A rough Pi approximation from coprime integers

In this short, plot all the points in the positive integer grid up to 101. If a coordinate (x,y) has numbers x and y that are relatively prime, we shade the dot blue; otherwise we shade the dot red. A famous result states that roughly 6/pi*pi of the dots will be blue. So we can use this image to get an approximation for 6/pi*pi and in turn use that to get an approximation for Pi. We get 3.12, which is not really good, but it’s kind of amazing that such a small grid gets us that close. Using a larger grid will give you a better approximation.
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[@Ninja20704]

For those who are wondering why, here’s a proof though a bit complicated. First look at the prob that two numbers share a prime factor. 1/2 of all numbers are divisible by 2, 1/3 of all numbers divisible by 3 and etc. In general, 1/p of all numbers are divisible by p. The chance that both numbers are divisible by p is 1/p*1/p=1/p^2. Thus, the chance that p does not divide both of them is (1-1/p^2). If the gcd of the two numbers are 1, then no prime is able to divide both of them. So the overall probability call it P is P=(1-1/2^2)(1-1/3^2)(1-1/5^2)… for every prime number. Now seemingly unrelated, consider 1/P. When we take the reciprocal of a product, we can take the reciprocal of each term then multiply. 1/P = [1/(1-1/2^2)][1/(1-1/3^2)][1/(1-1/5^2)]… Now the key is that each bracket is a geometric series 1/(1-x)=1+x+x^2+… So each bracket 1/(1-1/p^2) becomes 1 + 1/p^2 + 1/(p^2)^2 + 1/(p^3)^2 +… If we substitute this for each bracket and do all the distribution, we get the sum 1+1/2^2+1/3^2+1/4^2+1/5^2+… (Each bracket contains the prime to every possible power, and we are multiplying across all primes. So by FTA every whole number appears exactly once since we will have every possible combination of prime factorisations from the distribution) This sum is famously known to be pi^2/6. Thus, 1/P=pi^2/6 P=6/pi^2 In fact, if you chose n numbers, the probability that the n numbers are n-wise coprime is going to be 1/zeta(n) where zeta(n)=1+1/2^n+1/3^n+1/4^n+… This method is only an approximation because we cap x and y at 101. For this to be exact you have to allow x and y to be any arbitarily large size.

[@hugogonzalez1749]

Fascinating

[@adambrzeczyszczykiewicz7365]

Nice! ❤ 🎉

[@MichaelRothwell1]

Very nice, thanks for the explanation!

[@muditksingh]

thanks for the information!

[@Robin-Dabank696]

You lost me at the 7th paragraph... it's too complicated for me 😭😭😭

[@arlogodfrey1508]

I absolutely love this way of visualizing numbers. So much information about so many different aspects from such a simple map.

[@Blazingcharger]

This seems strangely simple

[@jblook777]

This seems familiar to zeta of 2 which is (pi)^2 /6. The zeta function relates to primes with analytic continuation, so naturally this similarity with the coprimes can relate with the Riemann Zeta Function.

[@bijipeter1471]

Thank you, sir

[@MinhBui-h4x]

interesting!! does it approach Pi when the number of cells increases?

[@MathVisualProofs]

Yes it does!

[@Sw3d15h_F1s4]

ive got a good way. first write down 3, since everyone knows that bit. then put the .14, and from there start randomly guessing numbers. close enough for engineering!

[@lamemelord]

We need a 3b1b video about this lol

[@CeresouslyAnimates]

Pi = approximately 3.1416, not approximately 3.1201

[@2048tile-s5y]

*3.14

[@yimphockany808]

My ocd💀💀💀💀💀💀💀

[@shruggzdastr8-facedclown]

What I find interesting about the decimalization of the fraction 6/pi^2 is how it roughly approximates phi-1

[@jasonzawtun]

I don't understand.

[@LamrabtiNaoufal]

it cornectded from the 5,000 balls of phi

[@LamrabtiNaoufal]

and phi is cornectded by 5,000 balls is just a bace 5,000 balls this is not 5,000 5,278

[@LamrabtiNaoufal]

that is bace 5,278 balls

[@JustinLe]

does this only work for 101 because if so it's kind of not that interesting

[@MathVisualProofs]

Approximation gets better as you take the grid size larger.

[@JustinLe]

​@@MathVisualProofs oh I didn't realized it worked for other N too 😮 then it's pretty cool now

[@MathVisualProofs]

@@JustinLe I should have snuck in at the end that the approximation gets better for larger grids ...

[@user-pr6ed3ri2k]

Sulfuric

[@hornkneeeee]

I am here to conclude yet again that this is extremely trivial as π is simply 3, i didnt know this proof but its quite fascinating it was able to get close to π to such a degree

[@officialalphabet]

Why?

[@daburak2559]

This is just basels formula written in a different way, when 101 becomes closer to infinity the value also becomes closer 6/pi²

[@philalethes216]

@@daburak2559 There needs to be a whole video on this relationship. This video is missing so much crucial information that it’s practically criminal.

[@deepeshr6025]

Early

[@Oldhogleg]

Here's an interesting way to approximate pi: Take the atomic number of carbon 666 and divide it by the boiling temperature of water 212. It's accurate to 4 decimal places.

[@dank.]

Try 355/113. It's the best approximation until the numerator/denominator get into the tens of thousands!

[@beniocabeleleiraleila5799]

Aint no way bro is using fahrenheit