Why you don't understand GREEN'S THEOREM -- Geometric Algebra, Calculus 3, Vector Calculus

In this video, we discuss the link between the fundamental theorem of calculus and Green's theorem. This offers an introduction to the exterior algebra, specifically, the wedge product. We discuss, albeit not explicitly, the de Rham pairing, which offers an enlightening interpretation of the fundamental theorem of calculus, bridging this theory with the theory of vector calculus. We discuss Stokes' theorem, verify properties of the wedge product, and more. This is part of a series on Geometric Algebra (c.f., • A Swift Introduction t... ).
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These videos are separate from my research and teaching roles at the Australian National University, University of Sydney, and Beijing University.
Hi, my name is Kyle and I'm currently doing my doctoral mathematics degree in complex differential geometry under the supervision of Professor Gang Tian and Professor Ben Andrews.

[@KyleBroder]

This audio is terrible....

[@such_a_cheatah4859]

fire your engineer

[@HilbertXVI]

@@such_a_cheatah4859 More like, get an engineer hehe

[@warisulimam3440]

The quality of the content does more than making up for it = ]

[@LetsbeHonest97]

Not a problem. The content is fantastic. I saved your reddit post some months ago. I was going through the saved posts today and decided to watch your videos. I have watched about 20 of them so far. The videos are absolutely brilliant. Thank you for putting in the effort in educating a lot of clueless undergraduates, myself included.

[@crehenge2386]

Extremely annoying music combined with crappy sound

[@gijsb4708]

You sure went to a special kindergarten!

[@neologicalgamer3437]

69th like

[@turtlecraft7996]

Me ∈{undergraduates} ⊆{Kindergarten}

[@danieljulian4676]

This is brilliantly concise. Edward Frenkel gave lectures for a semester of UC Berkeley's Math 53 (multivar calc) that were videotaped and put on YT, but only hints at this development without mentioning any defined concept (such as wedge product). Robert Ghrist even gives his online Calc 1 students a peek at the concept of boundaries, but doesn't note the next steps, not even in the Calc Blue series. It's not that I expect material offered openly to elaborate these next steps, but to come so close and then retreat seems a shame. And here you are with this wonderful presentation! I don't have enough math to go all the way there, but this is more than enough for me to see the outline of the advanced concepts. Beautifully done!

[@rhumblinesnavalactionchann5929]

In my undergraduate years at UNSW, I took courses in Vector Algebra, but the theorems seemed ad hoc. I never became aware of any deep insight in these theorems. Geometric Algebra is what I should have been taught. Thank you. There are right ways and wrong ways to present mathematical concepts. I think Geometic Algebra will take us into the next millenium.

[@thomasjefferson6225]

did you meet wildberger?

[@blipblap614]

In my undergrad vector calculus, 25 years ago, we got a flash in two lectures at the end: "there's a wedge product that unifies all this; you might see it in linear algebra." (We didn't.) Modern academia changes very slowly.

[@TheJara123]

What a clear cut presentation!! You made as clear as possible man...I am glad that I found your videos !!

[@KyleBroder]

Glad it was helpful!

[@abnereliberganzahernandez6337]

this was so simple yet so accurate and elegant I cant believe it!

[@peterhall6656]

Nice explanation. I'll check out your other work. Undergraduates are systematically short changed on these types of insights. It ’twas ever thus. Glimpses of some deeper structural reason for certain things are sometimes given but cannot be fully developed in the confines of a one semester course. And of course there is the tantalising expression: “The boundary of a boundary is zero”.

[@pseudolullus]

Undegrads who are actually interested in what they're studying will feel shortchanged, the others will complain about having to study stuff "outside the syllabus". After all, vector calc is taught to engineers, scientists etc, not just mathematicians, and the content is standardized. Full disclosure: I am a scientist and love this stuff

[@Rockyzach88]

Before watching the video, I just want to say that I don't understand Green's theorem because it's always taught at the end of the semester when you are trying to scramble to study for all your tests and get everything else done.

[@Jon.B.geez.]

I might argue that integration and derivatives are indeed opposites functionally, but that their is a geometric perspectives of being opposites that is simultaneous/equivalent to their relation, and that has to do with the relationship between a volume and it's boundary, as expressed by density function (and understanding that density function is a dual game between the Radon-Nikodyn derivative vs Reisz Representation). Moreover, I might argue that their is even a 3rd different perspective on what is "opposite to what" in this deep theorem, and that touches on homology vs cohomology.

[@Jon.B.geez.]

edit: This has a very insightful ending, bringing together wedges and the exterior derivative to derive Green's Theorem trivially. Well, perhaps I should have finished watching it, but he uses this derivation to showcase how Green's Theorem is a special case of Stoke's theorem, but damn, I really liked the derivation of Green's theorem itself okay, so he didn't derive Green's theorem, but implicitly, he showed that integrating a 1 form is equivalent to Green's theorem, thereby essentially deriving it

[@Nylspider]

Brilliant video, but one small correction at 3:07 - the right hand side would not be written as f dx, but rather it would just be written as simply f, because f dx is a one-form, which isn’t integrable over a zero-chain (which is the boundary of [a,b] in this case). If we write f with no dx, then we have a zero-form, which can be integrated over the given zero-chain and is in fact the correct statement of FTC using differential forms

[@null_s3t]

Nice catch!

[@KyleBroder]

Thanks for pointing this out :-)

[@alejandrocambraherrera8242]

And shouldn't the left-hand side integrand be df/dx dx? Where you could take both dx's out and be left with df, the integral thus being the sum of df's of your area function that make up the difference between boundary values when working with a definite integral. Which, now that I think about it... Wouldn't this be a proof of the fundamental theorem of calculus, given you could take a dx equal to the difference between boundaries and still be left with the correct result, since dx does not appear in the integrand anymore?

[@jacksonstenger]

Great video, best explanation of Green’s theorem I’ve seen so far

[@wonjunjang8623]

i think this might be my very first youtube comment. Thank you for making these videos. I just came across your channel and I'm looking forwards to watching all of your playlists.

[@a.v7998]

Nice Video man! You deserve a Million subs. As a Computer science student, I found this very useful!

[@KyleBroder]

Thanks, glad it is of some utility.

[@halneufmille]

"that we learned in kindergarten" I can't remember if we covered it after learning the square or the triangle.

[@Sidionian]

You're now an expert in Riemannian Geometry. Please post more of this stuff. In particular Ricci flows, curve shortening and surface minimization etc. I would like to learn more about Alexandrov Spaces. Please post more; you're great at explaining stuff.

[@visualgebra]

Please make videos on Abstract algebra:Group theory.

[@KyleBroder]

💪

[@nektariosorfanoudakis2270]

So differentiating the form is adjoint to taking the boundary, the "bilinear form" being integration.

[@KyleBroder]

👍 To support the channel, hit the like button and subscribe.

[@haipingcao2212_.]

No

[@haipingcao2212_.]

No

[@julianbruns7459]

3:20 If you denote the boundary of [a,b] as {a,b}, don't you lose the information of the orientation of the boundary? (since this would be equal to {b,a}) The manifold you integrate over must be oriented and the boundary must preserve this orientation.

[@Argoneui]

Yes, why does no one ever mention orientation? If you don't have one you can only integrate pseudo-n-forms aka densities. Which is often more appropriate anyway, after all the mass of a string shouldn't depend on which way we call right.

[@KyleBroder]

There's always a trade-off between precision and pedagogy. But you're right that orientation is an essential aspect of this stuff.

[@julianbruns7459]

@@KyleBroder yeah, maybe introducing new notation that does include orientation would be more confusing. Being pedantic is a luxury only the viewer has i guess.

[@q0x]

At 11:55 why do we write w= Pdx + Pdy and not w = P ^ dx + P ^ dy. Or why don't we write dw = dP ^ dx + dQ ^dy ? Why suddenly there is this implicit wedge where we assumed a multiplication? Is the multiplication just a special case for a scalar functions ?

[@null_s3t]

I’m not sure, but I interpret the differential multiplication as a geometric area as is defined by the wedge product. When we have iterated integrals we have to assume geometric area, thus I believe thinking about it as multiplication may have been the wrong way to think about it from the start!

[@yandrak6134]

This was so cool!! Thank you

[@briandwi2504]

That was so deep. Brilliant.

[@sahhaf1234]

masterfully explained...

[@KyleBroder]

Thanks!

[@Terieni-q7c]

Great video! Why should I understand dxdy as a wedge though? Is there any intuition? (this is coming for non-math major!)

[@yexiaorain]

very helpful thank you

[@Grateful92]

Thanks for posting this video 👍

[@zokalyx]

This is gold.

[@visualgebra]

And thank you for this video

[@milkwater1204]

One question: why is a wedge implied between dx and dy?

[@thomascancino1836]

i am also a bit confused with this assumption

[@__nog642]

You lost me at 3:02. Isn't f'(x) df/dx, not just df? I don't really understand what it means to have a single integral ∫ df dx.

[@ironsideeve2955]

Chain rule? D/dx f(x) = f’(x) . (x’) = df.dx

[@tonaxysam]

You can think about df like the way f changes for small values of x, and if you make a rectangle whose height is df and whose with is dx, then df dx is the area of that rectangle. The integral thing just means that you want to consider what happens to the sum of all the rectangles between a and b as dx approaches 0. If you think about it, yes, this área should be just f, and because you're adding up all the rectangles from a to b you should be adding all the values that f takes between a and b, right? But that blows up to infinity... This can't be right. The thing is, thinking about f' as a concrete fraction df/dx is wrong, because f' is not what df/dx is for a particular choice of dx, is Whatever the value of that fraction approaches as dx approaches 0. If you thought as f' like a literal fraction df/dx, then (df/dx) dx would indeed be df, and df is just f(x + dx) - f(x), the change in the function f. So when you add up the areas, you would get this: f(a + dx) - f(a) + f(a + 2dx) - f(a + dx) +... + f(b) - f(b + dx) = f(b) - f(a) Thus, the fundamental theorem in disguise. the thing here is that, the dx from df/dx may not change in the same way as the dx from the integral... Here let me show you what I mean: Consider dx = h in the first expression and h² in the second one. If h tends towards 0, Both dx will tend towards 0, hence, the value (df/dx) will tend towards the derivative of f, and the value \int f *dx* will tend towards the integral of f. However what happens if we combine these and interpret f' as a literal fraction df/dx? \int (df/dx) *dx* = \int df/h * h² = \int df * h This integral expression will be Whatever (f(b) - f(a))h approaches as h approaches 0. And that wasn't what we were looking for. That's why df is not exactly a fraction, is the limit of a series of fraction, and when df is inside the integral, its values at any particular point are already perfectly defined, no need to make a fraction. Formally, dx is a differencial 1-form, which means that its purpose in life is to be integrated under a 1-d region. It represents the with of a hypotetical rectangle whose height is... Whatever the value of the function to be integrated is. df is a notational short hand for the limiting process, not for a fraction. This is pretty confused since: f'(x) is Whatever a fraction whose numerador looks like a *difference in f* and whose denominator looks like *difference in x* approaches as dx approaches 0. But it gets a lot clearer in higher dimentions, because df tells you how a function changes over a given neighbourhood, and is not just checking how a tiny change in one variable affects the function. You need a higher diferencial form. A higher dimentional rectangle such that you can find is equivalent of area properly

[@pkskyutube]

Beautiful. But I wonder if the notion of the determinant as meaning volume is always represented so universally in teaching LA so as to call it’s understanding as “kindergarten”. The determinant in LA has a lot of other hard work to do. It was real eye opener to me after LA to hear this was true.

[@patrickpablo217]

Great video! One part that confused me though was when you were doing the derivation of dω and when you got to the step of distributing the dx and dy, in the next line the wedge product appeared. Why did it appear like that? (around time 12:30) I feel like I've seen the dx (and or dy) be distributed there with just some regular multiplication before, so that was a part where I got lost as to why that step happened that way.

[@KyleBroder]

Here, dx and dy are interpreted as 1--forms. There is no "multiplication" of 1--forms, the natural pairing of 1--forms is the wedge product. Check out the wedge product or the cotangent bundle, exterior algebra, on wikipedia 😄

[@patrickpablo217]

@@KyleBroder Cool 😊 probably what i had seen before was a "hand-wavey" version of this that just left out the wedge products so as not to "confuse" the students ..

[@patrickpablo217]

I looked into it some more but am still a bit confused about this part of your video. At this point where you start putting in the wedges, I hear you that they have been implied, but as a complete novice to this wedge stuff, I'm not clear about why they could be implied earlier in the derivation, but that at this point in the video they need to be make explicit and visible. Were the wedges invisible/implied all the way back to the beginning of the problem? Were they not there at all at the beginning, but then were there (but implied / not visible) at a later point, and then continued along implicitly until we get to this part of the video where they are revealed? I'm not trying to give you a hard time. I really like your video and appreciate you trying to explain things to us things most people don't bother trying to explain to us at all. I just know from my own attempts to teach people things I know but that they don't, that it's often hard for me to remember what it was actually like to not understand that thing. So sometimes I over-estimate what my student might know and sometimes I under-estimate, and often in surprising ways. So I'm acting in good faith here :) This is a great video showing what's going on behind the scenes here and how the explanations we got in school were "hand-wavey". However, as someone who has enough background to have seen those other explanations but not so much background that I already don't need this explained to me, I wanted to let you know that, to me, at this level of background, the wedges showing up at this point in the derivation felt "hand-wavey". But it also felt like probably they were not actually "hand-wavey" since that didn't seem at all like the sort of video you were trying to make. It seemed like there was a perfectly good explanation that you had just omitted because with your greater experience with the material, it didn't seem surprising at all that that is where the wedges would need to be made explicit. That's all :) [This is like way too long a reply for a non-fight lol 😂 but I wanted to let you know how much I appreciated your work and also wanted to give you the context for my question]

[@hydraslair4723]

He kinda "cheated" the definition of d(omega). The differential operator "d" is defined on 1-forms (and omega is a 1-form) as follows: if omega is written as Pdx + Qdy +..., then d(omega) = dP ^ dx + dQ ^ dy So the wedges are baked into the definition of how d acts on omega. There's also a way to make this into an even more rigorous argument by considering more complicated wedge products (for example, the wedge between zero terms) to generalise the definition of "d" to any form, not just 1-forms or functions.

[@Igdrazil]

OMG! Be carefull, your "transformed" version of the FTC is not correct. There is 2 mistakes, one on each side. The RHS should not have fdx (which is a 1-form), but instead simply f (which is function, i.e. a zero-form)! And the LHS should have, as integrand, the 1-form df (or more or less equivalently f'(x)dx), but not an horrible mixture of both : "dfdx" (which is a 2-form)! So you made quite a huge mess of notations and mathematical objects, in such a way that writing Int(df.dx)/D=Int(f.dx)/6D as you did (with D=[a,b]), is absolutely NOT correct. It should be instead the 1-dimensional form of...STOKE'S THEOREM : Int(df)/D=Int(f)/6D Moreover d and 6 ARE NOT equivalent operators, even in 1D! The differential operator d UPGRADES forms, and here for instance, upgardes zero-forms to 1-forms, and 1-forms to 2-forms. Whereas 6 is nothing more than the usual Lagrange dérivative f', simply extended to functions of several variables. Thus, in the rôle it has here, it doesn't upgrade anything! So you are not only making a simple "notation" mistake, but a fundamental one, illicitaly "identifying" different types of mathematical objects. Finaly your "derivation" of Green's theorem is not serious. You suddenly pull out of your hat some wedge products, magicaly springing from the middle of nowhere! This is because you knew the result your were seeking without knowing how to get there. And this comes from all your mistakes and confusions summerised up here. So your starting point is incorect and thus your "needed" wedge product are missing! Your mistake comes from your derivation of the 1-form w. The correct derivation is the folowing : w=P.dx+Q.dy (where P and Q are functions, i.e zero-forms of 2 real variables) dw=dP^dx+P.d2x+dQ^dy+Q.d2y (where dP and dQ are 1-forms. And their second derivatives : d2P and d2Q, are 2-forms, identicaly nul, by Poincaré's theorem). Thus the 2-form dw simplifies in : dw=dP^dx+dQ^dy And the rest follows immediately par expanding dP and dQ on the tangent space basis (dx,dy), and using the wedge product antisymmetry property. So finaly, all these falling stones confuse seriously some people and students who at least feel that some things are fishy here and there but don't always have the skill to identify your mistakes. But what about the rest who didn't even noticed your mistake and who are perhaps going to use such confusing crap maths for their exams???... Be carefull not to spread confusions and mistakes.

[@Tadesan]

I would have sworn it was because I've never heard of it before...

[@aosidh]

I was just talking with my friend about how my brain never internalized Greene's theorem like other concepts

[@NoNTr1v1aL]

Amazing video!

[@KyleBroder]

Thanks!

[@thomasjefferson6225]

I learned greens as an intergration by parts kind of idea, and its served me well in that fashion. What blew my mind was idea that the opposite of the dervative is the values of the boundry. Now that.... blew my fucking mind.

[@cmilkau]

Why can I write f' dx as df dx? I'd usually write this as just df *confused*

[@lucasm4299]

I am not familiar with the subject but f’(x) is a function so I guess df(x) acts like a function too?

[@null_s3t]

It should be just df on the RHS, as it should be zero-form

[@MehdiKaffash]

Nice similarity you have found, BUT quite screwed up explanation. We just eliminate do to power 2 as too small to effect the result.

[@billytheschmid]

Cool. Thanks man!

[@KyleBroder]

No problem!

[@zacklee5787]

Could you explain the first part where you rewrite the fundamental theorem of calculus? My first remark is why there is still a dx on the lhs? Also integration over the bounds would be summation not subtraction as that's what the integral is right? Thanks.

[@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ]

3:00 I'm 2 years too late, but why is that the case? Isn't f'(x)dx = df/dx * dx = df, rather than df*dx?

[@dontusethisname6693]

i think so too. He even talked about ∫ ω in region ∂S = ∫ dω in region S which i think would be a reason of that both sides of the equation don’t need dx . im japanese and i got no confidence in my English btw 😅

[@I-M-2.]

So derivatives are switchable per Schwartz theorem, but differentials are wedge multiplied? Very non intuitive. Thanks a lot for posting this derivation !

[@wellid2087]

Great video! But I got a problem at 12:15. Why dw=dPdx+dQdy not dw=dPdx+Pd^2x+dQdy+Qd^2y ?

[@KC_G4S]

I’m not positive but I believe this is because he defined dP and dQ to be the exterior derivative of P and Q respectively, which is different from differentiating each with respect to dx or dy.

[@DanielAnastasios]

What is the name of the book you show at 0:47?

[@underfilho]

isn't this in some sense the same as differential forms? I personally prefer to talk about differential forms, and i think its the most standard way to do calculus on manifolds

[@nahblue]

Why is the integral over ∂[a,b] not zero? In the real analysis sense it feels like integrating over just two points should be zero because the lengths of those subintervals are as good as zero. It's some kind of creative notation?

[@chri-k]

There's a typo in that integral, namely the dx shouldn't be there. This makes it a 0-dimensional integral, which is just a sum of f over the given points multiplied by the signs of the points ( in this case - for a and + for b )

[@RocketsNRovers]

at 3:03 , f'(x)dx = df and not dfdx ... so this is wrong from there on

[@RocketsNRovers]

this , or pls explain how thats correct , cause I am not able to follow

[@giangia8604]

i think he just writes f’ as df, abuse of notation i guess

[@djimms5644]

Does this theorem have anything to do with electrical theory? Like current and magnetic flux?

[@KyleBroder]

For applications to electricity and magnetism, I would check the standard physics-oriented books on vector calculus. I'm less familiar with the applications, admittedly.

[@djimms5644]

@@KyleBroder Are you a mathematician then?

[@eduardomontesinos5908]

excellent!!!!!

[@keyblade134679]

hi kyle. do you know any good text to learn differential forms for those that just finished a course on real analysis of several variables?

[@SVVV97]

A visual introduction to differential forms and calculus and manifolds by Fortney is superb

[@KyleBroder]

Lee's smooth manifolds, Jost's Geometric Analysis and Riemannian Geometry, Chern's book on differential geometry, Moroianu's lectures on Kähler geometry.

[@KyleBroder]

The wikipedia article is also nice to read.

[@CubingUniverse]

I just finished a course that used Hubbard's vector calc, linear algebra, and differential forms having a similar background as you in analysis and I thought it was great. It very nicely developed Stokes's theorem in full generality and included a proof in the appendix (which was about 10 pages long and quite complex :0)

[@sdsa007]

Im Stoked!

[@zhen_r_zyt1286]

Great video! btw the name of the university should be Peking Univ. instead of Beijing Univ.

[@mcpecommander5327]

3:40 Integral of dfdx doesnt make any sense. Its just integral of df, or df/dx * dx

[@dr_drw]

Love this video. Thank you!

[@jackozeehakkjuz]

Then you go to study about spaces without an orientation and realize that everything actually comes from Stokes' theorem for *pseudo*forms. It never ends, really.

[@abdulkadiryanalak1045]

THANK YOU! GOD WILL TAKE YOU TO HEAVEN

[@YazminAbat]

which is the book's title in 0:48 please

[@KyleBroder]

Lee's Introduction to smooth manifolds.

[@darkdevil905]

where is the geometric algebra in here? i just see the use of exterior algebra

[@KyleBroder]

The use of "geometric algebra" is not standardised.

[@Igdrazil]

Even the exterior algebra is not correct : OMG! Be carefull, your "transformed" version of the FTC is not correct. There is 2 mistakes, one on each side. The RHS should not have fdx (which is a 1-form), but instead simply f (which is function, i.e. a zero-form)! And the LHS should have, as integrand, the 1-form df (or more or less equivalently f'(x)dx), but not an horrible mixture of both : "dfdx" (which is a 2-form)! So you made quite a huge mess of notations and mathematical objects, in such a way that writing Int(df.dx)/D=Int(f.dx)/6D as you did (with D=[a,b]), is absolutely NOT correct. It should be instead the 1-dimensional form of...STOKE'S THEOREM : Int(df)/D=Int(f)/6D Moreover d and 6 ARE NOT equivalent operators, even in 1D! The differential operator d UPGRADES forms, and here for instance, upgardes zero-forms to 1-forms, and 1-forms to 2-forms. Whereas 6 is nothing more than the usual Lagrange dérivative f', simply extended to functions of several variables. Thus, in the rôle it has here, it doesn't upgrade anything! So you are not only making a simple "notation" mistake, but a fundamental one, illicitaly "identifying" different types of mathematical objects. Finaly your "derivation" of Green's theorem is not serious. You suddenly pull out of your hat some wedge products, magicaly springing from the middle of nowhere! This is because you knew the result your were seeking without knowing how to get there. And this comes from all your mistakes and confusions summerised up here. So your starting point is incorect and thus your "needed" wedge product are missing! Your mistake comes from your derivation of the 1-form w. The correct derivation is the folowing : w=P.dx+Q.dy (where P and Q are functions, i.e zero-forms of 2 real variables) dw=dP^dx+P.d2x+dQ^dy+Q.d2y (where dP and dQ are 1-forms. And their second derivatives : d2P and d2Q, are 2-forms, identicaly nul, by Poincaré's theorem). Thus the 2-form dw simplifies in : dw=dP^dx+dQ^dy And the rest follows immediately par expanding dP and dQ on the tangent space basis (dx,dy), and using the wedge product antisymmetry property. So finaly, all these falling stones confuse seriously some people and students who at least feel that some things are fishy here and there but don't always have the skill to identify your mistakes. But what about the rest who didn't even noticed your mistake and who are perhaps going to use such confusing crap maths for their exams???... Be carefull not to spread confusions and mistakes.

[@CertainlyNottt]

3:03 but if wouldnt we call f'(x) d(f(x))/dx rather than d(f(x))? This would make the integral Integral of df(x)) from a to b Integrating would give us f(b)-f(a)

[@rhumblinesnavalactionchann5929]

At 12:35, I don't get how the ()dx + ()dy = ()^dx + ()^dy.

[@hayekianman]

in kindergarten!

[@kibme5189]

I can't be the only one who found the background music annoying. Great video tho!

[@BongoFerno]

Got lost at the introduction of the wedge. No clue what it is doing geometrically.

[@junfour]

Why does this seem so much easier than vector calculus if it's more powerful too lol

[@KyleBroder]

It's often much more difficult to present differential forms to students. I succeeded in doing this in a second year undergraduate class, primarily targeted at engineers, but it was delicate and challenging from a teaching point of view.

[@junfour]

@@KyleBroder Anything in particular that was difficult?

[@markusantonious8192]

Way too much base in the sound production here.....

[@constantin2449]

That is stokes not greens

[@lilac624]

Handsome . 😊 💜

[@stuboyd1194]

Does this stuff have any real use?

[@KyleBroder]

"Real use" meaning?

[@declandougan7243]

@stuboyd1194 Honestly so much more than you know but when you ask an insulting question like that it doesn’t particularly inspire others to illuminate you.

[@stuboyd1194]

@@declandougan7243 Ok, so you choose to feel insulted by reading a question I wrote? Get over it snowflake.

[@sebastianmestre8971]

I think "Knuth's DP optimization" is a special case of Stoke's theorem

[@ianncarloalvim]

All names are made up, but Peter Petersen looks like is a made up kind of The Sims' NPC.

[@baxtermullins1842]

You know that we use these in undergraduate fluid dynamics - so, what so hard about understanding them? Maybe you should go back to kindergarten!

[@Whysicist]

Have you ever watched your video? Flashing equations for 1 to 2 seconds is a JOKE… LIKE THIS VIDEO. THERE ARE OTHER VIDEOS THAT ADDRESS THIS TOPIC! Boo!

[@eliotfan0118]

@Kyle Broder can you please stop saying things like "you learned in kindergarten" or "kindergarten level". It makes you come across as quite dismissive and arrogant and turns people off. None of this material is necessarily obvious or intuitive and to pretend that it is (even if it has become second-nature to you personally) is not the right attitude for an educator to have. Otherwise good video

[@KyleBroder]

Exactly, none of this material is necessarily obvious or intuitive. It is intentionally absurd to refer to it as "kindergarten".

[@jacksonstenger]

The phrase is actually helpful, since it indicates which statements are “fundamental statements” used at the beginning of the proof before he starts operating on them to arrive at a new result. It’s helpful for the student to differentiate the “basic” statements from the “new” statements in the context of a proof. Kindergarten just means a statement is nothing new, you’re already familiar with it, the new or derived statement that will be a little more unfamiliar is coming later. He’s not saying you’re dumb for not knowing something, since as all mathematicians know, everyone spends the vast majority of their time feeling dumb in some way or another since 99% of the time we don’t have the solution. Mathematicians have to grow thick skin and lose their egos to continue their pursuit. Long story short, referring to a statement as “kindergarten level” is a helpful semantic distinction for students, and it has no emotional or prescriptive meaning whatsoever. If you take it personally, that reflects on you and your ego more than it does on the teacher, who is teaching you for no reason other than because he’s passionate about Green’s theorem and wants to share it with people. He’s doing this for free, be grateful he makes videos at all

[@UniqueSundials]

While other kids were smearing ice cream over their faces and playing on swings you were doing differential equations.

[@hueheeuuehueuheuheuhueheeu1245]

this some kinder shit lol

[@Necrozene]

I don;t think the title is any good.

[@JW-ss8es]

nowhere geometric calculus is used in this video, only standard differential forms and wedge product (not geometric product). And this is exactly the confusing part that whether geometric calculus is needed given there is already differential form theory.

[@alexeyl22]

Less talking-head moments (or none at all) and more math please.

[@alexeyl22]

and you need to write (partial) derivatives of f in proper manner - df/dx, not just df. It is confusing what you did.

[@winstongludovatz111]

These are purely formal calculations with no geometric intuition -- i.e. worthless.

[@elquesohombre9931]

Geometric intuition is nice and all but seeing it calculated out is a lot better than just being handed the equation itself after being told why it should work. If you really want a geometric intuition, check out the khan academy article on greens theorem, but there is a lot of value in seeing it written out like this.

[@bucc5207]

Couldn't get past all the flexing in the first 90 seconds.

[@crehenge2386]

Funny how every math youtuber thinks they have the sole solution to teach students... Truly r/iamverysmart. But I guess you have to crap on others to motivate your own video, even though it sounds like you've barely read the wiki page... Also stealing others format ain't cool...

[@element118_5]

Great video!

[@KyleBroder]

🙏