March 2019 - mathfun@mailman.xmission.com

[math-fun] Pi s ~= nanocentury
by Bill Gosper 22 Mar '19

22 Mar '19

On 2019-03-22 01:15, Georg Dr. Fischer wrote: > It may be an old joke for you, but I did not know that pi seconds are > almost a nanocentury. - cf. < > https://en.wikipedia.org/wiki/List_of_humorous_units_of_measurement>. I.e., In[120]:= UnitConvert[Quantity[Pi, "Seconds"], "Nanocenturies"] Out[120]= Quantity[(1250000π)/3944619, "Nanocenturies"] In[121]:= N@% Out[121]= Quantity[0.995531080945268, "Nanocenturies"] Slightly more accurate: In[122]:= UnitConvert[Quantity[10^7 Pi, "Seconds"], "Years"] Out[122]= Quantity[(625π)/1971, "Years"] In[123]:= N@% Out[123]= Quantity[0.996192495430553, "Years"] But Mathematiica's sense of humor does not extend to UnitConvert[Quantity[Pi, "Decamegaseconds"], "Years"]: Quantity::unkunit: Unable to interpret unit specification Decamegaseconds. And In[127]:= UnitConvert[Quantity[Pi, "Decimegaseconds"], "Years"] Out[127]= Quantity[π/36500000, "Years"] is just weird. We'll have to settle for In[126]:= UnitConvert[Quantity[10 Pi, "Megaseconds"], "Years"] Out[126]= Quantity[(625 π)/1971, "Years"] —rwg

1 0

Re: [math-fun] How does this function manage to oscillate on the unit interval?
by Dan Asimov 22 Mar '19

22 Mar '19

I haven't been able to access the link provided, so would appreciate seeing the function definition in ascii. Thanks, Dan

2 1

[math-fun] How does this function manage to oscillate on the unit interval?
by Bill Gosper 22 Mar '19

22 Mar '19

https://www.wolframcloud.com/objects/d4fa673b-0423-42fb-b7b5-0f70a7c14c6e Sidarth Ghoshal has exactly determined the mean value and amplitude of an O(10^-6) ripple in a function whose definition appears nonoscillatory. It graphs like the error in a Remez minimax approximation, even unto the ripples Chebychummying at the endpoints. But unboundedly. Worse, Sid says there are O(10^-13) ripples on the O(10^-6) ripples. Whether or not he discovers their formula, I consider Sid's present formula a breakthrough. —rwg

1 1

[math-fun] Distribution of radii
by Dan Asimov 22 Mar '19

22 Mar '19

Suppose that inside a unit circle C (the edge the unit disk D) lie three smaller disks tangent to each other and to C. Suppose their radii are 0 < R, S, T < 1.* Call these radii "stage 1". The exterior of these disks now fall into four curvilinear triangles, inside each of which there is a unique largest disk, which will be tangent to all three sides. The radii of the three new disks are "stage 2". Iterating, we get an infinite tree of disks, each of maximal size in some curvilinear triangle lying between three previous disks. The nth stage will add 3^n new (disks and) radii. Now let X(n) denote the multiset of all radii 1, R, S, T, ..., through the nth stage, for a total of |X(n)| = 1 + 3 + ... + 3^n = (3^(n+1) - 1)/2 radii in X(n). Define the usual cumulative distribution function (cdf) for the data of X(n) via F_n : [0, 1] —> [0, 1] via F_n(x) = |{r in X(n) | r <= x}| / |X(n)|. ***> Question: What is the limiting distribution of F_n as n —> oo and how does it depend on R, S, T ??? —Dan ————— * From Frederick Soddy's poem "The Kiss Precise" we know that R, S, T can be any numbers satisfying (1 - (1/R + 1/S + 1/T)^2 = 2(1 + 1/R^2 + 1/S^2 + 1/T^2)

2 2

Re: [math-fun] Glass moon
by Keith F. Lynch 21 Mar '19

21 Mar '19

Tom Duff <td(a)pixar.com> wrote: > Keith F. Lynch <kfl(a)keithlynch.net> wrote: >> it's sudden changes in the refractive index that cause reflections. > I'd never thought about this before, but a quick Google suggests > that it's not true. Graded-index media have gradual Fresnel > reflection. Here's a paper that purports to explain the details: > https://pdfs.semanticscholar.org/9f8f/e29d292732c8a7fb1aa82f0ce5df94822077.… Thanks. My intuition was that as the distance over which the refractive index changed significantly became large compared to the wavelength of light, the reflectivity, while perhaps never reaching exactly zero, went down faster than linearly. And if I understand the formulas right, my intuition is correct. One abrupt transition of one part in a thousand has a reflection coefficient of 2.5E-7, but ten transitions of one part in ten thousand each have reflection coefficients of 2.5E-9 for a total reflection coefficient of 2.5E-8. So with more than a trillion transitions, each of less than one part in a trillion, the total reflections should be unmeasurably small. I can make no sense of the paper's distinction between electric and magnetic polarization, given that light is always both, with the two always at right angles to each other. Digression: The reason the first transatlantic telegraph cable (1858) worked so poorly is because its impedance varied widely, resulting in lots of internal reflections that smeared out the signals. That's also why no transatlantic *telephone* cable was attempted until 1956. Of course in those cases the relevant wavelengths are much longer than those of light. Eugene Salamin <gene_salamin(a)yahoo.com> wrote: > The Luneburg lens is a ball of radius R whose index of refraction > varies as ?2-(r/R)2.? A parallel beam of light is exactly focused, > in the approximation of geometric optics, to a point on the surface > opposite the direction of entry. Thanks, but I'm interested in the focal point being at the surface of *Earth*, not at the surface of the *moon*. An f110 lens. That requires a very low index of refraction, probably unrealistic for any known form of glass. > It looks like math-fun mangled the formula.? The index of > refraction is sqrt(2-(r/R)^2). The digest option replaces all non-ASCII characters with question marks, which makes some posts completely incomprehensible. Others can be decoded with effort. For instance I deduced that a recent mention of "?wave" in another thread, the ? was supposed to be the micro symbol. It's all Greek to me. Marc LeBrun <mlb(a)well.com> wrote: > I don't know, but it seems like a fun question for XKCD... Unfortunately, he stopped doing the what-ifs months ago. Henry Baker <hbaker1(a)pipeline.com> wrote: > Even if the Moon were made of 100% silicon glass, it would still > probably not have a uniform refractive index due to compression > effects of gravity at different depths. On the other hand, these > effects might be just what is needed to get proper focus. I was making several unrealistic assumptions. For instance that the glass moon is perfectly spherical and perfectly rigid, i.e. no tidal bulge or equatorial bulge. > I seem to recall that the human eye lens has different refractive > indices at different depths, but I don't know the profile. > You also have a very significant problem of different points of > focus for different colors/wavelenths. Designing a "Glass moon" > which corrected for this would be an interesting engineering > exercise. I designed it by fiat -- I chose a hypothetical kind of glass in which the refractive index is the same at all wavelengths. Also, it had better be *extremely* transparent, since otherwise the small amount of light absorption in the interior would soon melt the whole thing down. It's impressive how little chromatic aberration the human eye has. The only time I notice any is when looking at a purple Christmas light from a distance when it's dark out. At best it's either a red point embedded in a blue haze or vice versa. > BTW, even if your "Glass moon" weren't in proper focus, you could > still utilize it for its large light-gathering power, and use a > *corrective lens* on the camera at the Earth's surface to bring > things into better focus. This corrective lens might also be able > to have different corrections for different colors. I don't think that would work. But even without proper focus, the light gathering would certainly be useful. Astronomers can learn a lot about an object without resolving it. Similarly with using gravitational lenses to gather gravitational waves. Too bad the sun's focal length as a gravitational lens is so long. It would have been convenient had it been about the distance to the Earth.

2 1

[math-fun] Glass moon
by Keith F. Lynch 20 Mar '19

20 Mar '19

Decades ago, it was generally believed that no solid could be transparent to light for more than a few meters. Today it's known that non-transparency in transparent is due to a small handful of causes, including crystal boundaries and microscopic inclusions of opaque material. Today we've learned that extremely pure amorphous glass fibers can be transparent for tens or even hundreds of kilometers. That raises the question of what the moon would look like if it were perfectly transparent. Assume it's the same size it is now, in the same orbit, and is perfectly spherical. Assume it's immune to micrometeoroids, tidal stress, and radiation damage. Spheres don't make very good lenses, but that's only because they typically have a constant refractive index. If the refractive index varies smoothly with the distance from the center, it could have a perfectly sharp focus. How would the refractive index have to vary with depth for the moon to have a focal length equal to the distance to Earth's surface? (Is it practical to make (small) lenses on this principle today? If not, why not?) If the problem is underspecified, I'll further specify that the refractive index at the moon's surface should be 1, the same as the surrounding vacuum. That will prevent the surface from being reflective, since it's sudden changes in the refractive index that cause reflections. This would revolutionize astronomy, since it would turn the Earth- moon system into a giant telescope. You'd need to take very brief photos due to the speed of Earth's rotation. Either than or take photos (or naked eye observations) from a very fast jet that compensates for both Earth's and the moon's motions. If your first thought is that is that the moon would be a burning glass, leaving a trail of destruction during every solar eclipse, think again. In a week I will explain why that wouldn't happen unless someone beats me to it, which I hope they will.

5 5

[math-fun] Sugar + water paradox:
by Bill Gosper 20 Mar '19

20 Mar '19

Dump a couple of Tbsp of sugar into a quart or two jar of tepid water. Don't stir. 𝜇wave a couple minutes, hot but not boiling. Let it sit. Experimental claim: The total dissolved sugar is not an increasing function of time, even though a couple of Tbsp is << enough to saturate even cold water. —rwg

5 5

Re: [math-fun] Sugar + water paradox:
by Dan Asimov 20 Mar '19

20 Mar '19

Why wouldn't that be expected, once the jar is removed from the µwave? The heating has caused more sugar to dissolve than would normally dissolve at room temperature. So once it's taken out, as the temperature declines, less and less sugar can remain in solution (until equilibrium is reached). —Dan Bill Gosper wrote: ----- But for several minutes, at least, the amount of dissolved sugar actually declines. -----

2 1

[math-fun] Thm: You can never fully complete a lane change, even on a bicycle.
by Bill Gosper 20 Mar '19

20 Mar '19

https://en.wikipedia.org/wiki/Tractrix http://mathworld.wolfram.com/Tractrix.html http://mathworld.wolfram.com/PursuitCurve.html Unfortunately, https://www.its.caltech.edu/~mamikon/calculus.html appears to be broken. At any rate, to get your rear wheel traveling exactly straight in a new direction, you need to undercorrect your steering until the rear wheel is on the target line, and then continuously adjust the steering to keep it there. Of course, these adjustments will rapidly "incede" atomic diameters. —rwg

2 1

Re: [math-fun] Thm: You can never fully complete a lane change, even on a bicycle.
by Dan Asimov 20 Mar '19

20 Mar '19

Tractrices can be fun! * Rotate the graph of y = tractrixfunction(x) for x > 0 about the x-axis, and you get a smooth maximal surface among all such surfaces of constant Gaussian curvature = -1 that are embedded in 3-space. * Imagine, for each c >= 0, a rigid rod r(c) in the first quadrant of length = 1 with its lower endpoint lying on the x-axis such that a) r(0) is the segment from (0,0) to (0,1), and b) r(x_1) and r(x_2) are disjoint if x_1 ≠ x_2. Then among all arrangement of these unit segments according to a) and b): What is the least angle that the segment r(x) makes with the positive y-axis? That turns out to be the angle made by the tangent line to the tractrix. —Dan ----- https://en.wikipedia.org/wiki/Tractrix http://mathworld.wolfram.com/Tractrix.html http://mathworld.wolfram.com/PursuitCurve.html ... ... -----

1 0