Mobius' favorite function
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- Опубліковано 15 кві 2024
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I liked the part of the video where Michael said "it's mobin' time" and mob'd all of those formulas
I call square-free numbers primers. There’s more to them than meets the eye.
Question: Let f:ℕ→ℤ
Find all functions f such that
Σₖₗₙ f(k) = 1 if n=1
= 0 if n>1
Me: Since F(n)=Σₖₗₙ f(k) is multiplicative, morbius inversion can be applied, hence
f(n)=Σₖₗₙ F(k)μ(ⁿ/ₖ)=μ(n)
I used morbius to destroy morbius
I like that you are doing the proofs in a kind of inefficient but very enlightening way. Not explicitly using multiplicativity, dirichlet convolutions or such. Not that I think these tools are useless, but the video manages to create a feeling of a very down-to-earth understanding of the mobius mu without delving into more convoluted (haha) territory and obtuse things.
Note that the Euler's formula for Riemann's zeta function (product for all primes) appeared in the first proof. This is an awesome and important connection between calculus and number theory
It's called analytic number theory for a reason
'This is an awesome and important connection between calculus and number theory' - if u say so
michael never fails to do something i don't know
Typical GothamChess subscriber
Typical GothamChess subscriber.
@@dalibormaksimovic6399why hating for no reason bro? 😂 you must have issues in real life... sorry to hear
@mnokeee yeah I have issue called chess
@@dalibormaksimovic6399 I was about to make a similar joke.
Extremely beautiful use of Möbius function is its connection to the Riemann zeta function and prime numbers. Specifically, the Möbius function is intimately linked to the distribution of prime numbers through the Möbius inversion formula.
This connection between mu and zeta is why the Mertens conjecture (which was proved false) implied the Riemann hypothesis
Riemann Hypothesis? I thought it would have implied the Twin Prime Conj.
True, but I'm pretty sure RH would also be implied by M(x)=O(x^(1/2+ε)) for every epsilon, so not all hope is lost.
If Möbius were alive today his favorite math presenter would be Mike Penn.
Truly the most morb of all
This function is very nice.
10:07 It's Mewtwo!
That’s what Euler proved
Ah - the joys of concatenations
With a wonderful generalization by Rota (et al?) on locally finite partially ordered sets. (The partial order for the regular Möbius function is divisibility.)
I always kind of wondered; if I got my hair stylized and cut like Mike - would I be smart ?
Instead of using the C(m,k) choices pick one prime p pair each divisor of the form pd with the divisor d. Since pd and d differ by one prime factor they have opposite signs and since this exhausts all possible divisors they must sum to 0.
Michael can you do dirichlet function and explain it briefly
Take two functions f and g and a positive integer input n, and (just for illustrative purposes) create two rows of the divisors of n with one in ascending and one in descending order. For every column, put row 1’s entry through f(d) and row 2’s entry through g(n/d) [i.e. for a list of divisors, if a list entry is d, then the entry at the same index in the reversed list is n/d] and then multiply the two outputs. After you’ve taken a product for each column, sum all the products together, and you have (f * g) (n).
The ending is really nice
i found it in matrices with fractles
You know It's a good day when professor penn upload number theory!!
You named the video "Mobius favourite function" but you did not specify in which world is that 😁
I wish I knew when it was a good place to stop
Hi, I'm writing a term paper on prime numbers. Please tell me who is currently engaged in the study of prime numbers. Russian guys, mathematicians, they developed the program "The key to all doors". It seems like they moved either to London or New York. They are known for their research in the field of prime numbers, they found a justification that 1 is a prime number, and 2 is not. It's a huge breakthrough in mathematics there. Who are these guys anyway, is there any information on them? Thanks!
The "+ μ(n)" in the proof the second claim should actually be "+ … + μ(p₁p₂•…•pₘ)". The first ellipsis is needed because summands are being skipped. The expression in that final μ() is not necessarily n because we do not know n to be square-free.
Well, he had already explained at that point that if any of the divisors of n are not squarefree, then they'll make zero contribution to the sum. So it effectively suffices to consider only the 'squarefree part' of n, i.e. the product of all the distinct primes that divide n. That seems to be what he's doing, albeit it could have been better explained. (And you are of course right about the missing ellipsis!)
@@amritlohia8240: True. I watched the video in multiple sittings, and posted this comment at 4:00 in the morning, so I forgot about that explanation. Thanks!
Perhaps we could take Mike and twist him then join his head and feet to form a Möbius Mike.
One of my favorite functions. Maybe modular functions are my favorite, idk
That “filtering” process at 2:20 would have deserved more explanation- or rather an explanation at all. Not an ideal choice of variables and notation either. Would have helped me if the s-exponent was bracketed out, the natural numbers that are not divisible by 2 were labeled differently than “n” - say n’ - and if someone had explained why the sum of all 1/(2^s*n’) is exactly 1/n (without any overlap). Is all that really so intuitive? Obviously i’m just an amateur, but to me it sometimes feels as if parts of the reasoning that are more challenging to be put into words are much more likely to be glossed over whereas simple arithmetic gets dealt with in a a bit too much detail.
I dont know whats the idea behind this function but i am still looking for similarities with Moebius loop,
The only connection with the Moebius loop is that both were invented by the same mathematician.
Read Concrete Mathematics (by knuth).
g(n) = Σⁿₖ₌₁ f(ⁿ/ₖ)
where f(x)=f(⌊x⌋),
then
f(n)=Σⁿₖ₌₁ g(ⁿ/ₖ)μ(k)
iirc