mm-373 === Subject: : Re: Google's rankings> While it sucks for those of us who make our money on the net, its also> negative for the users of google. Have you tried finding something> lately? It seems most searches are quite skewed now. I find myself> having a harder time finding things on google now. And the funniest> part is, those sites that used to be top for terms like hardcore sex> are still at the top,I have to ask ... how would you know that?=== === Subject: : Re: More Accessible Algorithm-MazeThis puzzle has no solution; it is defective. The 11th step directsyou to skip the 11th step the second time you encounter it. You can'tskip the step if the step itself directs you to skip it.Also I'm lazy, and it was easier to stop the first time I got to the11th step.I'll try it again tomorrow, reading the 11th step as it was probablyintended.> Below is an abstract maze.> It is a bit more accessible/understandable and easier than my most > recently pos algorithm maze. > http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&threadm= b4be2fdf.0312081614.7770f73e%40posting.google.com&rnum=1&prev= > (For example, its only math is simple addition.)> So, perhaps it would be enjoyable for some children to try to solve> the 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 to> form the maze's solution-path.> By adjacent square, I mean a square immediately next> to 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 a> rook moves in Chess.)> And, importantly, you must not leave the grid or land> on any square more than once!> (But not every square is necessarily visi by the> solution'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 most> immediately previous number your path crossed before your> current 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 number> of [symbols] in the entire maze, where [symbol] is the> symbol in the square currently you are at > (which you moved to in previous step).> 11) Repeat the fifth step (step-5 above) twice, then repeat> steps (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??..)>=== === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et alContent-transfer-encoding: 8bit> Who's Madigin?Didn't Mozart write a piano concerto for her?-- Dot.=== === Subject: : Re: digital logicHi. Thanks for responding. I'm including some (working) Python code that mayhelp to clue you in wrt where I'm at regarding this problem. Arguably you'd beable to tell more from the code that doesn't work, but then you couldn't see itin action.Here are some basic questions that I hope you can help me with:1) Am I supposed to assume that the time required for a signal to travel along awire is negligible, in the sense that I can always assume it's instantaneous?2) Am I supposed to assume that JK-flipflops and D-flipflops (more generally,all types of flipflops) are time-equivalent? Here's what I mean by that:Suppose that t_0 is a moment of time (continuous time, the time of Newtonianphysics); let X be a JK-flipflop, and Y a D-flipflop, such that each input of Xand each input of Y stays the same (as itself) in some time period just beforet_0. If one or more of the inputs of each flipflop assume(s) a different valueat t_0, and the new inputs are, in each case, such that the output of theflipflop is going to change, then can I assume that X and Y will change theirrespective output values at the same moment?2') In case that still isn't clear, what I'm asking for is a yea or nayregarding the validity of the following argument:Hypotheses:i) X is a JK-flipflop with inputs A1...An and output Bii) Y is a D-flipflop with inputs C1...Cm and output Diii) There is a moment t_0 of time and a number e>0 such that, for (t_0-e) < tx< ty < t_0, Ai(tx)=Ai(ty) for 1<=i<=n and Bi(tx)=Bi(ty) for1<=i<=m.iv) There exists an i with Ai(t_0)=/=Ai(t) for (t_0-e) < t < t_0, and thereexists an i with Bi(t_0)=/=Bi(t) for (t_0-e) < t < t_0.v) The change of input values at t_0 causes both flipflops to change theirrespective output values.(Putative) Conclusion:If tX is the supremum of {t: B(T)=B(t_0) for t_0<=T<=t}, and tY is the supremumof {t: D(T)=D(t_0) for t_0<=T<=t}, then tX=tY.Peace, and thanks again. (Code follows.)MY LATEST FIDDLING, SHOWN HERE, IS A MOD-13 COUNTER.IT'S ESSENTIALLY OK, ALTHOUGH IT SHOWS 13 FOR HALF ACLOCK CYCLE BEFORE RESETTING TO ZERO########################################################## ## BOOLEAN FUNCTIONS:AND = lambda *bitVector: int(0 not in bitVector)OR = lambda *bitVector: int(1 in bitVector)XOR = lambda *bitVector: sum(bitVector)%2NAND = lambda *bitVector: int(0 in bitVector)NOR = lambda *bitVector: int(1 not in bitVector)NOT = lambda bit: int(not bit)########################################################## IF [d_n, ... ,d_0] IS A LIST OF BINARY DIGITS, THENnum([d_n, ... ,d_0]) IS THE NUMBER WHOSE BINARY NUMERALIS d_n...d_0 ; IF n IS THE NUMBER WHOSE BINARY NUMERALIS d_n...d_0 THEN bitVector(n) IS [d_n, ... ,d_0] .def num(bitVector): digits=list(bitVector) digits.reverse() return sum([digits[i]*pow(2,i) for i in range(len(digits))])def bitVector(num): twoExp = lambda x: x>=2 and twoExp(x/2)+1 digits=[(num>>i)%2 for i in range(twoExp(num)+1)] digits.reverse() return digits######################################################## #### NOT APPEARING IN THIS EPISODE:decoder = lambda *bitVector: [int(i==num(bitVector)) for i in range(len(bitVector))]mux = lambda *bitVector: lambda *dataVector: dataVector[num(bitVector)]def bus(*args): if sum([enabler for (driver,enabler) in args])!=1: raise bus error for (driver,enabler) in args: if enabler: return driver######################################################## ##NODES REPRESENT THE CONNECTIONS BETWEEN THE OUTPUT OF ONETHINGIE (OR OF A SWITCH OR A CLOCK) AND THE OUTPUT OFANOTHER THINGIE. IF X IS A NODE, THEN nodeVals[X] IS THECURRENT VALUE OF X; THAT IS, THE SIGNAL CURRENTLY BEINGCARRIED ALONG X. IF STRG IS A COMMA-SEPARA STRING OFWHICH IS IRRELEVANT, IS THEMSELVES); INITIALLY THEY ALLHAVE VALUE 0. toggle(node) DOES EXACTLY WHAT IT SAYS.IF transitionList IS A LIST OF INSTRUCTIONS FOR UPDATINGNODE VALUES, THEN doTransitions(transitionList) DOES EXACTLYWHAT IT SAYS; ITS PURPOSE IS TO TRANSLATE BOOLEAN FUNCTIONSINTO THE CORRESPONDING OPERATIONS ON nodeVals. counter DOESA SIMILAR TRANSLATION WRT num. T_edge ANSWERS TO AN EDGE-TRIGGEREDT-TYPE FLIPFLOP (WITH A RESET INPUT)nodeVals={}def declareNodes(strg): args=strg.split(,) for x in args: exec x + =' + x + ' in globals() nodeVals[x]=0def toggle(node): nodeVals[node]=NOT(nodeVals[node])def doTransitions(transitionList): global nodeVals newNodeVals=nodeVals.copy() for transition in transitionList: (assignee,funct,args)=transition newNodeVals[assignee]=funct(*[nodeVals[x] for x in args]) nodeVals=newNodeValscounter = lambda *nodeVector: num([nodeVals[x] for x in nodeVector])def T_edge(T,Q,RESET): if RESET: return 0 return (T and NOT(Q)) or (NOT(T) and Q)########################################################## EACH f_i REPRESENTS THE TRANSITION FUNCTION FOR THEFLIPFLOP WHOSE VALUE WILL BE THE i-TH DIGIT OF THENUMERAL SHOWN BY THE COUNTER. thirteen(t0,t1,t2,t3) IS1 IF num([t3,t2,t1,t0]) IS 13, AND 0 OTHERWISE.THE g_i'S ARE LIKE THE f_i'S, BUT INCORPORATE THE'RESET AT THIRTEEN' FUNCTIONf0 = lambda t0,t1,t2,t3,clk,reset: T_edge(clk,t0,reset)f1 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0),t1,reset)f2 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0,t1),t2,reset)f3 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0,t1,t2),t3,reset)thirteen = lambda t0,t1,t2,t3: AND(t0,NOT(t1),t2,t3)g0 = lambda t0,t1,t2,t3,clk: f0(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))g1 = lambda t0,t1,t2,t3,clk: f1(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))g2 = lambda t0,t1,t2,t3,clk: f2(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))g3 = lambda t0,t1,t2,t3,clk: f3(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))##################### ####################################### SCRIPT TO SHOW HOW THE ABOVE DEFINITIONS WORK:declareNodes(T0,T1,T2,T3,CLK)## LIST OF UPDATE-INSTRUCTIONS:mod_13_counter = [(T0,g0,[T0,T1,T2,T3,CLK]) ,(T1,g1,[T0,T1,T2,T3,CLK]) ,(T2,g2,[T0,T1,T2,T3,CLK]) ,(T3,g3,[T0,T1,T2,T3,CLK]) ]for i in range(30): print counter(T3,T2,T1,T0) toggle(CLK) doTransitions(mod_13_counter)## END|| > === Subject: : Digital Logic| >Level: Beginner| >|| >| >Essentially what I want to know is this: How can you tell how a circuit will| >behave without knowing the lengths of the gate delays? For example, how do| >simulation programs (eg: LogicWorks, Multimedia Logic) determine how circuit| >simulations should behave?| >| Others have given good answers but I will throw in my 2 cents worth| anyway. Simple state machines generally consist of flip-flops, which| may be either J-K or D input types, logic gates, and a clock. We might| denote the current state of a flip-flop by Q and its next state by Q+.| In the case of a JK flip-flop, for example, the outputs are immune to| the inputs when the clock is low. When the clock goes high the inputs| are locked in and read. On the transition from high to low the state| of the flip-flop changes from its current state Q to its next state| Q+. During this low clock period, the flip-flop is immune to the| changes in inputs. This gives time for all the logic gates to receive| their new inputs, including any fed-back values of the Q+ outputs| which now represent the new current state. All the gate delays have| time to settle to their new values and set the new J-K inputs, so the| flip flop is settled and ready to go when the clock goes back high and| the cycle repeats.|| This happens for all flip-flops in the circuit at the same time, so| what you get is a machine that proceeds from one state to the next| each time the clock goes from high to low for a JK flip flop.|| The boolean logic is used to specify the next state equations, that is| Q+ is a function of J and K, which are functions of the inputs and the| current state of the machine. Figuring out the logic equations for the| J and K inputs is where you usually get to dealing with Karnaugh maps.|| Machines of this type are called synchronous machines. They may appear| to respond immediately to input changes, but that is only because the| clocks are generally running very fast.|| Machines with no clock are called asynchronous machines and may indeed| have problems with gate delays and races between outputs and inputs| which must be carefully guarded against.|| Hope that helps.|| --Lynn|=== === Subject: : Re: digital logicNo, that's not right. As others have poin out, you're overlooking the fact that most flip-flops are clocked. They are designed so that they only change when the clock makes a transition. The flip-flop then takes up whatever new state it should have according to its inputs. The circuit sees _ONLY_ the values of the inputs at the clock transition, which can be thought of as a specific instant of time.Detailed simulations such as Pspice do consider gate delays, etc., but most simulations of circuits involving clocked logic assume idealized behavior in which clock transitions are instantaneous and all flip-flops change at the same instant.Jack> Hi. Thanks for responding. I'm including some (working) Python code that may> help to clue you in wrt where I'm at regarding this problem. Arguably you'd be> able to tell more from the code that doesn't work, but then you couldn't see it> in action.> Here are some basic questions that I hope you can help me with:> 1) Am I supposed to assume that the time required for a signal to travel along a> wire is negligible, in the sense that I can always assume it's instantaneous?> 2) Am I supposed to assume that JK-flipflops and D-flipflops (more generally,> all types of flipflops) are time-equivalent? Here's what I mean by that:> Suppose that t_0 is a moment of time (continuous time, the time of Newtonian> physics); let X be a JK-flipflop, and Y a D-flipflop, such that each input of X> and each input of Y stays the same (as itself) in some time period just before> t_0. If one or more of the inputs of each flipflop assume(s) a different value> at t_0, and the new inputs are, in each case, such that the output of the> flipflop is going to change, then can I assume that X and Y will change their> respective output values at the same moment?> 2') In case that still isn't clear, what I'm asking for is a yea or nay> regarding the validity of the following argument:> Hypotheses:> i) X is a JK-flipflop with inputs A1...An and output B> ii) Y is a D-flipflop with inputs C1...Cm and output D> iii) There is a moment t_0 of time and a number e>0 such that, for (t_0-e) < tx> < ty < t_0, Ai(tx)=Ai(ty) for 1<=i<=n and Bi(tx)=Bi(ty) for1<=i<=m.> iv) There exists an i with Ai(t_0)=/=Ai(t) for (t_0-e) < t < t_0, and there> exists an i with Bi(t_0)=/=Bi(t) for (t_0-e) < t < t_0.> v) The change of input values at t_0 causes both flipflops to change their> respective output values.> (Putative) Conclusion:> If tX is the supremum of {t: B(T)=B(t_0) for t_0<=T<=t}, and tY is the supremum> of {t: D(T)=D(t_0) for t_0<=T<=t}, then tX=tY.> Peace, and thanks again. (Code follows.)MY LATEST FIDDLING, SHOWN HERE, IS A MOD-13 COUNTER.> IT'S ESSENTIALLY OK, ALTHOUGH IT SHOWS 13 FOR HALF A> CLOCK CYCLE BEFORE RESETTING TO ZERO########################################################## > ## BOOLEAN FUNCTIONS:> AND = lambda *bitVector: int(0 not in bitVector)> OR = lambda *bitVector: int(1 in bitVector)> XOR = lambda *bitVector: sum(bitVector)%2> NAND = lambda *bitVector: int(0 in bitVector)> NOR = lambda *bitVector: int(1 not in bitVector)> NOT = lambda bit: int(not bit)> ##########################################################IF [d_n, ... ,d_0] IS A LIST OF BINARY DIGITS, THEN> num([d_n, ... ,d_0]) IS THE NUMBER WHOSE BINARY NUMERAL> IS d_n...d_0 ; IF n IS THE NUMBER WHOSE BINARY NUMERAL> IS d_n...d_0 THEN bitVector(n) IS [d_n, ... ,d_0] .> def num(bitVector):> digits=list(bitVector)> digits.reverse()> return sum([digits[i]*pow(2,i) for i in range(len(digits))])> def bitVector(num):> twoExp = lambda x: x>=2 and twoExp(x/2)+1> digits=[(num>>i)%2 for i in range(twoExp(num)+1)]> digits.reverse()> return digits> ##########################################################> ## NOT APPEARING IN THIS EPISODE:> decoder = lambda *bitVector: > [int(i==num(bitVector)) for i in range(len(bitVector))]> mux = lambda *bitVector: > lambda *dataVector: dataVector[num(bitVector)]> def bus(*args):> if sum([enabler for (driver,enabler) in args])!=1:> raise bus error> for (driver,enabler) in args:> if enabler:> return driver> ########################################################## NODES REPRESENT THE CONNECTIONS BETWEEN THE OUTPUT OF ONE> THINGIE (OR OF A SWITCH OR A CLOCK) AND THE OUTPUT OF> ANOTHER THINGIE. IF X IS A NODE, THEN nodeVals[X] IS THE> CURRENT VALUE OF X; THAT IS, THE SIGNAL CURRENTLY BEING> CARRIED ALONG X. IF STRG IS A COMMA-SEPARA STRING OF> WHICH IS IRRELEVANT, IS THEMSELVES); INITIALLY THEY ALL> HAVE VALUE 0. toggle(node) DOES EXACTLY WHAT IT SAYS.> IF transitionList IS A LIST OF INSTRUCTIONS FOR UPDATING> NODE VALUES, THEN doTransitions(transitionList) DOES EXACTLY> WHAT IT SAYS; ITS PURPOSE IS TO TRANSLATE BOOLEAN FUNCTIONS> INTO THE CORRESPONDING OPERATIONS ON nodeVals. counter DOES> A SIMILAR TRANSLATION WRT num. T_edge ANSWERS TO AN EDGE-TRIGGERED> T-TYPE FLIPFLOP (WITH A RESET INPUT)> nodeVals={}> def declareNodes(strg):> args=strg.split(,)> for x in args:> exec x + =' + x + ' in globals()> nodeVals[x]=0> def toggle(node):> nodeVals[node]=NOT(nodeVals[node])> def doTransitions(transitionList):> global nodeVals> newNodeVals=nodeVals.copy()> for transition in transitionList:> (assignee,funct,args)=transition> newNodeVals[assignee]=funct(*[nodeVals[x] for x in args])> nodeVals=newNodeVals> counter = lambda *nodeVector: num([nodeVals[x] for x in nodeVector])> def T_edge(T,Q,RESET):> if RESET:> return 0> return (T and NOT(Q)) or (NOT(T) and Q)> ##########################################################EACH f_i REPRESENTS THE TRANSITION FUNCTION FOR THE> FLIPFLOP WHOSE VALUE WILL BE THE i-TH DIGIT OF THE> NUMERAL SHOWN BY THE COUNTER. thirteen(t0,t1,t2,t3) IS> 1 IF num([t3,t2,t1,t0]) IS 13, AND 0 OTHERWISE.> THE g_i'S ARE LIKE THE f_i'S, BUT INCORPORATE THE> 'RESET AT THIRTEEN' FUNCTION> f0 = lambda t0,t1,t2,t3,clk,reset: T_edge(clk,t0,reset)> f1 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0),t1,reset)> f2 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0,t1),t2,reset)> f3 = lambda t0,t1,t2,t3,clk,reset: T_edge(AND(clk,t0,t1,t2),t3,reset)> thirteen = lambda t0,t1,t2,t3: AND(t0,NOT(t1),t2,t3)> g0 = lambda t0,t1,t2,t3,clk: f0(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))> g1 = lambda t0,t1,t2,t3,clk: f1(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))> g2 = lambda t0,t1,t2,t3,clk: f2(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))> g3 = lambda t0,t1,t2,t3,clk: f3(t0,t1,t2,t3,clk,thirteen(t0,t1,t2,t3))> ##########################################################> ## SCRIPT TO SHOW HOW THE ABOVE DEFINITIONS WORK:> declareNodes(T0,T1,T2,T3,CLK)> ## LIST OF UPDATE-INSTRUCTIONS:> mod_13_counter = [> (T0,g0,[T0,T1,T2,T3,CLK]) ,> (T1,g1,[T0,T1,T2,T3,CLK]) ,> (T2,g2,[T0,T1,T2,T3,CLK]) ,> (T3,g3,[T0,T1,T2,T3,CLK]) ]> for i in range(30):> print counter(T3,T2,T1,T0)> toggle(CLK)> doTransitions(mod_13_counter)> ## END> |> | > === Subject: : Digital Logic> | >Level: Beginner> | |> | | >Essentially what I want to know is this: How can you tell how a circuit will> | >behave without knowing the lengths of the gate delays? For example, how do> | >simulation programs (eg: LogicWorks, Multimedia Logic) determine how circuit> | >simulations should behave?> | | Others have given good answers but I will throw in my 2 cents worth> | anyway. Simple state machines generally consist of flip-flops, which> | may be either J-K or D input types, logic gates, and a clock. We might> | denote the current state of a flip-flop by Q and its next state by Q+.> | In the case of a JK flip-flop, for example, the outputs are immune to> | the inputs when the clock is low. When the clock goes high the inputs> | are locked in and read. On the transition from high to low the state> | of the flip-flop changes from its current state Q to its next state> | Q+. During this low clock period, the flip-flop is immune to the> | changes in inputs. This gives time for all the logic gates to receive> | their new inputs, including any fed-back values of the Q+ outputs> | which now represent the new current state. All the gate delays have> | time to settle to their new values and set the new J-K inputs, so the> | flip flop is settled and ready to go when the clock goes back high and> | the cycle repeats.> |> | This happens for all flip-flops in the circuit at the same time, so> | what you get is a machine that proceeds from one state to the next> | each time the clock goes from high to low for a JK flip flop.> |> | The boolean logic is used to specify the next state equations, that is> | Q+ is a function of J and K, which are functions of the inputs and the> | current state of the machine. Figuring out the logic equations for the> | J and K inputs is where you usually get to dealing with Karnaugh maps.> |> | Machines of this type are called synchronous machines. They may appear> | to respond immediately to input changes, but that is only because the> | clocks are generally running very fast.> |> | Machines with no clock are called asynchronous machines and may indeed> | have problems with gate delays and races between outputs and inputs> | which must be carefully guarded against.> |> | Hope that helps.> |> | --Lynn> |=== === Subject: : Re: digital logic>1) Am I supposed to assume that the time required for a signal to travel along a>wire is negligible, in the sense that I can always assume it's instantaneous?In practice it is negligible compared to the clock rate. If I werewriting a simulator I would sequentially:1. Read the states Q_i2. Read the inputs I_i3. Calculate J_i and K_i from the states and inputs.4. Evaluate the next states Q_i+5. Repeat, using the Q_i+ as the new Q_i>2) Am I supposed to assume that JK-flipflops and D-flipflops (more generally,>all types of flipflops) are time-equivalent? Here's what I mean by that:>Suppose that t_0 is a moment of time (continuous time, the time of Newtonian>physics); let X be a JK-flipflop, and Y a D-flipflop, such that each input of X>and each input of Y stays the same (as itself) in some time period just before>t_0. If one or more of the inputs of each flipflop assume(s) a different value>at t_0, and the new inputs are, in each case, such that the output of the>flipflop is going to change, then can I assume that X and Y will change their>respective output values at the same moment?I don't think so. Changing the J or K input just before the statetransition is a no-no. I would think a physical flip-flop probably hassome sort of mechanism built in to protect against this, but I don'tknow enough about hardware to tell you. I do know that good designrequires that inputs be synchronized with the low clock (for JK). Alsonote that D and JK flip-flops change state at different parts of theclock cycle so I don't think it is a good practice to mix them. Theycertainly don't change states at the same time. The idea is thatduring low clock is when the logic gates all settle down to their newvalues so everything is stable during high clock and for all practicalpurposes you get simultaneous state changes on the transition. Thingsare just the opposite for D types; they change on the rising clock.Disclaimer: I an not a EE person but have studied the topic and havedesigned and built several digital logic circuits. So anything that'sreally important you might want to check with the experts.>Peace, and thanks again. (Code follows.)Sorry, I don't know Python so I won't comment on your code. Are youwriting a simulator? Why don't you use logic works or somethingsimilar to test your logic and simulate your circuit?--Lynn=== === Subject: : some questions about a series of functionsLet f_n(x)=frac{1}{1+n^2 x}, and f(x)=sum_{n=1}^infty f_n(x).It is obvious to me that the sum is absolutely convergent for x>0, andeven so when x<-1. However, for -1<=x<=0 I am a little confused. Theseries definitely diverges for x=0, but for x=-1/3, for example, thereis a term in the sum that is 1/0... does that automatically make thesum undefined? So does this mean that f(x) is undefined wheneverx=-1/k for some integer k?I'm trying to answer some questions about this series (such ascontinuity, uniform convergence, and boundedness), but it is difficultwhen I do not understand how to handle these weirdness.Thanks=== === Subject: : Re: some questions about a series of functionsProblem 4 Chapter 7 Rudin eh?> Let f_n(x)=frac{1}{1+n^2 x}, and f(x)=sum_{n=1}^infty f_n(x).> It is obvious to me that the sum is absolutely convergent for x>0, and> even so when x<-1. However, for -1<=x<=0 I am a little confused. The> series definitely diverges for x=0, but for x=-1/3, for example, there> is a term in the sum that is 1/0... does that automatically make the> sum undefined? So does this mean that f(x) is undefined whenever> x=-1/k for some integer k?> I'm trying to answer some questions about this series (such as> continuity, uniform convergence, and boundedness), but it is difficult> when I do not understand how to handle these weirdness.Problem 4 Chapter 7 Rudin eh?The function sum( f_{n}(x) ) does not exist for the values -1/k^2, k =1,2,3...To show _uniform_ convergence on (-1/k^2,-1/(k+1)^2) proceed as such:Throw away the first __ terms of the series, so that the following makessense-1/k^2 < x < -1/(k+1)^2 if and only if1 - n^2/k^2 < 1 - n^2.x < 1 - n^2/(k+1)^2.Take recipricals. Note that k is fixed, apply the most applicable theorem ofthis chapter. You know, the one you've used like 5 times previously in thisexercise.Best, Andy=== === Subject: : Re: some questions about a series of functions> Let f_n(x)=frac{1}{1+n^2 x}, and f(x)=sum_{n=1}^infty f_n(x).> It is obvious to me that the sum is absolutely convergent for x>0, and> even so when x<-1. However, for -1<=x<=0 I am a little confused. The> series definitely diverges for x=0, but for x=-1/3, for example, there> is a term in the sum that is 1/0... does that automatically make the> sum undefined? So does this mean that f(x) is undefined whenever> x=-1/k for some integer k?I'm a little confused here. You have f_n(-1/3) = 1/(1 - n^2/3); forwhich n do you claim that this is 1/0?But, yes, that sum is defined only for thos valus of x such thatevery f_n(x) is defined.=== === Subject: : Re: some questions about a series of functions> Let f_n(x)=frac{1}{1+n^2 x}, and f(x)=sum_{n=1}^infty f_n(x).> It is obvious to me that the sum is absolutely convergent for x>0, and> even so when x<-1. However, for -1<=x<=0 I am a little confused. The> series definitely diverges for x=0, but for x=-1/3, for example, there> is a term in the sum that is 1/0... does that automatically make the> sum undefined? So does this mean that f(x) is undefined whenever> x=-1/k for some integer k?> I'm a little confused here. You have f_n(-1/3) = 1/(1 - n^2/3); for> which n do you claim that this is 1/0?> But, yes, that sum is defined only for thos valus of x such that> every f_n(x) is defined.> I'm sorry, a few minutes after I pos I realized my error, but Icouldn't get to a computer to correct. I mean to ask is f(-1/k^2)undefined whenever k is an integer, since a term in the sum would thenbe 1/0? Such as f(-1/9)=1/(1-1/9) + 1/(1-4/9) + 1/(1-9/9) + ...In hindsight of your reply, does that mean that the series isabsolutely convergent for all xin R, except for numbers of the form-1/k^2, for kin N ?Thanks=== === Subject: : Re: JSH: Past posters, consider facts> === Subject: : Re: JSH: Past posters, consider facts>Message-id: >A ring 'garantees' that if you add A and B, and A and B are on that>ring, the result A+B is also on that ring. A ring also 'garantees'>that A-B is on that ring (because every element has an additive>inverse), so substraction is covered too. Multiplication is also>'covered' (A*B is on the ring as well).>It was my understanding that subtraction is not covered.>>1-2 is not a Natural Number.>>No, the natural numbers are not a ring. A ring has an additive identity>(0) and additive inverses.> Ok, my understanding is wrong, but subtraction is still not a ring operation,> is it?In order to have a ring, you must first have an additive (Abelian) group, which mean that there must be an additive inverse, or negative, for every member of that group.Thus a - b is really a + (-b), and is _always_ defined in a ring.=== === Subject: : Re: JSH: Past posters, consider facts> === Subject: : Re: JSH: Past posters, consider facts>Message-id: > === Subject: : Re: JSH: Past posters, consider facts>Message-id: Okay, just so I get this...>>A ring 'garantees' that if you add A and B, and A and B are on that>ring, the result A+B is also on that ring. A ring also 'garantees'>that A-B is on that ring (because every element has an additive>inverse), so substraction is covered too. Multiplication is also>'covered' (A*B is on the ring as well).> It was my understanding that subtraction is not covered.> 1-2 is not a Natural Number.>It is covered. The Natural Numbers do not form a ring.>You can find some basic terminology defined on:> >( this shows also some difference in terminology between writers).>Also note that I use the word properties on that page. There are>texts around that call them axioms. That is a serious misnaming.Out of curiosity about how I had all this wrong, I went and looked upmy original source of the definition of a ring: the VNR Concise Encyclopedia of Mathematics. The definition of a ring makes no mention of the inverse or unit elements. What it _does_ say isUnder addition, R must be an Abelian groupOf course, it was my responsibility to go look up Abelian group(in a different section of the book about 300 pages away). So thismisunderstanding was due to careless use of reference materials.And unlike your page above, where rela material is all pulledtogether in one place, reading an encyclopedia isn't the best wayto try and learn something.I have no clue where I got the notion that the Natural Numbers were a ring. Can I blame Harris for that?>-- >dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,>+31205924131>home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/--=== === Subject: : Re: The Lost Proof of FermatB > AA^x + B^x[A^x + B^x]^[1/x]L'Hopital's Rule:Limit f(x)/g(x) = Limit f'(x)/g'(x)Take the natural logLn[A^x + B^x]^[1/x] = Ln[A^x + B^x]/x= f(x)/g(x)L'Hopital's Rule...Limit Ln[A^x + B^x]/x =[(A^x)*Ln[A] + (B^x)*Ln[B]]/[A^x + B^x] / 1= LnA/[1 + [B/A]^x] + LnB/[1+[A/B]^x]Take the limit= 0 + Ln[B]e^LnB = BLimitfor B > A[A^x + B^x]^[1/x] = B=== === Subject: : Re: The Lost Proof of Fermat> B > A> A^x + B^x> [A^x + B^x]^[1/x]> L'Hopital's Rule:> Limit f(x)/g(x) = Limit f'(x)/g'(x)Aha! What is lim_{x->1} x^2/(x + 1)?It is lim_{x->1} 2x/1 = 2. Very neat-- === === Subject: : Re: The Lost Proof of Fermat satisfy following constrains: f(a,b) = f(c,d) if and only if a=b, c=d or a=c, b=d.> for all positive integers a,b,c,d. ThanksDefine f(x,x) = 0 or f(x,y) = x + i*y, where i^2 = -1 and x <> y.=== === Subject: : Re: functions needed>> I need to construct such a family of functions which>> satisfy following constrains:>> f(a,b) = f(c,d) if and only if a=b, c=d or a=c, b=d.>> for all positive integers a,b,c,d.>> Thanks> Define f(x,x) = 0 or> f(x,y) = x + i*y, where i^2 = -1 and x <> y.It's supposed to be *commutative*; (a,b) -> a+bi is not,Nor is the earlier suggestion of (a,b) -> a.b.(I took it that f: Z+ x Z+ -> Z+ as well.)Try this ... f(a,b)=(2^(max(a,b)))*(3^(min(a,b)))=== === Subject: : Re: functions needed>> I need to construct such a family of functions which>> satisfy following constrains:>> f(a,b) = f(c,d) if and only if a=b, c=d or a=c, b=d.>> for all positive integers a,b,c,d.>> Thanks> Define f(x,x) = 0 or> f(x,y) = x + i*y, where i^2 = -1 and x <> y.> It's supposed to be *commutative*; (a,b) -> a+bi is not,> Nor is the earlier suggestion of (a,b) -> a.b.I do not see anywhere that either f(a,b) = f(b,a) is required,nor that f(f(a,b),c) = f(a,f(b,c)) is required, whichever you mean by commutative.My suggestion satisfies all the sta conditions.> (I took it that f: Z+ x Z+ -> Z+ as well.)> Try this ...> f(a,b)=(2^(max(a,b)))*(3^(min(a,b)))=== === Subject: : Re: hello...anaysis problem...>>so>> (int(|f(x)|^n,x in [0,1]))^(1/n) >= (b - a)^(1/n).N>>and this implies that>> lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= N.> You haven't shown the limit exists yet. >>Since this is true for each N < M,>> lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= M.> Same problem. M' < int(|f(x)|^n,x in [0,1]))^(1/n) <= Mfor every large enough n, and this proves that the limit exists and thatit is equal to M.> By the way, this is probably homework, so don't you think it would be > better to give a nice hint rather than the entire solution?You see, I did not *know* the answer; I had to think a bit beforeanswering and when that happens it's hard to resist to the temptationof giving the full answer.But OK, I guess that I should have just given a hint.=== === Subject: : Re: hello...anaysis problem...thank you very much...this problem is not homework.i am not going to the school.it's only studying doubt.thank you that concerned about my problem.=== === Subject: : Re: hello...anaysis problem...>so (int(|f(x)|^n,x in [0,1]))^(1/n) >= (b - a)^(1/n).N>>and this implies that lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= N.>> You haven't shown the limit exists yet. >Since this is true for each N < M, lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= M.>> Same problem.> M' < int(|f(x)|^n,x in [0,1]))^(1/n) <= M>for every large enough n, and this proves that the limit exists and that>it is equal to M.Yes. You might note that this issue, can't say anything about thelimit until we've shown the limit exists is one reason for the existence of the concepts lim sup and lim inf; any sequencehas a lim sup and a lim inf, and it has a limit if and only iflim sup = lim inf:You could make your original argument strictly correct bynoting that it's obvious that the lim sup is <= M, andthen the second half of the argument shows that thelim inf is >= M. Since lim inf <= lim sup it follows thatlim inf = lim sup = M, which implies lim = M.>> By the way, this is probably homework, so don't you think it would be >> better to give a nice hint rather than the entire solution?>You see, I did not *know* the answer; I had to think a bit before>answering and when that happens it's hard to resist to the temptation>of giving the full answer.>But OK, I guess that I should have just given a hint.>>************************=== === Subject: : Re: Resolvable Space <0401241647330.16637-100000@gandalf.math.ukans.edu>Sorry about the typoes. Here is a correc version.> A space S is resolvable when S is union two disjoint dense sets.> Are all spaces without open singletons resolvable?> I think not. Is there an example why not?An irresolvable T_1 space (X,t) with no isola points: Let X be aninfinite set. Let u be a free ultrafilter on X. Let t be the topologyon X consisting of the elements of u and the empty set.An irresovable Hausdorff space (X,t) with no isola points: Let X bean infinite set. Let t_0 be a Hausdorff topology on X with no isolapoints. Let T be the set of all topologies on X that extend t_0 andcontain no singletons. By Zorn's lemma, T has maximal elements. Let tbe a maximal element of T.> A space S is equally resolvable when S is union> two disjoint equinumerous dense sets.> Are all resovable spaces equally resovable?No, any finite multipoint resovable space with odd number of pts is> counterexample. Are there any Hausdorff or infinite counterexamples?No. Let X be an infinite topological space, say |X| = m. Suppose X isthe union of two disjoint dense sets A and B of different sizes, say|A| = m > |B|. Partition A into m disjoint sets A_i, each of size m.Assume for a contradiction that none of the sets AA_i is dense in X.Since B is dense in X, the closure of AA_i does not contain B. Choosea point b_i in B which is not in the closure of AA_i. Since |B| < m,there are two distinct indices i and j such that b_i = b_j. But thenb_i is not in the closure of A, a contradiction. Thus, for some i, Xis the union of the two disjoint dense sets AA_i and BuA_i, both ofsize m.=== === Subject: : Re: Resolvable Space <0401241647330.16637-100000@gandalf.math.ukans.edu> <0401250141230.17543-100000@gandalf.math.ukans.edu>Sorry about the typoes. Here is a correc version.Thanks for the courtesy of including the revisions in complete copy. >> A space S is resolvable when S is union two disjoint dense sets. >> Are all spaces without open singletons resolvable? >An irresolvable T_1 space (X,t) with no isola points: Let X be an >infinite set. Let u be a free ultrafilter on X. Let t be the >topology on X consisting of the elements of u and the empty set.An ultra filter space S is irresolvable becauseEvery open set is dense as an ultra filter space is hyperconnec.Dense sets are open. If a set dense isn't in the ultra filter, then as it intersects all open nonnul sets, it can be added to the ultra filter, oh my.No set and it's complement can both be dense as both can't be open.-- >An irresovable Hausdorff space (X,t) with no isola points: >Let X be an infinite set. >Let t_0 be a Hausdorff topology on X with no isola points.For example Q and R, for those cardinalities. >Let T be the set of all topologies on X that extend t_0 and >contain no singletons. By Zorn's lemma, T has maximal elements.Indeed, the sup topology of a chain C of topologies is the topologywith base union C. Thus it has no open singletons. >Let t be a maximal element of T.Now what? Show every t-dense set D has nonnul interior?-- >> A space S is equally resolvable when S is union >> two disjoint equinumerous dense sets. >> Are all resolvable spaces equally resolvable? >> No, any finite multipoint resolvable space with odd number of pts >> is counterexample. Are there any Hausdorff or infinite >> counterexamples? >No. Let X be an infinite topological space, say |X| = m. Suppose X >is the union of two disjoint dense sets A and B of different sizes, >say |A| = m > |B|. >Partition A into m disjoint sets A_i, each of size m. >Assume for a contradiction that none of the sets AA_i is dense in >X. Since B is dense in X, the closure of AA_i does not contain B. >Choose a point b_i in B which is not in the closure of AA_i. >Since |B| < m, there are two distinct indices i and j >such that b_i = b_j. >But then b_i is not in the closure of A, a contradiction.cl A = cl (A - A_i/A_j) = cl (AA_i / AA_j) = (cl AA_i) / (cl AA_j) >Thus, for some i, X is the union of the two disjoint >dense sets AA_i and BuA_i, both of size m.Well done cardinal proof.So the question of equally resolvable is barely interesting.----=== === Subject: : Re: Resolvable Space <0401241647330.16637-100000@gandalf.math.ukans.edu>[...]>An irresovable Hausdorff space (X,t) with no isola points:>Let X be an infinite set.>Let t_0 be a Hausdorff topology on X with no isola points.> For example Q and R, for those cardinalities.For any infinite cardinal m, you can get a trivial example by takingthe topological sum of m copies of Q.>Let T be the set of all topologies on X that extend t_0 and>contain no singletons. By Zorn's lemma, T has maximal elements.>Let t be a maximal element of T.> Now what? Show every t-dense set D has nonnul interior?That's clearly true if D = X, so assume that the complement of D, callit C, is nonempty. Since D is t-dense, C is not t-open. Maximality oft implies that there is a t-open set U whose intersection with C is asingleton {x}. Since {x} is closed, U{x} is a nonempty t-open subsetof D.[...]> So the question of equally resolvable is barely interesting.You can try to make it into a real question (I don't know howinteresting it is): Does every resolvable space X contain two disjointdense subsets A and B with |A| = |B| = density(X), where density(X) isdefined as the minimum cardinality of a dense subset of X? I guessit's false, but I don't know. It's trivially true for finite spaces.=== === Subject: : Re: A tiny puzzle in number theory> Is it always true that:> a) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n+z^n as n is any integer >0 ?> b) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n-z^n as n is any integer >0 ?> 3^n+6^n+9^n = 3^n(1+2^n+3^n) = 0 mod (18) for all n >03^n+6^n-9^n = 3^n(1+2^n-3^n) = 0 mod (18) for all n >0and other than (x,y,z)=(3,6,9) there are many other solutions.For example:x,y, z3,6,94,12,165,20,256,30,367,42,498,56,649,72,8110,15,2510,90,1 0014,35,4918,63,8121,28,4924,40,6430,70,10036,45,81=== === Subject: : Re: A tiny puzzle in number theoryCorrect! Tiny as the title indica. The point of problem was in the worddistinct, but not relative primes. ;-)Can you parameterize the result?> Is it always true that:> a) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n+z^n as n is any integer >0 ?> b) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n-z^n as n is any integer >0 ?> 3^n+6^n+9^n = 3^n(1+2^n+3^n) = 0 mod (18) for all n >0> 3^n+6^n-9^n = 3^n(1+2^n-3^n) = 0 mod (18) for all n >0> and other than (x,y,z)=(3,6,9) there are many other solutions.> For example:> x,y,z> 3,6,9> 4,12,16> 5,20,25> 6,30,36> 7,42,49> 8,56,64> 9,72,81> 10,15,25> 10,90,100> 14,35,49> 18,63,81> 21,28,49> 24,40,64> 30,70,100> 36,45,81=== === Subject: : Re: A tiny puzzle in number theory> Correct! Tiny as the title indica. The point of problem was in the word> distinct, but not relative primes. ;-)> Can you parameterize the result?How about (x,y,z) = (ai, a(a-i), a^2) with i < a-i?> Is it always true that:> a) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n+z^n as n is any integer >0 ?> b) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n-z^n as n is any integer >0 ?>>3^n+6^n+9^n = 3^n(1+2^n+3^n) = 0 mod (18) for all n >03^n+6^n-9^n = 3^n(1+2^n-3^n) = 0 mod (18) for all n >0and other than (x,y,z)=(3,6,9) there are many other solutions.For example:x,y,z> 3,6,9> 4,12,16> 5,20,25> 6,30,36> 7,42,49> 8,56,64> 9,72,81> 10,15,25> 10,90,100> 14,35,49> 18,63,81> 21,28,49> 24,40,64> 30,70,100> 36,45,81>=== === Subject: : Re: A tiny puzzle in number theory>> Is it always true that:>> a) there are distinct integers x,y,z >0 so that x+y+z always divides>> x^n+y^n+z^n as n is any integer >0 ?>> b) there are distinct integers x,y,z >0 so that x+y+z always divides>> x^n+y^n-z^n as n is any integer >0 ?>make |x+y+z|<=1 ?That's rather difficult for distinct positive integers...I'm a bit confused by the statements, . Do you mean there are distinct integers x,y,z > 0 so that for every integer n > 0, x+y+z divides x^n+y^n+z^n? Or do you mean for every n > 0, there are distinct integers x,y,z > 0so that x+y+z divides x^n + y^n + z^n?If n >= 3, 2^n + 6^n + 8^n = 0 mod 16.For n = 2, 1^2 + 2^2 + 4^2 = 0 mod 7.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: A tiny puzzle in number theory>> Is it always true that:>> a) there are distinct integers x,y,z >0 so that x+y+z always divides>> x^n+y^n+z^n as n is any integer >0 ?>> b) there are distinct integers x,y,z >0 so that x+y+z always divides>> x^n+y^n-z^n as n is any integer >0 ?>make |x+y+z|<=1 ?> That's rather difficult for distinct positive integers...Oh that? I thought it was some kind of weird alien smiley!-- Unpatched IE vulnerability: WebFolder data InjectionDescription: Injecting arbitrary data in the My Computer zoneReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/13.html=== === Subject: : Re: Past posters, consider facts Discussion, linux)> What if you were faced with arguments over the existence of sqrt(2),>> or sqrt(-1)?>> Imagine if as a group mathematicians had risen to protect the status>> quo, and rejec sqrt(-1)?>sqrt(-1), sqrt(2), and a host of other things we now use today at> one time were not accep by the mathematicians of the day. BUT, in> math, the truth always comes out eventually, so you have nothing to> worry about.James doesn't need more bad math history lessons. He creates enoughof them on his own.Can you please tell us, sir, when sqrt(2) [was] not accep by themathematicians of the day? (I don't personally know the history ofi, and so won't comment on it -- see that? If you don't knowsomething historical, you can refrain from comment. It's analternative to talking out your ass.)-- Jesse F. Hughes[Mathematical] society has evolved far enough away from mainstreamsociety that it has become rogue, and now is willing to push its needsagainst that of the majority. -- James S. Harris === === Subject: : Re: Past posters, consider facts> What if you were faced with arguments over the existence of sqrt(2),>> or sqrt(-1)?>> Imagine if as a group mathematicians had risen to protect the status>> quo, and rejec sqrt(-1)?>sqrt(-1), sqrt(2), and a host of other things we now use today at> one time were not accep by the mathematicians of the day. BUT, in> math, the truth always comes out eventually, so you have nothing to> worry about.> James doesn't need more bad math history lessons. He creates enough> of them on his own.> Can you please tell us, sir, when sqrt(2) [was] not accep by the> mathematicians of the day? (I don't personally know the history of> i, and so won't comment on it -- see that? If you don't know> something historical, you can refrain from comment. It's an> alternative to talking out your ass.)Jack Zamat's statement exposes some common misconceptions, so I thoughtit would be of interest to go into them.To even *say* sqrt(2) requires explaining what you mean, when you aretalking about mathematics in different time periods. If Jack Zamatmeant some geometrical object that the ancient Greeks would haveunderstood, like a line segment that has the property that building asquare on it gives a square that is twice the square build on a unitline segment, then clearly the ancient Greeks would have accepsqrt(2) exists; in fact, with this definition, the Greeks clearlyunderstood sqrt(2) could be construc by taking the diagonal of aunit square. Now of course, at some point in time (say, around the time of thePythagoreans), while sqrt(2) (as described above) was accep as ageometrical object, it was not accep as another arithmetic type ofobject. The Pythagoreans supposedly had problems accepting thatsqrt(2) was not commensurable to a unit segment. In modernterminology, we would say they were surprised that sqrt(2) was notrational. However, even with this, I'm hard pressed to understand how this couldbe interpre as sqrt(2) being not accep by the mathematicians ofthe day. In fact, Euclid mentions the differences between differentkinds of magnitudes (including commensurables) and these distinctionsare helpful for his mathematics. What I *can* say is that sqrt(2) was accep (certainly the ancientsused the diagonal of a square!) but to some mathematicians at some timein history it had counterintuitive properties. But to say it was NOTaccep, would to me imply that its use was somehow prohibi, whichwas not the case.The situation with i is different. The primary difference is that ipopped up as a result of symbolic manipulations and thus from the verystart was not attached to some object mathematicians at the time wouldhave considered real. The use of i as a way to enable useful symbolicmanipulations was considered suspect for some time. Not until later,when it was shown that these manipulations could be justified incertain frameworks (and so given meaning), did i become more accep.Jesse, it seems talking out of your ass has become a favorite expression of yours. ;-)=== === Subject: : Re: Past posters, consider facts>Jesse, it seems talking out of your ass has become a favorite >expression of yours. ;-)Up to cobordism, the expression simply states a banal fact;as Bob Stong used to point out (possibly incorrectly; see thethread on genus of the human body), one's mouth is cobordantto one's asshole.Lee Rudolph === === Subject: : Re: Past posters, consider facts>>Jesse, it seems talking out of your ass has become a favorite >>expression of yours. ;-)> Up to cobordism, the expression simply states a banal fact;> as Bob Stong used to point out (possibly incorrectly; see the> thread on genus of the human body), one's mouth is cobordant> to one's asshole.And you, sir, are a hnut!Gib=== === Subject: : Re: Past posters, consider facts <87ad4c4eiw.fsf@phiwumbda.org> Discussion, linux)> Jesse, it seems talking out of your ass has become a favorite> expression of yours. ;-)It's long been a favorite expression, because it's so useful. (Ithink I apply it reasonably fairly. I've referred to my own blatheras talking out my ass on many occasions -- more than I care toestimate.)Thanks for a much more careful history lesson. I'm not altogethersure I understand the statement, while sqrt(2) (as described above)was accep as a geometrical object, it was not accep as anotherarithmetic type of object. If I were to describe matters (and here,I talk out my ass a bit, but it's only a little fart), I'd say thatthe Pythagoreans doub that there were any incommensurable lengths,and the proof that square root of two is incommensurable to the unitlength contradic their beliefs.I honestly don't know to what extent they attemp to hold onto theirmistaken beliefs once the proof was found (apocryphal tales ofmathematicians overboard notwithstanding). Did they try to suppressthe proof? Did they deny that it was correct? Did they accept theproof and admit it caused their osophy deep difficulties?Also, didn't the proof cause more difficulties than just osophicalproblems for the Pythagoreans? I was under the impression that it wasa real mathematical crisis, since the proposition that every pair oflengths is commensurable had been implicit or explicit in many, manygeometrical proofs. If my failing memory is correct, Eudoxus (sp?)provided the solution to the crisis, but I'll bet that's a much tooglib account.The questions above are real questions. I don't know the answers tothem. Most of the sentences are good candidates for claims ofass-talking.-- 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 maimed. -- Lin Yutang=== === Subject: : Re: Biomathematics -> [...]> [...]> [...]> Nobody can usefully model turbulence> through space and over time.> [...]> True, but only because of an implicit> Falsehood.> No such thing as 'time' exists within> physical reality.> What's been referred to as time is> an Erroneous partial conceptualization> of the one-way flow of energy from> order to dis-order that is what's> =described= by 2nd Thermo [WDB2T].> When this Falsehood is elimina,> turbulence is easy.> In the limit of 3-D space -> 0, energy> will flow toward decreasing order.> Period.Whoever gets the Clay Millenium Navier-Stokes Prize will, necessarily, end with thissame Conclusion :-]K. P. Collins=== === Subject: : Proof on Cassinis identityI found such a proof athttp://austin.onu.edu/~mcaragiu1/saramillercapstone.ppt, slide 21. However,my question to the newsgroup is if there is any other, perhaps simpler wayto prove it.=== === Subject: : Re: Proof on Cassini's identityIn-reply-to: Robert Adams >I found such a proof at>http://austin.onu.edu/~mcaragiu1/saramillercapstone.ppt, slide 21. However,>my question to the newsgroup is if there is any other, perhaps simpler way>to prove it.An identity that makes this easier is a generalization of the inductivedefinition of the Fibonacci sequence: F = F F + F F [1] n k+1 n-k k n-k-1This is easily proven by induction. Since F_1 = F_2 = 1, the case k=1is verified by the inductive definition of the Fibonacci sequence: F = F + F [2] n n-1 n-2Suppose that [1] is true for some k, then F = F F + F F n k+1 n-k k n-k-1 = F ( F + F ) + F F k+1 n-k-1 n-k-2 k n-k-1 = F F + ( F + F ) F k+1 n-k-2 k k+1 n-k-1 = F F + F F [3] k+1 n-k-2 k+2 n-k-1Therefore, [1] is true for k+1.Cassini's Identity states 2 n F F - F = (-1) [4] n-1 n+1 nThe simplest proof seems to be by induction. The case n=2 is true: 2 2 F F - F = 1*2-1 = 1 = (-1) [5] 1 3 2Suppose that [4] is true for some n. By [1], both of the following aretrue 2 2 F = F + F [6a] 2n+1 n+1 n F = F F + F F [6b] 2n+1 n+2 n n+1 n-1Subtracting these gives 2 2 F F - F = -( F F - F ) n n+2 n+1 n-1 n+1 n n+1 = (-1) [7]Therefore, [4] is true for n+1.Rob Johnson take out the trash before replying=== === Subject: : Re: Proof on Cassini's identityI found such a proof [ that F_.F_ - F_n^2 = (-1)^n ] at> http://austin.onu.edu/~mcaragiu1/saramillercapstone.ppt, slide 21.> However, my question to the newsgroup is if there is any other,> perhaps simpler way to prove it.> An identity that makes this easier is a generalization of the inductive> definition of the Fibonacci sequence:> F = F F + F F [1]> n k+1 n-k k n-k-1> This is easily proven by induction. Since F_1 = F_2 = 1, the case k=1> is verified by the inductive definition of the Fibonacci sequence:> F = F + F [2]> n n-1 n-2I didn't have the patience to wait for the powerpoint presentation toload. But to prove the - ^2 relation above, it seemsmarginally easier to use F_n = (a^n - b^n)/(a - b) where a, b are(1 + SQRT(5))/2 and (1 - SQRT(5))/2 resp:Expanding G = (a^(n-1) - b^(n-1)).(a^(n+1) - b^(n+1)) - (a^n - b^n)^2gives immediately that G = - (a.b)^(n-1).(a - b)^2, and a.b = -1.Cheers----------------------------------------------------- ----------------------John R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen=== === Subject: : Re: Proof on Cassini's identityIn-reply-to: Robert Adams >I found such a proof at>http://austin.onu.edu/~mcaragiu1/saramillercapstone.ppt, slide 21. However,>my question to the newsgroup is if there is any other, perhaps simpler way>to prove it.>>An identity that makes this easier is a generalization of the inductive>>definition of the Fibonacci sequence:>> F = F F + F F [1]>> n k+1 n-k k n-k-1>>This is easily proven by induction. Since F_1 = F_2 = 1, the case k=1>>is verified by the inductive definition of the Fibonacci sequence:>> F = F + F [2]>> n n-1 n-2>>Suppose that [1] is true for some k, then>> F = F F + F F>> n k+1 n-k k n-k-1>> = F ( F + F ) + F F>> k+1 n-k-1 n-k-2 k n-k-1>> = F F + ( F + F ) F>> k+1 n-k-2 k k+1 n-k-1>> = F F + F F [3]>> k+1 n-k-2 k+2 n-k-1>>Therefore, [1] is true for k+1.>>Cassini's Identity states>> 2 n>> F F - F = (-1) [4]>> n-1 n+1 n>[snip proof]>Actually, Cassini's identity is a special case of Catalan's identity>which is proven just as easily using [1] above. Catalan's identity is> 2 n-k 2> F - F F = (-1) F [5]> n n-k n+k k>Note that when n=k, [5] is trivial since F_0 = 0. Suppose [5] holds for>some n. By [1], both of the following are true> 2 2> F = F + F [6a]> 2n+1 n+1 n> F = F F + F F [6b]> 2n+1 n-k+1 n+k+1 n-k n+k>Subtracting these gives> 2 2> F - F F = -( F - F F )> n+1 n-k+1 n+k+1 n n-k n+k> n-k+1 2> = (-1) F [7]> k>Thus, [5] holds for k+1.Make that: [5] holds for n+1.Rob Johnson take out the trash before replying=== === Subject: : Re: Fundamental basics of unilaterality> As an engineer, I do not need at all the fictitious notion of an>> infinitely small value zero. Its neighbours 0- and 0+ do the same job.>> Why not declaring the exact zero Cantors pipe dream?>>If you have 10 euros and then a mugger takes away those 10 euros>from you, how many euros have you left? 0+? 0-? or just 0?>>0+ if you were at least able to remember what he looked like>>when going to the police. Otherwise 0-.> Seriously, his question> d) Aren't 0+ and/or 0- quite sufficient?> I think he means: do there exist a pair of positive real numbers> that are so insanely close to zero that we will never be able> to measure physical quantities in macroscopic units that are> even closer to zero than these quantities? In other words,> if you have 10 euros and the mugger takes everything but leaves you> with 10^{-23} euros, didnt he just go ahead and take everything?> This question has been asked many, many times in different forms,> all of which are basically all of the form: do physical quantities> really not exist in the physical world just because they can't be> measured (simultaneously or acurately or or ...)?Yes, but his assumptions that there do exist physical quantitiesand that mathematics must be enslaved to those who study suchbeasts, must be challenged.-- === === Subject: : Re: Why is this a proof?>> You have complica notation. Consider a map f:M -> N of modules.>> We ask, how many such maps have image isomorphic to another module P.>> We count them according to their kernel and image.> Hmmm...do you say that the map f is completely determined by its kernel> and its image?!I didn't and I don't. Try reading the portion of the message that I snipped.-- === === Subject: : Evil numbersBelow are a few contributions to the concept of irrational depth.For each strictly positive number x, let's define:(1) rho(x) = PROD_{n>0} (FRAC(nx))^(1/n^2)(2) phi(x) = lim_{n --> infinity} [PROD_{k=1,...,n} FRAC(kx)]^(1/n)(3) psi(x) = lim_{n --> infinity} [n * inf_{k=1,...,n} FRAC(kx)]where FRAC(w) = w - INT(w) is the decimal part of w.---------------------------Bollinger-Billecart Theorem---------------------------(1) rho(x) defines the irrational depth of x, as rho(x) = 0 if x isrational.(2) phi(x) = 1/e for almost all positive x. However phi(x) = 0 forall rationals.(3) psi(x) = 1 for almost all positive x. However psi(x) = 0 for allrationals.------------------------Definition: evil numbers------------------------A positive x is an evil number iff x is not a rational number and psi(x)<> 1.-------------------------Bollinger-Kendall Theorem-------------------------There exist infinitely many evil numbers. Such numbers include e.g.SUM_{k=1,...,infinity} 10^{-a_k}where a_1=2 and a_{k+1} > 2*a_k.=== === Subject: : Re: Rational Limits permission for an emailed response.X-Tom-Swiftie: I like to draw pictures with lots of detail, Tom said graphically> Finding a particular uncomputable function is not, I think, easy given> only this background knowledge. Once you know *which* function one> ought to look at, it's pretty simple to prove it's uncomputable.> Discovering that function in the first place is not too simple.Well, maybe so. Ease-of-discovery and easy-of-explanation andease-of-understanding often do not correlate and this may be such acase. I think that (for people used to programming languages--adecidedly post-Turing set!) it is probable easies to understand whythe halting problem cannot be solved from my little program.> You can't tell me that discovering and proving that the halting> problem is uncomputable is comparably easy to *that*. No, I wouldn't do that. If it's about discovering and proving, Idunno. Turing is one of my personal heroes. I would never suggest that hiswork was easy to come up with. Many parts of it are easy today, butwere amazingly clever at the time--and indeed, they are only easytoday because Turing did the hard work.If Goedel and Turing hadn't done what they did, McCarthy et al wouldnever have been able to come up with Lisp, for example, which pinchesoff Goedel's and Turing's ideas at every turn.Thomas=== === Subject: : Re: Rational Limits <4eCdnaUn0MJfppLdRVn-sQ@comcast.com> <877jzkxzdf.fsf@phiwumbda.org> <87wu7hiygd.fsf@becket.becket.net> <8765f15z9o.fsf@phiwumbda.org> Discussion, linux)> I want to give a variation of Cantor's proof:> Let X be the set of all real numbers in [0,1].> Assume there exists a sequence, x_n, that> includes every member of X.> Construct two new sequences, a_n and b_n.> a_n = 0> b_n = x_m if x_m is of the form 1/k (k integer) and x_m < b_(n-1)> Is b_n a finite sequence?As Virgil says, this is incoherent. Let's try the definition of b_nagain. I assume you meant something like this:b_0 = x_m where m is the least number such that x_m > 0.b_{n+1} = x_m where m is the least number > n such that x_m = 1/k and x_m < b_n.Clearly, (b_n) is defined for all n in N.If you meant something else, then you need to write a coherentdefinition of it (or at least a description).> Assume b_n is infinite.> This is the basis of Cantor's proof.> If b_n is infinite, there are an infinite number of reals> not in the sequence x_n.> How does this relate to the intersection of (0,1/n)?> Proofs have been given showing the intersection of> (0,1/n) must be empty. Here is an example.> For any real x there is an n such that x > 1/n.> Therefore, there can be no x such that 1/n > x for all n.> Compare that to this:> For any natural, n, there is a real x such that 1/n > x.> Therefore, there can be no n such that x > 1/n for all x.Both pairs of claims look unobjectionable.> Let X be the set of rational number in [0,1].> Let x_n be a sequence that includes every member of X.> Many examples of x_n have been given in other threads.> Applying Cantor's algorithm, will b_n be a finite sequence?> Assume b_n is finite. This proves x_n does not contain any rationals> between (a_n, b_n) for some n.> Assume b_n is infinite. There must be an infinite number of> rationals not in the range of x_n.Whoa, Nelly. Why? Had a bit of a leap there, didn't we?> Saying the intersection of (0,1/n) is empty is equivalent to saying> the reals are countable.No, that's simply false. -- Jesse F. HughesMy baby don't allow me in the kitchen and I've come to love her decision. -- Bad Livers=== === Subject: : Re: Rational Limits> Let X be the set of rational number in [0,1].> Let x_n be a sequence that includes every member of X.> Many examples of x_n have been given in other threads.Applying Cantor's algorithm, will b_n be a finite sequence?> Assume b_n is finite. This proves x_n does not contain any rationals> between (a_n, b_n) for some n.Assume b_n is infinite. There must be an infinite number of> rationals not in the range of x_n.> Whoa, Nelly. Why? Had a bit of a leap there, didn't we?I make the same leap Cantor does.If x_n is a sequence of real numbers, Cantor assumes thatsince b_n is infinite there must be a number not in the rangeof x_n. How does my assumption differ form Cantor's.I know you are going to say that the missing numberin my proof must be irrational. Sure, there is such anirrational. There are also an infinte number of rationals.Cantor's proof works because b_n is an infinite set.> Saying the intersection of (0,1/n) is empty is equivalent to saying> the reals are countable.> No, that's simply false.You are saying that you can partition (0,1] into a countablenumber of intervals such that at least one interval is empty.Russell- 2 many 2 count=== === Subject: : Re: Rational Limits>> Let X be the set of rational number in [0,1].> Let x_n be a sequence that includes every member of X.> Many examples of x_n have been given in other threads.Applying Cantor's algorithm, will b_n be a finite sequence?> Assume b_n is finite. This proves x_n does not contain any rationals> between (a_n, b_n) for some n.Assume b_n is infinite. There must be an infinite number of> rationals not in the range of x_n.Whoa, Nelly. Why? Had a bit of a leap there, didn't we?> I make the same leap Cantor does.If you think that, you clearly do not understand what Cantor is saying.It is a property of the REALS, proved in both the Dedekind model an in the Cauchy model, that every strictly increasing but bounded sequence has a limit which is not a member of the sequence, and similarly for strictly decreasing sequences of reals.BUT the limits of such sequences of rationals need not exist WITHIN THE SET OF RATIONALS.Thus the set of rationals has gaps in it that the set of reals does not have.Cantor says that the absence of such gaps proves the reals uncountable but the presence of such gaps allows (but does not require) the rationals to be countable.> If x_n is a sequence of real numbers, Cantor assumes that> since b_n is infinite there must be a number not in the range> of x_n. How does my assumption differ form Cantor's.Not at all. He assumes no such thing. He PROVES it for the reals, based on other known properties of the reals. You assume it for the rationals based on nothing but arrogance. There is not, and cannot be, any such proof for the rationals, as the contrary has been proved here several times.But since you understand nothing of what is going on, all the important stuff has slid by you while you have focused in on irrelevancies and minor details. > I know you are going to say that the missing number> in my proof must be irrational. Sure, there is such an> irrational. There are also an infinte number of rationals.> Cantor's proof works because b_n is an infinite set.You have no clue as to why Cantor's proof works. It works because the reals always have real least upper bounds and real greatest lower bounds for suitably bonded non-empty sets of reals, whereas the rationals, for instance, do NOT ALWAYS have rational upper bounds nor rational lower bounds.If you do not understand why this is the important point in Cantor's proof, then you are totally ignorant of what is really going on here.But your past posts have made that abject ignorance evident to all.> Saying the intersection of (0,1/n) is empty is equivalent to saying> the reals are countable.No, that's simply false.>> You are saying that you can partition (0,1] into a countable> number of intervals such that at least one interval is empty.> Russell> - 2 many 2 count=== === Subject: : Re: Rational Limits <4eCdnaUn0MJfppLdRVn-sQ@comcast.com> <877jzkxzdf.fsf@phiwumbda.org> <87wu7hiygd.fsf@becket.becket.net> <8765f15z9o.fsf@phiwumbda.org> <877jzg4e3u.fsf@phiwumbda.org> <6fmdnQfiKtCYGo7dRVn-iQ@comcast.com> Discussion, linux)>> Let X be the set of rational number in [0,1].>> Let x_n be a sequence that includes every member of X.>> Many examples of x_n have been given in other threads.>> Applying Cantor's algorithm, will b_n be a finite sequence?>> Assume b_n is finite. This proves x_n does not contain any rationals>> between (a_n, b_n) for some n.>> Assume b_n is infinite. There must be an infinite number of>> rationals not in the range of x_n.>> Whoa, Nelly. Why? Had a bit of a leap there, didn't we?> I make the same leap Cantor does.> If x_n is a sequence of real numbers, Cantor assumes that> since b_n is infinite there must be a number not in the range> of x_n. How does my assumption differ form Cantor's.> I know you are going to say that the missing number> in my proof must be irrational. Sure, there is such an> irrational. There are also an infinte number of rationals.> Cantor's proof works because b_n is an infinite set.No, Cantor's proof works because R is gapless. Define a_n and b_n asin the proof of the theorem. They satisfy the following conditions:(1) (a_n) is strictly increasing and (b_n) strictly decreasing.(2) For every n, a_n < b_n.(Note: (1) and (2) are not really all that obvious from the text atwikipedia, but it's easy enough to make them explicit.)Define A = {x | (E n) x < a_n} and B = R A (so B = {x | (A n) x >=a_n}. Then A and B partition R. Moreover, every element of A is lessthan each element of B. Therefore, by the property of gaplessness, weconclude that there is a c such that for all x:* if c > x, then x in A;* if c < x, then x in B.We claim that c is not in (x_n). This claim requires proof, which Iomit here and will provide on request.The point is, quite clearly, the fact that (b_n) is an infinitesequence isn't what allows one to find the witness c. It is the factthat R is gapless, together with (1) and (2), that yields the witnessc.Q is not gapless. Your proof cannot work like Cantor's proof.>> Saying the intersection of (0,1/n) is empty is equivalent to saying>> the reals are countable.>> No, that's simply false.> You are saying that you can partition (0,1] into a countable> number of intervals such that at least one interval is empty.I'd never say that, since it is an abuse of the term partition. (Setsin a partition are non-empty by definition -- I think I omit thatfact when I defined A and B partition R earlier.) I suspect that youdon't mean partition here, but I won't guess what you mean.So the answer is definitely, no, I'm not saying that. Try toreformulate your claim so that I know what it means, and then I'lltell you whether that's what I'm saying.-- Jesse F. Hughes[blah blah blah] which you are trying to dispute, but I've found thatyou've used ***definitions*** rather than mathematical logic, often inyour posts. -- on the insufficiency of math. definitions.=== === Subject: : Regular LanguagesHi everybody,Is it possible to represent each regular language with a planar NFA graph?Behrang.=== === Subject: : Re: Regular Languages> Is it possible to represent each regular language with a planar NFA graph? Interesting question... my suspicion is no. Consider an alphabet S = {0,1,2,3,4} and a language L = {w in S* such that the sum of the characters of w is 0 mod 5} Then any machine would need states representing (at least) the five possible current sums (mod 5) and since any symbol (0 through 4) could move the action from one state to any other state, there has be some kind of path taking it there (be as creative as you like with epsilon transitions, such paths must still exist). This looks like a subdivision of the complete graph on five vertices, and so not planar. This of course isn't a complete proof... do you need one (for an assignment, or something?)J=== === Subject: : Re: Hilbert spaces in quantum mechanics permission for an emailed response.X-Tom-Swiftie: The Hurd is running, Tom said collectively>How come textbooks always gloss over these>contradictions as if they all make sense?> They're written by Physicists who don't care if the Mathematics are> right. They're written for students who don't have the Mathematical> background to understand a more complica approach. there isn't time> in the course to do it right and still cover the requisite Physics.> That's the way that it has always been taught. There are probably> other reasons as well.The authors of Structure and Interpretation of Classical Mechanics(Sussman and Wisdom) want to do a Quantum Mechanics book, if they canfigure out exactly how. Given the example of SICM and its carefulmathematical rigor, I think we could expect SIQM, should it appear, tobe similarly excellent.Thomas=== === Subject: : Re: Hilbert spaces in quantum mechanics permission for an emailed response.X-Tom-Swiftie: I've finished counting the horses, Tom said summarily> That's because there is a lot of hand waving in Physics. Some of what> they do can be put on a rigorous footing and some can't.It's worth pointing out that at least non-relativistic quantummechanics can be rigorously specified with good solid math.(*Including* renormalization. Just don't think of renormalization asa freaky fake way to get a sum from divergent integrals, but ratherignore the integrals, and think of renormalization as just aparticular functional operation.)Thomas=== === Subject: : Re: JSH: Push for Java>[...]>Could be. The other day you asked me whether I thought there>was a reason Macsysma(sp?) used Lisp. I was at www.python.org>just now, looking for something else, and I noticed this at the>top of the page:>'Python has been an important part of Google since the beginning, >and remains so as the system grows and evolves. Today dozens of >Google engineers use Python, and we're looking for more people >with skills in this language. said Peter Norvig, director of >search quality at Google, Inc.'>I wonder whether there's a reason they use Python there.>(I'm _temp_ to wonder whether Google counts as a>large system but I'm not going to, because no, of>course probably parts of Google use other languagesBut regardless, it _is_ hard to see how they can possibly>use Python for important _parts_ of the system, given>the instability of the language, the fact that it's such a>pain to learn the syntax, the fact that there's no>lexical closures, etc.)Not to reply to my own post just because it seems to havebeen missed, but there's a little joke that came uplast night:cosmology. They say something about how astronomersused to think that clusters of galaxies were the largeststructures in the universe - then a little later theydiscovered _large_ structures! Suddenly reminded meof this thread for some reason - if a _cluster_ of friggin_galaxies_ is not large then surely Google isn'tlarge either.(haha, no serious point intended).************************=== === Subject: : Re: JSH: Push for Java>> This is _Python_, an extremely fabulous language,>You might want to take a look at the following link>on another language called whitespace:Hey thanks - that looks like the language for me.> http://compsoc.dur.ac.uk/whitespace/> Hale************************=== === Subject: : Re: JSH: Push for Java>> If it's not available when someone starts a new large project>> that's irrelevant, because there's no reason the new project>> needs to use 1.5.2. The problem was supposed to be stability ->> if someone already has a large project using, say, 1.5.2,>> and he's concerned about stability there's no reason he>> can't just keep using 1.5.2 - it's going to remain available>> to him until he decides to delete it.>My point is that any version has bugs, always will. Not that this is>Python's fault; it's a big complex interpreter. And there's really>only one Python implementation out there.>So if you stick with version 1.5.2, you're stuck with the bugs it's>got, forever.Not that _I_ would do such a thing, but there's nothing stoppingyou from getting the source, finding the reason for the bug,and fixing it yourself. (Uh, probably pointing this out is a tacticalmistake, since a language that requires you to do this kind ofthing probably doesn't count as easy to use... )>If you upgrade to get bug fixes, you also get language changes to cope>with. >By contrast, if you stick with old-style C, you can upgrade your>compiler, and old-style C, being a stable thing, doesn't change. Even>when compilers star supporting ANSI C, it's still extremely common>for them to continue to support old-style C, so that upgrading your>compiler doesn't force you to use a new version of the language.But seriously: the vast majority of the changes they make inPython _are_ backwards compatible - it's proposed changesthat will break existing code that lead to the most headebates.>Thomas************************=== === Subject: : Re: JSH: Push for Java permission for an emailed response.X-Tom-Swiftie: I am writing lots of little verses, Tom said blankly> You Python bashers can laugh all you want, but David Ullrich is> right about Python being a fabulous language. It is the perfect fit> for what _I_ need in a programming language and that is all that> matters to _me_.Ah, well, fabulous to Mensanator (a name that sounds right out ofKids in the Hall) is fine; you can be the judge of that.> I, for one, appreciate guys like David Ullrich promoting their> the Usenet, I've learned about perl, Python, Java, Rexx, UBASIC,> Cygwin, .NET SDK and have downloaded and installed all of them.Have you spent any time with any really fabulous languages, like, say,Scheme?> I'll be the judge of whether a language is suitable for my needs.Well, sure. suitable for your needs is a measured statement, andwhile it's possible to be wrong about what is suitable for one'sneeds, I'm not interes in challenging *that*.Thomas=== === Subject: : Re: JSH: Push for Java> === Subject: : Re: JSH: Push for Java>Message-id: <874qukjlub.fsf@becket.becket.net> You Python bashers can laugh all you want, but David Ullrich is>> right about Python being a fabulous language. It is the perfect fit>> for what _I_ need in a programming language and that is all that>> matters to _me_.>Ah, well, fabulous to Mensanator (a name that sounds right out of>Kids in the Hall) is fine; you can be the judge of that.>> I, for one, appreciate guys like David Ullrich promoting their>> the Usenet, I've learned about perl, Python, Java, Rexx, UBASIC,>> Cygwin, .NET SDK and have downloaded and installed all of them.>Have you spent any time with any really fabulous languages, like, say,>Scheme?No, but I will make note of it for future reference. That's anotherlanguage I might add to my collection someday since you thinkso highly of it. I won't challenge your opinion that Scheme is _the_most fabulous language.But even if I were to find it so fabulous that I abandon Python andnever write another program in it, that wouldn't change my opinionthat Python is fabulous, I'd just say that Scheme is really fabulous.>> I'll be the judge of whether a language is suitable for my needs.>Well, sure. suitable for your needs is a measured statement, and>while it's possible to be wrong about what is suitable for one's>needs, I'm not interes in challenging *that*.Maybe wrong isn't the right word. Prior to learning about languagesprecision addition and simple multiplication using unlimi lengthstrings of 1s and 0s. Did it work? Yes. Did it meet my needs? Yes.But it was ridiculously slow. Now that I've got something better, there's no need for the perl solution. But perl's regular expressionhandling made the task a _lot_ easier than trying to do it in something like Visual Basic (assuming it could even be done).So I wouldn't say that I was wrong about perl being suitable for thetask, I would say I was wrong in my approach to the problem. If youhave a nail, then you're going to reach for a hammer. And when yourcarpentry collapses, you realise that you should have used screws.>Thomas--=== === Subject: : Re: JSH: Push for Java> Ok, I lied, you did mention one. But it says _nothing_ about why,> say, Python 1.5.2 is not suitable for large systems. A person> working on a large system is not required to upgrade as soon> as a new version comes out - in fact that Python on _my_> system has been 100% stable, it's exactly the same today> as it was many years ago when I first installed Python.Unfortunately, there are a number of reasons why stickingwith 1.5.2 is likely to be a problem.One is that the existing implementation of 1.5.2 is not maintained,and no-one is going to write a new one. Any complaint aboutbugs is going to be met with You need to upgrade.Even as 1.5.2 remains stable, the environment changes. At some point1.5.2 may cease to operate correctly (e.g. the operating system nolonger provides the libraries it needs, or moves them). The one that has caused us most trouble is that we have severalpackages using Python. When they use different versions ofPython problems occur. At a minimimum we and the users mustinstall different versions of python in parallel. Then thereare environment problems (which PYTHONPATH is correct, whereis the correct executable, where are the correct libraries. etc.)In practice upgrading is necessary. Fortunately this has notbeen hard. Unfortunately, this has not been trivial. - William Hughes=== === Subject: : Re: JSH: Push for Java>> Ok, I lied, you did mention one. But it says _nothing_ about why,>> say, Python 1.5.2 is not suitable for large systems. A person>> working on a large system is not required to upgrade as soon>> as a new version comes out - in fact that Python on _my_>> system has been 100% stable, it's exactly the same today>> as it was many years ago when I first installed Python.>Unfortunately, there are a number of reasons why sticking>with 1.5.2 is likely to be a problem.>One is that the existing implementation of 1.5.2 is not maintained,>and no-one is going to write a new one. Any complaint about>bugs is going to be met with You need to upgrade.>Even as 1.5.2 remains stable, the environment changes. At some point>1.5.2 may cease to operate correctly (e.g. the operating system no>longer provides the libraries it needs, or moves them). Well ok, never mind about the idea that a person can stick withone version. (_My_ os tends to remain as stable as my versionof Python, but it is going to happen at some point that I finallyhave to upgrade to a new version of Windows... not gonnahappen til I have a _really_ good reason - for example half therecent security holes in XP depend on new features.)>The one that has caused us most trouble is that we have several>packages using Python. When they use different versions of>Python problems occur. At a minimimum we and the users must>install different versions of python in parallel. Then there>are environment problems (which PYTHONPATH is correct, where>is the correct executable, where are the correct libraries. etc.)>In practice upgrading is necessary. Fortunately this has not>been hard. Unfortunately, this has not been trivial.> - William Hughes************************=== === Subject: : Re: JSH: Push for Java permission for an emailed response.> Even as 1.5.2 remains stable, the environment changes. At some point> 1.5.2 may cease to operate correctly (e.g. the operating system no> longer provides the libraries it needs, or moves them). Yes, I should have no this as well. This is one of the leadingsources of bit rot: the phenomenon that once-working andentirely-unchanged bits seem over time to just, well, rot.Thomas=== === Subject: : Re: JSH: Questioning my conclusions>> [...]>> Algebraically, to try and remain in the ring of algebraic integers,>> your only option is to have>> (5a_1(x)/7 + 1)(5b_2(x)/7 + 2) = 25x^2 + 30x + 2>That should be>(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2>and what's amazing to me now is how simple it all actually is.I don't know whether you've noticed this, but fyi everyone elsehas: It happens over and over that you're amused that nobodycan follow some utterly simple argument. Then it turns outthat the simple argument was simply _wrong_. Sometimesit's just a typo (that's the only sort of error you regularlyadmit to) but that's irrelevant: it still happens over andover that you're amused that we can't follow anincredibly simple _incorrect_ argument. Then youfix it and you're amused again we can't follow it.Then the fix turns out to be wrong...There's a _reason_ that you're the only person on theplanet able to prove all those things that nobody elsecan prove. And it's not because you're smarter thaneveryone else (one of my colleagues has a stockphrase for certain graduate students who areutterly out of their depth and always get everythingwrong: he can prove anything...)>[...]>The algebra is rigid here. There's just no other way it can be.>> and I know that it bothers you because it's possible to show that>> 5a_1(x)/7 in general is not an algebraic integers, but sorry Decker,>> that's what logically follows from the algebra.>One can consider this a test of your ability to accept what logically>follows, versus what you may wish to be true.>Physicists faced it with quantum mechanics. Maybe mathematicians are>weaker minded than physicists though.>[...]>> You're forced out, as the algebra clearly shows.>And I'd think that *some* mathematicians would actually accept>algebra!!!>************************=== === Subject: : Re: The gif ones> at 09:21 PM, j.schoenfeld@programmer.net (John Schoenfeld) said:>Quantum mechanics dictates discrete space,> No, it dictates continuous space-time. Discreteness comes in withSo then space-time is theoretically continuous but observably discrete?JS=== === Subject: : Re: The gif ones> at 09:21 PM, j.schoenfeld@programmer.net (John Schoenfeld) said:>Quantum mechanics dictates discrete space,> No, it dictates continuous space-time. Discreteness comes in with> So then space-time is theoretically continuous but observably discrete?No.=== === Subject: : Re: The gif ones> at 09:21 PM, j.schoenfeld@programmer.net (John Schoenfeld) said:Quantum mechanics dictates discrete space,No, it dictates continuous space-time. Discreteness comes in with> So then space-time is theoretically continuous but observably discrete?> No.Then, according to you, i can measure distances and time-intervalssmaller than h/2pi.JS=== === Subject: : Re: Action Device: Destination Alpha Centauri A> Look, until I plan my future life, I can not build this Action Device.> Alpha Centauri A is 4.35 ly away. So, please tell me, I am> accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How> much time it will take to cover 2.18 lightyears distance i.e. when I> have covered half the road?> Who holds the watch, idiot wog?> http://fourmilab.to/etexts/einstein/specrel/specrel.pdf> Uncle Al, I request you to unleash your wisdom on me by analysing this design..http://www.geocities.com/inertial_propulsion-Abhi.=== === Subject: : Re: Action Device: Destination Alpha Centauri AMessage-Id: <1075045528.11502.0@iris.uk.clara.netUncle Al, I request you to unleash your wisdom on me by analysing thisdesign..> http://www.geocities.com/inertial_propulsionI may not be Uncle Al but the error is obvious, youare still not showing how you keep the angle ABCfixed. This requires a force and the reaction tothat exactly cancels the desired unidirectionalforce. This shows the rod and springs version:http://www.dishman.me.uk/George/Abhi/abhi_rod.gif> All the people are still unaware. But what has happened that we have> technology through which we can reach to Mars in just one day.No, we don't, but ..> So I am going to build this Action Device ...Good, I suggest you start with the springs fromtwo biros and a small bit of wire and see if itcan lift just its own weight.George=== === Subject: : Re: Action Device: Destination Alpha Centauri A> Uncle Al, I request you to unleash your wisdom on me by analysing this> design..http://www.geocities.com/inertial_propulsion> I may not be Uncle Al but the error is obvious, you> are still not showing how you keep the angle ABC> fixed. This requires a force and the reaction to> that exactly cancels the desired unidirectional> force. This shows the rod and springs version:> http://www.dishman.me.uk/George/Abhi/abhi_rod.gifWelcome back, George. Happy news is that I have this angled rod andtwo springs. I am stretching these two springs and I find that angleABC still remains fixed!Obviously it requires force to keep angle ABC fixed. But as I am usingiron rod, intermoleculer forces at point B are so strong that due tospring forces acting at point B, angle ABC does not change. And alsothere are other ways to keep it fixed or make it in such a way that,if angle ABC is 60 degree, it changes from 60 degree to 0 degree. Forthis purpose, we can use another spring having restoring force atpoint B or attached in between middle points of spring AB and CB.BTW, I have noticed that you are still clinging to my spring designand not commenting on this magnet design..http://www.geocities.com/inertial_propulsionAnother happy news is that, take a elastic wooden stick of mass 'm'gram. Attach stone or any mass of '2m' gram. Bend this stick andrelease both ends. As the stick tries to straighten up, center pointof stick and hence stone gains momentum. Due to this momentum ofstone, stone and stick are propelled in only one direction. I did itwith bicycle spoke. Why don't you repeat it and place it on recordhere what exactly you observed?> All the people are still unaware. But what has happened that we have> technology through which we can reach to Mars in just one day.> No, we don't, but ..> So I am going to build this Action Device ...> Good, I suggest you start with the springs from> two biros and a small bit of wire and see if it> can lift just its own weight.> GeorgeWonder what will support weight of entire structure when it ishanging?Think about this. Imagine you are standing in between two walls just 2feet away from each other. With your back on one wall no.1, lift yourright leg and push wall no.2. Now also lift your left leg and pushagain wall no.2. You are hanging in between two walls. What issupporting your weight?And yes, it is very long hanging over this device. But I was reallyconfused why should I do this. I was not ready to get involved in myoffice work, business, theory development after this is over. I was insearch of new direction to my life. I always knew Magnitude but Inever knew direction of my life.-Abhi.-----------------------------------------har gha.Dii Khud se ulajhanaa hai muqaddar meraa mai.n hii kashtii huu.N mujhii me.n hai sama.ndar meraa kisase puuchhuu.N ki kahaa.N gum huu.N baraso.n se har jagah Dhuu.Ndhataa phirataa hai mujhe ghar meraa ek se ho gae mausamo.n ke chehare saare merii aa.Nkho.n se kahii.n kho gayaa ma.nzar meraa muddate.n biit gaii Khvaab suhaanaa dekhe jaagataa rahataa hai har nii.nd me.n bistar meraa aaiinaa dekhake nikalaa thaa mai.n ghar se baahar aaj tak haath me.n mahafuus hai patthar meraa -Nida Fazli.---------------------------------------------=== === Subject: : Re: Action Device: Destination Alpha Centauri A> Uncle Al, I request you to unleash your wisdom on me by analysing this> design..http://www.geocities.com/inertial_propulsion> I may not be Uncle Al but the error is obvious, you> are still not showing how you keep the angle ABC> fixed. This requires a force and the reaction to> that exactly cancels the desired unidirectional> force. This shows the rod and springs version:> http://www.dishman.me.uk/George/Abhi/abhi_rod.gif> Welcome back, George. Happy news is that I have this angled rod and> two springs. I am stretching these two springs and I find that angle> ABC still remains fixed!> Obviously it requires force to keep angle ABC fixed. But as I am using> iron rod, intermoleculer forces at point B are so strong that due to> spring forces acting at point B, angle ABC does not change. And also> there are other ways to keep it fixed or make it in such a way that,> if angle ABC is 60 degree, it changes from 60 degree to 0 degree. For> this purpose, we can use another spring having restoring force at> point B or attached in between middle points of spring AB and CB.> BTW, I have noticed that you are still clinging to my spring design> and not commenting on this magnet design..> http://www.geocities.com/inertial_propulsion> Another happy news is that, take a elastic wooden stick of mass 'm'> gram. Attach stone or any mass of '2m' gram. Bend this stick and> release both ends. As the stick tries to straighten up, center point> of stick and hence stone gains momentum. Due to this momentum of> stone, stone and stick are propelled in only one direction. I did it> with bicycle spoke. Why don't you repeat it and place it on record> here what exactly you observed?> All the people are still unaware. But what has happened that we have> technology through which we can reach to Mars in just one day.> No, we don't, but ..> So I am going to build this Action Device ...> Good, I suggest you start with the springs from> two biros and a small bit of wire and see if it> can lift just its own weight.> George> Wonder what will support weight of entire structure when it is> hanging?> Think about this. Imagine you are standing in between two walls just 2> feet away from each other. With your back on one wall no.1, lift your> right leg and push wall no.2. Now also lift your left leg and push> again wall no.2. You are hanging in between two walls. What is> supporting your weight?*Friction -- a force equal to the normal (perpendicular) force between your shoes and the wall, multiplied by the coefficient of friction.Try it with a slick wall and greasy shoes.earle*=== === Subject: : Re: Action Device: Destination Alpha Centauri AMessage-Id: <1075060385.33184.0@despina.uk.clara.netUncle Al, I request you to unleash your wisdom on me by analysing this> design..http://www.geocities.com/inertial_propulsionI may not be Uncle Al but the error is obvious, you> are still not showing how you keep the angle ABC> fixed. This requires a force and the reaction to> that exactly cancels the desired unidirectional> force. This shows the rod and springs version:http://www.dishman.me.uk/George/Abhi/abhi_rod.gif> Welcome back, George. Happy news is that I have this angled rod and> two springs. I am stretching these two springs and I find that angle> ABC still remains fixed!Excellent, try it and see is the best way to makeprogress, well done on taking the first step.Now tell me this, as you pull down on the springs,do you also have to hold the angled rod to stop itmoving? If not, which way are you pulling the springs,along the arms of the rod or are you pulling A in adirection directly away from C and vice versa?> Obviously it requires force to keep angle ABC fixed. But as I am using> iron rod, intermoleculer forces at point B are so strong that due to> spring forces acting at point B, angle ABC does not change.So far, so good.> And also> there are other ways to keep it fixed or make it in such a way that,> if angle ABC is 60 degree, it changes from 60 degree to 0 degree. For> this purpose, we can use another spring having restoring force at> point B or attached in between middle points of spring AB and CB.As long as you have a working method, go with it.It is easy to be distrac by alternatives.> BTW, I have noticed that you are still clinging to my spring design> and not commenting on this magnet design..> http://www.geocities.com/inertial_propulsionThe spring pulls the ends together and so do themagnets so from the point of view of the mechanicsthere is no difference. Since I already did thedrawing for the springs, I just reminded you of it.It applies equally well for the magnets.> Another happy news is that, take a elastic wooden stick of mass 'm'> gram. Attach stone or any mass of '2m' gram. Bend this stick and> release both ends. As the stick tries to straighten up, center point> of stick and hence stone gains momentum. Due to this momentum of> stone, stone and stick are propelled in only one direction. I did it> with bicycle spoke. Why don't you repeat it and place it on record> here what exactly you observed?There is an equal and opposite reverse force on yourfinger while you bend the stick. When you release thestone, it goes forward while your finger goes back. Ofcourse you have a mass much more than the stone andstick so the motion is not noticeable but you move atiny amount all the same. If you are standing firmlyon the ground it is the whole mass of the Earth thatmoves by an incredibly small amount, but the force isstill balanced.> So I am going to build this Action Device ...Good, I suggest you start with the springs from> two biros and a small bit of wire and see if it> can lift just its own weight.> Wonder what will support weight of entire structure when it is> hanging?> Think about this. Imagine you are standing in between two walls just 2> feet away from each other. With your back on one wall no.1, lift your> right leg and push wall no.2. Now also lift your left leg and push> again wall no.2. You are hanging in between two walls. What is> supporting your weight?Friction stops you sliding down the wall but your massis added to that of the wall so a worm under the wallwould feel just a little more force pressing down onhim. Again, the forces are balanced.> And yes, it is very long hanging over this device. But I was really> confused why should I do this. I was not ready to get involved in my> office work, business, theory development after this is over. I was in> search of new direction to my life. I always knew Magnitude but I> never knew direction of my life.Whether it works or not, I think you need to getthis out of your system so that you can thinkmore clearly about what you want to do next.George=== === Subject: : Re: Action Device: Destination Alpha Centauri ALook, until I plan my future life, I can not build this Action Device.I am precognitive -- I predic you would continue to post about it!Please provide a link to the post where you made this prediction.Minor Crank=== === Subject: : Re: Action Device: Destination Alpha Centauri AContent-transfer-encoding: 8bitLook, until I plan my future life, I can not build this Action Device.I am precognitive -- I predic you would continue to post about it!> Please provide a link to the post where you made this prediction.-=-=-=-=-== = === Subject: : Re: Action Device: Destination Alpha Centauri ALook, until I plan my future life, I can not build this Action Device.I am precognitive -- I predic you would continue to post about it!> Please provide a link to the post where you made this prediction.> Minor CrankHe predic it in Action Device Tragedy thread.Why he predic because of this...http://www.geocities.com/inertial_propulsion-Abhi.=== === Subject: : Re: Action Device: Destination Alpha Centauri AIn sci.math, Abhi<8d1edf6d .0401241427.2ccaf192@posting.google.com>:> Look, until I plan my future life, I can not build this Action Device.> Alpha Centauri A is 4.35 ly away. So, please tell me, I am> accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How> much time it will take to cover 2.18 lightyears distance i.e. when I> have covered half the road?> Please Note: Velocity Does Not Matter....> Thanks!> -Abhi.First, there's the issue of common units. 4.35 light years is4.12 * 10^16 m away. Half the distance to the goal line is2.06 * 10^16 m.Assuming Newtonian theory, which isn't quite right:d = 1/2 * a * t^2t = sqrt(2 * d / a)t = sqrt(2 * 2.06 * 10^16 m / 9.805 m/s/s) = 64822355 secondsor a smidge more than 2 years.Deceleration will take a similar amount of time.Since the final compu v = a * t = 9.805 * 64822355 =635583200 = 2.11c this isn't even close to being correctfrom a physical standpoint. However, I'd have to workout the Lorentz contraction, among other things.It should be possible to reach Alpha Centauri in about ayear (subjective), assuming sufficient fuel and a brutal10g trip, but the Twin Paradox still applies.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Action Device: Destination Alpha Centauri A> In sci.math, Abhi> :> Look, until I plan my future life, I can not build this Action Device.> Alpha Centauri A is 4.35 ly away. So, please tell me, I am> accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How> much time it will take to cover 2.18 lightyears distance i.e. when I> have covered half the road?> Please Note: Velocity Does Not Matter....> Thanks!> -Abhi.> First, there's the issue of common units. 4.35 light years is> 4.12 * 10^16 m away. Half the distance to the goal line is> 2.06 * 10^16 m.> Assuming Newtonian theory, which isn't quite right:> d = 1/2 * a * t^2> t = sqrt(2 * d / a)> t = sqrt(2 * 2.06 * 10^16 m / 9.805 m/s/s) = 64822355 seconds> or a smidge more than 2 years.Yes, it comes out to be around 750 days. It means total trip will takearound at least 8 years. Not possible for human being to spend 8 yearsof his/her life in just journey.All the people are still unaware. But what has happened that we havetechnology through which we can reach to Mars in just one day. We cantake trip of all the planets within month or two. But we just can'tleave this solar system. We are confined, for the moment, to our solarsystem.> Deceleration will take a similar amount of time.> Since the final compu v = a * t = 9.805 * 64822355 => 635583200 = 2.11c this isn't even close to being correct> from a physical standpoint. However, I'd have to work> out the Lorentz contraction, among other things.> It should be possible to reach Alpha Centauri in about a> year (subjective), assuming sufficient fuel and a brutal> 10g trip, but the Twin Paradox still applies.It is not possible for human being to sustain G-Force greater than 1Gfor long time, days and nights, for months. This action device canaccelerate the spaceship with acceleration far greater than 1g, say50g. But it will have to be Robotic mission first.For human being to travel in such spaceship accelerating with 50g, wewill have to neutralize 49g inside spaceship. Only way to do this isto create artificial gravity. Until and unless, me or someone elsesucceeds in creating artificial gravity, interstellar space voyage isimpossible.And even after creation of artificial gravity, we will not be able toleave this galaxy. Acceleration, decceleration is not useful to travelacross galaxies.(1) Now any human being, not necessarily only astronauts, can goanywhere in our solar system. But we are confined to our solar system.(2) Even after creation of artificial gravity, we can go to stars inour galaxy, but we can not leave this galaxy.(3) So to go to another galaxy, wait for another 300 years or so,until we find some new theory other than this Time Theory.So I am going to build this Action Device and my aim is to be firsthuman being to walk on Mars and take a trip of all the planets in oursolar system.-Abhi.=== === Subject: : Re: Action Device: Destination Alpha Centauri AIn sci.math, Abhi<8d1edf6d .0401250522.52c372db@posting.google.com>:>> In sci.math, Abhi>> <8d1edf6d.0401241427.2ccaf192@posting.google.com>:>> Look, until I plan my future life, I can not build this Action Device. Alpha Centauri A is 4.35 ly away. So, please tell me, I am>> accelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. How>> much time it will take to cover 2.18 lightyears distance i.e. when I>> have covered half the road? Please Note: Velocity Does Not Matter.... Thanks! -Abhi.>> First, there's the issue of common units. 4.35 light years is>> 4.12 * 10^16 m away. Half the distance to the goal line is>> 2.06 * 10^16 m.>> Assuming Newtonian theory, which isn't quite right:>> d = 1/2 * a * t^2>> t = sqrt(2 * d / a)>> t = sqrt(2 * 2.06 * 10^16 m / 9.805 m/s/s) = 64822355 seconds>> or a smidge more than 2 years.> Yes, it comes out to be around 750 days. It means total trip will take> around at least 8 years. Not possible for human being to spend 8 years> of his/her life in just journey.> All the people are still unaware. But what has happened that we have> technology through which we can reach to Mars in just one day. We can> take trip of all the planets within month or two. But we just can't> leave this solar system. We are confined, for the moment, to our solar> system.I'm not sure how you got the idea that Sojourner, Spirit,and Opportunity reached Mars in one *day*. A proper freeorbit trajectory from Earth to Mars would take about 9months or so, with a burn at the end to match the planet'sspeed, circularizing the orbit. Of course one can cheata bit here, as the burn at the end can bleed off some ofthe excess velocity in the wrong direction as well asincrease the velocity in the right direction.landed just the other day. That's about a 6 1/2 monthtravel time, give or take.>> Deceleration will take a similar amount of time.>> Since the final compu v = a * t = 9.805 * 64822355 =>> 635583200 = 2.11c this isn't even close to being correct>> from a physical standpoint. However, I'd have to work>> out the Lorentz contraction, among other things.>> It should be possible to reach Alpha Centauri in about a>> year (subjective), assuming sufficient fuel and a brutal>> 10g trip, but the Twin Paradox still applies.> It is not possible for human being to sustain G-Force greater than 1G> for long time, days and nights, for months. This action device can> accelerate the spaceship with acceleration far greater than 1g, say> 50g. But it will have to be Robotic mission first.Sure. Can it get off the ground? Will it spring up intospace orbit if you let it go, or just drop in a pile?You can theorize all you want but if it doesn't work in testingit's not worth jack .Your theoretical device reminds me for some reason of some veryold puppet toys. Basically, the toy has a string held in onehand; the other hand pulls down on the body of the toy, allowingit to lift its arms. Letting go of the body causes the arms todrop again.[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Action Device: Destination Alpha Centauri A[snip]> So I am going to build this Action Device and my aim is to be first> human being to walk on Mars and take a trip of all the planets in our> solar system.> -Abhi.Let us know when you get back.--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com=== === Subject: : Re: derivatives being in L^1>I think f is in L^2(0,2 pi), and the integration is performed over (0,2 pi).>And yes, f' and f'' (at least) are defined...Meaning that they both exist at _every_ point? I tend to doubt that that's what was really intended, but I could be wrong.In any case, right now the _statement_ of the problem doesn'teven make any sense: If f is in L^2(0, 2Pi) then there's no suchthing as f(x + y) in general, for x, y in (0, 2Pi). Probably youalso mean that f has period 2Pi? If so the statement makessense, but it's very easy to see that it's false (consideringan f that vanishes near 0, 2Pi and also near y.)I suspect you haven't sta the problem correctly.>> Fix y and define>> g_y (x) = [f(x + y) - f(y)] / [e^(ix) - 1].>> How do you show that>> INT |g'_y (x)| < infty iff INT |f ''(x)| < infty ?>> I see how to do this if you can let x --> 0 (and think of g_y as a sort>of>> Newton quotient for f at y, but, I don't see how to show this w/o that.>Can>> someone lead me down the right path?>> Come on. You're asking for free advice, how about some hypotheses? What>> kind of function is f? Where is it defined. Is it infinitely>> differentiable, two derivatives, what? What are the limits of integration?************************=== === Subject: : Re: general polyhedron> I *did* forget about thickness;> the problem of shingling can be mapped to good, ole 2D.> as for your novel conjecture> about the difference between a dodecasteron (icosahedron) and> a dodecaXIasteron (cuboctahedron),> How in Hell is that?> solid angle is only the same as a measure of area> for the sphere.Like, a tetrahedron and an octahedron are both spheres but the secondhas twice the area.If you are working in 2d, what's the problem with closing Noverlapping cones whose total angular deficit is 720 degrees?=== === Subject: : Re: general polyhedron> I *did* forget about thickness;> the problem of shingling can be mapped to good, ole 2D.Maybe you can provide an example for me. I don't get what the problem is. > as for your novel conjecture> about the difference between a dodecasteron (icosahedron) and> a dodecaXIasteron (cuboctahedron),> How in Hell is that?> solid angle is only the same as a measure of area> for the sphere.Polyhedron are all spheres. It is just that they are not all the same size. I don't get the XI notation. Please explain. I get the asteron part.Dick Fischbeck=== === Subject: : Linear Context Free GrammarA CFG is Linear if all production rules are in form of A->@B# inwhich, A,B are variables and @,# are members of T* .Consider language L={w in {a,b}* | number of a >= number of b}Is it Linear CF?Is there any relationship between Linear CF languages andDeterministic CF Languages?It would be appricia if anyone answers these questions,Behrang.=== === Subject: : Re: Socrates' Nothing is Everything> Nothing ventured, nothing gained and you can't make an> omlet [sic] without breaking eggs?> Please clarify this reponse.0 ventured, 0 gained, and 0/0. > (Also, since (0/0)^2 + 1 = 0 and (0/0)^2 - 1 = 0, it> follows that -1 = +1, leading to some practical issues.)Perhaps you might preach them.=== === Subject: : Re: Socrates' Nothing is EverythingIn sci.logic, Garry Denke<4e63857.0401250603.203a47e4@ posting.google.com>:>> Nothing ventured, nothing gained and you can't make an>> omlet [sic] without breaking eggs?>> Please clarify this reponse.> 0 ventured, 0 gained, and 0/0. I said clarify, not make completely useless. :-P>> (Also, since (0/0)^2 + 1 = 0 and (0/0)^2 - 1 = 0, it>> follows that -1 = +1, leading to some practical issues.)> Perhaps you might preach them.I would have thought the practical issues are obvious.Since -1 = +1, sqrt(-1) = i = sqrt(+1) = 1 or -1. Thereforei = -1 or +1. Therefore C does not exist. Thereforethe Fundamental Theorem of Algebra applies to real numbers only,although since -1 = +1, x = 1 can solve any equation, as one needmerely shift and scale: in x^2 + 4 * x + 4, x = -2 solves thatequation in normal math, but so doesx = N since x = ((N+2)/1)*(1) + (N-2)/2= ((N+2)/1)*(-1) + (N-2)/2 = -2 in Garry Denke math.I for one would think this verges on the ridiculous.Since many items -- the throwing of a baseball, the buildingof a bridge, the acceleration of a BMW from 0 to 60 mph -- canbe solved by quadratic equations, I for one see many problemswith this approach.(Not to mention everyone's probably laughing themselves sillyanyway.)-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : combinatorial algorithm neededI need an algo, which given an odd positive integer n,finds a permutation s in S_n such thatfor all primes p dividing nfor all j in {1..p} we have SUM [i=0..n/p-1] s(i*p+j) = n*(n+1)/(2*p)for example n=15:s(1..15)=2,13,4,11,1,7,6,8,10,9,15,5,12,3,14p=3: 2+11+6+9+12=13+1+8+15+3=4+7+10+5+14=40p=5: 2+7+15=13+6+5=4+8+12=11+10+3=1+9+14=24(this is needed to construct a pan-perfect magic hypercube of order n)=== === Subject: : Re: A problem >> I hope you can give me some advice in resolving the following>> question, which is a Number Theory problem. I've tried and tried but>> can't figure out the right way to solve it.>> The problem is:>> Let q(x,y) be a primitive (that is, with (a,b,c)=1) quadratic form>> ax^2+bxy+cy^2,>> and n a positive integer.>> Show the existence of two integers s,t , with (s,t)=1, such that>> (q(s,t),n)=1. Use Chinese remainder theorem to reduce to case n = p^r, p prime. Notice that is equivalent to n = p. Then one of (1,0), (0,1) and (1,1) works.> You mean, these are the values to assign to s and t?> I thought that the condition (a,b)=1 was defined just for a>1 and b>1,> not for a, b taking values 0 and-or 1.> And, the condition a,b,c coprime seems not necessary to me, if I solve> the problem in this way you told me.> Anyway, thank you for your advice!> I'm not sure I see what you are getting at.> I first say that one can always reduce to the case n = p prime.> Now if gcd(a,b,c) = 1, at least one of the three> values q(1,0) = a, q(0,1) = c and q(1,1) = a + b + c is> not a multiple of p.That's OK. This is perfectly clear to me. But the problem asked:Show the existence of two integers s,t , with s,t coprime, such thatq(s,t) and n are coprime. So, you say that s and t can assume the values 0 and 1 or 1 and 0 orboth 1s.The point is, 1 and 0 are considered coprime by definition? and 1 and1?Sorry if my post was a little hard to understand, i think it's amatter of language (i'm italian...!)Greetings, and thanks once againFrancesco=== === Subject: : Re: A problem> That's OK. This is perfectly clear to me. But the problem asked:> Show the existence of two integers s,t , with s,t coprime, such that> q(s,t) and n are coprime. I didn't spot the specification that s and t be coprime.Bu that's a non-issue... you can replace s and t by s/gcd(s,t)and t = gcd(s,t)/> So, you say that s and t can assume the values 0 and 1 or 1 and 0 or> both 1s.When n = p s prime.> The point is, 1 and 0 are considered coprime by definition? and 1 and> 1?Of course 1 and 0 generate the unit ideal in Z, as do 1 and 1-- === === Subject: : Re: A problem> That's OK. This is perfectly clear to me. But the problem asked:Show the existence of two integers s,t , with s,t coprime, such that> q(s,t) and n are coprime. > I didn't spot the specification that s and t be coprime.> Bu that's a non-issue... you can replace s and t by s/gcd(s,t)> and t = gcd(s,t)/Yes, but... (s,t) is one of (0,1), (1,0) and (1,1) ! And so gcd(0,1)doesn't make any sense ! That is the question. Maybe there's adifferent way to solve this problem?Francesco=== === Subject: : Re: A problemIn-reply-to: fscud@katamail.com (Francesco), Robin Chapman I hope you can give me some advice in resolving the following>> question, which is a Number Theory problem. I've tried and tried but>> can't figure out the right way to solve it.>> The problem is:>> Let q(x,y) be a primitive (that is, with (a,b,c)=1) quadratic form>> ax^2+bxy+cy^2,>> and n a positive integer.>> Show the existence of two integers s,t , with (s,t)=1, such that>> (q(s,t),n)=1.>> Use Chinese remainder theorem to reduce to case n = p^r, p prime.>> Notice that is equivalent to n = p.>> Then one of (1,0), (0,1) and (1,1) works.>You mean, these are the values to assign to s and t?>I thought that the condition (a,b)=1 was defined just for a>1 and b>1,>not for a, b taking values 0 and-or 1.>And, the condition a,b,c coprime seems not necessary to me, if I solve>the problem in this way you told me.>Anyway, thank you for your advice!You are given a, b, and c and looking for s and t. It only makes senseto assign values to s and t.For each prime p_k that divides n, if p_k divides a, c, and a+b+c, thenp_k divides (a,b,c). Thus, there must be at least one (s_k,t_k) from{(1,0), (0,1), (1,1)}, so that (q(s_k,t_k),p_k) = 1.Use the Chinese Remainder Theorem to find s and t so that for all p_kthat divide n s = s_k mod p_k t = t_k mod p_kSince q(s,t) = q(s_k,t_k) mod p_k, we must have that (q(s,t),p_k) = 1for all the p_k that divide n, and therefore (q(s,t),n) = 1.Rob Johnson take out the trash before replying=== === Subject: : Re: More Boolean Algebra> This isn't logic, it's algebra; so how do truth tables apply?If a statement of Boolean algebra always has value 1 in the two-valuedstructure, then it always has value 1 in any Boolean algebraicstructure.-- G.C.=== === Subject: : Question on basic analysisI don't really know how to solve the following problem:Let f:[a,b]?R be differentiable and let f'(a)?c?f'(b).Now I want to show that there is x element of [a,b], so that f'(x)=c.(It is not presupposed that f' is continuous.)I tried to start with c=0 and I know that f has a minimum and maximumbecause it is defined on a closed interval...Thank you,Marco=== === Subject: : Re: Question on basic analysis> I don't really know how to solve the following problem:> Let f:[a,b]?R be differentiable and let f'(a)?c?f'(b).> Now I want to show that there is x element of [a,b], so that f'(x)=c.> (It is not presupposed that f' is continuous.)> I tried to start with c=0 and I know that f has a minimum and maximum> because it is defined on a closed interval...> Thank you,> MarcoIf f is differentiable, then it must be continuous, right?=== === Subject: : Re: Question on basic analysisCharlie Johnson schrieb im NewsbeitragI don't really know how to solve the following problem:Let f:[a,b]?R be differentiable and let f'(a)?c?f'(b).> Now I want to show that there is x element of [a,b], so that f'(x)=c.> (It is not presupposed that f' is continuous.)>I tried to start with c=0 and I know that f has a minimum and maximum> because it is defined on a closed interval...>Thank you,Marco>> If f is differentiable, then it must be continuous, right?Yes, but it is f ' that's not presupposed to be continuous.=== === Subject: : Re: Question on basic analysis - correc verionI don't really know how to solve the following problem:Let f:[a,b]->R be differentiable and let f'(a)<=c<=f'(b).Now I want to show that there is x element of [a,b], so that f'(x)=c.(It is not presupposed that f' is continuous.)I tried to start with c=0 and I know that f has a minimum and maximumbecause it is defined on a closed interval... Thank you, Marco=== === Subject: : Re: Question on basic analysis - correc verion>I don't really know how to solve the following problem:>Let f:[a,b]->R be differentiable and let f'(a)<=c<=f'(b).>Now I want to show that there is x element of [a,b], so that f'(x)=c.>(It is not presupposed that f' is continuous.)Let g(x) = f(x) - cx and prove g has a minimum in (a,b).>I tried to start with c=0 and I know that f has a minimum and maximum>because it is defined on a closed interval...That's the general idea.=== === Subject: : Encrypt/Decrypt with java codejGpg, released under GNU GPL, a GnuPG java wrapperhttp://www.plmatrix.com/product/jgpg/index.html=== === Subject: : Re: Encrypt/Decrypt with java code> jGpg, released under GNU GPL, a GnuPG java wrapper> http://www.plmatrix.com/product/jgpg/index.htmlI don't see the source code available anywhere on that page. Where is it?-- === === Subject: : Susskind's Hologram not fundamental?Thanks Tony. I need to think about this.The area is fundamental in my model for how Einstein's curved metric field guv EMERGES as the phase ripples in the inflation vacuum coherence field from a globally flat unstable vacuum of local special relativistic quantum field theory like in the standard model. This in turn may emerge from spin foam where area is quantized. I guess the hologram idea may really boil down to the idea that there is only ONE independent spin foam quantum that is essentially Witten's alpha'?alpha' = (M theorystring tension)^-1 = (Loop Quantum of Area)/hcWitten's M theory and Loop Theory (or extended relativity in Castro's Clifford algebra formalism) are dual descriptions like in Cartan forms and projective geometry.Note, when you read Ch 3 of Lee Smolin's Three Roads to Quantum Gravity it becomes clear why1. signal locality from linearity + unitarity + Born probability sub-quantal equilibrium2. decoherence, consistent historiesmust break down in large scale physics. Smolin points out that for awhile he did not understand what Gell-Mann and Hartle were really talking about. It took a talk by Fay Dowker to show the very bizarre features of consistent histories like universes mixing different times as in Fred Hoyle's sci fi story Black Cloud. It gets down to the contingent sequence of quantum questions (projection operators in selecting out the mutually exclusive parallel world consistent histories. This is non-trivial and this is partly what Wheeler is alluding to in hisThe Question is: What is The Question?Much more on this in coming weeks.Jack, you mention ... the t'Hooft-Susskind hologram ...The quantum of 2D area is more fundamental thanthe quantum of 3D volume. ....The hologram/area picture may only be an artifact ofa semi-classical view of Hawking radiation,leading to the information paradox.In a fully quantum view,whether you adoptthe Chapline view - horizon as a quantum critical-opalescent condensatehttp://xxx.lanl.gov/abs/gr-qc/0012094orthe Cerf-Adami view - non-equilibrium thermodynamics of quantuminformation conditional entropyhttp://xxx.lanl.gov/abs/quant-ph/9904006orthe superstring fuzzball view (mentioned by Lubos Motl) of Mathur et alhttp://xxx.lanl.gov/abs/hep-th/0401115the end result is the same: the area = entropy relationshipis NOT fundamental, and there is no information paradox,so thatit seems to me that the t'Hooft-Susskind hologram programis not something that is part of a fundamental quantum theory,but only something that appears to be approximatelycorrect from a semi-classical viewpoint.As Mathur et al say in their above-ci paper:... the 'information paradox':Hawking radiation from the hole is crea bythe progressive dilation of vacuum modes near the horizon,and if the geometry here is determined only by the above conservedquantities ... mass, charge and angular momentum ...then the radiation will carry no details of the matter which went into make the hole.We thus get a violation of the unitarity of quantum mechanics...we propose that the interior of the horizon is not describedby the conventional picture ... where we have'empty space with a central singularity'.Ratherthe differences between the ... state information is distributhroughout the 'fuzzball' ... interior of the 'horizon' ...the horizon is just the boundaryof the region where the typical states differ from each other....consider the ... microstates that give the entropy of a black hole.Some attempts to locate the 'hair' have looked for smallperturbations near the horizon ...Other approaches would suggest that the microstates differ onlywithin a planck distance of the central singularity.In either caseeach microstate looks pretty much like ... a horizon with an area...But if each of the ... microstates has such a horizon,then it must itself represent different states.This makes no sense,since we wan the microstates to explain the entropy,not have further entropy themselves.We conclude that the ...microstates should have no horizons individually -the notion of a horizon should arise only after'coarse-graining' over these microstates....If we 'coarse-grain' by drawing a boundaryto enclose the region where these geometries differsignificantly from each other ... then from the area A of this boundarywe find S = A / 4G ...Thus the Bekenstein 'area entropy' arises directlyas a 'coarse-graining over hair'....Hawking's calculation of radiation showing information loss is sorobust because it uses no details of the physics at the planck scale.Resolving the information paradoxthus needs an explicit nonlocality over macroscopic distances.... thus the interior of the horizon is not just'empty space with a central singularity'.This makes it possible for radiation from the hole to pick upinformation from the 'hair' and avoid the information paradox....It is possible that infalling matter falls straight throughthe 'fuzzball' towards r = 0(as if it were falling through a conventional horizon),butover the Hawking evaporation time information is transferredto the 'light fractional modes' and into the radiation. ....Tony....=== === Subject: : Re: Practical problems with correlation dimension> , apparently tired of ranting to> those as cognitively diverse as he, chose to annoy newsgroups> typically free of idiocy like his, with:> } > 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!> } } > Karl> }> } What I will say has not yet been accep by others,> That's because you're about to prove to everyone not already aware of> the fact that you're a complete moron. Watch:> } You're missing some crucial data that cross-correlates> } your 'time' series to the cerebellar topology.> I couldn't possibly add anything to prove the point better.> Please keep your imaginary mathematical pollution confined to your> consciousness groups, although alt.usenet.kooks could use your> input.Far be it from me to provide succour for kooks or trolls but can I assumefrom your above statement that you consider all research or discussion intoconsciousness the sole 'domain' of kooks? If so, you are displaying 'kooktendencies yourself. In that you stubbornly cling to a belief which isdemonstrably wrong. Consciousness must be part of the study of the Universesince the Universe has apparently produced it as part of a natural processand must be accoun for in any valid scientific theory of 'everything'.I hope you were merely sloppy in your use of language and I am wrong in mydeductions.O'Ryan Wells.=== === Subject: : Re: MDS codes?[snip]> Some questions that I have are:> A. Given a fixed alphabet of cardinality q, can one find codes with> large n which are MDS, albeit codes that are not quite useful. As an> example, consider the cyclic code genera by x^1 + 1. The> polynomial will always be a factor of x^n + 1, thus a cyclic code> (n,2)C can be genera that will have a minimum distance of d_min => 2.> B. Are there known MDS codes whose length exceeds that of Reed-Solomon> codes?> (What is interesting to me is that the computational program MAGMA has> a function called MDSCode. This function gives the generator matrix> of a (q+1,k) code with d_min = q - k + 2. Is this a known code?> Any help, pointers to literature or suggestions will be greatly> apprecia> Jacoi recall vaguely that for any n <= q+1 and k where an MDS code is known toexist, there also exists a reed-solomon code with the same parameters.but to get it you may have to consider the extended reed-solomon codes.so i think that the real problem here is finding the appropriate finitefields F_q for large q, so that one can construct reed-solomon codes onthose finite fields.sorry, i can't remember any references for this.julius-- The most rigorous proofs will be shown by vigorous handwaving.http://www.mit.edu/~kusumaopinion of author is not necessarily of the institute=== === Subject: : Re: Algebra shows algebraic integer limitationIn sci.logic, Uncle Al<40131817.12671A08@hate.spam.net>:>> For about two years now I've been trying to explain an algebraic>> method for analyzing polynomials that relies on non-polynomial>> factors.> [snip]> For an equally long time a very large audience has empirically> demonstra by simple explicit evalution of your ravings that you are> a complete incompetent and a psychotic idiot.> Hey stooopid loud troll , put up or shut up,> http://www.rsasecurity.com/rsalabs/challenges/factoring/ faq.html> http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html> http://www.crank.net/harris.html> It's not every braying jackass that gets a whole page at crank.net> Is a $10,000 prize no questions asked too small to justify your> submission of two little prime numbers?Pedant point.RSA-576 has been factored. He'll have to go for the $20K.It's only a 640 bit number; with brute-force techniques itmight take 2.3 * 10^85 computer-years or so, assuming onefactorization attempt per millisecond and eliminating evennumbers and multiples of 3. Of course there are a numberof optimizations; I'll admit I don't know them.There's the little issue of power source, of course, as theheat death of the Universe is (AFAIK) currently estimaat about 5 * 10^10 year, and the death of the Sun at about5 * 10^9 year, which will probably vaporize anything on theEarth, if it doesn't swallow it whole. There are issues atabout 1 * 10^9 year in regards to the increased insolationcreating global warming which will make our current worrieslook very small.(Good luck, James. You'll need it!)> -- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Algebra shows algebraic integer limitation>> [...]>> What makes this result so surprising is that the necessary conclusion>> is that you cannot operate completely in the ring of algebraic>> integers if you divide both sides of>> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)>> by 7, as then *necessarily* from basic algebra, you're forced out of>> the ring as you then must have factors of 2 that are outside the ring>> of algebraic integers, as I demonstra with sqrt(7) as *necessarily*>> then the factors of 2 are sqrt(7) and 2/sqrt(7) which are not in the>> ring of algebraic integers.> But recall what Rick Decker found originally:>when x = 1, > P(x) = P(1) = 7*57.> Rick showed that P(1)/7 can be factored in the form> 57 = (5*sqrt(-2) +sqrt(7))*(-5*sqrt(-2) + sqrt(7)),>that is, there is a factorization in which all the>coefficients are algebraic integers. > You appear to be claiming this is impossible. How>do you resolve this?Well this is curious, Nora Baron. James has impliedrecently that the only reason people disagree with hisresults is that they don't like them - in snippage belowhe explains that some people just can't handle thetruth. But here it appears that you're _not_ just sayingyou don't like the result, you're giving a simple explanationof an error in the reasoning.I'm so confused. Surely James wasn't just _ignoring_this sort of reply when he explained that the onlyreason people disagree is that they don't _like_his results. He wouldn't do that.> ************************=== === Subject: : Re: Algebra shows algebraic integer limitation>I've just read this post and somehow I feel sorry for you.I've become really bothered by my posting, but I can't seem to stop. However, I realized that part of the sick cycle are some of thosepeople who reply to me so negatively, so in the hope that these peoplemight show mercy I thought I'd post something I found on the web, withlimi hope, I admit, that anything will change.Alternatively, the narcissist feels victimized by mediocrebureaucrats and intellectual dwarves who consistently fail toappreciate his outstanding - really, unparalleled - talents, skills,and accomplishments. Being haun by his challenged inferiorssubstantiates the narcissist's comparative superiority. Driven bypathological envy, these pygmies collude to defraud him, badger him,deny him his due, denigrate, isolate, and ignore him. The narcissistprojects onto this second class of lesser persecutors his owndeleterious emotions and transformed aggression: hatred, rage, andseething jealousy.But here are the scary quotes:Paranoid ideation - the narcissist's deep-roo conviction that heis being persecu by his inferiors, detractors, or powerfulill-wishers - serves two psychodynamic purposes. It upholds thenarcissist's grandiosity and it fends off intimacy.The paranoid delusions of the narcissist are always grandiose,cosmic, or historical. His pursuers are influential andformidable. They are after his unique possessions, out to exploit hisexpertise and special traits, or to force him to abstain and refrainfrom certain actions. The narcissist feels that he is at the center ofintrigues and conspiracies of colossal magnitudes.The paranoid narcissist ends life as an oddball recluse - derided,feared, and loathed in equal measures. His paranoia - exacerba byrepea rejections and ageing - pervades his entire life anddiminishes his creativity, adaptability, and functioning. Thenarcisstis personality, buffet by paranoia, turns ossified andbrittle. Finally, atomized and useless, it succumbs and gives way to agreat void. The narcissist is consumed.It looks like a person in the grips of this narcissism thing losesconnection with reality but somehow *feeds* on ANY attention,including negative attention.They called it narcissistic supply.It really is a twis illness that apparently can lead to evennastier illnesses. I am self-diagnosed (typical narcissisticbehavior) but it looks like it fits to me.However, I've decided that I don't want to succumb to mental woes, soI am fighting the dark narcissist within by making this informativepost, though it's probably a sign of the illness. In any event, maybeit will help in the long run as information is power.I will do my best to stop posting, but it will help if some of you tryto help out by not replying to me, or if you reply, by refraining fromthe attacks that feed my narcissistic side, and its irrational beliefsthat everyone is out to get me. If you have facts, sure, no problem. My inability to face facts presen without emotion will just be asign of the dark side.My only fear now is that some of you are sick as well, and act fromyour own illness, so that you can't stop attacking in your posting.But on this Christmas Day, I will act from hope, and hopefully thisNew Year will bring a change, and an escape from the madness. -- reverse my forename for mail! - saibot=== === Subject: : freeware nodal graphing software?Hi all, I have been looking for a freeware software package that will allowme to generate connections between nodes based on a matrix type inputsuch as[1 0 1 0 1][0 1 1 0 0][1 1 1 0 0][0 0 0 1 1][1 0 0 1 1]which would mean node 1 is connec to nodes 1, 3, and 5, node 2 isconnec to nodes 2 and 3, and so on.There seems to be a ton of programs out there that do this networkvisualization type of work (especially in the social network arena),but I dont know anything about any of them and would like to find aprogram that is simple and effective.Stephen=== === Subject: : Re: Implies> Let A be true and B be true then A implies B is true.> So if I'm asked to show that one statement implies another then surely> all I need to do is have them both true?> A = its raining> B = its summer.> Well lets say it is raining and that it is summer. In otherword, A is> true and B is true.> So I plug these truth values into the 'implies' operator and out comes> true. > Thus does that mean I can conclude that if its raining then its> summer?> Not quite. In order for A implies B (A => B) to hold, it must be the > case that whenever A is true, B is true. Now, if you happen to pick A > and B so that B is ALWAYS true (or A is always false), then A => B will > certainly be true, but not in general.It sound like you're misusing the word whenever. When there is someaddition variable in A and B, then A => B is just an abbreviation for(all x)A(x)=>B(x). That variable x may or may not be the time. Inthe original problem, the time is fixed, not a variable. There areno variables, and so A => B means just that. It means ~AvB which istrue.> In your example, assuming we're talking about Earth, A is sometimes > true, and B is sometimes false. If you had some way to show that > whenever it is raining is true, it is summer were also true, then A > => B would be true. But as experience shows, it frequently rains during > other seasons of the year (much to the chagrin of skiiers and > snowboarders)...so this example fails.> If you grew up on the Moon, where it never rains, then your example > would work -- it is raining is never true on the Moon, so whether or > not it is summer doesn't matter.> Similarly, if you lived on a planet where it was perpetually summer, > whether or not it is raining is irrelevant; whenever it DOES rain, it's > summertime (along with all the times when it's not raining, too).> -M=== === Subject: : Seeking relatively basic survey on TDOA analysisI am seeking a document which can be downloaded, and which contains arelatively basic introduction to Time-Delay-Of-Arrival analysis, ofthe kind that might be used for locationing in, say, a GPS, or similarsystem. I'm a tech writer; the company I work for is doing some workin this area, and I'd like to get a bit more background.I've done a Google on the subject, and there is no shortage ofdocuments out there; the problem is finding one at the right levelaudiences, and others which assume at least a Masters in EE or digitalsignal processing. I'm somewhere in between.I need something which:1. Is at roughly sophomore/junior level of engineering math.2. Has illustrations to accompany the math (I need a bit ofhand-holding with my math)3. Preferably covers at least two or three different algorithms.Any leads/links would be greatly apprecia.Steve O.Standard Antiflame Disclaimer: Please don't flame me. I may actually *be* an idiot, but even idiots have feelings.=== === Subject: : Re: Seeking relatively basic survey on TDOA analysis> I am seeking a document which can be downloaded, and which contains a> relatively basic introduction to Time-Delay-Of-Arrival analysis, of> the kind that might be used for locationing in, say, a GPS, or similar> system. I'm a tech writer; the company I work for is doing some work> in this area, and I'd like to get a bit more background. GPS Kindergarten & Up by Jerry Huang, ACME Services http://www.allgps.com/=== === Subject: : Re: Seeking relatively basic survey on TDOA analysis> I am seeking a document which can be downloaded, and which contains a> relatively basic introduction to Time-Delay-Of-Arrival analysis, of> the kind that might be used for locationing in, say, a GPS, or similar> system. I'm a tech writer; the company I work for is doing some work> in this area, and I'd like to get a bit more background.One of your search keywords should be geomatics- see:http://www.gps.gov.uk/additionalInfo/gpsLinks.asp=== === Subject: : Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication?> I have a square matrix A(NxN), and a column vector x(Nx1), > By computing y=A*x, I get another column vecotr y(Nx1). > But in implementation, due to some error, some small error was> introduced and actually I compu:> y=(A+E)*x, where E is a small error matrix(NxN), > now I want to analyze which element of y is impac most heavily, and> in what degree, and also want to have an order/ranking of the elements> of y being affec...This looks like the simplest application of something called backward error analysis.The important thing with this approach is that the right hand side of your equation is an exact mathematical equation, not a computational equation subject to numerical artifacts. The left side is the compu result. Since it is mathematics, distribution holds, so you can rewrite it asy = A*x + E*xWhere the RHS is still exact arithmetic. (A*x) is the exact answer to the matrix-vector product, (E*x) is the error.You can also apply backward error analysis to linear equation solutions (here y is given and x is the unkown), eigenvalue equations, polynomial solutions, and other matrix equation problems.$.02 -Ron Shepard=== === Subject: : Re: What kind of problem does this belong to? error analysis for matrix-vector multiplication?> I have a square matrix A(NxN), and a column vector x(Nx1), > By computing y=A*x, I get another column vecotr y(Nx1). > But in implementation, due to some error, some small error was> introduced and actually I compu:> y=(A+E)*x, where E is a small error matrix(NxN), > now I want to analyze which element of y is impac most heavily, and> in what degree, and also want to have an order/ranking of the elements> of y being affec...> I am not sure whether this kind of analysis can be made independent of> x or not... if it should be dependent on x, then what characteristics> of x combined with what kind of E gives most impact to which element> of y?I don't have an answer to your question. But I know a similar problemis addressed in numerical implementations of Gaussian Eliminationby the use of pivoting. So perhaps a search with keywords Gaussian,Elimination, and pivoting would suggest some ideas.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau=== === Subject: : Re: Dilemma with undergraduate real analysis> Anyway, to the original poster, > I say forget about your professor. It is more natural to learn > mathematics from books than from spoken lectures anyway. Things are written> more precisely and you can learn at exactly your own pace. While all books> are not equal, you can probably learn the analysis you need from just about> any half-way decent text. I recommend Elementary Classical Analysis by> Jerrold E. Marsden.> Yes he should forget about this particular dismal professor (who should > probably be fired), but I disagree about written versus a live learning > experience. A good teacher will supplement the text in important ways, > giving intuition and motivation, especially in working problems. The book > is there for all the details. One profits greatly from both.I never personally profit greatly from both, but I agree that somestudentscan. I find that a well-written text already gives intuition andmotivation.I also find that if I thoroughly read a good text, taking the time to understand each sentence, than I can solve more than 95% of theexercises,and I believe this would be true of most other people as well. Ithinkthe professor's main job is more to motivate and excite the students,and to keep them on a disciplined course of study. Also, it is theprofessor's job toselect material to be learned, topoint out errors in the text, give alternative proofs, and givecreativeassignments and tests. But a well-disciplined student can learn ALLof thematerial by reading a good text and occasionally going to the librarytosupplement the text with other points of view. A good student caneven findand correct textbook errors herself. I believe professors become toomuchof a crutch for most undergraduates. Students use them in order togetan abridged version of the material, thinking that if the professordoesn'tmention something than they don't have to learn it.Of course there are always exceptions. I'm sure that there are somegreatstudents out there learning the material very thoroughly fromtextbooks andgreatly benefitting from discussions with their professors. But thatis veryrare. I for one never learned any math from a professor that Icouldn'tfind in a text. However, I would encourage every student that readsthematerial and has a true desire to learn to start up dialogues withtheirprofessors and try to benefit from their years of experience.-Leonard=== === Subject: : Re: Dilemma with undergraduate real analysis>[...]>>What's wrong with the text for the class? (What book is it?)>Randy>>************************>The 'text' is a short workbook-style thing that the Professor made. We are not>using any standard textbook so he covers what he thinks is important.Well then of course it's impossible for any of us to judge whether thetext is really suitable or not. But as others have sugges, if you're worried about learning the material (which btw we all findvery refreshing - a lot of us are math teachers who are sick ofcomplaints about how things are too hard, a complaint aboutthings being too easy is unusual) you can always read a book.Rudin Principles of Mathematical Analysis is a classic. Somestudents find it too hard. I also like Ross Elementary Analysisa lot - it's a lot slower and gentler than Rudin.************************=== === Subject: : Re: Dilemma with undergraduate real analysis>> Anyway, to the original poster, >> I say forget about your professor. It is more natural to learn >> mathematics from books than from spoken lectures anyway. Things are written>> more precisely and you can learn at exactly your own pace. While all books>> are not equal, you can probably learn the analysis you need from just about>> any half-way decent text. I recommend Elementary Classical Analysis by>> Jerrold E. Marsden.>Yes he should forget about this particular dismal professor (who should >probably be fired), This is not clear from anything he said. His only comment on theprofessor was this:>To be>honest the professor is terrible and the way he goes about things >is to make the class so easy that even the worst students do great. That really doesn't tell us much. It could be that the professor isdismal and should be fired. Or it could be that he's perfectlycompetent, and stuck in a situation where if he taught thecourse the way it should be taught and gave the studentsthe grades they should get then more than half the classwould flunk, the students would complain, and he'd get in trouble for that!Yes, things like that happen. Yes, it happens thatadministrators judge competence _solely_ on thebasis of whether the students are complaining,with _no_ worries about whether the complaintsare justified. Hard to believe, but it _does_ happenthat faculty are told that complaints from studentsare bad and must not happen, and _any_ commentfrom the faculty member regarding the validity of thecomplaints is dismissed as irrelevant, by definition.The fact that this sometimes happens is not somethingI can prove mathematically, but trust me, it's somethingI know is true.(Note I also know that it does happen that the reason theprofessor makes the course so easy that the worststudents do well is that the professor is not competentto teach the class. But I don't see how you can concludethat here.)>but I disagree about written versus a live learning >experience. A good teacher will supplement the text in important ways, >giving intuition and motivation, especially in working problems. The book >is there for all the details. One profits greatly from both.That's all correct, of course.************************=== === Subject: : Re: Dilemma with undergraduate real analysis>Yes he should forget about this particular dismal professor (who should >probably be fired), > This is not clear from anything he said. His only comment on the> professor was this:>To be>honest the professor is terrible and the way he goes about things >is to make the class so easy that even the worst students do great. What could be clearer? He says the teacher is terrible. And this is from a student who thinks the class is way too easy. That's an evaluation worth consideration. He's in an upper level undergraduate math course. He's had other math profs to compare with. AND AGAIN, HE THINKS IT'S TOO EASY. I'm inclined to believe him. Besides, the main point here is not to determine if our OP is correct, but given his strong feelings on the matter, how best to advise him. Note that my firing comment is parenthetical.> It could be that the professor is> dismal and should be fired. Or it could be that he's perfectly> competent, and stuck in a situation where if he taught the> course the way it should be taught and gave the students> the grades they should get then more than half the class> would flunk, the students would complain, and he'd get in > trouble for that!Your argument seems to be that some outside agency could have caused the professor to be dismal. Fine, he's still dismal. And by trouble you mean what exactly? The stuff below? > Yes, things like that happen. Yes, it happens that> administrators judge competence _solely_ on the> basis of whether the students are complaining,> with _no_ worries about whether the complaints> are justified. Hard to believe, but it _does_ happen> that faculty are told that complaints from students> are bad and must not happen, and _any_ comment> from the faculty member regarding the validity of the> complaints is dismissed as irrelevant, by definition.> The fact that this sometimes happens is not something> I can prove mathematically, but trust me, it's something> I know is true.Sure, one can get complaints from above, ie, the sewer. But at any decent place you listen politely and then put them on ignore. And with tenure, you don't have much to worry about. Use yer academic freedom there mister perfesser.=== === Subject: : Re: Dilemma with undergraduate real analysis> 1. For myself, I never learned anything by reading books;Did you learn anything from writing books?-- === === Subject: : boolean algebra questionsI have several questions on boolean algebra I'm hoping someone cananswer. I'm not a mathematician, so please bear with me.Consider a boolean algebra of subsets B = where Pis the power set of A = {0, x, y, 1}, and where * will be deno byjuxtaposition.i.-----For B, does 0 = the empty set of A and 1 = Aii.-----What is the relationship between P and the following booleanexpressions/functions F --1. 02. xy3. xy'4. x5. x'y6. y7. xy' + yx'8. x + y9. x'y'10. (x + y')(x + y')11. y'12. x + y'13. x'14. x' + y15. x' + y'16. 1iii.-----Are the atoms and join irreducible elements of P 1, 2, 4, and 8? Ifnot, what are they each respectively? If so, is B an atomic booleanalgebra? Is it *not* a complete atomic boolean algebra?iv.-----For F, what are the:minterms;minimal sum-of-products expressions;complete sum-of-products expressions;prime implicants;sums of prime implicants?v.-----Is P a field of sets so that P, meaning the algebraic structure field? If so, how would P be characterized as a field? And then what wouldthe relationship be between its field characterization and ringcharacterization? I understand that all fields are rings, but amunsure of the nature of all the relationships here, presuming, ofcourse, they exist.vi.-----To my knowledge, the Hasse diagram for P with the edges removed can bediagrammed in the following manner: a b c d e f g h i j k l m n o pWhat I don't quite understand is how P would map to this diagram. Theanswer to ii should go some way towards illustrating this, I'd think.Any help would be much apprecia. Thanks in advance.Ross=== === Subject: : Re: Job options> Thanks to Pres. Bush I was laid off as a F-T community college math> instructor in August 2002 and have been unable to find another college> position since.could you elaborate?While I continue looking for another teaching job I was> wondering what other job options you nice folks can think of.go back to school and get another degree. math/math-ed degrees are> essentially worthless in begetting jobs outside academia.> Bah! and Humbug! They are liberal arts degrees. The two things business> needs are communication skills and problem-solving ability. (You can learn> the jargon and the established process as you go along.)question: how can that be accomplished?> If you have a math degree, you should know how to read and write (how to> communicate), and how to solve problems.of course - duh.> (The two procedures that work best> seem to be logic and lucky guesses. Keep in mind that even business majors> have trouble with delegation as a problem-solving technique, even though it> can be quite effective. It turns out that hiring a consultant solves the> political problem of how to sell your solution internally, and not of how> to solve the original problem.) So the problem of getting a job becomes> one of learning marketing techniques very quickly. You can get help in> this. A good place to start might be _What Color is Your Parachute?_. Or> might not.if your resume has math degrees and teaching experience, and is voidof other relevant experience, good luck accomplishing the above.> In other words, your *degree* doesn't get a job. *You* get hired -- and> hope to create a career out of it.some bright people with math degrees stand-alone await for your adviceon how to get hired...> Also, don't expect to do mathematics (unless you have a Ph.D.). You may> or may not use a mathematical approach to some problems in your career, but> you will do business. If that bothers you, what about the time you spend> cooking, or cleaning house, or bathing? Shouldn't you be doing> mathematics? Why didn't you get a Ph.D. in the first place? Or if you> did, why didn't you publish enough to be gran tenure?my sources tell me that math PhD's who do not graduate from top-rankedschools have serious difficulty getting research/teaching jobs. thosesources also tell me that there are exceptions, but significanly few.in addition to my sources, i have personally witnessed cases ofunsucessful, yet productive, math PhD's. some even from highly rankedschools.> Time spent getting another degree (unless you are genuinely interes) is> time was.if you know of a way of getting the formal qualifications in mannerthat is quicker and MORE effective in begetting non-math jobs, we thenwill consider your assessment positively.> Unless you are getting a professional degree, of course. But> even then, it's not a panacea (engineers get laid off, there are still more> architects gradua every year than currently practicing in the US, some> lawyers make very little money, etc., and there are some people with math> degrees working as engineers and analysts of various stripes). As nearly as> I can tell, the best road to riches (on average, in the US) is still to> become a doctor. But 85% report that they wouldn't do it again.i would venture to guess that the originator was not exactly askingfor advice on how to get hired as a marketer.> The key to marketing yourself is to start talking to people that do> something. Assess your abilities and interests and (surely) you know> someone who does something along those lines. Talk to them (1/2 hour, have> at least three questions that indicate you know something about what you're> talking about, and then end with Can you give me the names of two other> people that you know that I could talk to to get more information.) Then,> when you call the next two, start with, X referred me to you, I'm> interes in . . .. Because of the personal connection, you will get> talked to. If you've been honest about your abilities and interests,> eventually (since 2^n grows pretty fast), you'll eventuall meet someone in a> business with an interest and a need similar to your interests and> abilities.> Of course, you can use the more traditional means (resume floods, job fairs,> newspaper ads, etc.) as well.which applies to majority of the population. > Jon Miller=== === Subject: : Re: Recommend a Book?> Hi all,> I am a graduate - well 4 years ago anyway. I havnt been using maths to any> real standard and I don't want to forget the knowledge.> I wonder if anyone could recommend a good reference book - I would like the> book to contain details/tutorials etc on topics such as calculus, Fourier,> ODE/PDE - matrix ops... I know these topics are wide ranging - so I wonder> does a good book covering these topics decently exist?Dunford and Schwartz Linear Operators brings a lot of it together.-- G.C.=== === Subject: : ANN weight update algs / Vector calculus questionCONTEXT========With scalars, the Taylor expansion goes like this:y(x) = y(a) + y'(a) * (x-a) + (1/2) * y''(a) * (x-a) ^2 + (1/6) *y'''(a) * (x-a) ^ 3 + In artificial neural nets, many training algorithms use the Hessianmatrix to find the minimum of an error surface as a function of theweight vector.Let E = e(w)where w' is the weight vector space composed of N dimensionsE' is the scalar errore' is an error function of win particular, Ea' = e(wa)wherewa' is a given weight vector. Assume it is a column vector.Ea' is the scalar error at WaWhat I am trying to do here is to relate the Taylor series formula forscalars to the one that is applied in ANNs. The ANN literature usesthis approximation around wa:Ew = e(wa) + e'(wa) * (w-wa) + (1/2) * (w-wa)T * e''(wa) * (w-wa) +??????Note that e'(wa) is the gradient. It is a vector with N elements.e''(wa) is the Hessian matrix with NxN dimensions.(w-wa)T is row vector corresponding to the column vector (w-wa).QUESTIONS=========My questions are:1. How was the term (1/2) * (w-wa)T * e''(wa) * (w-wa) derivedaccording to Taylor's formula? The factor (1/2) is clear, but whatabout the Transposed Vector * Matrix * Vector?2. What would the 3rd order term be like? Since the 0th term is ascalar, the 1st term is a vector, the 2nd is matrix, the 3rd orderterm should naturally be something like a cubic matrix. Is there sucha thing?I am wondering if there is a way to understand this problem using thematrix notation. Perhaps this was all derived by doing the expansionsand later regrouped in matrix notation.Any pointer would be apprecia. G.=== === Subject: : Re: Rationals are UncountableIn sci.math, Virgil There are no values between zero and iota because that's iota's>> definition, a minimally positive value, next in the increasing>> sequence of ordered real numbers after zero. I could define iota as the leader of the little blue people who live>> in Ross's walls. That would not entitle me to proceed as if iota>> actually exis.>> A Google(tm) search for iota infinitesimal provides some references>> to the consideration of iota as an infinitesimal quantity,>> particularly with its quality of being the least real quantity.> The least real quantity, if it were to exist,would not be close to > zero, though its reciprocal might be.> And if iota should exist as the least real positive quantity, how big > is iota/2?Heh. And then there's the little issue that, if .999... != 1,then presumably 1 - iota = .999...., which means that1 - 10^N*iota is also .999... for any natural number N,by the simple proof:let k_n = 1 - 10^N*iota = .999... . Then compute the valuek_{n+1} = k_n * 10 - 9. For every digit in the expansion ofk_{n+1}, that digit is 9, just like k_n. Therefore,k_n = k_{n+1} = .999... .Unless one wants to give up the notion that the reals are a field,this way leads to madness. (I'm halfway there already... )-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Rationals are Uncountable permission for an emailed response.> The least real quantity, if it were to exist,would not be close to > zero, though its reciprocal might be.Hrm, or maybe the least real quantity (whatever that means) isperhaps any complex number with a zero real part? === === Subject: : Re: Rationals are Uncountable> The least real quantity, if it were to exist,would not be close to > zero, though its reciprocal might be.> Hrm, or maybe the least real quantity (whatever that means) is> perhaps any complex number with a zero real part? A least real, if such were to exist, would have to be a negative number of extremely large absolute value, and that absolute value would have to be a greatest real.A least positive real is a slightly different issue.=== === Subject: : Re: Rationals are UncountableLet X be the set of rational numbers in (0,1).> Let x_n be a sequence of rational numbers> that contains every member of X.> I have previously described such a sequence.Construct sequences a_n and b_n using> Cantor's algorithm.Are the sequences a_n and b_n finite?NOT if done as Cantor described.Assume they are finite.Why?> To eliminate the possibility.But that possibility is specifically banned by the conditions of the Cantor theorem that you are not yet understanding. > There exists an interval (a_i, b_i)> such that x_n contains no rationals> in this interval.Since we have star with a necessarily dense set, if we follow Cantor,> there cannot be any two members of that set with no members of the set> between them. This holds for the rationals.> Agreed.> Sequences A and B can never be finite if the rationals are dense.> Assume a_n and b_n are infinite.> There exists an infinite number of> rational numbers in X not in x_n.This is not necessarily true, as surjections from N to Q are known. if> the sequence x_n is one of those surjections onto Q, then there are NO> rationals not in the sequence x_n.>> The point of the proof is to show no such surjections exists.The point is to prove there are no surjections from the naturals to the REALS. There are known to be surjections from the naturals to the RATIONALS. I have construc a number of them myself, and you have read them, though clearly without understanding them if you can miss the point here by so much..> The wikipedia proof states:It states a lot of important stuff that you have left out, too.> The claim is that c cannot be in the range of the sequence x, and that is> the contradiction. If c were in the range, then we would have c = xi for> some index i. But then, when that index was reached in the process of> defining a and b, then c would have been added as the next member of one or> the other of those two sequences, contrary to the assumption that it lies> between their ranges.For the reals, by the least upper bound or gaplessness property of the reals, at least one 'c' must always exist which is not an 'x_n'. It is the always existing of 'c' that allows the throrem to guarantee the uncountablity of the realsFor X = Q, the rationals, c in Q need not always exist since such a c may have to be irrational. Examples like this have been construc here. That you do not appear to understand them is more a measure of your stupidity that of their complexity.> Nothing changes if we assume c is rational.> The assumption that the rationals> are dense is sufficient to prove c must exist.> Let a_n = x_i and b_n = x_j.> Rewrite the interval as (x_i, x_j)> For every x_j there exists an x_i, i such that no x_k, k<=j, lies in the interval> (x_i, x_j). This is true for all j.> There must be members of X that lie> in this interval if X is dense.> x_n can not contain every member of X.> Russell> - 2 many 2 count=== === Subject: : Re: Rationals are UncountableSince we have star with a necessarily dense set, if we followCantor,> there cannot be any two members of that set with no members of the set> between them. This holds for the rationals.Agreed.> Sequences A and B can never be finite if the rationals are dense.Assume a_n and b_n are infinite.> There exists an infinite number of> rational numbers in X not in x_n.This is not necessarily true, as surjections from N to Q are known. if> the sequence x_n is one of those surjections onto Q, then there are NO> rationals not in the sequence x_n.>The point of the proof is to show no such surjections exists.> The point is to prove there are no surjections from the naturals to the> REALS. There are known to be surjections from the naturals to the> RATIONALS. I have construc a number of them myself, and you have> read them, though clearly without understanding them if you can miss the> point here by so much..I have crea such surjections as well.No such surjection contains every rational.> The wikipedia proof states:> It states a lot of important stuff that you have left out, too.The claim is that c cannot be in the range of the sequence x, and thatis> the contradiction. If c were in the range, then we would have c = xi for> some index i. But then, when that index was reached in the process of> defining a and b, then c would have been added as the next member of oneor> the other of those two sequences, contrary to the assumption that itlies> between their ranges.> For the reals, by the least upper bound or gaplessness property of the> reals, at least one 'c' must always exist which is not an 'x_n'. It is> the always existing of 'c' that allows the throrem to guarantee the> uncountablity of the reals> For X = Q, the rationals, c in Q need not always exist since such a c> may have to be irrational.Yes. There exists an irrational that can be chosen to be c.There are also an infinite number of rationals that could be c.The assumption that the rationals are dense guarantees this.You agreed this is true earlier in the thread.> Examples like this have been construc here. That you do not appear to> understand them is more a measure of your stupidity that of their> complexity.Contructions have been presen that supposedly containno real number. This would mean the reals are not gapless.It also would mean the reals (and the rationals) are not dense.Make up your mind. Either the rationals are not dense orthe proof shows the rationals are uncountable.Russell- 2 many 2 count=== === Subject: : Re: Rationals are UncountableSince we have star with a necessarily dense set, if we follow> Cantor,> there cannot be any two members of that set with no members of the set> between them. This holds for the rationals.Agreed.> Sequences A and B can never be finite if the rationals are dense.> Assume a_n and b_n are infinite.> There exists an infinite number of> rational numbers in X not in x_n.This is not necessarily true, as surjections from N to Q are known. if> the sequence x_n is one of those surjections onto Q, then there are NO> rationals not in the sequence x_n.>The point of the proof is to show no such surjections exists.The existence of constructions of such surjections proves that they do exist.The point is to prove there are no surjections from the naturals to the> REALS. There are known to be surjections from the naturals to the> RATIONALS. I have construc a number of them myself, and you have> read them, though clearly without understanding them if you can miss the> point here by so much..> I have crea such surjections as well.> No such surjection contains every rational.What do you mean by surjection onto the rationals, if it is not to be onto ALL the rationals?> The wikipedia proof states:It states a lot of important stuff that you have left out, too.The claim is that c cannot be in the range of the sequence x, and that> is> the contradiction. If c were in the range, then we would have c = xi for> some index i. But then, when that index was reached in the process of> defining a and b, then c would have been added as the next member of one> or> the other of those two sequences, contrary to the assumption that it> lies> between their ranges.For the reals, by the least upper bound or gaplessness property of the> reals, at least one 'c' must always exist which is not an 'x_n'. It is> the always existing of 'c' that allows the throrem to guarantee the> uncountablity of the realsFor X = Q, the rationals, c in Q need not always exist since such a c> may have to be irrational.> Yes. There exists an irrational that can be chosen to be c.> There are also an infinite number of rationals that could be c.> The assumption that the rationals are dense guarantees this.> You agreed this is true earlier in the thread.The issue in the Cantor theorem is not whether there sometimes is a rational c, it is whether there must ALWAYS must be a rational c. Since, as you admit, there need not always be one, the Cantor theorem says nothing about the contability of the rationals.> Examples like this have been construc here. That you do not appear to> understand them is more a measure of your stupidity that of their> complexity.> Contructions have been presen that supposedly contain> no real number. By whom? Certainly not by anyone in this thread. Every construction of a sequence of nes intervals of reals that meets the requirements of the Cantor proof must contain such a c in all of them. Note that the sequence {(0,1/n)} does NOT satisfy those requirements, so it doesn not count.> This would mean the reals are not gapless.Except that it never happens.> It also would mean the reals (and the rationals) are not dense.Since it can never happen, it is irrelevant what it might mean if it shold happen.> Make up your mind. Either the rationals are not dense or> the proof shows the rationals are uncountable.Neither, since you arguments leading to this false dichotomy are multiply and hopelessly flawed. When, if ever, you can keep the whole of what that Cantor theorem says in your mind, even for the briefest fraction of a second, you will realize just how stupidly wrong you are being now.=== === Subject: : Re: Rationals are Uncountable> The point of the proof is to show no such surjections exists.> The existence of constructions of such surjections proves that they do> exist.The existence of constructions of surjections from N to Rproves they are surjections?> The point is to prove there are no surjections from the naturals tothe> REALS. There are known to be surjections from the naturals to the> RATIONALS. I have construc a number of them myself, and you have> read them, though clearly without understanding them if you can missthe> point here by so much..A number of people claim to have construc surjections from N to R.I am assuming you don't accept the claims of these people.> I have crea such surjections as well.> No such surjection contains every rational.> What do you mean by surjection onto the rationals, if it is not to be> onto ALL the rationals?I should have said there are no such surjections.> For the reals, by the least upper bound or gaplessness property ofthe> reals, at least one 'c' must always exist which is not an 'x_n'. It is> the always existing of 'c' that allows the throrem to guarantee the> uncountablity of the realsFor X = Q, the rationals, c in Q need not always exist since such a c> may have to be irrational.The wikipedia proof states:By the gaplessness of R, some point c must lie between them.We have agreed that gaplessness means the set in questionhas the LUB property. Earlier in this thread, Dave Seamanpoin out that the natural numbers have the LUB property.LUB does NOT guarantee that the interval (1,2) containsa natural number. We can assume c exists because thesets in question are DENSE.The rationals are dense and we can assume there is arational number, c, that lies in the intervals defined byCantor.The claim is that c cannot be in the range of the sequence x, and that isthe contradiction. If c were in the range, then we would have c = xi forsome index i. But then, when that index was reached in the process ofdefining a and b, then c would have been added as the next member of oneor the other of those two sequences, contrary to the assumption that itlies between their ranges.Assume no such rational c exists. If c were in the range of x_n we wouldhave c=x_i for some index i. When that index was reached in the processof defining a_n and b_n, c would would be added to A or B, contrary toour assumption that the rationals are dense.Russell- 2 many 2 count=== === Subject: : Re: Rationals are Uncountable> The point of the proof is to show no such surjections exists.The existence of constructions of such surjections proves that they do> exist.> The existence of constructions of surjections from N to R> proves they are surjections?Again you go to such extraordinary lengths to misundersand. The actual construction of a surjection from N to Q proves that such surjections can exist. Since you have acknowledged that such constructions have been made, you have acknowledged the countablility of the rationals. > The point is to prove there are no surjections from the naturals to> the> REALS. There are known to be surjections from the naturals to the> RATIONALS. I have construc a number of them myself, and you have> read them, though clearly without understanding them if you can miss> the> point here by so much..> A number of people claim to have construc surjections from N to R.> I am assuming you don't accept the claims of these people.I do not accept them. I would not accept any claim that runs contrary to a proof I accept without first carefully examining it. Those claims to have costruc surjections from N to R that I have examined have all been flawed.> I have crea such surjections as well.> No such surjection contains every rational.What do you mean by surjection onto the rationals, if it is not to be> onto ALL the rationals?> I should have said there are no such surjections.Then you would have been wrong. Again.> For the reals, by the least upper bound or gaplessness property of> the> reals, at least one 'c' must always exist which is not an 'x_n'. It is> the always existing of 'c' that allows the throrem to guarantee the> uncountablity of the realsFor X = Q, the rationals, c in Q need not always exist since such a c> may have to be irrational.> The wikipedia proof states:> By the gaplessness of R, some point c must lie between them.> We have agreed that gaplessness means the set in question> has the LUB property. Earlier in this thread, Dave Seaman> poin out that the natural numbers have the LUB property.But the Cantor proof requires that X be dense as well as gapless, which the naturals are not. You mind seems to be too small to contain more that one small thing at a time. Any compound ideas get fragmen by you into useless bits when you try to consider them. Dense AND gapless. A set with both must be uncountable. A set missing either need not be uncountable.The reals have both and so must be uncountable because of that theorem. The rationals do not have both and the naturals do not have both, and both are countable, as allowed by that theorem.The irrationals do not have both but are still uncountable, but not because of that theorem.> LUB does NOT guarantee that the interval (1,2) contains> a natural number. In the set of integers, (1,2) is empty. The LUB property does not say anything about empty sets.We can assume c exists because the> sets in question are DENSE.Wrong, again. You seem to have a great talent for being wrong.The c exists for reals because of the LUB and GLB properties of the reals, not because of density of the reals.> The rationals are dense and we can assume there is a> rational number, c, that lies in the intervals defined by> Cantor.You seem to be able, like the Red Queen in lookingglass land, to believe 10 impossible things before breakfast. But your belief does not make them true. And your ability to disbelieve true things even in the face of overwhelming proof puts your lookinglass skills way beyound those of the Red Queen. Given any strictly increasing sequence of rationals {a_n} and any strictly decreasing sequence of rationals which both converge to the same irrational number c, then there is one and only one real number in all of the intervals (a_n,b_n), and that real is the irrational c which is NOT rational. That c is very much like Russell, constitutionally incapable of being rational.> The claim is that c cannot be in the range of the sequence x, and that is> the contradiction. Not for X being the rationals, since the rationals have uncountaby many gaps, namely the irrationals.> If c were in the range, then we would have c = xi for> some index i. That assumes that there are no gaps in the rationals for c to be in, but each irrational is such a gap in the rationals, so there are lots of gaps for c to be in.[some garbage snipped]=== === Subject: : Re: Rationals are UncountableI know Virgil hates when I osophize.Of course, this is one of the reasons I do it.The following is not a proof.It is my take on what is happening.Let S be an unlimi set of symbolsand N be an unlimi set of natural numbers.I deliberately use the word unlimiinstead of infinite.I ask you to give me a function, f(), thatmaps an unique symbol in S to every member of N.After much thought you come up withsomething like this:{} = 0s{} = 1ss{} = 2...Let's call f() a numbering schema.You claim f() assigns an unique symbol toan infinite number of naturals and thereforef() assigns a symbol to every member of N.I take your f() and find a natural number thatf() doesn't assign a symbol to. I can do thisbecause N is unlimi.Not only can I find an unmatched member of N,I can find a symbol in S that f() does not use.S is also unlimi.So, I can find a natural in N that f() doesn'tassign a symbol to and a symbol in S thatf() doesn't use.This does not mean that S or N are finite,ie there is a largest natural number.It is f() that is finite where finite meansthere is a natural number that f() doesn'tassign a symbol to.It is easy to think that f() orders the symbols.After all, we never work directly with naturalnumbers, we always work with symbols.But, it is the natural numbers that are ordered,not the symbols.Early in this thread I gave an example of anatural number, x, that was not producedby a Russell Machine, RM.Once I knew the value of x it was easyto find a RM(i) = x where i >> x.This seems contradictory until we lookat what I did. I found a natural numberthat did not have a symbol and I randomlychose an unused symbol from S.Using the numbering schema, f(), x mustbe much smaller than the i in RM(i).But, I am not using the numbering schemaand the natural number that I am assigningx to is larger than any natural in the range of f().So, x represents an extremely large naturalnumber even though the numbering schema, f(),says x should represent a relatively smallnatural number.Russell- 2 many 2 count=== === Subject: : Re: Rationals are Uncountable <3c6b9c1e.0401191828.1bbb515b@posting.google.com> <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> Discussion, linux)> Assume no such rational c exists. If c were in the range of x_n we would> have c=x_i for some index i. When that index was reached in the process> of defining a_n and b_n, c would would be added to A or B, contrary to> our assumption that the rationals are dense.This is nonsense.First, you assume that the c here is not rational, and then you assumethat it's in the sequence (x_n) of rationals. Then you claim that itsmembership in A or B violates density. You are stringing random wordstogether in the vain hope that a cogent argument will result. I guessthat there are worse strategies, but they involve writing long lettersto world leaders and military honchos. Ask JSH how this gambit works.(After the first three lines were the same length by accident, thenext three took rather some effort.)-- It's an exercise in game theory[...] I've been brutally logical inmy analysis on this point.[...] My analysis indicates that theoptimal strategy for mathematicians is to acknowledge the resulttoday. --JSH gives practical reasons to accept his FLT proof=== === Subject: : Re: Rationals are Uncountable> Assume no such rational c exists. If c were in the range of x_n we would> have c=x_i for some index i. When that index was reached in the process> of defining a_n and b_n, c would would be added to A or B, contrary to> our assumption that the rationals are dense.> This is nonsense.> First, you assume that the c here is not rational, and then you assume> that it's in the sequence (x_n) of rationals.I said:Assume no such rational c exists.I never said c is irrational.> Then you claim that its> membership in A or B violates density.If c must be in set A or B there exists an interval(a_n, b_n) that contains no rational number.Are you claiming such an interval exists?Russell- 2 many 2 count=== === Subject: : Re: Rationals are UncountableAssume no such rational c exists. If c were in the range of x_n we would> have c=x_i for some index i. When that index was reached in the process> of defining a_n and b_n, c would would be added to A or B, contrary to> our assumption that the rationals are dense.This is nonsense.First, you assume that the c here is not rational, and then you assume> that it's in the sequence (x_n) of rationals.> I said:> Assume no such rational c exists.> I never said c is irrational.If c does not exist then the theorem does not apply to the rationals, and the rationals may be countable without violating anything.read what the theorem says, idiot.The density of the irrationals does not fill in those gaps in the rationals occupied by the irrationals, one of which is named Russell.> Then you claim that its> membership in A or B violates density.> If c must be in set A or B there exists an interval> (a_n, b_n) that contains no rational number.But c, if it has to exist, could be irrational, and then any non-empty open interval of rationals can still be infinite.And if a rational c does not exist, then the rationals can still be countable, at least accordingf to the theorem which you still do not understand.But you are the only one who seems to think that.> Are you claiming such an interval exists?I claim that the rationals are countable and that the irrationals are not. I have proved both to the satisfaction of everyone who can understand simple logic. You seem to want desparately to exclude yourself forever from this happy band.=== === Subject: : Re: Rationals are Uncountable <3c6b9c1e.0401191828.1bbb515b@posting.google.com> <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> Discussion, linux)> The wikipedia proof states:> By the gaplessness of R, some point c must lie between them.> We have agreed that gaplessness means the set in question> has the LUB property. Earlier in this thread, Dave Seaman> poin out that the natural numbers have the LUB property.> LUB does NOT guarantee that the interval (1,2) contains> a natural number. We can assume c exists because the> sets in question are DENSE.We can assume so, but we would be stupid to do so.Density is not gaplessness. Wikipedia was very explicit in whichproperty they used (though less explicit in how it applied to yieldc). Density is a different property.Look, if density suffices to give the existence of such a c, thenbloody well prove it.> The rationals are dense and we can assume there is a> rational number, c, that lies in the intervals defined by> Cantor.Bull. We can assume no such thing. Density does not yield theclaim.-- Jesse F. HughesWhat I represent is the unknowable future--the power of change. Inthat sense I'm a force of Nature, a force of the Universe, a livingemodiment of change itself. -- and his sense of humility=== === Subject: : Re: Rationals are Uncountable>> This is not necessarily true, as surjections from N to Q are known. if>> the sequence x_n is one of those surjections onto Q, then there are NO>> rationals not in the sequence x_n.>> The point of the proof is to show no such surjections exists.>> The point is to prove there are no surjections from the naturals to the>> REALS. There are known to be surjections from the naturals to the>> RATIONALS. I have construc a number of them myself, and you have>> read them, though clearly without understanding them if you can miss the>> point here by so much.. > I have crea such surjections as well.> No such surjection contains every rational.Then they aren't surjections.The definition of a surjection N->Q requires it to map to every rational.-Paul=== === Subject: : Re: Rationals are UncountableLet X be every rational in an open interval of the real line.Assume there is a sequence, x_1, x_2, ... such thatthe sequence contains every member of X.Claim:For every x_n there exists a rational number, c,such that c is a member of X and c is not a memberof (x_1, x_2, ..., x_n) and c < x_n.Proof:Let S be the set (x_1, x_2, x_3, ..., x_n)S must be a finite set.Assume x_n is the smallest member of S.Since X is an open set, there can be no smallestmember of X. Since x_n can not be the smallestmember of X there is a member, c, of X that issmaller than x_n and not in S.Assume x_n is not the smallest member of X.Create a set A that contains every member of Sthat is less than x_n.A is a finite set and must have a largest member.Let x_m be the largest member of A.There is no x_k, k<=n, such that x_klies in the interval (x_m, x_n).X must contain a member, c, in this intervalthat is not in S.Russell- 2 many 2 count=== === Subject: : Re: Rationals are Uncountable> Let X be every rational in an open interval of the real line.> Assume there is a sequence, x_1, x_2, ... such that> the sequence contains every member of X.Since any open interval of rationals can be mapped order-isomorphically onto any other such open interval, including the set of all rationals , there is no point in taking any interval or rationals smaller than all of Q here.> Claim:> For every x_n there exists a rational number, c,> such that c is a member of X and c is not a member> of (x_1, x_2, ..., x_n) and c < x_n.This follows trivially from the density of Q. No further proof is needed.> Proof:> Let S be the set (x_1, x_2, x_3, ..., x_n)> S must be a finite set.> Assume x_n is the smallest member of S.Unneeded, as there must be c in S smaller that the smallest of any finite umber of members sof X.> Since X is an open set, there can be no smallest> member of X. Since x_n can not be the smallest> member of X there is a member, c, of X that is> smaller than x_n and not in S.Russell, why do you keep swallowing camels then straining at gnats?> Assume x_n is not the smallest member of X.> Create a set A that contains every member of S> that is less than x_n.> A is a finite set and must have a largest member.> Let x_m be the largest member of A.> There is no x_k, k<=n, such that x_k> lies in the interval (x_m, x_n).> X must contain a member, c, in this interval> that is not in S.Since S is finite and a subset X which is an open interval in the rationals, there is at least one rational in x between any two members of S as well as below S and above S in X.You are arguiing in favor of a triviality. So what now?=== === Subject: : Re: Rationals are Uncountable> Since S is finite and a subset X which is an open interval in the> rationals, there is at least one rational in x between any two members> of S as well as below S and above S in X.> You are arguiing in favor of a triviality. So what now?This is true for all x_n.So we agree there is a member of X that is not inthe sequence (x_1, x_2, ...).Russell- 2 many 2 count=== === Subject: : Re: Rationals are Uncountable>> Since S is finite and a subset X which is an open interval in the> rationals, there is at least one rational in x between any two members> of S as well as below S and above S in X.You are arguiing in favor of a triviality. So what now?> This is true for all x_n.> So we agree there is a member of X that is not in> the sequence (x_1, x_2, ...).We agree only that for any finite subseqence of {x_n} there are values in X not yet in that finite subsequence, but if X is the set of ratiionals, then there are infinite sequences in X including every member of X. I have construc some, and you have admit their existence in another post.In that other post, you admit that there were surjections from N to Q, though you seemed not to understand what a surjection is. A function f from A to B is a *surjection* only if for every b in B there is at least one a in A for which f(a) = b.A function from the naturals, N, to an arbirary set B is called a sequence in B, and is surjective only if every member of B appears in the sequence.It appears that a part o your problem is that you are so unmathematically vague about the meanings of mathematically important words that you cannot understand what is really going on.You cannot think correctly about mathematical things if you misunderstand the meanings of those things. Which sort of misunderstanding you do. You seem to think that mathematical definitions can be as vague as your habitual though patterns and still mean something. You are wrong.> Russell> - 2 many 2 count=== === Subject: : Re: Rationals are Uncountable <87ptdjkk5r.fsf@phiwumbda.org> <3c6b9c1e.0401191828.1bbb515b@posting.google.com> <6badnbE5HvRzYIzdRVn-tw@comcast.com> <79CdnSQq58eWn47dRVn-iQ@comcast.com> <5Zmdnf5sKaz3Ho7dRVn-sA@comcast.com> Discussion, linux)>> Since S is finite and a subset X which is an open interval in the>> rationals, there is at least one rational in x between any two members>> of S as well as below S and above S in X.>> You are arguiing in favor of a triviality. So what now?> This is true for all x_n.> So we agree there is a member of X that is not in> the sequence (x_1, x_2, ...).(1) For all n, there exists x such that x is not in (x_1,...,x_n).(2) There exists x such that for all n, x is not in (x_1,...,x_n).(3) There exists x such that x is not in (x_1,...)Clearly, (2) implies (3). You have proven (1). But (1) does notimply (2). Get a new schtick. The whole commuting quantifiers thing is tiresome.For how many years have you been pursuing the same dead ends by makingthe same error of (A x)(E y)P => (E y)(A x)P? Why is it so difficultto learn that this is not a valid inference?This basic fact has nothing to do with infinity, by the way.Everybody loves somebody, but it is not the case that somebody isloved by everybody. The error you persist in repeating is just thatsimple and obvious.-- Jesse HughesMy experience indicates that the people who post on this newsgroupare about at the level of a 10 year old in the year 2060. -- More wisdom from , time traveler=== === Subject: : Re: sqrt(-1)=0/0> How can -1 be 0? The real numbers form a field and its trivial that> there can only be one number such that a * 0 = 0, for any a. But you> are saying -1 * 0 = 0. Contradiction.>I agree that the original poster lacks mathematical understanding in> this particular area. But calling someone an idiot, and then claiming> that -1 * 0 = 0 is a contradiction, does not reflect well on you today.> I imagine that was to read -1 * -1 == 0, he was trying to say there is only> one k for ak = 0, a = -1 and k = -1 is not it.Yes that is what I meant ... Jason=== === Subject: : Re: sqrt(-1)=0/0> How can -1 be 0? The real numbers form a field and its trivial that> there can only be one number such that a * 0 = 0, for any a. But you> are saying -1 * 0 = 0. Contradiction.*Someone who pos here a few months ago claimed that sqrt(-1) = 0/0 (which is the subject of this thread) because when he entererd sqrt(-1) on his calculator he got error. Then, when he entered 0/0 he also got error.Therefore they must be equal, since they are both equal to error.earle*BTW, the rationals are countable.Long ago, de Morgan showed that the ratio of log(-1) -- which is imaginary -- to the sqrt(-1) -- also imaginary -- is the same as the ratio of the circumference of a circle to its diameter.How about that?ej*=== === Subject: : Re: sqrt(-1)=0/0> Thus for Socrates, nothing = everything, This is highly contradictory,Nothing implies 0 state.Everything implies more than 0 states.0 != x where x > 0 AND x < 0, so Socrates was wrong or was using anundocumen system of logic.> for Denke, nothing/nothing = everything. There is insufficient justification for this statement, or anundocumen system of logic is being employed.If we accept that nothing is something, then you could argue thatEverything includes nothing because otherwise Since everythingdoes not include nothing, everything lacks something.In fact, I showed a proof on sci.math that with probability 1, allfinite sequences of integers occur infinitely in a countably infinitesequence of random integers. > As for myself, I know nothing, 0.If you knew nothing then how can you know that you know nothing? Thisis again contradictory or an undocumen system of logic is beingemployed.JS=== === Subject: : Re: sqrt(-1)=0/0> Thus for Socrates, nothing = everything, > This is highly contradictory,> Nothing implies 0 state.> Everything implies more than 0 states.> 0 != x where x > 0 AND x < 0, so Socrates was wrong or was using an> undocumen system of logic.> for Denke, nothing/nothing = everything. > There is insufficient justification for this statement, or an> undocumen system of logic is being employed.> If we accept that nothing is something, then you could argue that> Everything includes nothing because otherwise Since everything> does not include nothing, everything lacks something.> In fact, I showed a proof on sci.math that with probability 1, all> finite sequences of integers occur infinitely in a countably infinite> sequence of random integers.> As for myself, I know nothing, 0.> If you knew nothing then how can you know that you know nothing? This> is again contradictory or an undocumen system of logic is being> employed.> JS0/0 is every number, 0 is their placeholder.=== === Subject: : Re: sqrt(-1)=0/0> Thus for Socrates, nothing = everything, > This is highly contradictory,> Nothing implies 0 state.> Everything implies more than 0 states.> 0 != x where x > 0 AND x < 0, so Socrates was wrong or was using an> undocumen system of logic.> for Denke, nothing/nothing = everything. > There is insufficient justification for this statement, or an> undocumen system of logic is being employed.> If we accept that nothing is something, then you could argue thatEverything includes nothing because otherwise Since everything> does not include nothing, everything lacks something.> In fact, I showed a proof on sci.math that with probability 1, all> finite sequences of integers occur infinitely in a countably infinite> sequence of random integers.> As for myself, I know nothing, 0.> If you knew nothing then how can you know that you know nothing? This> is again contradictory or an undocumen system of logic is being> employed.> JS> 0/0 is every number, 0 is their placeholder.0/0=undefined, so 0 cannot be a place holder for all x in N, I or R. 0denotes the absence of quantity which is instrinsically a quantity(ie. nothing is something).Nothing has no constituent parts, and therefore cannot be divided byor multiplied against (10 times nothing is still nothing, half ofnothing is undefined).JS=== === Subject: : The Question is: What is The Question?http://www.holidays.net/passover/ question.htmlContextual background:Orthodox micro-quantum theory in the Gell-Mann/Hartle decoherent consistent histories interpretation, taken naively to the macro-realm as in conventional quantum cosmology and quantum gravity with the Wheeler-Dewitt timeless eq. HPSI(Universe) = 0 in Wheeler's superspace of 3D space-like geometries in the canonical 3+1 split (ADM), has many BIT thought-like parallel universes (attractor basins sans system point in Susskind's non-Bohmian Landscape that is subject to the Darwinian natural selection of the Weak Anthropic Principle (WAP) for the arbitrary parameters in the standard models of lepto-quarks and chaotic inflationary cosmology):These universes differ not only in the answers that the theory gives to questions, but by the questions that are asked. Everett's relative-state form of the theory, must therefore be supplemen by a theory of why what we observe corresponds to the answers of certain questions, and not to an infinite number of other questions. p. 43Dowker's and Kent's results showed that there had to be an infinite number of other worlds too. ... Even more disturbing there were worlds that were classical now that were arbitrarily mixed up superpositions of classical [worlds] at any point in the past. Dowker concluded [in 1995] that, if the consistent histories interpretation is correct, we have no right to deduce from the existence of fossils now that dinosaurs roamed the Earth one hundred million years ago. ... the people sitting near me were as shocked as I was ... Jim Hartle insis that the work he and Murray Gell-Mann had done on the consistent histories approach was not contradic by anything Fay Dowker had said. They were fully aware that their proposal imposed on reality a radical context dependence: one cannot talk meaningfully about the existence of any object or the truth of any statement without first completely specifying the questions that are to be asked. It's almost as if the questions bring reality into being. p.44It is ironic that Murray Gell-Mann in Chapter 11 of his The Quark and the Jaguar alludes to the Chickering letter to Under Secretary of Defense Richard De Lauer I helped to write in 1981 (see my book Destiny Matrix) that talks about signal nonlocality, which does violate micro-quantum theory. Gell-Mann's interpretation of micro-quantum theory is equally if not more bizarre than a sub-quantal non-equilibrium universe with signal-nonlocality that allows communication between the decoherent parallel BIT worlds and pulls the rug out from under the non-Boolean topos approach of Fotini Markopoulo-Kalamara (Ch 3 of Smolin's book). Of course, even with signal nonlocality allowing communication between parallel Hubble sphere worlds in Max Tegmark's Level 1 and Level 2, no single observer inside the relational multi-verse is ever able to have complete information. Therefore, some kind of Bayesean epistemology is needed. The problem here is that the boundary between epistemology and ontology is fuzzy.... there are other troubles with this conventional formulation of quantum cosmology. ... one is not free to ask any set of questions: instead these are constrained by having to be solutions to certain equations. ... It seems unlikely that this can ever be done. p. 45Thanks Tony. I need to think about this.The area is fundamental in my model for how Einstein's curved metric field guv EMERGES as the phase ripples in the inflation vacuum coherence field from a globally flat unstable vacuum of local special relativistic quantum field theory like in the standard model. This in turn may emerge from spin foam where area is quantized. I guess the hologram idea may really boil down to the idea that there is only ONE independent spin foam quantum that is essentially Witten's alpha'?alpha' = (M theorystring tension)^-1 = (Loop Quantum of Area)/hcWitten's M theory and Loop Theory (or extended relativity in Castro's Clifford algebra formalism) are dual descriptions like in Cartan forms and projective geometry.Note, when you read Ch 3 of Lee Smolin's Three Roads to Quantum Gravity it becomes clear why1. signal locality from linearity + unitarity + Born probability sub-quantal equilibrium2. decoherence, consistent historiesmust break down in large scale physics. Smolin points out that for awhile he did not understand what Gell-Mann and Hartle were really talking about. It took a talk by Fay Dowker to show the very bizarre features of consistent histories like universes mixing different times as in Fred Hoyle's sci fi story Black Cloud. It gets down to the contingent sequence of quantum questions (projection operators in selecting out the mutually exclusive parallel world consistent histories. This is non-trivial and this is partly what Wheeler is alluding to in hisThe Question is: What is The Question?Much more on this in coming weeks.=== === Subject: : Re: when NaturalNumbers = p-adics what alters in the Riemann Hypothesis Re: proof of the Riemann Hypothesis> (snips)I would have had to consider two extreme cases: (1) that the RH is> completely false just as FLT is completely false when NaturalNumbers => p-adics. If taking that extreme route would have had me find out what> was flawed in my claimed two proofs below.(2) the second extreme case is to say that something lies on the 1/2> Real line but not the NaturalNumbers of the illdefined notion of> finite-integers but rather instead the p-adics. Only the line is not a> straight line. And my second *alleged proof below* using a spiral sort> of touches or hints of a curved line.> (big snip)> TWO PROOFS OF THE RIEMANN HYPOTHESIS> [clip]> *> Mr. Plutonium: Have you ever submit a paper for publication?> earle> *For Archimedes Plutonium to submit any of his mathematical proofs to aMath Journal is akin to a Democrat in the USA Senate to submit aEarth Air Conditioner Bill espousing a research race to find achemical that can solve not only Global Warming but can also make thisplanet much more comfortable and much greener where humanity actuallybegins to climate control planet Earth.We all know what such a Democrat senator trying to solve GlobalWarming would face from a Republican controlled Congress and a BushGlobal Warming. Bush does not even accept that the American Academy ofSciences, that all the scientists have proven Global Warming is realand is the cause of our goofy weather.Just the other day I read a report that the Gulf Stream warm aircurrents will vanish as Global Warming increases which means that theEAst Coast USA will be more frigid in winters and that Europeextending to Russia will become freezer countries in the decadesahead.The Bush administration is anti-science. They are boneheads aboutGlobal Warming and about science. The mathematics journals editors areboneheads. They are ignorant of the idea that NaturalNumbers are notFiniteIntegers but are instead the P-adics. They are ignorantboneheads as to what NaturalNumbers are and why they are a ill-definedset, just as Bush is a ignorant bonehead about Global Warming.Is it any use for a Democrat to start a research program as to whatchemical we can put into the upper atmosphere to create a Global AirConditioner. No. We all know what the Republican controlled Congressand WhiteHouse will do-- reject reject and reject. They even rejecthe Academy of Scientists.Is it any use for Archimedes Plutonium to send two proof of theRiemann Hypothesis to any math journal editors in the world? Of coursenot. They are math boneheads, whose ignorant minds can only write--reject, reject, reject.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : JSH: A Looney Tune?expressed concern over their or someone else's morbidcuriosity with James' relentless pursuit of fame and fortunethrough mathematics. But maybe it isn't so morbid after all.Here's my theory.Consider: if Mathematical Truth is the Road Runner, then is a remarkable model for Wile E. Coyote. Heis constantly slamming into walls, sailing off cliffs orgetting entangled in his own traps. Readers delight indiscovering exactly where his latest contraption fails andcan plainly see the destructive path he insists on taking.Their efforts to help him surmount obstacles can be regardedas part of the good-natured fun. The level of entertainmentescalates as each failure is met with a new and less thanhalf-baked assault on his adversary.Once in a while he goes silent, perhaps waiting for thedelivery of the latest gadget from ACME Proof Works, afterwhich he wades back in with his newfangled thingamabob, onlyto smack into a new obstacle, or succumb to his ownill-conceived trap.Personally, I think the legacy of James' posts and thereaders responses resembles a Warner Bros. cartoon and amdeligh that he persists in providing this never-endingentertainment.--There are two things you must never attempt to prove: theunprovable -- and the obvious.----http://www.crbond.com=== === Subject: : Re: JSH: A Looney Tune?> [...]> Consider: if Mathematical Truth is the Road Runner, then> is a remarkable model for Wile E. Coyote.> [...]Not a bad analogy, but doesn't the Road Runner make strenuousefforts to avoid Wile E. Coyote, and don't Coyote's schemesoften have a certain ingenuity that would trap a less nimbleopponent?I can't help thinking a slightly better analogy has Bugs Bunnyas the truth, casually standing munching a carrot and not eventrying to escape, with the role of James played by Elmer Fudd.Cheers--------------------------------------------------- ------------------------John R Ramsden (jr@adslate.com)---------------------------------------------- -----------------------------Eternity is a long time, especially towards the end. Woody Allen