Divine high PHI: The power of AB=A+B (Mathologer masterclass)

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). 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/GoldenRa… 11. Something VERY pretty: An uncurling golden square spiral www.youtube.com/shorts/3hhonbIAKUU

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.

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...😂

Wunderbar, diese Wiederbelebung der fast in Vergessenheit geratenen Entdeckung von Steinbach! Wie immer, Vergnügen pur und besten Dank!

"Mathematicians, artists, wizards, and a lot of crazies." So... just crazies, then?

I always get a 'WOW!' when I watch your videos. Thanks for bringing this stuff to YouTube.

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

our monthly dose of underrated mathematics

When I saw the penrose tiling at the end I gasped! So interesting as always ❤

I find it astonishing how numbers align so well. Much better than stars. Beautiful video as always.

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.

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.

You're the only math youtuber that makes me audibly gasp every video, let alone multiple times. Keep up the amazing work!

Thank you Burkard. Another fine video with lots of interesting and surprising twists and turns!

These videos are so fun! Once I begin, I get so hooked that it's impossible for me to stop.

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 ;)

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 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

The videos are always worth the wait! ❤❤❤

The best Mathologer video I've ever seen (and the level of Mathologer videos is always very very veru very high...)