Clever Clairaut Proof
Vložit
- čas přidán 22. 11. 2020
- In this video, I give a very clever proof of Clairaut's theorem, which says that if the partial derivatives f_xy and f_yx are continuous at a point, then must be equal. Usually this is proved using difference quotients, but here I give a proof using double integrals. I also give a nice proof using Green's theorem. This is a must-see video for multivariable calculus lovers!
Clairaut Counterexample: • Clairaut Counterexample
Fubini Counterexample: • Fubini Counterexample
Green's Theorem: • Green's Theorem
Multivariable Calculus Playlist: • Multivariable Calculus
Partial Derivatives Playlist: • Partial Derivatives
Subscribe to my channel: / drpeyam
Check out my TikTok channel: / drpeyam
Follow me on Instagram: / peyamstagram
Follow me on Twitter: / drpeyam
Teespring merch: teespring.com/stores/dr-peyam
Since i'm a big fan of Green's theorem i find the second proof way more beautiful!
"It's-a me Fubini!" 3:55
The proof using differentiation under the integral sign is very nice too!
The first proof is more general, so I definitely do prefer this one.
Thank you Dr Peyam
the second one is waaaaaay better :P
this is cool, but I still like the rather technical proof via difference quotients, since it proves more. We only need to assume the existence and continuity of one of those 2nd order partial derivatives and the existence of the other is a conclusion.
Awesome! It also proves the Euler´s reciprocity condition, fundamental for exact differentials and thermodynamics.
I`m a big fan from Brazil.... greetings from here!
I needed this
Both the proofs are simple and interesting.
Dr. Peyam is fantastic! Big fan of yours, professor!!!
Thank youuuu
Never laughed so hard on Fubini theorem! Thanks alot!
You are very loco mi amigo. Saludos Dr. P...👍👍👍
The mustache in the thumbnail got me to click!
It's-a me, Dr Weselcouch!
@@matekichba6640 Hahaha the mustache makes him look like me!
I really like the part of putting a rectangle inside a small ball
I love that too, you can put a rectangle inside a ball and a ball inside a rectangle
Very nice
I liked the proof using integration.
-Very cool- *_Very clever_* 😄😄😄
Clean proofs. Thank you! But I think we do need f_xy and f_yx to be continuous in a region at least for the integration. Just having f_xy and f_yx continuous at a point won't allow the integration, I think.
I like both theorems
Thank you for the video. Since fxy=fyx in most examples in a Calculus TB, It'd be great to provide an example of an f with fxy>
Already done
This is awesome
Please do a proof of gauss's law some day
Too much physics, sorry
The first proof was really nice, but idk enough about double integrals to be certain I understand. The second seems beautiful, but I know even less about line integrals (but I can guess), and far far less about vector fields and Green's theorem. It still felt like it was intuitive though somehow, but I might be fooling myself into thinking I understood far more than I did.
Fantastic, thank YOU very much, greens better
Very nice. solving it two times!!!? each one " as clever as the otthe one. What can you ask more !!!!? Thx.👏
"Its a me! Fubini" that sound's so cuteee
great as always! I know this theorem under the name of Schwarz, it's the same I guess
Schwarz’s Theorem 💗
i like the greens theorem way
Dr π m
In step 2 you have 2 sign integrals
Remove one sign integral in step 2
Because already first integratef
Very very cool **Standing Ovation**
What is f_xy? I had never seen that notation before. Is it ẟ(ẟf/ẟx)/ẟy?
Exactly
@@drpeyam ... Thank you!
I was thinking, how would you solve the PDE
fxx=fyy
My initial solution would be
f(x,y) = c1*e^[c2(x+y)] + Axy+Bx+Cd+D.
How far off is my answer?
Far off actually. This is the wave equation and has a very deep theory, check out my PDE playlist
@@drpeyam oh yikes haha. I'll look into it!
@@drpeyam oh wow, so the general solution is any 2 1-D functions that are twice differentiable with a slight change in their inputs! That's really awesome.
Yep :)
Why didnt anyone tell me this!!
But can fubini’s theorem be proven without Clairiuts theorem.
Yes
@@drpeyam how though?
Green's Theorem proof seems to be the most elegant... but it has no Fubini. :-D
Dont we have that the functions are equal of k-cells and not everywhere?
Everywhere, I gave an outline of the proof
@@drpeyam Well almost everywhere. As far as I remember Fubini works if the function is continuous almost everywhere (meaning everywhere up to domains that measure zero) and if all integrals exist.
In that sense fxy and fyx only have to be continuous and equal almost everywhere.
I think we are going in with the assumption that fxy and fyx are continuous (everywhere). So they will be equal everywhere.
Fubini theorem is much more general and does not require continuity.
It's meme Fubini🤣😂
why are you so smart?
Doctor, something wrong happened at your left side hair. You should take a noon nap every day.
LOL, I literally woke up from a nap that day
Here's my proof:
f_xy = ddf/(dx*dy) = ddf/(dy*dx) = f_yx
It may not be rigorous, but this is honestly where more rigorous proofs start from anyways.
Peyam didn't give a counter example in the video (one is linked in the description though), however the assumption that the mixed derivatives are both continuous is a key ingredient to any proof of the theorem since the identity f_xy = f_yx doesn't hold for just any function. Sometimes ideas for proofs can run into roadblocks when the rubber meets the road. The devil really is in the details :)
@@elephantdinosaur2284 The requirement of continuity is implied by the infinitesimals dx, dy, and df. dx, dy, and df can't all approach zero if the function is discontinuous at the point they're supposed to approach. So naturally if you apply the equation to a discontinuous function you can get bad results. The reason I like this type of proof is because it translates more intuitively to a geometric visual. As I said, not rigorous, but immensely useful.
@@BlackEyedGhost0 Whoops. Sorry for the confusion. I agree that the function f has to be continuous for the reasons you mentioned, however I was referring to both the partial derivatives f_xy and f_yx being continuous which isn't always true.
The counter example Peyam gave in his other video is f(x,y) = xy(x^2-y^2)/(x^2+y^2) with f(0,0) = 0. This function is continuous and both the partial derivatives f_xy = ddf/(dx*dy) and f_yx = ddf/(dy*dx) exists everywhere, however they aren't continuous at the origin and in particular they aren't equal at the origin either.
If a result doesn't hold for a particular class of functions, a good proof will have exceptions buried somewhere in it to exclude them. I agree with you I much prefer proofs that can be visualised geometrically, however sometimes intuitions based on visualisations can give bad results (one of the reasons why this video is more than 2 minutes long xD).
@@elephantdinosaur2284 ...Yeah, I know. There was no confusion on my part.
First !,