"From any point?" "Eh,...no..." I'd love to see a compilation of Brady disarming professors with his ability to pose the most important questions that, unfortunately, are asked with a frequency inversely propionate to one's age.
Nah, if it starts at a corner and next hits a side at 45 degrees, that is the same path that would allow it to leave that side at 45 degrees and hit that corner.
If it starts in a corner (my DVD player starts in the upper left corner), then it will continue bouncing off sides until it hits another corner (in my case, the lower right), and then will reverse and eventually return to it's starting point hitting the upper left corner. And even if it starts on a side, it still might be able to hit a corner, unless the aspect ratio is perfectly square, which it never is. After all, every single point where it hit a side between hitting corners could have been a starting point that would result in those same corners being hit.
Luiz Sarchis If it hits a side after starting in a corner, and it WILL, then the reverse path is possible. If it started at that point on the side, it could hit a corner.
I don't expect formal proofs in these videos, but in contrast to Numberphile's excellent way of doing things over the years, no explanation or general formulation of any level is given in these set of two videos, just some results about rational angles without any logic behind it. Numberphile is usually fantastic in finding a middle road between popular science and math, but this time these couple of videos just felt like an anecdote collection.
I think with dynamical systems proofs or explanations are just so complex that you can't really fit them inside a video using language everbody is able to understand. But you're right Numberphile always did a great job!
Henry Segerman JANNIK well then those aren't really topics suitable for Numberphile videos, are they? There's lots of topics in math, and a large portion of those can't be covered in the format SeventyFive was talking about, but there's still a lot that can be, and maybe that's what this channel should focus on.
The square example reminds me of the dvd logo bouncing on the screen, and everyone wants it to go into a corner. Turns out, it can't, because then reflection doesn't mean anything. Like life.
Well, it can, because it usually starts from a corner. However, it CANNOT bounce into any corner, there are always only 2 corners the DVD logo can hit, unless it starts from somewhere else like in the center for instance.
when you are solving some problems with your students and you don't want to show them all at once so they aren't distracted by solving the difficult ones; however, when you switch off the projector, the students are even more distracted by the projector logo bouncing around. I was telling this from a teacher's perspective not to embarass myself :p
Actually it can bounce off the corner. If it goes in perfectly if till just bounce and come out the way it came in, reversing the "orbit". It's just treated as two separate reflections, one for each axis
There is a solution to the obtuse triangle problem!* *If you relax the notion of a periodic orbit. What Prof. Masur is implying is that there needs to exist a point for which the trajectory returns to that point _exactly_. If we only required that the trajectory returns to the point, but within an arbitrarily small radius of error, then the result is known to be true. In fact, this would be true for any shape billiard table, not just obtuse triangles. That is, provided that the billiard table has a finite area and all the points on the boundary of the billiard table are included. This is an application of Birkhoff's Recurrence Theorem, a result in Dynamical Systems.
For all practical purposes, it is not a different question, the idea of exact recurrence is an abstraction of mathematics and if we were to try to do verify the result with an experiment, we could never measure something exactly. As I pointed out, this is true for any crazy shaped billiard table provided it has a finite area Granted, exact recurrence is nicer, but we get a lot in return for this relaxation.
Why dividing the viedo in two, and make the second part hidden, reachable only through the link? Isnt better to have the video at full length, because it seems obvious that if one start watching this video and reaches the 9th minutes of it is actually interested and then it will keep watching these 6 minutes too, but dividing it in this manner... isnt less wise for the revenue too? usually much less ppl look at the 2nd part, but the 2nd part is interesting too...
That's true. I guess they don't want to make the videos too long and risk boring viewers. I'm pretty interested in most things on the channel, but occasionally I'll find myself getting bored toward the end of a particularly long video (usually when I'm tired). I personally prefer this method, which doesn't bombard subscribers with too many/too long videos, but allows those interested to find out more.
It’s easy to find a periodic orbit in any isosceles triangle - including obtuse irrational ones: Start at the middle of the base and drop a perpendicular to any of the other sides.
Do we have periodic orbits for a square given an irrational starting point and direction? What happens in a 3-D region? Do we always have some periodic path in a cube, given the starting point and direction are rational? Irrational? Can we extend this analogy to an n-dimensional region?
;) that I know, it is really a matter of representation. Some systems are better represented 2D and others 3D even when both systems by function are the same.
shure but it is propably way more comlicated to prove a 4 dimensional polygon (i don't think thats's caled a polygon if it isn't just substitute it for the word for 4 d shapes without curves.) has orbits and how much. w.itch is the old mathematicians ruling of amm fine but can we make this problem more complicated?
+Professor Howard Masur, what is so special about the distinction between rational and irrational angles? From my simplistic view, if I were to take an actual triangle with mirrors on the inside surface and use a laser to shine on the inside, you seem to say there are in infinite number of triangles that I could construct where the laser could be bounced all the way back to the starting point. Why then would there also be an infinite number of triangles in between any of those triangles?
If you have any obtuse triangle, you can divide it into two acute triangles, find the bisector pattern of each, and join the patterns to get it to repeat no?
Kram1032 That number is probably not mathematically special (as neither one hundred nor the unit of degrees is special). it probably resulted from bounding.
Quoting from the actual paper: "One might wonder whether 100 degrees is a natural cutoff for our result. It is not. We stopped at 100 degrees because it is a nice round number. With a lot more effort, we would perhaps get to 105 or 110 degrees." They do state, however, that 112.5 degrees (5/8 of Pi radians) is a "very hard barrier to pass" In the paper the problem is turned into one that can be done by a computer, and is thus limited by calculation time.
so if a polygon has all the angles a fraction of 180° then there is some point at some angle will have a periodic orbit does that apply to apeirogon (a polygon that has infinite sides) because an apeirogon with central angles add up like 180°(1+1/2+1/4+..) is a vaild apeirogon with all the angles a fraction of 180°
Just an observation with the periodic orbits of triangles, at least some isosceles triangle with one obtuse angle I believe has a periodic orbit.I tried it on one with the angles 120, 30 and 30. Do the method applied to the acute angled triangles, where you find a line that is perpendicular to each side but passes through the opposite angle (by extending the lines beyond the triangle). After that, connect up those right angles. In this case it forms an equilateral triangle (but there should at least always be two sides the same length). The segment of the two lines which hit the hypotenuse, the line opposite the obtuse angle should be a periodic orbit in at least this case.
Though corners mean an object bounce around unpredictably, what about a convex polygon without corners? Would that mean for any point on a rational polygon, a periodic orbit can be completed?
I came across this stuff making waveguide models for sound synthesis, and of course a circle (ie a drumhead) has an infinite number of orbits, tending towards infinity towards both the diameter and the tangent.
I wonder what the rule is for circles. My intuition tells me that any rational (measured in degrees not radians) angle of incidence will produce a periodic orbit "completed spirograph patten" and any irrational angle of incidence will and the line will not cross exactly the same path more than once. As for corners, you could define a rule for the infinitesimal corner region that it bounces out at the angle of reflection equal to the angle of incidence of the line that is the same angle away from each side that makes up the corner. For example on a square that would be the line 45 degrees from each side and touching the corner. The easier way to think of this is the line perpendicular to the centre point of the angle.
You mean a tangent? If the source IS the center of the circle, then it would bounce back to the center from the perimeter. The point at the perimeter would be the one a tangent to the circle would pass through.
A dynamical system is basically a system where the previous iteration (I don't recall the technical term) affects the current iteration. An example is a sequence x_n=(x_n-1)^2 : the value of x_n directly depends on the value of x_n-1. In the case of this video, the problem illustrated is a dynamical system because the location where the beam of light touches the wall after 5 rebounds depends on where it hit the wall on its fourth rebound. (My apologies if I'm being unclear :) ). Brady's vid on the Mandelbrot and Julia sets, which were published a few years ago and deal with dynamical systems, are one of the reasons that led me to major in math at university: I found them fascinating. As for the applications, there are many such as physics related problems (the position of a body at time t) and in biology (one can easily describe population growth using a dynamical system).
02:40 That's oxen excrements. Start at the corner, heading at 45° to the side, and you'll end up in the opposite corner of the square and you won't know what to do either.
This may help with an intuitive understanding of the problem: Rational angles usually work, because they involve a finite number of bounces; that number is an integer. With irrational angles, you might require an infinite number of bounces; that number might not be an integer... but a non-integer number of bounces would make no sense.
This reminds me so much of the halitng problem. May be you can represent Turing Machines with rational polygons and the periodic orbit problem is undecidable.
Well, the problem might not be solved for irrational obtuse triangles, but what about _specific_ irrational triangles? Are there any examples of specific triangles where they proved "this specific tringle has no periodic orbit" or "this specific triangle has orbits"?
Is there any chance that you could start another channel which contains only long edits of these interviews? They are fascinating to watch and I'm sure most of the audience would like to know more.
It seems odd that an arbitrarily small change in geometry can mean that suddenly we cant prove if there is a periodic orbit. Like if an internal angle is root(2) to 1 million decimal places we know there is a periodic orbit but if its exactly root(2) then we don't know. Surely there is nothing fundamentally different about the angle if it just happens to be irrational. Can an irrational angle actually exist?
Given angles are often measured in radians, aren't all rational angles in some sense irrational because they are ratios of pi (which is an irrational number)? Or does he mean that irrational angles are _irrational with pi_?
i didnt get the rational angle thing. Isnt 180 deg = pi rad? Why does it depends on the metric that you use? I have alwys thought that measuring angles in rads was the most "natural" and less arbitrary way. Could somenobe explain that for me?
The criterion here is that the angles all be rational multiples of π = 180º = a straight angle. Another way to say that, would be that all the angles are sums and differences of (a finite number of) those that occur in regular polygons. (!) That way of stating it, is independent of any choice of the unit of angle measure.
What I would really like to know is if there can be an orbit which doesn't contain the point of origin. In any shaped room at all? My intuition says no, but I've been wrong before...
why did you do trigonometric (don't know the english word, but i guess this is somthing like that) when you can just do this in very simple angle chasing ?
Is there a list of unsolved problems somewhere? I know about the millenial ones, and now these ones, and I'm sure I've heard of more, but I can't remember.
Extremely complex problem. Even for the simple case of any triangle it is very complex. But one can wrap his mind around it, I think. First of all, let's unwrap that closed trajectory (CT) over a straight line (further - direction line, named DL) with starting point laying on any side of the triangle and with certain angle of direction. Thus, we have the second point of CT by intersection of DL with one of the other two sides of the triangle. This particular side becomes an axis of symmetry over which we draw an mirror image (reflection) of our triangle so that one of its two remaining sides will certainly intersect DL and we will have the third point of CT, and so on. Now, if CT consists of k points, the k-th reflection of the triangle simply must be a parallel translation of the original triangle over DL. In other words, DL must intersect the k-th image of the triangle in the same point and with the same angle of direction, so that we have a cycle with lenght of k. Before continuing, we must reconsider the special case of DL intersecting not one of the sides of the triangle, but one of its vertices. Prof. Masur said in the video that in this particular case "we don't know what to do", but I think, that in order to avoid that kind of ambiguity we must state that DL reflects in line, perpendicular to bisectrix of the angle to which this vertex belongs. Then, instead of axis of symmetry we will have this vertex as center of rotational symmetry with angle of 180 degrees... after which we will continue with the "unwrapping" the CT over DL. So... every step of this procedure in fact tilts that side of the triangle (which has the first point of CT) with alpha, beta or gamma degrees (the masures of the three angles of the triangle) clockwise or counterclockwise... or with 180 degrees in the special case of rotational symmetry. Thus, we obtain the necessery (but insufficient) condition for CT: x*alpha + y*beta + z*gamma + w*180 = 0 ; where x,y,z are natural (positive or negative) numbers. (Here k=>abs(x)+abs(y)+abs(z)) This Diophantine equation has solution when alpha, beta and gamma are rational numbers; if they are irrational... there might be solution in very special cases, I think - when irrationalities will cancel out each other somehow. Besides of fulfilling the necessery condition - DL to intersect with k-th triangle with same angle - but it has to be in the same point, which will be sufficient for CT, as I mentioned beforehand. We can acommodate this procedure for any polygon, but resulting diophantine equations will have increasing complexity.
Hello I have a question, i got a riddle and i am unable to solve it is as follows: 1,3,5,7,9,11,13 and 15 we have these numbers and we have to arrange any three of them so that their sum is 30
shashank chandrashekar they defined rational angles as rational number multiples of 2pi, that is 180 degrees. So think about it as a rational number times 2pi radians
Samo Dimitrije Numberphile generally tried to provide access of higher level math to the common So I don't think a question would fit in here unless it's a really complicated one that no non mathematician can solve I would say you should post that question on math stackexchange first. Permutations is generally one of the easiest topics in math for me
Harinandan Nair so you have a box and in the box is a plastic tray with a 4×4 grid of rectangular waffle cookies.So you've been lazy and opened just a single corner of the foil over the tray so only one of the cookies can get out.The question is if you have the tray in the box and you can't see the arrangement of the cookies nor the number of them,how many moves and wich ones would you have to do to put them as much as you can in full columns and the leftover in the next column? It's probably supper simple but a great entry to the topic
How can a triangle have an irrational angle, if the sum of all angles must be 180? Are irrational angles not rationalized anyway, by actually constructing those angles, therefore cutting the irrational number somewhere?
Make one angle root(2), the second PI, and the third 180-(root(2) + PI). There you have it. Three irrational angles, adding up to 180. What do you mean with "rationalized" and "constructing"?
I was taking a more practical approach, obviously in theory you can just make a sketch of some triangle and say that the angles are root(2), etc. But if you actually wanted to draw that triangle, you would have trouble getting the angle right, because of how you can not possibly pin down the angle to be root(2), because as irrational number, the post comma numbers are never ending. Therefore, if you actually draw said triangle, you will have angles that can be pinned down to a rational number, maybe some dozen numbers after comma of some irrational number. But this approach questions the existence of "irrational triangles" because they, in fact, can not exist? IDK if that makes any sense. Probably, if not obviously, I am wrong here, else those scientists wouldn't bother with it, but then where am i wrong?
You're highlighting the distribution properties of rational and irrational numbers. Namely, that both are infinitely dense; that between any pair of distinct rational numbers, no matter how close together, there are infinitely many irrational numbers; and between any pair of distinct irrational numbers, no matter how close together, there are infinitely many rational numbers. So in the purely practical matter of constructing an irrational triangle, any degree of construction error, no matter how small, will defeat your attempt. This is a problem, not in practical construction, but of theoretical math.
Surprisingly, since there are only countable many rational numbers, but uncountable many irrationals, if you draw an arbitrary triangle, it is almost guarateed to have irrational angles. There is a subject in maths called "Constructible Numbers", you can find some vedos about it and I think even numberfiles made one. There it is pretty easy to construct (some) irrational numbers like root(2) (fun fact: PI is not constructible). But if you talk about physically constructing a triangle, you simply have the problem of measurement. You can not arbitrary precisely measure anything. So you would never know.
That "fractions of 180°" thing seems useless since you could always have 180 in the denominator, so it's really "fractions of degrees". - I suspect what he's really saying is that it's fractions of pi, i.e. half a circle (or, equivalently, fractions of tau, i.e. a full circle) - then that way of writing it makes a lot more sense "pi p/q" or "tau p/q".
I think what the p/q stands for is "any number that can be expressed by p/q", so "any rational number". Thus it is just a different way of expressing "a * 180° where a is element of the rational numbers".
Well, yes. p and q are integers. p/q is any rational number. But 180 is ALSO an integer and so can be absorbed into p (or alternatively, you can multiply q by 180 such that the 180 cancels). Adding the 180° remark is kind of pointless. What he's really saying is that we are talking about a rational part of a circle.
I guess some might think of it as clearer. To me it's actually less clear that way. I'd write it as π p/q or τ p/q (where τ = 2π), standing for a fraction of half a circle or one of a full circle respectively. Degrees are kind of arbitrary. They are simply chosen for being highly divisible. π and τ actually relate to the circle (circumference over diameter and circumference over radius respectively) But I guess it doesn't really matter in the end.
The irrational angle should come in pair. And we might says something like "the light need to cancel out all the irrationalities of the irrational angles in order the achieve a periodic path." Then, we can claim that "For any path, if one irrational angle is being cancelled, the pairwised angle cannot be cancelled with the same path". i.e. no periodic path. p.s. this is just an intuitive way for me to think of the question. no conclusion yet.
I find it deeply concerning that there's such a gap of knowledge between rationally-valued things and reals. Because the difference always seems to be infinitesimally small, so I have hard time figuring how an infinitesimal nudge in the angles might produce more than infinitesimal difference in the trajectory. And is it even about difference in the trajectories, or is is just that we don't have knowledge of them because the proof techniques don't yield to real analysis? But that's dynamical systems for you!
I'm always so grateful for the animations, thank you.
It's so satisfying to watch.
np dude..
4:21 when you remember you left the oven on but you are busy talking about polygons
Josh Alexander incredible
??.
"From any point?"
"Eh,...no..."
I'd love to see a compilation of Brady disarming professors with his ability to pose the most important questions that, unfortunately, are asked with a frequency inversely propionate to one's age.
Nice name, but I think you meant "proportionate"
??.
The maths behind the bouncing DVD logo...
wolverine96 Mathematical proof that it will never hit a corner if it starts at any side with 45 degrees angle
Nah, if it starts at a corner and next hits a side at 45 degrees, that is the same path that would allow it to leave that side at 45 degrees and hit that corner.
If it starts at a corner, by definition it didn't start at a side...
If it starts in a corner (my DVD player starts in the upper left corner), then it will continue bouncing off sides until it hits another corner (in my case, the lower right), and then will reverse and eventually return to it's starting point hitting the upper left corner. And even if it starts on a side, it still might be able to hit a corner, unless the aspect ratio is perfectly square, which it never is. After all, every single point where it hit a side between hitting corners could have been a starting point that would result in those same corners being hit.
Luiz Sarchis
If it hits a side after starting in a corner, and it WILL, then the reverse path is possible. If it started at that point on the side, it could hit a corner.
I don't expect formal proofs in these videos, but in contrast to Numberphile's excellent way of doing things over the years, no explanation or general formulation of any level is given in these set of two videos, just some results about rational angles without any logic behind it. Numberphile is usually fantastic in finding a middle road between popular science and math, but this time these couple of videos just felt like an anecdote collection.
Sometimes the proofs are too hard to even attempt to cover, even when the statement of the theorem is easily understandable.
I think with dynamical systems proofs or explanations are just so complex that you can't really fit them inside a video using language everbody is able to understand. But you're right Numberphile always did a great job!
Henry Segerman JANNIK well then those aren't really topics suitable for Numberphile videos, are they? There's lots of topics in math, and a large portion of those can't be covered in the format SeventyFive was talking about, but there's still a lot that can be, and maybe that's what this channel should focus on.
Or maybe we should remember we're watching videos on youtube.
Well I have to say even if I don't understand the maths behind it (yet hopefully haha) I found this video very interesting and intriguing.
The square example reminds me of the dvd logo bouncing on the screen, and everyone wants it to go into a corner.
Turns out, it can't, because then reflection doesn't mean anything. Like life.
Well, it can, because it usually starts from a corner. However, it CANNOT bounce into any corner, there are always only 2 corners the DVD logo can hit, unless it starts from somewhere else like in the center for instance.
when you are solving some problems with your students and you don't want to show them all at once so they aren't distracted by solving the difficult ones; however, when you switch off the projector, the students are even more distracted by the projector logo bouncing around.
I was telling this from a teacher's perspective not to embarass myself :p
Quess i shoulda read the comments first lol
Nobody puts Baby in the corner.
Actually it can bounce off the corner. If it goes in perfectly if till just bounce and come out the way it came in, reversing the "orbit". It's just treated as two separate reflections, one for each axis
I would LOVE to see videos made about the 4 other open problems with dynamical systems.
Dustin Rodriguez me too
This one is easy. I have discovered a truly remarkable proof which this margin is too small to contain.
+iSquared i see what you did there
@@Quantiad FERMET!!! haha.
??.
I love hearing about solved ones like these, it makes me feel like there's still some room left in this wonderful world of mathematics!
There is yet much to discover! :)
There is a solution to the obtuse triangle problem!*
*If you relax the notion of a periodic orbit.
What Prof. Masur is implying is that there needs to exist a point for which the trajectory returns to that point _exactly_. If we only required that the trajectory returns to the point, but within an arbitrarily small radius of error, then the result is known to be true.
In fact, this would be true for any shape billiard table, not just obtuse triangles. That is, provided that the billiard table has a finite area and all the points on the boundary of the billiard table are included.
This is an application of Birkhoff's Recurrence Theorem, a result in Dynamical Systems.
Please elaborate?
"there's an answer if you ask a different question"
Walt van Amstel - What an AWESOME comment - Obliged!!
xo
For all practical purposes, it is not a different question, the idea of exact recurrence is an abstraction of mathematics and if we were to try to do verify the result with an experiment, we could never measure something exactly.
As I pointed out, this is true for any crazy shaped billiard table provided it has a finite area
Granted, exact recurrence is nicer, but we get a lot in return for this relaxation.
It may eventually return to the exact same point as where it started but perhaps not on the same angle
I had the privilege of taking one of Dr. Katok's classes when I was at PSU. Very cool to hear him mentioned here.
Why dividing the viedo in two, and make the second part hidden, reachable only through the link? Isnt better to have the video at full length, because it seems obvious that if one start watching this video and reaches the 9th minutes of it is actually interested and then it will keep watching these 6 minutes too, but dividing it in this manner... isnt less wise for the revenue too? usually much less ppl look at the 2nd part, but the 2nd part is interesting too...
True.
Don't they usually also release the second part, just later.
That's true. I guess they don't want to make the videos too long and risk boring viewers. I'm pretty interested in most things on the channel, but occasionally I'll find myself getting bored toward the end of a particularly long video (usually when I'm tired).
I personally prefer this method, which doesn't bombard subscribers with too many/too long videos, but allows those interested to find out more.
they wanted to find out the amount of real people actually interested in their channel xD
It might be because the second part where uploaded before part one...
Would be nice to know a little bit about why proving for irrational angles has been so much harder and what methods the
You guys are masterminds! So amazing to watch and learn!
It’s easy to find a periodic orbit in any isosceles triangle - including obtuse irrational ones: Start at the middle of the base and drop a perpendicular to any of the other sides.
(EITHER of the other sides: there are only two, which in English requires "either", not "any". It's like "both" and "all".)
Mark Mandel ok
But can you find a periodical orbit inside a Parker square? :P
+Ilmar Bürck ha ha
The Parker square - a story about a man that tried the impossible (and sadly failed) xD
Ilmar Bürck Story now available as a lifetime movie.
Why did you do a parker?
+Adam Billman
It was scheduled to release as a lifetime movie, but they really made a parker square of the whole thing
Do we have periodic orbits for a square given an irrational starting point and direction?
What happens in a 3-D region? Do we always have some periodic path in a cube, given the starting point and direction are rational? Irrational?
Can we extend this analogy to an n-dimensional region?
what about orbits in higher dimensional shapes?
Or orbits that are either curved on a flat plane, or are on a curved plane?
;) that I know, it is really a matter of representation. Some systems
are better represented 2D and others 3D even when both systems by
function are the same.
shure but it is propably way more comlicated to prove a 4 dimensional polygon (i don't think thats's caled a polygon if it isn't just substitute it for the word for 4 d shapes without curves.) has orbits and how much. w.itch is the old mathematicians ruling of amm fine but can we make this problem more complicated?
"the old mathematicians ruling of amm fine" ???
2:48 What if I start at a 45degree angle from a CORNER of a square? I would hit the other corner an NOT bounce, wouldn't it?
At 3:08, he says if the 'triangles are acute', and captions show it as 'If the triangles are cute.'
TRIANGLES ARE ALWAYS CUTE.
Hemanth Pattem Why are triangles so cute you might ask? It's simple, they have :3 sides.
Women have very cute triangles built in
I love this guy's enthusiasm for the topic!
This guy was very at explaining this. Great video.
He was very.
+Professor Howard Masur, what is so special about the distinction between rational and irrational angles? From my simplistic view, if I were to take an actual triangle with mirrors on the inside surface and use a laser to shine on the inside, you seem to say there are in infinite number of triangles that I could construct where the laser could be bounced all the way back to the starting point. Why then would there also be an infinite number of triangles in between any of those triangles?
If you have any obtuse triangle, you can divide it into two acute triangles, find the bisector pattern of each, and join the patterns to get it to repeat no?
Why 100°? What's special about 100 that made this solvable?
Read the paper... I assume it's very complex, and way above our heads.
Kram1032 That number is probably not mathematically special (as neither one hundred nor the unit of degrees is special). it probably resulted from bounding.
Quoting from the actual paper:
"One might wonder whether 100 degrees is a natural cutoff for our result. It is not. We stopped at 100 degrees because it is a nice round number. With a lot more effort, we would perhaps get to 105 or 110 degrees."
They do state, however, that 112.5 degrees (5/8 of Pi radians) is a "very hard barrier to pass"
In the paper the problem is turned into one that can be done by a computer, and is thus limited by calculation time.
scowell, should have taken your own advice I guess ;)
That makes a lot of sense, thanks PanozGTR2
so if a polygon has all the angles a fraction of 180° then there is some point at some angle will have a periodic orbit
does that apply to apeirogon (a polygon that has infinite sides)
because an apeirogon with central angles add up like 180°(1+1/2+1/4+..) is a vaild apeirogon with all the angles a fraction of 180°
I would have appreciated seeing some of the actual maths. I hope you guys do a follow up of this.
Just an observation with the periodic orbits of triangles, at least some isosceles triangle with one obtuse angle I believe has a periodic orbit.I tried it on one with the angles 120, 30 and 30. Do the method applied to the acute angled triangles, where you find a line that is perpendicular to each side but passes through the opposite angle (by extending the lines beyond the triangle). After that, connect up those right angles. In this case it forms an equilateral triangle (but there should at least always be two sides the same length). The segment of the two lines which hit the hypotenuse, the line opposite the obtuse angle should be a periodic orbit in at least this case.
Though corners mean an object bounce around unpredictably, what about a convex polygon without corners? Would that mean for any point on a rational polygon, a periodic orbit can be completed?
I came across this stuff making waveguide models for sound synthesis, and of course a circle (ie a drumhead) has an infinite number of orbits, tending towards infinity towards both the diameter and the tangent.
I wonder what the rule is for circles. My intuition tells me that any rational (measured in degrees not radians) angle of incidence will produce a periodic orbit "completed spirograph patten" and any irrational angle of incidence will and the line will not cross exactly the same path more than once.
As for corners, you could define a rule for the infinitesimal corner region that it bounces out at the angle of reflection equal to the angle of incidence of the line that is the same angle away from each side that makes up the corner. For example on a square that would be the line 45 degrees from each side and touching the corner. The easier way to think of this is the line perpendicular to the centre point of the angle.
You mean a tangent? If the source IS the center of the circle, then it would bounce back to the center from the perimeter. The point at the perimeter would be the one a tangent to the circle would pass through.
Are there some results from triangles in other topologies? Like triangles at the surface of a sphere?
Now this is what I call a Periodic Video.
1:49 this sounds like a conversation the gf and I have sometimes... "Anywhere??" "NO NOT ANYWHERE!!!😲"
Now I want to know what dynamical systems are, and what their applications look like...
A dynamical system is basically a system where the previous iteration (I don't recall the technical term) affects the current iteration. An example is a sequence x_n=(x_n-1)^2 : the value of x_n directly depends on the value of x_n-1. In the case of this video, the problem illustrated is a dynamical system because the location where the beam of light touches the wall after 5 rebounds depends on where it hit the wall on its fourth rebound. (My apologies if I'm being unclear :) ). Brady's vid on the Mandelbrot and Julia sets, which were published a few years ago and deal with dynamical systems, are one of the reasons that led me to major in math at university: I found them fascinating. As for the applications, there are many such as physics related problems (the position of a body at time t) and in biology (one can easily describe population growth using a dynamical system).
Jakedesnake97 is the term you're looking for recursion?
See numberphile's video on -7/4
02:40 That's oxen excrements. Start at the corner, heading at 45° to the side, and you'll end up in the opposite corner of the square and you won't know what to do either.
"in the square no matter where I started, with a 45° angle it will be periodic"
*starts in middle, instant corner*
I like this dude. He knows stuff.
This may help with an intuitive understanding of the problem: Rational angles usually work, because they involve a finite number of bounces; that number is an integer. With irrational angles, you might require an infinite number of bounces; that number might not be an integer... but a non-integer number of bounces would make no sense.
So, of all prerequisites it's obtuseness and an irrational angle that are most likely to prevent one from getting into a rut?
Hi! I'm Earl, and this is my friend Lee!
Awesome, I was thinking this at the last video!
2:38 "What might happen is it might hit a corner then you don't know what to do." That's hilarious because I would say the same thing as a math major.
Might there have been any use of conformal mapping theory with the rational angle polygons?
Hi numberphile why when you press many times (ans+1)/ans its alaws egale to 1.618033989???? Pleas try it your self
This reminds me so much of the halitng problem. May be you can represent Turing Machines with rational polygons and the periodic orbit problem is undecidable.
For all equilateral triangles where A and B are the same length, then is C always irrational?
Well, the problem might not be solved for irrational obtuse triangles, but what about _specific_ irrational triangles? Are there any examples of specific triangles where they proved "this specific tringle has no periodic orbit" or "this specific triangle has orbits"?
So I have a question. Would this have any practical application? Could this aide in the designing of something?
No annotations for John Conway and Illumination Problem at the end :(
Is there any chance that you could start another channel which contains only long edits of these interviews? They are fascinating to watch and I'm sure most of the audience would like to know more.
what about a video (or ten) about chaotic billiards with all the math? would be great!
So does any solution to the Schrodinger equation in a rational polygon have scars ?
This guy has his math so figured out tho. i love it!
for bouncing on a corner, can't just use the bisector as the reflection line?
how does the orbit know whether an angle is rational or not? and why does it care?
It seems odd that an arbitrarily small change in geometry can mean that suddenly we cant prove if there is a periodic orbit. Like if an internal angle is root(2) to 1 million decimal places we know there is a periodic orbit but if its exactly root(2) then we don't know.
Surely there is nothing fundamentally different about the angle if it just happens to be irrational. Can an irrational angle actually exist?
Again, this reminds me of Lunar pool. I've fond memories on this game.
Daft question but if one angle of a triangle is irrational, will the others be too?
I study physics and today i had a geometry class just in time brady
Given angles are often measured in radians, aren't all rational angles in some sense irrational because they are ratios of pi (which is an irrational number)? Or does he mean that irrational angles are _irrational with pi_?
He means that the coeeficient beforw pi is irrational. For example 2*Pi is a rational angle while 1=(1/Pi)*Pi is an irrational angle
I have no idea what he said and I love it
but there is a possibility that the light goes everywhere to say they spread does this apply to the billard
i didnt get the rational angle thing. Isnt 180 deg = pi rad? Why does it depends on the metric that you use? I have alwys thought that measuring angles in rads was the most "natural" and less arbitrary way. Could somenobe explain that for me?
The criterion here is that the angles all be rational multiples of π = 180º = a straight angle.
Another way to say that, would be that all the angles are sums and differences of (a finite number of) those that occur in regular polygons. (!)
That way of stating it, is independent of any choice of the unit of angle measure.
ffggddss thanks!
What I would really like to know is if there can be an orbit which doesn't contain the point of origin. In any shaped room at all? My intuition says no, but I've been wrong before...
Hey. Whaif you've got a circle? Would angles matter
why did you do trigonometric (don't know the english word, but i guess this is somthing like that) when you can just do this in very simple angle chasing ?
Serraille Baptiste u haven't heard of trignometry yet u watch Numberphile videos?
:-\ :-\ :-\ :-\
If any one of the three angles in a triangle is known to be irrational, can it be said that all the angles are known to be irrational?
There are shelves upon shelves of books filled with questions like this.
the angle needs to be a rational fraction of pi ?
Won't any angle from any point in any shape cycle eventually given infinite time?
For all RATIONAL numbers, yes.
Is there a list of unsolved problems somewhere? I know about the millenial ones, and now these ones, and I'm sure I've heard of more, but I can't remember.
It follows that in the acute triangle the orbit is ortogonal to the heights. Why is that true?
Figured it out, the heights intersect at the center of the orbit triangle subscribed circle...
What is an accurate way to describe the star polygon(?) in the thumbnail of this video?
Halp?
Hey, sorry for the late response, but it's a 14/6-gon
Extremely complex problem. Even for the simple case of any triangle it is very complex. But one can wrap his mind around it, I think.
First of all, let's unwrap that closed trajectory (CT) over a straight line (further - direction line, named DL) with starting point laying on any side of the triangle and with certain angle of direction. Thus, we have the second point of CT by intersection of DL with one of the other two sides of the triangle. This particular side becomes an axis of symmetry over which we draw an mirror image (reflection) of our triangle so that one of its two remaining sides will certainly intersect DL and we will have the third point of CT, and so on.
Now, if CT consists of k points, the k-th reflection of the triangle simply must be a parallel translation of the original triangle over DL. In other words, DL must intersect the k-th image of the triangle in the same point and with the same angle of direction, so that we have a cycle with lenght of k.
Before continuing, we must reconsider the special case of DL intersecting not one of the sides of the triangle, but one of its vertices. Prof. Masur said in the video that in this particular case "we don't know what to do", but I think, that in order to avoid that kind of ambiguity we must state that DL reflects in line, perpendicular to bisectrix of the angle to which this vertex belongs. Then, instead of axis of symmetry we will have this vertex as center of rotational symmetry with angle of 180 degrees... after which we will continue with the "unwrapping" the CT over DL.
So... every step of this procedure in fact tilts that side of the triangle (which has the first point of CT) with alpha, beta or gamma degrees (the masures of the three angles of the triangle) clockwise or counterclockwise... or with 180 degrees in the special case of rotational symmetry.
Thus, we obtain the necessery (but insufficient) condition for CT:
x*alpha + y*beta + z*gamma + w*180 = 0 ; where x,y,z are natural (positive or negative) numbers. (Here k=>abs(x)+abs(y)+abs(z))
This Diophantine equation has solution when alpha, beta and gamma are rational numbers; if they are irrational... there might be solution in very special cases, I think - when irrationalities will cancel out each other somehow.
Besides of fulfilling the necessery condition - DL to intersect with k-th triangle with same angle - but it has to be in the same point, which will be sufficient for CT, as I mentioned beforehand.
We can acommodate this procedure for any polygon, but resulting diophantine equations will have increasing complexity.
Hello I have a question, i got a riddle and i am unable to solve it is as follows:
1,3,5,7,9,11,13 and 15 we have these numbers and we have to arrange any three of them so that their sum is 30
I know I'm 3 years late but... 3 * ( 1+9 ) = 30. You're welcome!
Suggestion: video about those 5 outstanding problems on dynamic systems.
you could make really a really cool screensaver based on this problem
Kek, i have been doing mandala like drawing based on that for years without knowing it was an actual thing in math.
why does the angle have to be in degrees and not in radians,
a rational degree angle can be an irrational radian angle and vice versa.
If you were in radians it would be [pi]*p/q instead of 180(degrees)*p/q. That's the only difference.
shashank chandrashekar they defined rational angles as rational number multiples of 2pi, that is 180 degrees. So think about it as a rational number times 2pi radians
pi
Do a Mathematics Nobel Prize Petition!
When you hit a corner, can't you bounce off both edges?
A corner is a single point, is has no edges.
Everyone watching this piece of video on the internet is an awesome guy. except the guy who disliked it ofc
anyone know i website that simulates these?
I've got an interesting idea for a video,a permutation problem.How can I reach out to you and discuss about it?
Samo Dimitrije Numberphile generally tried to provide access of higher level math to the common So I don't think a question would fit in here unless it's a really complicated one that no non mathematician can solve
I would say you should post that question on math stackexchange first.
Permutations is generally one of the easiest topics in math for me
Well it's a super simple but fun problem,something Cliff would explain,maybe a 6 minute video at most
Can u give me the question?
Harinandan Nair so you have a box and in the box is a plastic tray with a 4×4 grid of rectangular waffle cookies.So you've been lazy and opened just a single corner of the foil over the tray so only one of the cookies can get out.The question is if you have the tray in the box and you can't see the arrangement of the cookies nor the number of them,how many moves and wich ones would you have to do to put them as much as you can in full columns and the leftover in the next column? It's probably supper simple but a great entry to the topic
Anyone is there software to do this? Bouncing magnetic fields. 3 triangles pentagram? Then reverse. Can you show us that?
Aren't degrees an arbitrary thing? How is it that rational angles can create a periodic orbit?
Why does the 90 sqrt2 degree not "work"?
Do some Gödel videos and more videos on foundations in general.
How can a triangle have an irrational angle, if the sum of all angles must be 180? Are irrational angles not rationalized anyway, by actually constructing those angles, therefore cutting the irrational number somewhere?
Make one angle root(2), the second PI, and the third 180-(root(2) + PI). There you have it. Three irrational angles, adding up to 180.
What do you mean with "rationalized" and "constructing"?
I was taking a more practical approach, obviously in theory you can just make a sketch of some triangle and say that the angles are root(2), etc. But if you actually wanted to draw that triangle, you would have trouble getting the angle right, because of how you can not possibly pin down the angle to be root(2), because as irrational number, the post comma numbers are never ending.
Therefore, if you actually draw said triangle, you will have angles that can be pinned down to a rational number, maybe some dozen numbers after comma of some irrational number. But this approach questions the existence of "irrational triangles" because they, in fact, can not exist? IDK if that makes any sense. Probably, if not obviously, I am wrong here, else those scientists wouldn't bother with it, but then where am i wrong?
You're highlighting the distribution properties of rational and irrational numbers.
Namely, that both are infinitely dense;
that between any pair of distinct rational numbers, no matter how close together, there are infinitely many irrational numbers;
and between any pair of distinct irrational numbers, no matter how close together, there are infinitely many rational numbers.
So in the purely practical matter of constructing an irrational triangle, any degree of construction error, no matter how small, will defeat your attempt.
This is a problem, not in practical construction, but of theoretical math.
Surprisingly, since there are only countable many rational numbers, but uncountable many irrationals, if you draw an arbitrary triangle, it is almost guarateed to have irrational angles.
There is a subject in maths called "Constructible Numbers", you can find some vedos about it and I think even numberfiles made one. There it is pretty easy to construct (some) irrational numbers like root(2) (fun fact: PI is not constructible). But if you talk about physically constructing a triangle, you simply have the problem of measurement. You can not arbitrary precisely measure anything. So you would never know.
Now what does that mean for me? Is my approach wrong? Is it just close to reality? Is theoretical math not close to reality here?
Is there a 3D version of this topic?
Actually no, not all directions from any point inside a square lead to orbits, because some directions encounter corners.
You should make a video where to take everybody in number file and ask everyone their favorite shape, number and equation
1 angle can't be irrational, right? It has to be 2 or 3 angles?
Huh, I've actually pictured this sort of thing in my head a lot.
That "fractions of 180°" thing seems useless since you could always have 180 in the denominator, so it's really "fractions of degrees". - I suspect what he's really saying is that it's fractions of pi, i.e. half a circle (or, equivalently, fractions of tau, i.e. a full circle) - then that way of writing it makes a lot more sense "pi p/q" or "tau p/q".
I think what the p/q stands for is "any number that can be expressed by p/q", so "any rational number". Thus it is just a different way of expressing "a * 180° where a is element of the rational numbers".
Well, yes. p and q are integers. p/q is any rational number. But 180 is ALSO an integer and so can be absorbed into p (or alternatively, you can multiply q by 180 such that the 180 cancels). Adding the 180° remark is kind of pointless.
What he's really saying is that we are talking about a rational part of a circle.
Well I guess he just likes writing it like that, maybe for clarity? I have actually seen a few people do it that way.
I guess some might think of it as clearer. To me it's actually less clear that way. I'd write it as π p/q or τ p/q (where τ = 2π), standing for a fraction of half a circle or one of a full circle respectively.
Degrees are kind of arbitrary. They are simply chosen for being highly divisible. π and τ actually relate to the circle (circumference over diameter and circumference over radius respectively)
But I guess it doesn't really matter in the end.
In fact writing it as p/q*tau isn't better then p/q*180°... Both are just hiding the pi behind it...
If you start exactly in the middle of the quare and go in an 45° angle, there is no periodic orbit, because you hit exactly the edge.
The irrational angle should come in pair. And we might says something like "the light need to cancel out all the irrationalities of the irrational angles in order the achieve a periodic path." Then, we can claim that "For any path, if one irrational angle is being cancelled, the pairwised angle cannot be cancelled with the same path". i.e. no periodic path.
p.s. this is just an intuitive way for me to think of the question. no conclusion yet.
I find it deeply concerning that there's such a gap of knowledge between rationally-valued things and reals. Because the difference always seems to be infinitesimally small, so I have hard time figuring how an infinitesimal nudge in the angles might produce more than infinitesimal difference in the trajectory. And is it even about difference in the trajectories, or is is just that we don't have knowledge of them because the proof techniques don't yield to real analysis? But that's dynamical systems for you!
The difference is incredible big if you look at it from other perspectivesfor example, practcally all numbers are irrational
do a video on strange bases like √2!!
My next code project.