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Re: [math-fun] Miscounting in the Cheddar Gorge
by Guy Haworth 13 Nov '17

13 Nov '17

My feeling is that there is more to 'counting' and to the 'number exhibited' than merely its value. There is: a) the implied precision - which our various examples show can be overstated, b) the suggestion that the number could have been different: so, saying a chess-player has 7.5 points in a tournament suggests it might be 7.6 (not true btw) - '7½' is better c) the assurance that the number is correct: as an offshore yacht navigator, I would always right '01' rather than '1' to underline that we are genuinely looking at a 'one' The adjacent subject of units is also of interest. At least one space-mission tripped over this. A colleague suggested 'the unit' should be in the data-type definition at runtime so that, e.g., cm cannot be added to inches. Guy

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[math-fun] RMS paradox
by Keith F. Lynch 12 Nov '17

12 Nov '17

The root mean square (RMS) of a sine wave is always the peak value divided by the square root of 2. The same is true of the sums of multiple sine waves with different frequencies, phases, and amplitudes. But every repeating waveform is equal to the sum of sine waves with different frequencies, phases, and amplitudes. This includes the square wave, i.e. a function which always equals +X or -X, and never takes any other value. But obviously the RMS of that square wave is simply X, not X divided by the square root of 2. Explain. I learn a lot by coming up with such paradoxes then figuring out the solution.

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Re: [math-fun] RMS paradox
by Dan Asimov 12 Nov '17

12 Nov '17

It was kindly pointed out that in my last e-mail I neglected to square the A_j in RMS(f)^2 = lim_[T —> oo] (1/T) Integral_0 ^T (Sum (A_j)^2 sin(B_j x))^2 dx (now fixed). This did not affect anything else. —Dan

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Re: [math-fun] RMS paradox
by Dan Asimov 12 Nov '17

12 Nov '17

Good point, Gene. It's also true that since almost any n real numbers B_1, ..., B_n are linearly independent over the integers, almost any n sine waves of form f(x) = A sin(B x) will be pairwise orthogonal. Then RMS(f)^2 = lim_[T —> oo] (1/T) Integral_0 ^T (Sum A_j sin(B_j x))^2 dx the cross terms average to 0, giving RMS(f) = sqrt(1/2) (|A_1| + ... + |A_n|). It seems to me that in some appropriate sense this probably has a version that is true "almost always" for infinitely many terms as well — possibly requiring the condition that the set of periods is bounded. —Dan ----- Actually, the conjecture for sums of multiple sine waves is false. What is true is that the mean square MS = RMS^2 of the sum equals the sum of the mean squares of the component sine waves, with the proviso that the components are orthogonal. The peak value has no bearing on the matter. Consider sin(x)+sin(2x). Lazy guy that I am, I used a spreadsheet to find the peak value 1.76. This divided by sqrt(2) is 1.24. But the RMS of the sum is sqrt(1/2 + 1/2) = 1. On Sunday, November 12, 2017, 1:56:12 PM PST, Keith F. Lynch <kfl(a)KeithLynch.net> wrote: The root mean square (RMS) of a sine wave is always the peak value divided by the square root of 2. The same is true of the sums of multiple sine waves with different frequencies, phases, and amplitudes. But every repeating waveform is equal to the sum of sine waves with different frequencies, phases, and amplitudes. This includes the square wave, i.e. a function which always equals +X or -X, and never takes any other value. But obviously the RMS of that square wave is simply X, not X divided by the square root of 2. Explain. I learn a lot by coming up with such paradoxes then figuring out the solution. -----

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Re: [math-fun] Miscounting errors
by Henry Baker 12 Nov '17

12 Nov '17

Bingo! The "constant" speed is actually a red herring: it distracts one from the essence of the problem. The only real constraints are: 1. No backward motion, but continuous nondecreasing forward motion. 2. Constant *lap times*, not constant speed. 3. May need constant speed on the last (partial) lap. How about the following expression for swimmers' positions w.r.t. time: posx1(t)=(L/2)*(1-cos(%pi*t*S1/L)) posx2(t)=(L/2)*(1-cos(%pi*t*S2/L)) where L=25' (Hint: expand the forward & return swimming lanes into a *circle*; swimmers now swim constant speed around the circle, but we only worry about their x-coordinates.) Crossings are the zeros of posx1(t)-posx2(t): L*(cos(pi*t*S2/L)-cos(pi*t*S1/L))/2 which simplifies to the product: L*sin(pi*t*(S1-S2)/(2*L))*sin(pi*t*(S2+S1)/(2*L)) We can now count the zeros of the first factor plus the zeros of the second factor. We may have to "fix up" the last/partial lap. At 04:11 AM 11/12/2017, James Propp wrote: >So in fact one doesn't need to know that the swimmers swim at constant >speeds; if one knows how many laps each swimmer completes, and one knows >that the faster swimmer is faster throughout, and one knows that they never >reach a wall simultaneously (or, alternatively, if one agrees that such >occurrences count as double-passes), then the number of times the fast >swimmer passes the slow swimmer is a topological invariant.

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[math-fun] Miscounting in the Cheddar Gorge ...
by Guy Haworth 12 Nov '17

12 Nov '17

The Cheddar Gorge is 10,000,001 years old this year. I know this because last year ... Guy

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Re: [math-fun] Miscounting errors
by Henry Baker 12 Nov '17

12 Nov '17

I recall a question from my high school AP math or math SAT exam circa 1964: Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!) How many times do they pass one another in 30 minutes? This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer. I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test. At 05:06 AM 11/11/2017, James Propp wrote: >Anyone have a favorite puzzle in which miscounting plays a role? > >My favorite is the classic bookworm puzzle (see >https://math.stackexchange.com/questions/1271651/how-is-this-true-bookworm-… >). > >Another example: A man was born in 50 BC and died on the same day in 50 AD. >How old was he when he died? > >I'm especially interested in puzzles that lend themselves to solvers >committing fencepost errors, and off-by-one errors more generally. > >Jim Propp

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Re: [math-fun] Miscounting errors
by Dan Asimov 11 Nov '17

11 Nov '17

Henry's last comment reminds me of a project I was involved in that required optimizing a function over a certain rectangular box in some N-space. All the algorithms I could find (knowing nothing of optimization theory) were for optimizing on all of Euclidean space. No problem. (Or as they used to say, "You're welcome.") Just tessellate all of N-space by copies of that box, with the objective function extended by reflection (repeated if necessary) in the walls of the boxes. Now it's continuous on all of R^N with the same extrema as before. —Dan ----- Consider a plot of distance v. time. This is a long strip 25' wide and 30 minutes long. But you can *unfold* it -- e.g., make it 50' wide and 30 minutes long, which means that the wall doesn't exist and the swimmer keeps going. The elastic bounce has become a straight line. ... >Henry Baker writes: > > I recall a question from my high school AP math or math SAT exam circa 1964: > > > > Two swimmers are swimming laps of a 25' pool, but they swim at different > > (constant) speeds, say S1 & S2. (Assume that they bounce elastically off > > the pool ends!) > > > > How many times do they pass one another in 30 minutes? > > -----

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[math-fun] Tropical summation formulae
by Tom Duff 09 Nov '17

09 Nov '17

I got this from Dan Piponi (@sigfpe on twitter): sum (a+b)^-2 (c+d)^-2 (a+b+c+d)^-2 = 1/24 where the sum is over all integers (a,b,c,d) with 1<=a<=b, 1<=c<=d Paper: Tropical formulae for summation over a part of SL(2, Z) Nikita Kalinin, Mikhail Shkolnikov https://arxiv.org/abs/1711.02089 -- Tom Duff. Branches broken off trees.

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Re: [math-fun] Stupid steam train question
by Mike Beeler 08 Nov '17

08 Nov '17

I asked a train fan friend, who had this answer that seems to explain it. — Mike > Begin forwarded message: > > From: Claudio Topolcic <claudio(a)sunsetharrier.com> > Subject: Re: [math-fun] Stupid steam train question > Date: November 8, 2017 at 10:29:30 PM EST > To: Mike Beeler <mikebeeler2(a)gmail.com> > Cc: Claudio Topolcic <claudio(a)sunsetharrier.com> > > Yes, you guys are figuring it out. This was worked out in the early days of steam locomotives. The first design was more like a tea pot, where the hot gasses went around the boiler. This is very inefficient since most of the hot gasses don't come anywhere near the boiler. So they ran the hot gasses through tubes in the boiler, like your house furnace. But you want to force the cold air through the burning coal to burn faster. I think this is what tall smoke stacks are for in power stations. So you could use a fan, and I think this design was used. But there is a simpler design. The spent steam from the cylinders goes up in the "blast pipe" in the smoke box at the front of the boiler, and that pipe ends somewhat below the bottom of the stack. So the exhaust steam passing out of that pipe and to the stack causes a Venturi effect that sucks the hot gasses through the pipes in the boiler through the smoke box and out the stack. So yes indeed the puffing of the smoke and steam exhausted out the stack are in synch with the cylinders. > > I assume that below, Henry was seeing steam coming out of the cylinders, and this is what he was taking about. I assume he figured that was the spent steam being exhausted from the cylinders. That is not the case. There are various reasons why some steam would be exhausted directly from the cylinders. I don't know all, but an example that I do know is that there is a valve at the bottom of each cylinder that the engineer can open remotely. They open the valves when the cylinders are still cold because some of the steam condenses in the cold cylinders and if they left liquid water in the cylinders then that could break the pistons when they move to their extreme. But that is not the bulk of the exhausted spent steam, that goes up the stack. > > I don't know about steam ships, but I would expect that they do the same thing with the spent steam. But there is a lot of plumbing between the boiler and the funnel, and I don't know what happens to puffs. Also big steam ships have been using turbines for a while, with no puffs. But I don't know. > > Check out the Wikipedia page https://en.wikipedia.org/wiki/Steam_locomotive <https://en.wikipedia.org/wiki/Steam_locomotive>. It has this drawing, check out #12: [ I think math-fun is only text, so I removed the illustration. It shows the outside and a cut-away of a steam loco. Part # 12 is the “blast pipe”. ]

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