October 5

I'm pretty new to Wikipedia so I'm sorry if this is the wrong place for this. On the talk page for Bundle map, I previously asked a question about the definition of the term. The article doesn't cite any sources or provide references for additional information, and no one has answered the question on that page. I thought I would repost it here to see if I can get an answer, or perhaps someone could point me in the direction of resources that would answer the question for me. Let me know if it is inappropriate to ask the same question in two places, and where the question should have been asked initially.

I was following along with the lecture series "Lectures on the Geometric Anatomy of Theoretical Physics", given by Dr. Frederic P. Schuller and freely available on YouTube. In Lecture 6, when he introduces bundle maps (about 50 minutes in), he does not include the requirement that they are continuous. I also don't see that condition included here http://mathworld.wolfram.com/BundleMap.html, but the article includes ″continuous″ in the definition. I have been unable to find other information on the topic in my brief attempt at research (it doesn't help that the article lists no references), and was wondering if someone can clarify for me. Did Dr. Schuller simply forget to say the word continuous? Or is there some reasoning for intentionally leaving it out? Are both definitions used, and it simply depends on the author? --AlfonsoAnonymous (talk) 00:37, 5 October 2017 (UTC)[reply]

In topology, "map" is generally understood to mean continuous function. In the rare case where one might allow for discontinuous functions, then one will generally specifically use "function", and possibly even say something like "not necessarily continuous" for emphasis. --Deacon Vorbis (talk) 00:56, 5 October 2017 (UTC)[reply]

More generally, the meaning of a word like 'map' might include whatever assumptions are needed in the current context. So, for example, if you're talking about smooth manifolds then a map is assumed differentiable as well. This kind of thing would probably be caught and fixed in a textbook or research paper, but it's common in the slightly less formal setting of a lecture hall. Btw, I've seen Dr. Schuller series and it's a good introduction to much of the math you need for theoretical physics. --RDBury (talk) 16:37, 5 October 2017 (UTC)[reply]

@Deacon Vorbis: I figured it would be something like that, I just wanted to double check. He is usually pretty careful about explicitly stating assumptions, so that threw me off.

@RDBury: His lectures are great! Do you happen to know of any resources (textbook, online notes, etc) that would go along with the lecture series? My university unfortunately doesn't have anyone in theoretical physics, or who specializes in these ares of math.AlfonsoAnonymous (talk) 02:53, 6 October 2017 (UTC)[reply]

I don't don't know of any resources specific to the lecture series, but you might try The Road to Reality for a more detailed look at the subject. A warning though; it's a very dense book and finishing it is quite a feat in itself, not to imply that I've actually done it myself. --RDBury (talk) 12:43, 6 October 2017 (UTC)[reply]

I there may be an influence here from the idioms of category theory. In the category of topological spaces, continuous maps are the only morphisms. So other functions may not be considered "maps" at all. It's not necessarily sloppy — it's perfectly precise, as long as everyone understands that that's how the terminology is being used. --Trovatore (talk) 20:11, 7 October 2017 (UTC)[reply]

Quick challenge, take out your pocket calculator and compute i^i. Can your calculator do it? What's the answer and what's your model of calculator. 110.22.20.252 (talk) 05:36, 5 October 2017 (UTC)[reply]

My Casio from the 80's recognizes it as a series and lets me scroll through results: 0.20788, 3.8820e-04, 7.2495e-07, 1.3538e-09, 2.5281e-12, 4.7212e-15, 8.8165e-18, 1.6464e-20, 3.0746e-23, 5.7417e-26, 1.0722e-28, ... 209.149.113.5 (talk) 12:33, 5 October 2017 (UTC)[reply]

My trusty

TI-83 Plus gives me only 0.2078795764. shoy (reactions) 18:13, 5 October 2017 (UTC)[reply

]

My calculator can't do it. Staecker (talk) 18:33, 5 October 2017 (UTC)[reply]

The

HP-48 emulator (Droid48) on my phone: (.207879576351,0). -- ToE 19:52, 5 October 2017 (UTC)[reply

]

I don't have a calculator but I sometimes enter a sum into Google. Putting i^i into Google I get 0.20787957635, so it does a bit more than I expect - but I can't see how to enter complex numbers into the calculator that comes up.

Dmcq (talk) 14:02, 6 October 2017 (UTC)[reply

]

0.20787957635076193[1]

0.2078795763507619[2]

0.207879576350761908546955619834978770033877841631769608075... (using the principal branch of the logarithm for complex exponentiation)[3]

--Guy Macon (talk) 18:04, 7 October 2017 (UTC)[reply]

I had trouble with three parts of this word problem:

An outdoor decorative pond in the

ft, that water is pumped in at a rate of π ft³⁄min, and that the tank is initially empty. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant

of proportionality is k = 0.01.

1a. The rate of change dV/dt of the volume of the water at time t is a net rate. Use this net rate to determine a differential equation for the height h of the water at time t. The volume of the water is V = πRh² - 1/3πh³, where R = 10. Express the area of the surface of the water A = πr² in terms of h.

dh/dt =

1b. Solve the differential equation in part (a).

h(t) =

1c. What is h(2000/3)? Round your answer to three decimal places.

For problem 1c, I tried 9.9 ft, 660 ft, and 1,036.726 ft, which are all verified as incorrect. Any help would be appreciated. 147.126.10.148 (talk) 07:26, 5 October 2017 (UTC)[reply]

Sanity check - the volume of the hemisperical tank is 2000π / 3 ft³. So if we neglect evaporation, the tank will be full at time t=2000 / 3. So without evaporation h(2000 / 3) = 10 ft. When we include evaporation h(t) will be reduced, so h(2000 / 3) < 10 ft.

What equations did you get in part 1a for dV/dt and dh/dt ? Gandalf61 (talk) 10:26, 5 October 2017 (UTC)[reply]

That was my question—what is the value of h(2000/3) rounded to three decimal places? The equation dh/dt already accounts for evaporation. I don't have equations for either dV/dt or dh/dt—I didn't know where to start. 147.126.10.152 (talk) 16:19, 5 October 2017 (UTC)[reply]

_{(presumably ${\displaystyle k}$ has units of ft/min, and has been chosen to make ${\displaystyle \mathrm {flow} _{\mathrm {in} }-k\pi R^{2}=0}$, so that the integral is easy) --catslash (talk) 16:23, 5 October 2017 (UTC)}[reply]

The derivative of volume of water is:

${\displaystyle {\frac {dV}{dt}}=\pi -k\pi r^{2}=\pi -k\pi (2Rh-h^{2})}$.

On other hand it is

${\displaystyle {\frac {dV}{dt}}=\pi r^{2}{\frac {dh}{dt}}=\pi (2Rh-h^{2}){\frac {dh}{dt}}}$.

So, from the above two equations

${\displaystyle {\frac {dh}{dt}}={\frac {1}{2Rh-h^{2}}}-k}$.

You can solve it in the general case but you can also note that ${\displaystyle R=1/{\sqrt {k}}}$. In this case the equation is greatly simplified. So, omitting details the solution is

${\displaystyle h={\sqrt {{\frac {k^{2}t^{2}}{4}}+{\sqrt {k}}t}}-{\frac {kt}{2}}={\frac {2}{\sqrt {k}}}{\frac {1}{\displaystyle {\sqrt {1+{\frac {4}{k^{3/2}}}{\frac {1}{t}}}}+1}}}$.

So, ${\displaystyle \lim _{t\to \infty }h=1/{\sqrt {k}}=R}$ and then at ${\displaystyle t=2R^{3}/3=2/(3k^{3/2})}$ we have

${\displaystyle h={\frac {1}{3{\sqrt {k}}}}({\sqrt {7}}-1)={\frac {10}{3}}({\sqrt {7}}-1)\approx 5.49}$

Ruslik_Zero 19:17, 5 October 2017 (UTC)[reply]

Thank you so much for the help. 147.126.10.148 (talk) 00:23, 11 October 2017 (UTC)[reply]

I am covering both recursive functions and complex numbers in the same week. For complex numbers, every project has to do with AC electricity. Apparently, that is the only example of real-world complex number use that any textbook writer knows about. I've been googling (drowning in thousands of pages about AC electricity examples) trying to find a real-world example that uses complex numbers and recursion so I can cover both in one project. I've found nothing. Does anyone here know of any real-world examples where complex numbers and recursion are used together? 209.149.113.5 (talk) 12:25, 5 October 2017 (UTC)[reply]

Fractals are inherently recursive, and many fractal images use complex numbers. Computing the Mandelbrot set can easily be done using recursion. --Stephan Schulz (talk) 12:34, 5 October 2017 (UTC)[reply]

• See
  Linear difference equation#Solution of homogeneous case and particularly the subsection "Converting complex solution to trigonometric form". The solution of a difference equation of at least the second order involves complex numbers, which can be converted into trigonometric expressions, if some of the roots of the characteristic equation are complex. Loraof (talk) 02:00, 6 October 2017 (UTC)[reply
  ]

I agree with Loraof. There's also the explicit solution of degree 3 equations on reals, which involves complex numbers, and may have been the historically first application for them. – b_jonas 22:52, 8 October 2017 (UTC)[reply]

How many unique Eulerian (all edges visited) cycles are there on the Octahedron? (subject to rotation and mirror imaging). (vertices N, ABCD(at equator in that order clockwise)?Naraht (talk) 16:34, 5 October 2017 (UTC)[reply]

That sounds pretty rough to try to do by hand, but the number of possible paths to check is plenty small enough for a computer to brute force the solution quickly. --Deacon Vorbis (talk) 17:56, 5 October 2017 (UTC)[reply]

The problem is getting the computer to tell that two cycles are the same if one starts at position 6 of the other, and the octahedron has to be rotated and flipped to have one line up with the other. I'm trying to use the number and positions of "triangles" in the cycle and the number and positions of "over the tops" where the two vertices opposite of each other on the octahedron are in the cycle with only one vertex apart in the cycle.Naraht (talk) 20:03, 5 October 2017 (UTC)[reply]

@Naraht:, @Deacon Vorbis: There are 100 unique cycles. There's a cool trick for matching Eulerian cycles: since the cycles go through every edge, you can uniquely specify a starting edge and direction and then just compare the vertex sequence. Here's the full code. C0617470r (talk) 06:37, 14 October 2017 (UTC)[reply]

Just divide the number you get by 48 (the size of the octahedron's symmetry group), or by 8 if you start at a given vertex (the size of a vertex's stabilizer subgroup). --Deacon Vorbis (talk) 20:35, 5 October 2017 (UTC)[reply]

48 is because you can start at any of the 6, move to any of four and then left and right are simply flipped, right?Naraht (talk) 21:40, 5 October 2017 (UTC)[reply]

It works like that here, but in general it's a little more complicated (the size of the symmetry group of the n-

orthoplex

is ${\displaystyle 2^{n}n!}$). To explain the above a little more, the symmetry group of the octahedron

orbits of this action, which (since the action is free) is just the number of paths divided by the size of the group (48 in this case). --Deacon Vorbis (talk) 22:15, 5 October 2017 (UTC)[reply

]

Or maybe a little more simply, every unique path will be counted 48 times, once for each element in the symmetry group. And it will always be 48 because any two different symmetries applied to a particular path will always result in a different (but symmetric) path out. --Deacon Vorbis (talk) 22:27, 5 October 2017 (UTC)[reply]

OK, that makes sense.Naraht (talk) 22:44, 5 October 2017 (UTC)[reply]

I'm guessing yoiur vertices are N ABCD and S for north and south. You didn't name the 6th one. Due to mirroring, I presume it doesn't matter if ABCD is clockwise from above or from below, just that they are in sequence. ? -- SGBailey (talk) 06:39, 6 October 2017 (UTC)[reply]

Yes, missed the S. As pointed out here, mirroring doesn't matter. Just trying to designate N and S as opposites, A and C as opposites and B and D as opposites.Naraht (talk) 15:17, 6 October 2017 (UTC)[reply]

@Naraht:, By the way, I was bored and decided to refresh my Haskell a little, so I wrote a short program that brute forced it, and came up with 186 different paths. It's still running for the 4-orthoplex. I'm not especially confident that it's going to finish in a reasonable amount of time though. --Deacon Vorbis (talk) 01:33, 7 October 2017 (UTC)[reply]

@Deacon Vorbis:, so 186*48 and then divide by the symmetry group, or did you limit it in some way? (All paths start NA and then you divide by 2)???Naraht (talk) 01:57, 7 October 2017 (UTC)[reply]

186 actual unique paths (none of which are symmetries of each other). --Deacon Vorbis (talk) 02:23, 7 October 2017 (UTC)[reply]

Oh, and it finally finished overnight (a more efficient program probably would have done this a lot faster), and it turns out there are 19244800 115468800 unique paths for the 4-orthoplex. I doubt it would be able to find the next one in a reasonable amount of time though. --Deacon Vorbis (talk) 14:11, 7 October 2017 (UTC)[reply]

My concern is not only with rotations and reflections, but also how to make sure that two pathways that start in different places in the cycle are identified with each other.Naraht (talk) 16:28, 9 October 2017 (UTC)[reply]

Well, that's probably more complicated then. Presumably you want to consider reversed traversals as the same too. In that case, even if you start at a fixed vertex, each path could potentially be counted 4 times. Unfortunately, it's not as simple as dividing by 4 (since 186 isn't divisible by 4). That means that there are some circuits where some of these symmetries happen to coincide. Now you have two different groups acting on the set of circuits: the symmetry group of the octahedron, but also D₁₂ (the dihedral group of order 24). I'm not exactly sure how to proceed in general now other than just having a computer brute-force the remainder. --Deacon Vorbis (talk) 17:14, 9 October 2017 (UTC)[reply]

Consider the possible sets of edges unvisited by a Hamiltonian cycle along the edges of a Dodecahedron and the possible sets of edges unvisited by a Double Hamiltonian cycle (visiting all vertices twice along the path) on the edges of a icosahedron. Are these two sets in 1-1 relationship with each other? (IN Double Hamiltonian, I'm presuming that it doesn't make a difference if the Double Hamiltonian Cycle is one cycle followed by another or if is not).Naraht (talk) 16:37, 5 October 2017 (UTC)[reply]

According to my calculations there are 30 Hamiltonian cycles on the dodecahdron, which would be the number of possible sets of unvisited edges. The second set is essentially the number of matchings on an icosohedron for which the complement is Eulerian, assuming I understand your definition of double Hamiltonian correctly. But all the complements are connected so this is just the number of matchings (by the theorem that a connected graph with every vertex of even degree is Eulerian). I get 125 such matchings, and since 30 ≠ 125 there is no one-one correspondence. It might even things up a bit to include all matchings of the dodecahedron, not just the ones whose complements are cycles, but there doesn't seem to be any reason to suspect there would be a correspondence. --RDBury (talk) 23:53, 7 October 2017 (UTC)[reply]

PS. I went ahead and verified that there are 6 matchings of the dodecahedron whose complements aren't cycles, bringing the total matchings to 36 - still a log way from 125. It's clear without actually counting them that the number must be divisible by 3, eliminating 125 as a possible value. BTW, if you're counting up to symmetry, there are are 2 matchings of the dodecahedron, and 5 for the icosahedron. --RDBury (talk) 13:03, 8 October 2017 (UTC)[reply]