Modifiers of Cyclotomic Polynomials

[@winnablebtw459]

Nice use of an interrobang.

[@JayJynxTrash]

Did I understand this video....no did I watch it....all the way through

[@Vi51576]

I would have loved more abstract algebra at the end, it really helps solidity what has been shown IMO

[@Filup]

Last year I took two courses on algebra and number theory, and the vibes of group theory and cyclic/modular groups are coming back to me. It is interesting to see how these sorts of things are investigated without looking through the lens of pure abstraction.
Edit: Ok, I wrote this like, 2 minutes before the end of the video. At least my study didn't go to waste lmao

[@dylan7476]

Feels way too advanced for me, explained really well though.
What area of mathematics would you be studying for this? And does it have any real-world applications?

[@MariaBurns-o2x]

im not done with the video yet, but it reminds me of complex numbers!! i = i, i^2=-1 (by definition), i^3= -1 • i = -i, i^4 = -1 • -1= 1, and so on

[@TheGrayCuber]

I became interested in this when studying Galois theory. I’m not aware of any applications. And this does use the complex numbers! I didn’t want to focus too much on that since it’s not ‘necessary’ for an overview, but this is all using the complex numbers with magnitude 1

[@michaelboyd8546]

An introduction to these concepts would start with number theory and group theory. Group theory can be heavy though so take it slow and steady. While I'm not sure what this exact topic does for real world applications, group theory overall has many. For example, group theory can be used for mathematically solving Rubik's cubes, proving that there isn't a formula for 5th degree polynomials, and lots of stuff about permutations and rotations

[@michaelboyd8546]

​@@MariaBurns-o2x that's correct! "i" is a root of unity for x⁴-1 and is effectively a rotation of 90°. Group theory and complex numbers are closely (but carefully) intertwined in many ways

[@IdoF47]

Hi, I have a question.
At 10:58 you're multiplying 0 by "any polynomial h(z)", which makes sense since it's multiplied by 0.
But at 11:48, you're calculating a specific polynomial which h(z) should be.
Isn't that a contradiction?
Thanks for the video btw, very helpful

[@TheGrayCuber]

At 10:58, I'm saying that any two polynomials are equivalent if there exists some h that makes the statement true, regalrdless of what h is. At 11:48 we caculate what h would be for this case where those two polynomials are M_9 and x^6

[@IdoF47]

@@TheGrayCuber h was introduced to the equation by multiplying it by 0. The original logic of how it can be any polynomial due to that makes sense to me. I don't understand why once 0 gets replaced by something equivalent to 0, h loses that 'freedom' and has to be something specific

[@naturallyinterested7569]

One question after this Video. We can go from any Phi(x^k) to Phi(x^l) by doing first its respective I(x^k) modifier to Phi(x) and then the M(x^l) modifier to Phi(x^l). Or we could do I(x^k)*M(x^l) for the direct modifier. But polynomial multiplication is commutative. So we could first use the modifier M(x^l) and then I(x^k) and still end up at Phi(x^l). So there should be another shape outside the ones shown that is generated by applying M(x^l) to Phi(x^k) that generates Phi(x^l) when applying I(x^k)?

[@TheGrayCuber]

This is correct! For almost every n, there is an infinite family of shapes that we can reach using various combinations of the modifiers and their inverses

[@naturallyinterested7569]

@@TheGrayCuberCool! Thanks for the answer!

[@sakuhoa]

My dumbass thought this was a video about maimai💀