September 2018 - mathfun@mailman.xmission.com

[math-fun] music and math
by Cris Moore 11 Sep '18

11 Sep '18

For your enjoyment: https://portal.knme.org/video/majesty-of-music-and-math-egpyyf/ <https://portal.knme.org/video/majesty-of-music-and-math-egpyyf/> - Cris

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Re: [math-fun] Secure internet (OT)
by Henry Baker 11 Sep '18

11 Sep '18

Hi Dan: If you have reasonably current Apple gear, and *if it is configured properly* (the default configuration for Apple gear is "usually" pretty good), and if you use reasonable passwords, then you should be reasonably secure. If you have Apple gear that is so old that it is no longer receiving updates, then get new gear that is, because so many vulnerabilities have been found just in the past year or so, that firmware updates are mandatory. Apple is better than most other vendors at trying to keep their customers "reasonably" secure. An antivirus (AV) program is a government surveillance expert's dream: if you hand it a "definition" file, it is a shopping list of files the AV program is constantly pawing through your system trying to find. So the easiest way for a hacker to find something on your computer is to hack the AV program and ask *it* to look for the stuff. AV programs also tend to get the highest access "privileges", so once they've been compromised, you're toast. Nowadays, the biggest problems aren't the gear or its software, but clicking on "phishing" emails. If you click on what you *thought* was an email from your bank or your internet service provider, you will be in big trouble. This is how the Democratic National Committee was hacked -- it wasn't due to any HW/SW vulnerability. The other problem with individual home networks these days is that it takes only *one* vulnerable device to be hacked for someone to gain a foothold into your home network. After that, the hacked device can bang away incessantly 24x7 until it finds additional vulnerabilities. For example, a "smart" TV, a "smart" speaker, a "smart" home security camera, a "smart" thermostat, a "smart" fitness device, etc., can all be hacked. If any of these devices ask for your wifi password, simply use Nancy Reagan's rule: "just say no". If you have kids, or kids want to use your wifi, consider getting a separate wifi router that is used *only* for guests; they just aren't that expensive anymore (I purchased one for $35 new that works just fine, but it isn't Apple). There is one excellent rule for home routers: disable "plug-and-play" (UPnP), and don't allow any other devices -- e.g., game consoles -- to turn it back on. UPnP has been involved in too many vulnerabilities. Make sure that you use "WPA2" for your wifi passwords, and use relatively long WPA2 passwords. So-called "WEP" is totally broken & can be hacked in mere minutes. WPA3 will be even better when it finally comes out, as it provides for better protection against brute force attacks by someone in your neighborhood, or in a parked van. You might consider changing your DNS provider to Google (8.8.8.8), rather than relying on your own ISP's DNS service. While this gives Google access to every DNS lookup, Google is probably less likely to give you a fraudulent IP address from such a lookup. You can change the DNS server not only in your browser(s), but also in your computer, and in your router. Another rule that may be difficult for many people to agree to: turn off "SMB" file access on your home network. The Microsoft "SMB" protocols have been an infinite source of vulnerabilities. Another rule: completely remove Adobe Flash from every device that you own. It has also been a nightmare of vulnerabilities. Tor is just fine; I use it every day for general browsing. Because of the Tor Project's paranoia, they've done a lot to try to keep someone from the internet from messing up your machine(s). However, I wouldn't use Tor for accessing your bank accounts, because that would force your traffic to go overseas and back, and eliminates the possibility that your bank's own routing safeguards will protect you. There have been a number of articles on using VPN's (Virtual Private Networks). If you choose to go this route(!), make sure that you trust your VPN provider implicitly, because 1) they can see (and potentially manipulate) everything; and 2) you've now made it impossible for many other safeguards to operate -- e.g., firewalls. You should find out your own current IP address (there are a number of web sites that will tell you what it is just now; it can change from day to day or even hour to hour). Then run a *port scan* against your own IP address. If *any port is open*, then you should find out what it is and close it. Unless you are running your own web site or email server, you shouldn't have any ports open onto the main internet. At 07:09 PM 9/9/2018, Dan Asimov wrote: >I've had at least two computer breaches in the past two years, and further >reason to suspect my internet setup is not secure. > >So I wonder if math-fun folks who know about such things can suggest to me how to >set up a (relatively!) secure home internet from scratch. > >I'm a Mac shop, so will need Apple-compatible stuff. > >My concerns include the following: > >1) I'd like the router security to be reasonably unbreakable; > >I'm not sure how to achieve this. > >2) I'd like my own internet use to be private and reasonably un-break-in-able; > >This would involve at least some security software; I've used MacScan in the past >without evident problems; I don't trust Norton or McAfee. > >Maybe it also would require the Tor browser or something like it? I've avoided this >so far because I don't want to find myself on some gov't watchlist just because I use it. > >3) I would also like to be protected from someone in a windowless van down the street, >containing special equipment, eavesdropping on my internet use. > >Humor me for a moment no matter how unlikely you may think this scenario. > >Is there a not-too-expensive way to reasonably protect oneself from that kind of >eavesdropping? (Or at least a way to catch and have prosecuted the perpetrators?) > >All helpful comments will be much appreciated! > >Dan

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[math-fun] Pi: acceleration of the speed of convergence
by françois mendzina essomba2 11 Sep '18

11 Sep '18

Hello, dear all Here are two formulas of Pi, the first is convergence 2 * n + 1 And the second 2 * n + 3 The formulas converge to an odd multiple of Pi depending on the x [0] taken, for example x[0]=3 ; x[m+1]=x[m]+sum(2*(n!)^2*sin(k*x[m])/(k*(n+k)!*(n-k)!),k=1..n); x[m+1]=x[m]+((2*n+1)!)^2/(2^(4*n-1)*(n!)^2)*sum(cos((2*k+1)*x[m]/2)/((2*k+1)^2*(n-k)!*(n+k+1)!),k=0..n); The two formulas are particular cases of thousands of formulas of this form of which I have been able to highlight. Their transformation into linear convergence can allow me to reach 500 decimal places in each calculation. Best Regards... FME...

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[math-fun] Oligomorphic polytopes
by Allan Wechsler 10 Sep '18

10 Sep '18

I've gotten interested in enumerating symmetric convex polytopes in various dimensions, intrigued by the fact that in dimensions as low as 5, we don't yet have a proven-complete census of the uniform polytopes. Every polytope in N dimensions has a single N-cell, and then a graded collection of cells of dimensions from N-1 down to 0 (the vertices). The symmetry group of a polytope divides this collection of cells into transitivity classes, and I propose to focus on polytopes with a small number of classes of cells. Since symmetries never change the dimensionality of a cell, the number of classes is always at least one more than the dimension of the polytope as a whole. I've chosen to apply the term "oligomorphic" ("few-forms") to a polytope that is interesting because its cells fall into a small number of symmetry classes. The only monomorphic polytope is the trivial 0-dimensional isolated vertex. The only bimorphic polytope is a line segment; its two vertices clearly form one symmetry class, while the body of the segment provides the other. We can say it has a "profile" of 1:1 because it has 1 kind of 1-cell and 1 kind of 0-cell. The regular polygons form an infinite call of trimorphic polytopes with profile 1:1:1 (1 kind of 2-cell, 1 kind of 1-cell, 1 kind of 0-cell). There are no other trimorphic polytopes. When we come to consider tetramorphic polytopes, there are the 5 classical Platonic solids in 3 dimensions, with profile 1:1:1:1, but there are also some interesting 2-dimensional examples. A non-square rectangle and a non-square rhombus are examples of trimorphic polygons with signatures 1:2:1 and 1:1:2, respectively. There are similar examples with any even number of edges. (1:1:2 means 1 kind of 2-cell, 1 kind of 1-cell, and 2 kinds of 0-cells.) Up until this point I am pretty sure that I have enumerated all the possibilities. But when we get to pentamorphic polytopes, I quickly start to lose confidence. Here is my list, but except in four dimensions I do not know if it is complete: In four dimensions there are the six "classical" regular polychora, profile 1:1:1:1:1. In three dimensions the only ones I have been able to think of are: the cuboctohedron and the icosidodecahedron, with profile 1:2:1:1 (because they have two kinds of faces); their duals, the rhombic dodecahedron and the rhombic triacontahedron, 1:1:1:2 (because they have two kinds of vertices); and the "isoceles" tetrahedron, 1:1:2:1 (with two kinds of edge). Note that the duality operation reflects the profile, except for the boilerplate leading 1. The dual of an isoceles tetrahedron is a different isoceles tetrahedron. (In combinatorial algebraic topology, they often add on a single "-1-cell" with which every vertex is presumed to be incident, just to complete this kind of duality symmetry.) In two dimensions there is an interesting little menagerie of pentamorphic polygons. The simplest is the isoceles triangle, with two kinds of edges and two kinds of vertices, profile 1:2:2. This can be generalized into a family of polygons with 3k sides, with a vertex-edge pattern (AaAbBb)^k. (I hope it is clear what my silly notation means here). There is another family, of which the (non-rhombic) parallelogram is the simplest exemplar. Here, bigger examples have 2k sides, with a vertex-edge pattern (AaBb)^k. The profiles 1:1:3 and 1:3:1 are clearly impossible, since a single class of edge cannot be incident with three classes of vertex, and vice versa. It would absolutely not surprise me if someone were to come up with pentamorphic examples in 2 and 3 dimensions that were not covered by my sketched catalog above. Regarding hexamorphic polytopes, I have no confidence whatsoever that I have managed to exhaust all the possibilities, except that we know there are only three hexamorphic 5-dimensional polytopes. The isoceles trapezoids and their dual kites are 2-dimensional examples; right prisms and their dual right bipyramids are 3-dimensional ones; I cannot produce a single 4-dimensional one. Can anyone help complete the enumeration of hexamorphic types? The snub cube is octomorphic, with profile 1:3:3:1.

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[math-fun] Secure internet (OT)
by Dan Asimov 09 Sep '18

09 Sep '18

I've had at least two computer breaches in the past two years, and further reason to suspect my internet setup is not secure. So I wonder if math-fun folks who know about such things can suggest to me how to set up a (relatively!) secure home internet from scratch. I'm a Mac shop, so will need Apple-compatible stuff. My concerns include the following: 1) I'd like the router security to be reasonably unbreakable; I'm not sure how to achieve this. 2) I'd like my own internet use to be private and reasonably un-break-in-able; This would involve at least some security software; I've used MacScan in the past without evident problems; I don't trust Norton or McAfee. Maybe it also would require the Tor browser or something like it? I've avoided this so far because I don't want to find myself on some gov't watchlist just because I use it. 3) I would also like to be protected from someone in a windowless van down the street, containing special equipment, eavesdropping on my internet use. Humor me for a moment no matter how unlikely you may think this scenario. Is there a not-too-expensive way to reasonably protect oneself from that kind of eavesdropping? (Or at least a way to catch and have prosecuted the perpetrators?) All helpful comments will be much appreciated! Dan

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[math-fun] Oddity
by David Wilson 08 Sep '18

08 Sep '18

a = 12|[32876712]^k b = [32876712]^k|33 => a^2 + b^2 = a|b Examples 12^2 + 33^2 = 1233 1232876712^2 + 3287671233^2 = 12328767123287671233 &c

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[math-fun] Sum of digits in base -2 and base 4
by Keith F. Lynch 03 Sep '18

03 Sep '18

I just noticed that the sum of the digits of each of the numbers 0 through 13 was the same in base -2 and in base 4. Why should that be? There's no obvious pattern linking them. And the rule is broken with 14. 0 0 0 0 0 1 1 1 1 1 2 110 2 2 2 3 111 3 3 3 4 100 1 10 1 5 101 2 11 2 6 11010 3 12 3 7 11011 4 13 4 8 11000 2 20 2 9 11001 3 21 3 10 11110 4 22 4 11 11111 5 23 5 12 11100 3 30 3 13 11101 4 31 4 14 10010 2 32 5 Those digit sums are A027615 and A053737. A276328 and A276332 also start with those same 13 terms. Through 100, the base 4 sum of digits agrees with the base -2 sum of digits 81 times. Through 1000, they agree 560 times. Through 10000, they agree 4891 times. When they disagree, the latter always exceeds the former, and does so by a small multiple of 3 (at least through 10000, in which range the maximum difference is 15).

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[math-fun] Nested Radical
by françois mendzina essomba2 03 Sep '18

03 Sep '18

a nested radical for the odd alpha q=(q^alpha+k[1]*q^(p[1])-k[1]*(q^(alpha*p[1])+k[2]*q^(p[2])-k[2]*(q^(alpha*p[2])+k[3]*q^(p[3])-k[3]*(q^(alpha*p[3])+k[4]*q^(p[4])-k[4]*(q^(alpha*p[4])+k[5]*q^(p[5])-k[5]*(q^(alpha*p[5])+k[6]*q^(p[6])-k[6]*(1)^(1/alpha))^(1/alpha))^(1/alpha))^(1/alpha))^(1/alpha))^(1/alpha))^(1/alpha); q, p[i] and q[i] the real numbers for alpha = -1, we have a continuous fraction

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[math-fun] Identities for Pi (04)
by françois mendzina essomba2 03 Sep '18

03 Sep '18

to finish Pi/12=sum((1/(2*n+1))*(-sqrt(3)*sin(x)+sqrt(3*(sin(x))^2+1))^(2*n+1)*sin((2*n+1)*x),n=0..infinity); Pi/12=sum(((-1)^n/(2*n+1))*(sqrt(3)*sinh(x)-sqrt(3*(sinh(x))^2-1))^(2*n+1)*sinh((2*n+1)*x),n=0..infinity); (1/12)*Pi = sum((-1)^n/(2*n+1)*(-sqrt(3)*cosh(x)+(3*cosh(x)^2+1)^(1/2))^(2*n+1)*cosh((2*n+1)*x),n = 0 .. infinity); (1/6)*Pi = sum((-1)^n/(2*n+1)*(-(1/sqrt(3))*cosh(x)+((1/3)*cosh(x)^2+1)^(1/2))^(2*n+1)*cosh((2*n+1)*x),n = 0 .. infinity); With a general formula arctan(t) = f(t,x)

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[math-fun] integers that are either square or thrice a square, where do such beasts hide?
by Wouter Meeussen 02 Sep '18

02 Sep '18

do they occur anywhere naturally? in combinatorics or in number theory? I bumped into them by accident looking at bracelet-stuff. Wouter.

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