Level 5: Construct the set of real numbers from Dedekind-cuts, Tarski's axiomatization and Cauchy sequences. Prove that these produce isomorphic models.
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.
This was very VERY good. I have bs in pure math, so the ftc is pretty old news to me, and there were a shocking number of "aha!" moments in this video for me. Very well done. Subbed.
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.
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.
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).
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!
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.
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.
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.
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.
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.
@@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.
Level 4: proof using the generalized Stoke’s theorem
🤣🤣🤣
Level 5: Construct the set of real numbers from Dedekind-cuts, Tarski's axiomatization and Cauchy sequences. Prove that these produce isomorphic models.
@emanuellandeholm5657 can you please be my durg dealer😂
just curious is this supposed to be a joke?
@@niom9446 Yes.
Sandwich Theorem ♥
It's hard to understand 😢
@@user-ev2rj1bi4r Not really
@@user-ev2rj1bi4rsqueeze theorem 💪💪💪💪
You have to digest it properly
Cheeseburger theorem
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.
Thank you so much, that really means a lot!
This was very VERY good. I have bs in pure math, so the ftc is pretty old news to me, and there were a shocking number of "aha!" moments in this video for me. Very well done. Subbed.
@@kruksog Thanks so much! :)
And i implore you to make more of these wonderful proof videos!
Beautiful video, also quite relaxing! Well made!
Seeing expo marker on paper hurts, but your presentation is superb! Lovely demonstration.
The fundamental theorem was something that I could never quite grasp intuitively until now. Great video
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.
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.
Very clear video - nice job.
Great explanation... i like how you emphasize the importance behind each level of understanding... I hope you do more videos!
That's what I was looking for! Thanks for the video
Superb demonstration! I like how you used the construction paper as visual aids. Seeing the expo marker on paper hurt to watch haha
This is the first video I've unironically watched at speed 0.75 from beginning to end
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).
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!
Great video
Insightful! And the explainer is so attractive 😍
Beautiful.
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.
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.
Thank you!
Im gonna prove the fundamental theorem of calculus using the weight of the marker before and after coloring under the curve 😂
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.
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.
all about the rate of change of physics
It's only the first part of the theorem. Can you also make a video about the second part.
Done!
Is your table made of concrete?
Nope, it is just wood!
can dA be less than dx ? as when we get derivatives we get it at times less than one.
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.
@@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.
cool
When you can explain this. But can't land a front handspring front.
How large would you guess is the population of people that can do both?
That music made the video unwatchable. Seems this only applies to me.
What is the name of the music?
I composed it for this video! It doesn’t have a name.
Will you put on youtube, its really good!
@@paulostipanov7682
Thank you so much! I uploaded it as an unlisted video here: czcams.com/video/Bl5zXMEP_uM/video.html
Is your table made of concrete?