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.
so we could find that the sum of non-square free numbers i.e. any number that is divisible by a square number is (pi^4-90)/(6pi^2) approximately about 0.125
And also there are more square free numbers then primes (trivial since every prime is square free by definition), so the sum of the reciprocals must be bigger then the harmonic series sum which diverges.
@@nitayderei is there something I'm not getting right?? The harmonic series is the sum of the reciprocal of every integer, while the sum of the reciprocal of the square-free numbers is just a selection of integers, so shouldn't it be smaller, not bigger?
@@nitayderei How can the sum of the reciprocals of the primes (or of the square-free numbers) be bigger than the harmonic series if all three diverge, i. e. approach infinity? Judging from the corresponding partial sums, the reciprocals of the primes add up slower than the reciprocals of the square-free numbers which add up slower than all naturals since each of those contains smaller terms than the next, don't they?
@@tomctutor I know the harmonic series diverges, but he said that a sub-serie inside the harmonic series (the sum of the reciprocal of the square-free numbers) diverges because it is bigger than the harmonic series, which is not correct
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
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.
"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
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.
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
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?
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).
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
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 :-(.
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.
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)?
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...
Well I know that lim n -> ∞:P(n)= π (If we set the circle as unit diameter)? The perimeter P(n) = (n/√(2))√([1- cos(2π/n)] for unit diameter. (obtained that from the law of cosines on a segment) So you could check that ∀n≥3: P(n)
It converges to [e^π - e^(-π)]/2π, wich is about 3.676077... From eulers product (I Guess Thats How it's called): sin(πx)/πx = (1-x²/1²)(1-x²/2²)(1-x²/3²)... You can "plug" x=i; on the right hand all the minus change to plus; and on the left You need to use the complex definition on sin(z)=(e^iz - e^(-iz))/2i; and You get that. In general: sinh(πx)/πx = (1+x²/1²)(1+x²/2²)(1+x²/3²)... :^)
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.
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?)
No, what You got is the sum of the inverses of the not-square-free numbers squared. To help You understand: π²/6 is the sum of all the inverses of the square numbers. 15/π² is the sum of all the square free numbers squared. If you subtract them You are left with the squares of the not-square-free numbers (4,8,9,12,16,18...).
@@josevidal354 You are quite right, I would not call them the "not square free numbers" rather the "squared numbers", those numbers that contain at least the square of a prime number as a factor. Thankyou though for the correction. ∑ (n=1..∞) 1/(s_n)^2 = (π^4-90)/6π^2 ~ 0.12511 where s_n is n'th "squared" number.
This expression for the Riemann zeta function is only valid when the real part of the input is greater than 1 (else the expression does not converge). Both the trivial and nontrivial zeros are not in the region where the real part is greater than 1, so that is why it looks like the expression can never be 0.
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.
@@callmezinc4698 He said that already. This merely demands that particular terms from each factor be chosen; it does NOT demonstrate that such a choosing strategy is valid.
"It seems to me that there should be some left-over products" Well, he accounted for every product with finitely many terms that are not 1. If infinitely many terms are not 1 then the product evaluates to 0. So it doesn't look to me like he left anything out.
When trying to disprove something by counter-example, you have to actually provide (or at least construct) that counter example. So let's construct it: Suppose there is a such a product of left-over terms q1, q2, ... (along with the "1" from all the rest of the groups). Given that each 'q' is either prime or 1, then there is a single, unique integer Q that is their product (fundamental theorem of arithmetic). Since it is exists (and has ordinality), we can stick it into the arrangement you complained about in the appropriate place and since it's unique, it's not displacing anything. If none of the q1,q2,etc terms are 1, that's fine, too... just keep multiplying them out. This contradicts your hypothetical "left-over".
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)
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.
You can subtract one sequence from another (monotonic decreasing) if that other sequence is included in the first. You may still have to prove convergence I agree.
@@lexyeevee No, he left a ton of terms unaccounted for, he's just lucky that in this case they converge on less than (∞^2)/(2^∞) which is an infinitesimal. But it was not guaranteed by the technique he used, nor his explanation.
@@hurktang it's every combination of every number of prime factors (to the nth power), which will give you every natural number exactly once by the fundamental theorem of arithmetic. nothing is unaccounted for
@@lexyeevee You don't listen to what you say. It is an INFINITE products of INFINITE sums. Once you have every natural number EXACTLY ONCE, you still have an infinitesimally small sample of all terms. There is still an infinity-1 of infinite sums to account for.
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.
Math is even more dangerous than chess. One time, a young man used Cauchy's integral theorem on an integrand which was not analytic on the inside of the contour. He saw a bright flash of light after seeing the result and has yet to regain his sanity. The universe does not forgive such transgressions lightly.
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.
so we could find that the sum of non-square free numbers i.e. any number that is divisible by a square number is (pi^4-90)/(6pi^2) approximately about 0.125
If you’re interested, check out the Moebius function and Dirichlet series inversion.
it's just nice to see how intimately pi and hence circles are connected to the primes.
Fun fact: the density of the squarefree numbers is 6/π².
This means that the sum of the reciprocals of squarefree numbers diverges.
And also there are more square free numbers then primes (trivial since every prime is square free by definition), so the sum of the reciprocals must be bigger then the harmonic series sum which diverges.
@@nitayderei is there something I'm not getting right?? The harmonic series is the sum of the reciprocal of every integer, while the sum of the reciprocal of the square-free numbers is just a selection of integers, so shouldn't it be smaller, not bigger?
@@nitayderei How can the sum of the reciprocals of the primes (or of the square-free numbers) be bigger than the harmonic series if all three diverge, i. e. approach infinity?
Judging from the corresponding partial sums, the reciprocals of the primes add up slower than the reciprocals of the square-free numbers which add up slower than all naturals since each of those contains smaller terms than the next, don't they?
@@winoo1967 The harmonic series diverges end of story, there are many convergent series in a divergent series.
@@tomctutor I know the harmonic series diverges, but he said that a sub-serie inside the harmonic series (the sum of the reciprocal of the square-free numbers) diverges because it is bigger than the harmonic series, which is not correct
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
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.
"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
That's.... actually kinda neat!
It feels like there must be some relationship here, that 15->6 in this notation conversion and we just saw how 15/pi^2 is related to 6/pi^2.
@@BridgeBum that's probably a coincidence but I would be delighted if it isn't
4:09 Can you expand the expression when we look for ζ(-1) ? 👀
11:43 Good Place To Stop
Bruh
@@void7366 Brother
you naughty
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.
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
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.
@1:40 - the expansion of this product as an infinite sum converges for all n > 1.
Again a great video and thanks a lot for giving us this excellent UA-cam content! Happy holidays to you and best of wishes!
One of the best videos on this channel so far!
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?
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)
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).
Very nice Sir. Keep sharing a lot of good knowledge
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.
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
Hey michael merry christmas 😊
Man ..........this is way above my capabilities.
One of the most interesting videos I have seen on this channel!
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.
ikr. Great vid.
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 :-(.
The infinite sum of inverted square of square free numbers converges to a nice sum.Very interesting.
this is hypnotizing
I couldn't fall asleep but this video helped me get my rest. Thanks.
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.
30 does appear in the denominators
Up next: a closed form expression of 25.79435016... (π^3)/ζ(3). or, ζ(3) = π^3 divided by that number.
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)?
So, the sum of the reciprocals of the squares of the non-square free numbers is π^2/6 - 15/π^2.
Why doesn’t it converges when n=1
so a simple difference would then yield the sum of the the reciprocals of the squares of all non-square-free naturals
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)?
Does the sum of the reciprocals of the square-free numbers (without squaring in the denominator) diverge or converge?
Diverges because sum of reciprocal of primes diverges
@@shantanunene4389 oh yes I got it. Thank you!
Diverges, as the sum of reciprocal of primes diverges
Obviously diverges, as the square free numbers grow like O(n)
You are very serious person. Only time i have seen you laughing is in that video on learning chess.
How often do you laugh while teaching math? I love math but I don't do much more than smile.
Michael penn is a genius...amazing conceptual depth ...love the lectures !!
Cool 👌
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...
Well I know that lim n -> ∞:P(n)= π (If we set the circle as unit diameter)?
The perimeter P(n) = (n/√(2))√([1- cos(2π/n)] for unit diameter.
(obtained that from the law of cosines on a segment)
So you could check that ∀n≥3: P(n)
Does anyone know whether the infinite product of (1+1/n^2) for all natural numbers n converges and, if so, to what value?
It converges to [e^π - e^(-π)]/2π, wich is about 3.676077...
From eulers product (I Guess Thats How it's called):
sin(πx)/πx = (1-x²/1²)(1-x²/2²)(1-x²/3²)...
You can "plug" x=i; on the right hand all the minus change to plus; and on the left You need to use the complex definition on sin(z)=(e^iz - e^(-iz))/2i; and You get that.
In general:
sinh(πx)/πx = (1+x²/1²)(1+x²/2²)(1+x²/3²)...
:^)
pretty cool
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.
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?)
No, what You got is the sum of the inverses of the not-square-free numbers squared.
To help You understand:
π²/6 is the sum of all the inverses of the square numbers.
15/π² is the sum of all the square free numbers squared.
If you subtract them You are left with the squares of the not-square-free numbers (4,8,9,12,16,18...).
@@josevidal354 You are quite right, I would not call them the "not square free numbers" rather the "squared numbers", those numbers that contain at least the square of a prime number as a factor. Thankyou though for the correction.
∑ (n=1..∞) 1/(s_n)^2 = (π^4-90)/6π^2 ~ 0.12511 where s_n is n'th "squared" number.
so can you figure out Zeta(3)?
I believe there is no known solution for odd values.
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??
This expression for the Riemann zeta function is only valid when the real part of the input is greater than 1 (else the expression does not converge). Both the trivial and nontrivial zeros are not in the region where the real part is greater than 1, so that is why it looks like the expression can never be 0.
@@willnewman9783 Thank you! I'll have to delve into this more.
Wait. You could have continued.
Nice
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.
Fundamental theorem of Arithmetic
@@callmezinc4698 He said that already. This merely demands that particular terms from each factor be chosen; it does NOT demonstrate that such a choosing strategy is valid.
"It seems to me that there should be some left-over products"
Well, he accounted for every product with finitely many terms that are not 1. If infinitely many terms are not 1 then the product evaluates to 0. So it doesn't look to me like he left anything out.
When trying to disprove something by counter-example, you have to actually provide (or at least construct) that counter example. So let's construct it: Suppose there is a such a product of left-over terms q1, q2, ... (along with the "1" from all the rest of the groups). Given that each 'q' is either prime or 1, then there is a single, unique integer Q that is their product (fundamental theorem of arithmetic). Since it is exists (and has ordinality), we can stick it into the arrangement you complained about in the appropriate place and since it's unique, it's not displacing anything.
If none of the q1,q2,etc terms are 1, that's fine, too... just keep multiplying them out.
This contradicts your hypothetical "left-over".
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)
What is an irregular number?
He never told us.
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.
they are all taken eventually, he's just ordering them. and this is fine because all terms are positive
You can subtract one sequence from another (monotonic decreasing) if that other sequence is included in the first. You may still have to prove convergence I agree.
@@lexyeevee No, he left a ton of terms unaccounted for, he's just lucky that in this case they converge on less than (∞^2)/(2^∞) which is an infinitesimal. But it was not guaranteed by the technique he used, nor his explanation.
@@hurktang it's every combination of every number of prime factors (to the nth power), which will give you every natural number exactly once by the fundamental theorem of arithmetic. nothing is unaccounted for
@@lexyeevee You don't listen to what you say. It is an INFINITE products of INFINITE sums. Once you have every natural number EXACTLY ONCE, you still have an infinitesimally small sample of all terms. There is still an infinity-1 of infinite sums to account for.
Noice
6/pi^2 is also the probability of "2 random natural numbers are coprimed".
That is not a zeta, that is an S with some rho aspirations.
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.
Math is even more dangerous than chess. One time, a young man used Cauchy's integral theorem on an integrand which was not analytic on the inside of the contour. He saw a bright flash of light after seeing the result and has yet to regain his sanity. The universe does not forgive such transgressions lightly.
You are very serious person. Only time i have seen you laughing is playing chess.