- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] Partial fraction decomposition for the reciprocal of an infinite product
by Dan Asimov 22 Dec '17

22 Dec '17

I was recently reminded that the reciprocal of the famous infinite product sin(pi*z) / pi = z * Product_{1 <= n < oo} (1 - z^2 / n^2) can be expressed as pi / sin(pi*z) = Sum_{-oo < n < oo} (-1)^n / (z - n) (where the right side denotes the limit of the symmetrical finite sums). Cool, right? So I'm curious if there are any general rules for passing from an infinite product to the partial fraction decomposition of its reciprocal, or vice versa. Do people know other nice examples independent of the one above? —Dan

1 0

[math-fun] Why are so many states in two time zones?
by Dan Asimov 22 Dec '17

22 Dec '17

According to www.answers.com: ----- There are thirteen states in the USA that are divided into two time zones. Oregon and Idaho are split between the Mountain and Pacific time zones. Nebraska, Kansas, Texas, North and South Dakota are divided between Central and Mountain time zones. Florida, Michigan, Indiana, Kentucky, and Tennessee are split between Eastern and Central time zones. Also, Alaska is split between the Alaska time zone and the Hawaii-Aleutian time zone. ----- If you look at a map of it, like theone at http://www.icsanmartinoinpensilis.it/LearningObjects/The%20USA/Timezonesmap…, many states seem split between two time zones for no apparent reason. In many cases, the split causes a time zone to have either a more-easternmost or a more-westernmost piece than it probably would have had without the split. Does anyone know why so many states seem to have split time zones without apparent reason? —Dan

3 2

[math-fun] Phi-like numbers?
by Henry Baker 21 Dec '17

21 Dec '17

Is there a name for irrational numbers whose continued fractions eventually become all one's ?

2 1

[math-fun] Fwd: Cutting a pie to share equally among up to n people
by James Propp 21 Dec '17

21 Dec '17

This email sent to me by Warren Smith a few days ago may be of interest (especially the discussion of the inequalities that govern what Warren calls "variant 1"). ---------- Forwarded message ---------- From: *Warren D Smith* <warren.wds(a)gmail.com> Date: Friday, December 15, 2017 Subject: Cutting a pie to share equally among up to n people To: jamespropp(a)gmail.com On 12/15/17, James Propp <jamespropp(a)gmail.com> wrote: > Suppose we want to make cuts in a circular pie (the usual kind of cut, from > the center to a point on the perimeter) in such a way that, for every k > between 1 and n, the pieces can be divided equally among k people, each of > whom gets the same amount of pie. > > How many cuts are needed? --it is better to think of one the perimeter of the circle not its interior. Call your version variant 1. Variants of this problem: 2. suppose the k people all have different preferences (Amy likes pepperoni 3 times as much as she likes onions, Tom hates both but likes mushrooms...) and we want the divvying out of the pie to be equal in the views of all people simultaneously. And what if 3. we merely want it to be "envy free" i.e. each person regards own part as at least as good as any rival's? These variants especially #3, probably would be important if they could be solved; worth publishing. Variant 1: trivial upper bound is n!. Somewhat better upper bound is LCM(1,2,3,...,n) which grows like exp(n) roughly, while n! is superexponential. But neither bound is best possible: for n=3 4=3+1 cuts work, 6 are not needed. Ok, a much better upper bound is 1+2+...+n=(n+1)*n/2. Still better is (n-1)*n/2+1. Still better is n + SUM(k=2..n-1) [(k-1) if no multiple of k exists that is <=n, otherwise 0] which is equivalent to 1 + SUM(k=ceiling(n/2)..n) (k-1) which equals (n+2)(3n-4)/8+1 is n=even and 3(n-1)(n+1)/8+1 if n=odd. But this still is not best possible since with n=4 cutting into 3 pieces of measure 1/4, and 3 of measure 1/12 each, works. For variant 2, n=2, Amy "moves her knives" in such a way that knife #1 is at angle theta, while knife #2 provides a fair cut into 2 (in Amy's view) pieces. Bob calls out "now" whenever Amy's knives are also fair in his judgement. Then they cut. A solution must exist by some kind of intermediate value theorem, i.e. Bob will not remain silent. For variant 2, n=3, Amy's "knife space" is now a periodic curve in a 3-dimensional rather than 2-dimensional space of knife-angular positions. Hence it in general will not intersect Bob's curve. Hence variant 2 is unsolvable for any n>=3. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

2 1

[math-fun] Synchronizing 2 communicating processes
by Henry Baker 21 Dec '17

21 Dec '17

I'll have to beg your forgiveness about the sketchiness of this description. I'm trying to operate at a very high level of generality, so the details must necessarily be somewhat sketchy. We have 2 processes/devices A,B who want to "communicate" with one another, but must first "synchronize". Let's assume that the processes/devices are at a fixed constant distance d from one another, so there are no Doppler/relativity issues. If one process A "sends"/"emits" a sinusoidal signal, the other process B may "see"/"feel" the signal, and if there is some sort of *tuned circuit* (e.g., RLC "tank" circuit with a center frequency f and a Q of q) in the receiving process, then the receiver may see an amplification of the received signal as a result of the incoming energy being absorbed by the receiver *in phase*. Indeed, if the energy of the incoming sinusoidal signal is high enough, and it falls within the width of the frequency window of the RLC circuit, the receiver could actually explode! Think of a bell tuned to the sound of an incoming tone, and if the Q of the bell is high enough relative to the incoming energy and if the frequencies are matched well enough, the receiving bell could vibrate strongly enough to destroy itself. [I.e., the "R" in the RLC circuit loses less energy per cycle than the incoming energy per cycle, after which point the energy in the receiving RLC circuit "runs away".] Although the 2 processes/devices may have once been calibrated w.r.t. one another, it may have been a long time, so the two devices may not agree on either the absolute time or even the *rate* of time ("drift"), so for the two processes to communicate, they may first have to synchronize their clocks. By lowering the Q of the circuit, the receiver's sensitivity is reduced, but its frequency width is increased, so the receiver can "hear" incoming signals with greater difference between the sender's notion of frequency and the receiver's notion of frequency. So one strategy for the listener is to start with a low Q, and attempt to detect *any* energy increase due to an incoming signal. Let's assume that there is some mechanism to adjust the receiver's phase, so that the receiver can determine if its receiving frequency is too low or too high, and can therefore adjust the frequency in the correct direction for improved reception. The receiver then adjusts its receiving frequency by a small delta, and simultaneously increases the Q by a small amount. The improved center frequency and the improved Q should enable the receiver to more easily detect the incoming signal. If this process is *iterated*, then the receiver should be able to hone in on the correct frequency and reduce the *width* of the RLC frequency response. Of course, as noted above, the receiver needs to be careful to keep the Q low enough to avoid catastrophe. Question #1: What is the optimum search strategy for the receiver to hone in on the correct frequency of the sender as quickly as possible? Does the number of bits in the frequency precision grow linearly? Is there a strategy which allows quadratic increase in # of bits of frequency precision? So far, we have assumed that the sender unilaterally sends, and the receiver unilaterally receives. But in any real system, the sender and receiver are *symmetrical*: the receiver's RLC circuit also radiates its own signal which is "received" by the sender's RLC circuit. So both A & B are sending and receiving at the same time. So now, what strategy should be used by both A & B *simultaneously* to *converge* upon an extremely narrow frequency that they can both "agree" upon? Since only A & B want to communicate with one another, they don't really care about the *absolute* frequency, but only upon reducing the *difference* between their frequencies to the smallest amount possible, in terms of the number of bits of precision in the frequency. Question #2: What is the simultaneous & *symmetrical* optimal frequency "search" strategy? For the moment, we assume *classical* (non-quantum) physics, so we don't have to worry about discrete energy chunks.

3 4

[math-fun] why only the number 2
by François Mendzina Essomba 21 Dec '17

21 Dec '17

> why only the number 2. consider the following relation where x and y are functions of a given variable: x^n+y^n=1; if we drift, we get the differential equation: nx'x^(n-1)+ny'y^(n-1)=0 --- x'x^(n-1)+y'y^(n-1)=0 This equation has for solution: x'=-y^(n-1) et y'=x^(n-1) for n=2; x^2+y^2=1 x'=-y et y'=x we deduce: x=cos(x) et y=sin(x) for n=4; x^4+y^4=1 x'=-y^3 et y'=x^3 I defined by : x=qos(x) et y=qin(x) qos(x)=sum(a_(n)*(-1)^(n)*x^(4*n)/(4*n)!,n=0..infinity) qin(x)=sum(b_(n)*(-1)^(n)*x^(4*n+1)/(4*n+1)!,n=0..infinity) 1) why all mathematics seems to rest on the index n = 2; 2) I could not simplify the coefficients a_ (n) and b_ (n) despite all my efforts ... Can we put together a simplified algorithm? 3) for n = 0, what does the x and y functions return?

1 0

[math-fun] In[569]:= vertexSolidAngles["RhombicTriacontahedron"] // FullSimplify //Tally // tim
by Bill Gosper 21 Dec '17

21 Dec '17

During evaluation of In[569]:= 8.60579 secs, 2 components Out[569]= {{π, 12}, {3π/5, 20}} I.e., 12 vertices have solid angle π. True or false: Four RTs with disjoint interiors can intersect at a point and fill a neighborhood of it. Can five with disjoint interiors intersect at one point? Six?? 6×⅗π < 4π. Question 2: How many vertices, edges, and faces has the Stella Octangula? --rwg

3 4

[math-fun] Question suggest by solid angle issues
by Dan Asimov 20 Dec '17

20 Dec '17

Let S be the unit sphere in 3-space centered at (1, 0, 0). Now transform the x- and y-coordinates like the squaring operation (x + yi) |——> (x^2-y^2) + 2xyi, i.e., define a function f: R^3 —> R^3 via f(x, y, z) = (x^2-y^2, 2xy, z) and apply this to the surface S to get a new surface: X = {(x^2-y^2, 2xy, z) | (x-1)^2 + y^2 + z^2 = 1}. The origin 0 belongs to X. What is the solid angle SA(X; 0) subtended by the surface X at the point 0 ??? ((( Meaning: Let U be any open neighborhood of 0 in X. Let A(U) denote the area on the unit sphere of the set of all outward normal unit vectors to X at a point of U. Then we define the solid angle SA(X; 0) = inf {A(U) | U is an open neighborhood of 0 in X}. ))) —Dan

1 0

[math-fun] Cutting a pie ...
by Guy Haworth 20 Dec '17

20 Dec '17

The problem is usually expressed as one of dividing a cake (or pie) and distributing it so that no-one is in a position to complain about the size of their slice ... There's this ... https://www.scientificamerican.com/article/the-mathematics-of-cake-cutting/ ... and then there's this by Ian Stewart https://www.amazon.co.uk/How-Cut-Cake-Mathematical-Conundrums/dp/0199205906 However, the best solution I know of (which Ian Stewart confirmed he had not heard of) came from the 13-year-old son of a colleague of mine .... Everyone gets a knife and makes a cut in a circular cake which is standing on a turntable ... with the aim of making equal-size slices. The cake is then spun, and everyone gets the slice opposite them. [ In other words, the randomness takes the decision about who gets what away from the hungry mathematicians ] Guy

11 43

[math-fun] Fwd: Cutting a pie ...
by James Propp 20 Dec '17

20 Dec '17

More from Warren: ---------- Forwarded message ---------- From: Warren D Smith <warren.wds(a)gmail.com> Date: Wed, Dec 20, 2017 at 12:44 AM Subject: Re: [math-fun] Cutting a pie ... To: James Propp <jamespropp(a)gmail.com> I can't bring myself to believe F(n) obeys any linear upper bound; I think it must have a superlinear lower bound. Indeed F(2*n) >= 2*F(n)+n-1 holds for n=1,2,3,4 and perhaps 5, and it seems plausible F(2*n)>=2*F(n)+K*n holds for K>=0, with indeed K lower bounded by some positive constant on average... which would imply a n*log(n) style lower bound. So my conjecture is F(n) asymptotically is both superlinear and subquadratic, but I have no idea where it lies within that range. Very annoying! -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

1 0