mm-239 Yeah, there's just no way around it. I thought I had a proof of that> result, but now it's more interesting.*Two* iterations? That's *all*? First a claim, then a counterexample(accep without argument), a second claim and then admission of amistake? After *TWO*?Who are you and what === sha1:byvb+Bm2lEcbHQpVesJHCzG0Hro=> |One hallmark of new research is the ability to answer questions that> |others might not have even considered. I've been playing with a> |result that I've mentioned before, but became curious enough to post> |it again, and see if I'm wrong in assuming that what's easy for me to> |prove, is not so easy for contemporary mathematicians using> |contemporary mathematics!> |> |Following from my research on factoring polynomials into> |non-polynomial factors I have a rather simple result that given> |prime integef_1, f_2, and integer M, where> |> |M = f_1 f_2> |> |if you find integers x and y, where> |> |x^j = -y^j mod M> |> |where j is a positive odd integer, and abs(x)>1 and abs(y)>1,> |> |then it must be true that > |> |(x+y) = 0 mod f_1 or> |> |(x+y) = 0 mod f_2.> |> |I'd like to see someone prove that result, if you can.> |> |I consider this a rather easy test.> |> |And to answer another poster who replied in my previous thread, yes, I> |have the proof.> |> |The question here is, do any of you?> f1=7,f2=13,j=3,x=53,y=75Since I'm going around making everyone's reply for them tonight, letme reply for James.Following from my research on factoring polynomials intonon-polynomial factors I have a rather simple result that given primeintegef_1, f_2, and integer M, whereM = f_1 f_2if you find integers x and y, wherex^j = -y^j mod Mwhere j is a positive odd integer, and abs(x)>1 and abs(y)>1 AND x + y < 127 [...] Just curious if any of you are capable of proving this fact. This isa test. Of course, *I* have already proven it and it's trivial withmy methods.-- Many argue that its programmers have turned out shoddy programs, but[their] objective is to make profit, not superlative programs perse. By the profit criterion, Microsoft has been one of the greatestcompanies in the history of this country. -- ADTI === REVISED>where j is a positive odd integer, and abs(x)>1 and abs(y)>1 AND >x + y < 127 [...] I'm afraid it needs a little more refining...11^3 = -6^3 = 57 mod 91, but (11+6) = 3 mod 7 === REVISED <871xpunq84.fsf@phiwumbda.org> Discussion, is a positive odd integer, and abs(x)>1 and abs(y)>1 AND >>x + y < 127 [...] > I'm afraid it needs a little more refining...> 11^3 = -6^3 = 57 mod 91, but (11+6) = 3 mod 7 and (11+6) = 4 mod 13where j is a positive odd integer, and abs(x)>1 and abs(y)>1 AND x + y < 127 AND it's Tuesday [...](Pos Wednesday morning.)-- A set having three members is a single thing wholly constitu byits members but distinct from them. After this, the theologicaldoctrine of the Trinity as 'three in one' should be child's play. --Max Black, _Caveats and === <871xpunq84.fsf@phiwumbda.org> sha1:NJxqIIpZ7lzoGs62olOZOZxW+8I=>where j is a positive odd integer, and abs(x)>1 and abs(y)>1 AND >x + y < 127 [...] >> I'm afraid it needs a little more refining...>> 11^3 = -6^3 = 57 mod 91, but (11+6) = 3 mod 7 and (11+6) = 4 mod 13> where j is a positive odd integer, and abs(x)>1 and abs(y)>1 AND > x + y < 127 AND it's Tuesday [...]> (Pos Wednesday morning.)I missed my guess. The correct answer is...if there doesn't exist integers c and d, such that M - c^3 = d^3.-- Contrariwise, continued Tweedledee, if it was so, it might be, andif it were so, it would be; but as it isn't, it ain't. === Randow WalkWiener, Brownian and Random Walk - are all these essentially 3 differentnames for the same process? And, is this a special case of a white noiseprocess where the outcome === Brownian and Randow Walk> Wiener, Brownian and Random Walk - are all these essentially 3 different> names for the same process? And, is this a special case of a white noise> process say the first two are the same, continuous time, Gaussian,independent increments. But random walk could have discrete time andcould have non-gaussian increments.-- === Brownian and Randow Walk>Wiener, Brownian and Random Walk - are all these essentially 3 different>names for the same process? And, is this a special case of a white noise>process where the outcome has a gaussian distribution?A Wiener process and Brownian motion are the same thing. It is a special type of continuous-time random walk, which is more general. White noise refers to the derivative of Brownian motion: It is an uncountable collection of independent Gaussian r.v.'s.-- === the subject to myself right now. Answer wiener, brownian, andrandom walk are the same processW(t). Gaussian white noise is the formal derivative ofW(t) ie N(t)=dW/dt. Formal because although the paths of the wiener processare continuous they are not differentiable.White noise is said to be delta correla which isn't a good term, itscovariance function is the dirac delta function:E(N(t)N(s))=delta(t-s)E(W(t)W(s))=min(t,s)> Wiener, Brownian and Random Walk - are all these essentially 3 different> names for the same process? And, is this a special case of a white noise> process where the outcome has a === Randow Walk Actually wiener is a special case of brownian though its sometimes calledthat. Brownian refers to solns ofdX=aX+bdW> Just teaching the subject to myself right now. Answer wiener, brownian,and> random walk are the same process> W(t). Gaussian white noise is the formal derivative of> W(t) ie N(t)=dW/dt. Formal because although the paths of the wienerprocess> are continuous they are not differentiable.> White noise is said to be delta correla which isn't a good term, its> covariance function is the dirac delta function:> E(N(t)N(s))=delta(t-s)> E(W(t)W(s))=min(t,s)> Wiener, Brownian and Random Walk - are all these essentially 3 different> names for the same process? And, is this a special case of a white noise> process where the outcome has a gaussian === sha1:hhR1sYpgT7VCc4OELjcd9mNyq6E=> Wouldn't you know, I decided that it was wrong and saw what seems to> be an important condition I'd missed.I *knew* the real JSH wouldn't give up at just two missed guesses. Iknew it.Now, for sure, you've proved the new one. I'm with ya, buddy.-- All intelligent men are cowards. The Chinese are the world's worstfighters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby fighting one gets killed or === Algorithm-MazeBelow is an abstract maze.It is a bit more accessible/understandable and easier than my most recently pos algorithm maze. (For example, its only math is simple addition.)So, perhaps it would be enjoyable for some children to try to solvethe maze.(And I originally intended to make this maze accessible to children,but it ended up being a little harder than I had intended originally,even being too difficult for many *adults* to understand the RULES,let alone to get the solution.) Follow the rules in-order so as determine which path to take,from square (containing a symbol {ie. ascii-character})to adjacent square to adjacent square to...,forming a continuous path as the maze's solution.You must move from square to adjacent square so as toform the maze's solution-path.By adjacent square, I mean a square immediately nextto the square you are currently on,in the direction of above, below, left of, or right of;but *not* diagonal to.(So your pencil, or finger or mind, draws the path as arook moves in Chess.)And, importantly, you must not leave the grid or landon any square more than once!(But not every square is necessarily visi by thesolution's path.)And, if you cannot move to any adjacent square,given the rule associa with the move,then your path has reached a dead-end within the maze. (View with fixed-width font.)-------------------------| # | # | P | U | O | W |----+---+---+---+---+----| R | 3 | 7 | * | D | N |----+---+---+---+---+----| 2 | A | 4 | 3 | % | * |----+---+---+---+---+----| @ | 5 | 8 | + | @ | L |----+---+---+---+---+----| 1 | R | * | T | % | E |----+---+---+---+---+----| 3 | I | G | H | T | F |------------------------- Rules (follow in-order, similar to a computer-program):1) Start in upper-left square.2) Move to adjacent square with a letter.3) Move to adjacent square with a number.4) Move down one square.5) Move to adjacent square with a number.6) Add number in square you are currenty on and the mostimmediately previous number your path crossed before yourcurrent square, and move to adjacent square with this sum.7) Move to an adjacent square with a *.8) Move to adjacent square to its adjacent square to...so as to form a chain of squares, each containing a letter,spelling a word.9) Move ____[word spelled in previous step]___ one square.10) Move up or down the number of squares equal to the numberof [symbols] in the entire maze, where [symbol] is thesymbol in the square currently you are at (which you moved to in previous step).11) Repeat the fifth step (step-5 above) twice, then repeatsteps (6),(7),(8),(9) and (10) once each in-order.(Skip this {11th} step the second time you encounter it.) 12) Which square are you now on, and which symbol is in that square??(This maze === only has one solution, or does it??..)Leroy QuetSubject: Re: trisect a line segment, i.e. split it into> three equal portions, with just straighge and compass?Yes. Draw a ray such that it shares an endpoint (A) with the segment. Makethree points on the ray B, C, and D such that AB = BC = CD. Draw a linefrom D to the other endpoint of the segment (point E). Now draw a linethrough B parallel to DE. Repeat at point C.The details can be filled in by any high school geometry === CongruenceI must clarify that I myself have only proved that, generally,s(n,m) is divisible by (m+2n+1) for EVEN m's.(s(n,m) is as defined below.)So, mostly I am asking unanswered questions below.Leroy> This might be easy, but try to prove (which seems like it might be fun> {in a math-proving way} to me):> For m = EVEN positive integer,> (m+1) always divides> m! * sum{k=1 to m} F(m+1-k) /k ,> where F(n) is the n_th Fibonacci number.> (F(1) = 1, F(2) = 1, F(m+2) = F(m+1) +F(m))> In a private email to me, Robert G Wilson v claims that for many, but> not for all, of the ODD m's as well:(m+1) dividesm! * sum{k=1 to m} F(m+1-k) /k.(I will not say which odd m's...)> So, I guess the more general question is:For which positive integer m's does(m+1) divides(m) = m! * sum{k=1 to m} F(m+1-k) /k ?(I am referring to a mathematical means of determining if any m fits> the criteria in-general, not asking for a list of specific m's under> 100, say, unless the list {of odds, anyway} is finite.)> Or, even better, what is s(m) (mod {m+1}) ?> Or perhaps you would rather show, for which m does:(m+2n+1) divides(n,m) =m! (sum{k=1 to m} F(m+1-k) * H(n,k) ) ,in terms of n?> (H(n,k) is defined below.)> Or most generally, for the grand prize, show:What is s(n,m) (mod {m+2n+1}),for all nonnegative integers n and for all positive integers m?> And in general, for more advanced players...:> Let H(0,m) = 1/m,> for all positive integers m.> Let H(n,m) = sum{k=1 to m} H(n-1,k) ,> for all positive integers n.> Then, for all EVEN positive integers m, > and for all nonnegative integers n, > (m+2n+1) always divides> m! (sum{k=1 to m} F(m+1-k) * H(n,k) ).> For example,> (m+3) always divides, for even m,> m! *sum{k=1 to m} F(m+1-k) H(k),> where H(k) = 1 +1/2 +1/3 +..+ === Prove: Fibonacci-Number Sum Congruence> This might be easy, but try to prove (which seems like it might be fun> {in a math-proving way} to me):> For m = EVEN positive integer,> (m+1) always divides> m! * sum{k=1 to m} F(m+1-k) /k ,> where F(n) is the n_th Fibonacci number.> (F(1) = 1, F(2) = 1, F(m+2) = F(m+1) +F(m))In a private email to me, Robert G Wilson v claims that for many, butnot for all, of the ODD m's as well:(m+1) dividesm! * sum{k=1 to m} F(m+1-k) /k.(I will not say which odd m's...)So, I guess the more general question is:For which positive integer m's does(m+1) divides(m) = m! * sum{k=1 to m} F(m+1-k) /k ?(I am referring to a mathematical means of determining if any m fitsthe criteria in-general, not asking for a list of specific m's under100, say, unless the list {of odds, anyway} is finite.) Or, even better, what is s(m) (mod {m+1}) ?Or perhaps you would rather show, for which m does:(m+2n+1) divides(n,m) =m! (sum{k=1 to m} F(m+1-k) * H(n,k) ) ,in terms of n?(H(n,k) is defined below.)Or most generally, for the grand prize, show:What is s(n,m) (mod {m+2n+1}),for all nonnegative integers n and for all positive integers m?Leroy Quet > And in general, for more advanced players...:Let H(0,m) = 1/m,for all positive integers m.Let H(n,m) = sum{k=1 to m} H(n-1,k) ,for all positive integers n.Then, for all EVEN positive integers m, > and for all nonnegative integers n, > (m+2n+1) always divides> m! (sum{k=1 to m} F(m+1-k) * H(n,k) ).> For example,> (m+3) always divides, for even m,> m! *sum{k=1 to m} F(m+1-k) H(k),> where H(k) = 1 +1/2 +1/3 +..+ 1/k,> the kth harmonic number.> Leroy > === expressed as the sum of three squarestwo different ways, must N be a perfect cube or higher power? If anyhelp is given, so === natural numbeIf N can be expressed as the sum of three squares > two different ways, must N be a perfect cube or higher power? are quite a few numbers that can be expressed asthe sum of two squares in more than one way. 65 is the smallest.Adding any other square to it does not necessarily make it either acube or a higher power. So we have: 69 = 8^2 + 2^2 + 1^2 = 7^2 + 4^2 + 2^2.More striking: 4325 = 31^2 + 40^2 + 42^2 = = 10^2 + 16^2 + 63^2 = = 10^2 + 33^2 + 56^2 = = 6^2 + 8^2 + 65^2.-- === expressed as the sum of three squares>two different ways, must N be a perfect cube or higher power? If any>help is given, so think there are atleast a dozen numbers under 100 with this property (such as N=54) andonly one that's a perfect power (N=81), depending on whether you allowzero as a natural.Representations as sums of three squares are extremely common, so there'sabsolutely no reason to suspect that a number with two representationswould have to belong to such a tiny class as the cubes and higher powers.How ever did you come up === squares>In natural numbeIf N can be expressed as the sum of three squares>two different ways, must N be a perfect cube or counterexamples are very easy to find. I think there are at> least a dozen numbers under 100 with this property (such as N=54) and> only one that's a perfect power (N=81), depending on whether you allow> zero as a natural.Representations as sums of three squares are extremely common, so there's> absolutely no reason to suspect that a number with two representations> would have to belong to such a tiny class as the cubes and higher powers.> How ever did you come up with this hypothesis? -- ErickHello ErickLet me withdraw that previous question, and present a logical onerela thereto.Suppose 1) a+b+c = d+e+f is true.Then, if 2) a=e+f and 3) d=b+c are both true then 1) would be true. right?However, if 2) were true and 3) were false. or 2) were false and 3)were true, then 1) would be false. right?Now finally 4), If both 2) and 3) were false, could 1) still be true?Sure! If they are arithmetic statements about natural numbers(squares) and they are wrong by the same amount. (i.e., The errorscompensate)I don't know about logic statements though. I don't want to thinkabout that case right now.Anyway, my recast problem is: Prove that 2) and 3) can be wrong by thesame amount when they involve squares of a certain specific class ofnumbers.Here again, I can't reveal what that class is without giving === squares>In natural numbeIf N can be expressed as the sum of three squares>two different ways, must N be a perfect cube or counterexamples are very easy to find. I think there are at> least a dozen numbers under 100 with this property (such as N=54) and> only one that's a perfect power (N=81), depending on whether you allow> zero as a natural.Representations as sums of three squares are extremely common, so there's> absolutely no reason to suspect that a number with two representations> would have to belong to such a tiny class as the cubes and higher powers.> How ever did you come up with even though it (my question) wasnaive. And bless you for your patience. I realize now that it (myquestion) was poorly thought out (by me). I guess I was simply toolazy or too tired to think about it further. i.e., it's just easier toask you brilliant fellows. Although I have thought about it a lot inthe past, I had hoped that it (my hyp) was true. I'm sorry that Ican't risk explaining how I came up with it publically, until I'mready to fully disclose what might be a very significant discoveryconnec with it. Some of you are so smart that you might see theconnection with just the slightest clue. It's hard to dislose anythinghere with out giving the farm === aspirations to higher mathematicsMCKAY john> Follow Higher Mathematics in http://www.renyi.hu/:)Budapest and the R.8enyi Institute sound like real fun. And if I ever get togo, I will follow this eminently practical === those with aspirations to higher mathematics Discussion, Higher Mathematics in http://www.renyi.hu/Well, their Higher Mathematics dept. is more entertaining, but theiralgebra logic group has some impressive names.-- Radicals are interesting because they were considered 'radical' bymodern mathematics === those with aspirations to higher mathematics> http://www.renyi.hu/Huncut Mici (and cloned versions) seems === Synchronization Clocks in Relativity> Germaine to this 'dicussion' is the fact that contrary to all claims> that experiments have been performed which showed 'actually' that> light was independent of source motion, apparently only one has been> confirmed, AND TURNS OUT TO BE A CROCK!admitWhat on earth are you talking about?> How does one run a partcle accelerator up to speed?accellera to high speed-- OK now?, and they WERE travelling at c...........sounds good to> DHR's!What does that mean?> Could you please elaborate?Turs out, did it not, that the photon speed was not reallymeasured-- just an increased energy no which was not allowed toan increased velocity.> What they forgot to mention was that the SOURCE, at the precise> moment of emission of photons, STOPPED!!the pi0?> If not, please elucidate what you are talking about.> Please do so to reassure us that you are not just regurgitating some garbled> piece of half a story you heard somewhere.Might have been a caesium a for all I know!What I strongly suspect, is that the photon emit, and the remnantvelocity after the decay!If it is the pi0 decay experiment of which you speak, the following apply:They were most certainly not travelling at zero speed when they decayed.> They decayed in flight. This is borne out by the fact that the forward> going photons were shif up in energy in full accordance with the> relativistic Doppler effect and the backward going photons were shif> downwards in energy by the expec amount, for the same reason.> The experiment is a fraud (perhaps unknowingly)----the source was NOT> moving when the light was emit.You have said this twice now.> I have showed you why you are ill-informed. (just as a photon is not moving at> the surface of a mirrorThere is no analogy of any description whatsoever between the experiment> which I saw being conduc and a photon moving or not moving at the surface> of a mirror. Perhaps I have the wrong experiment in mind. That could be> cleared up easily by you telling us what the actual experiment is of which y> ou are speaking, and where it was done.> ....and that is ALL DHR's have to cling to.On the contrary, that experiment is but a tiny weapon in the armoury of> physicists. The really abundant evidence in favour of SR comes from the> designed in accordance with the principles of relativistic mechanics and> kinematics as accurately as can be measured.> All the rest involve> tweaked clocks and wavelength.Twaddle.Franz HeymannYou have no comprehension of what _causes_ Doppler. It is ALWAYS theresult of a past accelleration (decelleration). So if a Doppler shiftis observed, then the photon suffered said accelleration DUE TO theMOTION of its SOURCE.No comparison was done between the photon and source (remaining bits )right? The result just another self-serving the theory === Synchronization Clocks in Relativity> Germaine to this 'dicussion' is the fact that contrary to all claims>that experiments have been performed which showed 'actually' that>light was independent of source motion, apparently only one has been>confirmed, AND TURNS OUT TO BE A CROCK!>>admitWhat on earth are you talking about?> How does one run a partcle accelerator up to speed?> accellera to high speed-- OK now?, and they WERE travelling at c...........sounds good to>DHR's!What does that mean?> Could you please elaborate?> Turs out, did it not, that the photon speed was not really> measured-- just an increased energy no which was not allowed to> an increased velocity.You have quite obviously not read anything about that expariment. The timesbetween the birth and the death of the two photons were measured withextreme precision. These times agreed with the assertion that the speed ofthe photons were independent of the speed of the source, in spite of thefact that the source moved at around 99% of the speed of light.Additionally, the energies of the forward and backward photons weremeasured. These turned out to be in agreement with what was expec as aresult of the relativistic Doppler effect.I do sincerely recommend you to familiarise yourself with an experimentbefore you waffle about it. Please remember that there will be many readerswho are familiar with the experiment, and there might even be the occasionalone who was present when the experiment was done.>>What they forgot to mention was that the SOURCE, at the precise>moment of emission of photons, STOPPED!!Are you by any chance thinking of the experiments on the decay> the pi0?> If not, please elucidate what you are talking about.> Please do so to reassure us that you are not just regurgitating somegarbled> piece of half a story you heard somewhere.> Might have been a caesium a for all I know!It was not a caesuim a, so you do not know what you were talking about.> What I strongly suspect, is that the photon emit, and the remnant> velocity after the decay!Your srtong suspicion is unfounded. There was no remnant of whateverphotons themselves.If it is the pi0 decay experiment of which you speak, the followingapply:They were most certainly not travelling at zero speed when they decayed.> They decayed in flight. This is borne out by the fact that the forward> going photons were shif up in energy in full accordance with the> relativistic Doppler effect and the backward going photons were shif> downwards in energy by the expec amount, for the same reason.>The experiment is a fraud (perhaps unknowingly)----the source was NOT>moving when the light was emit.You have said this twice now.> I have showed you why you are ill-informed. (just as a photon is not moving at>the surface of a mirrorThere is no analogy of any description whatsoever between the experiment> which I saw being conduc and a photon moving or not moving at thesurface> of a mirror. Perhaps I have the wrong experiment in mind. That couldbe> cleared up easily by you telling us what the actual experiment is ofwhich y> ou are speaking, and where it was done.>.and that is ALL DHR's have to cling to.On the contrary, that experiment is but a tiny weapon in the armoury of> physicists. The really abundant evidence in favour of SR comes from the> designed in accordance with the principles of relativistic mechanics and> kinematics as accurately as can be measured.>>All the rest involve>tweaked clocks and wavelength.Twaddle.Franz Heymann> You have no comprehension of what _causes_ Doppler. It is ALWAYS the> result of a past accelleration (decelleration). So if a Doppler shift> is observed, then the photon suffered said accelleration DUE TO the> MOTION of its SOURCE.On the contrary. I know quite precisely how the Doppler effect occurs. Inthe experiment under consideration, it arose from the fact that the sourcewa smoving a a speed very close to that of light at the moment when the twophotons were born.> No comparison was done between the photon and source (remaining bits )> right?There were no remaining bits> The result just another self-serving the theory proves the> theoryYou have made it quite plain in your note that you do not have the foggiestidea of what the experiment of which you are talking, so that lastsentence of yours is totally without foundation, and merely expresseswishful thinking on your part.It was actually a most beautiful observation which convincingly vindicathe sourceindependence of the speed of photons, and as a bonus confirmed that therelativistic expression for the Doppler shift is correct, even at speeds inexcess of 99% of that of light.Sorry chum, you don't have a leg to stand === in Relativity...>>No, the light approaches at c relative to the mirror so will>>either be reflec at c or reset to c however you want to>>formulate the 'source dependency' part. There is no speed>>change relative to the mirrors or fibre (always c) or the>>lab (always c+v).>> Nah.>> ^ very small upward velocity>> A>> /------------> |>> | <---c+v-dv>> |>> -------------c+v-2dv-> C>> B>>The velocity of the mirrors is parallel to their surfaces>(I can't think of a way to add arrowheads but suppose the>table is rotating anticlockwise in your diagram) and also>applies to the source 'S' and detector 'C'. If the horizontal>component of v at S is v' then the light is moving at c+v'>to the left. The horizontal component at A is also v' so>it approaches at c relative. The vertical component at A>is also v' so the light moves down at c+v' and so on:v / A c+v' v> / /---------- |> |> | <---c+v'> |> |> ---------->C /> v B c+v' / v>>The other way round it is c-v' all the way.> You are ignoring the fact that the mirros rotate slightly while the lightis in> transit.It is taken into account in the calculation of v'. Forthe first leg from S to A, if the table were not rotatingthe angles of the mirror 'a' would be 45 degrees. Thesloping line is the surface of the mirror and also thedirection of motion: / /| / | /a | *<------------- /The source is moving in the same manner as the mirrorsbut the sloping line is the direction of motion of thesource: | | | a <-------------* Because the table is rotating the angle 'a' is increasedslightly. The drawings show the path of the light ashorizontal to make it easier to show the solution so youneed to think of light emit a little before the sourcereaches the 45 degree point.The vertical components would be v * sin(a) for both andRitzian theory says the light is emit at c relative tothe source. We are interes in light travelling alongthe S-A line so . c .' | v .' | .'b | a c + v' <-------------* c * sin(b) = v * sin(a)and c + v' = v * cos(a) + c * cos(b)The important point to note though is that a symmetricalpicture applies at the other end. though it's harder todraw. The light arrives at c+v' and call the angles a'and b'. By symmetry, a' = a and c * sin(b') = v * sin(a')so b' = bIf the speed of the light relative to the mirror is c',then c' * cos(b') = (c + v') - v * cos(a')and you should see that c' = c> The vertical component of A is not quite v'.The vertical component at S is v*sin(a) while at A itis -v*sin(a). The horizontal component adds to thespeed of the light at s and subtracts at A and they areequal.>> The rays SA and BC are not quite parallel and this causes the fringe>movement..v / A S' v> / /<----------S > | ^> | |> | |> | |> v |> ---------->/ /> v B C / v>>If the light is going the same way as the table, each angle>has to be slightly less than 90 degrees so that the light>gets to S' instead of S because S will have moved.>>Now consider the light going the other way round the table:v / A S' v> / /---------->S > ^ |> | |> | |> | |> | v> ---------->/ /> v B C / v>>This time the angles each have to be slightly more than>90 degrees for the light to reach S'.>>Since one angle increases and the other decreases, put the>two together to create fringes and the two beams meet at>exactly 90 degrees regardless of the motion of the table.>The only thing that affects the fringes is the time delay>and in Ritzian theory, there is none since the change of>speed in the lab frame exactly matches the change of path>length.> No! Every time there is a reflection from a (moving) 45 mirror, the lightgets> an extra kick in the 'forward' direction or is retarded slightly in the> 'backward' one.That would be true if the source was static since thelight would arrive at the mirror at a relative speed ofc-v' and be re-emit at c, but the source is moving inthe Sagnac experiment and when the horizontal componentsare equal, the relative incident speed is c so there isno change.> In Ritzian theory, light gains a velocity when reflec from a movingmirror.> If that is not well known, it is now - because I said it.Sure, but you forgot the source was moving.>This is obvious in the rotating frame where the speed is>always c and the path length is always the circumference>so the times taken are always equal.> Using the rotating frame is confusing. For instance, light doesn't move in> straight lines.True, but the curvature is constant so the change of angleat each end of the path is Light speed is NOT always c in this instance for the reasons I have justgiven.>>Consider the replacing the usual sine wave of the light>>with a narrow pulse waveform, the effect of the phase>>change is then obvious.>> In the c+v model, the two pulses would arrive at different times and be>> slightly displaced sideways. That's what happens.>>No, the pulses would arrive at the same time. Sideways>displacement is parallel to the wavefront so doesn't move>the fringes.> But the displacement is in opposite directions for the two beams.Draw out the paths, you'll find it is the same. It's a === Synchronization Clocks in RelativityExpires: 28 days>> You are ignoring the fact that the mirros rotate slightly while the light>is in>> transit.>It is taken into account in the calculation of v'. For>the first leg from S to A, if the table were not rotating>the angles of the mirror 'a' would be 45 degrees. The>sloping line is the surface of the mirror and also the>direction of motion:> /> /|> / |> /a |> *<-------------> />The source is moving in the same manner as the mirrors>but the sloping line is the direction of motion of the>source:> |> | > | a> <-------------*>Because the table is rotating the angle 'a' is increased>slightly. The drawings show the path of the light as>horizontal to make it easier to show the solution so you>need to think of light emit a little before the source>reaches the 45 degree point.>The vertical components would be v * sin(a) for both and>Ritzian theory says the light is emit at c relative to>the source. We are interes in light travelling along>the S-A line so> .> c .' | v> .' | > .'b | a> c + v' <-------------*> c * sin(b) = v * sin(a)>and> c + v' = v * cos(a) + c * cos(b)>The important point to note though is that a symmetrical>picture applies at the other end. though it's harder to>draw. The light arrives at c+v' and call the angles a'>and b'. By symmetry, a' = a and> c * sin(b') = v * sin(a')>so> b' = b>If the speed of the light relative to the mirror is c',>then> c' * cos(b') = (c + v') - v * cos(a')>and you should see that c' = c>> The vertical component of A is not quite v'.>The vertical component at S is v*sin(a) while at A it>is -v*sin(a). The horizontal component adds to the>speed of the light at s and subtracts at A and they are>equal.I'm afraid I find all that a bit hard to follow. It is too difficult to drawsuch things properly here. > The rays SA and BC are not quite parallel and this causes the fringe>>movement..>> v / A S' v>> / /<----------S >> | ^>> | |>> | |>> | |>> v |>> ---------->/ />> v B C / v>>If the light is going the same way as the table, each angle>>has to be slightly less than 90 degrees so that the light>>gets to S' instead of S because S will have moved.>>Now consider the light going the other way round the table:>> v / A S' v>> / /---------->S >> ^ |>> | |>> | |>> | |>> | v>> ---------->/ />> v B C / v>>This time the angles each have to be slightly more than>>90 degrees for the light to reach S'.>>Since one angle increases and the other decreases, put the>>two together to create fringes and the two beams meet at>>exactly 90 degrees regardless of the motion of the table.>>The only thing that affects the fringes is the time delay>>and in Ritzian theory, there is none since the change of>>speed in the lab frame exactly matches the change of path>>length.>> No! Every time there is a reflection from a (moving) 45 mirror, the light>gets>> an extra kick in the 'forward' direction or is retarded slightly in the>> 'backward' one.>That would be true if the source was static since the>light would arrive at the mirror at a relative speed of>c-v' and be re-emit at c, but the source is moving in>the Sagnac experiment and when the horizontal components>are equal, the relative incident speed is c so there is>no change.The source is rotating. Rotation is not a feature of source dependency. A rayof light from a rotating source will appear to follow a curved path.In the sagnac, the curvature is opposite for the rays traveling in oppositedorections. Thus they will be displaced sideways in opposite directions whenthe meet and cause movement of the fringes.>> In Ritzian theory, light gains a velocity when reflec from a moving>mirror.>> If that is not well known, it is now - because I said it.>Sure, but you forgot the source was moving.Moving AND rotating.>>This is obvious in the rotating frame where the speed is>>always c and the path length is always the circumference>>so the times taken are always equal.>> Using the rotating frame is confusing. For instance, light doesn't move in>> straight lines.>True, but the curvature is constant so the change of angle>at each end of the path is the same, is opposite for the two different rays. You are missing thispoint.>> Light speed is NOT always c in this instance for the reasons I have just>given.>Consider the replacing the usual sine wave of the light>with a narrow pulse waveform, the effect of the phase>change is then obvious.In the c+v model, the two pulses would arrive at different times and be> slightly displaced sideways. That's what happens.>>No, the pulses would arrive at the same time. Sideways>>displacement is parallel to the wavefront so doesn't move>>the fringes.>> But the displacement is in opposite directions for the two beams.>Draw out the paths, you'll find it is the same. It's a bit>surprising.The path length might be the same but the angle of incidence at the eyepiece isdifferent - as is the sideways displacement.>GeorgeHenri Wilson. === <1074713504.27040.0@despina.uk.clara.net>[snip my diagrams]> c * sin(b) = v * sin(a)>>and c + v' = v * cos(a) + c * cos(b)>>The important point to note though is that a symmetrical>picture applies at the other end. though it's harder to>draw. The light arrives at c+v' and call the angles a'>and b'. By symmetry, a' = a and c * sin(b') = v * sin(a')>>so b' = b>>If the speed of the light relative to the mirror is c',>then c' * cos(b') = (c + v') - v * cos(a')>>and you should see that c' = c> The vertical component of A is not quite v'.>>The vertical component at S is v*sin(a) while at A it>is -v*sin(a). The horizontal component adds to the>speed of the light at s and subtracts at A and they are>equal.> I'm afraid I find all that a bit hard to follow. It is too difficult todraw> such things properly here.I understand, I tried to do a sketch for the a' and b' anglesand it was hopeless, you can't get the resolution at all. Thekey point is simply that the motions are symetrical so lightincident on the mirror is then exactly c. If I get time I'lldraw it with Paint later.>> No! Every time there is a reflection from a (moving) 45 mirror, thelight>gets>> an extra kick in the 'forward' direction or is retarded slightly in the>> 'backward' one.>>That would be true if the source was static since the>light would arrive at the mirror at a relative speed of>c-v' and be re-emit at c, but the source is moving in>the Sagnac experiment and when the horizontal components>are equal, the relative incident speed is c so there is>no change.> The source is rotating.The source is both rotating and moving. Rotation does notaffect the launch speed but the movement does.> Rotation is not a feature of source dependency. A ray> of light from a rotating source will appear to follow a curved path.> In the sagnac, the curvature is opposite for the rays traveling inopposite> dorections.If the table is rotating anti-clockwise, the apparentcurvature of the paths in the rotating frame will beclockwise for both rays but you need to be careful whichway you draw it because the rays progress round the tablein opposite directions. As a result, one ray gets nearerto the centre when half way between the mirrors while theother gets further away. This didn't take long:http://www.briar.demon.co.uk/Henri/paths.gif> Thus they will be displaced sideways in opposite directions when> the meet and cause movement of the fringes.Sideways displacement doesn't cause fringe shift since youare moving parallel to the wavefront.>> In Ritzian theory, light gains a velocity when reflec from a movingmirror.>> If that is not well known, it is now - because I said it.>>Sure, but you forgot the source was moving.> Moving AND rotating.Yes, but it is the moving part that affects the speed andyour diagram even shows the light from the source movingat c+v, but you seem to have forgotten the effect of the+v when considering the relative speed of the lightincident on mirror A.>> Using the rotating frame is confusing. For instance, light doesn't movein>> straight lines.>>True, but the curvature is constant so the change of angle>at each end of the path is the same, hence the opposite for the two different rays. You are missing this> point.See the diagram. Curvature alters the path length as v^2(the path is shortest when v=0) and increases both pathsequally so does not produce a path length difference. Itis easiest to see in the fibre case which as you said isequivalent to an infinite series of mirrors so both pathsare just the circumference.>> But the displacement is in opposite directions for the two beams.>>Draw out the paths, you'll find it is the same. It's a bit>surprising.> The path length might be the same but the angle of incidence at theeyepiece is> different - as is the sideways displacement.Fringes are cused by the relative angle between the beamswhich stays the same as both rays rotate the same way. Thebeam has to be wide enough to fall on the eyepiece sosideways movement doesn't have an effect, the actual shiftis very small and much less than the === Synchronization Clocks in RelativityExpires: 28 days>> c' * cos(b') = (c + v') - v * cos(a')>>and you should see that c' = c>The vertical component of A is not quite v'.>>The vertical component at S is v*sin(a) while at A it>>is -v*sin(a). The horizontal component adds to the>>speed of the light at s and subtracts at A and they are>>equal.>> I'm afraid I find all that a bit hard to follow. It is too difficult to>draw>> such things properly here.>I understand, I tried to do a sketch for the a' and b' angles>and it was hopeless, you can't get the resolution at all. The>key point is simply that the motions are symetrical so light>incident on the mirror is then exactly c. If I get time I'll>draw it with Paint later.Yes. Putting complica diagrams on a webpage is the only satisfactory way.Trouble is it all takes TIME!> No! Every time there is a reflection from a (moving) 45 mirror, the>light>>gets> an extra kick in the 'forward' direction or is retarded slightly in the> 'backward' one.>>That would be true if the source was static since the>>light would arrive at the mirror at a relative speed of>>c-v' and be re-emit at c, but the source is moving in>>the Sagnac experiment and when the horizontal components>>are equal, the relative incident speed is c so there is>>no change.>> The source is rotating.>The source is both rotating and moving. Rotation does not>affect the launch speed but the movement does.>> Rotation is not a feature of source dependency. A ray>> of light from a rotating source will appear to follow a curved path.>> In the sagnac, the curvature is opposite for the rays traveling in>opposite>> dorections.>If the table is rotating anti-clockwise, the apparent>curvature of the paths in the rotating frame will be>clockwise for both rays but you need to be careful which>way you draw it because the rays progress round the table>in opposite directions. As a result, one ray gets nearer>to the centre when half way between the mirrors while the>other gets further away. This didn't take long:>http://www.briar.demon.co.uk/Henri/paths.gifThat's good.Two questions: How can you assume that both rays hit the same point on each mirror?Like I said, the apparatus rotates slightly while the light is in transit.Therefore the point where each ray hits will vary with rotation speed.Also, how can you say that the curves are identical and circular? I don't thinkthat is right.>> Thus they will be displaced sideways in opposite directions when>> the meet and cause movement of the fringes.>Sideways displacement doesn't cause fringe shift since you>are moving parallel to the wavefront.I'm sure it will have some kind of effect - the angle between the two beamsvaries with rotational speed.> In Ritzian theory, light gains a velocity when reflec from a moving>mirror.> If that is not well known, it is now - because I said it.>>Sure, but you forgot the source was moving.>> Moving AND rotating.>Yes, but it is the moving part that affects the speed and>your diagram even shows the light from the source moving>at c+v, but you seem to have forgotten the effect of the>+v when considering the relative speed of the light>incident on mirror A.Actually I am reaching the conclusion that source dependency or not make littledifference to the operation of a sagnac. The travel time of the light betweenmirrors is barely affec by c+v when v is extremely small. The standard analysis for source dependency in a ring gyro, using a rotatingframe, is wrong.It should treat the device as a 'four mirror interferometer with an infinitenumber of mirrors' rather than a fibre optic internal reflection thing - if youknow what I mean.> Using the rotating frame is confusing. For instance, light doesn't move>in> straight lines.>>True, but the curvature is constant so the change of angle>>at each end of the path is the same, hence the opposite for the two different rays. You are missing this>> point.>See the diagram. Curvature alters the path length as v^2>(the path is shortest when v=0) and increases both paths>equally so does not produce a path length difference. It>is easiest to see in the fibre case which as you said is>equivalent to an infinite series of mirrors so both paths>are just the circumference.But you have assumed both rays strike the mirrors at the same points. That isplainly wrong.> But the displacement is in opposite directions for the two beams.>>Draw out the paths, you'll find it is the same. It's a bit>>surprising.>> The path length might be the same but the angle of incidence at the>eyepiece is>> different - as is the sideways displacement.>Fringes are cused by the relative angle between the beams>which stays the same as both rays rotate the same way. The>beam has to be wide enough to fall on the eyepiece so>sideways movement doesn't have an effect, the actual shift>is very small and much less than the size of the beam.but the sideways movement is also indicative of an angular difference betweenthe two beams.>GeorgeHenri Wilson. === of Synchronization Clocks in Relativity>If it is the pi0 decay experiment of which you speak, the following apply:>>They were most certainly not travelling at zero speed when they decayed.>They decayed in flight. This is borne out by the fact that the forward>going photons were shif up in energy in full accordance with the>relativistic Doppler effect and the backward going photons were shif>downwards in energy by the expec amount, for the same reason.> Nah! The doppler shifts would have been in accordance with source dependency if> the pions hadn't stopped first..Another great demo.All experiments MUST support Henry Wilson's idea of reality,e.g.. source === Synchronization Clocks in RelativityExpires: 28 days>>If it is the pi0 decay experiment of which you speak, the following apply:>>They were most certainly not travelling at zero speed when they decayed.>>They decayed in flight. This is borne out by the fact that the forward>>going photons were shif up in energy in full accordance with the>>relativistic Doppler effect and the backward going photons were shif>>downwards in energy by the expec amount, for the same reason.>> Nah! The doppler shifts would have been in accordance with source dependency if>> the pions hadn't stopped first..>Another great demo.>All experiments MUST support Henry Wilson's idea of reality,>e.g.. source dependency.>Right?Well, Paul, Einstein obviously believed it was a fact of life.But until somebody directly measures the speed of light from a relativelymoving source I suppose you remain free to argue.Can I ask why hasn't an OWLS experiment been carried out when it is quitefeasible nowadays?>PaulHenri === Final Rout of Synchronization Clocks in Relativity>If it is the pi0 decay experiment of which you speak, the following apply:>>They were most certainly not travelling at zero speed when they decayed.>>They decayed in flight. This is borne out by the fact that the forward>>going photons were shif up in energy in full accordance with the>>relativistic Doppler effect and the backward going photons were shif>>downwards in energy by the expec amount, for the same reason.>> Nah! The doppler shifts would have been in accordance with source dependency if>> the pions hadn't stopped first..>>Another great demo.>>All experiments MUST support Henry Wilson's idea of reality,>e.g.. source dependency.>>Right?> Well, Paul, Einstein obviously believed it was a fact of life.I think this suffice, === in RelativityExpires: 28 days>>If it is the pi0 decay experiment of which you speak, the following apply:>>They were most certainly not travelling at zero speed when they decayed.>They decayed in flight. This is borne out by the fact that the forward>going photons were shif up in energy in full accordance with the>relativistic Doppler effect and the backward going photons were shif>downwards in energy by the expec amount, for the same reason.Nah! The doppler shifts would have been in accordance with source dependency if> the pions hadn't stopped first..>>Another great demo.>>All experiments MUST support Henry Wilson's idea of reality,>>e.g.. source dependency.>>Right?>> Well, Paul, Einstein obviously believed it was a fact of life.>I think this suffice, Henry.>PaulEinstein firmly believed that light moved at c relative to its source. Is thatnot a fact Paul?Henri Wilson. === of Synchronization Clocks in Relativity>> Paul B. Andersen.>>It is an indisputable fact that the MMX falsifies Michelsons ether theory.>>That's why it isn't dispu.>> The reason it isn't dispu is that its NULL result was easily manipulable to>> give the answer that the physics mafia wan.>>So in 1886 Michelson was part of a physics mafia that wan>to falsify the ether to prove that the establishment were wrong.>So that's why they manipula the highly unwan NULL result>to give the answer they didn't want, but which the mafia wan.>Right?> Wrong Paul. Physics was an honourable profession in the 19th century.Really? No mafia? No manipulation?What was you then talking about?:>> Null results prove very litle if anything.>> They usually reveal a flaw in the experiment.>> My demo shows that flaw. IF the michelson aether exis, then the 45 mirror>> would not remain at 45 after its length was contrac. The cross beam would be>> bent backwards as shown, thus exactly compensating for the presumed 'diagonal>> effect'..>>AH! Length contraction in Michelson's ether!.>So THAT's what your demo now shows.> Where is the joke? In aether theories, lengths DO physically contract. If the> 45 mirror contracts in the parallel direction, it obviously ends up sloping> greater than 45.> Even though my demo was not based on that fact, it clearly shows that it> happens.>>You are SO funny, Henry!> If one length contracts, then ALL lengths must contract. eh? Where is the joke?The joke is your ignorance.>>Every experiment MUST support [my idea of] reality, ie, source dependency.>>Thus even the experiments falsifying source dependency MUST support it!>>So there!>> There are NO experiments that falsify source dependency. De Sitter was wrong>> and there are no others.>>Every experiment MUST support reality, ie, source dependency>So the experiments that falisfies source dependency>simply CANNOT exist.>So there!> There is not one believeable bit of experimental evidence that === Rout of Synchronization Clocks in RelativityExpires: 28 days>> Germaine to this 'dicussion' is the fact that contrary to all claims>> that experiments have been performed which showed 'actually' that>> light was independent of source motion, apparently only one has been>> confirmed, AND TURNS OUT TO BE A CROCK!>> admit>What on earth are you talking about?>How does one run a partcle accelerator up to speed?>, and they WERE travelling at c...........sounds good to>> DHR's!>What does that mean?>Could you please elaborate?>> What they forgot to mention was that the SOURCE, at the precise>> moment of emission of photons, STOPPED!!>the pi0?>If not, please elucidate what you are talking about.>Please do so to reassure us that you are not just regurgitating some garbled>piece of half a story you heard somewhere.>If it is the pi0 decay experiment of which you speak, the following apply:>They were most certainly not travelling at zero speed when they decayed.>They decayed in flight. This is borne out by the fact that the forward>going photons were shif up in energy in full accordance with the>relativistic Doppler effect and the backward going photons were shif>downwards in energy by the expec amount, for the same reason.Nah! The doppler shifts would have been in accordance with source dependency ifthe pions hadn't stopped first..>> The experiment is a fraud (perhaps unknowingly)----the source was NOT>> moving when the light was emit.>You have said this twice now.>I have showed you why you are ill-informed.> (just as a photon is not moving at>> the surface of a mirror>There is no analogy of any description whatsoever between the experiment>which I saw being conduc and a photon moving or not moving at the surface>of a mirror. Perhaps I have the wrong experiment in mind. That could be>cleared up easily by you telling us what the actual experiment is of which y>ou are speaking, and where it was done.>> ....and that is ALL DHR's have to cling to.>On the contrary, that experiment is but a tiny weapon in the armoury of>physicists. The really abundant evidence in favour of SR comes from the>designed in accordance with the principles of relativistic mechanics and>kinematics as accurately as can be measured.>> All the rest involve>> tweaked clocks and wavelength.>Twaddle.>Franz HeymannHave you ever wondered why every supposed SR supportive experiment iscompletely shonky? Henri Wilson. === Egyptian-Fraction Expansions Of REALSAn obvious idea I have not seen *much* about before (which meansnothing...).If we have a positive REAL, possibly irrational, x,then we can find a sequence (many many sequences..., I bet) ofpositive integers{n(k)}such thatsum{k=1 to M} 1/n(k) = x,where M is infinity if x is irrational.For instance, after dealing with the integer-part differently, we canapply the Greedy-Algorithm to x.x = pi, as an example:pi = 3 + 1/8 + 1/61 +...Or x = (sqrt(5)+1)/2 = phi:phi = 1 + 1/2 + 1/9 + 1/145 +...(I am FAR from certain about the last terms in the above twoexpansions, since my calculator is low-precision.) We might want to, in order to be strict with our definition ofEgyptian-Fraction, rewrite the 3 in the pi-expansion as 1+1+1, to get:pi = 1 + 1 + 1 + 1/8 + 1/61(?) +...In any case, I do not believe these sequences are in the Encyclopediaof Integer Sequences, although this idea seems basic to me.I also wonder about alternative EF-expansions, both with all positiveterms and with the possibility of having negative terms.For instance, aside from:pi^2/6 = sum{k=1 to oo} 1/k^2,what other expansions exist?We could consider:pi^2/6 = sum{k=1 to oo} (4/3)/(2k-1)^2,but 4 does not divide (2k-1)^2, so we can rewrite this as:pi^2/6 =1/3 + 1/3 + 1/3 + 1/3 +1/27 + 1/27 + 1/27 + 1/27 +1/75 + 1/75 + 1/75 +1/75 +... So, therefore, if duplicate terms are allowed, then it is easy to seethat it is possible for an x to have multiple expansions.Anything more to be said about EFs of reals??(I am sure there *must* be more to === Expansions Of REALS>An obvious idea I have not seen *much* about before (which means>nothing...).>If we have a positive REAL, possibly irrational, x,>then we can find a sequence (many many sequences..., I bet) of>positive integers>{n(k)}>such that>sum{k=1 to M} 1/n(k) = x,>where M is infinity if x is irrational.Any positive real has continuum-many such strictly increasing sequences.Given positive real x and any sequence (d_j) of 1's and 2's, wecan construct a sequence (n_j) as follows. Let n_1 = ceil(1/x) + d_1,y_1 = x - 1/n_1, and for j > 1 given n_{j-1} and y_{j-1} letn_j = max(n_{j-1}, ceil(1/y_{j-1})) + d_j and y_j = y_{j-1} - 1/n_j.By construction, different sequences (d_j) produce different sequences (n_j). We have 0 < y_j < y_{j-1}. For any epsilon > 0, while y_j > epsilon we have n_j <= n_{j-1} + 1/epsilon + 3, and using the divergence of the harmonic series we must have eventually y_j < epsilon. So y_j -> 0 as j -> infinity, and thus sum_{j=1}^infinity 1/n_j = x. >For instance, after dealing with the integer-part differently, we can>apply the Greedy-Algorithm to x.>x = pi, as an example:>pi = 3 + 1/8 + 1/61 +...Or if you want all distinct fractions, 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/11+1/12+1/27+1/744+ 1/1173268+1/2586625801171+1/14348276635209672362238685+... >Or x = (sqrt(5)+1)/2 = phi:>phi = 1 + 1/2 + 1/9 + 1/145 +...1/1+1/2+1/9+1/145+1/37986+1/2345721887+1/ 26943815937041299094+1/811625643619814151937413504618770581764 +1/ 697120590223140234675813998970770820981012350673738243594006422 610850113672220+1/ 350525505629066374832579593963297873487088311690765674601194396 099875615389311387408232126549737069539300093331653997598972297 3498240764569878009748700790482+...Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia === Expansions Of REALS> If we have a positive REAL, possibly irrational, x, then we can find > a sequence (many many sequences..., I bet) of positive integers > {n(k)} such thatsum{k=1 to M} 1/n(k) = x,where M is infinity if x is irrational.x = pi, as an example:pi = 3 + 1/8 + 1/61 +...Or x = (sqrt(5)+1)/2 = phi:phi = 1 + 1/2 + 1/9 + 1/145 +...We might want to, in order to be strict with our definition of> Egyptian-Fraction, rewrite the 3 in the pi-expansion as 1+1+1, > to get:pi = 1 + 1 + 1 + 1/8 + 1/61(?) +...In any case, I do not believe these sequences are in the Encyclopedia> of Integer SequencesWell, something along these lines is. Go to the search page and instead of telling it to search for a particular sequence of numbers tell it to search for the word Egyptian. You'll get a long list of stuff, including %N A014015 Alternating Egyptian fraction expansion of e-2.%N A006525 Denominator of Egyptian fraction for e-2.%N A069139 Egyptian fraction for square root of 1/2.%N A069261 Denominators of the Egyptian fraction for Feigenbaum's constant, 4.6692...%N A006487 Egyptian fraction for square root of 2.%N A006526 Egyptian fraction for 1/e.%N A006524 Egyptian fraction for 1/ pi.%N A014013 Alternating Egyptian fraction expansion of pi.%N A001466 Denominator of Egyptian === Expansions Of REALS> If we have a positive REAL, possibly irrational, x,then we can find a sequence (many many sequences..., I bet) of> positive integers> {n(k)}> such thatsum{k=1 to M} 1/n(k) = x,where M is infinity if x is irrational.> I also wonder about alternative EF-expansions, both with all positive> terms and with the possibility of having negative terms.The difficulty is that there are at least a countable infinity ofthese expansions for any x. This makes many questions about them notvery interesting.This follows from the fact that the sum of 1/n diverges, starting fromany particular term. So to create an expansion for x, all we need todo is start with any particular sum of 1/n's that does not exceed x.Then use the greedy algorithm to append the appropriate further 1/n'sto cause the limit of the partial sums to be x.There are an uncountable infinity of subsets of sum{n} 1/n. I wonderif there is any real that is the sum of an *uncountable* number ofsubsets? But I suppose that any real must be, since while using thegreedy algorithm, we can exercise a countably infinite number ofchoices to skip over the next 1/n that could be used, and use the === REALS> An obvious idea I have not seen *much* about before (which means> nothing...).> If we have a positive REAL, possibly irrational, x,> then we can find a sequence (many many sequences..., I bet) of> positive integers> {n(k)}> such that> sum{k=1 to M} 1/n(k) = x,> where M is infinity if x is irrational.Since sum{n=1 to Infinity} 1/n diverges, it seems that there are an infinitenumber of such sequences. Another way to think of your sequence is toredefine it as the sequence b_k such that:sum{k=1 to M} b_k/n = x, where b_k in {0, 1}.Then for each integer n and irrational x, there is such a sequence with b_i= 0 for i < n. I think it follows (with some additional arguing) that thereare infinitely many such representations for x.Brian (too rusty to formalize === = 0 0 1 1 0 0 0 1 1 1. show that matrix A is diagonalizable or not diagonalizable on complex(C)2. find that nonsingular matrix P in Mat3(C) satisfied next(P^-1) A P = 1 0 0 0 w 0 0 0 w^2 ( 2w = -1 + (root3)i ) ------------------------------ i have a doubti find eigenvalue of this matrixit's 1, w = {-1+(root3)i}/2 , w^2 = {-1-(root3)i}/2thusi find minimum polynomial=p(x).because, p(x)=(x-1)(x-w)(x-w^2) <=> diagonalizable but p(A) is not zerobesides i fined eigenspace.but i think that all eigenspase is (0,0,0)thusi can't solve this problem.i am anxious about your thinking of === algebra problem......king-girl> let matrix> A => 0 0 1> 1 0 0> 0 1 1> 1. show that matrix A is diagonalizable or not diagonalizable oncomplex(C)> 2. find that nonsingular matrix P in Mat3(C) satisfied next> (P^-1) A P => 1 0 0> 0 w 0> 0 0 w^2> ( 2w = -1 + (root3)i )> ------------------------------> i have a doubt> i find eigenvalue of this matrix> it's 1, w = {-1+(root3)i}/2 , w^2 = {-1-(root3)i}/2I get the characteristic equationP = x^3 - x^2 - 1 = 0and the eigenvalues are different. But A is diagonalizable, since P has 3distinct complex roots. Also, you can check P(A)=0. I won't do the rest,because I'm afraid it would teach you some === === algebraic geometry textsAnyone know of any good ones?Shafarevich Basic Algebraic Geometry in two volumes from === geometry textsUndergraduate Algebraic Geometry by Miles Reidis a good undergrad book. Slightly idiosyncratic, but great fun to read.> Anyone know of any good ones?Shafarevich Basic Algebraic === introductory algebraic geometry texts> Anyone know of any good ones?Depends on your background. But, at the undergraduate level, you could lookat a book titled Ideals, Varities, and Algorithims by David Cox.http://www.amazon.com/o/ASIN/0387946802/102-4888135- 0200161?%5Fencoding=UTF8&coliid=I2ZCK1AT36F01E&colid= 7J7QGQP7I9K0I have looked through some of it, and it appears to be pretty good. I knowthat it gets high marks from a lot of === texts>Anyone know of any good ones?> Depends on your background. But, at the undergraduate level, you could look> at a book titled Ideals, Varities, and Algorithims by David Cox.> http://www.amazon.com/o/ASIN/0387946802/102-4888135-0200161?% 5Fencoding=UTF8&coliid=I2ZCK1AT36F01E&colid=7J7QGQP7I9K0I have looked through some of it, and it appears to be pretty good. I know> that it gets high marks from a lot of people.LurchAt a slightly higher level, you might look at Using Algebraic Geometry by Cox et. al.. I also enjoyed the book by Karen Smith, two Finns and a squid whose title escapes === on this, I thought of a way that you could get a similar result (ie, express any positive integer as a set of grouping symbols) without 0 being prime, although you can't express 0 in this way.You could start off the same way, express a number as it's prime factorization:1 = ()2 = (2(1)) ie. 2^14 = (2(2)) ie 2^25^10*7 = ((5(10))(7(1)))Now use recursion to get all the exponents in terms of their prime factors. Note the prime factors of 1 are ():5^10*7 = ((5((2())(5())))(7()))Now convert each prime number to it's prime index, ie which prime is iti 1 2 3 4 5 6 ...n 2 3 5 7 11 13 ...And at this time add a layer of ()'s to prevent multiple numbers with the same code.5^10*7 -> (((3)(((1)())((3)())))((4)()))prime factorize again, and then re-take index until every term becomes all parentheses. You get:1 ()2 ((())(()))3 (((()))(()))4 ((())((())))5 ((((())))(()))6 (((())(()))(((()))(())))7 ((((())((()))))(()))8 ((())(((()))))9 (((()))((())))10 (((())(()))((((())))(())))and 5^10 * 7 is (((((())))((((())(()))((((())))(())))))((((())((()))))(())))It seems like there may be a more compact way to store really large numbers in binary, but for most numbers this is alot longer than the usual binary encoding. But in any case this way doesn't require any separators between numbers.> Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt Your left with the ability to represent any number as a set of grouping> symbols after repea recursive prime decomposition. Don't know if this> is of any use at all, but its neat :)> === be a finite group. Show that the number ofelements x of G such that x^3=e is odd. Show that the number of elements x ofG such that x^2 is NOT equal to e is even.I'm having a hard time even understanding what he is asking. Is his use ofEnglish poor or am I just not reading this correctly? The answer given goes like this (which I also don't understand -- how hasanything been shown to be 'odd' or 'even'?):If x^3=e and x not equal to e, then (x^-1)^3=e and x is not equal to x^-1. Sononidentity solutions come in pairs. If x^2 is not equal to e then x^-1 is notequal to x and (x^-1)^2 is not equal to e. So solutions to === group problem> The problem states that Let G be a finite group. Show that the number of> elements x of G such that x^3=e is odd. Show that the number of elements> x of G such that x^2 is NOT equal to e is even.I'm having a hard time even understanding what he is asking. Is his use> of English poor or am I just not reading this correctly?It seems perfectly clear and unambiguous to me,and the English seems correct, and normal, too.The answer given goes like this (which I also don't understand -- how has> anything been shown to be 'odd' or 'even'?):If you show the solutions divide into paithe number will be even.The second question seems to me an order of magnitude more difficult than the first.You have to show that if the group has even orderthen the number of elements of order 2 is odd.I can only do it by choosing one element of order 2,and acting with this on the others.Is there a simpler solution?-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, === problem states that Let G be a finite group. Show that the number of>> elements x of G such that x^3=e is odd. Show that the number of elements>> x of G such that x^2 is NOT equal to e is even. I'm having a hard time even understanding what he is asking. Is his use>> of English poor or am I just not reading this correctly?>It seems perfectly clear and unambiguous to me,>and the English seems correct, and normal, too. The answer given goes like this (which I also don't understand -- how has>> anything been shown to be 'odd' or 'even'?):>If you show the solutions divide into paithe number will be even.>The second question seems to me an order of magnitude more difficult >than the first.>You have to show that if the group has even order>then the number of elements of order 2 is odd.>I can only do it by choosing one element of order 2,>and acting with this on the others.>Is there a simpler solution?It's easier if you consider the elements E that do not have order 2.There is the identity, but all others can be paired off with their inverses,so |E| is odd.Of course, if you prefer to use an ICBM to shoot a sparrow (to quote oneof my former lecturers), one of the stronger versions of Sylow's Theoremsays that the number of subgroups of a finite group G of order divisibleby p^n (p prime) is always congruent to 1 === problem states that Let G be a finite group. Show that the number of>> elements x of G such that x^3=e is odd. Show that the number of elements>> x of G such that x^2 is NOT equal to e is even. I'm having a hard time even understanding what he is asking. Is his use>> of English poor or am I just not reading this correctly?It seems perfectly clear and unambiguous to me,> and the English seems correct, and normal, too. The answer given goes like this (which I also don't understand -- how has>> anything been shown to be 'odd' or 'even'?):If you show the solutions divide into paithe number will be even.The second question seems to me an order of magnitude more difficult > than the first.> You have to show that if the group has even order> then the number of elements of order 2 is odd.> I can only do it by choosing one element of order 2,> and acting with this on the others.> Is there a simpler solution?The number of elements of order two is the cardinality of the union of all2-subgroups - 1 (omit the identity). which is clearly odd. if that's notactually that clear: the card of the union can be expressed as asigned sum of cards of intersections (inclusion-exclusion). eachintersection is a 2-group, so the terms in the sum are all === problemRandySW1983> The problem states that Let G be a finite group. Show that the number of> elements x of G such that x^3=e is odd. Show that the number of elementsx of> G such that x^2 is NOT equal to e is even.> I'm having a hard time even understanding what he is asking. Is his useof> English poor or am I just not reading this correctly?Well, the writing is a bit terse, IMO. I guess the author is assuming thatthe reader has previously seen some results and proofs of this sort.Take e.g. Show that the number of elements x of G such that x^3=e is odd.G is finite, so any subset has an odd or an even number of elements. Thesubset in question is the set of x in G such that xxx=e. e itself is onesolution. Suppose y is another. Then y^2 (call it z) is also a solution, forwe have z^3 = y^6 = y^3 y^3 = ee = e. But moreover, z^2 = y^4 = y y^3 = ye=y. Finally, we cannot have y=z unless y=e. See how the solutions y otherthan e thereby appear in pairs? But e appears alone, and so there are an oddnumber of solutions in === good introductory source (book or website)regarding Painleve analysis ultimately pertaining to the solution ofnonlinear systems of PDE's (keeping in mind its for a physicist -application orien).I have a few references, but i would like to know what other people thinkare good === bisector of angle ABC intersects segment AC at D. The bisector of angle ACB intersects segment AB at E. Segments BD and CE are congruent. True or false: triangle ABC is isosceles?-- === angle ABC intersects segment AC at D. The bisector > of angle ACB intersects segment AB at E. Segments BD and CE > are congruent. True or false: triangle ABC is isosceles?Well, with the p-adic norm, all triangles are isosceles, so...In Euclidean geometry, the statement is true (if I have the correct mentalpicture). If I remember, this isn't very easy to === Blackburn escribi.97:>> The bisector of angle ABC intersects segment AC at D. The>> bisector of angle ACB intersects segment AB at E. Segments BD>> and CE are congruent. True or false: triangle ABC is isosceles?> Well, with the p-adic norm, all triangles are isosceles, so...> In Euclidean geometry, the statement is true (if I have the correct> mental> picture). If I remember, this isn't very easy to prove.> -LeonardNot so ... . Consider the equivalent bisector of angle ACB intersects segment AB at E. Segments BD> and CE are congruent. True or false: triangle ABC is isosceles?False, of course.But if you meant Segments BD and BE are congruent...., it is true.The angle bisector divides opposite side in segments proportional tocontiguopus sides. ThenBE/AE = CB/CABD/CD = AB/ACAs BE = BD, you getBE*AC = AE*CB AB and the triangle is isosceles.-- Ignacio Larrosa Ca.96estroA Coru.96a === A plane geomerty problemIgnacio Larrosa Ca.96estro a .8ecrit:> Stephen J. Herschkorn escribi.97 at D. The>>bisector of angle ACB intersects segment AB at E. Segments BD>>and CE are congruent. True or false: triangle ABC is isosceles?> False, of course.no, true.But if you meant Segments BD and BE are congruent...., it is true.I think you read the post too quicly...The angle bisector divides opposite side in segments proportional to> contiguopus sides. ThenBE/AE = CB/CABD/CD = AB/ACno. D is on AC not an BC :Bisector of angle B intersects AC at D, notBisector of angle A intersects BC at D...[...]The question was just Steiner-Lehmus theorem.several proofs are known.For example first prove that if AB>=AC then BD>=CE(and of course if AC>=AB then CE>=BD)by === Larrosa Ca.96estro a .8ecrit:>> Stephen J. Herschkorn escribi.97 en el> The bisector of angle ABC intersects segment AC at D. The> bisector of angle ACB intersects segment AB at E. Segments BD> and CE are congruent. True or false: triangle ABC is isosceles?>> False, of course.> no, true.>> But if you meant Segments BD and BE are congruent...., it is>> true.> I think you read the post too quicly...>> The angle bisector divides opposite side in segments proportional to>> contiguopus sides. Then>> BE/AE = CB/CA>> BD/CD = AB/AC> no. D is on AC not an BC :> Bisector of angle B intersects AC at D, not> Bisector of angle A intersects BC at D...Yes, you are right. I put incorrectly the point D on my sketch in the sideBC. I apologize ...-- Ignacio Larrosa Ca.96estroA Coru.96a === A plane geomerty problem> Stephen J. Herschkorn bisector of angle ABC intersects segment AC at D. The>> bisector of angle ACB intersects segment AB at E. Segments BD>> and CE are congruent. True or false: triangle ABC is isosceles?> False, of course.Wrong. Look up the Steiner-Lehmus theorem. I proved this in highschool, but my proof was algebraic, not geometric. I didn't find outabout a geometric proof until years later.The correct conclusion is that AB = AC.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. How do you solve and ODE of the type:> P_1(x)x'' + P_2(x)x' + P_3(x)x+ P_4(x) = 0> where P_i are polynomials of x, (functions of x^i and constants).> Some very special cases are famous instance (Laguerre, Legendre and so> on...). But === with polynomials as coefficients> do you mean P_1(t) etc, with x' = dx/dt ?> if so, take x as a power series in t and equate coeff of t^nNo, I mean P_1(x(t)), for instance P_1(x)=x+3x^2(such and ODE could be:(1+x)x + (x^2+x)x' + x^3 === moth,A while ago you replied to a question of mine in sci.math, with the postbelow.I ve checked the expression you give (the last one, with the Besselfunctions)and it is correct save a factor 4*Pi.However, I ve been trying to go on from there to a form that I can usenumerically,but in vain. Indeed, as you state, one can integrate over k, yielding anellipticintegral, but that leaves the second integral over r', which I haven't beenable todeal with. It is given by:integrate(0,1) 2r' * K(4r'r/R^2) / R dr' withR=sqrt(z^2+r^2+r'2 - 2r'r)where K is the complete elliptic integral of the first kind.to this last integral in mind?Alexps i tried to email you personally about this but the email bounced----- Original Message ----- sci.math,sci.math.num-analysis,sci.math.symbolic, === sci.physics.electromagSubject: Re: integral over a disk> Here's a question that I already pos a while ago, to which I gotseveral> interesting answers> (thread 'definite integral' in sci.math and sci.math.num-analysis).> Nevertheless, I m still stuck with my practical problem, reason why I m> giving it another go, now including sci.math.symbolic and> sci.physics.electromag, and trying to specify> the problem and the kind of solution I m looking for more clearly:The problem (it's an electrostatics problem),> is to integrate the function 1/R over a flat, infinitely thin, circulardisk> where R represents the distance> from the integration point on the disk to an arbitrary point P in space.Mathematically:F = int_0^{2*pi} { int_0^1 { 1 over{ sqrt{ D^2 + r^2 + rho^2 -> 2*r*rho*cos{phi} }}} dr } dphiwhere the disk has radius 1, and it is centered at the origin of the> standard cylindrical coordinate-system> in the plane z = 0. The point P is at (r=rho, z=D, phi=0)This can be relatively easy reduced to a 1-D integral using Gauss'theorem:F = int_L { (rho*cos{phi} - 1)*R over Q^2 } dl,a line integral over the circumference L of the disk, and Q equals R forz => 0.> But then the problem starts. Mathematica gives me an expression of about13> pages of increasingly> complica elliptical functions, that becomes indeterminate at both the> integration limits.What I m looking for is a good approximation, that is, with a guaranteed> relative error> below, say 0.001, for any point P, which at the same time is not too> computationally demanding,> because it's part of a numerical simulation code that depends on speed.> Right now I m using numerical integration, and it's in advance,Alex> I presume you want to find the potential/field of a thin electrified isk -> if so you could refer threadm=bo9sej%24gr8%241%40merki.connect.com.au&rnum=4&prev=/ groups%3Fq%3Drancid%2Bmoth%26hl%3Den%26lr%3D%26ie%3DUTF-8% UTF-8&threadm=bo9srh%24gst%241%40merki.connect.com.au&rnum=5& prev=/groups%3Fq%3Drancid%2Bmoth%26hl%3Den%26lr%3D%26ie%3DUTF- 8%26scoring%3Dd> or you could start with fourier transforms of poissons equation andexpress> ...> phi(r,z) = 1/(8pi^3) * integrate(-oo,oo) Exp(i*z*k_z)/(k_r^2+k_z^2) dk_z> integrate(0,oo) k_r dk_r integrate(0,1) r'dr' ingegrate(0,2*pi)> Exp(-i*k_r*r'*cos(@'))d@' ingegrate(0,2*pi) Exp(-i*k_r*r*cos(@_k))d@_k> which caters for a disk at z=0 of radius 1, and a constant charge desnityof> 1.> from there you note that> ingegrate(0,2*pi) Exp(-i*k_r*r'*cos(@))d@ = 2*pi*J_0(r'*k_r)> where J_0(z) is the zeroth order bessel function of the first kind, andone> can easily evaluate> integrate(-oo,oo) Exp(i*z*k_z)/(k_r^2+k_z^2)> via contour methods and you should end up with> phi(r,z) = 1/2 * integrate(0,oo) exp(-k_r*|z|) *J_0(r'*k_r)*J_0(r*k_r)dk_r> integrate(0,1) r'dr> you can do these integrals - they turn out to be elliptic functions - they> can be expressed quite easily and will take up *one* line! (refer to the> posts === with correlation dimensionHallo!I want to calculate the correlation dimension of a time serie.What I have doneI calcula the correlation integral C(r) (number of point having adistance smaller than r) for different embedding dimensions. Takingthe slopes of the curve of log C(r) against log r for the differentembedding dimensions and plotting them against the embedding dimensionshould result in a limes of the slopes: the correlation dimension.My problemWhich slope shall I take?In examples I saw in text books there is a nice limit of the slopeswith higher embedding dimensions. In my data I do not know which slopeI should take because the slope of the curve varies. If I take theslope at a certain value of log r I can not get a limes.My curves (log C(r) against log r) can be seen inhttp://karlknoblich.4t.com/korrdim.jpgWhat to do? Does anybody knows such data and how to handle it?Hope somebody can === dimension> Hallo!> I want to calculate the correlation dimension of a time serie.> What I have done> I calcula the correlation integral C(r) (number of point having a> distance smaller than r) for different embedding dimensions. Taking> the slopes of the curve of log C(r) against log r for the different> embedding dimensions and plotting them against the embedding dimension> should result in a limes of the slopes: the correlation dimension.> My problem> Which slope shall I take?> In examples I saw in text books there is a nice limit of the slopes> with higher embedding dimensions. In my data I do not know which slope> I should take because the slope of the curve varies. If I take the> slope at a certain value of log r I can not get a limes.> My curves (log C(r) against log r) can be seen in> http://karlknoblich.4t.com/korrdim.jpg> What to do? Does anybody knows such data and how to handle it?> Hope somebody can help!> KarlThere is no guarantee that the limit exists.There may be different slopes on different scales with different widths.There is a huge difference between real life data and solution of low dim mathematical models without noise (as seen in books).-- Pavel PokornyMath Dept, Prague Institute of Chemical === Practical problems with correlation dimension> Hallo!> I want to calculate the correlation dimension of a time serie.> What I have done> I calcula the correlation integral C(r) (number of point having a> distance smaller than r) for different embedding dimensions. Taking> the slopes of the curve of log C(r) against log r for the different> embedding dimensions and plotting them against the embedding dimension> should result in a limes of the slopes: the correlation dimension.> My problem> Which slope shall I take?> In examples I saw in text books there is a nice limit of the slopes> with higher embedding dimensions. In my data I do not know which slope> I should take because the slope of the curve varies. If I take the> slope at a certain value of log r I can not get a limes.> My curves (log C(r) against log r) can be seen in> http://karlknoblich.4t.com/korrdim.jpg> What to do? Does anybody knows such data and how to handle it?> Hope somebody can help!> KarlWhat I will say has not yet been accep by others,so keep that in mind as you consider it.You're missing some crucial data that cross-correlatesyour 'time' series to the cerebellar topology.The cerebellum is a topographically-mapped subsystem.Any analysis must preserve, and incorporate, that mappingif the correlations are to be meaningful.And, then, to continue, one has to follow this mapping intothe rest of the brain.It's a =big= problem, but the mapping is mapped :-] throughthe efforts of Neuroscientists, and all one has to do is 'grind'through it.There a couple of other things that make your analysis Difficult.One is that the data is virtually always, itself, a transformation.The other is that the activation that occurs within the cerebellumis extremely-dynamic, with a =lot= of different inputs convergingand 'sliding' with respect to each other. There is such 'sliding'stuff with respect to every joint in the skelleton. [These enterinto the way that the nervous system maintains it's 'awareness'of the body's orientation in 3-D space [climbing fibers fromthe inferior olive].] And this is only one set of such 'sliding-field'stuff that occurs within the cerebellum. There are hundreds[perhaps thousands] more.So your analysis is Hard.ken [k. p. === Q_p(zeta)I am reading through the book Local fields by Cassels, and ran intoa problem I couldn't solve yet. It's an exercise I cannot fully solve,but which is needed later in a theorem. (Ex. in chapter 7, don't knowexact number at the moment.)== Prerequisites ==Let p > 2 be a prime, and let zeta =/= 1 be a p-th root ofunity. Consider the p-adic numbers Q_p and its extension Q_p(zeta).Let O denote the ring of integers in Q_p(zeta), i.e. all elementshaving value <= 1.The element lambda := 1 - zeta is a prime element of O.The galois auorphism sigma of Q_p(zeta), defined by sigma(zeta) = zeta^g for a primitive root g mod p, generates thegalois group of Q_p(zeta)/Q_p.By == I denote congruence. Since this extension is completelyramified, the residue class field O/(lambda) still is isomorphic toF_p = Z/pZ.I already solved the task to show* sigma(lambda) == g lambda (mod lambda^2)* p lambda^{-p+1} == -1 (mod lambda)Let a be a 1-unit in Q_p(zeta), i.e. a in O and a == 1 (modlambda). I already know that then there exists an integer m s.t. azeta^m == 1 (mod lambda^2).== My task ==* Let b be a 1-unit in Q_p(zeta). Then b == 1 (mod lambda^{p+1}),iff there exists an element a in Q_p(zeta) s.t. a^p = b.I proved one direction:For any a in Q_p(zeta) with a^p =: b a 1-unit, it holds thatb == 1 (mod lambda^{p+1}).For the other direction, a hint is given:Consider the polynomial g(Y) = (1 + lambda Y)^p - b.I know, that if for the given b an a with a^p = b exists, then a == 1 (mod lambda), so I can write a = 1 + lambda y. A zero y inO of g consequently leads to the a I'm looking for. But how toshow such a zero exists?I tried Hensel's lemma:g'(Y) = lambda p (1 + lambda Y)^{p-1}So |g'(y)| = |lambda p| = |lambda|^p, and i need a start approximation a0 with |g(a0)| < |lambda|^{2p} - which I cannot find.Any idea?Christian === Q_p(zeta)* Let b be a 1-unit in Q_p(zeta). Then b == 1 (mod lambda^{p+1}),> iff there exists an element a in Q_p(zeta) s.t. a^p = b.I proved one direction:> For any a in Q_p(zeta) with a^p =: b a 1-unit, it holds that> b == 1 (mod lambda^{p+1}).I wouldn't use Hensel. I'd note that as long as r > = 2,(1 + c lambda^r)^p = 1 + p c lambda^r (mod lambda^{p+r}).Take b and write it as 1 + p c_2 lambda^2for some integer c_2 in Q(zeta_p).Now b/(1 + c_2 lambda^2) = 1 (mod lambda^{p+2}) sowrite b/(1 + c_2 lambda^2)^p = 1 + p c_3 lambda^3.Keep going: for each j >= 2,b/(1 + c_j lambda^j)^p = 1 + p c_{j+1} lambda^{j+1}.Then b = a^p where a = === Data analysis software>> It plots and analyses any x-y data for peak location, peak height,> peak>> width, semi-derivative, derivative, integral, semi-integral,> convolution,>> deconvolution, curve fitting, and separating overlapped peaks and>> background.>> www.chemSoftware.comI keep racking my brain (it really has been abused enough -- guess> it's a slow day) but I just can't recall where I've read about this> before.OTOH, it helps us decide which data analysis software we definitelydon't want to buy!Barry in === and analyses any x-y data for peak location, peak height,> peak>> width, semi-derivative, derivative, integral, semi-integral,> convolution,>> deconvolution, curve fitting, and separating overlapped peaks and>> background.>> www.chemSoftware.comI keep racking my brain (it really has been abused enough -- guessit's a slow day) but I just can't === vs. propositionWhat is the difference between a theorem and a proposition?qualifications/disclaimers/comments:1) I mean these in the technical mathematical sense2) I feel like I've seen (mathematical) papers/books whichused both terms but I could not determine the distinguishing characteristics.3) I suspect that their primary content is identical, just their usage depends on context (like a corollary is akind of theorem, but a theorem is just a corollary of the 2nd to last item in a proof).4) Heath's commentary on Euclid discusses lots of terminologicalquestions but doesn't seem to touch on this one. Did I miss it? 5) Is this a problem of modern usage? how old are these two terms? In Heath, he explains that porism is a synonym of corollary, whereas (I thought) the modern usage is a theorem schema (a theorem whose construction produces many possible constructions/solutions(e.g. steiner's porism))5) Is there such a pair in other languages (in the original Greek or modern languages (in German it seems Satz is the common translation for both)-- Mitch Harris(remove q to === difference between a theorem and a proposition?A theorem is a === proposition> What is the difference between a theorem and a proposition?> qualifications/disclaimers/comments:> 1) I mean these in the technical mathematical sense> 2) I feel like I've seen (mathematical) papers/books which> used both terms but I could not determine the distinguishing> characteristics.> 3) I suspect that their primary content is identical, just> their usage depends on context (like a corollary is a> kind of theorem, but a theorem is just a corollary of the> 2nd to last item in a proof).> 4) Heath's commentary on Euclid discusses lots of terminological> questions but doesn't seem to touch on this one. Did I miss it?> 5) Is this a problem of modern usage? how old are these two terms?> In Heath, he explains that porism is a synonym of corollary,> whereas (I thought) the modern usage is a theorem schema (a theorem> whose construction produces many possible constructions/solutions> (e.g. steiner's porism))> 5) Is there such a pair in other languages (in the original Greek> or modern languages (in German it seems Satz is the common> translation for both)> -- > Mitch Harris> (remove q to reply)imo,All statements that are true or false are propositions.All theorems are provably true propositions. They are analytic and true.All propositions that are false or === propositionMitchFor a good definition of 'proposition' as used in osophy and logicstudied from a osophical viewpoint see:http://en.wikipedia.org/wiki/PropositionFor what looks like a better definition of 'proposition' as used inmathematics see:http://en.wikipedia.org/wiki/TheoremI hope this helps a little.Love and respectChris> What is the difference between a theorem and a proposition?> qualifications/disclaimers/comments:> 1) I mean these in the technical mathematical sense> 2) I feel like I've seen (mathematical) papers/books which> used both terms but I could not determine the distinguishing> characteristics.> 3) I suspect that their primary content is identical, just> their usage depends on context (like a corollary is a> kind of theorem, but a theorem is just a corollary of the> 2nd to last item in a proof).> 4) Heath's commentary on Euclid discusses lots of terminological> questions but doesn't seem to touch on this one. Did I miss it?> 5) Is this a problem of modern usage? how old are these two terms?> In Heath, he explains that porism is a synonym of corollary,> whereas (I thought) the modern usage is a theorem schema (a theorem> whose construction produces many possible constructions/solutions> (e.g. steiner's porism))> 5) Is there such a pair in other languages (in the original Greek> or modern languages (in German it seems Satz is the common> translation for both)> --> Mitch Harris> === proposition>What is the difference between a theorem and a proposition?> For a good definition of 'proposition' as used in osophy and logic> studied from a osophical viewpoint see:http://en.wikipedia.org/wiki/PropositionAh. didn't think of checking there (and I don't have a physical copy of a mathematical dictionary nearby)OK. Their 2nd paragraph talks about it with respect to Aristotelian logic, and that explanation seems tantalizingly close to ... > For what looks like a better definition of 'proposition' as used in> mathematics see:http://en.wikipedia.org/wiki/TheoremAh nice...mostly fits well except for...proposition: a result not associa with any particular theorem.This really doesn't do much for me. (I've pulled it out of context, but the context didn't === Re: theorem vs. proposition>What is the difference between a theorem and a proposition?For a good definition of 'proposition' as used in osophy and logic> studied from a osophical viewpoint see:http://en.wikipedia.org/wiki/Proposition> Ah. didn't think of checking there (and I don't have a physical copy of> a mathematical dictionary nearby)> OK. Their 2nd paragraph talks about it with respect to Aristotelian> logic, and that explanation seems tantalizingly close to ...> For what looks like a better definition of 'proposition' as used in> mathematics see:http://en.wikipedia.org/wiki/Theorem> Ah nice...mostly fits well except for...> proposition: a result not associa with any particular theorem.> This really doesn't do much for me. (I've pulled it out of context,> but the context didn't help me either)I sympathise Mitch - it didn't do much for me either at first reading. Ithink it makes more sense in context, which seems to say to methat the distinction is only one of degree, based on how interesting orimporant the result is. In summary I think the Wikipedia entry means:theorem:a statement which can be proven true within some logicalframework and which is interesting or important in some waylemma:a significant part of the proof of a theorem (and as such lessimportant than the theorem itself)corollary:also significant but very easily deduced from a theorem (and assuch less important than the theorem)proposition:proven true but not significant enough to be called a theorem or,even, a lemma or a corollaryI hope this helps, and represents the spirit of actual === and eulersI have the following problem:I have an object that has an orientation described with Euler angles andon which I can calculate the forces exer, thus finding the torques andangular accelerations wrt the object's axis. The problem is how tocompute the new angular position of the object.. In my case the euler angles describe rotations in the order X Y Z (worldaxis). I tried to convert these angles into a quaternion representation.The angular acceleration allows me in the object's frame gives me theangular velocity in the object's frame using simple Euler integration:w'(t+dt) ~= w'(t) + w''(t)*dtIf there are no forces exer on the object then w' is constant, i.e. itshould correspond to a rotation around a stationary axis. It should besimple to transform this axis from the object's to the world's frame ofreference... but I cannot seem to be able to do it..Maybe the problem lies in the fact that the euler angles describe aseries of rotations.. and thus I perhaps miss something when convertingfrom Euler to quaternion. Can anyone help?-- Christos DimitrakakisIDIAP === Isoperimetric ZeppHello sci.math,Consider the problem of maximizing the integral of some function F(x,y)over a simply connec region whose boundary C is allowed to varysubject to these two constraints:1) C contains the origin (0,0), and2) C has length 1.I've been corresponding with Rainer Rosenthal about this topic, whicharose on sci.math's German counterpart de.sci.mathematik. The caseF(x,y) = x was discussed in particular, yielding an optimal shape whichlooks like a Zeppelin, or a teardrop. The shape is pointy at the origin,with an apex angle of about 81.42 degrees.Here is some Mathematica code to plot it: xf[u_,m_] := Sqrt[1-m*Sin[u]^2]/m; yf[u_,m_] := ((m/2-1) EllipticF[u,m] + EllipticE[u,m])/m; u0 = 1.140659153420324; m0 = Csc[u0]^2; fac = 1 / Re[EllipticK[m0]]; ParametricPlot[fac {xf[u,m0], yf[u,m0]}, {u, -u0, u0}, AspectRatio -> Auatic];Rainer coined the name Zepp for this shape. Here's one physicalinterpretation:Phys 1: Hang a closed curve from a point, then fill it with water until it overflows.This works better if you live in a 2-d world, of course. The Zeppminimizes the potential energy of the water, so trying to add moreresults in an overflow. Now here's another interpretation:Phys 2: Take a rectangular piece of paper and gently bring two of its opposites edges together.Go ahead! Surely there's a piece of paper nearby! This one iseasier in 2-d as well, where you're just bending a line segment.The resulting shape is the one that minimizes Intg(s=0,1) kappa(s)^2 ds,where s is arc length and kappa(s) is curvature. Surprisingly,this shape turns out to be same as the Zepp!Surely such a wonderful shape cannot be new, and indeed Rainer hastraced it back to Euler. I'll let him tell you more about that.Maybe he will also explain why the two characterizations give thesame shape.I shall instead turn my attention to this: In general, the extremal curve C must satisfy F(x(s),y(s)) = alpha kappa(s), for some constant alpha.I derived this rather laboriously using the calculus of variations, butnow I realize that it's physically obvious: if we let F represent ascalar potential, it yields a normal force F at the boundary which iscounterbalanced by the tension force alpha*kappa.For example, the 3-d version of the Zepp would be an axisymmetricshape whose profile obeys the equation x(s) |y(s)| = alpha kappa(s)for some alpha. Can anyone provide more details or examples? In particular, it iscurious that A) Intg x dx dy is maximized when B) x(s) = alpha kappa(s), which also C) minimizes Intg(s=0,1) kappa(s)^2 ds.In general, then, how do we complete the following correspondence,if possible: A) Intg F(x,y) dx dy is maximized when B) F(x(s),y(s)) = alpha kappa(s), which also C) minimizes Intg(s=0,1) ??? ds.-Jim === be well-known, but not to me.Consider the sequence of ratios of neighboring Fibonaccinumbers: 1/1, 1/2, 2/3, 3/5, 5/8, etc.Let r_n be the nth rational on this list. What appearsto be true is this: r_{n+2} is the rational with the smallest denominator that is strictly between r_n and r_{n+1}I don't immediately see how to prove this, though.--Daryl === Fibonacci Numbers> This might be well-known, but not to me.Consider the sequence of ratios of neighboring Fibonacci> numbers: 1/1, 1/2, 2/3, 3/5, 5/8, etc.Let r_n be the nth rational on this list. What appears> to be true is this: r_{n+2} is the rational with the smallest denominator that> is strictly between r_n and r_{n+1}I don't immediately see how to prove this, though.I came up with a proof while arguing with a crank in another thread.There is an old result about the area of a polygon whose vertices arelattice points (i.e., their (x,y) coordinates are both integers): Thearea is one less than the number of lattice points in the interior plushalf the number of lattice points on the boundary. For example, thetriangle whose vertices are at (0,0), (4,0), (0,3) has 8 lattice pointson the boundary, 3 lattice points in the interior, and an area of 3.Similarly, the square with verticies at (+/- 1, +/- 1) has 8 latticepoints on the boundary, 1 lattice point in the interior, and an area of4.Ah, here's the name: Pick's Theorem.http://mathworld.wolfram.com/PicksTheorem.htmlhttp:// en.wikipedia.org/wiki/Pick%27s_theoremYou have each r_n = F_n/F_{n+1}Consider the point (x,y) where x/y is the lowest-terms representation ofthe rational with the smallest denominator that is strictly between r_nand r_{n+1}. You can show that (x,y) is an interior lattice point forthe triangle with vertices at (0,0), (3*F_n, 3*F_{n+1}), and (2*F_{n+1},2*F_{n+2}). You can also show that the only interior lattice point forthat triangle is (F_{n+2}, F_{n+3}). The desired result follows. -- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ === polynomial> Look at: http://sepwww.stanford.edu/oldsep/stew/descartes.pdfTheorem 1 (Lagrange) states:> If in p(x) the first negative coefficient is preceded by k coefficients > which are positive or zero, and if G denotes the greatest of the > magnitudes of the negative coefficients, then p(x) is always positive for > x > 1+ {G / p_n}^{1/k}, and so all real roots are less than that value.Use it for your P and for Q(x):=P(-x) (to get negative === Bound for roots of a polynomial> Hi all,> Let P(x) ba a polynomial with real coefficients. Is there> a simple way of finding a number R > 0 such that every real> root of P(x) is in [-R,R]? I'm thinking in terms of R being> obtained as a function of the coefficients of P(x).Yes, of course.If P(x) = x^n + a_1 x^{n-1} + .... + a_n,> take R = max(1, |a_1| + |a_2| + ... + |a_n|).Meanwhile, I decided not to be lazy and I got to the following formulafor R (using your notations):R = === is the meaning of imaginary numbers?>..imaginary numbers.. >Can anybody point me to any resource describing their TRUE physicalmeaning... Of course quantities can be applied to more than one physicalquality. An example:the addition of forces in a parallelogramm(Newton). That's a vector addition on a plane - yes, a complex numberis an ordinary 2D-vector, you can write a+i*b = (a,b) and i is thevector (0,1).And there is the multiplication, and the history and...See http://i-is-no-longer-imaginary.gmxhome.deor -the-same: http://i-z.eu.ttI pos this with sci.math in www.math-form.org === imaginary numbers?>..imaginary numbers.. >Can anybody point me to any resource describing their TRUE physical> meaning...> Of course quantities can be applied to more than one physical> quality. An example:the addition of forces in a parallelogramm> (Newton). That's a vector addition on a plane - yes, a complex number> is an ordinary 2D-vector, you can write a+i*b = (a,b) and i is the> vector (0,1).And there is the multiplication, and the history and...> See http://i-is-no-longer-imaginary.gmxhome.de> or -the-same: http://i-z.eu.ttI pos this with sci.math in www.math-form.org Have fun> HeroMultiplication by i represents a counterclockwise === polyhedronCan you think of any reason why N stiff, flexible, conical,vertex elements would have a problem joining by overlap to produce apolyhedron whose Euler number is 2 where each cone has an angulardeficit alpha and each edge of the polyhedron is arbitrary, andalpha=720degrees/N?Dick === SpamAssassin (score=-22.8, required 5, EMAIL_ATTRIBUTION, QUO_EMAIL_TEXT, REFERENCES, REPLY_WITH_QUOTES, USER_AGENT) Here's another example of a Galois group that physicists should like. >> Let C(z) be the field of rational functions in one complex variable z ->> in other words, functions like f(z) = P(z)/Q(z) where P and Q are polynomials in z with complex coefficients. You can >> add, subtract, multiply and divide rational functions and get other >> rational functions, so they form a field. And they contain C as a >> subfield, because we can think of any complex number as a *constant* >> function. So, we can ask about the Galois group of C(z) over C.>> What's it like? It's the Lorentz group! To see this, it's best to think of rational functions as functions not>> on the complex plane but on the Riemann sphere - the complex plane >> together with one extra point, the point at infinity. The only >> conformal transformations of the Riemann sphere are fractional linear>> transformations:Could you please explain why an element of the Galois group must> be a conformal transformation? At first sight one would think> it can be any rational substitution that is bijective.> So why does bijective imply conformal?Hmm, let met try. Any holomorphic (analytic) function gives you aconformal transformation (and I think in 2D every conformal transformationgives you a holomorphic function).All rational functions are holomorphic (except for the zeros of thedenominator, but since we define the functions on the Riemann sphere, thiscould be glossed over).Every rational function is surjective (fundamental theorem of algebra).If the denominator has more than one root, the function is notinjective (it's zero at more than one place).If the denominator has more than one root, the function is notinjective either (more than one point gets mapped to the point at infinity).This leaves just f(z) = (a z + b)^n/(c z + d)^m. But if n and m areanything other than 1, this function is not injective either f(z) = cwould have max(m,n) solution.I guess this just leaves fractional linear transformationsf(z) = (a z + b)/(c z + d).Hope I didn't miss === small program for array inversions.Can someone give me exaples of matrix 2x2 3x3 and 4x4 - how it should look after inverse (best - some easy examples - with small numbers)Or maybe program for linux that will do such thing?-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~l-.~~~~~~~~~~~~~~~~~~~GG- 1175498 ____| ]____, Rafal 'Raf256' Maj X-( * === )Rafal(at)Raf256(dot)com ,---------- Subject: Re: Matrix small program for array inversions.> Can someone give me exaples of matrix 2x2 3x3 and 4x4 - how it should> look after inverse (best - some easy examples - with small numbers)> Or maybe program for linux that will do such thing?[3, 5; -1, 4]^(-1) = [4/17, - 5/17; 1/17, 3/17][3, 6, -1; 4, 0, 5; 1, 3, 8]^(-1) = [5/73, 17/73, - 10/73; 9/73, - 25/219,19/219; - 4/73, 1/73, 8/73][1, 0, 1, 0; 0, 1, 1, 0; 1, 1, 0, 1; 1, 1, 1, 1]^(-1) = [1, 0, 1, -1; 0, 1,1, -1; 0, 0, -1, 1; -1, -1, -1, 2](Elements are delimi by ',', rows by ';')-- Ignacio Larrosa CanestroA Coruna === Matrix inverse examples>i'm testing my small program for array inversions.>Can someone give me exaples of matrix 2x2 3x3 and 4x4 - how it should look >after inverse (best - some easy examples - with small numbers)??? Doing matrix multiplication is simple. Just calculate AA^(-1).Also I think it would also be useful if such a program was known towork for matrices with === course and they have a linear algebra problem here. Mylinear algebra skills are a bit rusty and I don't have the book here to takea look. Can someone give me a hint as to how I would show that c1 + c2sin^2(x) + c3cos^2(x) forms a vector space? === University of Montana.>> I am taking a pde course and they have a linear algebra problem here. My>linear algebra skills are a bit rusty and I don't have the book here to take>a look. Can someone give me a hint as to how I would show that c1 + c2sin^2>(x) + c3cos^2(x) forms a vector space? c1, c2 and c3 are constants.You mean, show that the set of all functions of the formc1 + c2*sin^2(x) + c3*cos^2(x)with c1, c2, c3 (arbitrary real) constants forms a vector spacepresumably over the real numbeand presumably with the 'usual'operations?First, you need to remember out the vector space operations (vectoraddition and scalar multiplication), but that is not hard: given twosuch functions, which are the vectors,v1 = c1 + c2*sin^2(x) + c3*cos^2(x)v2 = d1 + d2*sin^2(x) + d3*cos^2(x)then their sum isv1 + v2 = (c1+d1) + (c2+d2)*sin^2(x) + (c3+d3)*cos^2(x)which is again of the same form.And given v1 and a scalar k, the scalar product k*v1 isk*v1 = (k*c1) + (k*c2)*sin^2(x) + (k*c3)*cos^2(x)Once you have these, you need to prove the following:(1) The vector sum is associative and commutative (this is easy, since it is just the sum of functions).(2) There is a vector, called 0, such that for all vectors v, v+0=0+v=v (HINT: show that the 0 function, which maps everything to 0, is one of the elements of the set, by choosing c1, c2, and c3 suitably).(3) For each vector v, there is a vector w such that v+w = 0 (HINT: multiply everything by -1)(4) Scalar multiplication is associative: for any two scalars k,m and any vector v, k*(m*v) = (k*m)*v. Again, this is easy, since this is true for any functions.(5) 1*v = v for all vectors v.(6) Scalar products distribute over vector sums: for all scalars a,b and all vectors v and w, a*(v+w) = (a*v) + (a*w) (a+b)*v = (a*v) + (b*v).If you prove these properties, that means the collection === sum? sum_{i=1}^{infty}frac{mu^iln{i!}}{i!}or this one? sum_{i=1}^{infty}frac{mu^iiln i}{i!}i've given === dx->0 of g(x,x+dx)Given a funtion of 2 variables g(x,y), how does one find the limitfunction f(x) for the situation in which y=x ? (Having no Greekcharactelet dx represent delta x in the following.) Asked moreelegantly, how does one findf(x) = lim g(x,x+dx) dx->0 In one hemodialysis paper (call it paper 1) that I am studying,it states that / K ( 1 - -- ) Qb | Qd |Ko*A = ------ * ln| ------- | eqn1 Qb | K | 1 - -- ( 1 - -- ) Qd Qb /when Qd <> Qband KKo*A = ------ eqn2 K 1 - -- Qbwhen Qd = Qbwhere: Ko is the mass transfer coefficient of the dialyzer A is the mass transfer area of the dialyzer Ko*A is the mass transfer area coefficient of the dialyzer Qb is the blood flow rate through the dialyzer Qd is the dialysate flow rate through the dialyzer K is the urea clearance coefficient of the dialyzerwithout any physical or mathematical proof. Another hemodialysis paper (call it paper 2), which did notmention Iwas able to follow the derivation of eqn1. *If* eqn2 is true, possibly without knowing it, in essence, paper 1was stating that the limit of eqn1 when Qd = Qb is eqn2. Via numericalmethods, I have found that this appears to be true by settingQd = Qb + 1 ml/min, etc. Obviously, eqn1 has a singularity at Qb = Qd where it (eqn1) isindeterminate because its 1st term tends toward infinity and its 2ndterm tends toward 0. As a result, this singularity plane must beapproached via limiting. Thinking of eqn1 and eqn2 as / K ( 1 - -- ) Qb | Qd |g(Qb,Qd) = ------ * ln| ------- | eqn3 Qb | K | 1 - -- ( 1 - -- ) Qd Qb /and Kf(Qb) = ------ eqn4 K 1 - -- Qb Since both eqn3 and eqn4 are equal to Ko*A, in essence, paper 1 wasstating thatf(Qb) = lim g(Qb,Qb+dQ) dQ->0where Qd = Qb + dQ How does one find the limit function of eqn1 as Qd goes to Qb? Orfrom another point of view, how does one prove that eqn2 is === lim dx->0 of g(x,x+dx)> Given a funtion of 2 variables g(x,y), how does one find the limit> function f(x) for the situation in which y=x ? (Having no Greek> charactelet dx represent delta x in the following.) Asked more> elegantly, how does one findf(x) = lim g(x,x+dx)> dx->0> In one hemodialysis paper (call it paper 1) that I am studying,> it states that / K > ( 1 - -- )> Qb | Qd |> Ko*A = ------ * ln| ------- | eqn1> Qb | K |> 1 - -- ( 1 - -- )> Qd Qb /when Qd <> Qband K> Ko*A = ------ eqn2> K> 1 - --> Qbwhen Qd = Qbwhere: Ko is the mass transfer coefficient of the dialyzer> A is the mass transfer area of the dialyzer> Ko*A is the mass transfer area coefficient of the dialyzer> Qb is the blood flow rate through the dialyzer> Qd is the dialysate flow rate through the dialyzer> K is the urea clearance coefficient of the dialyzerwithout any physical or mathematical proof. Another hemodialysis paper (call it paper 2), which did not> mention was able to follow the derivation of eqn1. *If* eqn2 is true, possibly without knowing it, in essence, paper 1> was stating that the limit of eqn1 when Qd = Qb is eqn2. Via numerical> methods, I have found that this appears to be true by setting> Qd = Qb + 1 ml/min, etc. Obviously, eqn1 has a singularity at Qb = Qd where it (eqn1) is> indeterminate because its 1st term tends toward infinity and its 2nd> term tends toward 0. As a result, this singularity plane must be> approached via limiting.> Thinking of eqn1 and eqn2 as / K > ( 1 - -- )> Qb | Qd |> g(Qb,Qd) = ------ * ln| ------- | eqn3> Qb | K |> 1 - -- ( 1 - -- )> Qd Qb /and K> f(Qb) = ------ eqn4> K> 1 - --> Qb Since both eqn3 and eqn4 are equal to Ko*A, in essence, paper 1 was> stating thatf(Qb) = lim g(Qb,Qb+dQ)> dQ->0where Qd = Qb + dQ> How does one find the limit function of eqn1 as Qd goes to Qb? Or> from another point of view, how does one prove that in advance,> Kevin BrewerYou will learn these methods when you study calculus.In this case, ln(1)=0 and 1-Qb/Qb=0, so we have an indeterminate form0/0. Methods for this are in your === interesting paper by Paul Garrett on cycloicpolynomials.This will be familiar to number theorists, but I will post it anyway.I also recall doing several problems from Hungerford like thefollowing;if p is prime, n > 1 and odd;F_(p^k)(x) = F_p(x^p^(k-1)) ; F_2n(x) = F_n(-x) when n is odd.(I call F_n the cycloic polynomials).For those not used to them, they come up as irreducible polynomialsassocia with the roots of unity, i.e., x^n - 1 = 0.There are 2 expressions for F_n that are very useful;F_n = (x - a_1)(x - a_2) ... (x - a_m) ; where m = phi(n) is theEuler fcn. of n,and the a_i are primitive roots of unity. If we are looking atx^n - 1 = 0, then the complex solns. are exp[i*2*pi*k/n], which isprimitiveiff (k,n) = 1, i.e. if k is in Z*_n. In this case x^n = 1 but x^(n/d)!= 1 (d|n ; d < n).The other useful formula isF_n(x) = [x^n - 1]/D ; where D = prod(d < n; d|n) F_d(x)This allows calculation of F_n from the F_d where d|n and d < n.Its a good exercise to calculate all of them up to about 15 or 20.F_1 = x - 1 ; F_2 = x + 1 ; F_3 = x^2 + x + 1 =( x^3 - 1)/(x - 1) ;F_4 = x^2 + 1, etc.Based on fairly extensive hand calculations, one might suspect thatallcoefficients of all cycloic polynomials are either +1, 1, or 0, but this is not true. It is true for n prime, and for n having at most2 distinctprime factobut not generally.The smallest n where F_n has coefficients not +/- 1, 0, isn = 105 = 3 x 5 x 7 (there must be at least 3 odd primes).One gets F_105 = [x^105 - 1]/D ; D = F_1 * F_3 * F_5 * F_7 * F_15 *F_21 * F_35= F_3 * F_15 * F_21 *(x^35 - 1) ; then(x^105 - 1)/(x^35 - 1) = x^70 + x^35 + 1 ;continuing like this, and multiplyingtop and bot by appropriate F_d's and x^d - 1, noting that deg(F_105) = phi(105) = 2.4.6 = 48, and that the symmetry of thebinomial coefs.implies symmetry of the coeffs. of F_105 (coefs of decreasing powersare the sameas read for increasing powers of x), one gets a formula forF_105 = N/D ; where both N and D are products of several F_d.Garrett then assumes that assumes that |x| < 1, and expands thingslike1/(1 - x^21) = 1 + x^21 + x^42 + x^63 + ..., and gets all the coefsup to the x^24, since the coefs of x^25 to x^48 will be the sameas x^23 down to x^0 = 1.Finally, he gets a term -2*x^7 (I have not checked all hiscalculations.)which means also a -2*x^41 term.He also has === More Boolean AlgebraI have a couple more boolean algebra expressions that need minimized.I have the answers but I don't know exactly how to get to them:1) X + Y' + X'.Y + (X+Y').X'.Yanswer: 12) ((P'.Q)'.(P.R)')'answer: P'.Q + P.RCan someone show me the steps to get these answers === writing a Master's thesis on ggt and some applications to Riemannian geometry. My base text seems to be de la Harpe's Topics in GGT. The problem is that I need a punch-line; a big, interesting theorem to work towards. I know there's Gromov's theorem on polynomial growth, the only problem is that the proof is really way out of the scope of what I have in mind. At least, that is, according to the proof in Kapovich's Notes on GGT and my cursory look over Gromov's original paper. My problem seems to boil down to Gromov's theorem is easy to state, probably won't be provable in my thesis and doesn't seem to have any geometric consequences that interest me. Some more geometric results like Mostow rigidity or Geometrization don't really use any of the background I want to cover, as detailed in de la Harpe.To cut a longer post a bit shorter: does anyone have any suggestions for major theorems for me to === message ):> But there it is, a single proof where recompressing already compressed > data saves space.What I'm wondering is this:Suppose we have a file F. We have two compression algorithms, A and B. A is > better than B.Will it always be true that...> sizeof(A(F)) <= sizeof(A(B(F)))> in the case where> sizeof(A(F)) <= sizeof(B(F))In other words, does applying a crappy algorithm and then a good one always > result in a LARGER file than just applying the good one?I think the answer is NO, because different algorithms will have different > strengths. But I'm not really sure.I tried thinking about the problem in terms of Kolmogorov complexity,but that seems to be a dead end without some further information. Anyideas on how to decide this?(cross-pos to sci.math. I hope nobody minds.)'cid === Acid Pooh [CapitalEth][EDoubleDot][Micro] this:Suppose we have a file F. We have two compression algorithms, A and B. Ais> better than B.Will it always be true that...> sizeof(A(F)) <= sizeof(A(B(F)))> in the case where> sizeof(A(F)) <= sizeof(B(F))In other words, does applying a crappy algorithm and then a good onealways> result in a LARGER file than just applying the good one?No.Assume n = 2*3*k wlog. (so we don't have to deal with floor functions andsuch)Let A, B compress n consecutive identical bytes using run-length encoding,as follows:A(xxx...x) = (n/2)x xxx...x [(n/2) bytes left uncompressed, for a total sizeof (n/2)+2 bytesB(xxx...x) = (n/3)x xxx...x [n-n/3=2n/3 bytes left uncompressed, for atotal size of 2n/3+2 bytesBy definition, sizeof(A(xxx...x)) < sizeof(B(xxx..x)), on identical equalsized data.A(B(xxx...x)) = A((n/3)x xxx...x) = (n/3)x ((2n/3)/2)x xxx...x = (n/3)x(n/3)x xxx...x, with (n/3) x's left, for a total of (n/3)+4 < (n/2) +2, <=>n/3+2 < n/2,which holds for n > 12.> I think the answer is NO, because different algorithms will havedifferent> strengths. But I'm not really sure.> I tried thinking about the problem in terms of Kolmogorov complexity,> but that seems to be a dead end without some further information. Any> ideas on how to decide this?> (cross-pos to sci.math. I hope nobody minds.)> 'cid 'ooh--------------------------------------------Eventually, === prepare this>device. I have sent my patent application no.8 and now what is holding>me back, apart from my personal problems, is that all these 8>applications might not be enough to secure this device.>So, you have filed seven patents? You should have gotten reference>numbers for this, right? What are they?> (8) Unknown> I couldn't find the manywhere. Are they know what is the source of all my problems?It is lack of Motivation. theory) just to fill vacuum, dipression crea in my mind after thatBig Bang, Glimpses of World History episode way back in 1988. But nowI have reached to my Manzil, aim(which was just to achieve somethingin life); I am just confused what to do next. All the most importantthings of my life are left behind.Without motivation, I just can't stand up to build this device. Sosomehow, I will have to find new motivation to keep me going.I want 180 degree change in circumstances. I am moving to Switzerland when this drama will be over, of course ifI get some money.I think, Switzerland will motivate me to === tried to prepare this>> device. I have sent my patent application no.8 and now what isholding>> me back, apart from my personal problems, is that all these 8>> applications might not be enough to secure this device.>So, you have filed seven patents? You should have gotten reference>numbers for this, right? What are they?>>(8) Unknown> I couldn't find the Volker> Volkar, do you know what is the source of all my problems?> It is lack of Motivation.> theory) just to fill vacuum, dipression crea in my mind after that> Big Bang, Glimpses of World History episode way back in 1988. But now> I have reached to my Manzil, aim(which was just to achieve something> in life); I am just confused what to do next. All the most important> things of my life are left behind.> Without motivation, I just can't stand up to build this device. So> somehow, I will have to find new motivation to keep me going.> I want 180 degree change in circumstances.> I am moving to Switzerland when this drama will be over, of course if> I get some money.> I think, Switzerland will motivate me to build this device.> -Abhi.Actually, your answer is closer to home.Go into your garden and find a small stone. The kind is really irrelevant.Return to your workshop and place the stone atop your head. Now, place yourright thumb into your right ear, and your hook your left thumb into yourNow, quietly wait one minute. Slowly open your eyes, remove the stone fromthe top of your head and place it into your left pocket. with your righthand. Put down your right leg.Lift your left leg, lean your head back and holler IMMAY IMMAY IMMAY IMMAYagain at the top of your lungs. Again slowly open your eyes and take a deepbreath.After another quiet moment, pick up your tools, springs, levehinges,etc. and build your machine. If it then works, it was meant to be and youare the chosen one. If it doesn't, then you are just another ordinaryperson like the rest of us. If that is the case, get over it and get onwith your life. DO NOT - repeat - DO NOT waste another seventeen yearsagonizing over this once you've done the freedom ritual. Your life willbe your own and you can === circumstances. He trained me>to execute Logic. If switch is ON, it just can't be OFF at>exactly same moment. Yes # No.>I am talking about generation of unidirectional force which will>change course of history, physics and will open gateway to entire>universe in future.>Now we make a statement that this device does work.>(1)Above statement is TRUE statement. >Then looking at the simplicity of this device, any person who know>physics can see that mechanism of this device does not need my theory.> Here's a logical mistake:> The simplicity of the device does not depend on it working.V-shaped streched spring is not simple!!! Nevertheless, the simplicity (or Logical mistake number two: Why are you always posting long explanations> of how and why it has to work? Right now I could take any of those,> build it and sell it. Right. Answer is circumstances. Never underestimate awesome power of> circumstances. It can crash you and make very simple task impossible.Feeble excuse. What circumstance? What did you gain >Till this date, I have not prepared or even tried to prepare this>device. I have sent my patent application no.8 and now what is holding>me back, apart from my personal problems, is that all these 8>applications might not be enough to secure this device.> So, you have filed seven patents? You should have gotten reference> numbers for this, right? What are they?(8) UnknownI couldn't find the manywhere. Are they === transformationshelloit's problem of auaticobject model is:x'=Ax+buy=Cx+Dyour intresting matrixes A[n x n]ib[nx1]where A [n x n] is any matrixwhat algorithm must I use to transform matrix A form free form to formcanonic control (?) which means: