Grant Sanderson

Grant Sanderson

@3blue1brown.com

Math videos

20.4K followers18 following54 posts

Happy Pi Day! In a certain sense, π is not a constant, but a variable. Using our usual Euclidean distance, it is 3.14159… but applying other L^p norms on ℝ², half the unit circle's perimeter will give other values. For instance, at p=1 (taxicab geometry), “π” = 2√2. At p ≈ 2.2, it's 3.20.

Ah! Correction, as several have pointed out, it should be 4 in the taxicab metric. And the same correction, i.e., measuring the circle in the appropriate metric, the relevant date-shifting-joke-value to be closer to 2.6

Happy Pi Day! In a certain sense, π is not a constant, but a variable. Using our usual Euclidean distance, it is 3.14159… but applying other L^p norms on ℝ², half the unit circle's perimeter will give other values. For instance, at p=1 (taxicab geometry), “π” = 2√2. At p ≈ 2.2, it's 3.20.

Anyway, the video I was hoping to have out this day will be out closer to the 20th. Some call it “missing your deadline”, but I prefer to think of it as giving the L_{2.2} norm a little love.

Well, math terminology being what it is, something like this was bound to happen eventually. (If you're curious about why these balls are so puny, the full talk is up on YouTube)

In the fifth and final of a series of guest videos I've been posting, @BenSyversen delves into a question anybody who has had to do ruler and compass constructions in a geometry class may have wondered: What's the point?

For context, I knew I'd want to take some time away this year (paternity leave!), so I reached out to a few other creators whose work I respect and asked if they'd be interested in me commissioning a guest video during my absence. It's a pretty good lineup coming!

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It's known you cannot do better than O(√N), which is certainly not as earth-shattering as an exponential speed-up would be, and questionably useful given the enormous overheads of quantum computing. Nonetheless, it's thought-provoking that such a thing is possible!

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What do they do then? This video builds up to Grover’s algorithm, a general method in quantum computing for finding solutions to any NP problem, i.e., anything where you have a quick way to verify solutions, even if finding them in the first place may be hard.

As a generic stand-in for the kind of problem it solves, suppose you have a function acting on {1, ..., N} which returns True on one and only one value in this set. If all you can do with this function is try it out on numbers, then it takes an average of (1/2)N steps to find the answer.

A common misconception about quantum computers is that they would solve hard problems by trying all possible solutions in parallel. This vaguely gestures at something true, but the reality is more subtle.

I have a pi-day challenge for all the physics students among you (or anyone willing to set up an experiment). If you share your results with me by March 10th, I may feature them in a video, depending on how good the results are and how many I get.

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More generally, with a mass ratio of N-to-1, the number of collisions is around π / arctan(1 / sqrt(N)). So any big mass ratio gives you an approximation of pi by multiplying the number of collisions by arctan(1/sqrt(N))

I have a pi-day challenge for all the physics students among you (or anyone willing to set up an experiment). If you share your results with me by March 10th, I may feature them in a video, depending on how good the results are and how many I get.

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Also, it's a wildly inefficient way to compute pi. To even get "3.14" you'd need this to work with a 10,000-to-1 mass ratio and have a way to count all 314 collisions. Matt Parker and I actually gave this a go, and the results were...okay, but could definitely have been improved :)

Note, there's no reason to restrict yourself to powers of 100. For example, you could use powers of 4 to compute pi in binary. A mass ratio of 64-to-1 should give 25 collisions, which is 11001 in binary, and pi looks like 11.001...

I have a pi-day challenge for all the physics students among you (or anyone willing to set up an experiment). If you share your results with me by March 10th, I may feature them in a video, depending on how good the results are and how many I get.

The original puzzle assumes zero friction and zero energy loss in collisions, so obviously there are limits to how far you can get. I can tell you the real limiting factor is energy lost in collisions, more so than friction. The hardest part is energy lost in collisions, more so than friction.

Well, perhaps not “new” new. Books have a notion of a second edition, and while YouTube videos usually don’t, I wanted to make a new edition of one of the earliest videos on the channel, as this is one of my favorite pieces of math.

Aside from animating this beautiful piece of math better, I had an itch to address new research that’s happened since, and to pull in numerous other mind-bending connections.