3 Levels of Proving the Fundamental Theorem of Calculus |

  Рет қаралды 20,502

Trivial

Trivial

Күн бұрын

This video contains three proofs of the Fundamental Theorem of Calculus at increasing levels of rigor, as well as a more general argument for why intuitive and rigorous proofs enhance each other.
Chapters:
0:00 Introduction
0:26 Review of the theorem
1:17 Proof One
2:30 Proof Two
4:34 Proof Three
7:22 Conclusion
This video is an entry to the third Summer of Math Exposition contest, hosted by ‪@3blue1brown‬.
The content and visuals of this video were all made by me. I also composed the soundtrack using MuseScore 4.
Thanks for watching. I hope you enjoy it!

Пікірлер: 53
@hubertmasson7550
@hubertmasson7550 9 ай бұрын
Level 4: proof using the generalized Stoke’s theorem
@academyofuselessideas
@academyofuselessideas 8 ай бұрын
🤣🤣🤣
@emanuellandeholm5657
@emanuellandeholm5657 3 ай бұрын
Level 5: Construct the set of real numbers from Dedekind-cuts, Tarski's axiomatization and Cauchy sequences. Prove that these produce isomorphic models.
@precociousdeathdealer202
@precociousdeathdealer202 2 ай бұрын
​@emanuellandeholm5657 can you please be my durg dealer😂
@niom9446
@niom9446 2 ай бұрын
just curious is this supposed to be a joke?
@emanuellandeholm5657
@emanuellandeholm5657 2 ай бұрын
@@niom9446 Yes.
@LB-qr7nv
@LB-qr7nv 11 ай бұрын
Sandwich Theorem ♥
@user-ev2rj1bi4r
@user-ev2rj1bi4r 11 ай бұрын
It's hard to understand 😢
@TU7OV
@TU7OV 2 ай бұрын
@@user-ev2rj1bi4r Not really
@pyrotechnicalbirdman5356
@pyrotechnicalbirdman5356 Ай бұрын
​@@user-ev2rj1bi4rsqueeze theorem 💪💪💪💪
@oxbmaths
@oxbmaths Ай бұрын
You have to digest it properly
@primenumberbuster404
@primenumberbuster404 Ай бұрын
Cheeseburger theorem
@instrumental7809
@instrumental7809 5 ай бұрын
I've been reading about proofs that are essentially based on the same ideas as the 3rd proof in this video but I have been failing to understand exactly how they worked until I watched this video. The way you explained how the squeeze theorem comes into play made it so easy to grasp. You sir have gained a new subscriber.
@trivial-math
@trivial-math 5 ай бұрын
Thank you so much, that really means a lot!
@tristinbell
@tristinbell 9 ай бұрын
And i implore you to make more of these wonderful proof videos!
@user-br5hj4oj9i
@user-br5hj4oj9i 11 ай бұрын
Beautiful video, also quite relaxing! Well made!
@GhostyOcean
@GhostyOcean 7 ай бұрын
Seeing expo marker on paper hurts, but your presentation is superb! Lovely demonstration.
@newtona8798
@newtona8798 9 ай бұрын
That's what I was looking for! Thanks for the video
@tedsheridan8725
@tedsheridan8725 9 ай бұрын
Very clear video - nice job.
@academyofuselessideas
@academyofuselessideas 8 ай бұрын
Great explanation... i like how you emphasize the importance behind each level of understanding... I hope you do more videos!
@zaccandels6695
@zaccandels6695 3 ай бұрын
The fundamental theorem was something that I could never quite grasp intuitively until now. Great video
@kellystevens6464
@kellystevens6464 11 ай бұрын
Thank you!
@isavenewspapers8890
@isavenewspapers8890 6 ай бұрын
Beautiful.
@HPTopoG
@HPTopoG 8 ай бұрын
Neat, but there’s an implicit assumption you’ve made without mentioning it! The function needs to be sufficiently continuous! The Cantor function on [0,1] has integral 1/2 and is even uniformly continuous, but it has derivative 0 almost everywhere. So it can’t satisfy the FTC. It’s probably more than this video calls for, but I think if you make a follow up video it might be a good idea to include at least some mention of different continuity strengths.
@trivial-math
@trivial-math 7 ай бұрын
I think you're confusing the two parts of the theorem! Though its true that the second fundamental theorem of calculus fails for some continuous functions like the Cantor function, the first fundamental theorem holds for any continuous function that is Riemann integrable.
@azorbarros3308
@azorbarros3308 3 ай бұрын
Great video
@GhostyOcean
@GhostyOcean 7 ай бұрын
Superb demonstration! I like how you used the construction paper as visual aids. Seeing the expo marker on paper hurt to watch haha
@aidansunbury9341
@aidansunbury9341 11 ай бұрын
Insightful! And the explainer is so attractive 😍
@pedro-z1z
@pedro-z1z 15 күн бұрын
This is the first video I've unironically watched at speed 0.75 from beginning to end
@Sstevewong36
@Sstevewong36 2 ай бұрын
all about the rate of change of physics
@alexanderkotnik2625
@alexanderkotnik2625 6 ай бұрын
Is your table made of concrete?
@elomensch9566
@elomensch9566 7 ай бұрын
cool
@instrumental7809
@instrumental7809 5 ай бұрын
Amazing video, but one thing I simply cannot understand in these proofs is the step at 3:40 where you take the limit of both sides which makes the left hand side A'(x) but how do we conclude the limit as delta x approaches zero of f(x) is equal to f(x)? I would appreciate it greatly if you could explain that to me.
@trivial-math
@trivial-math 5 ай бұрын
When finding this limit, we are only changing delta x; x is staying constant. Since x is constant, f(x) is constant regardless of the value of delta x, so the limit is just f(x). It's like how lim_{b -> 0} (a) = a. What's really happening "under the hood" is that the error between both sides of the approximation goes to zero, so we get true equality in the limit. I'm just not explicitly writing it out.
@sachinrath219
@sachinrath219 4 ай бұрын
can dA be less than dx ? as when we get derivatives we get it at times less than one.
@trivial-math
@trivial-math 4 ай бұрын
Yes! dA is equal to f(x)dx, so dA is less than dx whenever f(x) is less than 1. Try looking at the graph of y = 0.5 and see how the area function grows at 1/2 the rate of x.
@sachinrath219
@sachinrath219 4 ай бұрын
@@trivial-math thanks a lot, my confusion was asking both are represented by the letter d what stands for near to zero, so both are always equal i e beyond comparison, so dA can be greater, lesser or equal depending on situation, pl reply, thanks.
@Avighna
@Avighna 4 ай бұрын
3:48 My problem with this proof is not necessarily the lack of rigor, but more that you've implicitly assumed that Δx > 0. So when you take the limit as Δx approaches 0, you have shown that the one-sided limit (specifically the Δx -> 0+) is equal to A'(x), but not that the other (0-) limit approaches A'(x) as well. This can easily be fixed since lim h -> 0 (f(x) - f(x-h)) / h is also a perfectly valid definition for f'(x). So do the same thing for x-Δh instead of x+Δh (if you take both cases, taking Δx > 0 is completely fine), and say that A(x) - A(x-Δx) ≈ f(x) Δx, and continue the same way. I suppose this is technically a complaint about rigor in a way.
@trivial-math
@trivial-math 4 ай бұрын
This proof actually still works even if Δx approaches from below 0! When Δx is negative, both sides of the approximation A(x + Δx) - A(x) ≈ f(x)Δx get their signs flipped. If f(x) is positive for example, then f(x)Δx is negative and A(x + Δx) < A(x), so A(x + Δx) - A(x) is also negative. After dividing both sides by Δx, the sign flips cancel each other out and we get A(x + Δx) - A(x)/Δx ≈ f(x). The real lack of rigor is when the approximately equals sign turns into an equals sign.
@davethesid8960
@davethesid8960 11 ай бұрын
It's only the first part of the theorem. Can you also make a video about the second part.
@trivial-math
@trivial-math 7 ай бұрын
Done!
@eugene1317
@eugene1317 Ай бұрын
Im gonna prove the fundamental theorem of calculus using the weight of the marker before and after coloring under the curve 😂
@neilmcguire5153
@neilmcguire5153 4 ай бұрын
I wouldn’t call these very rigorous. They are definitely ways to explain the concepts and prove them from a layperson’s point of view. But if you’re a senior undergraduate math major or a graduate student in math these proofs won’t fly. You need more real analysis and you need to prove both versions of the FTC. The first version with the definite integral being the difference of the antiderivatives at a and b and the second version involving the indefinite integral with basepoint a. The first version states given a finite set E and functions f, F: [a,b]-> R such that: (a). F is continuous on [a,b], (b). F’(x) = f(x) for x ∈︎ [a,b] \ E, (c). f belongs to R[a,b]. Then we have ∫︎ f = F(b) - F(a).
@trivial-math
@trivial-math 4 ай бұрын
This video is only about the first fundamental theorem! I have a seperate video on this channel about the second part of the theorem, which is what you are describing in your comment. This video proves that, if f is a continuous real-valued function, a is a constant in the domain of f, and A(x) = the integral f from a to x, then A'(x) = f(x). The first two proofs in this video are not meant to be rigorous, but the third proof is fully rigorous. Please let me know if there are any specific steps of the third proof you think are incorrect!
@alexanderkotnik2625
@alexanderkotnik2625 6 ай бұрын
When you can explain this. But can't land a front handspring front.
@graf_paper
@graf_paper 4 ай бұрын
How large would you guess is the population of people that can do both?
@openyard
@openyard 2 ай бұрын
That music made the video unwatchable. Seems this only applies to me.
@paulostipanov7682
@paulostipanov7682 9 ай бұрын
What is the name of the music?
@trivial-math
@trivial-math 9 ай бұрын
I composed it for this video! It doesn’t have a name.
@paulostipanov7682
@paulostipanov7682 9 ай бұрын
Will you put on youtube, its really good!
@trivial-math
@trivial-math 9 ай бұрын
@@paulostipanov7682 Thank you so much! I uploaded it as an unlisted video here: kzfaq.info/get/bejne/eNJlrbt-qrbDpYE.html
@APaleDot
@APaleDot 9 ай бұрын
Is your table made of concrete?
@trivial-math
@trivial-math 9 ай бұрын
Nope, it is just wood!
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