Fun fact: Phi (1.618) is really close to the ratio between miles and kilometers (1.609) which means you can use adjacent Fibonacci numbers to quickly mentally convert between them. For instance: 89 miles is nearly 144 km (it's actually 143.2), or 21 kilometers is roughly 13 miles (13.05). You can even shift orders of magnitude to do longer distances! e.g., 210 miles is around 340 km (multiplying 21 and 34 by 10) which is close to the actual answer of 337.96 km.
I love the way this channel give us puzzles which they dont have the answer yet. This is like "I trust you to go beyond" And this is simply awesome. I wish our schools had more of that.
Wow, there are a lot of mean comments in here. Please don't fall into the trap of simplifying this channel too much because of commenters. Even though I didn't fully understand what was going on in the previous video I really appreciate there being a maths channel that's willing to approach complex topics.
Found this video to be much better than recent ones. I think because it involved somewhat more accesable maths whilst still being suprising and less well known information. I think episodes like this, where you dont try to attempt complex proofs but instead introduce interesting ideas in an intuitive way and give people a direction and material for further reading if they are interested, have the best format. In cases where you want to explain more complex ideas and give long proofs maybe make them more detailed by break them up over episodes and/or look for proofs with good graphical representions (like 3Blue1Brown) perhaps. Mirroring the way PBS spacetime makes videos, with each providing interesting new information whilst building the knowledge base to allow for more complex topics would be good. Not to be too critical tho, you guys do a great job!
This answered questions I've had for a while but didn't know how to ask. The golden ratio reappears with any seed numbers or if you divide powers of the seed numbers too. This makes it obvious why those patterns emerge. Thanks guys!
Next observation: By d_(m,k) > d_(n,k) for m>n and d_(n,k)>d_(n,j) for j < k < ceil(n/2)-1 it is possible to proof (!) that sigma_n is not any diagonal-ratio of any regular polygon (for a given n>2). Indeed that is much easier than in the paper. Now for a given n you can easily proof with a machine that sigma_n is not such a numbers, you have to only calculate very few numbers for each n. I can give the details if anybody is interested. That does not give us the answer for all n, but it is maybe a way to go.
It's nice to see a video about the other metallic ratios but I think the golden ratio is more famous because it's simpler. Especially since the Fibonacci sequence doesn't involve multiplication unlike the other metallic sequences. There are other ways to generalize as well, such as the Tribonacci sequence and its constant (which I often like to call "Tri" but is more commonly referred to as just "the Tribonacci constant".) The Tribonacci rule is adding 3 numbers together instead of 2.
Easy numerical proof (really a proof), that sigma_3 does not occur as a diagonal in any n-regular polygon (with site length = 1): You have d_(m,k) > d_(n,k) for m>n and d_(n,k)>d_(n,j) for j < k < ceil(n/2)-1 The first n-regular polygon which can produce a diagonal with length sigma_3 = 3.3027756 ... is n = 11 (because for n < 11 the radius is too small). Next we check, which diagonal in the 11-polygon is the biggest, but smaller than sigma_3. It is the 3rd diagonal d_{11,3} = 3.228707... < sigma_3 and d_{11,4} = 3.51333 So with the second above inequality the 11-polygon is ruled out. We try the 12-polygon and find: d_{12,2} < sigma_3 < d_{12,3} Now see that d_{n,k} < k+1 and together with the first inequality above ALL other regular n-polygons are ruled out. So sigma_3 never occurs as a diagonal-ratio. Q.E.D. Similar you can easily check any other sigma_m, but of course with higher m you will have more numbers to check, which is a easy task for a script. At least we can now state the following: For any given m we can give an algorithm which proofs (pretty fast) that sigma_m does not occur as diagonal-ratio in ANY regular polygon, IF that is true. Of course the same algorithm will find any occurences, like d_{5,1} = sigma_1 and d_{8,2} = sigma_2, too.
I looked up 5, 8 on the Online Encyclopedia of Integer Sequences. The first sequence I found that started with 5, 8 was numbers that are the sum of two successive primes. If this is the pattern for what polygons have the metallic ratios as the diagonal:side ratio, then the next polygon should be the dodecagon. I assume you have probably already tried the doddcagon though.
I know people like shitting on Tae, but I was so excited to see an accessible RUclips channel not focused solely on numerology. Number theory and stuff can be ok, but I can always turn to numberphile for that. Please keep bringing us the super deep stuff she was giving. She's taking feedback very well and fixing her presentation skills, plus the topics are a lot cooler.
It should be noted that a real number r>1 is a metallic ratio if and only if r * (r - floor(r)) = 1 (equivalent to definition). So sigma_n -> n with n -> infinity (side note). Hence you can check for a given n all diagonal-ratios d_j of the regular, if the equation d_j if d_j*(d_j-floor(d_j)) = 1 holds. I checked for all n0 some n, k and m with d_(n,k) - sigma_m < epsilon (like you would expect from a somewhat random distribution)
Except that pulling out the big guns didn't help me in any way. I am sure she proved SHE understands it...but I never doubted that, and I don't feel any closer. :(
Bottom line: I hope she goes with a bit simpler topic next time. Sometimes (like today) I can fully dig in and fully grok what's going on. Other times, I can understand enough to enjoy the ideas without getting the details. (See: probably 2/3 of previous topics) That last one just lost me and I never got within hailing distance of the basic idea. It made for a much less fun experience. Today's response didn't help. Not because she's not smart (she's obviously brilliant) but because I am not mathematically sophisticated enough to even enjoy the show. I hope she takes folks like me into account when she chooses her next topic.
I liked Diffie-Hellman key exchange more, but that may be the case because I program a bit. It was actually kinda fun to write a program determining whether an Mod-N group is cyclic or not and if so, what numbers are its generators.
i am sad to not see more episodes of infinite series, but much like the completion of a supertask, the work was done at such an accelerating rate, that an infinite amount of enjoyment was produced in a finite amount of time.
Might be useful to know: Let distance from 0th vertex of an n-gon to the m th vertex be D(m,n). Ratio D(a,n)/D(b,n) = sin(aπ/n)/sin(bπ/n). So ratio of D(a,n) to side length is sin(aπ/n)/sin(π/n). For Φ, σ_1, a=2, n=5, For σ_2, a=3, n=8, For σ_m, a>m else D(a,n)
sadly the world was just not ready for this video or this channel for that matter. I hope the future generations will stumble upon these relic videos and spawn a new age of enlightenment
First time watching this series... Glad to see you (Gabe) on another PBS show... And I am also happy that you are speaking slower than you did in the SpaceTime videos. Time is short, and we should make the most of it! Keep it going! and going....
I spent about 6 hours at work today taking a crack at the Copper Ratio and seeing if it fit into any polygon's diagonals. Allow me to take a stab at this as I believe I have proved it to not occur. My approach was to find a formula for, first, the triangles formed by connecting diagonals of any regular polygon. Assuming that the triangle is of sides 1-1-r, then r is 2*sin(theta/2) where theta is the angle between the equal sides. The properties of triangles put an upper limit on r of 2, so we can immediately disqualify every Metallic Ratio save for the Golden Ratio, and substituting that for r and solving for theta easily shows we must have a regular pentagon. Next I took the possible trapezoids formed from essentially slicing them off a regular polygon. This gave a formula for the diagonal of r = 1 - 2 * cos(theta). Once again, the properties of quadrilaterals put an upper limit of 3 on r, and setting r equal to the Silver Ratio gives us a theta that is the interior angle of an octagon. Similarly, setting r to the Golden Ratio still gives us the interior angle of the pentagon as this is still perfectly valid. So of course next was the possible pentagons we could form. The equation I found here was r = -4 * cos(theta) * sin(theta/2) after some trigonometric substitution, and this very much works. r cannot be greater than or equal to 4 this time, and testing any polygon below the octagon gives a diagonal formed by a simpler shape that we have already tested. While this formula can be used to find a pentagon that has the Copper Ratio for r, this does not correspond with a theta matching the interior angle of any polygon. From the octagon to the dodecagon, the diagonals formed are roughly 2.613, 2.879, 3.078, 3.229, and 3.346, and we have now surpassed the Copper Ratio and so if it exists it must be in another diagonal we have not seen yet. Next we have the hexagon with r = 1 - 2*cos(theta) + 2*cos(2*theta). We aren't interested in the diagonals of anything smaller than a decagon this time, and once again, testing up to the 11-gon gives us values for r of 3.236 and 3.513, and once more we miss the Copper Ratio. I also tested to see if the 4th Metallic Ratio could fit here as our limit for r is to be smaller than 5, and tested up to the 16-gon, by which we passed and missed it. For brevity I will exclude the values, they are simple enough to calculate following this formula. Finally I tested heptagons, and which point my formula gets unwieldy and for some reason when I attempt to simplify it, it stops working, so perhaps I was either tired or have forgotten something essential to trigonometric identities. r = -4*cos(theta)*sin(theta/2) + 2*cos(pi/2 - 5*theta/2). I should be able to rewrite this entirely in terms of sin functions but for whatever reason it just breaks when I try and I do not know why. Anyhow, we are not interested in any diagonals of any shapes smaller than the dodecagon, as we have already tested them. The diagonal given for the dodecagon is r = 3.864, which is already larger than the Copper Ratio. Now, I believe I have exhausted all the possible hiding spots for the Copper Ratio. My reasoning is that, excuse me if I don't know how to put this into formal mathematical terms, but I have tested every shape that could be sliced off of a regular polygon and when the equations became too unwieldy for me to know how to solve offhand, which was almost immediately, tested the smallest value for our interior angle that could be used that did not also form a smaller shape when sliced off the required polygon, and tested each possible interior angle as we sliced our shape off of the next-sided polygon until it became obvious that we had missed the Copper Ratio, for of course as the interior angles increased so too must the missing side of our shape, approaching the limit for that shape as the interior angles approached infinity. This is enough to prove, I think, that the Copper Ratio cannot be formed using that particular shape. So, we go to the next number of sides, and repeat the process, until we reach a point where the number of sides for a shape sliced off a polygon, and the interior angle required, demands a value larger than the Copper Ratio. I look forward to someone doing this far more succinctly than I have. Also, I did test to see if the heptagons could produce the 4th or 5th Metallic Ratio. They surpass the 4th at the 14-gon and surpass the 5th at the 21-gon, and the limit for r is 6 so no larger one can appear. I don't know to what degree I'd have to go to see if the 4th Metallic Ratio does not appear. I similarly look forward to someone showing whether this behavior of appearing in the diagonals of regular polygons is limited to the Golden and Silver Ratios, personally, I suspect but cannot prove that this is the case. Obviously this brute force method is inelegant, but personally I am more proud of these formulas for these almost-regular polygons. Additionally, checking wolfram alpha, and indeed for a pentagon of these qualities to have the Copper Ratio as its missing side, it would need an angle of x ~ 4 (pi * n + 0.650013) which yields an interior angle of 2.600052 radians, which does not correspond to any regular polygon. I come back to this particular case merely to make certain that the pentagon did in fact miss it, as some theories here seem to believe it would not.
Re the heptagon, the polynomial for the ratio between two of the diagonals is given as x^3 =x^2 + 2x +1 = 0 ruclips.net/video/ciXwHMBUMgg/видео.html frame 6:44 My maths isn't good enough to know if that can ever be re-expressed as one of the metallic ratios, but surely they aren't the only game in town anyway. Maybe we should go beyond them too.
It's been 30 months, or 2 and a half years, that I don't touch deeply any mathematics, since high school. I was among the best in class because I learn easily, but it's depressing how months ago I tried to watch this video, my first on the channel, but I couldn't keep up with it, and saved it for later. Some days ago I watched the numberphile video on this, okay, not bad, and now I came back here. Geez, I understood most, but superficially, barely understanding what he meant with it, having to pause. Except some things in the end that I didn't even get at all. It's just sad. Some weeks from now I'll enter college for a science degree, so it gets even more embarrassing and depressing. But great channel, like as always from PBS. I watch PBS Space Time sometimes and it's awesome. You can't just live by very superficial educational channels. Thanks.
I have indeed found some "equally cool analogs in the rest of the metallic family", though regretfully not polygonal ones yet. Each number in each metallic ratio formula, such as (2+sqrt8)/2 for the Silver, contains a term in, or derivable from, the sequence associated with that formula. But we're talking about the cooler "Lucas" version of that sequence. You probably know that what distinguishes the Lucas sequence from the standard Fibonacci is that while they share the same Golden ratio constant 1.618..., the terms in the Lucas approximate much more closely to 1.618...^n where n is the position or index. For example the 10th term in Lucas is 123, and 1.618..^10 ~ 122.996..., whereas in Fibonacci the 10th term is 55. Another interesting difference is that if you take the product of any pair of terms separated by one, such as 34 and 89, in Fibonacci, the absolute difference with the square of the term between is always 1. So for example (34*89)-(55^2)=1, but in Lucas it's always 5, the square root of which crops up in the formula for the Golden ratio, (1+sqrt5)/2. Lastly Lucas begins with the two terms 2 1, which also occur in the denominator and numerator respectively of the golden ratio formula. All this seems to hold for the other metallic ratios as listed in that table too, something which may have escaped observation if nobody has thought of "Lucasating" the associated sequences. Try it with the Silver. Begin it with the two terms 2 2 and continue, using its recurrence a(n) = 2*a(n-1) + a(n-2).
I noticed a pattern between two methods of calculating the ratio including your open question: σ_n = (n+sqrt(n^2 + 4))/2 = the ratio of the nth diagonal of the regular (n^2 + 4)th polygon to that polygon's side. In the video you showed this works for σ_1 and σ_2. I ran through some code and found that this does not generalize to n>2, but it does nonetheless approximate the true ratios rather well, maxing out at 8.0689% error for σ_6. I was also wondering if the properties would extend to the "trivial" case, σ_0 = 1, and, checking back through the video, I think most if not all of them do!
Ok so i tired every possible arc of every polygon up the 1000-gon and saw if it was equale to any silver ratio and the result is negartive appart from the 2 already known exemples giving by Gab. But more regourously speaking, the problem can be stated in this equation that i couldn't solve : sin(pi*i/m)/sin(pi/m) = (n + sqrt(n^2+4))/2 with m,n,i being natural numbers and i
a little python script if you wanna play with it : import numpy as np silver_ratios, p = [ (n+(n**2+4)**(1/2))/2 for n in range(1,320) ], [ ] def arc_ratios_for_ngone(n): ia = np.pi/n return [ np.sin(i*ia)/np.sin(ia) for i in range(2,n//2+1) ] for i in range(3,1000): A.extend(arc_ratios_for_ngone(i)) for k in A : for j in silver_ratios: if np.abs(j-k) < 1e-6 : print(j)
My first thought is this: A pentagon is a line that turns 360/n degrees. It does this n times. n = 5 for a pentagon. You are drawing a diagonal by 'skipping' a turn. This gives you sigma-1. An octagon is n=8. You are skipping two turns. This gives you sigma-2. I thought maybe this could generalize: a 13-gon where n=13. You skip three turns. This gives you sigma-3. I wrote a program to test this (and generalized for sigma-4 in a 21-gon with 4 turns and sigma-5) and... uh, I'm not sure if the results are correct. The measurements look good, but it was a pretty sloppy empirical test (and I have no reason why this would be theoretically sound math). I can double-check but, before I do; has anyone else tried this? Is it a dead-end? It looks promising, but I'd rather not chase this rabbit into a brick wall.
He said in other comments that he tested the 13-gon already, and it doesn't work. It was actually the first one he tested. Somebody else posted a link to a paper claiming to prove that sigma(3) doesn't continue the pattern. I haven't had time to read it yet (it's late in my part of the world), but you may want to check that out.
Yes. This was my intuition as well. I even started up a theoretical process to prove it, but then, alas, I read the aforementioned article. The article gives a correct proof, by the way, that the Bronze Ratio does not continue in this pattern, nor any other such n-gon/diagonal type relationship for that matter. I wonder, however, if some sort of variation of this pattern persists in higher dimensions, i.e. Platonic solids or beyond, as Leach pointed out ... .
For anyone that still cares, this is a repost from a previous thread: "After some work it turns out that the following equation has to be true to get the right ratio: sin[ b*pi / n ] / sin[ a*pi / n ] = s + (4+s^2)^.5/2 where "a" is the first point along the n-gon, "b" is the second point along the n-gon, "n" are the amount of sides, and "s" is the sigma used. (test this with the pentagon and octagon if you want.) I could supply the proof for this, but it's really just number manipulation. There may be a taylor formula or lesser known trig identities that can be used to break this into fractions that can be tested... but I've spent a half day on this already and I'm done for now." I played around with it a bit more since then, but to no avail... maybe someone else can solve for a,b, n and s
I had a lot of fun working out all the math to build up a spreadsheet to check out to the 12th Phi and polygons up to 70 sides. Now that I have a generalized case for everything, I could throw this all into a python program and run it to infinity...but I see someone else has already done that out to ludicrous values. Oh well, I did all the parts that would teach me stuff anyway. Great topic and well-explained! Thanks!
OH MAN. When I was in High School I explored these ratios in quite a bit of depth. I should have my notes lying around somewhere. I'm afraid I didn't look at polygons, but I remember this was so much fun to look at. I'm going to go dig up my notes, might find something interesting.
My fav thing about the Fibonacci Sequence is that if you take the digital roots of each number in the sequence it actually produces a simple repeating pattern that is 24 steps in total. And what's even cooler is that if you take the first 12 over the second 12, they all add to 9s. So Digital Root (for anyone who doesn't know what that is) is summing the digits of a multi digit number until they are a single digit. Ex. 15 = 1+5 248 = 2+4+8 = 14 = 1+4 = 5) . So now... 1-1-2-3-5-8-13-21-34-55-89-144 8= 8-7-6-5-4-3-2-1-9 7= 7-5-3-1-8-6-4-2-9 6= 6-3-9-6-3-9-6-3-9 5= 5-1-6-2-7-3-8-4-9
whoa whoa whoa... WHOA!!!!, its this guy again!!!!!!... havent seen him forever!! i hope he sticks around again, but dosnt replace the guy who replaced him on Space TIme....lets keep them both!
I love Matt over at Space Time. Honestly, he and the writing staff over there are really knocking it out of the park. But damn it, I missed Gabe. It's truly great to see him again.
i think this could be a solution: Golden ratio =(1+sqrt(5))/2 => satisfied by five sided figure (1st diagonal) (sigma 1) Silver ratio =(1+sqrt(8))/2 satisfied by 8 sided figure (second diagonal) (sigma 2) so for any sigma n: polygon (n^2 +4 sides) nth diagonal : side = sigma n
The ratios (algebraic ratios of linear functions) lead to very interesting mathematics ; including elliptic curves. These ratios are a generalization of the "Golden Ratio". I am writing a paper on the subject.
I really wish there was more visual demonstrations or examples of the ratio use , great work though, keep up the good work guys ♡ Greeting from #Baghdad
pentagon (5) then octogon (8), maybe it would follow the Fibonacci sequences, and so be the ratio of the height of a 13-gon to it's side ? Also, it could go one dimension higher so maybe we should consider looking at solids..
What I have done is written a formula given two inputs that could POTENTIALLY generate metallic ratios. it uses the same logic as the polygons, given x number of vertexes on a circle, what is the ratio of the chord of two nonadjacent vertexes over the chord of two adjacent vertexes. This can be expressed simply as sin(C*pi/S)/sin(pi/S). where C is the nth vertex away (2 for sigma 2, 3 for sigma 3) and S is the number of sides to a polygoon First thing I noticed: C must be at least the next whole number larger than sigma-n because it is increasing with increasing numbers of S and the limit as S goes to infinity is C Second thing: for sigma-1 and sigma 2, when C=2, and S is half the number of sides their polygon has, the result is sigma-n minus one. Third thing: You can input any two numbers where S=C+3 and find a solution that satisfies sin(Cpi/S)/sin(pi/S)-sin(4pi/S)/sin(2pi/S)=1, but only two of those equations are metallic ratios, because those are the only two ratios whose values are less than 3, and the max value for sin(Cpi/S)/sin(pi/S) with the constraint S=C+3 is 3. Conclusion: For a ratio to be metallic, it must fit the rules discussed in the video, but for that ratio to be expressed in terms of a polygon, the polygon MUST have 3 more sides than the nth vertex, which means the highest value the ratio can be is 3, and all values for sigma-n where n>2 are greater than 3, therefore no other metallic ratio can be expressed in terms of two lengths of a polygon.
10:01 I can't believe PBS gets away with monetization RUclips channel, has on camera sponsorship and is not for profit 501c3. Great content. Transparency answers where the #moneygoes 💰
I suspect that if there are no further metallic numbers with corresponding regular polygon representations, it may be to do with the fact that starting with the bronze ratio, they are all greater than pi
The silver ratio is very closely related to the ratio between the sides of A-series paper (and B- and C-series, though B-series isn't used all that much and C-series is mostly used for envelopes), which is the standard for paper sizes in much of the world.
It's 9 minutes of video on new topic, 1 minute add for Brilliant and 5 minutes of explaining questions to previous video on 'Brouwer's fixed point theorem'.
3:41 Fun fact: This series produces an infinite list of numbers from the most irrational to the least as all continued fractions with period 1 with coefficients 1 to infinity are covered. As you progress through it, numbers can be approximated faster and faster with rationals.
If it helps you really just need to solve the following: sin[ b*pi / n ] / sin[ a*pi / n ] = s + (4+s^2)^.5/2 where a is the smaller length, b is the larger length, n is the sides of the n-gon and s is the sigma index. for example on your first entry, d_{5,1,0} = sigma_1, a = (0)+1 b = (1)+1, n = (5) and s = (1)
I've made some progress The side of a regular n-gon inscribed in a circle of radius 1 is: >b=2sin(pi/n) the diagonal of a regular n-gon inscribed in a circle of radius 1 skipping m sides is: >a=2sin(m*pi/n) Thus, if the metallic ratios appear in any n-gon, they will be: >f=sin(m*pi/n)/sin(pi/n) When this happens this expression should be true: >f-/sigma_k=0 try to solve for m, we get: > m = (n/pi)*arcsin[/sigma_k*sin(pi/n)] Now, I suspect that f is as close as possible to a metallic ratio if and only if: n=k^2+4 if we put this in the expression for m then whenever we get whole numbers for m (as happens with k=1 and 2) then we will know that that n-gon has a metallic ratio Thus it is only necessary to prove that the closest ratio will happen at n=k^2+4 to solve this open problem
Given a polynomial y^n = a^yn-1 + by^n-2 + ... + z, you can create an associated series f(n+m) = af(n+m-1) + bf(n+m-2) + ... + zf(m). If the ratio of successive terms in the series converges, it converges to a root of the original polynomial. As an example, if the polynomial is x^2 = x + 1, the associated series is f(n+2) = f(n+1) + f(n) (The standard Fibonacci rule) and the ratios of successive terms converge to a root of x^2 = x + 1 which is (1+ SQRT(5))/2.
Find a/b = phi_n for an s-sided regular polygon with side length b, and m-th diagonal length a. Using a side length of 1 (b = 1), you then have a/1 = a = phi_n. Therefore, the question can be restated as: given an s-sided regular polygon with side length 1, what are the lengths of the diagonals (the line between two corner points on the polygon). Then, which of those diagonal length are equal to a metallic ratio phi_n? If you can find a formula or method for calculating those diagonal lengths, you can then check which ones match a given phi_n.
I wrote a program to run through lots of regular polygons for each possible diagonal checked against each metallic ratio from 1 to 50. I've found some very close but none that are exact yet. Tomorrow I am going to try to use the function I found to find the diagonal length to prove or disprove this overall since I;m not finding any solutions.
Small note: simple continued fractions (all the numerators are 1) can be unique, but the video doesn't give enough conditions. 1) Finite continued fractions of non-integers never end in 1; 2) no terms except possibly the first term in the continue fraction are 0; and, 3) either all terms are negative, or only the first term is negative. With these three conditions, simple continued fractions are indeed unique. (1) is something like the zero-point-nine repeating problem for decimal numbers, (2) is like the leading- or trailing-zero problem for decimal numbers, and (3) is doesn't really have an analog for numbers as we usually write them.
Well-said, and keep including comments like this. Almost any time we omit details like this, it's either b/c of "flow" in the script or b/c we have pretty tight length constraints on the videos. We'd prefer to cover every logical base, obviously, but we feel less bad about it since we consistently see commenters (like you) fill in the gaps for those who may be unable to do so for themselves. So please, keep comments like this coming! We rely on them.
very nice part about the metallic ratio constants family. if n=3 doesn't work with planar polygons, perhaps in a space with a different metric it works. maybe there have to be more degrees of freedom
such as diagonals of polyhedra. thanks for the video again, and there is another real nice video recently by mathologer. "way beyond the golden ratio.." ruclips.net/video/cCXRUHUgvLI/видео.html
The association of the main numbers in the field of mathematics with each other, reflects numerical sequences that correspond to the dimensions of the Earth, the Moon, and the Sun in the unit of measurement in meters, which is: 1' (second) / 299792458 m/s (speed of light in a vacuum). Ramanujan number: 1,729 Earth's equatorial radius: 6,378 km. Golden number: 1.61803... • (1,729 x 6,378 x (10^-3)) ^1.61803 x (10^-3) = 3,474.18 Moon's diameter: 3,474 km. Ramanujan number: 1,729 Speed of light: 299,792,458 m/s Earth's Equatorial Diameter: 12,756 km. Earth's Equatorial Radius: 6,378 km. • (1,729 x 299,792,458) / 12,756 / 6,378) = 6,371 Earth's average radius: 6,371 km. The Cubit The cubit = Pi - phi^2 = 0.5236 Lunar distance: 384,400 km. (0.5236 x (10^6) - 384,400) x 10 = 1,392,000 Sun´s diameter: 1,392,000 km. Higgs Boson: 125.35 (GeV) Phi: 1.61803... (125.35 x (10^-1) - 1.61803) x (10^3) = 10,916.97 Circumference of the Moon: 10,916 km. Golden number: 1.618 Golden Angle: 137.5 Earth's equatorial radius: 6,378 Universal Gravitation G = 6.67 x 10^-11 N.m^2/kg^2. (((1.618 ^137.5) / 6,378) / 6.67) x (10^-20) = 12,756.62 Earth’s equatorial diameter: 12,756 km. Phi (1.618) & Pi (3.141) • (3.141^1.618) x (10^3) = 6,371.70 Earth's average radius: 6,371 km. The Euler Number is approximately: 2.71828... Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2. Golden number: 1.618ɸ • (2.71828 ^ 6.67) x 1.618 x 10 = 12,756.23 Earth’s equatorial diameter: 12,756 km. Planck’s constant: 6.63 × 10-34 m2 kg. Circumference of the Moon: 10,916. Gold equation: 1,618 ɸ • (((6.63 ^ (10,916 x 10^-4 )) x 1.618 x (10^3)= 12,756.82 Earth’s equatorial diameter: 12,756 km. Planck's temperature: 1.41679 x 10^32 Kelvin. Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2. Speed of Sound: 340.29 m/s • (1.41679 ^ 6.67) x 340.29 - 1 = 3,474.81 Moon's diameter:: 3,474 km. Cosmic microwave background radiation 2.725 kelvins ,160.4 GHz, Pi: 3.14 Earth's polar radius: 6,357 km. • ((2,725 x 160.4) / 3.14 x (10^4) - (6,357 x 10^-3) = 1,392,000 The diameter of the Sun: 1,392,000 km. Numbers 3, 6 & 9 - Nikola Tesla One Parsec = 206265 AU = 3.26 light-years = 3.086 × 10^13 km. The Numbers: 3, 6 and 9 • ((3^6) x 9) - (3.086 x (10^3)) -1 = 3,474 The Moon's diameter: 3,474 km. Now we will use the diameter of the Moon. Moon's diameter: 3,474 km. • (3.474 + 369 + 1) x (10^2) = 384,400 The term L.D (Lunar Distance) refers to the average distance between the Earth and the Moon, which is 384,400 km. Moon's diameter: 3,474 km. • ((3+6+9) x 3 x 6 x 9) - 9 - 3 + 3,474 = 6,378 Earth's equatorial radius: 6,378 km. Orion: The Connection between Heaven and Earth eBook
5:56 I was once messing around with Phi (because I have an adoring fascination with it) and accidently discovered this and freaked out because I thought it was so cool. P.S. You can also create the Lucas Sequence with Phi^x+(-Phi)^-x. I also figured that one out on my own randomly while messing with Phi. And yes, I'm weird.
I found the lucas sequence by accident once (I only know it's called the lucas sequence now because I recognised the number sequence from this video and checked back to see what I had done). I got the numbers by plotting a graph of: - y = x/2 + 45 = arctan(G^n), where x is in degrees and G is the Golden Ratio. where y when mapped onto a sphere viewed from above appears to give a curve that looks the same as the curve constructed from the golden ratio. so x = 2[ arctan(G^n) - 45] degrees. Then find tan(x) for every G^n I got a value of tan(x) of: - G^0 -> 0 G^1 -> 0.5 G^2 -> 1 x (G - 0.5) G^3 -> 2 G^4 -> 3(G - 0.5) G^5 -> 5.5 G^6 -> 8(G - 0.5) After that I did the following also (*indicates appearance in list above): - G^0 - G^0 = 0 G^0 + G^0 = 2 G^1 - G^-1 = 1 = (2 x *0.5) G^2 + G^-2 = 3 G^3 - G^-3 = 4 = (2 x *2) G^4 + G^-4 = 7 G^5 - G^-5 = 11 = (2 x *5.5) G^6 + G^-6 = 18 so those are the Lucas numbers which fit with 2 times every other number in the list above it (indicated by *) While the fibonacci sequence 1, *1, 2, *3, 5, *8, 13,.. fits with 1 times the remaining numbers (indicated by *) Don't ask me why I did the above? I have no idea. As I say - I had never even come across the Lucas sequence before. Even did a similar thing to get fractions made of prime numbers. Anyway, on with the video...
For sigma-3, you can do it using the fact that 3 < sigma-3 < 4. As a polygon gets increasingly many sides, those sides approach a straight line. Thus, after a certain point, the line connecting 4 consecutive line segments will be more than sigma-3. The line connecting three consecutive line segments will always have length less than 3, so we can just ignore them. Using the incredibly powerful theoretical tool of a calculator, we can come up with a function of the interior angle of the polygon that gives us the distance between the ends of 4 consecutive lines, and we get an angle of about 133.8 degrees. Translating this to the number of sides of a polygon, we get that the theoretical number of edges for this maximal polygon as 7.792 sides. Rounding down, we that such a diagonal length would only be possible in a polygon with less than or equal to 7 sides. Once again using the powerful calculator, we can just try all the possibilities by hand, and get... impossible! This ration doesn't exist as any diagonal of any regular polygon.
Of course by handyou can't do that reasonably, but a simple script can do that, and I did write it. The not-a-diagonal-conjecture holds for any sigma_m with 2
If we created a sequence satisfying the recurrence relation with pi and e it would be wonderful. Also, if you replace 1 by phi^2-phi in the Euler's equation in becomes quite interesting.
OMG IT'S THE GUY that used to be on the pbs space time. i really like you and i hope to see more and more of you Also nice hoodie laces in the first half of the video
The script proved that sigma_3, sigma_4, ..., sigma_250 are not diagonal-length in any regular polygon (with site length =1). It took 10 min to prove that and for sigma_250 is had 4323 polygons to check. Output of the last lines: Proven: sigma_248 is no diagonal-ratio. Number of checked polygons = 4268 Proven: sigma_249 is no diagonal-ratio. Number of checked polygons = 4295 Proven: sigma_250 is no diagonal-ratio. Number of checked polygons = 4323
First, good video on both parts of it. Thanks! (I may have to listen to the last bit a few more times to make sure I properly got it). A small nitpick though, (sorry!) with the proof that the ratios of successive terms in the sequences converge to the golden (or appropriate metallic ratio) You demonstrated that _if_ it does converge, that's what the limit would equal, but, near as I can tell, you omitted the part of the proof that for all starting points it does, in fact, converge, toward anything at all. Only mentioning this because you made it sound like it wasn't just proving "here's what it converges to" but in fact was "here's a proof that it does, in fact, converge".
All true, and I know that. As Tai-Danae said, our issue is a somewhat strict total length constraint, so some stuff ends up getting cut from the script if I can't find a way to say it in very few words without generating more confusion. Sometimes, we have to make an executive decision to get just clear enough that the viewer should be able to fill in any gaps. It's always a tough decision on our part, because on the one hand we don't want to leave *any* logical gaps (I'm fully aware of the one you just brought up), but on the other we have a length constraint. Welcome to my world.
Yeah, I have no problem with that. As I said, I wouldn't have mentioned it at all if the phrasing hadn't made it sound like it was, in fact, a proof of convergence. But again, just a small nitpick, and overall I found the video quite good, the exploring of the metallic numbers was interesting.
Very Interesting and stimulating. I see that ing geometric shapes you have started with the Pentagon and then on to an Octagon. Now, if you carry on with this to a polygon of infinite numbers, you will end up with Pi !!!
Did this series ever cover the fundamentals of topology? You have to explain functor, group homomorphism, composition, inclusion map. And all that was in one sentence. I'm sure it can be built up, but before understanding advanced new concepts we need lots of examples of basic stuff.
the last question, work with regular polygons with number of sides that are equal to numbers appearing in the fibonacci sequences. eventually you get to a circle; and then the get a/b with is the diameter of the the circumference... PI
Woo! Open problems! Even if they've been closed in the annals of some esoteric journal, good on you for asking something original without a bread and butter solution! =D
One thing I noticed right off the bat is that for the first two at least, the number of sides is the value of the radicand in the radical definition of the metallic ratios. So I would assume some ratio in a (regular) 13-gon would be the bronze ratio, some ratio in a regular 20-gon would be sigma 4, and so on. Maybe the answer can be found here? I'll look in to it and report back here if I see a pattern or if I find that it isn't true for all metallic ratios.
Matthew Sylvester, big up Bro. I would like to soppose that's a beautiful coincidence but the probability that sigma showed up two time in the diagonal of an n-regular polygons with n=8 is pretty low. It's reasonable to think there's a pattern to be discovered.
It's possible that the core pattern is misidentified. Maybe the ratio isn't the 1st diagonal for the golden ratio. It's actually the 2nd (it just so happens to also be the 1st but that's dumb luck). And the Silver ratio isn't the 3rd diagonal, it's the 5th (again, dumb luck it had a conjugate). So consider this pattern. Phi(1) = (1 + fib(5))/2 and Phi(2) = (2 + fib(6))/2 and Phi(3) = (3 + fib(7))/2. We can generalize to say Phi(n) = (n + fib(4+n))/2. The syntax here is that fib(n) represents the nth fibonacci number. A ratio shows up in a regular polygon of fib(n) sides between the length of a side and the kth diagonal where k = fib(fib(n-1)). So for n = 5 you have fib(5) = 5 and it's kth diagonal. k = fib( fib (5-1) ) = fib ( fib (4) ) = fib ( 3 ) = 2 For n = 6 you have fib (6) = 8 and it's kth diagonal. k = fib (fib (6-1) ) = fib ( fib (5) ) = fib ( 5 ) = 5 But when you do n = 7 you see why it falls apart For n = 7 and fib (7) = 13 but the kth diagonal is k = fib ( fib (7-1) ) = fib ( fib (6) ) = fib (8) = 21... so the ratio will only work with the 21st diagonal of the polygon with 13 sides and it all falls apart. Note: This generalization (like many of the others) works for triangles n = 4, fib (4) = 3. kth diagonal. k = fib ( fib (4 - 1) ) = fib ( fib (3) ) = fib (2) = 1 and we know that Phi(0) = 1. So this is why Phi(3) and indeed all values greater than 2 fall apart. Going to type up a slightly more formal paper and send it in.
Following the logic of: one side of the pentagon equals the length of the line between corner 1 and 3 dividet by phi AND one side of an octagon equals the length of a line between corner 1 and 4 dividet by phi2 FOR phi3 might be interesting: a 13 sidet geometric form where one side equals the length of a line between corner 1 and 5 dividet by phi3 Why 13 sidet? 1 1 2 3 "5" "8" "13" 3 sides correspond to: x/(x/phi)=0,618033... 2 sides (a line) correspond to: just 0, or division by 0 1 says origin or singularity (that's deep :) ) go check it perhaps? I could do so, but i guess you can calculate much better than me ;)
I personally found that there exist an n-polygon that satisfy the condition (distance between (k+1)-th and first point)/(side length)=sigma_m only if (and probably if) n satisfy a polynomial equation in cos(2pi/n) dipendent from m (but not from sigma_m) of degree 2k. I didn't found the explicit form of the polynomial cause its complexity, but it can be obtained with recurrency probably. Anyway replacing cos(2pi/n) with x in the polynomial one can use something like geogebra to see if there exist a root of the polynomial in (-/2,1), at this point the last thing to check is to see if the equation cos(2pi/n)=x_0 correspond to a an n natural. In the end, i did my little work, but honestly don't seem that i really simplified the problem. For any n significant for the problem, the polynomial don't have roots that can be obtained analitically. Even checking the case n=5 for sigma_1 would be very complex and it can be done only with approximations.
Did anyone end up doing the problem? I'd definitely wack at it, the connection with the pentagon and phi is good motivation. have you examined edge ratios on polyhedra? phi is definitely found on the icosahedron because of the pentagonal faces, but now there are more diagonals so maybe one of those could be a metallic ratio?
I wrote a program to test all the unique diagonals of all the polygons up to 'n' (2048 was as high as I bothered) against sigma up to 'm' (1024 was overkill, but fine, since you can early out if you hit a sigma higher than that diagonal). Even using double precision, and checking if the values were within a margin of error (to account for any precision error), I don't feel like I found any new solutions. Using an error threshold of 5E-7 percent, I got 7 results out of 321100023 pairs checked. However, after 8, the next polygon "found" is the 1016-gon, diagonal 361, sigma 292, 4.82779E-7% off, but this just makes me think it's a precision issue. I'm really doubting that there are any others. It was a fun excercise though, and a good chance to flex my coding skills.
I use to hate math, so please correct me if I 'm saying something wrong. Take a regular octagon A, B, C, D,... With sides of length 1, and its second diagonal AD. Now project B and C on AD and find the points B' and C'. Now AD is the sum of B'C'=BC=1, and 2AB' AB is the diagonal of a square and its length is 1, therefore AB'=AB/sqr2=1/sqr2. Given that AD=2(1/sqr2)+1. Therefore σ2=1/(2(1/sqr2)+1) Am I right?
The ratio of the third diagonal in a tridecagon is approximately sigma 3 but is further away than the analogue of Phi or sigma 2. I think this pattern would continue with (Fibonacci sequence)-gons but each one less accurate than the last
Fun fact: Phi (1.618) is really close to the ratio between miles and kilometers (1.609) which means you can use adjacent Fibonacci numbers to quickly mentally convert between them.
For instance: 89 miles is nearly 144 km (it's actually 143.2), or 21 kilometers is roughly 13 miles (13.05).
You can even shift orders of magnitude to do longer distances! e.g., 210 miles is around 340 km (multiplying 21 and 34 by 10) which is close to the actual answer of 337.96 km.
This should be on top. I can finally understand when in movies they talk with miles
I can't believe I never knew this before. That's awesome! :)
Fuck imperial system
fine addition → my collection
That's cool : )
I love the way this channel give us puzzles which they dont have the answer yet. This is like
"I trust you to go beyond"
And this is simply awesome. I wish our schools had more of that.
Our school gives us shitty education "trusting" we're forced to go beyond. So they indeed fulfill your requirement xD
Wow, there are a lot of mean comments in here. Please don't fall into the trap of simplifying this channel too much because of commenters. Even though I didn't fully understand what was going on in the previous video I really appreciate there being a maths channel that's willing to approach complex topics.
Found this video to be much better than recent ones. I think because it involved somewhat more accesable maths whilst still being suprising and less well known information. I think episodes like this, where you dont try to attempt complex proofs but instead introduce interesting ideas in an intuitive way and give people a direction and material for further reading if they are interested, have the best format. In cases where you want to explain more complex ideas and give long proofs maybe make them more detailed by break them up over episodes and/or look for proofs with good graphical representions (like 3Blue1Brown) perhaps. Mirroring the way PBS spacetime makes videos, with each providing interesting new information whilst building the knowledge base to allow for more complex topics would be good. Not to be too critical tho, you guys do a great job!
I found this video very satis-phi-ing.
KungFuBlitzKrieg I see what u did there
Ha ha ha but I found 4 and 20
Are you sure it wasn’t satis-pi-ing?
Me too!
Nice Pun! Hi-Phive! 😃🖐
Now *this is exactly why I subscribe to 'Infinite Series'!
Cheers.
sparhopper true that, this was my favorite video in a while.
This answered questions I've had for a while but didn't know how to ask. The golden ratio reappears with any seed numbers or if you divide powers of the seed numbers too. This makes it obvious why those patterns emerge. Thanks guys!
You guys are great. Thanks for the education! I'm almost 60 but I never stop learning!
Next observation:
By d_(m,k) > d_(n,k) for m>n and d_(n,k)>d_(n,j) for j < k < ceil(n/2)-1 it is possible to proof (!) that sigma_n is not any diagonal-ratio of any regular polygon (for a given n>2). Indeed that is much easier than in the paper. Now for a given n you can easily proof with a machine that sigma_n is not such a numbers, you have to only calculate very few numbers for each n. I can give the details if anybody is interested.
That does not give us the answer for all n, but it is maybe a way to go.
Brilliant, I'm happy you guys talked more about the gaps in the theory of the last video! Keep it up!
The ratio of his uneven hood drawstrings is equal to the golden ratio.
I was thinking the same except I thought it was closer to the silver ratio... I found it distracting...
Wait what
This is the type of mathematics that I love, please keep doing videos of this kind.
It's nice to see a video about the other metallic ratios but I think the golden ratio is more famous because it's simpler. Especially since the Fibonacci sequence doesn't involve multiplication unlike the other metallic sequences. There are other ways to generalize as well, such as the Tribonacci sequence and its constant (which I often like to call "Tri" but is more commonly referred to as just "the Tribonacci constant".) The Tribonacci rule is adding 3 numbers together instead of 2.
Easy numerical proof (really a proof), that sigma_3 does not occur as a diagonal in any n-regular polygon (with site length = 1):
You have d_(m,k) > d_(n,k) for m>n and d_(n,k)>d_(n,j) for j < k < ceil(n/2)-1
The first n-regular polygon which can produce a diagonal with length sigma_3 = 3.3027756 ... is n = 11 (because for n < 11 the radius is too small).
Next we check, which diagonal in the 11-polygon is the biggest, but smaller than sigma_3. It is the 3rd diagonal
d_{11,3} = 3.228707... < sigma_3
and d_{11,4} = 3.51333
So with the second above inequality the 11-polygon is ruled out. We try the 12-polygon and find:
d_{12,2} < sigma_3 < d_{12,3}
Now see that d_{n,k} < k+1 and together with the first inequality above ALL other regular n-polygons are ruled out. So sigma_3 never occurs as a diagonal-ratio. Q.E.D.
Similar you can easily check any other sigma_m, but of course with higher m you will have more numbers to check, which is a easy task for a script.
At least we can now state the following:
For any given m we can give an algorithm which proofs (pretty fast) that sigma_m does not occur as diagonal-ratio in ANY regular polygon, IF that is true. Of course the same algorithm will find any occurences, like d_{5,1} = sigma_1 and d_{8,2} = sigma_2, too.
I looked up 5, 8 on the Online Encyclopedia of Integer Sequences. The first sequence I found that started with 5, 8 was numbers that are the sum of two successive primes. If this is the pattern for what polygons have the metallic ratios as the diagonal:side ratio, then the next polygon should be the dodecagon. I assume you have probably already tried the doddcagon though.
I know people like shitting on Tae, but I was so excited to see an accessible RUclips channel not focused solely on numerology. Number theory and stuff can be ok, but I can always turn to numberphile for that. Please keep bringing us the super deep stuff she was giving. She's taking feedback very well and fixing her presentation skills, plus the topics are a lot cooler.
I really like the topology videos, even when I can't keep up. Mathologer is pretty good at satisfying anyone's phi, e, or pi needs.
It should be noted that a real number r>1 is a metallic ratio if and only if r * (r - floor(r)) = 1 (equivalent to definition).
So sigma_n -> n with n -> infinity (side note).
Hence you can check for a given n all diagonal-ratios d_j of the regular, if the equation d_j if d_j*(d_j-floor(d_j)) = 1 holds. I checked for all n0 some n, k and m with
d_(n,k) - sigma_m < epsilon
(like you would expect from a somewhat random distribution)
Someone should show this video to the creators of all those other videos where they pretend that Φ is some mystical and sacred number.
Joseph Noonan I loved Matt Parker's Numberphile video where he similarly points out that the Fibonacci Sequence isn't that cool
it is mystical, its even on your face, you have 1 mouth, 1 nose and 2 eyes. Fibonacci!
Not sure if joking, but that order and the choice of those specific organs are arbitrary.
ijfharvey Yeah they were joking. The joke is that finding something that holds the pattern 1, 1, 2 is arbitrarily easy and meaningless
Yes, that's tru
It's all fun and games until Tai-Danae pulls out the big guns.
I know, right? She answered a question for me yesterday about polynomials over finite fields in like 5 seconds. I was like, "whoa...".
Except that pulling out the big guns didn't help me in any way. I am sure she proved SHE understands it...but I never doubted that, and I don't feel any closer. :(
Bottom line: I hope she goes with a bit simpler topic next time. Sometimes (like today) I can fully dig in and fully grok what's going on. Other times, I can understand enough to enjoy the ideas without getting the details. (See: probably 2/3 of previous topics) That last one just lost me and I never got within hailing distance of the basic idea. It made for a much less fun experience. Today's response didn't help. Not because she's not smart (she's obviously brilliant) but because I am not mathematically sophisticated enough to even enjoy the show. I hope she takes folks like me into account when she chooses her next topic.
I liked her last video. The visuals help. When it's all just words like her follow-up in this video it's harder to follow.
Best video yet since the host change
I'm really excited for the math heavy videos coming up. This was a really great, down to earth video tho.
kindlin Agreed.
I liked Diffie-Hellman key exchange more, but that may be the case because I program a bit. It was actually kinda fun to write a program determining whether an Mod-N group is cyclic or not and if so, what numbers are its generators.
Wait is this the first guy from pbs spacetime?
Yep, it's me. Totes back.
Thanks for that bit at the end on Brouwer's theorem, I really appreciate it!
i am sad to not see more episodes of infinite series, but much like the completion of a supertask, the work was done at such an accelerating rate, that an infinite amount of enjoyment was produced in a finite amount of time.
Might be useful to know:
Let distance from 0th vertex of an n-gon to the m th vertex be D(m,n).
Ratio D(a,n)/D(b,n) = sin(aπ/n)/sin(bπ/n).
So ratio of D(a,n) to side length is
sin(aπ/n)/sin(π/n).
For Φ, σ_1, a=2, n=5,
For σ_2, a=3, n=8,
For σ_m, a>m else D(a,n)
sadly the world was just not ready for this video or this channel for that matter. I hope the future generations will stumble upon these relic videos and spawn a new age of enlightenment
First time watching this series... Glad to see you (Gabe) on another PBS show... And I am also happy that you are speaking slower than you did in the SpaceTime videos. Time is short, and we should make the most of it! Keep it going! and going....
Good to see you back, Gabe! Still the best SpaceTime host.
I spent about 6 hours at work today taking a crack at the Copper Ratio and seeing if it fit into any polygon's diagonals. Allow me to take a stab at this as I believe I have proved it to not occur. My approach was to find a formula for, first, the triangles formed by connecting diagonals of any regular polygon. Assuming that the triangle is of sides 1-1-r, then r is 2*sin(theta/2) where theta is the angle between the equal sides. The properties of triangles put an upper limit on r of 2, so we can immediately disqualify every Metallic Ratio save for the Golden Ratio, and substituting that for r and solving for theta easily shows we must have a regular pentagon.
Next I took the possible trapezoids formed from essentially slicing them off a regular polygon. This gave a formula for the diagonal of r = 1 - 2 * cos(theta). Once again, the properties of quadrilaterals put an upper limit of 3 on r, and setting r equal to the Silver Ratio gives us a theta that is the interior angle of an octagon. Similarly, setting r to the Golden Ratio still gives us the interior angle of the pentagon as this is still perfectly valid.
So of course next was the possible pentagons we could form. The equation I found here was r = -4 * cos(theta) * sin(theta/2) after some trigonometric substitution, and this very much works. r cannot be greater than or equal to 4 this time, and testing any polygon below the octagon gives a diagonal formed by a simpler shape that we have already tested. While this formula can be used to find a pentagon that has the Copper Ratio for r, this does not correspond with a theta matching the interior angle of any polygon. From the octagon to the dodecagon, the diagonals formed are roughly 2.613, 2.879, 3.078, 3.229, and 3.346, and we have now surpassed the Copper Ratio and so if it exists it must be in another diagonal we have not seen yet.
Next we have the hexagon with r = 1 - 2*cos(theta) + 2*cos(2*theta). We aren't interested in the diagonals of anything smaller than a decagon this time, and once again, testing up to the 11-gon gives us values for r of 3.236 and 3.513, and once more we miss the Copper Ratio. I also tested to see if the 4th Metallic Ratio could fit here as our limit for r is to be smaller than 5, and tested up to the 16-gon, by which we passed and missed it. For brevity I will exclude the values, they are simple enough to calculate following this formula.
Finally I tested heptagons, and which point my formula gets unwieldy and for some reason when I attempt to simplify it, it stops working, so perhaps I was either tired or have forgotten something essential to trigonometric identities. r = -4*cos(theta)*sin(theta/2) + 2*cos(pi/2 - 5*theta/2). I should be able to rewrite this entirely in terms of sin functions but for whatever reason it just breaks when I try and I do not know why. Anyhow, we are not interested in any diagonals of any shapes smaller than the dodecagon, as we have already tested them. The diagonal given for the dodecagon is r = 3.864, which is already larger than the Copper Ratio.
Now, I believe I have exhausted all the possible hiding spots for the Copper Ratio. My reasoning is that, excuse me if I don't know how to put this into formal mathematical terms, but I have tested every shape that could be sliced off of a regular polygon and when the equations became too unwieldy for me to know how to solve offhand, which was almost immediately, tested the smallest value for our interior angle that could be used that did not also form a smaller shape when sliced off the required polygon, and tested each possible interior angle as we sliced our shape off of the next-sided polygon until it became obvious that we had missed the Copper Ratio, for of course as the interior angles increased so too must the missing side of our shape, approaching the limit for that shape as the interior angles approached infinity. This is enough to prove, I think, that the Copper Ratio cannot be formed using that particular shape. So, we go to the next number of sides, and repeat the process, until we reach a point where the number of sides for a shape sliced off a polygon, and the interior angle required, demands a value larger than the Copper Ratio.
I look forward to someone doing this far more succinctly than I have. Also, I did test to see if the heptagons could produce the 4th or 5th Metallic Ratio. They surpass the 4th at the 14-gon and surpass the 5th at the 21-gon, and the limit for r is 6 so no larger one can appear. I don't know to what degree I'd have to go to see if the 4th Metallic Ratio does not appear. I similarly look forward to someone showing whether this behavior of appearing in the diagonals of regular polygons is limited to the Golden and Silver Ratios, personally, I suspect but cannot prove that this is the case.
Obviously this brute force method is inelegant, but personally I am more proud of these formulas for these almost-regular polygons. Additionally, checking wolfram alpha, and indeed for a pentagon of these qualities to have the Copper Ratio as its missing side, it would need an angle of x ~ 4 (pi * n + 0.650013) which yields an interior angle of 2.600052 radians, which does not correspond to any regular polygon. I come back to this particular case merely to make certain that the pentagon did in fact miss it, as some theories here seem to believe it would not.
Re the heptagon, the polynomial for the ratio between two of the diagonals is given as x^3 =x^2 + 2x +1 = 0 ruclips.net/video/ciXwHMBUMgg/видео.html frame 6:44 My maths isn't good enough to know if that can ever be re-expressed as one of the metallic ratios, but surely they aren't the only game in town anyway. Maybe we should go beyond them too.
It's been 30 months, or 2 and a half years, that I don't touch deeply any mathematics, since high school.
I was among the best in class because I learn easily, but it's depressing how months ago I tried to watch this video, my first on the channel, but I couldn't keep up with it, and saved it for later. Some days ago I watched the numberphile video on this, okay, not bad, and now I came back here.
Geez, I understood most, but superficially, barely understanding what he meant with it, having to pause. Except some things in the end that I didn't even get at all. It's just sad. Some weeks from now I'll enter college for a science degree, so it gets even more embarrassing and depressing.
But great channel, like as always from PBS. I watch PBS Space Time sometimes and it's awesome. You can't just live by very superficial educational channels. Thanks.
Wow, cool to see Gabe back doing videos since he left Spacetime
I have indeed found some "equally cool analogs in the rest of the metallic family", though regretfully not polygonal ones yet.
Each number in each metallic ratio formula, such as (2+sqrt8)/2 for the Silver, contains a term in, or derivable from, the sequence associated with that formula. But we're talking about the cooler "Lucas" version of that sequence. You probably know that what distinguishes the Lucas sequence from the standard Fibonacci is that while they share the same Golden ratio constant 1.618..., the terms in the Lucas approximate much more closely to 1.618...^n where n is the position or index. For example the 10th term in Lucas is 123, and 1.618..^10 ~ 122.996..., whereas in Fibonacci the 10th term is 55. Another interesting difference is that if you take the product of any pair of terms separated by one, such as 34 and 89, in Fibonacci, the absolute difference with the square of the term between is always 1. So for example (34*89)-(55^2)=1, but in Lucas it's always 5, the square root of which crops up in the formula for the Golden ratio, (1+sqrt5)/2. Lastly Lucas begins with the two terms 2 1, which also occur in the denominator and numerator respectively of the golden ratio formula.
All this seems to hold for the other metallic ratios as listed in that table too, something which may have escaped observation if nobody has thought of "Lucasating" the associated sequences. Try it with the Silver. Begin it with the two terms 2 2 and continue, using its recurrence a(n) = 2*a(n-1) + a(n-2).
I noticed a pattern between two methods of calculating the ratio including your open question: σ_n = (n+sqrt(n^2 + 4))/2 = the ratio of the nth diagonal of the regular (n^2 + 4)th polygon to that polygon's side. In the video you showed this works for σ_1 and σ_2. I ran through some code and found that this does not generalize to n>2, but it does nonetheless approximate the true ratios rather well, maxing out at 8.0689% error for σ_6.
I was also wondering if the properties would extend to the "trivial" case, σ_0 = 1, and, checking back through the video, I think most if not all of them do!
Fascinating... so for sigma_3, the answer is approximately the 3rd diagonal of a triskaidecagon, for sigma_4, the 4th diagonal of an icosagon, etc...
Ok so i tired every possible arc of every polygon up the 1000-gon and saw if it was equale to any silver ratio and the result is negartive appart from the 2 already known exemples giving by Gab. But more regourously speaking, the problem can be stated in this equation that i couldn't solve : sin(pi*i/m)/sin(pi/m) = (n + sqrt(n^2+4))/2 with m,n,i being natural numbers and i
a little python script if you wanna play with it :
import numpy as np
silver_ratios, p = [ (n+(n**2+4)**(1/2))/2 for n in range(1,320) ], [ ]
def arc_ratios_for_ngone(n):
ia = np.pi/n
return [ np.sin(i*ia)/np.sin(ia) for i in range(2,n//2+1) ]
for i in range(3,1000):
A.extend(arc_ratios_for_ngone(i))
for k in A :
for j in silver_ratios:
if np.abs(j-k) < 1e-6 :
print(j)
My first thought is this:
A pentagon is a line that turns 360/n degrees. It does this n times. n = 5 for a pentagon. You are drawing a diagonal by 'skipping' a turn. This gives you sigma-1.
An octagon is n=8. You are skipping two turns. This gives you sigma-2.
I thought maybe this could generalize: a 13-gon where n=13. You skip three turns. This gives you sigma-3.
I wrote a program to test this (and generalized for sigma-4 in a 21-gon with 4 turns and sigma-5) and... uh, I'm not sure if the results are correct. The measurements look good, but it was a pretty sloppy empirical test (and I have no reason why this would be theoretically sound math).
I can double-check but, before I do; has anyone else tried this? Is it a dead-end? It looks promising, but I'd rather not chase this rabbit into a brick wall.
He said in other comments that he tested the 13-gon already, and it doesn't work. It was actually the first one he tested.
Somebody else posted a link to a paper claiming to prove that sigma(3) doesn't continue the pattern. I haven't had time to read it yet (it's late in my part of the world), but you may want to check that out.
Yeah, I see that here. Reading it now! Thanks for the heads up!
Darn. I was going to check 13. Oh well.
Yes. This was my intuition as well. I even started up a theoretical process to prove it, but then, alas, I read the aforementioned article. The article gives a correct proof, by the way, that the Bronze Ratio does not continue in this pattern, nor any other such n-gon/diagonal type relationship for that matter. I wonder, however, if some sort of variation of this pattern persists in higher dimensions, i.e. Platonic solids or beyond, as Leach pointed out ... .
For anyone that still cares, this is a repost from a previous thread:
"After some work it turns out that the following equation has to be true to get the right ratio:
sin[ b*pi / n ] / sin[ a*pi / n ] = s + (4+s^2)^.5/2
where "a" is the first point along the n-gon, "b" is the second point along the n-gon, "n" are the amount of sides, and "s" is the sigma used. (test this with the pentagon and octagon if you want.) I could supply the proof for this, but it's really just number manipulation.
There may be a taylor formula or lesser known trig identities that can be used to break this into fractions that can be tested... but I've spent a half day on this already and I'm done for now."
I played around with it a bit more since then, but to no avail... maybe someone else can solve for a,b, n and s
I had a lot of fun working out all the math to build up a spreadsheet to check out to the 12th Phi and polygons up to 70 sides. Now that I have a generalized case for everything, I could throw this all into a python program and run it to infinity...but I see someone else has already done that out to ludicrous values. Oh well, I did all the parts that would teach me stuff anyway. Great topic and well-explained! Thanks!
Old Spacetime guy is back! Awesome!
OMG this explains actually the hint to Sigma and Phi's relationship in ZTD.
You're welcome ;)
OH MAN. When I was in High School I explored these ratios in quite a bit of depth. I should have my notes lying around somewhere. I'm afraid I didn't look at polygons, but I remember this was so much fun to look at. I'm going to go dig up my notes, might find something interesting.
My fav thing about the Fibonacci Sequence is that if you take the digital roots of each number in the sequence it actually produces a simple repeating pattern that is 24 steps in total. And what's even cooler is that if you take the first 12 over the second 12, they all add to 9s.
So Digital Root (for anyone who doesn't know what that is) is summing the digits of a multi digit number until they are a single digit.
Ex. 15 = 1+5
248 = 2+4+8 = 14 = 1+4 = 5) .
So now...
1-1-2-3-5-8-13-21-34-55-89-144
8= 8-7-6-5-4-3-2-1-9
7= 7-5-3-1-8-6-4-2-9
6= 6-3-9-6-3-9-6-3-9
5= 5-1-6-2-7-3-8-4-9
Hey man, I miss you on PBS Space Time... Where have you been?
whoa whoa whoa... WHOA!!!!, its this guy again!!!!!!... havent seen him forever!!
i hope he sticks around again, but dosnt replace the guy who replaced him on Space TIme....lets keep them both!
I love Matt over at Space Time. Honestly, he and the writing staff over there are really knocking it out of the park. But damn it, I missed Gabe. It's truly great to see him again.
i think this could be a solution:
Golden ratio =(1+sqrt(5))/2 => satisfied by five sided figure (1st diagonal) (sigma 1)
Silver ratio =(1+sqrt(8))/2 satisfied by 8 sided figure (second diagonal) (sigma 2)
so for any sigma n:
polygon (n^2 +4 sides)
nth diagonal : side = sigma n
Where did I see you before? SpaceTime?
Yes
The ratios (algebraic ratios of linear functions) lead to very interesting mathematics ; including elliptic curves. These ratios are a generalization of the "Golden Ratio". I am writing a paper on the subject.
How did it go?
I really wish there was more visual demonstrations or examples of the ratio use , great work though, keep up the good work guys ♡
Greeting from #Baghdad
Thumb up for the detailed respond. And even more so for the last sentence.
pentagon (5) then octogon (8), maybe it would follow the Fibonacci sequences, and so be the ratio of the height of a 13-gon to it's side ? Also, it could go one dimension higher so maybe we should consider looking at solids..
What I have done is written a formula given two inputs that could POTENTIALLY generate metallic ratios. it uses the same logic as the polygons, given x number of vertexes on a circle, what is the ratio of the chord of two nonadjacent vertexes over the chord of two adjacent vertexes. This can be expressed simply as sin(C*pi/S)/sin(pi/S). where C is the nth vertex away (2 for sigma 2, 3 for sigma 3) and S is the number of sides to a polygoon
First thing I noticed: C must be at least the next whole number larger than sigma-n because it is increasing with increasing numbers of S and the limit as S goes to infinity is C
Second thing: for sigma-1 and sigma 2, when C=2, and S is half the number of sides their polygon has, the result is sigma-n minus one.
Third thing: You can input any two numbers where S=C+3 and find a solution that satisfies sin(Cpi/S)/sin(pi/S)-sin(4pi/S)/sin(2pi/S)=1, but only two of those equations are metallic ratios, because those are the only two ratios whose values are less than 3, and the max value for sin(Cpi/S)/sin(pi/S) with the constraint S=C+3 is 3.
Conclusion: For a ratio to be metallic, it must fit the rules discussed in the video, but for that ratio to be expressed in terms of a polygon, the polygon MUST have 3 more sides than the nth vertex, which means the highest value the ratio can be is 3, and all values for sigma-n where n>2 are greater than 3, therefore no other metallic ratio can be expressed in terms of two lengths of a polygon.
10:01 I can't believe PBS gets away with monetization RUclips channel, has on camera sponsorship and is not for profit 501c3. Great content. Transparency answers where the #moneygoes 💰
I suspect that if there are no further metallic numbers with corresponding regular polygon representations, it may be to do with the fact that starting with the bronze ratio, they are all greater than pi
I often to use "sexy" to describe interesting mathematical concepts and theorems. Now that the new guy has done it, he's officially accepted.
The silver ratio is very closely related to the ratio between the sides of A-series paper (and B- and C-series, though B-series isn't used all that much and C-series is mostly used for envelopes), which is the standard for paper sizes in much of the world.
Hmmm.. 10 minutes of content for 5 minutes of ads. The golden ratio of youtube.
It's 9 minutes of video on new topic, 1 minute add for Brilliant and 5 minutes of explaining questions to previous video on 'Brouwer's fixed point theorem'.
3:41 Fun fact: This series produces an infinite list of numbers from the most irrational to the least as all continued fractions with period 1 with coefficients 1 to infinity are covered. As you progress through it, numbers can be approximated faster and faster with rationals.
We can generalize the hypothesis in considering not only the ratio between a diagonal and the side of the polygon but ratios between all diagonals. We find then many occurences of sigma_1 and sigma_2 but no sigma_3 or higher again.
Let d_{n,d1,d2} = Quotient of the d1-diagonal and the d2-diagonal in the n-polygon. The 0-diagonal is the side-length of the n-polygon ( = 1).
Output of the searching:
d_{5,1,0} = sigma_1
d_{5,2,0} = sigma_1
d_{8,2,0} = sigma_2
d_{10,3,1} = sigma_1
d_{15,5,2} = sigma_1
d_{16,5,1} = sigma_2
d_{20,7,3} = sigma_1
d_{24,8,2} = sigma_2
d_{30,4,2} = sigma_1
d_{30,8,4} = sigma_1
d_{30,11,5} = sigma_1
d_{32,11,3} = sigma_2
d_{40,14,4} = sigma_2
d_{40,15,7} = sigma_1
d_{48,17,5} = sigma_2
d_{60,9,5} = sigma_1
d_{60,17,9} = sigma_1
d_{60,23,11} = sigma_1
d_{64,23,7} = sigma_2
d_{80,29,9} = sigma_2
d_{80,31,15} = sigma_1
d_{85,33,16} = sigma_1
d_{96,35,11} = sigma_2
d_{120,19,11} = sigma_1
d_{120,35,19} = sigma_1
d_{120,44,14} = sigma_2
d_{120,47,23} = sigma_1
d_{128,47,15} = sigma_2
d_{136,50,16} = sigma_2
d_{160,59,19} = sigma_2
d_{160,63,31} = sigma_1
d_{170,67,33} = sigma_1
d_{192,71,23} = sigma_2
d_{240,39,23} = sigma_1
d_{240,71,39} = sigma_1
d_{240,89,29} = sigma_2
d_{240,95,47} = sigma_1
d_{255,101,50} = sigma_1
d_{256,95,31} = sigma_2
d_{272,101,33} = sigma_2
d_{320,119,39} = sigma_2
d_{320,127,63} = sigma_1
d_{340,135,67} = sigma_1
d_{384,143,47} = sigma_2
d_{408,152,50} = sigma_2
d_{480,79,47} = sigma_1
d_{480,143,79} = sigma_1
d_{480,179,59} = sigma_2
d_{480,191,95} = sigma_1
If it helps you really just need to solve the following:
sin[ b*pi / n ] / sin[ a*pi / n ] = s + (4+s^2)^.5/2
where a is the smaller length, b is the larger length, n is the sides of the n-gon and s is the sigma index.
for example on your first entry, d_{5,1,0} = sigma_1, a = (0)+1 b = (1)+1, n = (5) and s = (1)
How do you think I found those values? :)
Fair enough :)
I've made some progress
The side of a regular n-gon inscribed in a circle of radius 1 is:
>b=2sin(pi/n)
the diagonal of a regular n-gon inscribed in a circle of radius 1 skipping m sides is:
>a=2sin(m*pi/n)
Thus, if the metallic ratios appear in any n-gon, they will be:
>f=sin(m*pi/n)/sin(pi/n)
When this happens this expression should be true:
>f-/sigma_k=0
try to solve for m, we get:
> m = (n/pi)*arcsin[/sigma_k*sin(pi/n)]
Now, I suspect that f is as close as possible to a metallic ratio if and only if: n=k^2+4
if we put this in the expression for m then whenever we get whole numbers for m (as happens with k=1 and 2) then we will know that that n-gon has a metallic ratio
Thus it is only necessary to prove that the closest ratio will happen at n=k^2+4 to solve this open problem
I miss this channel
Given a polynomial y^n = a^yn-1 + by^n-2 + ... + z, you can create an associated series f(n+m) = af(n+m-1) + bf(n+m-2) + ... + zf(m). If the ratio of successive terms in the series converges, it converges to a root of the original polynomial. As an example, if the polynomial is x^2 = x + 1, the associated series is f(n+2) = f(n+1) + f(n) (The standard Fibonacci rule) and the ratios of successive terms converge to a root of x^2 = x + 1 which is (1+ SQRT(5))/2.
Gabe is *such* a great presenter
Find a/b = phi_n for an s-sided regular polygon with side length b, and m-th diagonal length a. Using a side length of 1 (b = 1), you then have a/1 = a = phi_n.
Therefore, the question can be restated as: given an s-sided regular polygon with side length 1, what are the lengths of the diagonals (the line between two corner points on the polygon). Then, which of those diagonal length are equal to a metallic ratio phi_n?
If you can find a formula or method for calculating those diagonal lengths, you can then check which ones match a given phi_n.
I wrote a program to run through lots of regular polygons for each possible diagonal checked against each metallic ratio from 1 to 50. I've found some very close but none that are exact yet. Tomorrow I am going to try to use the function I found to find the diagonal length to prove or disprove this overall since I;m not finding any solutions.
Love this video! Thank you for bringing this information to the everyone!
Small note: simple continued fractions (all the numerators are 1) can be unique, but the video doesn't give enough conditions. 1) Finite continued fractions of non-integers never end in 1; 2) no terms except possibly the first term in the continue fraction are 0; and, 3) either all terms are negative, or only the first term is negative. With these three conditions, simple continued fractions are indeed unique.
(1) is something like the zero-point-nine repeating problem for decimal numbers, (2) is like the leading- or trailing-zero problem for decimal numbers, and (3) is doesn't really have an analog for numbers as we usually write them.
Well-said, and keep including comments like this. Almost any time we omit details like this, it's either b/c of "flow" in the script or b/c we have pretty tight length constraints on the videos. We'd prefer to cover every logical base, obviously, but we feel less bad about it since we consistently see commenters (like you) fill in the gaps for those who may be unable to do so for themselves. So please, keep comments like this coming! We rely on them.
My man! How have you been. We've missed you at space time. Hella cool to see you.
very nice part about the metallic ratio constants family. if n=3 doesn't work with planar polygons, perhaps in a space with a different metric it works. maybe there have to be more degrees of freedom
such as diagonals of polyhedra. thanks for the video again, and there is another real nice video recently by mathologer. "way beyond the golden ratio.." ruclips.net/video/cCXRUHUgvLI/видео.html
The association of the main numbers in the field of mathematics with each other, reflects numerical sequences that correspond to the dimensions of the Earth, the Moon, and the Sun in the unit of measurement in meters, which is: 1' (second) / 299792458 m/s (speed of light in a vacuum).
Ramanujan number: 1,729
Earth's equatorial radius: 6,378 km.
Golden number: 1.61803...
• (1,729 x 6,378 x (10^-3)) ^1.61803 x (10^-3) = 3,474.18
Moon's diameter: 3,474 km.
Ramanujan number: 1,729
Speed of light: 299,792,458 m/s
Earth's Equatorial Diameter: 12,756 km. Earth's Equatorial Radius: 6,378 km.
• (1,729 x 299,792,458) / 12,756 / 6,378) = 6,371
Earth's average radius: 6,371 km.
The Cubit
The cubit = Pi - phi^2 = 0.5236
Lunar distance: 384,400 km.
(0.5236 x (10^6) - 384,400) x 10 = 1,392,000
Sun´s diameter: 1,392,000 km.
Higgs Boson: 125.35 (GeV)
Phi: 1.61803...
(125.35 x (10^-1) - 1.61803) x (10^3) = 10,916.97
Circumference of the Moon: 10,916 km.
Golden number: 1.618
Golden Angle: 137.5
Earth's equatorial radius: 6,378
Universal Gravitation G = 6.67 x 10^-11 N.m^2/kg^2.
(((1.618 ^137.5) / 6,378) / 6.67) x (10^-20) = 12,756.62
Earth’s equatorial diameter: 12,756 km.
Phi (1.618) & Pi (3.141)
• (3.141^1.618) x (10^3) = 6,371.70
Earth's average radius: 6,371 km.
The Euler Number is approximately: 2.71828...
Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2. Golden number: 1.618ɸ
• (2.71828 ^ 6.67) x 1.618 x 10 = 12,756.23
Earth’s equatorial diameter: 12,756 km.
Planck’s constant: 6.63 × 10-34 m2 kg.
Circumference of the Moon: 10,916.
Gold equation: 1,618 ɸ
• (((6.63 ^ (10,916 x 10^-4 )) x 1.618 x (10^3)= 12,756.82
Earth’s equatorial diameter: 12,756 km.
Planck's temperature: 1.41679 x 10^32 Kelvin.
Newton’s law of gravitation: G = 6.67 x 10^-11 N.m^2/kg^2.
Speed of Sound: 340.29 m/s
• (1.41679 ^ 6.67) x 340.29 - 1 = 3,474.81
Moon's diameter:: 3,474 km.
Cosmic microwave background radiation
2.725 kelvins ,160.4 GHz,
Pi: 3.14
Earth's polar radius: 6,357 km.
• ((2,725 x 160.4) / 3.14 x (10^4) - (6,357 x 10^-3) = 1,392,000
The diameter of the Sun: 1,392,000 km.
Numbers 3, 6 & 9 - Nikola Tesla
One Parsec = 206265 AU = 3.26 light-years = 3.086 × 10^13 km.
The Numbers: 3, 6 and 9
• ((3^6) x 9) - (3.086 x (10^3)) -1 = 3,474
The Moon's diameter: 3,474 km.
Now we will use the diameter of the Moon.
Moon's diameter: 3,474 km.
• (3.474 + 369 + 1) x (10^2) = 384,400
The term L.D (Lunar Distance) refers to the average distance between the Earth and the Moon, which is 384,400 km.
Moon's diameter: 3,474 km.
• ((3+6+9) x 3 x 6 x 9) - 9 - 3 + 3,474 = 6,378
Earth's equatorial radius: 6,378 km.
Orion: The Connection between Heaven and Earth eBook
Additionally, irrational numbers' continued fractions are periodic for algebraic numbers, and non periodic for transcendental numbers.
might have been able to pay attention if his hood strings were equally dangling
Hey you're back!! Nice!!
5:56 I was once messing around with Phi (because I have an adoring fascination with it) and accidently discovered this and freaked out because I thought it was so cool.
P.S. You can also create the Lucas Sequence with Phi^x+(-Phi)^-x. I also figured that one out on my own randomly while messing with Phi. And yes, I'm weird.
I found the lucas sequence by accident once (I only know it's called the lucas sequence now because I recognised the number sequence from this video and checked back to see what I had done). I got the numbers by plotting a graph of: -
y = x/2 + 45 = arctan(G^n), where x is in degrees and G is the Golden Ratio. where y when mapped onto a sphere viewed from above appears to give a curve that looks the same as the curve constructed from the golden ratio.
so x = 2[ arctan(G^n) - 45] degrees. Then find tan(x)
for every G^n I got a value of tan(x) of: -
G^0 -> 0
G^1 -> 0.5
G^2 -> 1 x (G - 0.5)
G^3 -> 2
G^4 -> 3(G - 0.5)
G^5 -> 5.5
G^6 -> 8(G - 0.5)
After that I did the following also (*indicates appearance in list above): -
G^0 - G^0 = 0
G^0 + G^0 = 2
G^1 - G^-1 = 1 = (2 x *0.5)
G^2 + G^-2 = 3
G^3 - G^-3 = 4 = (2 x *2)
G^4 + G^-4 = 7
G^5 - G^-5 = 11 = (2 x *5.5)
G^6 + G^-6 = 18
so those are the Lucas numbers which fit with 2 times every other number in the list above it (indicated by *)
While the fibonacci sequence 1, *1, 2, *3, 5, *8, 13,.. fits with 1 times the remaining numbers (indicated by *)
Don't ask me why I did the above? I have no idea. As I say - I had never even come across the Lucas sequence before.
Even did a similar thing to get fractions made of prime numbers.
Anyway, on with the video...
Can there be imaginary metallic ratios? Like sigma i where it is defined as i plus one over i plus one over i and so on upto infinity? And so on?
Yes, but it forms a loop that repeats after 12 iterations
For sigma-3, you can do it using the fact that 3 < sigma-3 < 4. As a polygon gets increasingly many sides, those sides approach a straight line. Thus, after a certain point, the line connecting 4 consecutive line segments will be more than sigma-3. The line connecting three consecutive line segments will always have length less than 3, so we can just ignore them.
Using the incredibly powerful theoretical tool of a calculator, we can come up with a function of the interior angle of the polygon that gives us the distance between the ends of 4 consecutive lines, and we get an angle of about 133.8 degrees. Translating this to the number of sides of a polygon, we get that the theoretical number of edges for this maximal polygon as 7.792 sides. Rounding down, we that such a diagonal length would only be possible in a polygon with less than or equal to 7 sides.
Once again using the powerful calculator, we can just try all the possibilities by hand, and get... impossible! This ration doesn't exist as any diagonal of any regular polygon.
Of course by handyou can't do that reasonably, but a simple script can do that, and I did write it. The not-a-diagonal-conjecture holds for any sigma_m with 2
If we created a sequence satisfying the recurrence relation with pi and e it would be wonderful. Also, if you replace 1 by phi^2-phi in the Euler's equation in becomes quite interesting.
OMG IT'S THE GUY that used to be on the pbs space time.
i really like you and i hope to see more and more of you
Also nice hoodie laces in the first half of the video
The script proved that sigma_3, sigma_4, ..., sigma_250 are not diagonal-length in any regular polygon (with site length =1).
It took 10 min to prove that and for sigma_250 is had 4323 polygons to check.
Output of the last lines:
Proven: sigma_248 is no diagonal-ratio.
Number of checked polygons = 4268
Proven: sigma_249 is no diagonal-ratio.
Number of checked polygons = 4295
Proven: sigma_250 is no diagonal-ratio.
Number of checked polygons = 4323
First, good video on both parts of it. Thanks! (I may have to listen to the last bit a few more times to make sure I properly got it).
A small nitpick though, (sorry!) with the proof that the ratios of successive terms in the sequences converge to the golden (or appropriate metallic ratio)
You demonstrated that _if_ it does converge, that's what the limit would equal, but, near as I can tell, you omitted the part of the proof that for all starting points it does, in fact, converge, toward anything at all. Only mentioning this because you made it sound like it wasn't just proving "here's what it converges to" but in fact was "here's a proof that it does, in fact, converge".
All true, and I know that. As Tai-Danae said, our issue is a somewhat strict total length constraint, so some stuff ends up getting cut from the script if I can't find a way to say it in very few words without generating more confusion. Sometimes, we have to make an executive decision to get just clear enough that the viewer should be able to fill in any gaps. It's always a tough decision on our part, because on the one hand we don't want to leave *any* logical gaps (I'm fully aware of the one you just brought up), but on the other we have a length constraint. Welcome to my world.
Yeah, I have no problem with that. As I said, I wouldn't have mentioned it at all if the phrasing hadn't made it sound like it was, in fact, a proof of convergence. But again, just a small nitpick, and overall I found the video quite good, the exploring of the metallic numbers was interesting.
GABE! Miss you on Spacetime
Very Interesting and stimulating. I see that ing geometric shapes you have started with the Pentagon and then on to an Octagon. Now, if you carry on with this to a polygon of infinite numbers, you will end up with Pi !!!
Did this series ever cover the fundamentals of topology? You have to explain functor, group homomorphism, composition, inclusion map. And all that was in one sentence. I'm sure it can be built up, but before understanding advanced new concepts we need lots of examples of basic stuff.
The Best Explanation
I miss you on PBS Spacetime
the last question, work with regular polygons with number of sides that are equal to numbers appearing in the fibonacci sequences. eventually you get to a circle; and then the get a/b with is the diameter of the the circumference... PI
I didn't think about this very long so I might be wrong
The polygons have ratios that converge to integer n. So with infinities involve yes. t^2-nt-1=0
This. Video. Kicks. ASS
My brain just poured out my left earhole, but I learned a lot!!
Woo! Open problems! Even if they've been closed in the annals of some esoteric journal, good on you for asking something original without a bread and butter solution! =D
Silver ratio ... sounds interesting
One thing I noticed right off the bat is that for the first two at least, the number of sides is the value of the radicand in the radical definition of the metallic ratios. So I would assume some ratio in a (regular) 13-gon would be the bronze ratio, some ratio in a regular 20-gon would be sigma 4, and so on. Maybe the answer can be found here? I'll look in to it and report back here if I see a pattern or if I find that it isn't true for all metallic ratios.
Matthew Sylvester, big up Bro. I would like to soppose that's a beautiful coincidence but the probability that sigma showed up two time in the diagonal of an n-regular polygons with n=8 is pretty low. It's reasonable to think there's a pattern to be discovered.
It's possible that the core pattern is misidentified. Maybe the ratio isn't the 1st diagonal for the golden ratio. It's actually the 2nd (it just so happens to also be the 1st but that's dumb luck). And the Silver ratio isn't the 3rd diagonal, it's the 5th (again, dumb luck it had a conjugate).
So consider this pattern.
Phi(1) = (1 + fib(5))/2 and Phi(2) = (2 + fib(6))/2 and Phi(3) = (3 + fib(7))/2. We can generalize to say Phi(n) = (n + fib(4+n))/2. The syntax here is that fib(n) represents the nth fibonacci number.
A ratio shows up in a regular polygon of fib(n) sides between the length of a side and the kth diagonal where k = fib(fib(n-1)).
So for n = 5 you have fib(5) = 5 and it's kth diagonal. k = fib( fib (5-1) ) = fib ( fib (4) ) = fib ( 3 ) = 2
For n = 6 you have fib (6) = 8 and it's kth diagonal. k = fib (fib (6-1) ) = fib ( fib (5) ) = fib ( 5 ) = 5
But when you do n = 7 you see why it falls apart
For n = 7 and fib (7) = 13 but the kth diagonal is k = fib ( fib (7-1) ) = fib ( fib (6) ) = fib (8) = 21... so the ratio will only work with the 21st diagonal of the polygon with 13 sides and it all falls apart.
Note: This generalization (like many of the others) works for triangles
n = 4, fib (4) = 3. kth diagonal. k = fib ( fib (4 - 1) ) = fib ( fib (3) ) = fib (2) = 1 and we know that Phi(0) = 1. So this is why Phi(3) and indeed all values greater than 2 fall apart. Going to type up a slightly more formal paper and send it in.
Following the logic of:
one side of the pentagon equals the length of the line between corner 1 and 3 dividet by phi
AND
one side of an octagon equals the length of a line between corner 1 and 4 dividet by phi2
FOR phi3 might be interesting:
a 13 sidet geometric form where one side equals the length of a line between corner 1 and 5 dividet by phi3
Why 13 sidet? 1 1 2 3 "5" "8" "13"
3 sides correspond to: x/(x/phi)=0,618033...
2 sides (a line) correspond to: just 0, or division by 0
1 says origin or singularity (that's deep :) )
go check it perhaps? I could do so, but i guess you can calculate much better than me ;)
I miss this channel.
will you make a sort of periodic table of metallic ratios? i what about the titanium ratio? the lead ratio? bismuth?
I personally found that there exist an n-polygon that satisfy the condition (distance between (k+1)-th and first point)/(side length)=sigma_m only if (and probably if) n satisfy a polynomial equation in cos(2pi/n) dipendent from m (but not from sigma_m) of degree 2k. I didn't found the explicit form of the polynomial cause its complexity, but it can be obtained with recurrency probably. Anyway replacing cos(2pi/n) with x in the polynomial one can use something like geogebra to see if there exist a root of the polynomial in (-/2,1), at this point the last thing to check is to see if the equation cos(2pi/n)=x_0 correspond to a an n natural. In the end, i did my little work, but honestly don't seem that i really simplified the problem. For any n significant for the problem, the polynomial don't have roots that can be obtained analitically. Even checking the case n=5 for sigma_1 would be very complex and it can be done only with approximations.
Did anyone end up doing the problem? I'd definitely wack at it, the connection with the pentagon and phi is good motivation. have you examined edge ratios on polyhedra? phi is definitely found on the icosahedron because of the pentagonal faces, but now there are more diagonals so maybe one of those could be a metallic ratio?
Great video, I really liked it!
I wrote a program to test all the unique diagonals of all the polygons up to 'n' (2048 was as high as I bothered) against sigma up to 'm' (1024 was overkill, but fine, since you can early out if you hit a sigma higher than that diagonal). Even using double precision, and checking if the values were within a margin of error (to account for any precision error), I don't feel like I found any new solutions.
Using an error threshold of 5E-7 percent, I got 7 results out of 321100023 pairs checked. However, after 8, the next polygon "found" is the 1016-gon, diagonal 361, sigma 292, 4.82779E-7% off, but this just makes me think it's a precision issue.
I'm really doubting that there are any others. It was a fun excercise though, and a good chance to flex my coding skills.
I use to hate math, so please correct me if I 'm saying something wrong.
Take a regular octagon A, B, C, D,... With sides of length 1,
and its second diagonal AD.
Now project B and C on AD and find the points B' and C'.
Now AD is the sum of B'C'=BC=1, and 2AB'
AB is the diagonal of a square and its length is 1, therefore AB'=AB/sqr2=1/sqr2.
Given that AD=2(1/sqr2)+1.
Therefore σ2=1/(2(1/sqr2)+1)
Am I right?
The ratio of the third diagonal in a tridecagon is approximately sigma 3 but is further away than the analogue of Phi or sigma 2. I think this pattern would continue with (Fibonacci sequence)-gons but each one less accurate than the last