Irregular numbers and the coolest sum I have ever seen! 

interesting how the sum of reciprocals of **all** squares has a pi^2 in the numerator, and the sum of reciprocals of **only square-free** composed squares has a pi^2 in the denominator.

This is a special case of the identity sum(n=1 to infinity) of |μ(n)|/n^s = zeta(s)/zeta(2s). The Mobius function captures the whole square free thing very nicely

Fun fact: the density of the squarefree numbers is 6/π². This means that the sum of the reciprocals of squarefree numbers diverges.

"Square-free numbers provide an alternate mapping of binary numbers: the exponent of the nth prime is the digit representing 2^n, where 2 is the pth prime." As an example: 15 = 3x5 -> 0b110 = 6 14 = 2x7 -> 0b1001 = 9 This is also to say: here's how i would map N onto square-free N

i love how you can show some products that are extremely fascinating and beautiful and just be like "so im sure you've all seen these before" in mathematics. this subject is so beautiful

You can continue this idea to find the sum of the reciprocals of the cube-free numbers squared as well. It comes out to 315/(2pi^4) if I’ve done it correctly. I used zeta(6) and the difference of cubes formula instead.

1:51 I think you meant to say it does not converge if n equals 1. Very nice video about a splendid and accessible result. I discovered a love for the Möbius Mu function last year, and we can write this result in the neat way using this function: the infinite sum indexed over all natural numbers of the terms (u(n)/n)² equals 15/π², where u(n) is the Möbius function.

I first discovered squarefree integers as a way to represent sets of natural numbers compactly. Very neat equation here.

That is wonderful Michael - loved it.

Again a great video and thanks a lot for giving us this excellent RU-vid content! Happy holidays to you and best of wishes!

That was a lot of fun. I've been enjoying your videos, but I'm enjoying the ones I can actually follow even more

4:09 Can you expand the expression when we look for ζ(-1) ? 👀 11:43 Good Place To Stop

One of the best videos on this channel so far!

Very nice Sir. Keep sharing a lot of good knowledge

One of the most interesting videos I have seen on this channel!

Very nice Christmas gift. Merry Christmas

Reposting and slight editing of recent mathematical ideas into one post: Split-complex numbers relate to the diagonality (like how it's expressed on Anakin's lightsaber) of ring/cylindrical singularities and to why the 6 corner/cusp singularities in dark matter must alternate. The so-called triplex numbers deal with how energy is transferred between particles and bodies and how an increase in energy also increases the apparent mass. Dual numbers relate to Euler's Identity, where the thin mass is cancelling most of the attractive and repulsive forces. The imaginary number is mass in stable particles of any conformation. In Big Bounce physics, dual numbers relate to how the attractive and repulsive forces work together to turn the matter that we normally think of into dark matter. The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass. In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards. Mediants are important to understanding the Big Crunch side of a Big Bounce event. Matter has locked up, with particles surrounding and pressuring each other. The matter gets broken up into fractions of what it was and then gets added together to form the dark matter known from our Inflationary Epoch. Sectrices are inversely related, as they deal with all stable conformations of matter being broken up, not added like the implosive "shrapnel" of mediants. Ford circles relate to mediants. Tangential circles, tethered to a line. Sectrices: the families of curves deal with impossible arrangements. (The Fibonacci spiral deals with how dark matter is degenerated/broken up and with supernovae. The Golden spiral deals with how the normal matter, that we usually think of, degenerates, forming black holes.) The Archimedean spiral deals with matter spiraling in upon itself, degenerating in a Big Crunch. The Dinostratus quadratrix deals with the laminar flow of dark matter being broken up by lingering black holes. I'm happily surprised to figure out sectrices. Trisectrices are another thing. More complex and I don't know if I have all the curves available to use in analyzing them. But, I can see Fibonacci and Golden spirals relating to the trisectrices. General relativity: 8 shapes, as dictated by the equation? 4 general shapes, but with a variation of membranous or a filament? Dark matter mostly flat, with its 6 alternating corner/cusp edge singularities. Neutrons like if a balloon had two ends, for blowing it up. Protons with aligned singularities, and electrons with just a lone cylindrical singularity? Prime numbers in polar coordinates: note the missing arms and the missing radials. Matter spiraling in, degenerating? Matter radiating out - the laminar flow of dark matter in an Inflationary Epoch? Connection to Big Bounce theory? "Operation -- Annihilate!", from the first season of the original Star Trek: was that all about dark matter and the cosmic microwave background radiation? Anakin Skywalker connection?

Merry Christmas to you, Michael. Thanks for your video. Note: The "Euler Product" for zeta(s) is even true for Re(s)>1. For zeta(n) with n a natural that surely means: n>=2, because there is no natural number n that satisfied 1

So...the sum of inverse square squares is pi^2 / 6 - 15/pi^2. Which is quite close to 1/8. Ratio - the sum of inverse square squares / the sum of inverse squares - is 1 - 90 / pi^4 or around 7.6%, the inverse of which is roughly 13 (close to 10 + pi + 0.005).

Hey michael merry christmas 😊

An interesting problem to do : calculate ζ(4) or ζ(2n) as a function of π^2n and bernoulli number (it may be a little bit longer)

We can even describe it more succinctly by saying sum from 1 to infinity of (u(n) / n) ^ 2 = 15 / pi ^ 2 where u(n) is the mobius function.

Man ..........this is way above my capabilities.

I really liked this video as a tour through some concepts in math that have felt mystifying for some time now, in a way that felt natural.

this is hypnotizing

Yes Michael! Especially interesting sum you will get if instead of the squares of the square-free numbers use first powers of them, because zeta(2*z)/zeta(z) -> 0 as z -> 1. So that the sum of the inverse of the square-free’s will be 0 instead :-(.

I couldn't fall asleep but this video helped me get my rest. Thanks.

@1:40 - the expansion of this product as an infinite sum converges for all n > 1.

I love square free numbers! Nice identity

I know that the infinite product diverges when n=1, but do you know of any generalization of the finite terms that approach the product as p_m -> oo? Specifically, is there a good way to think about (or describe) the analytic continuation of the function f(n)=Π from i=1 to i=n of 1/(1-1/p_i), or of that of 1/f(n)?

pretty cool

Up next: a closed form expression of 25.79435016... (π^3)/ζ(3). or, ζ(3) = π^3 divided by that number.

The infinite sum of inverted square of square free numbers converges to a nice sum.Very interesting.

You are very serious person. Only time i have seen you laughing is in that video on learning chess.

Is there not a much simpler way to do that : Every integer is in only one way the product of a square free and a square. The sum of the inverses of the squares of the integers is then the product of the sum of the squares of the inverses of the squarefree and the sum of the inverses of 4th power of the integers. So the sum we're interested in is the ratio of pi²/6 divided by pi^4/90.

Cool 👌

Nice

Does anyone know whether the infinite product of (1+1/n^2) for all natural numbers n converges and, if so, to what value?

So, the sum of the reciprocals of the squares of the non-square free numbers is π^2/6 - 15/π^2.

so a simple difference would then yield the sum of the the reciprocals of the squares of all non-square-free naturals

Can we prove that if we inscribe two regular polygons to the same circle with one of them having n sides and the other n+1 sides(n≥3), then P(n+1)>P(n), where P(k) is the perimeter of a regular polygon inscribed to a circle with k sides. I can't prove that, please can we prove it...

Does the sum of the reciprocals of the square-free numbers (without squaring in the denominator) diverge or converge?

I wonder if my logic is correct: since we want the sum of 1 over the square of square-free numbers, can't we just use zeta(2) - zeta(4)?

Either I did not get it or there is a mistake in the argument. The set of square free numbers is different from from the set of numbers in the denominators because the latter only have nbers with 2 prime factors. 30 (2x3x5) has 3 prime factors, is square free (no prime more than once) but does not show in the denominators.

Why doesn’t it converges when n=1

So we can conclude that: ∑ (n=1..∞) 1/(p_n)^4 = (π^4-90)/6π^2 ~ 0.12511 where p_n is n'th prime number (I just subtracted your result from the Basel solution π^2/6, would like someone to verify that?)

also known as 60/τ² :)

Here's something that might be fun to explore. Say we have a Pythagorean rectangle (that is, a rectangle whose diagonal is an integer measure) ABCD (clockwise from top left) with a Pythagorean right triangle BEC glued to its right side and not overlapping. The height of both (AD, BC) is a. The rectangle has width (AB, CD)=b, the triangle has base (BE)=c. Together they form a quadrilateral ABED with angles at A and D being right and angle E is the larger acute angle of triangle BEC. Can we choose a combination of Pyth rectangle and Pyth triangle where not only is side BE an integer, not only the quadrilateral's (and the rectangle's) diagonal BD an integer, but *also* the other diagonal of the quadrilateral, AE? Put another way: if all variables that follow are natural numbers, can all these statements be true? a^2+b^2=n (rectangle diagonal) a^2+c^2=m (triangle hypotenuse) a"2+(b+c)^2=k (quadrilateral long diagonal)

so can you figure out Zeta(3)?

The Riemann product has complex exponents but I still don't see how this could ever be zero, even at the trivial roots. What am I not seeing??

Wait. You could have continued.

Noice

6/pi^2 is also the probability of "2 random natural numbers are coprimed".

At 3:40, you did NOT demonstrate that rearranging those infinite sums (picking and choosing which terms to multiply) is valid in this case. It seems to me that there should be some left-over products, but I cannot give an example.

That is not a zeta, that is an S with some rho aspirations.

What is an irregular number? He never told us.

Is this really the coolest sum you have ever seen? That's seem like an overstatement.

ya u lost me

Chess is more dangerous than math. In chess game we cannot undo things after we make a wrong move. Even if a move is very simple and a player misses it there is no way he can undo things. Chess is unforgiving.

the error is at 2:58 , you cannot cherry pick a fraction of an infinite sequence and leave the rest behind. You have to ensure every fractions are taken. You cannot take more 2s then 7s ... Math error.