July 2013 - mathfun@mailman.xmission.com

Re: [math-fun] math-fun Digest, Vol 125, Issue 17
by Warren D Smith 17 Jul '13

17 Jul '13

>> Question: what about a combined photovoltaic plus heat-engine/thermal >> based solar system? Couldn't that achieve greater efficiency than >> either alone, and why is it not heavily used? > > That's an interesting question. A photovoltaic only uses about 10% of the > photons because > they have to be in the right frequency band. The others are mostly just > absorbed, since > the panels look black or dark blue. But maybe the absorbtion is at the > electrodes. A > mesh electrode and a reflective backing might allow most of the photons that > are not > absorbed by the photovoltaic to be reflected to a boiler. --to me the most obvious scheme is a "hotbox" (insulated box with black interior, glass front, gets hot inside due to greenhouse effect) with photovoltaic cells inside. The PV cells would have to be ones which can work when hot or not, and you can get hot water via pipes passing thru the box. Then you generate both electricity and hot water instead of just one. But should be greater efficiency and lower cost. Why is this not done? Another scheme which works best at high altitudes (direct sun, not scattered light) involves cylindrical shaped concentrator solar cells at the focus of a parabolic trough mirror; then pass fluid thru the pipe. It is expensive to fabricate cylinder-shaped solar cells. And tracking parabolic mirrors, while not exactly high-tech, are also not trivial to build using supplies from hardware store. I think this second scheme has been built (?).

1 0

[math-fun] stupid solar mirror tricks, and solar energy
by Warren D Smith 17 Jul '13

17 Jul '13

The TV program "mythbusters" did a show about Archimedes / Clarke ship-roasting legend. The participants aimed their mirrors with the aid of a net 1-2 meters in front of them (by shining thru the net, and seeing the lit-up part of the net, they could aim decently) However, I believe their conclusion was, it was not an effective weapon. They could char the ship a bit, and annoy the sailors with the light and heat, at best. For solar energy, mirror-based systems are best used in non-cloudy high altitude locations while photovoltaics are better in cloudy or lower-altitude locations since the latter can use scattered light. Question: what about a combined photovoltaic plus heat-engine/thermal based solar system? Couldn't that achieve greater efficiency than either alone, and why is it not heavily used?

4 3

[math-fun] von Neumann universe packing puzzle
by Adam P. Goucher 17 Jul '13

17 Jul '13

What's the minimum area of a bounding rectangle capable of accommodating a graphical representation of the nth iteration of the von Neumann universe? For n <= 3, the problem is reasonably trivial. For n = 5, the solution would be enormous (since there are 65536 sets). For n = 4, I've managed to get it down to 75 * 79, hoping for 75 * 75: http://cp4space.wordpress.com/2013/07/17/von-neumann-universe/ It should be evident that the dimensions are necessarily congruent to 3 (modulo 4). Sincerely, Adam P. Goucher

1 0

[math-fun] Turning tetrahedra inside out
by Henry Baker 17 Jul '13

17 Jul '13

I've been playing with a mathematical model of physical tetrahedrons: The four vertices are joined by edges which are Hookean springs with non-zero rest lengths. The vertices are 'pins'; i.e., the edges can move freely into different angles -- i.e., the vertices exert no forces other than axial forces along the edges. Each of the springs can have a different rest length and a different spring constant (non-regular tetrahedra). Define 'failure' by a tetrahedron turning itself inside out. Interestingly, Buckminster Fuller played with this problem of inverting tetrahedrons; he called the process of turning inside-out 'dimpling'. Also, modern computer graphics uses such models for 'deformable' surfaces -- e.g., human skin. However, computer graphics goes to great lengths to make sure that the tetrahedra never turn themselves inside out (never 'dimple'). So the major question is: how much energy is required to invert/'fail' the tetrahedron. Notice that you have 4 choices over which vertex to 'push through'. I want to find the minimum 'failure energy' in closed form. Does it even matter which vertex is pushed through? --- I assume that this problem has been investigated by people other than architects & computer graphics folks.

2 4

Re: [math-fun] one crazy idea
by Henry Baker 17 Jul '13

17 Jul '13

Google Arthur C Clarke's "A Slight Case of Sunstroke". I think that Clarke's idea has been analyzed in the scientific literature, as well. I see several problems: 1. It's difficult enough to point a mirror (or a laser) a few hundred feet and keep it on target. An analogous problem is keeping high-powered binoculars focused on a target -- it's basically impossible without a tripod, at least. 2. How can you tell if _your_ beam is hitting the target? 3. Atmospheric problems. Modern telescopes use lasers to test the atmosphere & correct using 'adaptic optics'. That's a bit expensive for every single person to have with their mirror. 4. Various refractive & diffractive effects. Since these mirrors aren't lasers, they have a multitude of wavelengths, so your 'focus' 'point' would be a pretty big volume. Volume grows cubically, while surface area grows quadratically, so your spaceship might be able to 'hide' in plain sight. 5. Cosine atmosphere problem. The further you are from the closest point on the Earth's surface, the more atmosphere you have to go through. This would remove most of the UV, at least. 6. The laser ignition project that the U.S. has spent billions on already may be shut down due to lack of results. And they're only trying to hit a target right here in California! At 02:07 AM 7/17/2013, Simon Plouffe wrote: >Hello, > > I was thinking at how we could procuce energy in a simple >way, like having solar ovens with mirrors and such devices. >This lead me to Archimedes and the way he used to 'toast' >roman warships from the Syracuse citadel, apparently that >idea was somewhat validated once, they found a way to >toast a pile of plywood resembling a roman ship with plates >of metals used at the time. > >Anyhow which lead me to think : what about today ?, suppose >we would hypothetically having the same kind of threat >but from space : an alien warship threatening to zap >humans from existence. > >What about doing the same thing ? Billion of people using >mirrors directly aimed at (more or less precisely) at this >alien ship : what would happen ? > >More precisely, what would be the effect in energy having >let's say 100 miilion mirrors aimed at one point in space >toward one of those things, would it do the same thing ? >Is this feasible, we would have one shot at it. > >This may sound naive, well is it ? > >Can someone calculate what output of energy this >represent ? > >Best regards, >Simon PLOUFFFE

2 1

[math-fun] Symbolic null space solutions
by Scott Fenton 15 Jul '13

15 Jul '13

Hi all, I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ] I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this? Thanks, Scott

3 5

[math-fun] multiply tangent lines to polyomials
by rcs＠xmission.com 15 Jul '13

15 Jul '13

a post from Erich Friedman <efriedma(a)stetson.edu> --rich --- greetings funsters, consider the problem of which monic 4th degree polynomials x^4 + a x^3 + b x^2 + c x + d have the property that some line is tangent to the graph at 2 different places. there are some hard ways to think about this, equating slopes with average slopes, but the solution is obvious once you realize that these are the polynomials with 2 inflection points, so take 2 derivatives and look for 2 real roots. turns out the condition is 8b < 3a^2. the fact that c and d don't matter should be clear because of horizontal and vertical shifts. all that is background for my real question, one i am having trouble answering. which monic 6th degree polynomials x^6 + A x^5 + B x^4 + C x^3 + D x^2 + Ex + F have a line that is tangent to the graph at 3 different places? if such a line y = U x + V exists, then the polynomial x^6 + A x^5 + B x^4 + C x^3 + D x^2 + Ex + F - Ux - V is tangent to the x-axis at 3 points, which means it is of the form (x - r1)^2 * (x - r2)^2 * (x - r3)^2 for some distinct r1, r2, r3, but which A, B, C, and D accomplish that? i can't seem to solve the resulting equations. is there an easier way of doing this? erich friedman

5 8

[math-fun] erich friedman polynomial problem
by Warren D Smith 14 Jul '13

14 Jul '13

>a post from Erich Friedman <efriedma(a)stetson.edu> --rich greetings funsters, consider the problem of which monic 4th degree polynomials x^4 + a x^3 + b x^2 + c x + d have the property that some line is tangent to the graph at 2 different places. there are some hard ways to think about this, equating slopes with average slopes, but the solution is obvious once you realize that these are the polynomials with 2 inflection points, so take 2 derivatives and look for 2 real roots. turns out the condition is 8b < 3a^2. the fact that c and d don't matter should be clear because of horizontal and vertical shifts. --I did not understand that. My way of solving it (which maybe Erich Friedman disagrees with and is wrong, but I'll say it anyhow?): Let the quartic polynomial be Q(x). Let the line be Ax+B. Then Q(x)-Ax-B is another monic quartic, which is tangent to the x-axis at 2 places. It would seem to me, these are precisely the quartics ((x-J)*(x-K))^2. And that is the solution of the problem: there must exist A and B such that Q(x)-Ax-B has this squared-quadratic form. >all that is background for my real question, one i am having trouble answering. which monic 6th degree polynomials x^6 + A x^5 + B x^4 + C x^3 + D x^2 + Ex + F have a line that is tangent to the graph at 3 different places? if such a line y = U x + V exists, then the polynomial x^6 + A x^5 + B x^4 + C x^3 + D x^2 + Ex + F - Ux - V is tangent to the x-axis at 3 points, which means it is of the form (x - r1)^2 * (x - r2)^2 * (x - r3)^2 for some distinct r1, r2, r3, but which A, B, C, and D accomplish that? i can't seem to solve the resulting equations. is there an easier way of doing this? erich friedman --similarly, let the sextic be S(x) and let the line be Ax+B. The solution to the problem is to demand there exist A and B such that S(x)-Ax-B adopts the squared-cubic form ((x-J)*(x-K)*(x-L))^2. Now I don't know whether you consider my solutions "solutions" but let me say this. I have given a parameterized description (parameters being J,K,L,A,B) of every monic sextic, and it yields only the monic sextics, that do this. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

1 0

Re: [math-fun] Symbolic null space solutions
by Bill Gosper 14 Jul '13

14 Jul '13

ES>Work in n-dimensional space with basis e1, ... , en. You are given n-1 vectors, the rows of your matrix, by hypothesis linearly independent. The 1-dimensional null space is generated by a vector orthogonal to these. Use the n-dimensional generalization of the cross product. Append to your matrix the row [e1, ... , en]. Now you have an nxn matrix. Calculate its determinant, and express it as A1 e1 + A2 e2 + ... + An en. Then [A1, ... , An] is a basis for the null space. -- Gene ________________________________ From: Scott Fenton <sctfen(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Saturday, July 13, 2013 4:48 PM Subject: [math-fun] Symbolic null space solutions Hi all, I'm not sure if Google's and my linear algebra education are failing me, but I've got a matrix problem that I have no idea how to solve. The basics go like this. I've got an n by (n-1) matrix Q = [ q[1,1] q[1,2] .... q[1,n] ] [ . ] [ . ] [ . ] [ q[n-1,1] .... q[n-1,n] ] I can guarantee that all (n-1)x(n-1) matrix minors are non-singular. I'm looking for the general solution of this matrix for a proof I'm working on. Any ideas how to go about doing this? Thanks, Scott -------------------- Isn't this just the vector of the determinants of the minors, times an alternating sign? In[183]:= {{a, b, c, d}, {e, f, g, h}, {i, j, k, l}}; In[184]:= Table[Drop[Transpose[%], {z}], {z, 4}] Out[184]= {{{b, f, j}, {c, g, k}, {d, h, l}}, {{a, e, i}, {c, g, k}, {d, h, l}}, {{a, e, i}, {b, f, j}, {d, h, l}}, {{a, e, i}, {b, f, j}, {c, g, k}}} In[185]:= Det /@ % Out[185]= {-d g j + c h j + d f k - b h k - c f l + b g l, -d g i + c h i + d e k - a h k - c e l + a g l, -d f i + b h i + d e j - a h j - b e l + a f l, -c f i + b g i + c e j - a g j - b e k + a f k} In[186]:= FullSimplify[#*Denominator[#[[1, 1]]] &@NullSpace[%%%]] Out[186]= {{-d g j + c h j + d f k - b h k - c f l + b g l, d g i - c h i - d e k + a h k + c e l - a g l, -d f i + b h i + d e j - a h j - b e l + a f l, c f i - b g i - c e j + a g j + b e k - a f k}} --rwg

3 2

[math-fun] Reviews of the HoTT book ?
by Guy Haworth 14 Jul '13

14 Jul '13

Would like to be pointed at reactions to the Adv.Inst.Princeton HoTT book which can be appreciated by the relative layman, especially reactions which put it in the context of Human, Automated and Assisted Theorem Proving. Thanks in advance, Guy

1 0