A couple of more fun/interesting bits and pieces that did not get mentioned in the video and/or popped up in the comments so far (I'll update these and also collect them in the description of the video until I hit the word limit :) 1. Nice log troll: log(r+s)=log(r)+log(s) 2. T-shirt design is called Fibonightmare. 3. e-mail from Marty commenting on my 2 × 2 = 2 + 2 remark at the beginning: When I was doing my PhD, I read a paper by my supervisor, Brian White. It was on “surfaces mod 4”. So, it was a way to make surfaces multiple valued, turned the world of them into a ring, and mod 4 meant the obvious thing: four copies of a surface summed to 0. Anyway, I read the paper, and was trying to figure out why his theorem worked specifically for mod 4. And I asked: “Does it boil down to 2 x 2 = 2 + 2”? “Yep, that’s it." link.springer.com/article/10.1007/BF01403190 4. (1+i)(1-i)=(1+i)+(1-i) 5. The locus of points with xy=x+y is the hyperbola xy=1 shifted up and right 1. 6. B(m,n) = A(m+1,n-1), in other words the r component grid is also just a translated s component grid. And so r^m s^n = A(m,n)s+A(m+1,n-1)r+A(m,n-1). Another super nice and super symmetric way of of summarising the relationships between A(m,n), B(m,n), and C(m,n): A(m,n)=C(m,n+1), B(m,n)=C(m+1,n), C(m,n)=(1*(r/1)^m*(s/1)^n+r²(-s/r)^m*(1/r)^n+s²(1/s)^m*(-r/s)^n)/(1+r²+s²) The same viewer who pointed this out also noted: When I am working with the heptagon ratios, I prefer using the three side lengths of a heptagon with the smallest side negated: a=2sin(τ/7), b=2sin(2τ/7), c=2sin(4τ/7) (1 : r : s = sin(π/7) : sin(2π/7) : sin(4π/7)) Using these values, any polynomial f(x,y,z) such that f(a,b,c)=0 also has f(b,c,a)=0 and f(c,a,b)=0. It also has the property that ab+ca+bc=a+b+c+abc=a²+b²+c²-7=0 7. r³ = r² + 2r - 1 and s³ = 2s² + s - 1 and (sr⁻¹)³ = -(sr⁻¹)² + 2(sr⁻¹) + 1. These translate into higher-order Fibonacci like recursion relations. E.g. A(m,n)=2 A(m-1,n) + A(m-2,n) -A(m-3,n). The cubic equations are also the characteristic equations of the matrices R, S and SR⁻¹. The roots of the first two cubic equations are the six special numbers that pop up in my "Binet's formula" for A(m,n). 8. A nice extension to the Φ calculator trick: first show what happens when you push the squaring button and then follow this up by showing what happens when you push the 1/x button: Φ²=Φ+1, 1/Φ=Φ-1. 9. F(n) can be written as a geometric sum by iterating Φⁿ=F(n)Φ+F(n-1). E.g. F(4) = Φ³ - Φ¹ + Φ⁻¹ - Φ⁻³, F(5) = Φ⁴ - Φ² + Φ⁰ - Φ⁻² + Φ⁻⁴. 10. The golden ratio in a manga. The golden spin :) www.rwolfe.brambling.cdu.edu.au/pt7stands/GoldenRatio.html 11. Something VERY pretty: An uncurling golden square spiral czcams.com/users/shorts3hhonbIAKUU 12. You can generalise the idea of the golden ratio to matrices and ask what n by n matrix equals its inverse plus the identity matrix. The solution involves the Catalan series multiplied by phi. 13. Check out the analysis of the 9-gon case by @charlottedarroch in the comments below (long so I won't reproduce it here). 14. r=s=2 and r=s=0 may be the only solutions to r+s=rs in N, the naturals, but it is an open question whether or not there are any other solutions in βN, the Stone-Čech compactification of N. If there were another solution, it would immediately solve the open Ramsey theory problem: If you colour the natural numbers with a finite set of colours, can you guarantee that two distinct numbers x+y, xy have the same colour, where x,y are natural numbers? See, Pretend numbers in Chalkdust. 15. A recent VERY comprehensive paper On the Construction of 3D Fibonacci Spirals by Mariana Nagy, Simon R. Cowell and Valeriu Beiu www.mdpi.com/2227-7390/12/2/201 16. (by Andrew) The other nice heptagon fact is the construction by neusis, based on the same cubic. A fun problem may be to find a natural connection between this geometry and the equations for r and s. There will be a purely algebraic one, but is there a natural way to use these lengths of the heptagon’s diagonals in a neusis construction? .. or demonstrate that in one of the neusis constructions (or origami constructions) of the regular heptagon, the values r & s occur naturally in a way where rs=r+s etc are obvious. 17. The heptagonal triangle with sides 1, r and s has a lot of interesting properties and has its own wiki page: en.wikipedia.org/wiki/Heptagonal_triangle 18. 1, r and s satisfy the optic equation: en.wikipedia.org/wiki/Optic_equation 19. Heptagonal Fibonacci numbers" (e.g. 1, 1, 3, 6, 14 etc) "Multinacci rijen" by Jacques Haubrich in the maths magazine Euclides archive.org/details/nvvw-euclides-074_1998-99_04 An instance of our new number sequences occurring in nature in terms of numbers of different reflection paths through a number of glass plates. I've translated this to English www.qedcat.com/multifibonacci.pdf There is also this V. E. Hoggatt Jr. and M. Bicknell-Johnson, Reflections across two and three glass plates, Fibonacci Quarterly, volume 17 (1979), 118-142. 20. Check out a bunch more references at the Encyclopedia of integer sequences oeis.org/A006356 (1, 3, 6, 14, 31, 70, ... s-components of rᵐ sⁿ m=0) Number of distributive lattices; also number of paths with n turns when light is reflected from 3 glass plates and other incarnations. oeis.org/A006054 (1, 2, 5, 11, 25, 56, ... r-components of rᵐ sⁿ m=0) Also check out the cross references of closely related sequences listed on these pages. 21. If R=(x+y)/x and S = (x+y)/y, then RS=R+S. Choosing x and y to be integers gives rise to all rational solutions of RS=R+S. 22. Worth a look: Ptolemy's Theorem and Regular Polygons, L. S. Shively, The Mathematics Teacher, MARCH 1946, Vol. 39, No. 3 (MARCH 1946), pp. 117-120 talks about the basic equations of the regular heptagon, nonagon and 17-gon. 23. Extending the Fibonacci sequence to the negative numbers gives: ...-8,5,-3,2,-1,1,0,1,1,2,3,5,8... Then F(n)/F(n-1) → φ for n → +∞ and → -1/φ for n->-∞. 24. For matrices AB=A+B implies AB=BA. Proof: (*) AB = A + B implies AB - A - B = 0. We add the identity matrix and get: I -A - B + AB = I. This means that:(I-A)(I-B) = I ....and now the 🎩 MAGIC happens ✨ : (I-B)(I-A) = I. We conclude that I - B - A+ BA = I , i.e. (#) BA = A + B. The conclusion AB = BA follows from (*) and (#). @yyaa2539 25. We know that Φ is one solution of x² = x + 1. Dividing this equation by x² gives 1 = 1/x + 1/x². Therefore 1/x² = -1/x + 1 and (-1/x)² = (-1/x) + 1. Therefore -1/Φ is the second solution of this quadratic equation.
Hi I have a question about a previous video, the anti-shapeshifter one The question being is cedar tree (like the one on the Lebanese flag) anti-shapeshifter in there own way, i.e they grow but they keep the same proportions
@@zakariah_altibi The Koch spiral? As they say "it can be continually magnified 3x without changing shape." So, yes, there some "exponential/logarithmic" in all this :) Also have a look at logarithmic spirals. There rotation around the center is the same as scaling.
@@Mathologer Okay You seem open to some discussion! There is a linguistic theory of the word logarithm to stem from the word log, the log of the tree, meaning "to stay static, to stay the same, to maintane the same shape" i.e "anti shape shifting" Somthing like the curse of io, the curse of agelessness, to stay forever the same So, that's that
Constructable or turns out to not play any role in this. Except perhaps in creating some sort of bias that prevents some people from seriously exploring "non-constructible monsters" like regular heptagons :)
@@Mathologer heptagons always seem like the "cursed" one, especially when it's surrounded by hexagons and pentagons and octagons, so huge thanks for giving it the well-deserved limelight this time
What amazes me is just how accessible the Mathologer videos are. I would rate myself as 'clever' up through High School mathematics (pre-calculus), but also a bit lazy and underperforming. I am not a 'math guy.' I remember the general idea of trig but only the most basic identities, a handful of geometric proofs, some of the definitions in calculus but none of the methods or shortcuts. And yet, excepting a very few, I almost always manage to watch these explanations and follow along - delighted and inspired the whole way through. The man is truly a very gifted teacher and a master of this "you-tube" way of presentation. Thank you, Dr. Polster, for leading us on this wonderful journey.
You can generalise the idea of the golden ratio to matrices and ask what n by n matrix equals its inverse plus the identity matrix. The solution involves the Catalan series multiplied by phi.
Yes!! This is part of what I started talking about at the end of the video. The counterpart of the 3x3 matrices that I mention there for Fibonacci numbers is the 2x2 matrix {{1,1},{1,0}}.
For the curious, r and s are solutions to cubic equations, so WolframAlpha present them with formulas involving complex numbers, even when the final result is real.But when asked to simplify, it found a closed form using trig functions: (1 + Sqrt[7] Cos[ArcTan[3 Sqrt[3]]/3] + Sqrt[21] Sin[ArcTan[3 Sqrt[3]]/3])/3
@@Mathologer I think you mistyped those values. r and s are greater than 1. cos(pi/7) and cos(pi/7)^2-2 are less than 1. I find r = 2*cos(pi/7) And the r^2 = s+1 formula directly gives s= r^2 - 1.
thank you for always having captions :) auto-generated aint toooo terrible these days fortunately but the extra effort to have your own is noticed and appreciated
Luckily captions are no longer as much of a pain as they used to be. Since all my videos are fully scripted I can just take my script, clean it a bit and CZcams takes care of the rest automatically.
Thank you for the mention, Burkard! And thank you for bringing Steinbach's work back into the light. Some years ago I tried some exposition on the same topic, seen from a very different angle, in a triplet of iPython (now Jupyter) notebooks. Apparently I cannot include the link in this comment.
With a little bit of trigonometry on the heptagon, the set of equations for r and s can be used to show that cos(π/7) is the root of a cubic polynomial (8x³-4x²-4x+1=0). This immediately shows why the heptagon cannot be constructed (cube roots are not constructible). It's the same reason that one cannot construct a cube of volume 2. Using "origami math" (look it up, it's cool) one can construct roots of quartic polynomials. Geretschläger (1997) used this to construct a heptagon starting from just a quadratic sheet of paper (no straightedge or ruler required), but where you can make paper folds as in origami. The proof is given in the form of an origami folding instruction.
The other nice heptagon fact is the construction by neusis, based on the same cubic. A fun problem may be to find a natural connection between this geometry and the equations for r and s. There will be a purely algebraic one, but is there a natural way to use these lengths of the heptagon’s diagonals in a neusis construction? .. or demonstrate that in one of the neusis constructions (or origami constructions) of the regular heptagon, the values r & s occur naturally in a way where rs=r+s etc are obvious.
Check out these earlier videos czcams.com/video/O1sPvUr0YC0/video.html and czcams.com/video/IUC-8P0zXe8/video.html (the second one has the origami among many other things)
@@Mathologer I have, thank you! This just led me down the rabbit hole of A055034 in the Online Encyclopedia of Integer Sequences (degree of minimal polynomial of cos(π/n) over the rationals). This is equal to phi(2n)/2, where phi is Euler's totient function. As a consequence of the known lower bounds on phi, n=7 and n=9 are the only cases where that degree is 3. The heptagon and nonagon are the only regular polygons where origami help with the construction.
@@renerpho Origami can be used to solve higher degree polynomials, apparently. I mainly know this from a casual discussion with an expert in this, but he told me that there is no limit to the polynomial, if you allow sufficiently sophisticated systems of folds that simultaneously place certain points on certain lines or circles (at least, that is what I understood of it). If you google “origami to solve quartic polynomials” you’ll find some examples of fairly general quartics solved where the origami operation is not too extreme.
I was introduced to the Steinbach ratios 8 years ago. When I am working with the heptagon ratios, I prefer using the three side lengths of a heptagon with the smallest side negated: a=2sin(τ/7), b=2sin(2τ/7), c=2sin(4τ/7) Using these values, any polynomial f(x,y,z) such that f(a,b,c)=0 also has f(b,c,a)=0 and f(c,a,b)=0 It also has the property that ab+ca+bc=a+b+c+abc=a²+b²+c²-7=0 Also, to expand on your formulas at 43:00, A(m,n)=C(m,n+1), B(m,n)=C(m+1,n), and C(m,n)=(1*(r/1)^m*(s/1)^n+r²*(-s/r)^m*(1/r)^n+s²*(1/s)^m*(-r/s)^n)/(1+r²+s²)
plugging real numbers into the Fibonacci function and using complex numbers gives a nice exponential helix: the center follows exp(x) while the radius follows exp(-x) (or really, exponents of phi).
Thank you for introducing us to Peter Steinbach's work; your mission is accomplished! This has to be one of the most beautiful math lectures I've experienced. My eyes welled up when we got to the powers of phi, and I had to pause the video and go for a walk when I realized there's a Pascal triangle coming. Math (for me at least) can be overwhelmingly beautiful at times.
I have not verified this myself, but I was told that the fact that R^n has exactly one smooth structure for all n except n=4 (where it has uncountably infinitely many) is related to the fact that 2*2 = 2+2 = 4
Not aware of anything like this. If you find a reference please let me know. After proofreading the script for this video my friend Marty sent me this e-mail: : When I was doing my PhD, I read a paper by my supervisor, Brian White. It was on “surfaces mod 4”. So, it was a way to make surfaces multiple valued, turned the world of them into a ring, and mod 4 meant the obvious thing: four copies of a surface summed to 0. Anyway, I read the paper, and was trying to figure out why his theorem worked specifically for mod 4. And I asked: “Does it boil down to 2 x 2 = 2 + 2”? “Yep, that’s it." link.springer.com/article/10.1007/BF01403190
30:15 phi can be written like (1+sqrt(5))/2, which is what you get from the positive solution. for the negative solution the square root is subtracted, so we end up with (1-sqrt(5))/2, and doing some algebra leads to: -(1+sqrt(5))/2 + 1 which is -phi + 1. at this point i would try find a correlation between the two solutions (or try finding a mistake in my calculations), but i've known for quite some time now that 1/phi = phi-1, and i will try prooving it now: if you have a golden rectangle with sidelengths 1 and phi, and then remove a square, the new sidelengths are going to be 1 and phi-1. but since the ratio is the same, we conclude that 1/phi-1 = phi/1, which (hopefully) proves phi-1 is the reciprocal of phi. 31:50 for puting real numbers there is already a video on youtube online, it basically just results in complex numbers because of the exponents. for the negatives, it just goes backwards and switches between positive and negative, which, now that i think about it, makes the sequence even better! when taking the ratio of two consecutive positive fibonacci numbers, their ratio approaches phi as you go bigger and bigger. if you go in the opposite direction, the ratios approach -1/phi, which is the second solution to the previous equation! i find that funny as it seems as such a clear fact, the equation we use to get the fibonacci numbers uses -1/phi afterall. 41:35 my guess is that it will produce the mixed r and s identities.
Top shelf stuff Burkard. Extending the Fib S "leftwards" we obtain: ...-8,5,-3,2,-1,1,0,1,1,2,3,5,8... and note that just as F(n)/F(n-1)->φ for n->+∞ it also->-1/φ for n->-∞.
I independently stumbled onto the Fibonacci cascade when I was in high school… ended up using the name as my gamer tag and my account here… still find it really cool, and sometimes I wish I went down an academic path in mathematics, but had no idea how to go down that road. Still don’t. Oh well, cool videos though, keep up the good work (and I want pretty much all your t-shirts, man)
@@briancooke4259 All equations of the form axy = bx+cy are hyperbolas. This can be derived with some algebraic manipulation (we are essentially solving for y in terms of x). axy = bx + cy can be rearranged as axy - cy = bx. Factoring y on the left hand side gives y(ax-c) = bx and isolating y gives y = bx/(ax-c). Note that we may write bx as bx - bc/a + bc/a, which becomes (b/a)(ax-c) + bc/a. Substituting this back into our equation gives y = b/a + bc/(a(ax-c)). This is the classic y = 1/x scaled vertically by bc/a, shifted right by c/a, and shifted up by b/a, all of which preserve the hyperbolic shape of y = 1/x. To answer your actual question, the desired hyperbola is y = φ + φ^2/(x-φ) according to the general formula or more simply y = φx/(x-φ), where φ is the golden ratio (sqrt(5)+1)/2.
I love your videos so much because they help me a lot to learn new stuffs and invites me to delve deeper into mathematics. Thanks a lot. As a student and a regular viewer of your videos, I request you to create a video on Matrix, their determinants for higher order like 4 by 4 or 5 by 5, ..., their properties, decomposition of matrices into factors and related stuffs. These are very common topics, but students really need these.
High School Maths teacher here, so glad I happened in this video! Researching how to bring life into functions for my Precalculus class. I think I found it! These connections will live through students and maybe future mathematicians.
Truly beautiful. There is endless room for exploration here, like seeing how the ratios behave, and possibly proliferate in n-gons as n approaches infinity, which should spit out Ptolemy's Theorem. Also, how does this behave in 3 or more dimensions, when ratios of area, volume and various hyper-volumes come into play? IIRC there are only 2 regular hyperhedra(?) when we go above 4-d. The world to be explored is bigger than the observable universe, and just as rich!
47m20s - "What's next 3 1 4 5 9 ? :)" Did you forget something? No, apparently, π forgot something! Thanks for another enjoyable safari through some mathematical curiosities! Fred
Stumbled across that one while playing with different Fibonacci seed values for this video. Have to make this into a t-shirt. Also did you notice that the next number is 14?
@@Mathologer Yes I did, actually! But then it goes 23, 37, 60, ... Oh well . . . On one of his first two record albums*, ca 1960, Bob Newhart did a short bit on the famous monkeys-and-typewriters principle, playing the part of the guy who's walking up and down the line of primates reporting their output to the experiment's conductor, when he remarks: "Oh wait. I think we may have something here. 'To be or not to be - that is the gzornmplat ...' " * "The Button-Down Mind of Bob Newhart" and "The Button-Down Mind Strikes Back"
While playing with Excel one rainy afternoon, I discovered on my own how any two random numbers can seed a Fibonacci-ish sequence in which the ratio of adjacent elements converges to phi. I thought at first I'd done something wrong, then much later saw a hand-wavy Numberphile video that explained it. Amazing stuff. There's beauty to be found in all these dusty little corners...
Amazing, awesome video, thanks. I'd like to add that moving into 3D is also good with golden triangles. Yes, the pentagon is filled with phi, but also the icosahedron. In 2008 I discovered a way to use icosahedral geometry to create structures with some amazing properties which one can see via search for towerdome. The golden ratio contributes (I believe) to a unique routing of stress forces to make it perhaps the most efficient way of holding weight above ground with multiple floors.
@@Mathologer Thanks, and back to your video, your ability to display math/geometry with CG is so good, that I wish that students could have curricula that use these techniques and content. For a long time I've been distressed that so few people seem to know that the Pythagorean Theorem is "just" a special case of Ptolemy's Theorem. The latter seems so powerful that I always wondered how it could be used, and your video with the inscribed pentagon and heptagon and connections to phi ratio and the "golden R/S" is truly mind-blowing.
Glad you got something out of all this. You are of course familiar with zome tool (and vZome?). Check out the note on vZome in the description of this video. Maybe some towerdome action lurking there :)
This is nice: r = 2 * cos(π/7), (and s = r^2 - 1). I saw that Mathologer typed something similar somewhere, but it disappeared :-) And oh, we could also write s as: s = 2 * cos(2π/7) + 1 . Loving it!
The dodecahedron has diagonals That are sqr(2)Phi, sqr(3)phi and (phi)^2 the geometry demonstration is beautiful by showing the square, cube, and large pentagon that lays on the surface of the dodecahedron - love your videos makes me happy to see other math enthusiasts even though I am not a mathematician
First thing I would think of is the number fields Q[r, s] (and Q[phi]). Q[phi] is the splitting field of x^2-x-1, in particular Q[phi] is isomorphic to Q[x]/(x^2-x-1). Q[r, s] is isomorphic to Q[x, y]/(xy-x-y, x^2-y-1, y^2-x-y-1). We can look at the ring of algebraic integers inside these number fields. For Q[phi], that would be Z[phi]. If I recall correctly, this is a unique factorisation domain. Perhaps it would be interesting to check if the ring of algebraic integers in Q[r, s] is a unique factorisation domain (else we may consider computing class numbers, it might be interesting too). Since we get them from polygons, I think these lie in cyclotomic extensions, so Q[phi] is inside Q[z] where z^5=1. It should be clear that Q[r, s] is inside Q[z] where z^7=1, and for the infinite family of polygons I would expect something similar. (Proof: identify the vertices of the polygon with the complex numbers 1 to z^6. Then r^2, s^2 are clearly |1-z^2|^2/|1-z|^2=|1+z|^2 and |1-z^3|^2/|1-z|^2=|1+z+z^2|^2 respectively. Since s^2=s+r+1 and r^2=s+1, we can solve for r and s from there.) I believe it should be clear that [Q[r, s]:Q]=3 since all powers of r and s can be written as Ar + Bs + C.
Total nostalgia! I discovered all of this back in 1992. Not the geometrical visualizations, but ALL of the algebra ... simply by virtually swimming in the beauty of the number sequences. In the desire to find the algebraic expressions of r and s, I ended up with x³-7x-7=0, where x=rs-1, for which there are three real solution, from none of which I was able to eliminate the imaginary part of their algebraic expressions. Major disappointment! 😅😅.. And when I finally gave up, and completely emptied my mind of all attempts at algebraic manipulation toward a non-complex expression, and instead demanded my mind to directly visualize the solution, it in fact came up with an approximation that was so close to the true value of rs-1 that my calculator fooled me to believe I actually hit jackpot! But I ran to my computer to compare with the real value of rs-1, only to find a discrepancy just a few points down the line of decimals 🙄😢😢..
Great video, thanks. After my recent geometric explorations of polyhedra, I reached the conclusion that somehow sqrt(2) and phi are part of a special kind of irrational numbers regarding geometric ratios.
I’ve tackled the geometry and trigonometry behind the heptagonal triangle, which is the triangle that has sides r, s and 1. That is worth exploring on its own in a video series, not just one video. Wikipedia has a whole article dedicated to it and there are a number of papers dedicated to it as well.
If Aₙ is the n'th length in the circle, A₀=0, A₁=1, ..., then the product of two of these can be calculated as Aₙ₊₁Aₘ₊₁=Aₘ₋ₙ₊₁+Aₘ₋ₙ₊₃+...+Aₘ₊ₙ₊₁ Note that this sequence has to become negative for the identity to hold. This makes all the diagonals look really nice in the multiplication table, since two diagonally adjacent multiplications only differ by a single factor (which also follows from Ptolemy's theorem, and this implies the general result because it holds if n=1). In a pentagon: φ=A₂=A₃, A₂A₂=A₁+A₃, so φ²=φ+1. The only other identities involve linear dependence, like A₂+1=A₄ in a nonagon, although you can see this because the expansions of A₃A₄=A₂+A₃+A₄ is equal to the sum of the expansions of A₃A₁=A₃ and A₃A₂=A₂+A₄.
Steinbach's ratios are extremely pretty, thanks for showcasing them! I'm going to have to look at the ratios α, β, and γ for the enneagon. I'm surprised you didn't mention that A, B and C, the Fibonacci-like recurrent functions are actually only a single function. In particular, taking their definition r^m s^n = A(m,n)s+B(m,n)r+C(m,n) and multiplying through by r and applying the relations for r and s give recurrences for A(m+1,n), B(m+1,n) and C(m+1,n), and similarly multiplying through by s gives recurrences for A(m,n+1), B(m,n+1), and C(m,n+1). And one obtains that C(m+1,n) = B(m,n) and C(m,n+1) = A(m,n). Therefore not only is C a shifted copy of A, but so is B, as B(m,n) = A(m+1,n-1). So r^m s^n = A(m,n)s+A(m+1,n-1)r+A(m,n-1). That can make building the values up by recurrence slightly awkward, but you need to build them from a set of initial values anyway, so it's not any less practical. And this means you only really need a Binet type formula for A(m,n), and the formula for B(m,n) is just a re-indexing of that.
@@Mathologer Most things work out similarly nicely for the enneagon, with one exception which is particularly curious. There's still a little I haven't quite figured out, so I'll explain what I do know. In the paper Steinbach uses α,β and γ for the ratios in increasing size, but I'll be using a, b and c. Applying Ptolemy's theorem we obtain the following 6 relations, after simplifying a few things: a^2 = b+1 ab = a+c ac = b+c b^2 = b+c+1 bc = a+b+c c^2 = a+b+c+1 Just as in the case with the pentagon and heptagon, we can use these relations to reduce any non-negative integer powers of a, b and c to an integer linear combination of 1, a, b and c. However, unlike the pentagon and heptagon, we don't get all integer powers in general. The relations alone can be used to obtain 1/a = a+b-c-1 and 1/c = c-b, however 1/b cannot be obtained from the relations alone, for reasons we shall see later. We can then define maps P, Q, R, S: ZxNxZ->Z, where Z and N are the integers and the non-negative integers respectively, such that: a^i b^j c^k = P(i,j,k)c+Q(i,j,k)b+R(i,j,k)a+S(i,j,k) for all i, k in Z and j in N. Multiplying through by a, b and c gives forward recurrences for P,Q,R and R in the i, j and k indices, and dividing by a and c gives reverse recurrences for P, Q, R and S in the i and k indices. Like with the pentagon and heptagon, we see that Q, R and S are just shifted copies of P. In particular: Q(i,j,k) = P(i,j+1,k-1), R(i,j,k) = P(i+1,j,k-1) and S(i,j,k) = P(i,j,k-1) for all i, k in Z, j in N. I do wonder if it's perhaps better to write P, Q and R in terms of S, as then P, Q and R are positive shifted versions of S in a fairly symmetric way. To understand why we don't get 1/b from the relations, consider that there are multiple solutions to the relations. Any triple (x,y,z) satisfying the relations of (a,b,c), must also have: x^i y^j z^k = P(i,j,k)z+Q(i,j,k)y+R(i,j,k)x+S(i,j,k) for all i, k in Z and j in N. What are these solutions? Well we can use the relations to show that if (x,y,z) is a solution, x, y and z must be roots of the polynomials x^4-x^3-3x^2+2x+1, y^4-3y^3+3y and z^4-2z^3-3z^2+z+1 respectively. Each has four real roots which match up to form 4 distinct solutions and can be expressed as (a,b,c), (1,0,-1), (-c/a,b/a,-1/a) and (-1/c,-b/c,a/c). We can now see the problem with deriving 1/y from the relations; the solution (1,0,-1). If 1/y could be derived from the relations, we'd be able to express negative powers of y as a linear combination of 1, x, y and z. But negative powers of y in the solution (1,0,-1) correspond to dividing by 0. We shall see this isn't all to the story though. Now that we have our solutions to the relations, we can form a matrix equation A = MB, where: A = {a^i b^j c^k, (-1)^k δ_j, (-c/a)^i (b/a)^j (-1/a)^k, (-1/c)^i (-b/c)^j (a/c)^k} B = {P(i,j,k), Q(i,j,k), R(i,j,k), S(i,j,k)} M = {{c, b, a, 1}, {-1, 0, 1, 1}, {-1/a, b/a, -c/a, 1}, {a/c, -b/c, -1/c, 1}} Note that 0^j for j in N is just δ_j the Kronecker Delta at j, where we're using the combinatorial convention that 0^0 = 1. We can then invert M and obtain B = M^(-1)A, where: M^(-1) = 1/9 {{c-b+1, -3, -a+1, b}, {-c+2b-1, 0, c-b+a, -a-b+1}, {a-1, 3, -b, -c+b-1}, {-b+3, 3, -a+b+1, c+1}} So we do indeed get Binet-like formulae for P,Q,R and S, for i, k in Z and j in N. But the story doesn't quite end there. What would happen if you try to use these formulae for negative values of j? Do we just get nonsense? We no longer get integer outputs for P, Q, R and S, we get 3-adic rationals in general, but they do indeed allow you to express negative powers of y as a linear combination of 1, x, y and z for three of the four solutions (x,y,z). And the formula actually works correctly! I don't currently have a complete explanation as to why, but I did observe one thing. If y =/= 0, we can obtain a new relation z = x+1. We see this relation is indeed satisfied by three of the four solutions, but is not satisfied by the solution (1,0,-1), so perhaps this additional relation is enough to show that the Binet formulae indeed works for three of the four solutions for negative values of j.
Very interesting and thanks for sharing :) Maybe also have a look at the collection of additional insights that I pinned to the top of the top of the comment section. There was one about a super simple way of writing the heptagon Binet's formula. Also, maybe e-mail me any other comments to make sure that things don't get lost. burkard.polster@monash.edu
At first blush, rs = r+s appears to violate dimensionality, but with your proof, using dimensions of length, there are "hidden" unit lengths in the equation, so, if our units are meters, we have rs m^2 = r*1 m^2 + s*1 m^2. Wonder if considering the dimensionality for some of the other development might yield more interesting insights?
With regards to time stamp 1:20. Works for all x and y when r = (x+y)/x and s = (x+y)/y This is also the rational set of solutions when x and y are integers.
The transformation of the pentagon ratios to a golden rectangle and the heptagon ratios to a golden 3-d box seems to suggest that the ratios present in a golden n-dimensional hyperbox (probably the wrong word) are present in a (2n+1)-gon. At a quick glance, Steinbach's paper doesn't seem to address this, so is it true?
As far as I can tell nobody has really looked at any of these things in detail beyond the heptagon. Having said that and based on what is known in general, I'd also expect a lot of the things I talk about in terms of geometry to carry over into higher dimensions.
Great to see! A few avenues not explored: The four triangles and their trigonometry. Including THE heptagonal triangle (with sides 1:r:s or angles pi/7, 2pi/7, 4pi/7) r satisfies r^3-r^2-2r+1=0, and s is a root of the reciprocal polynomial. Recurrence relations, similar to those relating to the Tribonacci constant. Rational approximations to r (and s), starting with 1,2,9/5,182/101 (and 2,9/4,182/101)
Step 1. You Tube recommendation Step 2. What a nice title, A+B=AB ---> Golden Equation... Step 3. Liked 👍 and Saved in "Watch Later" folder. Step 4. Post this comment😅 Step 5. Enjoying the whole video...now....
Back in college I did my senior project involving some number theory-esque shenanigans and I came across the imaginary golden ratio by complete accident. Simply phi^2=phi-1. Lot of cool stuff with that guy too.
@@Mathologer - I'm replying to your suggestion in the video to find the equivalents to phi, r&s for the nonagon. I found the 6 equations for products of r, s and t, then calculated their real values and then .. .. oh and while you're there - at the time you released this video I was playing with the extended integers a+phi*b and had just found a fun integer sequence. I'm trying to get it into the OEIS at the moment. If I can get the pair of sequences approved they will be A375484 + phi*A347493. I've had some helpful feedback from the reviewers so far, but do you know of other pairs of sequences in the form a+phi*b in the OEIS?
30:15 The phi is equal to (sqrt(5) + 1) / 2 the other solution to the equation is (-sqrt(5) + 1) / 2. The difference between them is sqrt(5). If we rearrange the phi = (sqrt(5) + 1) / 2 we get sqrt(5) = 2 * phi - 1. So the other solution in terms of phi is: phi - (2 * phi - 1) which simplifies to: - phi + 1 = - (phi - 1) and since phi - 1 = 1 / phi, -(phi - 1) = - 1 / phi.
Jeez! Lovely video! In addition to the gold rush in finding all the great stuff about r and s (rho and sigma), those ratios from all the higher odd-number sided regular polygons seem like another gold rush waiting to happen.
like 2 years ago when i first heard of the fact that the golden ratio was the diagonal of the pentagon i thought "hmm, that's interesting, what about other shapes", and since i already knew the diagonals of the hexagon i jumped to the heptagon. i independently figured out pretty much everything in this video, and because of that 1.801 and 2.247 became my favorite numbers, and i cannot express the euphoria i got when i saw the letters r and s at the beginning and realizing where it was going. thank you so much i am so giggity right now
Thanks Mathologer. Amazing and high quality presentation as usual. BTW, for 9-gons the relations are t = r+1, r+t = r*s, s+t = r*t, for 11-gons are r^2 = s+1, r+t = r*s, s+u = r*t, t+u = r*u.
11:58 One small correction. The golden spiral is _not_ made up of quarter-circles. It _is_ the logarithmic spiral with growth factor φ. The succession of quarter-circles approximates the golden spiral. A Fibonacci spiral also approximates the golden spiral, with the approximation improving as the number of squares increases.
24:27 I was curious about the area of the thin green rectangle that's leftover. For a 5x8 rectangle it's about 0.721 units. Interestingly, as you scale up to higher Fibonacci rectangles, that leftover rectangle gets thinner and thinner, and the rate it gets thinner cancels out the rate that the Fibonacci rectangle grows, so that the area converges to a fixed value of 0.7236. After some fiddling around, I found that the exact value that the area of the leftover rectangle converges to is (phi^2 + 1)/5 or (1 + (1/sqrt(5))/2
Watching this I had a few thoughts/questions: 1. Looking at diagonals, there was a "jump" from the pentagon to the heptagon - two shapes with odd numbers of sides. Are there any similar ratios/relationships for the hexagon or octagon? What about the square - too trivial? 2. If not, is there some reason why the even sided shapes do not exhibit this behaviour? 3. If the ratios of segments in a pentagon bear a connection to the golden rectangle, and those in a heptagon a connection to the "golden box", can we hypothesise that the ratios of line segments in a nonagon have a connection to a 4-dimensional golden object? Can we generalise to higher dimensions? Something about those diagonals is hinting at graphs and vertices but I can't make the connection. These are just thoughts - I don't think I have the maths to effectively investigate, maybe other viewers have some ideas 🙏🏼
A few months ago was was playing around with phi^n and 1/(phi^n). There are some interesting recursion relations in there. Another great video from Prof Mathologer!
I’m still on this video ( a week later ) Been chipping away at it bit by bit. It’s like a little candy I can enjoy at the end of each day after I’m done with all my school and work related math.
Another nice thing about fibo fractions a/b, so that a>b. For both a/b and b/a, let's multiply the denominator by 2. When both a=1 and b=1, both a+2b and b+2 = 3. After that, a+2b -> Lucas numbers and b+2a -> Fibonacci numbers.
20:20 I wonder what the set of all possible limit points looks like. Without having given much thought to that yet, I guess it would look like some kind of fractal?
Yes it is a Cartesian product of three Cantor sets: all points (x,y,z) where x,y and z all lie in a different Cantor set. Rather than the standard ratio of 1/3 in the cantor set construction, the ratios vary and, well, are some other ratios that probably have nothing to do with the rest of this video. 😉 If you consider the equivalent thing with the rectangle and φ, then the cut points in each segment are 1/φ and 1/φ² , so the Cantor set is full of golden ratios. This is fun to look at in base φ.
This was very interesting to watch. I was thinking about making some video animation about the golden ratio and pentagons, but this video also gave me some new food for thoughts. It's like the discoveries around this ratio is endless.
Lovely video as always, well presented and many fun insights! The 2+2=2•2 is also 2^2. This also reminded me of a different equality: x^y = y^x, which has the obvious solutions x=y, but also another curve of solutions with only 1 integer solution: 2^4 = 4^2. Just a fun curiosity I once thought about while trying to go to sleep. 😂
Some number curiosities that I've included in videos so far include 2x2=2+2, 1x2x3=1+2+3 and 3^3+4^3+5^3=6^3. Those other curiosities you mention are definitely also bouncing around in the back of my mind waiting for their time to shine :)
A little late to comment, but I actually read the paper this is from a few months back! I was browsing the Bridges Archive in my free time, and it is so satisfying to see it covered here.
I always roll my eyes when people talk about galaxies and nautilus shells in relation to phi, it's not true, but also just observing patterns is far less interesting than understanding them. This meant I had always had a "resistant" feeling towards phi. Mathologer has well and truly turned me into a phi appreciator, first with the continued fraction representation explanation, and now this, I'd somehow never seen Binet's formula, or at least if I had, I had forgotten it, but the derivation and wider connections was absolutely wonderful. I do want to note that as a colour blind viewer, I do struggle with some of the coloured lines in some videos (at the start), it's not an easy thing to solve, I know I tend to prefer dashed, dotted, double, ect. lines instead of colour indicators but that can get messy fast, it's just something maybe worth considering.
If you're interested I've linked to the original phi debunking articles in the description of this video. Also, have you seen Mario Livio's book? That one does an excellent job in separating fact from fiction :)
Very cool! When you gave your formula for generating each triple from the previous triple, that technique has a clear higher-dimensional analogue. If you generate n-tuples using this rule, I would imagine that would correspond to a golden n-dimensional hypercube, which would have a corresponding golden spiral. I’d be curious to know if these higher dimensional analogies had any interesting properties, applications, or divergences from their lower dimensional counterparts.
You answered a question I had a long time ago in undergrad when the professor used a shift operator on the recurrence relation that leads to the golden ratio. Great video as always! I'm in Greece rn on holidays so a little rho and sigma would go nice :)
What a wonderful journey through mathematics land! :) I really didn't see that rectangular grid of equations coming at 38:10. I'm now assuming that if you start with a 9-gon, that there will be 3 constants that, together with 1, satisfy 4 equations. Super interesting video, which scratches my itch for integer sequences as well :D
Maybe also check out my comment pinned to the top of this comment section for a summary of interesting additional details that I did not mention in the video and/or that popped up in this comment section.
Excellent video. Thank you. Folks should also check out: What's 1/2 Doing in Φ? - the Golden Ratio What's inside Φ? - the Golden Ratio on the Imaginary Angle YT channel.
I have a special relationship to Ptomelys Inequality because it's the first theorem where I came up with an original and beautiful proof of it. And it was actually because of you :D Back then, I was re-watching some video about proofs of Pythagoras and one of the proofs involved scaling the original triangle by the factors a, b and c and then rearranging the parts in a clever way. You can do the same thing with Ptolemys Inquality! Take any quadrilateral with side lengths a, A, b, B, c, C like in your video but make sure that c is "in between" a and b. Take three copies of the quadrilateral and scale it by a, b and c respectively. Notice that two of your copies now have a side with length ab and diagonals ac and bc respectively. Join them on the side ab. Now notice that you can fit the third copy perfectly on the diagonals of the other two, as the third copy has two side lengths ac and bc too and the angle also matches. Put it in place and look carefully. You will have formed a triangle with side lengths aA, bB and cC. Ptolemys Inquality therefore follows. And the equality will hold exactly when the point lands exactly on the cC side. Which will happen exactly when the opposite angles of the original quadrilateral add up to 180 degrees which is exactly the case when it is circular. qed I stand by my opinion that this proof is absolutely marvelous and the best proof of Ptolemys Inequality ever. And it's perfect for an animation so if you ever want to show it, I absolutely allow you to do so :)
I agree, that proof is super nice. Someone actually animated it on the Wiki page for Ptolemy's theorem :) en.wikipedia.org/wiki/Ptolemy An earlier version of the slideshow for this video actually also included an animation of this proof.
@@Mathologer Ah cool I didn't see that before :) I also see it "only" animates the equality part for circular quadrilaterals so I guess my version is pretty much a generalized version that also covers the inequality part. And the animation is based on work from 2012? It's always so interesting how we constantly find new beautiful proofs of things that has been known for centuries
@@Supremebubble Just in case I end up covering your proof would be good to know your name :) Can you send me an e-mail please? burkard.polster@monash.edu
A couple of more fun/interesting bits and pieces that did not get mentioned in the video and/or popped up in the comments so far (I'll update these and also collect them in the description of the video until I hit the word limit :)
1. Nice log troll: log(r+s)=log(r)+log(s)
2. T-shirt design is called Fibonightmare.
3. e-mail from Marty commenting on my 2 × 2 = 2 + 2 remark at the beginning: When I was doing my PhD, I read a paper by my supervisor, Brian White. It was on “surfaces mod 4”. So, it was a way to make surfaces multiple valued, turned the world of them into a ring, and mod 4 meant the obvious thing: four copies of a surface summed to 0. Anyway, I read the paper, and was trying to figure out why his theorem worked specifically for mod 4. And I asked: “Does it boil down to 2 x 2 = 2 + 2”? “Yep, that’s it." link.springer.com/article/10.1007/BF01403190
4. (1+i)(1-i)=(1+i)+(1-i)
5. The locus of points with xy=x+y is the hyperbola xy=1 shifted up and right 1.
6. B(m,n) = A(m+1,n-1), in other words the r component grid is also just a translated s component grid. And so r^m s^n = A(m,n)s+A(m+1,n-1)r+A(m,n-1). Another super nice and super symmetric way of of summarising the relationships between A(m,n), B(m,n), and C(m,n):
A(m,n)=C(m,n+1), B(m,n)=C(m+1,n), C(m,n)=(1*(r/1)^m*(s/1)^n+r²(-s/r)^m*(1/r)^n+s²(1/s)^m*(-r/s)^n)/(1+r²+s²)
The same viewer who pointed this out also noted: When I am working with the heptagon ratios, I prefer using the three side lengths of a heptagon with the smallest side negated: a=2sin(τ/7), b=2sin(2τ/7), c=2sin(4τ/7) (1 : r : s = sin(π/7) : sin(2π/7) : sin(4π/7)) Using these values, any polynomial f(x,y,z) such that f(a,b,c)=0 also has f(b,c,a)=0 and f(c,a,b)=0. It also has the property that ab+ca+bc=a+b+c+abc=a²+b²+c²-7=0
7. r³ = r² + 2r - 1 and s³ = 2s² + s - 1 and (sr⁻¹)³ = -(sr⁻¹)² + 2(sr⁻¹) + 1. These translate into higher-order Fibonacci like recursion relations. E.g. A(m,n)=2 A(m-1,n) + A(m-2,n) -A(m-3,n). The cubic equations are also the characteristic equations of the matrices R, S and SR⁻¹. The roots of the first two cubic equations are the six special numbers that pop up in my "Binet's formula" for A(m,n).
8. A nice extension to the Φ calculator trick: first show what happens when you push the squaring button and then follow this up by showing what happens when you push the 1/x button: Φ²=Φ+1, 1/Φ=Φ-1.
9. F(n) can be written as a geometric sum by iterating Φⁿ=F(n)Φ+F(n-1). E.g. F(4) = Φ³ - Φ¹ + Φ⁻¹ - Φ⁻³, F(5) = Φ⁴ - Φ² + Φ⁰ - Φ⁻² + Φ⁻⁴.
10. The golden ratio in a manga. The golden spin :) www.rwolfe.brambling.cdu.edu.au/pt7stands/GoldenRatio.html
11. Something VERY pretty: An uncurling golden square spiral czcams.com/users/shorts3hhonbIAKUU
12. You can generalise the idea of the golden ratio to matrices and ask what n by n matrix equals its inverse plus the identity matrix. The solution involves the Catalan series multiplied by phi.
13. Check out the analysis of the 9-gon case by @charlottedarroch in the comments below (long so I won't reproduce it here).
14. r=s=2 and r=s=0 may be the only solutions to r+s=rs in N, the naturals, but it is an open question whether or not there are any other solutions in βN, the Stone-Čech compactification of N. If there were another solution, it would immediately solve the open Ramsey theory problem: If you colour the natural numbers with a finite set of colours, can you guarantee that two distinct numbers x+y, xy have the same colour, where x,y are natural numbers? See, Pretend numbers in Chalkdust.
15. A recent VERY comprehensive paper On the Construction of 3D Fibonacci Spirals by Mariana Nagy, Simon R. Cowell and Valeriu Beiu www.mdpi.com/2227-7390/12/2/201
16. (by Andrew) The other nice heptagon fact is the construction by neusis, based on the same cubic. A fun problem may be to find a natural connection between this geometry and the equations for r and s. There will be a purely algebraic one, but is there a natural way to use these lengths of the heptagon’s diagonals in a neusis construction? .. or demonstrate that in one of the neusis constructions (or origami constructions) of the regular heptagon, the values r & s occur naturally in a way where rs=r+s etc are obvious.
17. The heptagonal triangle with sides 1, r and s has a lot of interesting properties and has its own wiki page: en.wikipedia.org/wiki/Heptagonal_triangle
18. 1, r and s satisfy the optic equation: en.wikipedia.org/wiki/Optic_equation
19. Heptagonal Fibonacci numbers" (e.g. 1, 1, 3, 6, 14 etc) "Multinacci rijen" by Jacques Haubrich in the maths magazine Euclides archive.org/details/nvvw-euclides-074_1998-99_04 An instance of our new number sequences occurring in nature in terms of numbers of different reflection paths through a number of glass plates. I've translated this to English www.qedcat.com/multifibonacci.pdf There is also this V. E. Hoggatt Jr. and M. Bicknell-Johnson, Reflections across two and three glass plates, Fibonacci Quarterly, volume 17 (1979), 118-142.
20. Check out a bunch more references at the Encyclopedia of integer sequences
oeis.org/A006356 (1, 3, 6, 14, 31, 70, ... s-components of rᵐ sⁿ m=0) Number of distributive lattices; also number of paths with n turns when light is reflected from 3 glass plates and other incarnations.
oeis.org/A006054 (1, 2, 5, 11, 25, 56, ... r-components of rᵐ sⁿ m=0)
Also check out the cross references of closely related sequences listed on these pages.
21. If R=(x+y)/x and S = (x+y)/y, then RS=R+S. Choosing x and y to be integers gives rise to all rational solutions of RS=R+S.
22. Worth a look: Ptolemy's Theorem and Regular Polygons, L. S. Shively, The Mathematics Teacher, MARCH 1946, Vol. 39, No. 3 (MARCH 1946), pp. 117-120 talks about the basic equations of the regular heptagon, nonagon and 17-gon.
23. Extending the Fibonacci sequence to the negative numbers gives: ...-8,5,-3,2,-1,1,0,1,1,2,3,5,8... Then F(n)/F(n-1) → φ for n → +∞ and → -1/φ for n->-∞.
24. For matrices AB=A+B implies AB=BA. Proof: (*) AB = A + B implies AB - A - B = 0. We add the identity matrix and get: I -A - B + AB = I. This means that:(I-A)(I-B) = I ....and now the 🎩 MAGIC happens ✨ : (I-B)(I-A) = I. We conclude that I - B - A+ BA = I , i.e. (#) BA = A + B. The conclusion AB = BA follows from (*) and (#). @yyaa2539
25. We know that Φ is one solution of x² = x + 1. Dividing this equation by x² gives 1 = 1/x + 1/x². Therefore 1/x² = -1/x + 1 and (-1/x)² = (-1/x) + 1. Therefore -1/Φ is the second solution of this quadratic equation.
Hi
I have a question about a previous video, the anti-shapeshifter one
The question being is cedar tree (like the one on the Lebanese flag) anti-shapeshifter in there own way, i.e they grow but they keep the same proportions
@@zakariah_altibi Have a look at this en.wikipedia.org/wiki/Self-similarity
@@Mathologer
Thank you for the response
Another question if i may, is this self similarity can be expressed with a logarithmic function?
@@zakariah_altibi The Koch spiral? As they say "it can be continually magnified 3x without changing shape." So, yes, there some "exponential/logarithmic" in all this :) Also have a look at logarithmic spirals. There rotation around the center is the same as scaling.
@@Mathologer
Okay
You seem open to some discussion!
There is a linguistic theory of the word logarithm to stem from the word log, the log of the tree, meaning "to stay static, to stay the same, to maintane the same shape" i.e "anti shape shifting"
Somthing like the curse of io, the curse of agelessness, to stay forever the same
So, that's that
I'll have to watch 3 times. It always takes me at least 3 times because I constantly want to pause, and go explore something that was said...😂
Constructable or turns out to not play any role in this. Except perhaps in creating some sort of bias that prevents some people from seriously exploring "non-constructible monsters" like regular heptagons :)
@@Mathologer heptagons always seem like the "cursed" one, especially when it's surrounded by hexagons and pentagons and octagons,
so huge thanks for giving it the well-deserved limelight this time
14 😅
Same here. I decided to try the nonagons and see how far I would get.
@@nanamacapagal8342I’ve always been particularly partial towards heptagons and dodecagons
What amazes me is just how accessible the Mathologer videos are. I would rate myself as 'clever' up through High School mathematics (pre-calculus), but also a bit lazy and underperforming. I am not a 'math guy.' I remember the general idea of trig but only the most basic identities, a handful of geometric proofs, some of the definitions in calculus but none of the methods or shortcuts.
And yet, excepting a very few, I almost always manage to watch these explanations and follow along - delighted and inspired the whole way through. The man is truly a very gifted teacher and a master of this "you-tube" way of presentation. Thank you, Dr. Polster, for leading us on this wonderful journey.
Same here. Am a teacher myself, but not of this level - not by a long shot. Burkhard and his team are something else 😊
That's great, glad these videos work so well for you :)
Calculus is so easy, though.
"Mathematicians, artists, wizards, and a lot of crazies." So... just crazies, then?
Mostly crazies ... :)
@@Mathologerthe crazies are the best lol
surely regulars wouldn't reveal any hidden identities. :P Even if you square them
Except for the wizards, I think- they’re the sanest of the bunch.
I always wanted to be a wizard when I was a kid. Settled for being a mathematician in the end. As close as you can get :)
I always get a 'WOW!' when I watch your videos. Thanks for bringing this stuff to CZcams.
That's great :)
our monthly dose of underrated mathematics
Wunderbar, diese Wiederbelebung der fast in Vergessenheit geratenen Entdeckung von Steinbach!
Wie immer, Vergnügen pur und besten Dank!
You're the only math youtuber that makes me audibly gasp every video, let alone multiple times. Keep up the amazing work!
Glad you like them!
You can generalise the idea of the golden ratio to matrices and ask what n by n matrix equals its inverse plus the identity matrix. The solution involves the Catalan series multiplied by phi.
Good one, thanks ❤
Yes!! This is part of what I started talking about at the end of the video. The counterpart of the 3x3 matrices that I mention there for Fibonacci numbers is the 2x2 matrix {{1,1},{1,0}}.
I find it astonishing how numbers align so well. Much better than stars.
Beautiful video as always.
These videos are so fun! Once I begin, I get so hooked that it's impossible for me to stop.
Perfect, mathematical magic in action :)
For the curious, r and s are solutions to cubic equations, so WolframAlpha present them with formulas involving complex numbers, even when the final result is real.But when asked to simplify, it found a closed form using trig functions:
(1 + Sqrt[7] Cos[ArcTan[3 Sqrt[3]]/3] + Sqrt[21] Sin[ArcTan[3 Sqrt[3]]/3])/3
Here are two easier ways of writing r and s: r = cos(pi/7) and s= r^2 - 2 :)
@@Mathologer
cos(π/7) is transcendental or not
@@conanedojawa4538 No because it is a root of polynomial equation r^3-r^2-2r+1=0
@@Mathologer I think you mistyped those values.
r and s are greater than 1.
cos(pi/7) and cos(pi/7)^2-2 are less than 1.
I find r = 2*cos(pi/7)
And the r^2 = s+1 formula directly gives s= r^2 - 1.
Yes, that's right :)
When I saw the penrose tiling at the end I gasped! So interesting as always ❤
A lot more to be said about all this but 50 minutes was of course pushing it already :)
Thank you Burkard. Another fine video with lots of interesting and surprising twists and turns!
thank you for always having captions :) auto-generated aint toooo terrible these days fortunately but the extra effort to have your own is noticed and appreciated
I sometimes wonder whether anybody actually uses them. Good to know that somebody does and actually thinks they are worth the effort :)
@@Mathologer Captions are also sometimes helpful to get the exact spelling of the name of something e.g. the mathematician Simon Jacob you mentioned
Luckily captions are no longer as much of a pain as they used to be. Since all my videos are fully scripted I can just take my script, clean it a bit and CZcams takes care of the rest automatically.
Thank you for the mention, Burkard! And thank you for bringing Steinbach's work back into the light.
Some years ago I tried some exposition on the same topic, seen from a very different angle, in a triplet of iPython (now Jupyter) notebooks. Apparently I cannot include the link in this comment.
Yes, CZcams's comment environment is pretty hopeless. Can you perhaps e-mail me those notebooks? burkard.polster@monash.edu
Ha, echo of Fermat. "I have a link to a notebook proving this, which this comment field is incapable of accepting". 😁
With a little bit of trigonometry on the heptagon, the set of equations for r and s can be used to show that cos(π/7) is the root of a cubic polynomial (8x³-4x²-4x+1=0). This immediately shows why the heptagon cannot be constructed (cube roots are not constructible). It's the same reason that one cannot construct a cube of volume 2.
Using "origami math" (look it up, it's cool) one can construct roots of quartic polynomials. Geretschläger (1997) used this to construct a heptagon starting from just a quadratic sheet of paper (no straightedge or ruler required), but where you can make paper folds as in origami. The proof is given in the form of an origami folding instruction.
The other nice heptagon fact is the construction by neusis, based on the same cubic. A fun problem may be to find a natural connection between this geometry and the equations for r and s. There will be a purely algebraic one, but is there a natural way to use these lengths of the heptagon’s diagonals in a neusis construction? .. or demonstrate that in one of the neusis constructions (or origami constructions) of the regular heptagon, the values r & s occur naturally in a way where rs=r+s etc are obvious.
Check out these earlier videos czcams.com/video/O1sPvUr0YC0/video.html and czcams.com/video/IUC-8P0zXe8/video.html (the second one has the origami among many other things)
@@Mathologer I have, thank you!
This just led me down the rabbit hole of A055034 in the Online Encyclopedia of Integer Sequences (degree of minimal polynomial of cos(π/n) over the rationals). This is equal to phi(2n)/2, where phi is Euler's totient function. As a consequence of the known lower bounds on phi, n=7 and n=9 are the only cases where that degree is 3.
The heptagon and nonagon are the only regular polygons where origami help with the construction.
Ah, yes, that's right :)
@@renerpho Origami can be used to solve higher degree polynomials, apparently. I mainly know this from a casual discussion with an expert in this, but he told me that there is no limit to the polynomial, if you allow sufficiently sophisticated systems of folds that simultaneously place certain points on certain lines or circles (at least, that is what I understood of it).
If you google “origami to solve quartic polynomials” you’ll find some examples of fairly general quartics solved where the origami operation is not too extreme.
The videos are always worth the wait! ❤❤❤
I feel like a kid who just found a $100 bill on the roadside. He will now burn his weekend to think about all the toys he can get ;)
Have fun :)
I was introduced to the Steinbach ratios 8 years ago. When I am working with the heptagon ratios, I prefer using the three side lengths of a heptagon with the smallest side negated:
a=2sin(τ/7), b=2sin(2τ/7), c=2sin(4τ/7)
Using these values, any polynomial f(x,y,z) such that f(a,b,c)=0 also has f(b,c,a)=0 and f(c,a,b)=0
It also has the property that ab+ca+bc=a+b+c+abc=a²+b²+c²-7=0
Also, to expand on your formulas at 43:00, A(m,n)=C(m,n+1), B(m,n)=C(m+1,n),
and C(m,n)=(1*(r/1)^m*(s/1)^n+r²*(-s/r)^m*(1/r)^n+s²*(1/s)^m*(-r/s)^n)/(1+r²+s²)
That's nice ! :)
What do these look like as colours?
Its really cool to learn that mathematics I am familiar with often turns out to be generalizations of even more powerful mathematics
you are the mathematical superman, saving mathematical discoveries from oblivion!
mathman :)
plugging real numbers into the Fibonacci function and using complex numbers gives a nice exponential helix: the center follows exp(x) while the radius follows exp(-x) (or really, exponents of phi).
Yes, some nice mathematics hiding there :)
I'm only 7 minutes in and this is my favorite video of yours by far
That's great :)
I say that for every one of his videos! This one was special - I easily followed most of it 1st watching.
:)
The maths is cool but your genuine enthusiasm is the hook that keeps me watching. Thanks
Thank you for introducing us to Peter Steinbach's work; your mission is accomplished! This has to be one of the most beautiful math lectures I've experienced. My eyes welled up when we got to the powers of phi, and I had to pause the video and go for a walk when I realized there's a Pascal triangle coming. Math (for me at least) can be overwhelmingly beautiful at times.
Glad it worked so well for you :)
fascinating, I could watch you for days!
I have not verified this myself, but I was told that the fact that R^n has exactly one smooth structure for all n except n=4 (where it has uncountably infinitely many) is related to the fact that 2*2 = 2+2 = 4
Not aware of anything like this. If you find a reference please let me know. After proofreading the script for this video my friend Marty sent me this e-mail: : When I was doing my PhD, I read a paper by my supervisor, Brian White. It was on “surfaces mod 4”. So, it was a way to make surfaces multiple valued, turned the world of them into a ring, and mod 4 meant the obvious thing: four copies of a surface summed to 0. Anyway, I read the paper, and was trying to figure out why his theorem worked specifically for mod 4. And I asked: “Does it boil down to 2 x 2 = 2 + 2”? “Yep, that’s it." link.springer.com/article/10.1007/BF01403190
These videos should really be called 'Mathologer Masterpieces' instead. Thanks to everyone who contributes to making them.
Glad you think so :)
30:15 phi can be written like (1+sqrt(5))/2, which is what you get from the positive solution. for the negative solution the square root is subtracted, so we end up with (1-sqrt(5))/2, and doing some algebra leads to:
-(1+sqrt(5))/2 + 1 which is -phi + 1. at this point i would try find a correlation between the two solutions (or try finding a mistake in my calculations), but i've known for quite some time now that 1/phi = phi-1, and i will try prooving it now: if you have a golden rectangle with sidelengths 1 and phi, and then remove a square, the new sidelengths are going to be 1 and phi-1. but since the ratio is the same, we conclude that 1/phi-1 = phi/1, which (hopefully) proves phi-1 is the reciprocal of phi.
31:50 for puting real numbers there is already a video on youtube online, it basically just results in complex numbers because of the exponents. for the negatives, it just goes backwards and switches between positive and negative, which, now that i think about it, makes the sequence even better! when taking the ratio of two consecutive positive fibonacci numbers, their ratio approaches phi as you go bigger and bigger.
if you go in the opposite direction, the ratios approach -1/phi, which is the second solution to the previous equation! i find that funny as it seems as such a clear fact, the equation we use to get the fibonacci numbers uses -1/phi afterall.
41:35 my guess is that it will produce the mixed r and s identities.
Very good :)
Top shelf stuff Burkard.
Extending the Fib S "leftwards" we obtain:
...-8,5,-3,2,-1,1,0,1,1,2,3,5,8...
and note that just as F(n)/F(n-1)->φ for n->+∞ it also->-1/φ for n->-∞.
Actually F(n)/F(n-1) -> -1/φ for n->-∞ is a very nice observation (which I myself have never made :)
I independently stumbled onto the Fibonacci cascade when I was in high school… ended up using the name as my gamer tag and my account here… still find it really cool, and sometimes I wish I went down an academic path in mathematics, but had no idea how to go down that road. Still don’t. Oh well, cool videos though, keep up the good work (and I want pretty much all your t-shirts, man)
The locus of points with AB=A+B is the hyperbola XY=1 shifted up and right 1.
This gives nice pairs like (3/2,3), (4/3,4), (1+1/n,n+1).
That's amazing! Would there be a hyperbola that gave AB = phi*(A + B)
?
Good point :)
@@Utesfan100 combo class had that right?
@@Mathologer pun intended?
@@briancooke4259 All equations of the form axy = bx+cy are hyperbolas. This can be derived with some algebraic manipulation (we are essentially solving for y in terms of x).
axy = bx + cy can be rearranged as axy - cy = bx. Factoring y on the left hand side gives y(ax-c) = bx and isolating y gives y = bx/(ax-c).
Note that we may write bx as bx - bc/a + bc/a, which becomes (b/a)(ax-c) + bc/a. Substituting this back into our equation gives y = b/a + bc/(a(ax-c)). This is the classic y = 1/x scaled vertically by bc/a, shifted right by c/a, and shifted up by b/a, all of which preserve the hyperbolic shape of y = 1/x.
To answer your actual question, the desired hyperbola is y = φ + φ^2/(x-φ) according to the general formula or more simply y = φx/(x-φ), where φ is the golden ratio (sqrt(5)+1)/2.
Love your work, you're rigorous and funny! Thank you.
I love your videos so much because they help me a lot to learn new stuffs and invites me to delve deeper into mathematics. Thanks a lot. As a student and a regular viewer of your videos, I request you to create a video on Matrix, their determinants for higher order like 4 by 4 or 5 by 5, ..., their properties, decomposition of matrices into factors and related stuffs. These are very common topics, but students really need these.
I have loved each depth of unveiling the mathematical concepts ❤
Thank you, mathologer, for building concepts deeply always😊
High School Maths teacher here, so glad I happened in this video! Researching how to bring life into functions for my Precalculus class. I think I found it! These connections will live through students and maybe future mathematicians.
Glad you found it then :)
The best Mathologer video I've ever seen (and the level of Mathologer videos is always very very veru very high...)
Glad you think so :)
Truly beautiful.
There is endless room for exploration here, like seeing how the ratios behave, and possibly proliferate in n-gons as n approaches infinity, which should spit out Ptolemy's Theorem. Also, how does this behave in 3 or more dimensions, when ratios of area, volume and various hyper-volumes come into play?
IIRC there are only 2 regular hyperhedra(?) when we go above 4-d.
The world to be explored is bigger than the observable universe, and just as rich!
There really is a lot of room for exploration. Hopefully there will be a gold rush :)
I SWEAR I JUST DISCOVERED THE FORMULA MYSELF YESTERDAY!!!
Just amazing how mathologer brings less famous ideas into life and give them a whole new life altogether!
47m20s -
"What's next
3 1 4 5 9 ?
:)"
Did you forget something? No, apparently, π forgot something!
Thanks for another enjoyable safari through some mathematical curiosities!
Fred
Burkard the troll strikes again. 😆
Stumbled across that one while playing with different Fibonacci seed values for this video. Have to make this into a t-shirt. Also did you notice that the next number is 14?
@@Mathologer Yes I did, actually! But then it goes 23, 37, 60, ... Oh well . . .
On one of his first two record albums*, ca 1960, Bob Newhart did a short bit on the famous monkeys-and-typewriters principle, playing the part of the guy who's walking up and down the line of primates reporting their output to the experiment's conductor, when he remarks:
"Oh wait. I think we may have something here. 'To be or not to be - that is the gzornmplat ...' "
* "The Button-Down Mind of Bob Newhart" and "The Button-Down Mind Strikes Back"
@@ffggddss Fibonacci sequence with initial values a(0) = 3 and a(1) = 1
3,1,4,5,9,14,23,37,60,97,157,254,411,665,1076,1741,2817,4558,7375,11933,19308,31241,50549,81790,132339,214129,346468,560597,907065,1467662,2374727,3842389,6217116,10059505,16276621,26336126,42612747,68948873,111561620,180510493,292072113,472582606,764654719,1237237325,2001892044,3239129369,5241021413,8480150782,13721172195,22201322977,35922495172,58123818149,94046313321,152170131470,246216444791,398386576261,644603021052,1042989597313,1687592618365,2730582215678,4418174834043,7148757049721,11566931883764,18715688933485,30282620817249,48998309750734,79280930567983,128279240318717,207560170886700,335839411205417,543399582092117,879238993297534,1422638575389651,2301877568687185,3724516144076836,6026393712764021,9750909856840856,15777303569604876,25528213426445732,41305516996050610,66833730422496340,108139247418546940,174972977841043260,283112225259590200,458085203100633500,741197428360223700,1199282631460857300,1940480059821081000,3139762691281938400,5080242751103019000,8220005442384957000,13300248193487976000,21520253635872930000,34820501829360910000,56340755465233840000,91161257294594750000,147502012759828600000,238663270054423360000,386165282814251960000,624828552868675300000
getting warmer and warmer to nature, which is numberless.
I loved the way we go from equations to matrices
While playing with Excel one rainy afternoon, I discovered on my own how any two random numbers can seed a Fibonacci-ish sequence in which the ratio of adjacent elements converges to phi. I thought at first I'd done something wrong, then much later saw a hand-wavy Numberphile video that explained it. Amazing stuff. There's beauty to be found in all these dusty little corners...
Amazing, awesome video, thanks. I'd like to add that moving into 3D is also good with golden triangles. Yes, the pentagon is filled with phi, but also the icosahedron. In 2008 I discovered a way to use icosahedral geometry to create structures with some amazing properties which one can see via search for towerdome. The golden ratio contributes (I believe) to a unique routing of stress forces to make it perhaps the most efficient way of holding weight above ground with multiple floors.
Very interesting. Just checked out your website :)
@@Mathologer Thanks, and back to your video, your ability to display math/geometry with CG is so good, that I wish that students could have curricula that use these techniques and content. For a long time I've been distressed that so few people seem to know that the Pythagorean Theorem is "just" a special case of Ptolemy's Theorem. The latter seems so powerful that I always wondered how it could be used, and your video with the inscribed pentagon and heptagon and connections to phi ratio and the "golden R/S" is truly mind-blowing.
Glad you got something out of all this. You are of course familiar with zome tool (and vZome?). Check out the note on vZome in the description of this video. Maybe some towerdome action lurking there :)
👏👏👏
So excellent lecture.
Thank you, Doctor, your scientists and your colleagues.
and
With luck and more power to you.
hoping for more videos.
This is nice: r = 2 * cos(π/7), (and s = r^2 - 1). I saw that Mathologer typed something similar somewhere, but it disappeared :-)
And oh, we could also write s as: s = 2 * cos(2π/7) + 1 . Loving it!
The dodecahedron has diagonals
That are sqr(2)Phi, sqr(3)phi and (phi)^2 the geometry demonstration is beautiful by showing the square, cube, and large pentagon that lays on the surface of the dodecahedron
- love your videos makes me happy to see other math enthusiasts even though I am not a mathematician
First thing I would think of is the number fields Q[r, s] (and Q[phi]). Q[phi] is the splitting field of x^2-x-1, in particular Q[phi] is isomorphic to Q[x]/(x^2-x-1). Q[r, s] is isomorphic to Q[x, y]/(xy-x-y, x^2-y-1, y^2-x-y-1).
We can look at the ring of algebraic integers inside these number fields. For Q[phi], that would be Z[phi]. If I recall correctly, this is a unique factorisation domain. Perhaps it would be interesting to check if the ring of algebraic integers in Q[r, s] is a unique factorisation domain (else we may consider computing class numbers, it might be interesting too).
Since we get them from polygons, I think these lie in cyclotomic extensions, so Q[phi] is inside Q[z] where z^5=1. It should be clear that Q[r, s] is inside Q[z] where z^7=1, and for the infinite family of polygons I would expect something similar. (Proof: identify the vertices of the polygon with the complex numbers 1 to z^6. Then r^2, s^2 are clearly |1-z^2|^2/|1-z|^2=|1+z|^2 and |1-z^3|^2/|1-z|^2=|1+z+z^2|^2 respectively. Since s^2=s+r+1 and r^2=s+1, we can solve for r and s from there.) I believe it should be clear that [Q[r, s]:Q]=3 since all powers of r and s can be written as Ar + Bs + C.
Steinbach also has a few things to say about all this in his papers :)
Total nostalgia! I discovered all of this back in 1992. Not the geometrical visualizations, but ALL of the algebra ... simply by virtually swimming in the beauty of the number sequences.
In the desire to find the algebraic expressions of r and s, I ended up with x³-7x-7=0, where x=rs-1, for which there are three real solution, from none of which I was able to eliminate the imaginary part of their algebraic expressions. Major disappointment! 😅😅.. And when I finally gave up, and completely emptied my mind of all attempts at algebraic manipulation toward a non-complex expression, and instead demanded my mind to directly visualize the solution, it in fact came up with an approximation that was so close to the true value of rs-1 that my calculator fooled me to believe I actually hit jackpot! But I ran to my computer to compare with the real value of rs-1, only to find a discrepancy just a few points down the line of decimals 🙄😢😢..
That's great :)
Great video, thanks. After my recent geometric explorations of polyhedra, I reached the conclusion that somehow sqrt(2) and phi are part of a special kind of irrational numbers regarding geometric ratios.
28:13 my high school teacher showed this to us once, and then he proved it to explain how proofs by induction work!!
You are in luck, you've got a good maths teacher :)
What can I say, this is just a gem ! thank you !
I’ve tackled the geometry and trigonometry behind the heptagonal triangle, which is the triangle that has sides r, s and 1. That is worth exploring on its own in a video series, not just one video. Wikipedia has a whole article dedicated to it and there are a number of papers dedicated to it as well.
Good point :)
Thank you. This video was very nice and the tessalation at the end was very interesting!
Glad you enjoyed it :)
If Aₙ is the n'th length in the circle,
A₀=0, A₁=1, ..., then the product of two of these can be calculated as
Aₙ₊₁Aₘ₊₁=Aₘ₋ₙ₊₁+Aₘ₋ₙ₊₃+...+Aₘ₊ₙ₊₁
Note that this sequence has to become negative for the identity to hold.
This makes all the diagonals look really nice in the multiplication table, since two diagonally adjacent multiplications only differ by a single factor (which also follows from Ptolemy's theorem, and this implies the general result because it holds if n=1).
In a pentagon:
φ=A₂=A₃,
A₂A₂=A₁+A₃, so
φ²=φ+1.
The only other identities involve linear dependence, like
A₂+1=A₄ in a nonagon, although you can see this because the expansions of
A₃A₄=A₂+A₃+A₄ is equal to the sum of the expansions of A₃A₁=A₃ and A₃A₂=A₂+A₄.
Very good. Check out the details in Peter Steinbach's paper: Golden fields: a case for the heptagon: www.jstor.org/stable/2691048
Thank you for everything you do. What a great teacher!!
Thanks !
hi, always nice to see your videos.
Steinbach's ratios are extremely pretty, thanks for showcasing them! I'm going to have to look at the ratios α, β, and γ for the enneagon. I'm surprised you didn't mention that A, B and C, the Fibonacci-like recurrent functions are actually only a single function. In particular, taking their definition r^m s^n = A(m,n)s+B(m,n)r+C(m,n) and multiplying through by r and applying the relations for r and s give recurrences for A(m+1,n), B(m+1,n) and C(m+1,n), and similarly multiplying through by s gives recurrences for A(m,n+1), B(m,n+1), and C(m,n+1). And one obtains that C(m+1,n) = B(m,n) and C(m,n+1) = A(m,n). Therefore not only is C a shifted copy of A, but so is B, as B(m,n) = A(m+1,n-1).
So r^m s^n = A(m,n)s+A(m+1,n-1)r+A(m,n-1). That can make building the values up by recurrence slightly awkward, but you need to build them from a set of initial values anyway, so it's not any less practical. And this means you only really need a Binet type formula for A(m,n), and the formula for B(m,n) is just a re-indexing of that.
Very good :) If you write up anything for the enneagon please share it here/with me.
@@Mathologer Most things work out similarly nicely for the enneagon, with one exception which is particularly curious. There's still a little I haven't quite figured out, so I'll explain what I do know. In the paper Steinbach uses α,β and γ for the ratios in increasing size, but I'll be using a, b and c. Applying Ptolemy's theorem we obtain the following 6 relations, after simplifying a few things:
a^2 = b+1
ab = a+c
ac = b+c
b^2 = b+c+1
bc = a+b+c
c^2 = a+b+c+1
Just as in the case with the pentagon and heptagon, we can use these relations to reduce any non-negative integer powers of a, b and c to an integer linear combination of 1, a, b and c. However, unlike the pentagon and heptagon, we don't get all integer powers in general. The relations alone can be used to obtain 1/a = a+b-c-1 and 1/c = c-b, however 1/b cannot be obtained from the relations alone, for reasons we shall see later.
We can then define maps P, Q, R, S: ZxNxZ->Z, where Z and N are the integers and the non-negative integers respectively, such that:
a^i b^j c^k = P(i,j,k)c+Q(i,j,k)b+R(i,j,k)a+S(i,j,k) for all i, k in Z and j in N.
Multiplying through by a, b and c gives forward recurrences for P,Q,R and R in the i, j and k indices, and dividing by a and c gives reverse recurrences for P, Q, R and S in the i and k indices. Like with the pentagon and heptagon, we see that Q, R and S are just shifted copies of P. In particular:
Q(i,j,k) = P(i,j+1,k-1), R(i,j,k) = P(i+1,j,k-1) and S(i,j,k) = P(i,j,k-1) for all i, k in Z, j in N.
I do wonder if it's perhaps better to write P, Q and R in terms of S, as then P, Q and R are positive shifted versions of S in a fairly symmetric way.
To understand why we don't get 1/b from the relations, consider that there are multiple solutions to the relations. Any triple (x,y,z) satisfying the relations of (a,b,c), must also have:
x^i y^j z^k = P(i,j,k)z+Q(i,j,k)y+R(i,j,k)x+S(i,j,k) for all i, k in Z and j in N.
What are these solutions? Well we can use the relations to show that if (x,y,z) is a solution, x, y and z must be roots of the polynomials x^4-x^3-3x^2+2x+1, y^4-3y^3+3y and z^4-2z^3-3z^2+z+1 respectively. Each has four real roots which match up to form 4 distinct solutions and can be expressed as (a,b,c), (1,0,-1), (-c/a,b/a,-1/a) and (-1/c,-b/c,a/c).
We can now see the problem with deriving 1/y from the relations; the solution (1,0,-1). If 1/y could be derived from the relations, we'd be able to express negative powers of y as a linear combination of 1, x, y and z. But negative powers of y in the solution (1,0,-1) correspond to dividing by 0. We shall see this isn't all to the story though.
Now that we have our solutions to the relations, we can form a matrix equation A = MB, where:
A = {a^i b^j c^k, (-1)^k δ_j, (-c/a)^i (b/a)^j (-1/a)^k, (-1/c)^i (-b/c)^j (a/c)^k}
B = {P(i,j,k), Q(i,j,k), R(i,j,k), S(i,j,k)}
M = {{c, b, a, 1},
{-1, 0, 1, 1},
{-1/a, b/a, -c/a, 1},
{a/c, -b/c, -1/c, 1}}
Note that 0^j for j in N is just δ_j the Kronecker Delta at j, where we're using the combinatorial convention that 0^0 = 1.
We can then invert M and obtain B = M^(-1)A, where:
M^(-1) = 1/9 {{c-b+1, -3, -a+1, b},
{-c+2b-1, 0, c-b+a, -a-b+1},
{a-1, 3, -b, -c+b-1},
{-b+3, 3, -a+b+1, c+1}}
So we do indeed get Binet-like formulae for P,Q,R and S, for i, k in Z and j in N. But the story doesn't quite end there. What would happen if you try to use these formulae for negative values of j? Do we just get nonsense? We no longer get integer outputs for P, Q, R and S, we get 3-adic rationals in general, but they do indeed allow you to express negative powers of y as a linear combination of 1, x, y and z for three of the four solutions (x,y,z). And the formula actually works correctly! I don't currently have a complete explanation as to why, but I did observe one thing.
If y =/= 0, we can obtain a new relation z = x+1. We see this relation is indeed satisfied by three of the four solutions, but is not satisfied by the solution (1,0,-1), so perhaps this additional relation is enough to show that the Binet formulae indeed works for three of the four solutions for negative values of j.
Very interesting and thanks for sharing :) Maybe also have a look at the collection of additional insights that I pinned to the top of the top of the comment section. There was one about a super simple way of writing the heptagon Binet's formula. Also, maybe e-mail me any other comments to make sure that things don't get lost. burkard.polster@monash.edu
Thanks my friend. I always watch you early-anticipated videos on a nice Sunday afternoon here in Ontario Canada, bi a big mug of coffee. :)
That's great :)
At first blush, rs = r+s appears to violate dimensionality, but with your proof, using dimensions of length, there are "hidden" unit lengths in the equation, so, if our units are meters, we have rs m^2 = r*1 m^2 + s*1 m^2. Wonder if considering the dimensionality for some of the other development might yield more interesting insights?
With regards to time stamp 1:20. Works for all x and y when
r = (x+y)/x and s = (x+y)/y
This is also the rational set of solutions when x and y are integers.
Good point :)
The transformation of the pentagon ratios to a golden rectangle and the heptagon ratios to a golden 3-d box seems to suggest that the ratios present in a golden n-dimensional hyperbox (probably the wrong word) are present in a (2n+1)-gon. At a quick glance, Steinbach's paper doesn't seem to address this, so is it true?
As far as I can tell nobody has really looked at any of these things in detail beyond the heptagon. Having said that and based on what is known in general, I'd also expect a lot of the things I talk about in terms of geometry to carry over into higher dimensions.
Great to see!
A few avenues not explored:
The four triangles and their trigonometry. Including THE heptagonal triangle (with sides 1:r:s or angles pi/7, 2pi/7, 4pi/7)
r satisfies r^3-r^2-2r+1=0, and s is a root of the reciprocal polynomial.
Recurrence relations, similar to those relating to the Tribonacci constant.
Rational approximations to r (and s), starting with 1,2,9/5,182/101 (and 2,9/4,182/101)
Yes, lots and lots of nice stuff I did not get around to talking about in this video.
Step 1. You Tube recommendation
Step 2. What a nice title, A+B=AB ---> Golden Equation...
Step 3. Liked 👍 and Saved in "Watch Later" folder.
Step 4. Post this comment😅
Step 5. Enjoying the whole video...now....
Back in college I did my senior project involving some number theory-esque shenanigans and I came across the imaginary golden ratio by complete accident. Simply phi^2=phi-1. Lot of cool stuff with that guy too.
yeah, the nonagon is fun the way it unexpectedly collapses. Just when it's looking neat it suddenly gets even simpler. Makes a very satisfying doodle.
Sounds interesting what comment are you replying to here?
@@Mathologer - I'm replying to your suggestion in the video to find the equivalents to phi, r&s for the nonagon. I found the 6 equations for products of r, s and t, then calculated their real values and then ..
.. oh and while you're there - at the time you released this video I was playing with the extended integers a+phi*b and had just found a fun integer sequence. I'm trying to get it into the OEIS at the moment. If I can get the pair of sequences approved they will be A375484 + phi*A347493. I've had some helpful feedback from the reviewers so far, but do you know of other pairs of sequences in the form a+phi*b in the OEIS?
30:15
The phi is equal to (sqrt(5) + 1) / 2 the other solution to the equation is (-sqrt(5) + 1) / 2. The difference between them is sqrt(5). If we rearrange the phi = (sqrt(5) + 1) / 2 we get sqrt(5) = 2 * phi - 1. So the other solution in terms of phi is: phi - (2 * phi - 1) which simplifies to: - phi + 1 = - (phi - 1) and since phi - 1 = 1 / phi, -(phi - 1) = - 1 / phi.
That's it :)
Jeez! Lovely video! In addition to the gold rush in finding all the great stuff about r and s (rho and sigma), those ratios from all the higher odd-number sided regular polygons seem like another gold rush waiting to happen.
All the same humongous gold rush :)
Wow, this is really a gold mine, and very inspiring! Thank you so much! 😀
Glad you enjoyed this one :)
like 2 years ago when i first heard of the fact that the golden ratio was the diagonal of the pentagon i thought "hmm, that's interesting, what about other shapes", and since i already knew the diagonals of the hexagon i jumped to the heptagon. i independently figured out pretty much everything in this video, and because of that 1.801 and 2.247 became my favorite numbers, and i cannot express the euphoria i got when i saw the letters r and s at the beginning and realizing where it was going. thank you so much i am so giggity right now
Thanks Mathologer. Amazing and high quality presentation as usual. BTW, for 9-gons the relations are t = r+1, r+t = r*s, s+t = r*t, for 11-gons are r^2 = s+1, r+t = r*s, s+u = r*t, t+u = r*u.
11:58 One small correction. The golden spiral is _not_ made up of quarter-circles. It _is_ the logarithmic spiral with growth factor φ. The succession of quarter-circles approximates the golden spiral. A Fibonacci spiral also approximates the golden spiral, with the approximation improving as the number of squares increases.
0:21 "mathematicians, artists, wizards, and a lot of crazies" ... It feels like you've said "mathematicians" four times!
Well, the intersection of those four sets of people is not insignificant :)
24:27 I was curious about the area of the thin green rectangle that's leftover. For a 5x8 rectangle it's about 0.721 units. Interestingly, as you scale up to higher Fibonacci rectangles, that leftover rectangle gets thinner and thinner, and the rate it gets thinner cancels out the rate that the Fibonacci rectangle grows, so that the area converges to a fixed value of 0.7236. After some fiddling around, I found that the exact value that the area of the leftover rectangle converges to is (phi^2 + 1)/5 or (1 + (1/sqrt(5))/2
That's also quite a fun and surprising observation :)
Watching this I had a few thoughts/questions:
1. Looking at diagonals, there was a "jump" from the pentagon to the heptagon - two shapes with odd numbers of sides. Are there any similar ratios/relationships for the hexagon or octagon? What about the square - too trivial?
2. If not, is there some reason why the even sided shapes do not exhibit this behaviour?
3. If the ratios of segments in a pentagon bear a connection to the golden rectangle, and those in a heptagon a connection to the "golden box", can we hypothesise that the ratios of line segments in a nonagon have a connection to a 4-dimensional golden object? Can we generalise to higher dimensions? Something about those diagonals is hinting at graphs and vertices but I can't make the connection.
These are just thoughts - I don't think I have the maths to effectively investigate, maybe other viewers have some ideas 🙏🏼
A few months ago was was playing around with phi^n and 1/(phi^n). There are some interesting recursion relations in there. Another great video from Prof Mathologer!
Glad you enjoyed this one :)
this super cool not gonna lie, i never thought about the golden ratio in so depth
I’m still on this video ( a week later ) Been chipping away at it bit by bit. It’s like a little candy I can enjoy at the end of each day after I’m done with all my school and work related math.
Glad you are enjoying this video so much :) Recent subscriber, where have you been all those years?
Another nice thing about fibo fractions a/b, so that a>b. For both a/b and b/a, let's multiply the denominator by 2. When both a=1 and b=1, both a+2b and b+2 = 3.
After that, a+2b -> Lucas numbers and b+2a -> Fibonacci numbers.
Maybe also check out my video on the Fibonacci and Tribonacci numbers. It's also got all sorts of fun stuff like this about the Lucas numbers :)
20:20 I wonder what the set of all possible limit points looks like. Without having given much thought to that yet, I guess it would look like some kind of fractal?
Yes it is a Cartesian product of three Cantor sets: all points (x,y,z) where x,y and z all lie in a different Cantor set. Rather than the standard ratio of 1/3 in the cantor set construction, the ratios vary and, well, are some other ratios that probably have nothing to do with the rest of this video. 😉
If you consider the equivalent thing with the rectangle and φ, then the cut points in each segment are 1/φ and 1/φ² , so the Cantor set is full of golden ratios. This is fun to look at in base φ.
Thanks for doing some of the answering of questions :)
@@Mathologer can’t help myself. :-)
Some interesting generalizations. I like it.
This was very interesting to watch. I was thinking about making some video animation about the golden ratio and pentagons, but this video also gave me some new food for thoughts. It's like the discoveries around this ratio is endless.
I've made quite a few golden ratio themed videos already. Check out some of the other ones linked to from the description of this video :)
@@Mathologer One step ahead of you, already watched those videos :) But I'll probably need to watch one of those again in the future.
I love ab=a+b. It's also closely related to my favorite pair of rational functions, y=1-1/x and y=1/(1-x).
Lovely video as always, well presented and many fun insights!
The 2+2=2•2 is also 2^2.
This also reminded me of a different equality: x^y = y^x, which has the obvious solutions x=y, but also another curve of solutions with only 1 integer solution: 2^4 = 4^2.
Just a fun curiosity I once thought about while trying to go to sleep. 😂
Some number curiosities that I've included in videos so far include 2x2=2+2, 1x2x3=1+2+3 and 3^3+4^3+5^3=6^3. Those other curiosities you mention are definitely also bouncing around in the back of my mind waiting for their time to shine :)
A little late to comment, but I actually read the paper this is from a few months back! I was browsing the Bridges Archive in my free time, and it is so satisfying to see it covered here.
For fun you should try to apply Ptolemy's theorem to the hexagon in your logo :)
I always roll my eyes when people talk about galaxies and nautilus shells in relation to phi, it's not true, but also just observing patterns is far less interesting than understanding them. This meant I had always had a "resistant" feeling towards phi.
Mathologer has well and truly turned me into a phi appreciator, first with the continued fraction representation explanation, and now this, I'd somehow never seen Binet's formula, or at least if I had, I had forgotten it, but the derivation and wider connections was absolutely wonderful.
I do want to note that as a colour blind viewer, I do struggle with some of the coloured lines in some videos (at the start), it's not an easy thing to solve, I know I tend to prefer dashed, dotted, double, ect. lines instead of colour indicators but that can get messy fast, it's just something maybe worth considering.
If you're interested I've linked to the original phi debunking articles in the description of this video. Also, have you seen Mario Livio's book? That one does an excellent job in separating fact from fiction :)
Very cool!
When you gave your formula for generating each triple from the previous triple, that technique has a clear higher-dimensional analogue. If you generate n-tuples using this rule, I would imagine that would correspond to a golden n-dimensional hypercube, which would have a corresponding golden spiral. I’d be curious to know if these higher dimensional analogies had any interesting properties, applications, or divergences from their lower dimensional counterparts.
Let's hope somebody gets curious enough to nut out the details. Shouldn't be too hard :)
También hubiéramos visto la Súper proporción áurea.
Buen video.
You answered a question I had a long time ago in undergrad when the professor used a shift operator on the recurrence relation that leads to the golden ratio. Great video as always! I'm in Greece rn on holidays so a little rho and sigma would go nice :)
That's great :)
@@Mathologer :)
Sir your laugh in between seems you read some viewers minds 😅
I can predict quite reliably what many viewers will be thinking (not hard :)
Reminds me some work I've done with Phi as a numerical base :D
The dimensionality expansion of all this is... staggeringly amazing !
Yes, very beautiful, isn't it ? :)
Time for golden tesseract 😎
Amazing display of the art of mathematics, brilliant video ! Many thanks
What a wonderful journey through mathematics land! :) I really didn't see that rectangular grid of equations coming at 38:10. I'm now assuming that if you start with a 9-gon, that there will be 3 constants that, together with 1, satisfy 4 equations. Super interesting video, which scratches my itch for integer sequences as well :D
Maybe also check out my comment pinned to the top of this comment section for a summary of interesting additional details that I did not mention in the video and/or that popped up in this comment section.
As a Phi fan, I loved this video especially. Great work. Thanks.
Great :)
This is GOLD
Excellent video. Thank you.
Folks should also check out:
What's 1/2 Doing in Φ? - the Golden Ratio
What's inside Φ? - the Golden Ratio
on the Imaginary Angle YT channel.
I have a special relationship to Ptomelys Inequality because it's the first theorem where I came up with an original and beautiful proof of it. And it was actually because of you :D Back then, I was re-watching some video about proofs of Pythagoras and one of the proofs involved scaling the original triangle by the factors a, b and c and then rearranging the parts in a clever way. You can do the same thing with Ptolemys Inquality!
Take any quadrilateral with side lengths a, A, b, B, c, C like in your video but make sure that c is "in between" a and b. Take three copies of the quadrilateral and scale it by a, b and c respectively. Notice that two of your copies now have a side with length ab and diagonals ac and bc respectively. Join them on the side ab. Now notice that you can fit the third copy perfectly on the diagonals of the other two, as the third copy has two side lengths ac and bc too and the angle also matches. Put it in place and look carefully. You will have formed a triangle with side lengths aA, bB and cC. Ptolemys Inquality therefore follows. And the equality will hold exactly when the point lands exactly on the cC side. Which will happen exactly when the opposite angles of the original quadrilateral add up to 180 degrees which is exactly the case when it is circular. qed
I stand by my opinion that this proof is absolutely marvelous and the best proof of Ptolemys Inequality ever. And it's perfect for an animation so if you ever want to show it, I absolutely allow you to do so :)
I agree, that proof is super nice. Someone actually animated it on the Wiki page for Ptolemy's theorem :) en.wikipedia.org/wiki/Ptolemy An earlier version of the slideshow for this video actually also included an animation of this proof.
@@Mathologer Ah cool I didn't see that before :) I also see it "only" animates the equality part for circular quadrilaterals so I guess my version is pretty much a generalized version that also covers the inequality part. And the animation is based on work from 2012? It's always so interesting how we constantly find new beautiful proofs of things that has been known for centuries
I'll also make an animation of your proof in the next couple of days :)
@@Supremebubble Just in case I end up covering your proof would be good to know your name :) Can you send me an e-mail please? burkard.polster@monash.edu
@@Mathologer I sent you one :)