Beyond the Golden Ratio | Infinite Series

You know the Golden Ratio, but what is the Silver Ratio? Learn through active problem-solving at Brilliant:

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Polygons, Diagonals, and the Bronze Mean by Antonia Redondo
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Cut a line segment into unequal pieces of lengths a and b such that the ratio a to b is the same as the ratio (a + b) to a --- that is, so that big over medium equals medium over small. This is how you construct the golden ratio Phi. If a rectangle has an aspect ratio of Phi, you can subdivide it forever into a square and another golden rectangle, and make fun logarithmic spirals by connecting the corners.

Written and Hosted by Gabe Perez-Giz
Produced by Rusty Ward
Graphics by Ray Lux
Assistant Editing and Sound Design by Mike Petrow and Meah Denee Barrington

Special thanks to Roman Pinchuk for supporting us on our Converse level on Patreon.

Along with thanks to Matthew O'Connor, Yana Chernobilsky, and John Hoffman who are supporting us on Patreon at the Identity level!

And thanks to Nicholas Rose, Jason Hise, Thomas Scheer, Marting Sergio H. Faester, CSS, and Mauricio Pacheco who are supporting us at the Lemma level!

[robspiess]

Fun fact: Phi (1.618) is really close to the ratio between miles and kilometers (1.609) which means you can use adjacent Fibonacci numbers to quickly mentally convert between them.

For instance: 89 miles is nearly 144 km (it's actually 143.2), or 21 kilometers is roughly 13 miles (13.05).
You can even shift orders of magnitude to do longer distances! e.g., 210 miles is around 340 km (multiplying 21 and 34 by 10) which is close to the actual answer of 337.96 km.

[matheuscardoso]

I love the way this channel give us puzzles which they dont have the answer yet. This is like

"I trust you to go beyond"

And this is simply awesome. I wish our schools had more of that.

[KungFuBlitzKrieg]

I found this video very satis-phi-ing.

[ronanstephens]

Found this video to be much better than recent ones. I think because it involved somewhat more accesable maths whilst still being suprising and less well known information. I think episodes like this, where you dont try to attempt complex proofs but instead introduce interesting ideas in an intuitive way and give people a direction and material for further reading if they are interested, have the best format. In cases where you want to explain more complex ideas and give long proofs maybe make them more detailed by break them up over episodes and/or look for proofs with good graphical representions (like 3Blue1Brown) perhaps. Mirroring the way PBS spacetime makes videos, with each providing interesting new information whilst building the knowledge base to allow for more complex topics would be good. Not to be too critical tho, you guys do a great job!

[docopoper]

Wow, there are a lot of mean comments in here. Please don't fall into the trap of simplifying this channel too much because of commenters. Even though I didn't fully understand what was going on in the previous video I really appreciate there being a maths channel that's willing to approach complex topics.

[sparhopper]

Now *this is exactly why I subscribe to 'Infinite Series'!
Cheers.

[Actuarium]

Next observation:

By d_(m, k) > d_(n, k) for m>n and d_(n, k)>d_(n, j) for j < k < ceil(n/2)-1 it is possible to proof (!) that sigma_n is not any diagonal-ratio of any regular polygon (for a given n>2). Indeed that is much easier than in the paper. Now for a given n you can easily proof with a machine that sigma_n is not such a numbers, you have to only calculate very few numbers for each n. I can give the details if anybody is interested.

That does not give us the answer for all n, but it is maybe a way to go.

[josealvim]

Brilliant, I'm happy you guys talked more about the gaps in the theory of the last video! Keep it up!

[plasmaballin]

I looked up 5, 8 on the Online Encyclopedia of Integer Sequences. The first sequence I found that started with 5, 8 was numbers that are the sum of two successive primes. If this is the pattern for what polygons have the metallic ratios as the diagonal:side ratio, then the next polygon should be the dodecagon. I assume you have probably already tried the doddcagon though.

[plasmaballin]

Someone should show this video to the creators of all those other videos where they pretend that Φ is some mystical and sacred number.

[ryanburnette]

This answered questions I've had for a while but didn't know how to ask. The golden ratio reappears with any seed numbers or if you divide powers of the seed numbers too. This makes it obvious why those patterns emerge. Thanks guys!

[billrussell]

You guys are great. Thanks for the education! I'm almost 60 but I never stop learning!

[jkun]

The ratio of his uneven hood drawstrings is equal to the golden ratio.

[Actuarium]

Easy numerical proof (really a proof), that sigma_3 does not occur as a diagonal in any n-regular polygon (with site length = 1):

You have d_(m, k) > d_(n, k) for m>n and d_(n, k)>d_(n, j) for j < k < ceil(n/2)-1

The first n-regular polygon which can produce a diagonal with length sigma_3 = 3.3027756 ... is n = 11 (because for n < 11 the radius is too small).

Next we check, which diagonal in the 11-polygon is the biggest, but smaller than sigma_3. It is the 3rd diagonal

d_{11, 3} = 3.228707... < sigma_3
and d_{11, 4} = 3.51333

So with the second above inequality the 11-polygon is ruled out. We try the 12-polygon and find:

d_{12, 2} < sigma_3 < d_{12, 3}

Now see that d_{n, k} < k+1 and together with the first inequality above ALL other regular n-polygons are ruled out. So sigma_3 never occurs as a diagonal-ratio. Q.E.D.

Similar you can easily check any other sigma_m, but of course with higher m you will have more numbers to check, which is a easy task for a script.

At least we can now state the following:

For any given m we can give an algorithm which proofs (pretty fast) that sigma_m does not occur as diagonal-ratio in ANY regular polygon, IF that is true. Of course the same algorithm will find any occurences, like d_{5, 1} = sigma_1 and d_{8, 2} = sigma_2, too.

[Actuarium]

It should be noted that a real number r>1 is a metallic ratio if and only if r * (r - floor(r)) = 1 (equivalent to definition).
So sigma_n -> n with n -> infinity (side note).

Hence you can check for a given n all diagonal-ratios d_j of the regular, if the equation d_j if d_j*(d_j-floor(d_j)) = 1 holds. I checked for all n<100.000 : no other solution was found besides the one mentioned in the video.

But there is no obvious argument just by distances alone, because numerical experiments suggest that you can find for all given epsilon>0 some n, k and m with

d_(n, k) - sigma_m < epsilon

(like you would expect from a somewhat random distribution)

[TaiFerret]

It's nice to see a video about the other metallic ratios but I think the golden ratio is more famous because it's simpler. Especially since the Fibonacci sequence doesn't involve multiplication unlike the other metallic sequences. There are other ways to generalize as well, such as the Tribonacci sequence and its constant (which I often like to call "Tri" but is more commonly referred to as just "the Tribonacci constant".) The Tribonacci rule is adding 3 numbers together instead of 2.

[escobasingracia]

This is the type of mathematics that I love, please keep doing videos of this kind.

[Lucroq]

It's all fun and games until Tai-Danae pulls out the big guns.

[AdventuresWithPCS]

I noticed a pattern between two methods of calculating the ratio including your open question: σ_n = (n+sqrt(n^2 + 4))/2 = the ratio of the nth diagonal of the regular (n^2 + 4)th polygon to that polygon's side. In the video you showed this works for σ_1 and σ_2. I ran through some code and found that this does not generalize to n>2, but it does nonetheless approximate the true ratios rather well, maxing out at 8.0689% error for σ_6.

I was also wondering if the properties would extend to the "trivial" case, σ_0 = 1, and, checking back through the video, I think most if not all of them do!

[chrisg]

I have indeed found some "equally cool analogs in the rest of the metallic family", though regretfully not polygonal ones yet.

Each number in each metallic ratio formula, such as (2+sqrt8)/2 for the Silver, contains a term in, or derivable from, the sequence associated with that formula. But we're talking about the cooler "Lucas" version of that sequence. You probably know that what distinguishes the Lucas sequence from the standard Fibonacci is that while they share the same Golden ratio constant 1.618..., the terms in the Lucas approximate much more closely to 1.618...^n where n is the position or index. For example the 10th term in Lucas is 123, and 1.618..^10 ~ 122.996..., whereas in Fibonacci the 10th term is 55. Another interesting difference is that if you take the product of any pair of terms separated by one, such as 34 and 89, in Fibonacci, the absolute difference with the square of the term between is always 1. So for example (34*89)-(55^2)=1, but in Lucas it's always 5, the square root of which crops up in the formula for the Golden ratio, (1+sqrt5)/2. Lastly Lucas begins with the two terms 2 1, which also occur in the denominator and numerator respectively of the golden ratio formula.

All this seems to hold for the other metallic ratios as listed in that table too, something which may have escaped observation if nobody has thought of "Lucasating" the associated sequences. Try it with the Silver. Begin it with the two terms 2 2 and continue, using its recurrence a(n) = 2*a(n-1) + a(n-2).