3Blue1Brown could you do simplectic geometry in classical mechanics ? Maybe a topic from Arnold's or mardsens book on classical mechanics. Those are graduate text to mathematics, but they cover physics

3Blue1Brown If i^2 = -1, does i^4 (4x90° rotation) equal 1?

EV4 Gaming yes. Multiply i by itself 4 times.

Max Yurievich Okay, thank you

This is no proof, i really can't answer a teacher this way can I? You need to get mathematical with intuition.

Thanks so much John!

John! You watch this channel?! I didn't know there were other nerdfighters around here!

But he didn't really prove anything whatsoever!

aa That's not the point of the channel. Anyone who has done a maths degree has seen the proofs. This channel is to show maths to people who haven't done a maths degree

Hey nerdfighter!!!

you are a genius.

Brilliant!

@19:07 - now it's time to get funky

that comment is under rated AHAHAHAHAHAHAHA

Lmao can someone explain

still alive?

@@slinkywhite9255bruh

@@ryan-tabar I mean he's legitimate 70

@@slinkywhite9255💀

same

which uni?

Same!!!

Of course it's india...

@@aurelia8028 no, he just got a shitty teacher. There are good and bad teachers everywhere. I'm in high school and my teacher introduced me to this channel. He has taught so many things that are out of my syllabus because he has real passion of teaching.

@@notsoclearsky you are lucky man , we got teachers who just want marks

+Tim H. Wow, thanks so much! That's kind of you on so many levels, I'm glad you enjoyed.

His videos have the same effect on me!

Ohh i thought i was the only weird guy who cried over mathematics videos 😂

Wow, could have said it better myself. I never thought I'd cry over math---with happy tears.

if there was a bit of sad music i probably would've cried too

not the same as a reflection, only the same as reflecting the point at 1+0i using the imaginary axis as a mirror line, note that such an action does however reflect the real axis across the "mirror point" of 0 because the real line is 1 dimensional and therefore has no orientation to help distinguish reflection and rotation

It is the same as the central reflection centered on 0, but not as the axial reflection along the imaginary axis

The one thing I didn't understand is what is the rule according to which the points on the vertical axis get mapped to the points on the circle at the end

@@randomname7918 you take the imginary Number shown on the left and that's your x, then you raise the esponent (2, then 5 then e) to the x and that Is your result shown on the right

Very cool way to interpret it geometrically.

Yeah this hurt me too

@@MatthewWroten What hurts me is that my comment didn't get a heart. I want to assume that Grant didn't have a chance to read it! :)

thanks for point out error❤

Archibald Belanus it is.

I intend to, I agree that it's a really nice topic. For quotient groups, it helps a lot to contemplate of tightening up on the structure you care about for the thing being acted on, so to speak. For example, think of the map from D_8 -> Z_2, where for each square symmetry, you consider how it shuffles the two diagonal lines. That each, a symmetry maps to 1 if those diagonal lines are exchanged, and 0 otherwise. The Kernel of this map (which is a quotient D_8/Z_2) can be seen in all actions of D_8 that don't exchange those two diagonal lines, which consists of four actions, isomorphic to K_4. That would be better with some animations, but hopefully it helps a bit.

3Blue1Brown Algebraic Geometry is a really interesting and visual discipline for Group Theory, Rings, Ideals, as well as concepts in Topology and Linear Algebra.

the quotient group G/H is a generalisation of "congruence modulo n". take 2 elements, a,b in G. in G/H, these elements represent the cosets aH and bH, which are equal if and only if ab^-1 is in H (or a+H = b+H iff a-b in H if you use additive notation). notice that this is basically the definition of congruence mod n: integers a,b are congruent mod n iff a-b is a multiple of n. example: suppose G = Z, H = 5Z. if a,b are elements in G, then the elements a+H, b+H are equal in G/H iff a-b is in H, i.e. if a-b is divisible by 5, i.e. if a=b mod 5. so G/H is the group of integers mod 5. another way to think about it is that to find the group G/H, you take G and replace all elements of H by the identity. so in the example Z/5Z, take the integers and replace all multiples of 5 with zero. consequently you have to replace all numbers of the form 5n+1 with 1, etc. so you are just left with 0,1,2,3,4 in the group. this idea is the same for quotients of rings by ideals. an example i like is R[x]/(x^2+1) where R[x] is the ring of real polynomials and (x^2+1) is the ideal generated by x^2+1, i.e. all real polynomials that have x^2+1 as a factor. the quotient ring R[x]/(x^2+1) is what you get when you take R[x] and replace all instances of x^2+1 by 0, so this ring is isomorphic to C. if f(x) in R[x] is a polynomial, you can write f(x) = (x^2+1)q(x)+r(x) where r has degree 0 or 1 by euclidean division, then x^2+1 becomes zero so f(x) is the same as r(x) in the quotient. example: take f(x) = 9x^3 + 2x^2 - x - 7. then f(x) = (9x+2)(x^2+1) + (-9-10x). replacing x^2+1 with 0, you see that f(x) is the same as -9-10x in the quotient, and indeed, f(i) = -9-10i

Yes please! That's probably the closest he will ever get to number theory.

I absolutely love this I'm a chemist, and I love how this video in particular gets me thinking about how group theory works Some day I want to teach group theory for chemists too chemists so this mathematical understanding is really awesome

I would definitely like to do an "Essence of real analysis" one day, and the topics you mention would be right on there.

This would be amazing.....I have thought the hardest part of math is analysis and you're just the guy to do it.

Evi1M4chine I study math and most of the people I know don't like to memorize something. We try to understand. And this counts for every single one of my lecturers and their assistances. Sometimes this goes so far down the rabbit hole, that we nearly commit an entire lecture, just to understand that one thing. Sure there are theorems you might only need to know exist in the first place, like stuff from Frobenius, fudge this dude and his proofs. Also there was a time in mathematics, where it wasn't accepted to use graphical assistance to help the understanding of a topic. Just look at most of Euler's and Gauß's proofs. Those feel like they had a completely different understanding for the topic, but every evidence of sketches or graphical explanation got deleted, since it was a 'no go' in the era. But this isn't the case anymore. Not to say that everything get's easier with graphical understanding of a simplified problem, but either way you have to put effort and a lot of work in understanding simple theorems to explain them in the way 3Blue1Brown does. And if it takes 10 hours just to get one idea, that's okay. And most of the fellow students I share my time with, love mathematics. Like two friends of mine stood in front of a chalkboard, trying to understand why you need certain properties in a theorem, if it would be enough to use more global statements. They spent 2 hours (after a complete day of lectures) just to understand this one bit. This is what studying mathematics is for. Understanding and expanding your knowledge on the topic of your choosing. BUT Wikipedia is a page of knowledge, not understanding. If you want to understand, you are welcomed to study mathematics. Get books about a topic or just search it up on Math Stackexchange. Wikipedia is to fast look up a certain theorem and it's not built for mathematicians, it's for everyone. To much information will sometimes lead to struggle, if certain people are unwilling to put in the said effort and time. They don't need an explanation how it works or why it works, but only that it will work and achieve the output needed. Nothing else. They simply don't have the time and energy to do so. Don't try to force them to understand those things.

Wikipedia decided long ago to be an _encyclopedia,_ that is, to collect knowledge supported by reputable sources. It's not the place to add your own explanations for complicated concepts. A blog would be a better place; or an homepage provided by your institution or company; or a KZfaq channel like this one; or you could write a book if you have enough material. Either way, if and when your material gains acceptance and recognition, in a way that can be measured by that "cabal" of editors you despise, then it will be gladly accepted into Wikipedia.

Zaknafein Do'Urden A Zin-Carla in the comment section? Didn't thought you could type...

I am in the same situation, feeling the same gratitude.

Jovana Krstevska Hi

Jovana Krstevska I can relate. That is very true. Lesser and lesser imagination and visualization as we try to learn more. Kinda kills the fun. Fortunately, you have been saved. I am happy for you.

Prvpat gledam makedonec da komentira na vakvo video lol

No offense but I find it really surpising a master's student would say this. Don't get me wrong, I'm up late watching this for the hell of it because I like the teaching elements, but I'm just shocked someone in a graduate program needs to be taught to imagine things.

The crazy thing is even in math programs (at least at my undergraduate institution), a lot of what we had to do was take math on faith. We were basically given the definition of a group as though they were handed to us on stone tablets like Mosses on Mount Sinai. Never asking "why the hell are we using this as our definition of a group?" or "what even IS a group and what is the broader concept it's supposed to represent?" It would be like teaching a six year old: multiplication is the process of repeating the addition property an indicated number of times. Sure it's correct, but you aren't really learning multiplication at a deeper level and it becomes harder to later generalize to multiplying by zero, fractions, or God forbid, negative numbers. Visuals of rotating a 4x5 grid to show that 4x5 = 5x4 are a much better way of understanding the fact that multiplication is commutative than simply being told it's an axiom and just something you have to accept. These 3Blue 1 Brown videos are a great example of the rotating the grid type of visuals that give one a much deeper understanding of what these concepts actually represent as opposed to some vague abstract construct with which to work.

Ur mom gay

@@mohsinuddin7049 no u

@@mohsinuddin7049 Lol!

Well, i is DEFINED by being the square root of one. it would be the same thing as saying: "Wow! way to prove that the ratio between a circles perimiter and its diameter is exactly pi!"

It wasn't a conclusion as such, but more of a process in the context of group theory of how i*i is -1. Well, i*i=-1 is the very equation by which i is defined, so he didn't prove anything new, but rather showed us the process of how it works. So people who would ask what's x*i or essentially, "what does multiplying by i do?" would find this illustration useful. Just saying i*i=-1 is a given is one answer, but more of a declarative one, and it raises plenty more questions than it solves. Giving an animation to explain the process helps people explore complex numbers. So it was the same thing as saying, "Just saying i*i=-1 is not satisfying. Here is something that you might find satisfactory and it will help you explore complex numbers as well."

Trust me when I say this: do NOT make spaghetti in the toaster. Don't ask me how I know, it's not important. Just don't do it.

But i^i is not - 1, it's ≈ 0.2

Nevermind, I mistook the * for a ^

@Sophisticated Coherence but he does not teach everything else

@@jsutinbibber9508 What stage are you at in your education? Because these videos become absolutely indispensable at some point and fourth (not gonna get ya to the finish line - bad analogy - but gives some ideas about stamina (for example).

University has always been a joke. A scam

Im sorry to hear you didnt go to a good one. In europe its very cheap and right now im studying at University of Twente in systems and control, it is very good. The teachers are competent and caring, our facilities are good. The campus promotes sports and social associations that help you develop skills outside of the core classes. I might be lucky but it is not a scam.

They really aren't a scam, lol ANY university in the world teach this kind of thing in mathematics related courses.

The one I hear often is 'those who can, do. those who can't, teach' which is kinda funny. sorry about the four year old reply

Ooh! Good catch, thanks for that. Silly me, mixing up my sides.

Yup. Thought the same thing, immediately! :D

Omg you're right, i measured it with my protractor (because that's quiqer than calculating it) and I got exactly that

Lol it's just arrcos(2/sqrt(5)). Awesome video

is arrcos the pirate version of arccos?

e^(i*tau) = 1 is the same as multiplying by 1. As for e^(i*pi) = -1, it is the same as e^(1/2*i*tau) = --1 or rotation of half turn. So, yes pi obscures its true meaning.

If e^(i*tau) is 1, doesn't that mean i * tau = 0?

@@linus6718 No, because the exponential function in this case is periodic. You see, if e^0 = 1 then e^i*tau = 1 too because the exponential is periodic with period i*tau (the same as saying one rotation by 360 degrees does nothing), so: e^(k+tau*i) = e^k for every number k

but you could say i*tau is congruent to 0 mod(i*tau)

@Linus Yes, it is. x^0 = 1 for all x's that are real numbers. The reals are rational, irrational and transcendental numbers (think any point you can choose on a continuous numberline). π is a transcendental number so it's included. e too (también!). So, yep, i*tau = 0. Another way to verify is with e^iπ = -1 or e^i*tau/2 = -1 Square both sides and you get e^i*2π = (-1)^2 = 1 or e^i*(tau/2)*2 = e^i*tau = 1. Thus e^i*tau = 1 = e^0 Now take the natural log of both sides In(e^i*tau) = ln(e^0) i*tau * ln(e) = 0 * ln(e) and ln(e) = 1 so i*tau = 0 Booyah! @Daniel Lopes actually states the generalized formula for any value k an integer perfectly but denies his own equation for the valid k = 0. cred: I have an MS in Physics '85 and a Math minor, 30 years of scientific work experience, am left-handed, and have spoken with Doug Hofstadter personally about "Göedel, Escher, Bach, an Eternal Golden Braid."

Good for you! I got my degree in geophysics from Berkeley, and I am so proud and grateful that I went there.

Agreed! The video is great, but this one step is not clear enough I think. The explanation at 21:06 kinda touches on it, because it hints that the derivative of different exponentials would mean a different rate of change around the unit circle, meaning that Euler's formula only works for e. That said, he still didn't elaborate on it enough I think. Edit: If you do the derivative with 2^x instead of e^x, you get i(ln(2)2^(i*t)) which might be the explanation we were looking for! Because the factor in the derivative is converted into a factor in the velocity around the circle. This explains the angle we get if we use 2^x instead of e^x, as well as e^x is special.

I'd love that too. There are some textbook like introductions on youtube (check out Sarada Herke), but nothing to bridge jumping to any of Babai's lecture videos, and nothing really conveys to me the core difficultly they are fighting against. Linear in time for almost all graphs, I had no idea the gap was that big! I'd love to see a visualization of what that core difficulty is that somehow escapes almost all graphs. So mysterious, I bet there is something beautiful in there when explained well!

I think it has nothing to do with group theory, but instead just computer science. It's because the problem is NP-Complete as you can reduce the satisfaction problem to it somehow.

Completely agreed. It's so stupid how we just jump to that conclusion and then make more assumptions based on that with no explanations given.

Fun fact: The transformations of the multiplicative group of th ecomplex numbers represents exactly those 2D lienar transformatiosn which are not just real linear transformations but even Complex linear transformations (meaning they are the linear transformations commuting with multiplication by complex numbers)

It is intuition, but it isn’t proof. Both should be considered.

Is that a category theory joke? xd

Patrick hehehehe The Joy of Cats.

Hey. At least it's locally small!

@@nasajetpropulsionlaborator8727 Lmao, projecting much?

How dare you make me read that pun with my own two i’s

@@sophiap.8859 i cannot unsee that

Did you pass bro?

@@boston0086 Hahah In the end I didn't even do it. The corona quarantine had just started and so I just got a passing grade. So yeah, I guess.. Sorta passed :)

@@garlic7099 Congrats bro:) the most important thing is that you you were interested in it! The topic about complex analysis is so much, you will always learn new stuff about that :D

I agree, sometimes you just miss having the opportunity to show that extra gratitude. However, you could show monetary appreciation at patreon.

Just make sure that the total number of clicks on the like button is uneven

Lol

I desperately need the love button for this video!

share all of his videos on all your social networks and to your friends on messages

Wow thats a pretty good catch. Indeed if you look back at the group of rotations of the square, this group is the same as the integers modulo 4, and the circle group (group of all rotations) is just the Real numbers modulo 2*Pi. The loss of information you describe comes from the fact that exp(2*Pi*i*k)=1 for all integer k, so multiple points are mapped to 1 (adn thus multiple points are mapped to any point). In general, "loss of information" like this are so important they are described through the fundamental theorem on homomorphisms. the most important theorem of grouptheory

I have a similar question. Would love to see if anyone responds to this.

Me 3! At 19:40, he said "it happens to be 0.693 radians", which is the moment that made me really want to now *why*! The explanation at 21:06 kinda touches on it, because it hints that the derivative of different exponentials would mean a different rate of change around the unit circle, meaning that Euler's formula only works for e. That said, he still didn't elaborate on it enough I think. Edit: If you do the derivative with 2^x instead of e^x, you get i(ln(2)2^(i*t)) which might be the explanation we were looking for! Because the factor in the derivative is converted into a factor in the velocity around the circle. This explains the angle we get if we use 2^x instead of e^x, as well as e^x is special.

+

You're telling me. I just kept thinking "2 years? That can't be right, surely that's not right".

for me it was more like: _only_ 2 years? this channel feels like such a classic and is has so many great things in it already.. :D

Odd how your vocalization has slowed slightly over two years. Is the aging process exponential or linear? Will your voice be twice as slow in another 2 years or slowed to a barely audible rumble? ;-)) Wish we had vids this back in the Sixties. Nice!

This really interests me as a student studying Computer Science. How can group theory be used to optimize machine code?

I actually found that quite genius of him. The point was that positive real numbers are pure stretchings while negative numbers also include rotations, which are a complex thing (and notice that the real exponential function only puts out positve numbers, so the real exponential function only gives you a relation between the additive real numbers and the multiplicative positve numbers)

And he did spell it out when he explained the complex multiplication and how it relates to rotation ;)

uhm.... it's genius of him because frankly multiplication by real negative number is just stretching and flipping through 0, it's not really a complex thing. In 1D, there is no need for rotation to involve. It happens that in complex, "flipping" through y-axis is 180 degree rotation.

Exactly. If he had talked about multiplying by negative numbers in 1D, it could have looked like flipping. But that would have created a misconception when extending to the complex plane because multiplying by negative 1 is not flipping in the complex plane but just rotation.

It's 'cause only the positive reals are in the range of the exponential map e^x (where x is real), but all complex numbers other than zero are in the range of the exponential map e^x (where x is complex).

Mohith S N Raju all it means is that given the two actions, the result is the same for these two groups (so in this case -1 and 2 give the same result (but different paths to get to them, which is of course to be expected))

agreed, got stuck at the same point

Me too, got stuck there. Why does exponentiation by complex number correspond to rotation of the plane? It can be verified mathematically, but I'm not getting the intuition behind it.

You can think of exponentiation by reals as a way of turning a discrete function into a continuous one - it still has to go through the same discrete points. That can be one starting point towards understanding. It turns out that when you do a series expansion of a real exponential then it usually works just fine in the complex domain, and you can then look up the sine and cosine in the same series expansion and you’ll see the Euler’s formula that way as well. Again: that’s just one of many approaches. But going from discrete to continuous exponentials requires a better definition of an exponential than the elementary one - whether you look at a series expansion or something involving logarithms doesn’t quite matter.

He didn't actually explain it, he just asserted it. At 18:49 he said "So wouldn't you agree that it would be *reasonable* for this new dimension of additive actions, slides up and down, to map directly to this new dimension of multiplicative actions, pure rotations?" I completely understand how multiplying by 2^(real number) will stretch the space, but he never said why multiplying by 2^i would rotate it, he just made a comparison to how multiplying by i made the space rotate. I agree with @Mithilesh Hinge, I can see it algebraically, using Euler's formula ( e^ix = cos( x ) + i sin( x ) ), but I don't get the same intuitive feel as Grant/3B1B provided for how multiplying by a complex number can rotate a space in the complex plane. Can someone give an intuitive, geometric reason as to why 2^i causes a rotation? Maybe there isn't one, which is acceptable, but I was hoping for something as elegant as his earlier explanation of rotations due to multiplying by complex numbers. And, as some others have mentioned, even if 2^i = 0.769 + 0.639 i, that looks like a stretch and a rotation, which is not the "pure rotation" he mentioned. Otherwise, though, this video is wonderful. 😊

@@tonylopez5937 if you multiply the complex number z= r*exp(it) by the complex number exp(ix), you get r*exp(i(t+x)). You are just rotating z by x radians because the angles add. The author of the video is not explaining what is really happening here. He is assuming Euler’s formula and the polar form of a complex number. He is just saying that the image of pi*i under the exponential map (homomorphism) is -1 because Euler’s formula is true. His graphics are nice, but he is not using group theory to explain why Euler’s formula is true. So the title of this video is a bit misleading.

Three years later, but here's how I understood it. I don't think there's necessarily a direct relation between the operations themselves, meaning a way to deduce the operation that must be performed below by seeing the operation performed above, but what happens is: On the top, you slide by whatever the power is. So if the power is 3, you slide by 3. If it's -1, you slide by -1. On the bottom, you squish by 2 to the respective power. So if it's -1, you stretch by 2^(-1)=0.5. If it's 3, you stretch by 2^3=8. This is because exponentiation is repeated multiplication, so if you raise 2 to the third power, it's the same as stretching by 2, then again by 2, then by 2 once more.

Isn't it trying to say that both additive and multiplicative actions have similar result for the same expression? Sliding 1 unit left and then 2 unit right lands on the number 1. Squish by 0.5 and then stretch by 4 lands on number 1 also right?

@@cliffordwilliam3714 If you start the arrow on 0, squishing and stretching by any amount shouldn't move the arrow away from 0. I don't understand at all what he's trying to get at in that part of the video.

Well you should stop waiting now

@@eccentricOrange lmao