX/dt = dN/dy - dM/dz. > and he Žnds > 1/V dX/dt¹ = d/dy¹(bN - bv/V Y) - d/dz¹(bM + bv/V Z). > I¹ve made the computation but I can¹t obtain such a result. Can someone > show me? 1/V dX/dt¹ = 1/V dX/dt dt/dt¹ + 1/V dX/dx dx/dt¹ + 1/V dX/dy dy/dt¹ + 1/V dX/dz dz/dt¹ = (dN/dy - dM/dz) dt/dt¹ + 1/V (-dY/dy - dZ/dz) dx/dt¹ + 0 + 0 = (dN/dy¹ - dM/dz¹) b + 1/V (-dY/dy¹ - dZ/dz¹) bv = d/dy¹( b( N - v/V Y) ) - d/dz¹( b( M + v/V Z ) ) crucial steps: dX/dx + dY/dy + dZ/dz = div(E) = 0 for empty space and t = b(t¹+vx¹/V^2) x = b(x¹+vt¹) y = y¹ z = z¹. so dt/dt¹ = b dx/dt¹ = bv dy/dt¹ = 0 dz/dt¹ = 0 and d/dy = d/dy¹ d/dz = d/dz¹ hth Dirk Vdm === Subject: Re: Einstein 1905 Nicola > 1/V dX/dt¹ > = 1/V dX/dt dt/dt¹ + 1/V dX/dx dx/dt¹ > + 1/V dX/dy dy/dt¹ + 1/V dX/dz dz/dt¹ > = (dN/dy - dM/dz) dt/dt¹ + 1/V (-dY/dy - dZ/dz) dx/dt¹ + 0 + 0 > = (dN/dy¹ - dM/dz¹) b + 1/V (-dY/dy¹ - dZ/dz¹) bv > = d/dy¹( b( N - v/V Y) ) - d/dz¹( b( M + v/V Z ) ) > > crucial steps: > dX/dx + dY/dy + dZ/dz = div(E) = 0 for empty space > and > t = b(t¹+vx¹/V^2) > x = b(x¹+vt¹) > y = y¹ > z = z¹. > so > dt/dt¹ = b > dx/dt¹ = bv > dy/dt¹ = 0 > dz/dt¹ = 0 > and > d/dy = d/dy¹ > d/dz = d/dz¹ > > hth > > Dirk Vdm > > === Subject: Re: in the sequel? Adjunct Assistant Professor at the University of Montana. >> Yes. Sequel comes from the Latin sequella, meaning to follow. In >> papers, sometimes below is used instead of in the sequel, but if >> you are refering to something which is some way ahead (e.g., in >> books), below is not normally used. >Latin sequella actually that which follows. >The verb is sequor (inŽnitive, sequi). rather as an adverb (that which is ->to follow<-). But you are absolutely correct. -- ============================================================= ========= It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================= ========= Arturo Magidin magidin@math.berkeley.edu === Subject: Re: in the sequel? permission for an emailed response. X-Tom-Swiftie: I can¹t drink alcohol, Tom said spiritually > > > >> Yes. Sequel comes from the Latin sequella, meaning to follow. In > >> papers, sometimes below is used instead of in the sequel, but if > >> you are refering to something which is some way ahead (e.g., in > >> books), below is not normally used. > > > >Latin sequella actually that which follows. > > > >The verb is sequor (inŽnitive, sequi). > > rather as an adverb (that which is ->to follow<-). But you are > absolutely correct. Oh, duh! English is such a wonderfully žexible language. === Subject: Re: Hilbert space related ... > H a Hilbert space, K a closed linear subspace. Let x0 be any pt in H. > > w.t.s. Min{||x0-y|| : y in K} = Max{|| : y in K^perp, ||y||=1} > > > I¹m getting stuck but should be pretty easy :( > > know from given info that ||x0 -y|| >= ||x0|| for all y in K > Also that there exists a y in K, z in K^perp s.t. x0 = y + z > > so ||x0 - y|| = ||z|| > = ||x0|| ... I think I¹m going wrong way with this. > This result is false as stated. If K = H then the min is 0; but the sup of the empty set is -inŽnity. (The max doesn¹t exist.) Always watch the boundary cases... --Ron Bruck === Subject: Re: best calculus book by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJ5SqN20250; > at 07:48 PM, victorfrankenstein2@juno.com (Benjamin) said: >>Why is it that the best calculus books seem to have been written in >>the pre-modern math days? Is there anyone from the modern math >>generation who has authored high quality calculus books? >I¹ve heard good things about Apostol¹s Calculus. I¹ve got a copy of >his Analysis, and it¹s excellent. >-- > Shmuel (Seymour J.) Metz, SysProg and JOAT Is Apostol from the baby boom generation or is he a generation Xer? === Subject: Re: best calculus book > > at 07:48 PM, victorfrankenstein2@juno.com (Benjamin) said: > > > >>Why is it that the best calculus books seem to have been written in > >>the pre-modern math days? Is there anyone from the modern math > >>generation who has authored high quality calculus books? > > > >I¹ve heard good things about Apostol¹s Calculus. I¹ve got a copy of > >his Analysis, and it¹s excellent. > > > >-- > > Shmuel (Seymour J.) Metz, SysProg and JOAT > > > Is Apostol from the baby boom generation or is he a generation Xer? Tom Apostol Professor Emeritus B.S. in Chemical Engineering, University of Washington, Seattle, 1944 M.S. in Mathematics, University of Washington, Seattle, 1946 Ph.D. in Mathematics, University of California, Berkeley, 1948 === Subject: Re: Žnding circle radius from other measurements by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJ5StP20312; >My wife wanted to frame a project she is working on and she had a >picture of a sample mat. I Žgured I could take measurements from the >picture of the original and create a plan that would match it very >closely. Everything was easy until I tried to Žgure out the radius of >a circular cut. The way it was cut, it is the top portion of a circle >(less than half) with about 2/3 of another circle covering part of the >top. So, I can only measure two small arcs of the original circle. I >made a picture showing the relevant parts here: >http:// wwwcsif.cs.uc davis.edu/~brownjb/cir1.jpg >I am after the diameter of the larger circle. The smaller circle is >irrelevant for the problem I¹m trying to solve and its diameter was >easy to measure anyway. I took some measurements and made the >following diagram of the larger circle: >http:// wwwcsif.cs.uc davis.edu/~brownjb/cir2.jpg >b is the measurement from one end of the arc to the other end. I >connected the ends of the arcs and continued the lines to where they >met. The distance from the edge of the circle to the intersection >point is a. I also measured the angle between the lines, c. Now, I had >a, b, and c. I Žgured this was more than enough information to Žnd >r, the radius of the circle, in terms of a, b, and c. But, I could >never Žgure it out. I spent more time than really should have been >necessary and ran it by someone else who also should have been able to >solve it and we never came up with a solution. >I don¹t need an answer to this for the mat any more, but the problem >has really been bugging me. It seems like there should be a >straightforward, quick solution, but I¹m starting to think there >isn¹t. To Žnd the center of a circle (or circular arc) draw any two chords--the longer the better. Then erect a perpendicular bisector on each chord. The two bisectors meet a the center of the circle. phil O / ----o o === Subject: Re: out-of-kilter algorithm by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJ5Sso20295; http://wwwpub.utdallas.edu/~chandra/documents/networks/net4. pdf >Try http://citeseer.nj.nec.com >>Hello! >>I was just wondering if there is someone who could help me in Žnding some >>usefull site with explanation of out-of-kilter algorithm and examples of >>problems solved with this algorithm. >>Mimmy === Subject: Re: Complex logs by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJ5Ss120306; >> Could someone give the integral formula for the natural log of a complex number ? In another group , it was mentioned to be the integral from 1 to z of 1 / z , z = complex number . >> Is that the correct and complete deŽnition ? >> What is the best way to relate that to : >> ln z = ln (modulus) + i * ( argument +- 2npi ) , n=integer >> This seems to be the best math group for such a query. >> Gary >Well here¹s a start: >z = exp [ln z] = exp [ln (modulus)] + i * ln (argument)] Gary === Subject: Weaker Version of AC by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJBVDk13929; Is there a weaker version of AC, or at least a set-theoretic statement, that is equivalent to the fact, that every Želd has an alegraically closed extention? === Subject: Re: Reposting (was: Basic ring theory (query reposted)) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJBVA313867; But, isn¹t it true that a nerwly posted message goes to the top of the thread? >This does not make much sense. Where your post appears within a thread >depends on the particular newsreader someone uses. >Marc === Subject: Re: Reposting > But, isn¹t it true that a nerwly posted message goes to the top of the thread? Perhaps some web-based forums arrange them in the order top=last. The reason for this could be, that readers have to see _every_ posted message. But if you read news via a newsreader, it is up tou you, how you want to have the message displayed. Moreover, you can conŽgure the newsreader, such that within a thread the Žrst unread message is selected anyway, so it does not make much of a difference. Also, your old message remains visible, so there is simply no need to repost. Marc === Subject: Re: Final Rout of Synchronization Clocks in Relativity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJDo3q23742; >> ABSTRACT. The synchronization of clocks in Relativity has speculative >> chatter, and this speculative chatter about Synchronizations of >> clocks in Relativity has not ACTUAL TECHNICAL EMBODYING In CONCRETE >> TECHNICAL DEVICES. >The deductive Analysis of Surprising paradox of mythical so-called >pseudo of synchronization of clocks in the Relativity is given below: >> >> > Sometimes claimants misquote or exaggerate to further their >> > own agendas. It is best to keep an open opinion until you have heard >> > from both sides of any story. -|Tom|- >> > >> > >> > Tom Van Flandern - Washington, DC - see our web site on replacement >> > astronomy research at http://metaresearch.org Plato¹s Time: ....He planned to make a movable image of Eternity, .... .....set in order the Heaveen.... ...HE MADE AN ETERNAL IMAGE MOVING ACCORDING TO NUMBER, ...Time was created along with the Heaven,in order to disolve concurrently with it,IF EVER A DISSOLUTION OF THEM WILL TAKE PLACE [Timaeus 38B] Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: Final Rout of Synchronization Clocks in Relativity > >> ABSTRACT. The synchronization of clocks in Relativity has speculative > >> chatter, and this speculative chatter about Synchronizations of > >> clocks in Relativity has not ACTUAL TECHNICAL EMBODYING In CONCRETE > >> TECHNICAL DEVICES. > > > >The deductive Analysis of Surprising paradox of mythical so-called > >pseudo of synchronization of clocks in the Relativity is given below: > > > > > >> > >> > Sometimes claimants misquote or exaggerate to further their > >> > own agendas. It is best to keep an open opinion until you have heard > >> > from both sides of any story. -|Tom|- > >> > > >> > > >> > Tom Van Flandern - Washington, DC - see our web site on replacement > >> > astronomy research at http://metaresearch.org > > Plato¹s Time: > ....He planned to make a movable image of Eternity, .... > .....set in order the Heaveen.... > ...HE MADE AN ETERNAL IMAGE MOVING ACCORDING TO NUMBER, > ...Time was created along with the Heaven,in order to disolve > concurrently with it,IF EVER A DISSOLUTION OF THEM WILL > TAKE PLACE [Timaeus 38B] > Panagiotis Stefanides > http://www.stefanides.gr Please, Panagiotis Stefanides look at: Ron Knott, Fibonacci Numbers and Golden sections in Nature; http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/ Žbnat.html#pinecones http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/Žbnat2 .html http://www.spirasolaris.ca OVERVIEW Bode¹s Flaw Bode¹s Law - more correctly the Titius-Bode relationship - was an ad hoc scheme for approximating mean planetary distances that was originated by Johann Titius in 1866 and popularized by Johann Bode in 1871. The law later failed in the cases of the outermost planets Neptune and Pluto, but it was žawed from the outset with respect to distances of both MERCURY and EARTH, as Titius was perhaps aware. II The Alternative Describes an alternative approach to the structure of the Solar System that employs logarithmic data, orbital velocity, synodic motion, and mean planetary periods in contrast to ad hoc methodology and the use of mean heliocentric distances alone. III The Exponential Order The constant of linearity for the resulting planetary framework is the ubiquitous constant Phi known since antiquity. Major departures from the theoretical norm are the ASTEROID BELT, NEPTUNE, and EARTH in a resonant synodic position between VENUS and MARS. THE SPIRAL FORM AND UBIQUITOUS GOLDEN SECTION IV Spira Solaris Archytas-Mirabilis Most suitably represented in terms of exponential growth and a complex equiangular spiral, the Phi-series based planetary model appears to be new in one sense and yet quite ancient in another. IVd2 Spira Solaris and The Middle Ages Ostensively the translation of Aristotle¹s De Caelo from medieval latin to French, Nicole Oresme¹s Le Livre du ciel et du monde [ ca.1375 CE] was more than a translation and a commentary. The numerous references in this work to the insights of the Arab scholar Ibn Rushd [Averroes,1128-1198 CE] lead back to Plato¹s Republic, Archimedes, Pythagoras, and the Golden Section in early Alchemical contexts. === Subject: Re: Complex logs by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJDdN622697; > Could someone give the integral formula for the natural log of a complex number ? In another group , it was mentioned to be the integral from 1 to z of 1 / z , z = complex number . > Is that the correct and complete deŽnition ? > What is the best way to relate that to : > ln z = ln (modulus) + i * ( argument +- 2npi ) , n=integer > This seems to be the best math group for such a query. > Gary ln z = ln [{modz}e^i(theta)]=ln[mod z] +i(theta) for z=e^[theta/90] ,then ln z = theta/90 =[2theta rads/pi rads] ref:http://www.stefanides.gr/why_logarithm.htm http://www.stefanides.gr/nautilus.htm http://www.stefanides.gr/logarithm.htm === Subject: displacement by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJITX012945; hello my name is Kris Straus. I have a chance to get extra credit in math and I have to Žnd out why s is the symol for displacement. I was wondering if you knew === Subject: Re: ait - meaning by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBJJiIx18783; >I¹m trying to quantify a ghost concept of algorithmic information >theory, that is the meaning. Of course meaning of a string is >relative to the receiver. And in some way a byte can contain more >meaning for a receiver that a žow of data. a bit can means british >encyclopedia... But the receiver have in this case to contain a copy >of B.E. for decode it. >So we can deŽne the meaning of a string the minimal lenght of the >message plus the minimal length of UTM that describe the receiver. >(that is in some way objective). >In short I think that receiver is not a passive entity, but an UTM >which decription can be linked algorithmically to message for lead us >to something that is similar to meaning. I think also the >transmitter is an UTM, which produce the message. I think that is a >step (small) to shannonian statistic receiver and transmitter, and is >near to Algorithmic Information Theory. >help in this sense. >blind Here¹s something to think about. Suppose that if the last bit of a message is a 1 the receiver is to ignore the preceding message, and if it is 0 then ignore the last bit. How much meaning does the last bit convey? How much information? (hint: maybe negative information?) phil === Subject: Re: ait - meaning > >I¹m trying to quantify a ghost concept of algorithmic information > >theory, that is the meaning. Of course meaning of a string is > >relative to the receiver. And in some way a byte can contain more > >meaning for a receiver that a žow of data. a bit can means british > >encyclopedia... But the receiver have in this case to contain a copy > >of B.E. for decode it. > >So we can deŽne the meaning of a string the minimal lenght of the > >message plus the minimal length of UTM that describe the receiver. > >(that is in some way objective). > > > >In short I think that receiver is not a passive entity, but an UTM > >which decription can be linked algorithmically to message for lead us > >to something that is similar to meaning. I think also the > >transmitter is an UTM, which produce the message. I think that is a > >step (small) to shannonian statistic receiver and transmitter, and is > >near to Algorithmic Information Theory. > >help in this sense. > >blind > > Here¹s something to think about. Suppose that if the last bit of > a message is a 1 the receiver is to ignore the preceding message, > and if it is 0 then ignore the last bit. How much meaning does > the last bit convey? How much information? (hint: maybe negative > information?) > > phil a very interesting point of view. The answer may be: The information , or better the meaning is distributed in all message, or better in structure of message. best wishes blind === Subject: Probability without countable additivity I recall attending a lecture long ago where the instructor (James Pittman at U.C. Berkeley, I think) presented an example of an oddity that can occur if one requires only Žnite (as opposed to countable) additivity. I have forgotten the what the example was. Could anyone here provide one? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Shape of closest packed things. that was cool, how pi came-up ... in approximation, of course. as for the Wolframites, I recommend the review of his magnum opus in *Skeptic* magazine (it was funny that they called him the Jedi Knight of math, or wome thing, but didn¹t note that he *was* ka-nighted, a sir David, by Betty Dos ... probably because she dug the program, not that there weren¹t others, or that he deserved it). now, Bucky¹s whole spiel on pi is just sophistry, as I have argued on some of the maillists/newsgroups -- without much of a contradiction, either, by others. > The simplest circle that I know is based on the fact that d/dx Sin is > Cos and d/dx Cos is -Sin. > > I used this quite often in my various graphics. > > Let X be a two-byte number (set it to 0 for the top) and let Y be > another two-byte number (set it to FFFF). > > For a 256-dot scale, use the top half (byte) of X and Y to plot a dot. > > (loop) > Subtract the top byte of X (extended to double) from the whole of Y, > whilst adding the top old byte of Y (extended to double) to the whole > of X. > > Plot another dot and keep going back to (loop). > > Surprisingly, this gives not 4 times 256 plots but PI times 256 - and > you never even mentioned PI in the algorithm. --Give the Gift of Dick Cheeny -- out of ofŽce, at last! http://www.benfranklinbooks.com/ http://www.wlym.com/pages/music.html http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: For anyone who has intro to topology and modern analysis by simmons on P. 20, problem 6. The wording of the question is confusing to me: Let X,Y be nonempty sets, and f:X->Y, if A is a subset of X, B a subset of Y then prove: (a) ff^-1(B) subset of B, and ff^-1(B) = B <=> f is onto; now my question is, let p : ff^-1(B) subset of B q : ff^-1(B) = B r : f is onto am i trying to show p&q <=> r, or something else? because since q => p, it seems strange to phrase it this way. why not just say prove q <=> r ? this strange phrasing makes me wonder if i am reading the question wrong. === Subject: Re: For anyone who has intro to topology and modern analysis by simmons The book is asking you to Žrst show that f f^{-1}(B) subset B Once you¹re done with that, show that f f^{-1}(B) = B if and only if f is onto. > on P. 20, problem 6. > > The wording of the question is confusing to me: > > Let X,Y be nonempty sets, and f:X->Y, if A is a subset of X, B a > subset of Y then prove: > > (a) ff^-1(B) subset of B, and ff^-1(B) = B <=> f is onto; > > now my question is, > let p : ff^-1(B) subset of B > q : ff^-1(B) = B > r : f is onto > > am i trying to show p&q <=> r, or something else? > > because since q => p, it seems strange to phrase it this way. > why not just say prove q <=> r ? > > this strange phrasing makes me wonder if i am reading the question > wrong. === Subject: Re: For anyone who has intro to topology and modern analysis by simmons > on P. 20, problem 6. > The wording of the question is confusing to me: > Let X,Y be nonempty sets, and f:X->Y, if A is a subset of X, B a > subset of Y then prove: > (a) ff^-1(B) subset of B, and ff^-1(B) = B <=> f is onto; > now my question is, > let p : ff^-1(B) subset of B > q : ff^-1(B) = B > r : f is onto > am i trying to show p&q <=> r, or something else? > because since q => p, it seems strange to phrase it this way. > why not just say prove q <=> r ? > this strange phrasing makes me wonder if i am reading the question > wrong. ********************************************************** The question is asking you to prove two things: 1. ff^-1(B) subset of B and 2. ff^-1(B) = B <=> f is onto Eric J. Wingler Department of Mathematics and Statistics Youngstown State University === Subject: Re: GCD > 1 sequence > Consider this sequence... > > 2, 4, 6, 3, 8, 9, 10, 5, 12, 14, 7, 15, 16, 18, 20,... > > > a(1) = 2; > a(m) = lowest unpicked positive integer which is *not* coprime with at > least one previous term of the sequence. > > By unpicked, I mean the integer is not among {a(1),...,a(m-1)}. > And by not coprime with at least one previous term, I mean that at > least one prime dividing a(m) also divides at least one element of > {a(1),...,a(m-1)}. > (But these deŽnitions are obvious, I trust.) > > (It might be advantageous to deŽne a(0) = 1 for whatever reason. {I > am > sure there is a good reason for doing such, but I am unsure now why}.) > > It was pointed out to me that this sequence is similar to, but not the > same as, the ŒEKG sequence¹: > http://www.research.att.com/projects/OEIS?Anum=A064413 > > http://www.sciencenews.org/20020406/mathtrek.asp > > http://mathworld.wolfram.com/EKGSequence.html > > http://arXiv.org/abs/math.NT/0204011/ > > I wonder if any of the mentioned EKG-sequence conjectures/theorems > have > analogs with my sequence. > (If I did not err, the fact that my sequence is a permutation of all > integers >= 2 was much easier to prove than proving the same for the > EKG > sequence, however.) > > Proof, my sequence is permutation of integers > 2: > I am errrorr-prone lately, > but I would guess every even integer appears, because of the > pigeon-hole > principle (at least because of the PHP as loosely deŽned). > (We can always pick an even integer, because of 2 {and 4 and 6 > and...}, > and we WILL pick the lowest unpicked even because otherwise we would > only be picking the odds {which would skip the unpicked even > eventually, > which contradicts the deŽnition of the sequence}.) > > And so, every 2*p, p =prime, is picked, so EVERY prime is eventually > picked (p immediately following 2p). > So, the lowest still unpicked multiple of p WILL eventually be picked, > because otherwise we would skip it, which contradicts the deŽnition > of > the sequence. > > Right??? (never can be *too* sure...) > > > Any more that can be said about this sequence??? > > > Leroy > Quet First, my sequence is now at http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A089088 Well, if we let a(1) = n, integer >= 2 {and not necessarily 2}, we can get the sequence {a_n(m)}. If we have n = odd prime, a_n(m) = a_2(m-2) for all m where 3<=m<=M, where M is deŽned by a_2(M-1) = 2n. So, the limit of {a_n(m)}, as n = primes -> oo, is {n,2n,a_2(m-2)}. But I have not discovered anything about the sequence if n is composite. Leroy Quet === Subject: Minimizing/Maximizing A DeŽnite Integral Let f be any real -> real integrable function which is strictly monotonically increasing, f(0) = 0 and f(1) = 1, and where integral{0 to 1} f(x) dx = 1/2. Let g be the inverse of f, f(g(x)) = x for each x from 0 to 1. What f(x)(Œs) has the minimum, and which the maximum, for integral{0 to 1} f(x)*g(x) dx ?? Actually, the max is easy. f(x) = x = g(x) works, and gets the integral¹s value = to 1/3. ;) Leroy Quet === Subject: Re: Minimizing/Maximizing A DeŽnite Integral >Let f be any real -> real integrable function which is strictly >monotonically increasing, f(0) = 0 and f(1) = 1, and >where integral{0 to 1} f(x) dx = 1/2. >Let g be the inverse of f, f(g(x)) = x for each x from 0 to 1. >What f(x)(Œs) has the minimum, and which the maximum, for >integral{0 to 1} f(x)*g(x) dx ?? >Actually, the max is easy. >f(x) = x = g(x) works, >and gets the integral¹s value = to 1/3. How do we know this is the maximum? Use the calculus of variations. Let Df be a variation of f. To Žnd the corresponding variation of g, consider f(g(x)) = x. The variation of a composition of two functions is given by D(f(g(x))) = (Df)(g(x)) + f¹(g(x))Dg(x). Since D(f(g(x))) is 0, we get that Dg(x) = -(Df)(g(x))/f¹(g(x)). Now to Žnd a critical value of |1 | f(x) g(x) dx [1] | 0 we must have that the variation is 0: 0 |1 |1 = | Df(x) g(x) dx + | f(x) Dg(x) dx | 0 | 0 |1 |1 = | Df(x) g(x) dx - | f(x) (Df)(g(x))/f¹(g(x)) dx | 0 | 0 |1 |1 = | Df(x) g(x) dx - | f(f(x)) Df(x) dx [2] | 0 | 0 The last step follows from the substitution x -> f(x). For [2] to be 0 for all variations of f, we must have g(x) = f(f(x)), which means f(f(f(x))) = x. f is strictly monotonically increasing, so if f(x) > x for any x then f(f(f(x))) > x; likewise for f(x) < x. Thus, f(x) = x for all x in [0,1]. As for the minimum, the inŽmum of the integral is 0, but that cannot be realized within the restrictions given. One family of functions which tends to this limit is f_n(x) = x^n, for which [1] < 1/(n+1). Rob Johnson take out the trash before replying === Subject: A Family Of Permutations of the Positive Integers Let a(1) = 1; a(2) = n, n = integer >= 2; a(m+2) = the |a(m+1) -a(m)|th highest yet-unpicked positive integer. By yet-unpicked, I mean an integer that is not among {a(1),a(2),...,a(m+1)}. So, for each n >=2, we get a permutation of the + integers. Now, the n =2 case is uninteresting, just giving the + integers in their own order. But it seems that for all sufŽciently large n¹s, the terms of the sequence fall into a speciŽc pattern. And, for all sufŽciently high n¹s, every 3rd term forms a sequence of constants. So, if a(n,m) = the m_th term of the sequence with a(n,2) = n (and a(n,1) = 1), then, as n -> oo, we can get the sequence: A(m) = a(n,3m-2), which begins (perhaps...): A(m) -> 1, 2, 5, 6, 9(?),... Generally, the a-sequence, for all sufŽciently high n¹s, is(??).. 1, n, n+1, 2, n+3, n+6, 5, n+8, n+11, 6, ... Anything interesting anyone can add to the knowledge of these permutations?? Leroy Quet === Subject: Re: Probability question > > > My understanding of probability mathematics is virtually zero, > > Then you should probably not be making any claims (such as not > random) at all. Very helpful. I take it that you think 78 occurrences out of 80 is random then? Or 96 out of 96 occurrences? Or that you have to be a mathematical genius like yourself to conclude otherwise (who, by the way, was unwilling - or unable - to answer the question). === Subject: Energy of Gravity Vacuum bcc bcc Part III PZ: Why make it look superŽcially as if the physical effects of acceleration do not mark off inertial frames if in fact they do? You cannot throw away differential geometry. JS: Huh? Standard GR has a difference between LIFs and LNIF¹s, it¹s simply that the local Želd equations do not depend on the difference. It¹s tensors all the way Jose. PZ: The point is that general covariance alone is not actual general relativity -- although it can look superŽcially like general relativity in a formally general covariant theory. ... PZ: Einstein wished to extend the special principle of relativity into a general theory in which the inertial frames were not marked off as physically special. JS: Obviously. PZ: This is not at all obvious. And it turns out it didn¹t actually work, since there is in fact no Einsteinian strict equivalence. JS: I do not understand why you keep saying this? I am not sure where the miscommunication is. The g-force is locally equivalent to an inertial force. The LOCAL and non-metricity of course. PZ: So, Jack, is this really what you think Einstein meant by equivalence? Isn¹t this also trivially true in Newtonian physics? JS: Sure, they have that in common. Newtonian gravity is a limiting case of Einstein gravity, which has a richer mathematical structure and additional physical predictions like event horizons and gravimagnetism for example. Scale by Steven Weinstein Naive Quantum Gravity that demolishes basis for Yilmaz & Puthoff¹s really a žat background spacetime. In contrast, the curvature tensor local relative tidal acceleration separation shrinks to zero (neglecting Lp effect) need not vanish PZ: That¹s what? Einstein equivalence? Or do you mean that¹s all one can really say? JS: The latter. All you can say is: 1. No g-force on timelike geodesics (for non rotating extended systems) in Einstein gravity as well as Newton¹s (The painter on the falling ladder Einstein story and žoating astronauts.) 2. Test for local curvature via timelike geodesic deviation in relative 3. What to do in LNIF say on surface of Earth is more complicated since one must correct for electrical reaction forces to get a pure gravity curvature measurement. 4. Special relativity works to good approximation in LIFs for all non-gravity forces i.e. mainly electrical-magnetic & light. PZ: This is all just as true in Newtonian theory. Nothing speciŽcally Einsteinian about it. JS: Correct, until you look for post-Newtonian effects including gravimagnetism, bending of light with factor of 2 not found in Newton. Also Grad^2V(stuff) = 4piGrho(1 + 3w) w = 0 in Newton w = Pressure/Energy Density of stuff Also gravity waves not found in Newton¹s math. PZ: Einstein¹s idea of extending the special principle of relativity, and what is wrong with this, is explained very clearly in Roberto Torretti¹s highly regarded Relativity and Geometry (Dover 1996): While Einstein¹s description of his strong Principle of Equivalence... as Œa natural extrapolation of one of the most universal empirical statements of physics¹ is not unjustiŽed, his original presentation of it as a generalization of the Relativity Principle is completely misleading. The physical equivalence of reference frames at rest in a homogeneous gravitational Želd and uniformly accelerated frames does not obliterate the physical inequivalence between the latter and inertial frames... It simply entails that a reference frame at rest in a gravitational Želd cannot be inertial. JS: What does obliterate mean? PZ: The same as Landau & Lifz¹s annihilate. That is at the core of Einstein¹s idea: that the *entire physical reality* of the gravitational Želd goes out of existence in an LIF. JS: Only in the approximate heuristic correspondence sense of Wheeler, MTW obviously. PZ: Apparently you are having a very hard time understanding what all these people are worried about. JS: Yes. It¹s a non-problem. Of course Einstein knew there is a physical difference between LIF¹s and LNIF¹s mutually coincident at same point event P. PZ. Then how do you think Einstein could imagine that he was extending the Relativity Principle to accelerated frames? JS: I see no problem here? Where is the conžict? General coordinate transformations that are local at P are not identical to global Lorentz and translational transformations. Tensors of the general local coordinate transformations include the accelerated frames missing in the global Lorentz symmetry group of special relativity. Tensors of the global Poincare group are not the same as local LNIF GR tensors. One can use the local tetrad map to go from LNIF GR tensors to LIF local Lorentz sub-group tensors. Note that they are not global translation group tensors anymore. The global translational group is locally gauged to get the guv LNIF metric Želd of the base space. All laws of physics (in GR forgetting quantum theory) are LOCAL tensor laws with respect to the given base space group Diff(4) and the given tangent Žber group O(1,3). The tetrad maps between base space and tangent Žber. EEP is formally in that LOCAL map. All this neglects Lp^2 = hG/c^3 effects. LIF observers are weightless, LNIF observers feel weight from non-gravity electrical reaction forces putting them off a geodesic world line through P. The point is that there is a local tetrad map eu^a(P) and inverse from LIF to LNIF in which the local geometrodynamic Želd equation has same form PZ: But that in itself does not extend the Relativity Principle as Einstein thought. That¹s Torretti¹s point. JS: I do not understand what you just said. I do not see why it is important for anything interesting? Give examples. It¹s too abstract and vague with no context for my mind. i,e, tuv(Marble Geometry) + Tuv(Wood Matter) = 0 in LNIF at P t..(Geometry) = (String Tension)G..(Einstein) G..(Einstein) = R..(Ricci) - (1/2)R(Ricci)g.. gab = Minkowski LIF metric at P guv = LNIF metric at same P in non-exotic vacuum where /zpf = 0 corresponding to a large critical VACUUM COHERENCE BTW http://qedcorp.com/APS/EmergentGravity.pdf PZ: Fine. Who disagrees with this? JS: The Question is: Who is even aware of this new way to look at what was thought was understood. Who agrees with /zpf = Lp^-2(Lp^2|Vacuum Coherence|^2 - 1) ? I would say none of the Pundits are thinking this way at all. Vacuum Coherence? Who ordered that? they would say at least today. Also String Tension in an Einstein gravity formula? They would scratch their heads. Huh? Yes, a local frame at rest in a gravity Želd is LNIF. PZ: And then you experience the *true* strength of the real gravitational force. JS: Nonsense. What is this true? Means nothing in this context. G-force is a frame dependent quantity. DeŽnitely it is physically real and very measurable. But it is purely contingent. The way you use true here is not physically important IMHO. Whenever there is a g-force you are in a LNIF. PZ: Again, the term g-force is ambiguous, since it ignores the distinction between inertial and gravitational effects. JS: This is your basic illusion here. There is no such distinction! electrical reaction forces! G-force + electrical rection force = 0 In free žoat electrical reaction force = 0 EACH in FREE FALL with zero G-Force on each. It¹s GEODESIC DEVIATION. So it¹s apples and oranges at different levels with different operational procedures to deŽne the concepts. You can split the connection Želd into a tensor part and a non-tensor part, but that is I think not a unique split. It changes as the LNIF changes. And in LIF tensor part + nontensor part = 0. PZ: You can get g-forces without any gravitational Želds. JS: Obviously. So what? First ask HOW do you measure the tidal curvature YOU HAVE NOT ASKED THE RIGHT QUESTION! You measure components of the Ruvwl curvature tensor using PAIRS of That is you Žrst ELIMINATE G-Force on your PAIR and THEN measure CURVATURE. I mean that¹s the direct procedure, the simplest. If there are G-Forces on them it¹s a more complicated procedure. The geodesic deviation equation has no ELECTRICAL-MAGNETIC FORCES in it. That adds noise to the signal. As a practical matter however we need an analysis of that. I have not seen one. Have you? But it is clear to me Paul that you did not think this far operationally and therefore I think your idea is not sound, it not self-consistent, is not well-posed. There is no there there as far as I can understand your obsession on all this? feeling ZERO g-force, to measure their intrinsic curvature relative tidal acceleration if one is there. PZ: You don¹t need two test bodies. You can use a water droplet. JS: That counts as two. PZ: You can use all kinds of simple devices to measure Riemann curvature -- some of which do NOT scale down in sensitivity to curvature with the size of the neighborhood. JS: Give examples of last remark? I do not understand it. PZ: So even the MTW EEP is in trouble. JS: Justify with examples. Otherwise I don¹t believe what you say. PZ: Far from generalizing the Principle of Relativity, [the equivalence principle] drastically alters its meaning and restricts its scope. (Relativity and Geometry, Section 5.2 pp 135-136 ) JS: Much Ado About Nothing IMHO. PZ: : Because you don¹t understand Einstein equivalence -- which is OK, since it¹s wrong! JS: You are correct that I do not understand what you think it is you are understanding. ;-) PZ: But don¹t try and tell me that this strict equivalence was not the core of Einstein¹s theory. It was. JS: I do not know. Even if Einstein early on overstated something informally so what? That happens all the time. PZ: And here is Ohanian and RufŽni: Unfortunately, Einstein¹s statement [Einstein (1916)] has often been generalized to sweeping assertions about all laws of physics about all laws of physics being the same in a laboratory freely falling in a gravitational Želd and in another laboratory far away from any Želd... JS: O & R do not say Einstein did that. BTW Hal Puthoff does do that in his PV Tables on K. OR: Such generalizations are unwarranted since... even quite simple devices will signal the presence of a true gravitational Želd by their sensitivity to tidal forces and will therefore permit us to discriminate between a gravitational Želd and the pseudo-force of acceleration. JS: Exactly what I have said all along. PZ: You have not said that there are tidal measurement devices that are scale- insensitive. JS: It¹s obvious. PZ: The point is that we can always tell the difference between a gravitational Želd and an inertial Želd essentially everywhere in an accelerated frame, and the magnitude of the measured effects does not necessarily scale down with spacetime volume. JS: This is purely a semantic issue on how to use informal language precisely and clearly. There is no new physics here. Einstein understood all this. PZ: Of course he knew about Riemann curvature and tidal forces. But he did not consider them essential to the deŽnition of the gravitational Želd. That¹s the point. JS: Again you never are clear enough here what you mean by Želd? G-Force or tidal acceleration? PZ: Why do you think all these authors -- including your old buddy Dick Feynman -- have made such a song and dance about this if there is no issue here? JS: What did Feynman say that you think applies here? This statement [of the Einstein equivalence principle] is true only in a limited sense. Gravitation and acceleration are equivalent only as far as the translational motion freedom of the motion of the masses are taken into consideration, then the equivalence fails. JS: Of course, I have repeatedly said the same thing. PZ: OK. So then you agree with me that Einsteinian strict equivalence, as I have deŽned it, JS: Yes you set up a Straw Man. PZ: is untenable, and with Ohanian and RufŽni that the EEP as formulated by Wheeler is not actually empirically valid? JS: I need to read the exact Wheeler statement. PZ: Gravitation and Spacetime, Second Edition (Norton (1995), Section 1.9) JS: I do not have a copy here in lobby of Bay Club. PZ: So there is at this point no physical reason to take Einstein¹s belief in the complete physical equivalence of gravitational and inertial Želds seriously, except in the sense of a *limited heuristic analogy*. JS: Depends what one means by vague complete. PZ: Einstein meant *complete*. That means: NO fundamental physical distinction between gravitational and inertial Želds, PERIOD. One and the same. JS: If you mean by Želd G-FORCE, then YES no fundamental difference LOCALLY! If you mean by Želd tidal curvature , then NO there is a fundamental difference LOCALLY! This is not a big deal. Everyone understands what OR say and agrees with it. It¹s all in MTW. PZ: That you can *always* operationally discriminate between a physical gravitational Želd and an inertial force, no matter how small the neighborhood? JS: Here you garble tidal curvature with G-Force. PZ: *z- Where? The how can this EEP support the argument on p 476 that the physical stress- energy of the gravitational Želd *must* vanish at at least one point in an LIF, because the connections (gammas) also vanish at the same point? JS: This is a logically independent problem from what we were just discussing. I showed a tensor for that which is zero in non-exotic vacuum , but is NOT zero in the exotic vacuum. Classical GR does not have quantum idea of exotic vacuum. That is NEW PHYSICS. PZ: Which is what I have been arguing for at least a year. JS: No one ever disagreed with that. You have been very obscure about this getting it mixed up with the Yilmaz issue. PZ: It is directly related to the Yilmaz issue. You have now agreed (and you say you always did agree) that the gravitational Želd is fundamentally different from a Žctitious inertial Želd. JS: NO! Because I never know what you mean by Želd? Sometimes you mean G-Force and sometimes you seem to mean CURVATURE. PZ: Now, I ask you again: (1) Does the vacuum stress-energy of the gravitational Želd self-gravitate? JS: In my deŽnition YES. My deŽnition is tuv(vacuum gravity = (String Tension)Guv(Einstein) = -(String Tension)/zpfguv The only vacuum recognized in CLASSICAL GR is non-exotic where /zpf = 0. Dark energy (i.e. /zpf > 0) and dark matter (i,e. /zpf < 0) exotic vacua not known to Einstein or to anyone before 2002! This is REALLY NEW PHYSICS. PZ: (2) Does the vacuum stress-energy of the inertial Želd self-gravitate? JS: Meaningless question unless you show me the formula for it. PZ: (3) Is the vacuum stress-energy of the inertial Želd even deŽned? If not, why not? JS: I don¹t think it is. Maybe it is in Yilmaz¹s theory. You tell me. PZ: By self-gravitate I mean act as its own source. JS: In GR ALL stress-energy density is its own source! Every important idea in physics is a limited heuristic analogy IMHO. There are no absolute truths in science. PZ: OK, in the context of discovery. But need we concern ourselves with dreams of snakes biting tails in order to understand the ring structure of benzene? JS: Ring structure is itself a concept of limited utility and meaningfulness. PZ: If so, then there is no reason to expect inertial forces to contribute to or alter the physical stress-energy of the gravitational Želd, and thus no deep reason to demand that this physical stress-energy density be annihilated *anywhere* in an LIF -- contrary to the bold Einsteinian assertions of Charles Misner et al. JS: This does not follow logically from what was said before. PZ: OK -- then are you now saying that there *is* a deep reason to expect inertial forces to contribute to or alter the physical stress-energy of the gravitational Želd? JS: No, I am saying I see NO deep reason to try to build gravity stress-energy from G-Forces at all! Apparently Yilmaz does? I don¹t do it that way at all! Again my formula is very different from that. It is tuv(Geometry) = (String Tension)Guv(Einstein) = -(String Tension)/zpfguv PZ: Which is it? It has to be one or the other. JS: NO there is NO deep reason IMHO. Indeed, I show that when quantum zero point energy is included in the exotic w = -1 vacua of both dark energy and dark matter as LOCAL Želds, which in FRW case limit to Einstein¹s Cosmological Constant that tuv(Marble Geometry Vacuum) = - tuv(Exotic Vacuum) tuv(Exotic Vacuum) = (String Tension)/zpfguv /zpf = 0 in non-exotic vacuum /zpf > 0 repulsive dark energy exotic vacuum /zpf < 0 attractive dark matter exotic vacuum /zpf = Lp^-2[Lp^3|Vacuum Coherence|^2 - 1] guv = (Minkowski)uv + du,v + dv,u du = Lp^2(arg Vacuum Coherence),u ,u = ordinary partial derivative Lp^2 = hG/c^2 = 1 Bekenstein BIT on surface of Susskind¹s World Hologram that has the 3D projection image that is ordinary uncompactiŽed space. When Vacuum Coherence = 0 we have globally žat Minkowski space-time, but with a huge cosmological constant. This is unstable and is why we have inžation IMHO. PZ: Interesting, but this does not answer the question I posed, which is: Does the stress-energy of the pure inertial Želd, with no matter sources (if it is even deŽned) self-gravitate, i.e., itself act as a source of the gravitational Želd? JS: stress energy of the pure inertial Želd has no meaning to me unless you can produce a GR formula for it. Z. * /zpf = 0 is like Lenny Susskind¹s supersymmetry limit where the vacua on the Landscape of Megalopolis (Francis Ford Coppola¹s new movie) do not depend on the moduli of the extra hyperspace dimensions of Calabi-Yau. === Subject: A (New?) Sequence Transform Let {a(k)} be an inŽnite sequence of positive integers, and where {a} contains an inŽnite numbers of terms equal to 1. Here is a simple(well...) transform which converts {a} into another inŽnite sequence {b(k)} of positive integers, and where {b} also has an inŽnite number of 1¹s. Let {c(k)} be a permutation of the positive integers, where c(1) = 1, and c(m+1) = the a(m)_th yet-unpicked positive integer. (By yet-unpicked, I mean that c(m) is not among c(1),c(2),c(3),...c(m-1).) Let {d(m)} be the inverse of {c}. (ie. c(d(m)) = m for all m.) Apply the reverse of the {a}->{c} step to {d} to get {b}. In other words, b(m) = the order among + integers not in {d(1),d(2),d(3),...d(m)} of d(m+1). So, this simpler-than-it-must-appear-to-you transform seems like it must have some uses, but I do not know. An example of it being used: Sequence A001222 of EIS: (from a(2) on) a -> 1,1,2,1,2,1,3,2,2,1,3,... c -> 1,2,3,5,4,7,6,10,9,11,8,14,... d -> 1,2,3,5,4,7,6,11,9,8,10,... b -> 1,1,2,1,2,1,4,2,1,1,... (amazing!...) Aside from if a -> b implies b -> a, what additionally can be said about this transform on sequences? Leroy Quet === Subject: Re: sqrt(8n^2+1) is integer > Is there a way to Žnd the values of n > for which sqrt(8n^2+1) is an integer? See methods for solving the Pell equation given in Solving the generalized Pell equation - PDF File at http://hometown.aol.com/jpr2718/ to solve x^2 - 8y^2 = 1. Others have posted correct solutions. John Robertson === Subject: Re: sqrt(8n^2+1) is integer this) > See methods for solving the Pell equation given in > Solving the generalized Pell equation - PDF File > at > http://hometown.aol.com/jpr2718/ > to solve x^2 - 8y^2 = 1. > Others have posted correct solutions. > John Robertson === Subject: Re: sqrt(8n^2+1) is integer >Is there a way to Žnd the values of n for which sqrt(8n^2+1) is an >integer? Or is testing n=1,2,3,4... the only way? Starting with a_0 = 0 and a_1 = 1, use a_n = 6 a_{n-1} - a_{n-2}. 0,1,6,35,204,1189,6930,40391,235416,1372105,etc. This should give you all of them. This is derived from the continued fraction for sqrt(8) = (2,1,4,1,4,1,4,...). In fact the sequence above is the sequence of denominators of every second partial quotient. It can be shown that if p/q - sqrt(8) < 1/(2q^2) then p/q is a partial quotient of the continued fraction for sqrt(8). Rob Johnson take out the trash before replying === Subject: Re: sqrt(8n^2+1) is integer > > Hello. > Is there a way to Žnd the values of n for which sqrt(8n^2+1) is an > integer? Or is testing n=1,2,3,4... the only way? Look at http://www.research.att.com/projects/OEIS?Anum=A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2). 8*a(n)^2 + 1 is a perfect square. 0,1,6,35,204,1189,6930,40391,235416,1372105,7997214,.... with lots of formulas, links and references. Hugo Pfoertner === Subject: Re: sqrt(8n^2+1) is integer Charlie Johnson a .8ecrit: > >>Hello. >>Is there a way to Žnd the values of n for which sqrt(8n^2+1) is an >>integer? Or is testing n=1,2,3,4... the only way? > > Mathematical Induction. > > Lurch > Pell Equation m^2 - 8*n^2 = 1 (search Google for details) This one is used to Žnd the square = triangle numbers. n=1,6,35,204 ... n_(i+2)=6*n_(i+1) - n_i with n_0 = 1 and n_1 = 6 or if you prefer getting directly n_i with m + n*sqrt(8) = (3 + sqrt(8) )^i Otherwise, just Mathematical Induction is not enough... -- chephip at free dot fr === Subject: Re: sqrt(8n^2+1) is integer >> Hello. >> Is there a way to Žnd the values of n for which sqrt(8n^2+1) is an >> integer? Or is testing n=1,2,3,4... the only way? >Mathematical Induction. >Lurch Induction???!! You are kidding of course. I know it isn¹t responsive to your question, but I ran it through MAPLE and up through 10,000, only n = 1, 6, 35, 204, 1189, 6930 work. Not too many that¹s for sure. --Lynn === Subject: Re: Secret Santa probability question >Last month my wife and Žve of her girlfriends randomly and secretly >selected names in a Secret Santa gift exchange. On Sunday they >exchanged their gifts and discovered that whenever lady A had chosen >lady B, lady B had also chosen lady A. In other words, there were >three pairs of ladies who had chosen each others names. What is the >probability of this happening? (Note: The ladies were not allowed to >pick their own name. If they did, they threw it back into the hat.) I didn¹t see the answer I expected in anyone else¹s post so: It seems to me that you are assuming that you have a standard Secret Santa arrangement among 6 people, and you want to know how surprising it is that the 6 people are paired off. As was noted in a couple of posts, a SS arrangement is called a derangement, and it can¹t necessarily be created by name-drawing. But your question seemed to assume that the SS thing _had_ in fact been set up. So for this part of the problem it may be sufŽcient to just assume that an Oracle had selected a derangement of the n names at random. Then your question is, OK, now that we¹ve done that, how likely is it that the contestants are exactly paired? The answer is Pretty darn unlikely! when n is odd... For n even, the derangements which precisely pair the participants can be counted as follows. Take any permutation f of the numbers from 1 through n. Write the numbers f(1), ..., f(n) in order on a piece of paper. Group these numbers into pairs {f(1), f(2)}, {f(3), f(4)}, ... There¹s an exact-pairing derangement for you! And in fact, every such pairing will show up for some permutation f, and once you have found one f that works, you can Žnd other ones too: you can swap any or all of the n/2 pairs, and you can scramble the pairs, as groups, in any way you like. So in fact there are 2^(n/2) (n/2)! permutations which will give rise to any particular pairing. But that means the number of pairings is the total number of possible f¹s (which is n! ) divided by the number which lead to a single pairing (which is 2^(n/2) (n/2)! ) . You can work this out; it¹s equal to the produt (n-1) (n-3) (n-5) ... (5) (3) (1), which is sometimes written as (n-1)!! (!). For example, when n=6 there are 5!!=5.3.1=15 pairings; 15 of the derangements will leave the 6 participants buying presents in pairs. (Of course there are other ways of deciding that the number of pairing is (n-1)!! .) So the probability that a SS exercise involving an even number n of participants will result in everone buying their own Santa¹s present is (n-1)!!/D_n . You can compute this without much trouble for small n. For larger n it can be estimated using D_n ~ (1/e) n! so that the probability is about e / (2^(n/2) (n/2)!) . Obviously this tends to zero with n ; less obviously, it¹s approximately (e/n)^(n/2) (where I¹ve opted for a simple-to-write estimate rather than a particularly accurate one...) When n=8, the odds agains this happening are about 140-to-1, and the odds against group an order of magnitude when n=10, and jump even faster when n=12. dave === Subject: Trigonometry textbook I don¹t currently have access to a university library, or I¹d browse the stacks to Žnd what I¹m looking for. I am in need of a trigonometry textbook. I am not interested in anything published in the last 30 years. Does anyone know of a good, solid, old-school trig book with the four-place table of sine/cosine/tan in the back, & a discussion of spherical as well as plane trig? To re-iterate: anything Brett === Subject: Re: Trigonometry textbook If you can Žnd it, the old trig text book by Welchons, Krickenberger and Pearson sounds exactly like what you are looking for. > I don¹t currently have access to a university library, or I¹d browse > the stacks to Žnd what I¹m looking for. I am in need of a > trigonometry textbook. I am not interested in anything published in > the last 30 years. Does anyone know of a good, solid, old-school trig > book with the four-place table of sine/cosine/tan in the back, & a > discussion of spherical as well as plane trig? To re-iterate: anything > Brett === Subject: Re: solutions to a^2 + b^2 = x (mod p) >I think there exists a somehow stronger (related) result: >for every a, x and for every prime p, there exists b such that: >either: a - b ^ 2 = x (mod p) >or: a + b ^ 2 = x (mod p) This is only true when p=2 or p == 3 (mod 4). For instance, there¹s no solution for p=5, a=1, x=3. -- Erick === Subject: Re: solutions to a^2 + b^2 = x (mod p) Yes, this is true. Here¹s a quick proof. Theorem: Let p be prime. For every value of x, there are integers a and b so that a^2 + b^2 = x (mod p). Proof: Look at the set S = { a^2 (mod p) : 0 <= a <= (p-1)/2 } The set S contains (p+1)/2 different elements, since a nonzero square k^2 has the two square roots mod p, namely k and p-k, and exactly one of them is in the interval between 1 and (p-1)/2. Now Žx any value for x and look at the set T = { x - b^2 (mod p) : 0 <= b <= (p-1)/2 }. The same reasoning shows that T contains (p+1)/2 distinct elements. Thus S and T each contain (p+1)/2 distinct elements mod p. But there are only p different numbers mod p, so by the pigeonhole principle, S and T have an element y in common. That element y is equal to A^2 for some A, since y is in S, and y is equal to x-B^2 for some B, since y is in T. Therefore A^2 = x - B^2, which gives you your desired solution. By the Chinese Remainder Theorem, this implies that a^2 + b^2 = x (mod n) always has a solution if n is a product of _distinct_ primes. So for your second question, primes that appear in n to the Žrst power don¹t affect whether or not there is a solution. That may explain some of what you¹ve found. JoeS > Is it true that, for every prime p and every 0<=x solution a,b such that a^2+b^2=x (mod p) ? > > I¹ve determined that it is not necessarily true if p is not a prime, but > what about when it is? It seems that since a prime p has (p+1)/2 > quadratic residues, the probability of choosing any a at random and > being able to solve b = sqrt(x-a^2) is .5, so there ought to be lots of > solutions. But how can you prove this? > > Also, some experimentation has convinced me that when p is not prime, > the chances of having some x¹s that can¹t be expressed as a sum of > squares is higher when p has lots of small prime factors, then when it > has some large factors. Is there any criterion for this? === Subject: real analysis puzzlers hi, i have been thinking as hard as i can for the last week on these two real analysis problems. can anybody help me out? these are from birkhoff¹s book (which is the same as royden really) I¹m totally stumped. Let T: X -> Y be a linear map from banach space to banach space Y, with T continuous in the weak topology on X to the norm topology on Y. Show that the image T(X) must have Žnite dimension. === Subject: Re: real analysis puzzlers >hi, >i have been thinking as hard as i can for the last week on these two >real analysis problems. can anybody help me out? these are from >birkhoff¹s book (which is the same as royden really) I¹m totally >stumped. > Let T: X -> Y be a linear map from banach space to banach space Y, >with T continuous in the weak topology on X to the norm topology on Y. > Show that the image T(X) must have Žnite dimension. It follows from the deŽnitions that there exist Žnitely many L_1, ... L_n in X* such that |Tx| <= |L_1 x| + ... |L_n x| for all x in X. (So if L_j x = L_j y for 1 <= j <= n then ...) ************************ David C. Ullrich === Subject: Re: real analysis puzzlers >hi, >i have been thinking as hard as i can for the last week on these two >real analysis problems. can anybody help me out? these are from >birkhoff¹s book (which is the same as royden really) I¹m totally >stumped. > Let T: X -> Y be a linear map from banach space to banach space Y, >with T continuous in the weak topology on X to the norm topology on Y. > Show that the image T(X) must have Žnite dimension. > Pretend for a moment that X is just R^inŽnity with the product topology. Fix a product neighbourhood U of 0 such that T[U] sits in the unit ball of Y. Say U=U_1 times U_2 times ... times U_n times R times R times ... Now prove that every x with x_1=x_2=...=x_n=0 belongs to the kernel of T. After this special case the general proof is like this: a weak neighbourhood is determined by Žnitely many functionals and the above argument shows that the intersection of their kernels sits inside the kernel of T. KP === Subject: Re: Please solve this Diophantine Problem. > To. John R Ramsden (jr@adslate.com) > Is there any one can solve and correct proof. this problem !!! ??? > Someone ! someone !!!! I proved it, without a computer, using 6th grade Algebra. Has anyone tried one of those automated theorem provers? Hint: neither 32942845 nor 4059 are squares. Nathan the Great Age 16 === Subject: All of them presented at one time! >Message 1 in thread === >Subject: Jack Sarfatti¹s important physics ideas > > > >Post a follow-up to this message >Message 2 in thread === >Subject: Re: Jack Sarfatti¹s important physics ideas > > > >Post a follow-up to this message >Message 3 in thread === >Subject: Re: Jack Sarfatti¹s important physics ideas > > > >Post a follow-up to this message And that about sums it up for Dr. Jack Sarfatti¹s important physics ideas! === Subject: Re: Riemann Surfaces in Analysis >If you¹d stated that you¹re > happier with branch cuts than with the deŽnition in terms of > continuations along paths nobody would have argued. interested in techniques which is where cuts are useful. >The > bone of contention was your statement that there is necessarily some > arbitrariness in the construction of the Riemann surface > for a function. That statement is simply false. Having thought about it I can understand that applying analytical continuation on its own does produce a unique surface. Where I went wrong was in using cuts and then retaining their images when they are disposable. Anyway I have learned one thing: the way to attract attention is to make a mistake. === Subject: Re: Riemann Surfaces in Analysis >>If you¹d stated that you¹re >> happier with branch cuts than with the deŽnition in terms of >> continuations along paths nobody would have argued. >interested in techniques which is where cuts are useful. >>The >> bone of contention was your statement that there is necessarily some >> arbitrariness in the construction of the Riemann surface >> for a function. That statement is simply false. >Having thought about it I can understand that applying analytical >continuation on its own does produce a unique surface. Where I went >wrong was in using cuts and then retaining their images when they are >disposable. Anyway I have learned one thing: the way to attract >attention is to make a mistake. Well it sounds like you _have_ learned something about analytic continuation, which is a good thing. For future reference, regarding that last thing you say you¹ve learned: Making a mistake on sci.math is a good way to attract _corrections_. You could have learned what you¹ve learned about analytic continuation a lot faster if you¹d simply tried to avoid being certain you were right and everyone else was wrong, _even though_ (as you _said_) you did not understand the construction that others were talking about. ************************ David C. Ullrich === Subject: Re: calculus - tan limit question >>> >>>> So you know that tan¹ pi/4 = whatever. >>>> Now what does that *mean*? >>>> (Can you recall the *deŽnition* of derivative?) >>> >>> It means that you can apply l¹Hopital¹s rule without knowing in >>> advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is. >>Try again! That¹s no deŽnition of derivative. >>How do you use L¹H¹s rule to calculate this limit without >>already knowing what it is? > > It might be clearer to point out that the limit we are trying to > evaluate is > > f(x) - f(a) > lim ----------- > x->a x - a > > where f(x) = tan(x) and a = pi/4. The problem in that form might jog > some memories. But alas, not Virgil¹s :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: calculus - tan limit question > > > > >> So you know that tan¹ pi/4 = whatever. > >> Now what does that *mean*? > >> (Can you recall the *deŽnition* of derivative?) > > > > It means that you can apply l¹Hopital¹s rule without knowing in > > advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is. > > Try again! That¹s no deŽnition of derivative. > > How do you use L¹H¹s rule to calculate this limit without > already knowing what it is? The deŽnition of derivative of tan(x) may be deduced from the derivatives of sin(x) and cos(x) and the quotient rule for derivatives, the derivative of the numerator and denominator of (tan(x) - 1)/(x - pi/4) can be found without being aware that it is a difference quotient. If f(x) = tan(x) - pi/4 = sin(x)/cos(x), then f¹(x) = (cos(x)*cos(x) - sin(x)(-sin(x))/cos(x)^2 + 0 = 1/cos(x)^2 and if g(x) = x -pi/4 then g¹(x) = 1 Since f(pi/4) = g(pi/4) = 0 but g¹(pi/4) and f¹(pi/4) are continuous and nonzero at x = pi/4, L¹Hopital applies. === Subject: Re: calculus - tan limit question > >> >> > >> >> So you know that tan¹ pi/4 = whatever. >> >> Now what does that *mean*? >> >> (Can you recall the *deŽnition* of derivative?) >> > >> > It means that you can apply l¹Hopital¹s rule without knowing in >> > advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is. >> >> Try again! That¹s no deŽnition of derivative. >> >> How do you use L¹H¹s rule to calculate this limit without >> already knowing what it is? > > The deŽnition of derivative of tan(x) may be deduced from the > derivatives of sin(x) and cos(x) and the quotient rule for > derivatives, the derivative of the numerator and denominator of > (tan(x) - 1)/(x - pi/4) can be found without being aware that > it is a difference quotient. > > If f(x) = tan(x) - pi/4 = sin(x)/cos(x), then > f¹(x) = (cos(x)*cos(x) - sin(x)(-sin(x))/cos(x)^2 + 0 > = 1/cos(x)^2 Excellent: you can compute derivatives. Alas you seem to have forgotten what they are. :-( You have (d/dx)(tan x - pi/4) = 1/cos^2 x [dunno what the pi/4 is doing, but it¹s irrelevant anyway]. Now what does that mean? (Heavy hint: apply the deŽnition of derivative.) > and if g(x) = x -pi/4 then g¹(x) = 1 Any point to this? > Since f(pi/4) = g(pi/4) = 0 but g¹(pi/4) and f¹(pi/4) are > continuous and nonzero at x = pi/4, L¹Hopital applies. But why waste time using L¹H? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: JSH: Consider Dik Winter > > You probably knew this already [if it is right], though if > you did, I don¹t see why you would say ... neither I do know.. > > Theorem: Algebraic integer units are dense in the reals. > > Proof: First show that algebraic integers are dense in the > reals. This is clear from consideration of the function > > s(n) = sqrt(n) - [sqrt(n)], > > where [.] denotes the greatest-integer function. Sorry to be obtuse, but could you expand on this? To show that the algebraic integers are dense in the reals one has to take an arbitrary real and demonstrate that there are algebraic integers which are arbitrary close. I don¹t see how your function shows this, what have I missed? > > Next, consider the function > > f(x) = sqrt(x + 1) - sqrt(x) > > where x > 0. This is a continuous function from (0, 1] to > (0, 1]; f(0) = 1 and f(x) --> 0 as x --> inŽnity. Restricting > f(x) to the algebraic integers and using the fact that the > algebraic integers are dense in the reals shows that the > range of f(a), a = algebraic integer, is dense in (0, 1]. > f(a) is a unit for each algebraic integer a since > > f(a) * (sqrt(a + 1) + sqrt(a)) = 1. > > Finally, consider the function g(x) = 1/x: This is continuous > on (0, 1] and has range (0, inŽnity]. If u is a unit, certainly > g(u) = 1/u is a unit also. Therefore the set U = {g(u), u an alg. > int. unit in (0, 1]} is dense in the positive real numbers. > Thus U + (-U) is dense in the reals. > > > Nora B. === Subject: Re: JSH: Consider Dik Winter > > > > You probably knew this already [if it is right], though if > > you did, I don¹t see why you would say ... neither I do know.. > > > > Theorem: Algebraic integer units are dense in the reals. > > > > Proof: First show that algebraic integers are dense in the > > reals. This is clear from consideration of the function > > > > s(n) = sqrt(n) - [sqrt(n)], > > > > where [.] denotes the greatest-integer function. > > Sorry to be obtuse, but could you expand on this? To show that > the algebraic integers are dense in the reals one has to take > an arbitrary real and demonstrate that there are algebraic > integers which are arbitrary close. I don¹t see how your function > shows this, what have I missed? > There are a couple of ways to see this, one graphical, the other analytic - 1. Graph the function s(x) = sqrt(x), a smooth curve with slope that goes to 0 as x --> inŽnity. Put points on the graph corresponding to x = n = positive integer. Note that the values of s(n) on the y-axis get closer & closer together as n --> inŽnity. This shows that S = {s(n) - [s(n)]} is dense in (0, 1). Then add arbitrary integers to each point in (0, 1) to conclude that S is dense in the reals. 2. Consider two large consecutive perfect squares, say, 1024 and 1089. The numbers 0, sqrt(1025) - 32, sqrt(1026) - 32, ..., 33 - 32 are all in [0, 1], and the distance between consecutive such numbers is no larger than sqrt(1025) - 32. Thus no number in (0, 1) is farther away than .5*(sqrt(1025) - 32) from one of these numbers. Generalize to large n, noting that .5*(sqrt(n^2 + 1) - n) goes to 0 as n --> inŽnity, to conclude again that the set S is dense in [0, 1]. Then proceed as above. Nora B. > > > > Next, consider the function > > > > f(x) = sqrt(x + 1) - sqrt(x) > > > > where x > 0. This is a continuous function from (0, 1] to > > (0, 1]; f(0) = 1 and f(x) --> 0 as x --> inŽnity. Restricting > > f(x) to the algebraic integers and using the fact that the > > algebraic integers are dense in the reals shows that the > > range of f(a), a = algebraic integer, is dense in (0, 1]. > > f(a) is a unit for each algebraic integer a since > > > > f(a) * (sqrt(a + 1) + sqrt(a)) = 1. > > > > Finally, consider the function g(x) = 1/x: This is continuous > > on (0, 1] and has range (0, inŽnity]. If u is a unit, certainly > > g(u) = 1/u is a unit also. Therefore the set U = {g(u), u an alg. > > int. unit in (0, 1]} is dense in the positive real numbers. > > Thus U + (-U) is dense in the reals. > > > > > > Nora B. === Subject: Re: Alternatives to axiomatic systems Can you imagine math without Euclid? It would be very hard > to think of western math without axioms. Even more: mathematical reasoning is exact, and involves _steps_. So, no matter what you do, these steps can be analyzed and then formalized as an axiom system. However, there are forms of reasoning that are not based on steps or are not Œexact¹ enough to be so easily formalized. (For example: everyday life reasoning). Attempts to formalize these are sometimes done using Žxpoint constructions, and other tricks. You might be interested to Google for non-monotonic logic. Especially the argumentation-theoretic approach (Phan Minh Dung, Henry Prakken) and inheritance networks (John Horty). Hope this brings you further. Good luck and cheers, Herman Jurjus === Subject: how to study math i am taking some proving classes should i memorize all the theorems and their proofs? === Subject: Re: how to study math > i am taking some proving classes What¹s a proving class? A class where they teach you how to prove? > should i memorize all the theorems and their proofs? Likely the important part is where you prove yourself. === Subject: Re: how to study math >> i am taking some proving classes >Likely the important part is where you prove yourself. I thought Descartes did this already? === Subject: Re: how to study math > > > > >> i am taking some proving classes > >> > >Likely the important part is where you prove yourself. > > I thought Descartes did this already? LOL! === Subject: Re: how to study math <7kb8uv8m6gqmvh9p2159oc196kvonucg6r@no.spam> > >> i am taking some proving classes > >Likely the important part is where you prove yourself. > I thought Descartes did this already? ;-) Even so, his proof is equivalent to I think I am. === Subject: Re: maths spreadsheet modelling > I often use spreadsheets for prototyping methods before programming > them up directly (or even never leaving their prototyped stages!) > > I Žnd where they really get useful is for small scale Markov chain, > dynamic programming, simulation and general optimization problems. > > Obviously by nature such models will never be computationally very > efŽcient but this is often offset by the rapidity with which they can > developed. > > It often strikes that commonly used spreadsheets (that is in 99% > of cases, Excel!) have a whole load of functionality that is > irrelvant for this kind of application (e.g. lots of formatting, > web stuff) whereas by tightening up certain other aspects they > could be considerably more potent. For example: > > - better handling for array formulas (these are what really makes > use of spreadsheets in this way possible, in my opinion) > > - more intelligent execution plans > > - perhaps even partial complilation of certain sets of cells > > and to a lesser extent > > - support for linear algebra (Excel seems to allows multuplication, > inversion and determinant calculation only for matrices of > dimension less than 70 or so) > > My question is, has anyone ever written a spreadsheet application with > this type of use in mind? I gather use of spreadsheets is popular with > cellular automata researchers, are there any specialized tools out > there? > > > Tom In terms of general math modelling with Excel, I have found Excel to be good for solving PDEs using the Žnite difference method. There are plenty of web-links explaining how to do this - most of them use Laplace¹s equation as an example, but it is easier to solve more complicated equations (eg I have used Excel to solve the Reynolds¹ equation (this equation appears in problems to do with hydrodynamically lubricated contacts). In the Žnite difference method you don¹t need to invert any matrices, you just need to iterate the cell¹s value depending on the values of cells surrounding it - and to speed the process up you can always use successive over relaxation. Of course, Excel is nowhere near as fast as a decent numerical programming language (ie FORTRAN), but you can always call FORTRAN routines from Excel (if you compile them as dll¹s) Hope this helps Ian Taylor === Subject: Re: maths spreadsheet modelling X-message-žag: Outlook is rather hackable, isn¹t it? X-Home-Page: http://www.cbbrowne.com/info/ X-Affero: http://svcs.affero.net/rm.php?r=cbbrowne Clinging to sanity, Gus Gassmann mumbled into her beard: >> Oops! Harlan Grove was seen spray-painting on a wall: >> > .. >> >>- better handling for array formulas (these are what really makes >> >>use of spreadsheets in this way possible, in my opinion) >> > Details! Excel does a pretty good job with them already, but much of >> > its array semantics are undocumented except in newsgroup threads, >> > and then only empirically. >> Hmm? I¹m not sure there¹s a useful way to use =TRANSPOSE(ARRAY)... > I would say the following is quite useful: > =mmult(transpose(array1),mmult(array2,array1)) The problem with the array formulae is that there¹s no way of passing back arrays; all that you get is the top-left-most scalar element. Not much good if I actually wanted to get 15 values back... -- If this was helpful, rate me http://www.ntlug.org/~cbbrowne/postgresql.html DSK: STAN.K; ML EXIT -- FILE NOT FOUND === Subject: Re: maths spreadsheet modelling >...Gus Gassmann ... >> I would say the following is quite useful: >> =mmult(transpose(array1),mmult(array2,array1)) >The problem with the array formulae is that there¹s no way of passing >back arrays; all that you get is the top-left-most scalar element. >Not much good if I actually wanted to get 15 values back... Meaning you don¹t like having to preselect a range large anough to hold your array results before pressing [Ctrl]+[Shift]+[Enter] (in the Excel context)? Now in the limited case of array formulas that return 1-by-1 arrays, I agree it¹s a PITA to array-enter them rather than expect them to be treated as scalars, but that¹s an infrequent pain. === Subject: Confused by what should be a fairly simple inequality problem! :*( Let a, b, x be reals and suppose we know that a < x < b. What can we infer, inequality-wise, about 1/x? One is tempted to say 1/a > 1/x > 1/b... This is true if a,b, and x are positive. But what about the true, more general answer, where some or all may be negative?? Can we infer anything that *doesnt* have to do with taking absolute values or the sgn function or etc. of the things involved? it isn¹t homework === Subject: Re: Confused by what should be a fairly simple inequality problem! :*( > Let a, b, x be reals and suppose we know that a < x < b. > What can we infer, inequality-wise, about 1/x? > One is tempted to say 1/a > 1/x > 1/b... > This is true if a,b, and x are positive. > But what about the true, more general answer, where some or all may be > negative?? > Can we infer anything that *doesnt* have to do with taking absolute > values or the sgn function or etc. of the things involved? > it isn¹t homework The larger the denominator the smaller the number, and vice versa. Lurch === Subject: Re: Confused by what should be a fairly simple inequality problem! :*( >Let a, b, x be reals and suppose we know that a < x < b. >What can we infer, inequality-wise, about 1/x? >One is tempted to say 1/a > 1/x > 1/b... >This is true if a,b, and x are positive. >But what about the true, more general answer, where some or all may be >negative?? >Can we infer anything that *doesnt* have to do with taking absolute >values or the sgn function or etc. of the things involved? If a and b have the same sign, then 1/a > 1/x > 1/b. However, if a < 0 and b > 0, we can¹t even infer that 1/x exists since x could be 0. If x <> 0 and a < 0 and b > 0, then we can infer that either 1/a > 1/x or 1/x > 1/b, but 1/a < 1/b, so we cannot infer both. Rob Johnson take out the trash before replying === Subject: Re: lots of balls = 0 balls In response to Lewis Mammel , >> Any Žfth grader would not hesitate to answer: inŽnity . >This is why Žfth graders aren¹t generally known to be accomnplished >Set Theorists. >InŽnity is not an acceptable answer, as there is no number n, which >when multiplied by 10, gives you inŽnity. As each label number n >must be a successor of some previous label number, inŽnity is never >actually attained. Then, by that same reasoning, 12:00 is never actually attained. Suppose we add a tally each time the balls in the bucket are changed. How many tallys are there at 12:00? Certainly, no Žnite number. We have to allow ourselves to use inŽnity, at least in the transŽnite guise of w (omega), the Žrst limit ordinal. The time per event is 0 at 12:00, so if we are allowed to proceed to 12:00, any number of events could happen during that instant. That is, at 12:00, balls numbered w to w+9 are added and the ball numbered w is removed, but also at the same time, balls numbered w+10 to w+19 are added and ball w+1 is removed. Exactly how many events happen? It is as indeterminate as 0/0. So, although there are deŽnitely no balls with a Žnite index in the bucket, what is preventing any multiple (Žnite or otherwise) of 9 balls (with transŽnite index) from being in the bucket? Part of the trouble with this problem is that we are trying to pass from the Žnite to the transŽnite with rules only for successor ordinals; no rule is given for limit ordinals. To deŽne a transŽnite process, we must have a rule for limit ordinals as well as successor ordinals. Rob Johnson take out the trash before replying === Subject: Re: lots of balls = 0 balls > In response to Lewis Mammel , >>> Any Žfth grader would not hesitate to answer: inŽnity . >>This is why Žfth graders aren¹t generally known to be accomnplished >>Set Theorists. >>InŽnity is not an acceptable answer, as there is no number n, which >>when multiplied by 10, gives you inŽnity. As each label number n >>must be a successor of some previous label number, inŽnity is never >>actually attained. > Then, by that same reasoning, 12:00 is never actually attained. Suppose > we add a tally each time the balls in the bucket are changed. How many > tallys are there at 12:00? Certainly, no Žnite number. We have to > allow ourselves to use inŽnity, at least in the transŽnite guise of w > (omega), the Žrst limit ordinal. He didn¹t say we were disallowed from considering inŽnity altogether. He merely pointed out that no ball labelled inŽnity is ever placed in the bucket. > The time per event is 0 at 12:00, so if we are allowed to proceed to > 12:00, any number of events could happen during that instant. That is, > at 12:00, balls numbered w to w+9 are added and the ball numbered w is > removed, but also at the same time, balls numbered w+10 to w+19 are > added and ball w+1 is removed. Exactly how many events happen? It is > as indeterminate as 0/0. And if I take 2 apples from 5 apples, the result is 17 apples. If I am allowed to make use of secret information that was not given in the problem statement, then anything is possible and there is no point in discussing any word problem whatsoever. > So, although there are deŽnitely no balls with a Žnite index in the > bucket, what is preventing any multiple (Žnite or otherwise) of 9 balls > (with transŽnite index) from being in the bucket? The fact that no such balls were ever placed in the bucket? > Part of the trouble with this problem is that we are trying to pass from > the Žnite to the transŽnite with rules only for successor ordinals; > no rule is given for limit ordinals. To deŽne a transŽnite process, > we must have a rule for limit ordinals as well as successor ordinals. There is nothing in the problem that mentions successors. All we are given is what happens at step n for each n. Note that one of those is not a successor. The same goes for the transŽnite subway. We are given a rule to apply at each station (each ordinal). The rule does not distinguish between successor ordinals and limit ordinals. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: lots of balls = 0 balls _ === Subject: Re: lots of balls = 0 balls > > >> Cool. Zeno didn¹t understand a millennia ago. Real Cool. > > >> I don¹t recall mentioning a Žnite time interval. > > > > > >What do you mean a millennia? One can have a millennium, but one > > >cannot have less then two millennia. But Zeno understood better than > > >you seem to. > > > > Haven¹t you ever learned that the punctuation is to be included > > inside the quotes. What in the world?! only periods and commas go in quotes, question and exclamation marks _are_ supposed to remain outside of the quotation marks! === Subject: Re: lots of balls = 0 balls > > > >> Cool. Zeno didn¹t understand a millennia ago. Real Cool. > > > >> I don¹t recall mentioning a Žnite time interval. > > > > > > > >What do you mean a millennia? One can have a millennium, but one > > > >cannot have less then two millennia. But Zeno understood better than > > > >you seem to. > > > > > > Haven¹t you ever learned that the punctuation is to be included > > > inside the quotes. > > What in the world?! only periods and commas go in quotes, question > and exclamation marks _are_ supposed to remain outside of the > quotation marks! I was asked for a reference for what I just said. I remember it from grade school, truthfully, but googling for punctuation and following the Žrst hit returned http://owl.english.purdue.edu/handouts/grammar/g_quote.html , it¹s at the very bottom. I¹ll quote for simplicity: Put commas and periods within closing quotation marks, except when a parenthetical reference follows the quotation. He said, I may forget your name, but I never remember a face. History is stained with blood spilled in the name of civilization. was to do nothing (27). Put colons and semicolons outside closing quotation marks. Williams described the experiment as a deŽnitive step forward; other scientists disagreed. Benedetto emphasizes three elements of what she calls her Olympic journey: family support, personal commitment, and great coaching. Put a dash, question mark, or exclamation point within closing quotation marks when the punctuation applies to the quotation itself and outside when it applies to the whole sentence. Philip asked, Do you need this book? Does Dr. Lim always say to her students, You must work harder? Sharon shouted enthusiastically, We won! We won! I can¹t believe you actually like that song, If You Wanna Be My Lover! === Subject: Re: lots of balls = 0 balls > > What in the world?! only periods and commas go in quotes, question > > and exclamation marks _are_ supposed to remain outside of the > > quotation marks! > I was asked for a reference for what I just said. I remember it from > grade school, truthfully, but googling for punctuation and following > the Žrst hit returned > http://owl.english.purdue.edu/handouts/grammar/g_quote.html , it¹s at > the very bottom. I¹ll quote for simplicity: I consider to be a more logical way of handling the punctuation. My revisions are in the lines which begin with asterisks. Some other comments of mine are given in brackets. David Cantrell ------------------------------------------------- Punctuation with Quotation Marks Use a comma to introduce a quotation after a standard dialogue tag, a brief introductory phrase, or a dependent clause, for example, He asked, She not show the use of a comma!] As D. H. Nachas explains, The gestures used for greeting others differ greatly from one culture to another. * ... from one culture to another.. [Perhaps that looks ugly. But it seems logically correct, assuming that an entire sentence was quoted.] Put commas and periods within closing quotation marks, except when a parenthetical reference follows the quotation. He said, I may forget your name, but I never remember a face. * ... remember a face.. History is stained with blood spilled in the name of civilization. * ... in the name of civilization. Put a dash, question mark, or exclamation point within closing quotation marks when the punctuation applies to the quotation itself and outside when it applies to the whole sentence. Philip asked, Do you need this book? * Philip asked, Do you need this book?. Does Dr. Lim always say to her students, You must work harder? * Does Dr. Lim always say to her students, You must work harder.? Sharon shouted enthusiastically, We won! We won! * Sharon shouted enthusiastically, We won! We won!. === Subject: [OT] Punctuation: Was: Re: lots of balls = 0 balls >[...] > >> >>> >>>> >>>> >>>>>Cool. Zeno didn¹t understand a millennia ago. Real Cool. >>>>>I don¹t recall mentioning a Žnite time interval. >>>>> >>>>> >>>>What do you mean a millennia? One can have a millennium, but one >>>>cannot have less then two millennia. But Zeno understood better than >>>>you seem to. >>>> >>>> >>>Haven¹t you ever learned that the punctuation is to be included >>>inside the quotes. >>> >What in the world?! only periods and commas go in quotes, question >and exclamation marks _are_ supposed to remain outside of the >quotation marks! > Do you have a reference for that? On the web, maybe? (This is a request for info, not rhetorical questions.) -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: lots of balls = 0 balls Hehehhe, I forgot I had a part 2! > certainly is not: That there is only a Žnite amount of matter in the > universe preventing this problem from existing is not paradoxical, > merely physically impossible. The same goes with putting the balls in > the bucket at speeds exceeding the speed of light, the aggregate > mathematical standpoint, there is a single answer as well; perhaps > counter-intuitive, but still not a paradox. OK, what happens if we make ball n weigh 2^-n? The number of balls in the bucket always increasing proponent would almost certainly also accept a mass of balls in the bucket always decreasing argument. Their therefore the count tends to inŽnity becomes therefore the mass decreases to zero. And Žnally the therefore there are an inŽnite number of balls at the end becomes therefore there are no balls at the end. Can someone try this out on someone who¹s not seen the problem before (I¹ve posed the highest/lowest/random variations to most people I know, now) pose it not in terms of _balls_, but in terms of _masses_? Phil -- Unpatched IE vulnerability: dragDrop invocation Description: Arbitrary local Žle reading through native Windows dragDrop invocation. Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0302/12.html Exploit: http://kuperus.xs4all.nl/security/ie/xŽles.htm === Subject: Re: lots of balls = 0 balls > > I think the physical imagery of the balls creates the paradox. It creates > > a block against the realization of the actual inŽnite sequence of Žnite > > operations. > > There really is no paradox here. While it may be mathematically explainable in fairly simple terms to an intelligent Žfth grader, and leads to no logical paradox, I think it¹s a bit arrogant of modern day mathematicians to claim entire ownership of the word Œparadox¹. The description of problem, unless you¹ve seen similar situations before, _does_ invite different approaches which lead to contradictory conclusion. Therefore, classically, there is what would be called a paradox. The fact that 20th century (late 19th?) logic can resolve the situation and show that one of the approaches is mathematically invalid does not cancel, for me, the paradoxical setup of the problem. If you throw out this one, then you¹ve got to throw out half of the paradoxes created by the people who invented the word. And if so, you¹ll be telling the physicists to not call atoms atoms either! (Yes, slippery slope, I know. But it makes good rhetoric, I Žnd.) > perhaps > counter-intuitive, but still not a paradox. Paradox, amongst its meanings, means counter-intuitive, coming from Gk. /paradoxon/ meaning contrary to expectation. TomAHto, tomAYto, but my version¹s 3 syllables, yours is 6. Phil -- Unpatched IE vulnerability: DNSError folder disclosure Description: Gaining access to local security zones Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html === Subject: elliptic modular function lambda, product representation At http://functions.wolfram.com/EllipticFunctions/ModularLambda/ 02/ on the mathematica website, the following formula is given for the elliptic modular function lambda where q = e^{pi i z}: lambda(z) = 16q prod_{k ge 1} ( frac{1+q^{2k}}{1+q^{2k-1}} )^8 Does anybody know a reference for this formula (or where I should look for one)? Any help appreciated, Richard. === Subject: Developing the cosine of a cosine which is the fastest (i.e. more rapidly converging) way to develop the cosine of a cosine into an harmonic series? I¹m aware of the developement through Bessel functions, but I was wondering if there are alternative ways to achieve this. francesco === Subject: Re: Perplexing Patterns of Square Numbers ... > >>>> and a scintillating Perplex-9 pattern ... A few more Perplex-9¹s, reading across - 27261 21046 18089 13875 25123 15542 10937 14568 12712 10591 11986 15238 12723 25879 30764 14723 29904 12842 16191 25256 16688 13577 21641 29316 21307 28884 12842 15869 23486 10908 29313 17279 22894 10723 26394 12842 23161 19906 25746 20696 18581 13654 31392 15204 12908 18042 15722 23972 22737 13629 17782 26966 26242 21093 21146 20431 27918 12028 22842 18822 19437 12762 24788 29466 26426 29761 15412 20766 23194 11907 23926 25719 30746 20596 23361 17818 13924 25616 22873 10596 25719 20627 16841 23091 21479 26823 19868 11065 16983 30186 11477 18073 12839 27718 15293 11822 23229 14908 31488 Eg, squared out the Žrst one looks like: 743162121 442934116 327211921 192515625 631165129 241553764 119617969 212226624 161594944 and the last one like 131721529 326633329 164839921 768287524 233875849 139759684 539586441 222248464 991494144 -jiw === Subject: Re: Perplexing Patterns of Square Numbers > Now is the turn of forming TWO more Perplex-8 patterns out of 2783 eight digit nonzero square numbers.Try to form out these patterns and get perplexed with their exploration. > HINT : These patterns have 3408^2 and 4116^2 respectively,in their last row/column. Reading across for each solution - 7511 7909 6729 3969 7275 3535 5335 3334 6421 4079 5184 4871 9918 4782 6674 3408 7591 8921 8096 5331 5882 4718 9174 3408 8279 9129 7316 6579 4282 9318 6674 3408 6079 7918 9843 7734 6667 5261 6516 3869 8519 5326 7343 8728 6081 5718 8215 4116 7371 6604 6023 5636 3556 8279 6426 4119 7918 5141 8034 9615 7042 8839 5341 6454 9732 6813 8609 3337 4056 9479 5183 7014 3444 4341 9418 8057 3854 3567 6184 7842 9066 4875 4182 9843 5094 6014 7228 8093 3336 4044 3721 5046 9139 9477 9665 8134 -jiw === Subject: Re: Perplexing Patterns of Square Numbers > >After a lapse of two months, I am disclosing details of the SEVEN Perplex-7 patterns. Expand the following to arrive at the actual patterns and get perplexed : > > 1913^2 2057^2 1457^2 1311^2 2723^2 1057^2 1524^2 > 2636^2 1526^2 1284^2 2708^2 2044^2 1223^2 1942^2 > 2333^2 1813^2 1557^2 1172^2 1335^2 1414^2 1668^2 > 3134^2 1366^2 1686^2 2887^2 2173^2 2756^2 1459^2 > 2343^2 1643^2 2937^2 2694^2 2816^2 1654^2 2308^2 > 2643^2 2183^2 2133^2 1639^2 1524^2 2072^2 2762^2 > 3114^2 3114^2 3144^2 1204^2 3108^2 3157^2 2538^2 In your third set 3144 is a typo for 3114. I get the following 13 solutions (with each set of 7 numbers below reading across rather than down) 1311 2708 1172 2887 2694 1639 1204 2261 1292 1278 1713 1106 1639 1204 1069 1212 2162 1687 2816 2539 1204 1188 2043 1313 1157 1974 2113 2236 1065 1109 1811 2236 1696 1682 2262 1524 1942 1668 1459 2308 2762 2538 3094 2314 2744 1581 2925 1984 2581 2284 1637 1339 2631 2606 2372 2629 2723 2044 1335 2173 2816 1524 3108 1913 2636 2333 3134 2343 2643 3114 1457 1284 1557 1686 2937 2133 3114 2057 1526 1813 1366 1643 2183 3114 1057 1223 1414 2756 1654 2072 3157 === Subject: Re: Perplexing Patterns of Square Numbers ... > >>The THIRTEEN Perplex-6 patterns can be obtained by expanding the following sets of 6 squares each : > >1) 865^2 2) 885^2 3) 965^2 4) 379^2 5) 939^2 6) 369^2 7) 761^2 8)479^2 > > 668^2 918^2 582^2 694^2 944^2 631^2 891^2 486^2 > > 932^2 568^2 432^2 565^2 335^2 829^2 969^2 984^2 > > 476^2 526^2 526^2 814^2 844^2 335^2 435^2 642^2 > > 472^2 472^2 472^2 658^2 482^2 815^2 535^2 704^2 > > 738^2 738^2 738^2 407^2 407^2 334^2 334^2 408^2 > > > >9) 981^2 10)715^2 11)965^2 12)344^2 13)661^2 > > 806^2 409^2 627^2 357^2 615^2 > > 544^2 424^2 372^2 935^2 828^2 > > 608^2 477^2 462^2 585^2 962^2 > > 796^2 536^2 478^2 585^2 478^2 > > 408^2 719^2 771^2 834^2 393^2 ... I think there might be a couple more ending in 407 and another 738 -- 389 744 365 594 482 407 389 756 335 594 482 407 385 668 932 476 472 738 -jiw === Subject: Re: Rationality test 2, math > I am going to try to follow the convention of addressing the assembly > as a whole, not just the poster to whom I am replying. I think this > practice improves the ratio of light to heat in a discussion. > > > > Given, where x is in the ring of algebraic integers, I¹ve shown the > > factorization > > > > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > > > where b_3(x) = a_3(x) - 3 and the a¹s are roots of > > > [Polynomial P] > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > > > so when x=0, a_1(0) = a_2(0) = b_3(0) = 0. > It took me two tries, even with Mathematica, but this computation > checks out OK. > > > > Now consider the factorization shown again, but with the 49 multiplied > > through: > > > > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > > > 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > > > > and since a_1(0)= a_2(0) = b_3(0) = 0, it¹s not surprising that the > > values thus shown to be constant in the factors on the left side i.e. > > 7, 7 and 22 are in fact factors of what¹s constant on the right side > > i.e. 1078. > Sustitute 0 for x and the equation reduces to > 7 * 7 * 22 = 1078. Yes. > > > > Now if I divide both sides by 49, I end up with a change where now I > > have constant factors 1, 1, and 22 on the left which are still factors > > of 22 on the right. > Well, let¹s see if I understand. The left side could just be written > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)/49. > But one can say, Well, let¹s merge the factor of 1/49 into the other > factors. The way that looks nicest is to multiply each of the Žrst > two factors by 1/7. Then we have this: > > [Equation A] > ((5/7) a_1(x) + 1)((5/7)a_2(x) + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > > > > Does that fact tell you that *two* of the factors on the left, the > > ones that have 7 as a constant factor were each divided by 7, or does > > it tell you *nothing* at all? > It tells me a little bit, but I don¹t think it tells me as much as Mr. > Harris wants. I am not sure, but I think he wants to infer that the > three factors in Equation A will be algebraic integers when x is a > non-zero algebraic integer. > That is incorrect. I¹ve noticed several posters making that claim. One poster after repeated correction *still* made that claim. I¹ve noted that *in general* a_1(x)/7 and a_2(x)/7 are NOT algebraic integers! James Harris === Subject: Re: Rationality test 2, math > Hi James. Wrong again! Maybe you should stop posting. Note post under alias with intent not really clear. What is this poster *really* after? Obviously attention is still a factor, but what is the desired attention if a person doesn¹t provide their name? My guess is that such posters represent people who feel they are part of a given social order that their social order is under attack and that they are providing a general service to their society! Of course, what is the society of the sci.math and sci.logic newsgroups? Or better yet, what society does this person *think* they represent? Unfortunately, it¹s harder to follow-up on posters such as this, as they tend to come and go, but it¹s worth considering them at least briežy. James Harris === Subject: Re: condensation points in R > > > Let S be an uncountable set in R. I want to prove: > > > > (1) If P is the set of bilateral condensation points of S, then P > > intersection S is uncountable > > > > (2) The set of condensation points of S that don¹t belong to S is > > countable. > > > > To prove (1), I tried to prove that the set of unilateral > > condensation points of S is countable. Since it¹s known the set of all > > condensation points of S is uncountable, this gives the conclusion. > > > > But I could`t get through yet. > > > > PS.: x is a bilateral condensation point of S if the points of S > > condensate to the left and to the right of x, that is, for every > > eps>0, the intervals (x-eps, x) and (x, x+eps) contain uncountably > > many elements of S. > > For (1), let V be the union of all open intervals whose intersection with S > is countable. Then V intersect S is countable. Write V as the pairwise > disjoint union of open intervals (an,bn). If you put V* = U [an,bn], then > V* intersect S is still countable and it looks to me like every point of S > V* is a bilateral condensation point of S. Yes, I think you¹re right. The numbers a_n a and b_n are unilateral condensation points of S. (the elements of S condensate to the left of the a_n¹s and the the right of the b_n¹s. If P is the set of all condensation points of S, then S is a union of pairwise disjoint closed intervals. If x is in P, then there is a natural n such that x is in [b_n, a_n+1], which is an interval whose elements are condensation popints of S. (b_n may be -inf and a_n+1 may be inf.) Suppose b_n < x < a_n+1. If the elements of S don¹t condensate to the right of x, then there¹s eps>0 such that (x, x+eps) is in (x, a_n+1) and intersects S in only countably many elements. Therefore, no element of (x, x+eps) is a condensation point of S, which is a contadiction. Similarly, we see the elements of S also condensate to the left of x, and so we concluse x is a bilateral condensation point. With this we have proved that: (a) The set B of all bilateral condensation points of S is open and, so,uncountable. (b) the set U of the unilateral condensation points of s is {a1, b1, a2,b2.....}, therefore countable. (c) Since the set P of all condensation points of S is closed, it follows U is closed, because U = P / B (d)The set U intersect S is clearly countable. (e)We have P intersect S = (B intersect S) Union (U intersect S). Since P intersect S is uncountable, it follows from (d) that B intersect S is uncountable. Now it looks to me that B intersect S is open and U intersect S is closed, but I¹m not sure. Amanda === Subject: Re: condensation points in R === Subject: condensation points in R >Let S be an uncountable set in R. I want to prove: >(1) If P is the set of bilateral condensation points of S, >then P intersection S is uncountable >(2) The set of condensation points of S that >don¹t belong to S is countable. Remove from S all maximal closed intervals [a,b], a < b for which [a,b] / S is countable / intersect [a,b] is maximal when for all (c,d) containing [a,b] (c,d) is uncountable. There are at most countable many of these pairwise disjoint closed intervals. Thus removing all of them from S, leaves S unchanged cardinally. The points in C, the remainder of S, are bi-sided condensending. If x in C wasn¹t, then for some a, either [x,a] or [a,x] would be countable which contradicts the construction of C. For every so constructed maximal [a,b], the endpts a,b, are mono-sided condensending and the interior points at best, are mundane accumulators. ---- === Subject: Re: condensation points in R > > > Let S be an uncountable set in R. I want to prove: > > > > (1) If P is the set of bilateral condensation points of S, then P > > intersection S is uncountable > > > > (2) The set of condensation points of S that don¹t belong to S is > > countable. > > > > To prove (1), I tried to prove that the set of unilateral > > condensation points of S is countable. Since it¹s known the set of all > > condensation points of S is uncountable, this gives the conclusion. > > > > But I could`t get through yet. > > > > PS.: x is a bilateral condensation point of S if the points of S > > condensate to the left and to the right of x, that is, for every > > eps>0, the intervals (x-eps, x) and (x, x+eps) contain uncountably > > many elements of S. > > For (1), let V be the union of all open intervals whose intersection with S > is countable. Then V intersect S is countable. Here, V must be the union of a countable collection of open intervals that form a _countable_ topological basis for R, right? For example, the collection of all open intervals centered at rational elements and with rational radius. This is essential to ensure V intersect S is given by a countable union of countable sets, so that V is countable. Right? Write V as the pairwise > disjoint union of open intervals (an,bn). If you put V* = U [an,bn], then > V* intersect S is still countable and it looks to me like every point of S > V* is a bilateral condensation point of S. It¹s clear U (a_n, b_n) is the set of all real numbers that are not condensation points of S. Therefore, all of the a_n¹s and the b_n¹s are condensation points. But for every eps satisfying, 0< eps Let f(x) be a function from the set of non-negative integers to itself. > > Any suggestions as to how to Žnd functions Q(x), such that Q(f(x)) = > Q(x). > > I¹ve found, empirically, that x % p, p prime, is such a function for > f(x) = x^k, where k = p + (p-1)n, and also that if x % p and x % q are > such functions for x^k, then x % p*q is also such a function. Any _constant_ function Q will do the trick, for every f. Without further information about f, these may be the only ones. For example in the case f(x) = x+1 the above condition forces Q(x+1)=Q(x) for every x, hence Q must be constant. More generally, you can look at the orbits of f i.e., the subsets of the form {f^k(c) | k=0,1,2,...} for a Žxed c. (Here f^k is f applied k times). A function Q as above has to be constant on those orbits. You can also build new such functions: if Q1 and Q1 have the above property with respect to f, then Q1+Q2 and Q1*Q1 also have this property. Marc === Subject: Title: Re: Shannon defeats Cantor = single inŽnity type |-|erc says... >> Look at the following two statements: >> 1. d is on the list >> <-> exists n, forall i, >> digit number i of d = digit number i of number n >> 2. d can be approximated to any accuracy by numbers on the list >> <-> forall i, exists n, >> forall jSame statement. No, the Žrst statement is false, but the second statement is true. >change forall jStatement 2 2nd part -> >exists n, forall i, >forall j=i, digit j of d = digit j of n No, you can¹t switch the quantiŽers forall i and exists n. That¹s a logical error. >a formula is a type of algorithm No, it¹s not. Here¹s a formula: n is the largest prime number such that n+2 is also a prime number What algorithm is there to compute n? -- Daryl McCullough Ithaca, NY === Subject: Re: C++ Simulator of a Universal Turing Machine > --------------------------------- <^> <()> <^> ----------------------------------- > > C++ Simulator of a Universal Turing Machine can be downloaded at : > > * http://alexvn.freeservers.com/s1/utm.html > > * http://sourceforge.net/projects/turing-machine/ > > > > > > The program simulates a Universal Turing Machine (UTM). > > > > The UTM used in the Simulator is three-tape Turing Machine: > > * Tape#0 contains transition table and initial instantaneous description > > of a Particular Turing Machine (TM); > > * Tape#1 and Tape#2 are working UTM-tapes. > > > > The UTM can simulate the behavior of a Multitape TM. > > > > The package consists of two executable Žles : > > * t2u - compiler TM-to-UTM > > which translates description and input of TM to UTM-language; > > t2u generates several output Žles, one of them is used as input of the utm. > > * utm - the Simulator itself. > > > > Detailed log Žle is generated. > > Resources used (input size, output size, UTM-space, UTM-time) are computed as well. > > > > > > Testsuites. Two Turing Machines (TM-1 and TM-2) are used to create inputs for UTM. > > Each of them is an addition program which adds two numbers: > > * TM-1 is one-tape TM, > > * TM-2 is two-tape TM. > > > > > Has the UTM been encoded itself as a TM. I.e. have you tested > UTM(UTM(testfunction)) > If so what is the size of the encoded UTM? > I¹m interested if there is much of a space saving to a single tape UTM. > Herc -- ===================================== Alex Vinokur mailto:alexvn@connect.to http://mathforum.org/library/view/10978.html ===================================== === Subject: Re: Book For a First Analysis Course I would like to add one more to the list. Introduction to Calculus and Analysis by Richard Courant. Lurch === Subject: Re: Book For a First Analysis Course John Harrison > I have been working through Herstein¹s Topics in Algebra and Rudin¹s > Principles of Mathematical Analysis on a course of independent study. > So far, my progress with Herstein has been good ( at least, I think > so). I have grasped the theorems and solved most of the exercises. The > theory seems to develop naturally. Above all, I am enjoying learning > the material. My progress with Rudin seems much slower, however. In > particular, I have much more trouble mastering the theorems and their > proofs. ... Just a tip based on a lot of experience in self study... Analysis is intrinsically rigour heavy in a way that algebra isn¹t. The main theorems in Herstein, or a similar course, are essentially structure theories, e.g. about solvable groups and Galois extensions. But in analysis, what are called theorems are often more in the nature of technicalities, made necessary by the inŽnite limiting processes that are always present. E.g. the Jordan curve theorem, although intuitively obvious, requires a substantial proof. Something like Dini¹s theorem is intuitively plausible, and doesn¹t seem to say much on its own, but needs the epsilon-delta treatment too. It might be well to look ahead at how the theorems in Rudin are used. There are lots of books but I¹m a bit partial to Friedman _Foundations of Modern Analysis_, which is back in print as a cheap Dover paperback. LH === Subject: Re: question... In sci.math, A N Niel >> a and b are non-zero positive integers > [...] >> >> a/b < 0 See previous post. :-) -- #191, ewill3@earthlink.net It¹s still legal to go .sigless. === Subject: Re: question... In sci.math, Tom <3fe34847$0$9096$80265adb@spool.cs.wisc.edu>: >> In sci.math, Rustom >> >> >>>I¹m trying to prove that for any real number m<1, there is at least >>>one number of the form (3^a)/(2^b) , a and b are non-zero positive >>>integers, between m and 1, and i¹m trying to do so without using >>>supremums and inŽmums. does anyone has an idea about how to proceed ? >> >> >> 0 < m < 3^a/2^b < 1 if and only if log(m) < a*log(3)/b*log(2) < 0, >> or log(2) * log(m) / log(3) < a/b < 0 (since log(2) and log(3) > 0). >> >> Since the rationals are dense around any point, you shouldn¹t >> have a problem. :-) > That¹s wrong. The if and only if shold be > log(m) < a*log(3) - b*log(2) < 0 > and then you have to solve the simultaneous inequalities. Whoopsie...you¹re right. This could be trickier than I thought... :-) However, it¹s not all bad; m is an arbitrary variable and therefore vanishes in the analysis (yeah, I know that sounds weird, but replace m with epsilon and you might get the idea); and one merely needs to Žnd a Cauchy sequence such that a_n*log(3) < b_n*log(2) for every term in the sequence, with the limit such that a_n/b_n = log(2)/log(3). (Since log(2)/log(3) is a real number, obviously there is a general Cauchy sequence. I don¹t know if it necessarily follows that there¹s a Cauchy sequence that is forced to the left.) -- #191, ewill3@earthlink.net It¹s still legal to go .sigless. === Subject: Re: question... > I¹m trying to prove that for any real number m<1, there is at least > one number of the form (3^a)/(2^b) , a and b are non-zero positive > integers, between m and 1, and i¹m trying to do so without using > supremums and inŽmums. does anyone has an idea about how to proceed ? What integers a, b lies between m and 1 and is non-zero positive, if m < 1? Lurch === Subject: Re: question... In sci.math, Charlie Johnson : >> I¹m trying to prove that for any real number m<1, there is at least >> one number of the form (3^a)/(2^b) , a and b are non-zero positive >> integers, between m and 1, and i¹m trying to do so without using >> supremums and inŽmums. does anyone has an idea about how to proceed ? > What integers a, b lies between m and 1 and is non-zero positive, if m < 1? > Lurch REAL*4 m :-) -- #191, ewill3@earthlink.net -- insert random FORTRAN here It¹s still legal to go .sigless. === Subject: Re: Largest ball in a convex polyhedron X-Authenticated: @ Hi again to the replies in this thread, I also received a mail from Brian Borchers suggesting linear programming for solving the problem. Another mail I received presented a different take on the subject, and is quoted below for completeness. Sren Nielsen QUOTE === Subject: Re: Largest ball in a convex polyhedron Hi Soren, I¹m e-mailing because I don¹t have access to post to newsgroups right now. (You can post this e-mail if for some reason that helps.) I thought of a fairly simple algorithm, but you¹ll have to Žll in the details yourself. The important thing to realize is that in any convex set S, if B0 and B1 are two balls that Žt in S (individually, that is, not together), then B(t) = (1-t) B0 + t B1 can be Žt in S for any 0 <= t <= 1. The center and radius of B(t) are x(t) = (1-t) x0 + t x1 and r(t) = (1-t) r0 + t r1, respectively, where x0, x1, r0, and r1 are the centers and radii of B0 and B1. Let v0 = x¹(t) = x1 - x0, and w0 = r¹(t) = r1 - r0. We can then write B(t) as x(t) = x0 + v0 t, and r(t) = r0 + w0 t. For the algorithm, you need to code 3 key functions: (1) Given a ball B0 = (x0,r0) in S, does there exist (v0,w0) with w0 > 0 such that B(t) is in S for some t > 0? (2) If the answer to (1) is yes, provide some (v0,w0). (3) If the answer to (1) is yes, and given the (v0,w0) from (2), Žnd the largest t for which B(t) is in S. B0 is maximal if and only if the answer to (1) is no. Intuitively it seems like the answer is yes if and only if there exists a closed hemisphere of B0 which does not intersect the surface of S. The algorithm, then, is to loop over asking (1), and if yes, then to perform (2) and (3) to get a B(t) of larger radius, and then to set B0 = B(t). This algorithm gives a sequence of larger and larger spheres. It should terminate in a Žnite number of steps when S is a polyhedron (I think). I don¹t have time to work out the details, but I¹m conŽdent that this idea is a good way to go. Maybe it reduces to applying the simplex method for (David Eppstein¹s formulation of the problem as) a linear program, but this seems more intuitive to me. (Then again, I dislike linear programming.) -Jim | Jim Ferry | http://www.uiuc.edu/ph/www/jferry/ | +-----------------------+ jferry@[delete_this]uiuc.edu | | Center for Simulation | (217) 333-8985 (OfŽce) | | of Advanced Rockets | (217) 333-1910 (Fax) | QUOTE === Subject: Re: Homology of join of spaces Originator: grubb@lola >Is there a formula (exact sequence, spectral sequence...) that >expresses the homology of the join of spaces in terms of the homology >of such spaces? In most cases, the pair (X,Y) is excisive in the join, so you can just use Mayer-Vietoris. --Dan Grubb === Subject: cubic equations I would like to know a bit more about how to Žnd reasonably neat and tidy expressions for the zeroes of third degree polynomials. For the past two or three years I have been fascinated with third degree polynomials and methods of Žnding their zeroes. This all started when, in October 2001, someone posted as follows: Given the polynomial: P(x) = x^3 - x - 1 how can one arrive at its zero of 1.324717957...? The reply given was to use the recurrence relation: x_n+1 = sqrt(1 + x_n) that is directly derived from P(x) = 0. Since reading that reply, I rediscovered Newton¹s method (which converges to the desired result faster than the recurrence relation above). I also discovered what I believe is referred to as Cardan¹s method, which involves deriving an associated quadratic equation, Žnd its two roots, and using them to determine the one real and two complex or reqal zeroes of any cubic of the form P(x) = x^3 - ax - b An interesting feature of this method is that when the cubic has three real zeroes, the quadratic has complex zeroes, and when the quadratic has real zeroes, then the cubic has only one real zero and a pair of complex ones. None of my math books include anything about Žnding zeroes of cubics, and the mathematical landscape over which I travelled in my undergrad days as a physics major was restricted to well-travelled paths like calculus and differential equations. Although I have close at hand the standard analysis tools I generally used as a high-school math & physics teacher, like the rational roots theorem, Descartes¹ Sign Change Test, they are only a beginning, and don¹t really help much. Can the above solution of P(x) = x^3 - x - 1 be expressed as a simple algebraic expression with radicals - possibly one of the form: m + p sqrt(q) ? I am guessing that maybe it is, provided the pair of complex zeroes contains some similarly complicated radical expressions, but I am interested in the perspectives of those who have navigated a lot more mathematical terrain than I have. Bob Lindsay === Subject: Webpage for a dead friend http://www.geocities.com/mswiegandrip === Subject: Re: JSH: Equation has no memory > > : Given, where x is in the ring of algebraic integers, I¹ve shown the > : factorization > > : (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > : 49(300125 x^3 - 18375 x^2 - 360 x + 22) > : where b_3(x) = a_3(x) - 3 and the a¹s are roots of > : a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > : so when x=0, a_1(0) = a_2(0) = b_3(0) = 0. > : Now you can divide both sides by 49. > > : Some posters have claimed that when you divide by 49, what results > : varies depending on what value x has. > : Do you understand? > > I don¹t understand what you mean by the phrase what results. > > Please consider my question in the following step-by-step: > > 1) You get your factorization. I agree. > 2) 49 divides right-hand side. I agree. > 3) 49 divides left-hand side. I agree. > 4) So when x=0 we get > > (7)(7)(22) = 49(22) I agree. > > 5) But when x!=0 it¹s not immediately clear what results from this > equation. Simply because 7 divides the Žrst two terms when x=0 does not > imply anything about when x!=0. Are you suggesting that 7 divides the > Žrst two terms when x!=0? If so, could you please explain why? > > : I¹m curious to know if you will admit understanding now, or if you > : will continue to ask questions, so I ask again: > : Do you understand? > > I will certainly admit understanding after you clear up my question. JSH does not answer questions. Gib