August 2013 - mathfun@mailman.xmission.com

Re: [math-fun] quantum theory foundational issues, my theory of how they should be resolved
by Adam P. Goucher 05 Aug '13

05 Aug '13

> David Makin wrote: > > > But, equivalently, we could write the same binary input stream on its initial tape, so this self-contained Turing machine behaves exactly the same as your `computer with a method of sensing the rest of existence'. > > No because neither you, nor anyone (or anything) else could predict what that binary stream should be *until the time when that state exists* (true based on either uncertainty in the current physics model or based on my own belief that existence is fractal in nature). There's no need to predict the future. Equally well one could record the past, and feed that into the machine. The machine wouldn't know that it's being given a recording of past events. Sincerely, Adam P. Goucher

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Re: [math-fun] quantum theory foundational issues, my theory of how they should be resolved
by Adam P. Goucher 05 Aug '13

05 Aug '13

> David Makin wrote: > > > I disagree strongly., why are "Turing machines" obviously not consdcious ? Turing machines are mathematical concepts, not materialistic entities. When programmed correctly, they are capable of miraculously sophisticated behaviour -- probably greater than that of human minds, potentially, due to the unlimited storage -- but they can't possibly be conscious. Here is a thought experiment: Suppose a Turing machine X with initial tape T_0 is `conscious'. Then, by definition of consciousness, it must be aware of its existence. Now, I suppose that cannot be ruled out in the following situation: 1. An electronic computer simulates X with the initial tape T_0, and is therefore conscious. But the Platonic abstraction of `X with the initial tape T_0' has always existed as a mathematical abstraction, so even before anyone built an electronic computer it must have been conscious. Even before the universe existed, `X with the initial tape T_0' has always had a predetermined set of future states. Going in the other direction, what about the following? 2. Two electronic computers each simulate X with the initial tape T_0. Are you suggesting that the `consciousness' therefore inhabits both electronic computers simultaneously? Finally, if Turing machines are conscious, then so are (by emulation) Diophantine equations. > Turing machines *with no method of sensing the outside world* may well be "obviously not conscious" but give the computer the programmed ability to self-evolve and sense the rest of existence - what then ? Suppose you give the computer a method of sensing the rest of existence. Without loss of generality, it may as well receive a binary input stream through a USB port (or whatever). But, equivalently, we could write the same binary input stream on its initial tape, so this self-contained Turing machine behaves exactly the same as your `computer with a method of sensing the rest of existence'. Sincerely, Adam P. Goucher

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[math-fun] "The" q-Fibonacci numbers
by Bill Gosper 04 Aug '13

04 Aug '13

On Sun, Aug 4, 2013 at 3:41 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > Replacing the Binomials with QBinomials in the Fibonacci identities > below, and throwing in a factor of q^k^2, we get the "q-Fibonacci numbers": > In[47]:= > Expand[FunctionExpand[Table[Sum[q^k^2*QBinomial[n-k,k,q],{k,0,n}],{n,{0,1,2,3,4,5,6,19}}]]] > Out[47]= {1,1,1+q,1+q+q^2,1+q+q^2+q^3+q^4,1+q+q^2+q^3+2 > q^4+q^5+q^6,1+q+q^2+q^3+2 q^4+2 q^5+2 q^6+q^7+q^8+q^9,1+q+q^2+q^3+2 q^4+2 > q^5+3 q^6+3 q^7+4 q^8+5 q^9+6 q^10+7 q^11+9 q^12+10 q^13+12 q^14+14 q^15+17 > q^16+19 q^17+23 q^18+25 q^19+29 q^20+32 q^21+37 q^22+40 q^23+46 q^24+50 > q^25+56 q^26+61 q^27+68 q^28+73 q^29+81 q^30+86 q^31+94 q^32+100 q^33+108 > q^34+113 q^35+122 q^36+127 q^37+135 q^38+140 q^39+148 q^40+152 q^41+160 > q^42+163 q^43+169 q^44+172 q^45+177 q^46+177 q^47+182 q^48+181 q^49+183 > q^50+181 q^51+182 q^52+178 q^53+178 q^54+172 q^55+170 q^56+164 q^57+160 > q^58+152 q^59+148 q^60+139 q^61+133 q^62+124 q^63+118 q^64+108 q^65+102 > q^66+92 q^67+85 q^68+76 q^69+70 q^70+61 q^71+56 q^72+48 q^73+43 q^74+36 > q^75+32 q^76+26 q^77+23 q^78+18 q^79+15 q^80+12 q^81+10 q^82+7 q^83+6 > q^84+4 q^85+3 q^86+2 q^87+2 q^88+q^89+q^90} > In[48]:= %/.q->1 > Out[48]= {1,1,2,3,5,8,13,6765} > (6765 was the console phone # at MIT AI.) These polynomials > approach a limit, q-Fibonacci(∞) which is the Rogers-Ramanujan function: > In[55]:= Series[1/Product[(1-q^(5*n-4))*(1-q^(5*n-1)),{n,∞}],{q,0,36}] > Out[55]= 1+q+q^2+q^3+2 q^4+2 q^5+3 q^6+3 q^7+4 q^8+5 q^9+6 q^10+7 q^11+9 > q^12+10 q^13+12 q^14+14 q^15+17 q^16+19 q^17+23 q^18+26 q^19+31 q^20+35 > q^21+41 q^22+46 q^23+54 q^24+61 q^25+70 q^26+79 q^27+91 q^28+102 q^29+117 > q^30+131 q^31+149 q^32+167 q^33+189 q^34+211 q^35+239 q^36+O[q]^37 > (The sum (with its upper limit infinite, with or without reflecting the > q-binomial) > is not obviously the usual one in the Rogers-Ramanujan identity.) > The coefficient sequence A003114 <https://oeis.org/A003114>has a bunch > of interesting > combinatorial interpretations, enriching the property list of 239 (if you > accept its index 36 as somehow fundamental). > --rwg > > This is apparently the original definition of q-fibonacci, dating back to [9] I. Schur, Gesmmelte Abhandungen, Vol. 2, Springer-Verlag, Berlin, 1973, pp. 117-136 according to http://arxiv.org/pdf/math/0508546v2.pdf . But just eight years ago, when I could pass Turing's test, I mentioned here an attractive alternative, qfb, to the above, which I called qib. OEIS wisely disinguishes qfb as "Fibonacci polynomial": ----------------------- [Massive headercide] Date: Sun, 11 Sep 2005 01:09:10 -0700 (PDT) From: "R. William Gosper" <rwg(a)osots.com> Message-Id: <200509110809.j8B89AXH055413(a)m130.osots.com> To: math-fun(a)mailman.xmission.com Many of you may remember the spectacularly ill-conditioned "Filbert" matrix pointed out to me by Peter ("Feathers") Groot in the late 60s: E.g., (c89) (genmatrix(lambda([i,j],1/fib(i+j+9)),3),%% = float(%%)) [ 1 1 1 ] [ -- --- --- ] [ 89 144 233 ] [ ] [ 0.01124 0.00694 0.00429 ] [ 1 1 1 ] [ ] (d89) [ --- --- --- ] = [ 0.00694 0.00429 0.00265 ] [ 144 233 377 ] [ ] [ ] [ 0.00429 0.00265 0.00164 ] [ 1 1 1 ] [ --- --- --- ] [ 233 377 610 ] Contrary to appearance, this is good to six figures: (c90) 1/%; [ 89 144 233 ] [ 89.0 144.0 233.0 ] [ ] [ ] (d90) [ 144 233 377 ] = [ 144.0 233.0 377.0 ] [ ] [ ] [ 233 377 610 ] [ 233.0 377.0 610.0 ] But (c91) float(map('invert,d89)) [ 1.0019e+11 2.62304e+11 - 6.86715e+11 ] [ ] (d91) [ 2.62304e+11 6.86691e+11 - 1.79782e+12 ] = [ ] [ - 6.86715e+11 - 1.79782e+12 4.70676e+12 ] [ 5.01351e+7 1.44257e+8 - 3.64668e+8 ] [ ] [ 1.3994e+8 3.67339e+8 - 9.60738e+8 ] [ ] [ - 3.57682e+8 - 9.72036e+8 2.50923e+9 ] The disagreement is even *more* dramatic if the floating numbers on the rhs of d89 are first converted to exact rationals, and the inversion performed exactly! The reason is that the matrix is far closer than 10^-6 to being singular, and truncation to six figures pushes it way over the brink. The true inverse is symmetric and integer. I tried substituting a couple of candidates for q-Fibonaccis to see if inverting made huge q-polynomials. It is tempting to define qib(0):=0, qib(1):=1, n - 2 qib(n) = qib(n - 1) + q qib(n - 2)), for then n 1 floor(- - -) 2 2 ==== 2 \ k qib(n) = > qbn(n - k - 1, k) q / ==== k = 0 where qbn is the q-binomial coefficient (q; q) n qbn(n, k) = -------------------. (q; q) (q; q) k n - k Then, thru qib(11) 0 1 1 1 + q 2 1 + q + q 2 3 4 1 + q + q + q + q 2 3 4 5 6 1 + q + q + q + 2 q + q + q 2 3 4 5 6 7 8 9 1 + q + q + q + 2 q + 2 q + 2 q + q + q + q 4 6 2 3 4 5 6 (1 + q + q ) (1 + q + q + q + q + q + q ) 4 2 3 4 5 6 7 8 9 10 (1 + q ) (1 + q + q + q + q + q + 2 q + 2 q + 2 q + 2 q + q 11 12 + q + q ) 2 3 4 2 3 4 (1 - q + q - q + q ) (1 + q + q + q + q ) 4 5 6 7 8 9 10 11 12 (1 + q + q + q + q + q + q + q + q + q + q ) 2 3 4 5 6 7 8 9 10 1 + q + q + q + 2 q + 2 q + 3 q + 3 q + 4 q + 5 q + 5 q 11 12 13 14 15 16 17 18 + 5 q + 6 q + 6 q + 6 q + 6 q + 6 q + 5 q + 5 q 19 20 21 22 23 24 25 + 4 q + 4 q + 3 q + 2 q + q + q + q and qib(oo) is defined and equal to inf ==== \ A003114(n) 1 > q = ----------------------------------- / oo ==== /===\ n = 0 | | 5 n + 1 5 n + 4 | | (1 - q ) (1 - q ) | | n = 0 oo 2 ==== n \ q = > -------, / (q; q) ==== n n = 0 the famous Rogers-Ramanujan function. But this qib definition produces messy denominators in the Filbert matrix, and worse, fails to distribute over gcd, f(gcd(n,k)) = gcd(f(n),f(k)), unlike f:= vanilla fib. Defining instead qfb[0] := 0, qfb[1] := 1, qfb[n] := qfb[n-1] + q qfb[n-2] (A011973) answers these objections: 0 0 1 1 2 1 3 1 + q 4 1 + 2 q 2 5 1 + 3 q + q 6 (1 + q) (1 + 3 q) 2 3 7 1 + 5 q + 6 q + q 2 8 (1 + 2 q) (1 + 4 q + 2 q ) 2 3 9 (1 + q) (1 + 6 q + 9 q + q ) 2 2 10 (1 + 3 q + q ) (1 + 5 q + 5 q ) 2 3 4 5 11 1 + 9 q + 28 q + 35 q + 15 q + q 2 12 (1 + q) (1 + 2 q) (1 + 3 q) (1 + 4 q + q ) 2 3 4 5 6 13 1 + 11 q + 45 q + 84 q + 70 q + 21 q + q 2 3 2 3 14 (1 + 5 q + 6 q + q ) (1 + 7 q + 14 q + 7 q ) Note that the degree of qfb[n] is floor((n-1)/2), and qfb[2 p] = qfb[p] times a polynomial of equal degree which is 1 mod p. We also have the pleasant addition formula [ qfb qfb ] [ qfb qfb ] [ k + 1 k ] [ 1 0 ] [ n + 1 n ] [ ] . [ ] . [ ] = [ qfb qfb ] [ 0 q ] [ qfb qfb ] [ k k - 1 ] [ n n - 1 ] [ qfb qfb ] [ n + k + 1 n + k ] [ ]. [ qfb qfb ] [ n + k n + k - 1 ] [error in A011973, since corrected, but F(n+1, x) = F(n, x) + x*F(n-1, x)) still has an extra close paren!-] And the q-Filbert inverse contains satisfyingly huge polynomials divided by mere monomial powers of q. (It looks possible to guess the general form, which will probably require q-"superfactorials", i.e. q-extensions of 1^1 2^2 3^3 ... n^n or 1! 2! 3! ... n!.) [...] --rwg PS: 4 2 %i %pi k ==== ---------- \ 1/5 5 > sqrt(8 - 7 %e ) / ==== k = 0 1 1/5 3/5 1 1/5 2/5 2 (------- + 1) 7 (7 + 3) (1 - -------) (7 + 2) 7 sqrt(5) sqrt(5) 2 = ------------------------------- + ----------------------------- + - sqrt(5) sqrt(5) 5 ~ 14.1420530099223 --------------------------- Note that qib(6) is irreducible. I failed to mention that qfb also has a pretty little Binet formula: In[60]:= F[n+1,x]==F[n,x]+x*F[n-1,x]/.F[k_,x]->qfb[k]/.x->q Out[60]= qfb[1+n]==q qfb[-1+n]+qfb[n] In[61]:= RSolve[{%,qfb[0]==0,qfb[1]==1},qfb[n],n] Out[61]= {{qfb[n]->-((2^-n ((1-Sqrt[1+4 q])^n-(1+Sqrt[1+4 q])^n))/Sqrt[1+4 q])}} which A011973 could mention. --rwg WDS>``BUT gravitons cannot be shielded even in principle. When you emit them, they fly away right thru the box walls. Bingo -- you've been measured by an "external environment."'' Then, can gravitons be detected?

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Re: [math-fun] quantum theory foundational issues, my theory of how they should be resolved
by Adam P. Goucher 04 Aug '13

04 Aug '13

> Dan wrote: > > Adam, what is the connection between emulating human behavior (or the vaguely defined Turing test) on the one hand, and Turing completeness on the other? If any digital computer can emulate human behaviour, then so could any Turing-complete system. > Mike wrote: > > > Also, computers are provably not conscious, since they can be emulated by > Turing machines, which are obviously not conscious. > > It's not obvious to me. Instead of thinking of the Turing machine as an electronic machine (which could indeed be consicous), think of it as a long tape of paper, with erasable symbols, and a reference book (the `rule table'). Every day, a different volunteer walks into the room, reads a single symbol, obeys the instruction given in the reference book (e.g. replace the symbol with another symbol, move left and change state), and leaves the room. An inanimate book cannot be conscious. Neither can the tape. Hence, if there is any consciousness, it must lie in the individual volunteers (assuming they're human, they definitely will be conscious). But they can be replaced with much simpler machines (say, mechanical automata made of LEGO), with no memory, which only exist for the few minutes it takes to enter the room, obey the instruction and leave. Now, these LEGO automata are *certainly* not aware of their own existence. And there is only one alive at any time, so we can't argue that they are forming some `super-brain'. The automata themselves cannot make decisions, nor be conscious, since they're far too simple. But the overall system I've described is a Turing machine, which advances its computation each day. Admittedly it's about 10^17 times slower than the average desktop computer, and will take billions of years to emulate just a single second of a large neutral network, but it's still capable of emulating one, nonetheless. Sincerely, Adam P. Goucher

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[math-fun] Is Adam Goucher conscious?
by Warren D Smith 04 Aug '13

04 Aug '13

How do I know you are aware of your own existence? Because you say you are? Well, a Turing machine could print out a statement saying it is, too. In principle a Turing machine presumably could/should exist which would completely emulate Adam Goucher's every utterance and action. This seems to prove Adam Goucher is not a conscious being (rest of proof supplied by AG himself previously). Probably explains why I only am aware of him thru the internet.

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Re: [math-fun] Is Adam P. Goucher conscious?
by Adam P. Goucher 04 Aug '13

04 Aug '13

(I've corrected the subject title by inserting the non-optional middle initial. Apologies go to anyone whose e-mail clients interpret this as a completely unrelated message.) > Warren Smith wrote: > > How do I know you are aware of your own existence? Because you say you are? Exactly, you don't know that I'm conscious. You don't know that anyone is conscious, with the (possible!) exception of yourself. Indeed, for a very long time people believed that I was a fictional being invented by Dr. Geoff Smith in an attempt to motivate people training for the British IMO team. > Well, a Turing machine could print out a statement saying it is, too. Correct. > In principle a Turing machine presumably could/should exist which > would completely emulate Adam Goucher's every utterance and action. Indeed. > This seems to prove Adam Goucher is not a conscious being (rest of > proof supplied by AG himself previously). That is an epic case of ignoratio elenchi. I asserted that machines incapable of consciousness are capable of human-like behaviour, and not that humans themselves are incapable of consciousness. Sincerely, Adam P. Goucher

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Re: [math-fun] quantum theory foundational issues, my theory of how they should be resolved
by Marc LeBrun 04 Aug '13

04 Aug '13

>="Dan Asimov" <dasimov(a)earthlink.net> [...] > There are two striking aspects of the universe that are so difficult to > address that physics can't touch them (so far): a) the flow of time, and b) > conscious awareness. These two things must be very closely related. But very > mysterious. Maybe I'm just being too densely mechanistic, but I find the popular idea that consciousness is somehow closely related to deep physics puzzlingly unmotivated. Other smart folks (eg Roger Penrose) also assert things along these lines, but I can't begin to follow these arguments because conscious awareness just seems to me a natural consequence of animals evolving the ability to create and manipulate models. Sometimes we just happen to build and manipulate reflective models of ourselves that capture aspects of our own model-processing--but that doesn't seem fundamentally different from many other internal-model driven behaviors--no quantum weirdness required. Perhaps our own consciousness feels mysterious to us because access to our self-models is necessarily vague, murky, incomplete and sketchy--that is, well-adapted to the natural limitations of the jellyware platform we run on. We're self-aware, but through a gloss, darkly. (This is probably a good thing, lest we be vulnerable to pinwheeling recursive nostalgia!) We don't really come with a full introspective debugger that can dump the code that implements our reflexes and instincts for our inspection, analysis, optimization, refactoring or reimplementation. For our augmented and automated descendents it may be a different story. Isn't that outcome of natural evolution wondrous enough, without conjuring ghosts in the machines?

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[math-fun] consciousness, N-body, heat
by Warren D Smith 04 Aug '13

04 Aug '13

I'm actually the one who did N-body turing machines: http://rangevoting.org/WarrenSmithPages/homepage/church.pdf While I agree there is a sense in which a small number of point bodies under Newton laws can "simulate a Turning machine running forever" the paper shows this is untrue for more-realistic modified Newton's laws, and also it seems likely that the initial conditions causing this are of measure 0 (although this is not known). Re consciousness (what is that?), I made a paper trying to give a definition of "intelligence", see http://rangevoting.org/WarrenSmithPages/homepage/iq.pdf (too long version) http://rangevoting.org/WarrenSmithPages/homepage/iqshort.pdf (too short version) My definition of intelligence turned out to have been basically the same as one invented by Marcus Hutter a bit earlier (and he published a book); the long paper contrasts me vs Hutter near the end. The long version contains a short section (only a few paragraphs) "Some speculation about what consciousness is" Various posters here have non-explicitly noted that every known life form is a heat engine which depends for survival on energy input and dumping heat to some external cold bath. True. There then seems to be this notion that if there were no energy source & cold bath combination, we'd all die, and "time would not exist" (for us, anyhow). I don't think that is about time, that is about survival. For you time continues even if I die. Would you contend "time does not exist without free oxygen?" Nonsense. Anyhow, consider a tank full of gas, just sitting there in thermal equilibrium very boringly. Does it "experience time?" Well, facts about that gas get "measured" by the "external environment" in particular via gravitons which inherently cannot be shielded. This happens continually, and no thermal equilibrium is ever reached in that respect. As a result the gas "decoheres" preventing it from getting too quantum-weird. This happens continually. The gas therefore "experiences time" in the same time-direction as the rest of the universe, e.g. you.

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Re: [math-fun] generalization of 'trapezoid/trapezium' to 3D ?
by Henry Baker 04 Aug '13

04 Aug '13

'Prismatoid' ! Wow, that's a word that I _never_ learned in school. Thanks! In case anyone cares, the most general fundamental step in the 2D 'Quickhull' algorithm uses a _trapezoid/trapezium_ (Note to Wikipedia: NOT a triangle). Analogously, the most general fundamental step in the 3D 'Quickhull' algorithm uses a 'prismatoid' (NOT a tetrahedron). If you organize your Quickhull algorithm correctly, you don't have to worry about degenerate cases at all; e.g., you don't have to worry about repeated vertices, collinear vertices, etc. The correct 2D 'Quickhull' produces the hull points in cyclic order. This is important to the functioning of 3D 'Quickhull', where the 2 parallel convex polygons of a prismatoid must be 'merged' in the same cyclic order. --- Very cool insight by KQ Brown in 1979: you can compute 2D Voronoi diagrams with 3D convex hulls. Hint: stereographic projection! Brown, K.Q. "Voronoi diagrams from convex hulls". Info. Proc. Letters 9,5 (Dec. 1979), 223-228. This paper screams for a follow-on using quaternions for 3D/4D, but I haven't been able to find it. At 08:06 PM 8/3/2013, Bill Gosper wrote: >Prismatoid! I learned it from my grandmother's old math text. >Its volume formula h*(Area(bottom)+4Area(midsection)+Area(top))/6 >works for just about all the familiar solids (cones, spheres,...) >--rwg. >hgb> > >What is the generalization of a (2D) trapezoid/trapezium to 3D? > >I'm thinking of a convex polyhedron with 2 parallel faces, >whose only other edges run between these 2 parallel faces. > >I.e., we have the convex hull of 2 parallel planar polygons, >not necessarily congruent or similar, and not necessarily >with the same number of sides. > >Is there a _name_ for such an object? (Hint: it _isn't_ >a frustum, but some of the frustum formulae still work.)

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Re: [math-fun] Atrocious function plots
by Bill Gosper 04 Aug '13

04 Aug '13

Fred> A short while back, I described a purported plot of the classical Beta function at http://mathworld.wolfram.com/BetaFunction.html --- or maybe it was the the very similar offering at https://en.wikipedia.org/wiki/Beta_function --- as "passable identikit". Given that identikit reconstructions are notorious for bearing at best only a remote resemblance to their subjects, the verdict may well have been more appropriate than I could known at the time. The sole noteworthy feature of these hilarious travesties proves to be merely a lattice of vertical planar artifacts, resulting from a misguided attempt to plot across lines of discontinuity, and completely obliterating any structural detail of the actual surface. In the course of drafting an account of the recent discussion of binomial coefficients in this list, I took up Gosper's challenge to produce an accurate plot of the associated function of two real variables; and while I was at it knocked out a plot of the (closely related) Beta function, which I have now posted at https://www.dropbox.com/s/rzcojf8wibd0mpw/beta5.png Compare the competing versions; and weep. Fred Lunnon -------------- I hope Fred will soon announce his lovely new pdf that combines enhancements of these plots with compelling mathematics supporting Binomial[-n,-k]:=0 . Here's one I just thought of. For nonnegative integer n In[12]:= Sum[Binomial[n - k, k], {k, 0, u}] Out[12]= Fibonacci[1 + n] for Floor[n/2] ≤ u ≤ n . For u>n, we crash into the 2 o'clock wedge and tack on (-1)^n times an interesting integer depending on how far we transgress. For u=∞, the sum diverges. But we can dispel all this ugliness by merely left-right flipping the sum trajectory: In[12]:= Sum[Binomial[n - k, n-2*k], {k, 0, u}] Out[12]= Fibonacci[1 + n] for all u ≥ Floor[n/2], including ∞, because the 10 o'clock wedge (which Fred calls the Bermuda Triangle) isn't there. Playing with this in Mma, I encountered such a swarm of bugs that I don't know where to begin reporting them. E.g., I could get √5 as a finite sum of integers. The situation is nearly irreparable due Mma's inconsistent and incomplete Sum semantics, and to the depressing fact that pFq notation itself is broken. E.g., Hypergeometric1F1[1-n,-n,n]==Sum[n^(i-1)*(n-i)/i!,{i,0,∞}]==0 unless n is a nonnegative integer, when Hypergeometric1F1[1-n,-n,n]==Sum[n^(i-1)*(n-i)/i!,{i,0,n}]==n^n/n! In other words, there is the insane convention that the hypergeometric series terminates if a random term happens to vanish, regardless of ensuing terms. Thus In[25]:= FunctionExpand[Hypergeometric1F1[1-n,-n,n]] Out[25]= 0 is improvident, because In[28]:= Hypergeometric1F1[1-n,-n,n]/.n->2 Out[28]= 2 Oh well, another reason to switch to matrix products. --rwg

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