November 2017 - mathfun@mailman.xmission.com

[math-fun] Draft of my November 2017 blog post
by James Propp 13 Nov '17

13 Nov '17

Hi, I started writing a new draft titled "Impaled on a Fencepost" and would love to get people's feedback. I plan on publishing it on the 17th. Some questions and misgivings that remain in my mind: Am I missing some fun or important examples of fencepost error? Should I cite linguistics literature on the ambiguity of the word "between" (assuming it exists)? I feel I'm forgetting some famous movie or TV show featuring the fail-to-count-yourself error. Is "third, fifths, etc." terminology really called "ordinal" terminology for intervals? (I just made that up.) Was it really invented during the Renaissance? (I just made that up too.) Am I correct in what I write about legal matters? I forgot to remind everyone that if you think your comment is one that might elicit a response from me (e.g., "You're right, my wording was confusing; is this better?"), or if you want your contribution to be acknowledged, then please sign your response; otherwise I won't know who you are. Please leave your feedback here: https://mathenchant.wordpress.com?p=1998&shareadraft=5a0912d2a2e4a Thanks, Jim Propp

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Re: [math-fun] Miscounting in the Cheddar Gorge ...
by Keith F. Lynch 13 Nov '17

13 Nov '17

Guy Haworth <g.haworth(a)reading.ac.uk> wrote: > The Cheddar Gorge is 10,000,001 years old this year. > I know this because last year ... That's a precision fail, not a counting fail. It's perfectly valid to add one year to the age of something each year, no matter how old it is. Except that the age of very old things is seldom known to the nearest year. And gorges, unlike, say, individual dinosaurs, don't have precise birthdays anyway. (That joke is more commonly told about dinosaurs, with a museum guard answering that a particular specimen is is 100,000,003 years old, or whatever.) Also, for very high precision or for extremely long distances or times, the age depends on the trajectory of the clock. For instance how old is Mount Everest? I haven't been able to find that number anywhere. Wikipedia says it's 8848 meters tall and that the Himalayas are growing by 5 millimeters per year. That would imply an age of 1769600 years. If, at the instant the mountain started its climb from sea level someone had placed a perfect clock where it would ride up the peak and another perfect clock where it would remain at sea level, today the two clocks would disagree by 24 seconds as to the age of the mountain, due to general relativity. You'd get far more variance on how long ago the Big Bang was, since there are so many choices of "reasonable" trajectories for a clock that formed that very instant and conveniently lands in from of you just when you want to know the age of the universe. Precision fails are very common. For instance: Naturally, this means that it's ridiculously hot on the planet's surface, with temperatures reaching 2000 degrees Celsius (3632 Fahrenheit), ... https://www.outerplaces.com/science/item/16744-astronomers-planet-titanium-…

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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 13 Nov '17

13 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 13 Nov '17

13 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 13 Nov '17

13 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 13 Nov '17

13 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 12 Nov '17

12 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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