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[math-fun] Can a polyhedron
by R. William Gosper 31 Dec '02

31 Dec '02

with its vertices on a sphere and its edge lengths all equal have more than 92 faces? I'm looking at one (a doenneacontahedron?) with 12 regular pentagons (pentaga?), 60 equilateral triangles, and 20 non-regular (but equilateral) hexagons. There are 90 vertices, all tetravalent, so maybe there is a C_90 fullerane (as opposed to fullerene). I also made a pentacontahedral candidate for C_48 fullerane with 30 squares, 8 equilateral triangles, and twelve rhombi. --rwg

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[math-fun] Re: Integers with many 0s in base 2 and base 3 (fwd)
by Helger Lipmaa 28 Dec '02

28 Dec '02

---------- Forwarded message ---------- Date: Sat, 28 Dec 2002 12:06:13 -0500 From: Robert Harley <harley(a)argote.ch> To: NMBRTHRY(a)LISTSERV.NODAK.EDU Subject: Re: Integers with many 0s in base 2 and base 3 Andreas Weingartner wrote: >Prove or disprove: >There exists an epsilon>0, such that no natural number has the >property that in base 2 as well as in base 3, at most (epsilon)*100% >of the digits are nonzero. Heuristically one might expect (except possibly for some small examples) such an eps for bases a and b at the solution of: (2*eps-1)*log(a)*log(b) = (eps*log(eps)+(1-eps)*log(1-eps))*(log(a)+log(b)) For bases 2 and 3, this is about 0.104939... I ran a search up to 2^64 for small examples: 6: 0.666667 9: 0.5 18: 0.4 162: 0.375 261: 0.333333 4376: 0.307692 19712: 0.3 32805: 0.25 65610: 0.235294 131220: 0.222222 4785156: 0.217391 9570312: 0.208333 272629962: 0.206897 1208614932: 0.2 2542645806624: 0.190476 154206526918656: 0.1875 2348694485729280: 0.181818 9341451062288388: 0.176471 18049789376104448: 0.171429 451521135633235968: 0.169492 Regards, Rob. .-. .-. / \ .-. .-. / \ / \ / \ .-. _ .-. / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / `-' `-' \ / \ / \ \ / `-' `-' \ / `-' `-'

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[math-fun] symmetric venn diagrams/Geometry Junkyard
by Shel Kaphan 27 Dec '02

27 Dec '02

I just happened across this and thought some of you might like it. Includes some nice pictures and an interesting bit about symmetric gray codes. http://www.theory.csc.uvic.ca/~cos/venn/VennSymmEJC.html Found while browsing "The Geometry Junkyard" http://www.ics.uci.edu/~eppstein/junkyard/topic.html which has some other fun and odd stuff.

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[math-fun] Proof of Uncertainty Principle
by David H Bailey 27 Dec '02

27 Dec '02

I personally find the proof of the uncertainty principle quite amusing. Here it is, if any of you are interested. DHB Let us consider a real, continuously differentiable, $L^2$ function $f(t)$, which further satisfies $f(t) \sqrt{t} \rightarrow 0$ as $|t| \rightarrow \infty$. For convenience, we will also assume here that $f(-t) = f(t)$, as this insures that the Fourier transform $\hat{f}(x)$ of $f(t)$ is purely real, although in general this condition is not necessary. Define \begin{eqnarray*} E(f) &=& \int_{-\infty}^\infty f^2(t) \, dt \\ V(f) &=& \int_{-\infty}^\infty t^2 f^2(t) \, dt \\ \hat{f}(x) &=& \int_{-\infty}^\infty f(t) e^{-i t x} \, dt \\ Q(f) &=& \frac{V(f)}{E(f)} \cdot \frac{V(\hat{f})}{E(\hat{f})} \end{eqnarray*} \vspace{1ex}\noindent {\bf Theorem} With the above assumptions and definitions, $Q(f) \geq 1/4$, with equality if and only if $f(t) = a e^{-(b t)^2/2}$ for real constants $a$ and $b$. \vspace{1ex}\noindent {\bf Proof.} By applying Schwarz' inequality to the functions $t f(t)$ and $f'(t)$, we can write \begin{eqnarray*} \left| \int_{-\infty}^\infty t f(t) f'(t) \, dt \right|^2 &\leq& \left[\int_{-\infty}^\infty t^2 f^2(t) \, dt \right] \left[\int_{-\infty}^\infty (f'(t))^2 dt \right] \end{eqnarray*} \noindent Furthermore, \begin{eqnarray*} \int_{-\infty}^\infty t f(t) f'(t) \, dt &=& \frac{1}{2} \int_{-\infty}^\infty t \frac{d f^2(t)}{dt} \, dt \; = \; -\frac{1}{2} E(f) \end{eqnarray*} \noindent by applying integration by parts. Let $g(t) = f'(t)$. By noting that $\hat{g}(x) = - i x \hat{f}(x)$, and by applying Parseval's identity \cite[pg. 65]{papoulis1977} to $f(t)$ and $g(t)$, we obtain, respectively, \begin{eqnarray*} \int_{-\infty}^\infty |f(t)|^2 \, dt &=& \frac{1}{2 \pi} \int_{-\infty}^\infty |\hat{f}(x)|^2 \, dx \\ \int_{-\infty}^\infty |f'(t)|^2 \, dt &=& \frac{1}{2 \pi} \int_{-\infty}^\infty x^2 |\hat{f}(x)|^2 \, dx \end{eqnarray*} \noindent Combining these results, we obtain our desired inequality. Recall that equality in the Schwarz inequality occurs only when the two functions ($t f(t)$ and $f'(t)$ in this case) are linear scalings of each other, i.e. $f'(t) = c t f(t)$ for some $c$. By solving this elementary differential equation, we conclude that the minimum value $1/4$ is achieved if and only if $f(t) = a e^{-(b t)^2/2}$ for constants $a$ and $b$. By the way, it is worth noting that $Q(f)$ is unaffected by a linear scaling of either the function argument $t$ or the function value $f(t)$.

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RE: [math-fun] Items from the most recent AMS journals
by Eugene Salamin 27 Dec '02

27 Dec '02

Yes, "Uncertainty Principle" sounds good, as long as it's made clear that it's not Heisenberg's Uncertainty Principle. I didn't want to quibble over exactly what constant appears in the RHS, but a precise statement is that if f(x) and F(p) are normalized so that int(|f(x)|^2 dx) = int(|F(p)|^2 dp) = 1, if |f(x)|^2 and |F(p)|^2 are interpreted as probability densities in their respective spaces, and if DeltaX and DeltaP are the respective standard deviations, then DeltaX * DeltaP >= hbar/2. A Gaussian function achieves the minimum uncertainty product. The ordinary uncertainty principle relates position with spatial frequency (or wavelength) or time with frequency. The Heisenberg uncertainty principle relates position with momentum or time with energy. The identification of spatial frequency with momentum or of frequency with energy is possible only within quantum mechanics. Gene Salamin __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

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[math-fun] Items from the most recent AMS journals
by Eugene Salamin 27 Dec '02

27 Dec '02

The January 2003 issue of the Bulletin of the AMS has a review by Steven Krantz of "A new kind of science". This has been thoroughly discussed within math-fun, so I will make only two remarks. Footnote 2 of the review claims: "Wolfram formerly had a contract with Addison-Wesley to publish his book. One sticking point in that relationship was that he demanded that any reviewer sign a nondisclosure agreement and promise not to study math or physics for the ensuing ten years." Footnote 7 repeats an error that I have seen one time too many. If f(x) and F(p) are Fourier transforms of each other, then int(x^2 f(x)^2) * int(p^2 F(p)^2) >= 1 is called a statement of the Heisenberg uncertainty principle (HUP). Does this theorem (with suitable hypotheses appended) have a name? It is not the HUP. Calling this theorem the HUP would falsely imply, for example, that if I am given the exact electric and magnetic fields in an electromagnetic pulse, that I could not calculate precisely both its centroid and total momentum. But this is not so. The energy and momentum densities are given by the field components; there is no need to determine frequencies or wavelengths. This example should make it clear that the HUP is strictly a principle of quantum theory; it does not appear in any manner in classical physics. For the case of an electromagnetic pulse, in order for the Fourier transform theorem above to be applicable, I need to be concerned with individual photons; but then this is quantum theory. The January 2003 issue of the Notices of the AMS has a review of "Indra's Pearls". Severe ongoing constraints on my time have prevented me from reading this book, but it is my intention to do so eventually, and report on it to math-fun. Apparently this book was 20 years in the making. It seems incredible with these two books having been released at about the same time, that any competent mathematician would choose to devote the time to reading Wolfram's tome. Gene Salamin __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

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[math-fun] more solid angles
by R. William Gosper 26 Dec '02

26 Dec '02

In my version of Macsyma's Geofuncs package is now (c1) solid_angle(a,b,c) 2 (cos(c) + cos(b) + cos(a) + 1) (d1) acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) where a, b, and c are the ordinary angles at the trihedral vertex. It seems too nice to be new, but an hour or so of (low bandwidth) Googling failed to turn it up. Here are some playings with it. A sector with wedge angle v will have solid angle v/(2 pi) 4 pi = 2 v. But we can split it perpendicular to its straight edge to make two trihedrals with angles pi/2, pi/2, and v: (c2) 2*solid_angle(%pi/2,%pi/2,v) (d2) 2 v We can dissect a regular n-gonal pyramid of height h and base circumradius r into n congruent tetrahedra with trihedral components v, v, and 2 asin(sin(v) sin(pi/n)), where v = atan(r/h), i.e., half the apex angle. This gives, after much simplification, two expressions for the solid angle of the apex: 2 %pi 2 %pi 2 %pi 2 cos (---) (2 sin (---) cos(v) + cos(-----)) n n n (d3) n acos(--------------------------------------------- 2 %pi 2 2 %pi sin (---) cos (v) + cos (---) n n 2 %pi - cos(-----)) n n = acos((- 1) n - 1 /===\ 2 %pi 2 | | %pi %pi k 2 2 cot (---) n | | (cos(v) + cot(---) cot(-----)) n | | n n k = 1 (1 - ------------------------------------------------------)) 2 2 %pi n (cos (v) + cot (---)) n Letting n -> oo, the solid angle of a cone is 2 pi (1 - cos v), where, again, tan v = (base radius)/height. Exercise: what is the black/white (pentagons/hexagons) area ratio of a soccer ball (spherical truncated icosahedron)? Thanks to Ed Hudson for the loan of the laptop on which this was derived, amid numerous calls to the formerly great technical support organization of COMPAQ ("taken from you by HP").

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Re: [math-fun] more solid angles
by Henry Baker 25 Dec '02

25 Dec '02

This looks suspiciously like a quaternion product in drag... (Quaternions are very nice for spherical geometry.) At 11:22 AM 12/24/02 -0800, R. William Gosper wrote: >In my version of Macsyma's Geofuncs package is now >(c1) solid_angle(a,b,c) > 2 > (cos(c) + cos(b) + cos(a) + 1) >(d1) acos(-------------------------------------- - 1) > (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) > >where a, b, and c are the ordinary angles at the trihedral vertex. >It seems too nice to be new, but an hour or so of (low bandwidth) >Googling failed to turn it up. Here are some playings with it. > >A sector with wedge angle v will have solid angle v/(2 pi) 4 pi = 2 v. >But we can split it perpendicular to its straight edge to make two >trihedrals with angles pi/2, pi/2, and v: > >(c2) 2*solid_angle(%pi/2,%pi/2,v) > >(d2) 2 v > >We can dissect a regular n-gonal pyramid of height h and base >circumradius r into n congruent tetrahedra with trihedral components >v, v, and 2 asin(sin(v) sin(pi/n)), where v = atan(r/h), i.e., >half the apex angle. This gives, after much simplification, two >expressions for the solid angle of the apex: > > 2 %pi 2 %pi 2 %pi > 2 cos (---) (2 sin (---) cos(v) + cos(-----)) > n n n >(d3) n acos(--------------------------------------------- > 2 %pi 2 2 %pi > sin (---) cos (v) + cos (---) > n n > > 2 %pi > - cos(-----)) > n > > n > = acos((- 1) > n - 1 > /===\ > 2 %pi 2 | | %pi %pi k 2 > 2 cot (---) n | | (cos(v) + cot(---) cot(-----)) > n | | n n > k = 1 > (1 - ------------------------------------------------------)) > 2 2 %pi n > (cos (v) + cot (---)) > n > >Letting n -> oo, the solid angle of a cone is 2 pi (1 - cos v), where, >again, tan v = (base radius)/height. > >Exercise: what is the black/white (pentagons/hexagons) area ratio >of a soccer ball (spherical truncated icosahedron)? > >Thanks to Ed Hudson for the loan of the laptop on which this was >derived, amid numerous calls to the formerly great technical support >organization of COMPAQ ("taken from you by HP").

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[math-fun] Re: Your discrete log question
by Mike Stay 24 Dec '02

24 Dec '02

Sorry, you're right! I forgot to say that the p's are primes. OK, let me try again with less cumbersome notation (although my question was mostly about how to make the notation better in the first place...) Pick p,q primes > some composite n. Make sure (p-1) and (q-1) are smooth so that one can use the Pohlig-Hellman algorithm to calculate discrete logs. Also choose them so they both have g as a generator; that is, the powers of the element g give every number mod p and mod q except 0. Let lg_p, p prime, be discrete log base g over p; that is, find x given y,p in the equation y=g^x mod p. Then x=lg_p(y). Let a(x,p) = g^(lg_p(n) - lg_p(x)) mod p Then find x != 1,n such that a(x,p) == a(x,q) The idea is that only factors of n satisfy both sides simultaneously. (I haven't proven that.) It's a strange problem, I think, because it's easy (i.e. polynomial time) to solve either side of the problem, but seems to be very hard to solve both at once. One problem is the notation: usually one writes "equivalent to" with the triple-bar equals and says that both sides of the equation are in the same equivalence class. Here I'm saying that the bits match when one uses the customary "mod" function in your favorite language. a is self-inverse for fixed p: x == a(a(x,p),p) One can rewrite the above problem as: Find x != 1,n such that x == a(a(x,p),q). -- Mike Stay staym(a)clear.net.nz

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[math-fun] Most efficient cone puzzle
by asimovd＠aol.com 24 Dec '02

24 Dec '02

For an open cone (ice cream cone shape) of fixed area, find the proportions (ratio of its height to the radius of its base) giving it the maximum volume. --Dan

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