Why We Never Actually Learn Riemann's Original Definition of Integrals - Riemann vs Darboux Integral

2024 ж. 9 Мам.
82 671 Рет қаралды

We typically credit Riemann for his discovery of integrals. However, in school, we never actually learn the actual Riemann Integral as he invented. Instead we learn the "Riemann" Integral, which is actually the Darboux Integral. Why is that? In this video, we uncover the truth behind the two different definition of integrals.
Links:
Formal Definition of Supremum and Infimum
en.wikipedia.org/wiki/Infimum...
Why Sup and Inf Always Exist
en.wikipedia.org/wiki/Least-u...
Proof That the Lim of Upper Sum = Inf of Upper Sum
www.overleaf.com/read/dynwhfq...
Integrability of Monotone and Continuous Functions
www.kau.edu.sa/Files/0003377/...
Chapters:
00:00 Intro
01:24 Rigorous Foundations of Calculus
02:06 Different Types of Integration
02:48 Generalized Riemann Sum
04:08 Riemann Integrability
04:49 Failure of Limit
05:24 Non-Integrable Function
06:16 Riemann Integrability of x^3
11:11 Upper and Lower Sum
13:01 Redefining Riemann integrals
13:29 Darboux Integrability
14:49 Darboux Integrability of x^3
16:59 Fatal Shortcomings of the Riemann Integral
17:32 Outro
Corrections:
15:12 The denominator for the sum of cubes is 4

Пікірлер
  • Riemann's definition generalizes to the Gauge Integral, where the max width δ isn't the same for all intervals, but depends on the tag, so it's really a function δ(x), called a gauge. All Lebesgue- and indefinite-Riemann-integrable functions are gauge integrable. It has apparently found use in evaluation of some path integrals in Quantum Mechanics.

    @waarschijn@waarschijn2 ай бұрын
    • Thank you for decimating this clickbait slide reel in 3 sentences.

      @jaredroussel@jaredroussel2 ай бұрын
    • isn't it the Henstock-Kurzweil theory of integral?

      @eliacampigotto2632@eliacampigotto26322 ай бұрын
    • @@eliacampigotto2632 Yes, that's a different name.

      @waarschijn@waarschijn2 ай бұрын
    • @@jaredroussel I'm not "decimating" anything. The video is informative and the title is appropriate: I also learned the Darboux version first.

      @waarschijn@waarschijn2 ай бұрын
    • @@waarschijn Not talking about the title. Talking about the slide reel about the original Riemann Integrating sucking.

      @jaredroussel@jaredroussel2 ай бұрын
  • I think the important thing about the reiman integral is it "feels" more general and natural. So proving that the Darboux integral is well founded and equivalent to the reiman integral is an important logical step, even though once the proof is done, you dont ever want to look at it again. The knowledge that they are equivalent is in itself valluable.

    @varno@varno2 ай бұрын
    • It's historically interesting, but I don't think it's any more natural. "The infimum of all upper sums" seems like a clean and straightforward definition, compared to a limit as the norm of some infinitely-dimensioned space of partitions and tags approaches zero. An upper sum is independent of "tags" entirely and only depends on the partition, which makes it much more natural to reason about. The fact that you often want to use some limit to bound it from above falls out naturally as you try to compute the value by sandwiching it, instead of being baked into the definition.

      @OMGclueless@OMGclueless2 ай бұрын
  • 1) Riemann never intended to use his definition for calculations. This is one of those typically theoretical definitions, very generell and useful for existence proofs and such. 2) In 1868 Riemann was already 2 years dead. He found his legendary definition around 20 years earlier as a young student in Göttingen.

    @markborz7000@markborz70002 ай бұрын
  • when you understand the construction of the riemann integral, you understand the construction of the lebesgue integral when you yearn the basics in measure theory

    @FreeGroup22@FreeGroup222 ай бұрын
  • We learn Riemanns definition of integrals in Calculus 1 mate... And then immideately learn Darboux integral. Then in any further analysis course, you learn about Lebesgue integrals.

    @Eknoma@Eknoma2 ай бұрын
  • Maybe highschool, but not in uni. My analysis class taught the Riemann integral first, then the darboux integral. We have some useful theorems to prove, almost never used original Riemann definition, just like you said.

    @NaHBrO733@NaHBrO7332 ай бұрын
    • my AP class in high school just taught riemann integral

      @maxthexpfarmer3957@maxthexpfarmer39572 ай бұрын
  • Yet another "you never learned this in school" topic that we DID learn in HS... Thank you Bélabá, you were the best math teacher we ever had!

    @HA7DN@HA7DNАй бұрын
  • Nice video! You should do a follow up on Lebesgue integration next, it would be epic!

    @fedebonons8453@fedebonons84532 ай бұрын
  • There's a misprint on 15:12 - there must be "4" instead of "3" in the denominator of an expression for sum of cubes. I actually learned both integrabilities in university and can say that this is a really cool and interesting video! Keep it up 👍

    @infas0tka693@infas0tka6932 ай бұрын
  • This is absolutely outstanding.

    @edwardperry5041@edwardperry50412 ай бұрын
  • Please do a Video about Lebesgue integrals :)

    @maxfred1696@maxfred16962 ай бұрын
  • Very nice video! To my mind, a strong advantage of the Darboux variant is that you need only check one specific sequence of partitions. In contrast to that, in the Riemann variant you had to keep the partitions (and tags) arbitrary. Just wanted to mention this since you used it (of course) but didn't actively point it out.

    @Mau-vz8qo@Mau-vz8qo2 ай бұрын
    • Using the squeeze theorem, they are easily seen to be equivalent.

      @cparks1000000@cparks10000002 ай бұрын
    • ​@@cparks1000000 Yes, the equivalence is kind of the video's point. I was referring to how you can apply the different formulations.

      @Mau-vz8qo@Mau-vz8qo2 ай бұрын
  • Love this channel! One of the best videos I’ve ran into!

    @iammikesimon@iammikesimon2 ай бұрын
  • Great video! Would you consider covering the Ito and Stratonovich integrals? They are very interesting topics!

    @williamdavis3658@williamdavis36582 ай бұрын
  • Nice one 🎉 Good topic to discuss...But I think need more like this

    @5Stars49@5Stars492 ай бұрын
  • I actually did dadboux integrability proofs in my freshman calc class. My math dept was on some crazy shir lmao

    @Basilisk-tu7ud@Basilisk-tu7ud2 ай бұрын
  • There is flexibility in Riemann integral when integrating from the definition. For instance you can integrate sqrt(x) from 0 to 1 by choosing x_i = (i/n)^2 for i = 1, ..., n as the partition points. Now, Δx_i = (2i + 1)/n and sqrt(x_i) = i/n. So, this non-regular partition results in a sum that can be evaluated. The Darboux definition on a regular partition results in evaluations points for which the sum cannot be evaluated directly because we have no general exact formula of sum of sqrt(i).

    @jonathanewebster@jonathanewebster2 ай бұрын
    • Actually you have it backward, the Darboux integral does not require uniform partitions and indeed it is the Darboux integral that allows you to check just a single convenient partition.

      @TheEternalVortex42@TheEternalVortex422 ай бұрын
    • Darboux integral is solvable with any partition of choice while Riemann requires you to show that all partitions work. However there are theorems that allow you to take a limit as the number of intervals goes to infinity of say for example a uniform partition.

      @ccbgaming6994@ccbgaming699410 күн бұрын
  • Very informative video bro!!👍👍

    @66sbjaygoti80@66sbjaygoti802 ай бұрын
  • So why do I hear Darboux‘ name the first time today if their method is more powerful?

    @maxfred1696@maxfred16962 ай бұрын
    • Power is not fame. Although in this case it is, but because Riemann has his hands in more than just integration.

      @hybmnzz2658@hybmnzz26582 ай бұрын
    • You should open any real analysis book and you will find it.

      @joelwillis2043@joelwillis20432 ай бұрын
    • his*

      @NoNameAtAll2@NoNameAtAll22 ай бұрын
    • ​@@joelwillis2043 but Riemann did appear in highschool and Darboux didn't.

      @dekippiesip@dekippiesip2 ай бұрын
    • Have you have heard about the lebesgue integral?

      @youtubepooppismo5284@youtubepooppismo52842 ай бұрын
  • Defining integrals based on a definition would be a bit unmathematical anyway, as you always want to use an axiomatisation instead. Darboux has the first rigorous definition, equivalent to the following three/four properties. 1. The integral of a constant function is... 2. The integral of f(x) from a to b plus from b to c is the integral of f(x) from a to c. 3a. If f(x)≥0 for a≤x≤b, the the integral of f(x) from a to b is greater or equal to 0. 3b. If f(x)≤0 for a≤x≤b... The cool part about axiomatisation with inequalities is that you don't have any limits. I guess it cool, but you would want to use limits in practise anyways, since you basically only care about cases of equality.

    @caspermadlener4191@caspermadlener41912 ай бұрын
    • There’s not really a difference between a definition an an axiomatisation. Not in this case anyways

      @biblebot3947@biblebot39472 ай бұрын
    • Describing constructions as “unmathematical” is a bit silly for a variety of reasons. Tons of things in math are described via constructions, and integrals are certainly one of them. There are pros and cons to both approaches. For example, people often define the real numbers to be “the unique complete ordered field”. This is all well and good until you realize that you have no idea if such a thing exists. Indeed, in order to rigorously prove that this definition is usable, you need to A: Construct a model of the reals. B: Show that it is a complete ordered field. C: Show that all complete ordered fields are isomorphic to it. This is just straight up objectively more work than constructing the real numbers, which is the step in part A. A similar thing happens for most axiomatic definitions: To create an axiomatic definition you need to first list the axioms and then prove that they have a unique solution, but the fastest way to do that second part almost always involves constructing the thing anyway, so most mathematicians don’t bother. I realize that I said there were pros and cons but have listed only cons, so just for the record, the reason some mathematicians like axiomatic definitions is that there’s a certain conciseness to them that makes them “prettier”. Like, “the unique complete ordered field” is a much prettier definition than [insert your favorite construction of the reals], even though it takes more work to prove that the definition works.

      @TheBasikShow@TheBasikShow2 ай бұрын
    • Just those properties? That doesn't seem sufficient, unless I'm missing something; I suspect you need also that it's a linear functional?

      @Stirdix@Stirdix2 ай бұрын
    • @@Stirdixseems like those axioms only characterize integrals of piece-wise constant functions.

      @Deguiko@Deguiko2 ай бұрын
  • This is great and interesting man.. But can you make a video or two on Lebesgue integral? Since it is subtle and blends sets, measures and calculus ideas unlike others.. I am still looking for good videos on these topics!?

    @TheJara123@TheJara1232 ай бұрын
  • Can you please do a video on Lebesgue Integral ?

    @SayakKolay@SayakKolay2 ай бұрын
  • I don't know where you're coming from saying we probably never learned the Riemann integral in college. Every elementary calculus textbook I've ever seen teaches some variation of the Riemann integral. By contrast, I've never seen Darboux's definition in anything short of an upper-level text.

    @DrCorndog1@DrCorndog1Ай бұрын
  • excellent video!

    @schweinmachtbree1013@schweinmachtbree10132 ай бұрын
  • I've actually learnt about both Riemann and Darboux integration, although the arbitrary partitioning method was attributed to Darboux, not Riemann. We've also formalized it using the Lower and Upper sums and their convergence to the same number from the get go. Never seen the arbitrary rectangle point/height method before.

    @bigshrekhorner@bigshrekhorner2 ай бұрын
    • That's the Darboux integal that you're describing.

      @cparks1000000@cparks10000002 ай бұрын
  • Quite right 👍 the Darboux version is technically easier and more natural to prove results with. For first year calculus it won't matter but later on it's preferable to use Darboux.

    @hanks.9833@hanks.98332 ай бұрын
  • Hi EpsilonDelta, I came accross your channel from the SOME3 playlist. Appreciate what you do here. Is there an email address I can contact you at?

    @elinmalikzade@elinmalikzade2 ай бұрын
  • So in my Analysis class we did start with the upper/lower integral versions, but also saw the Riemann version and proved it's equivalence, so the title did confuse me

    @TheLuckySpades@TheLuckySpades2 ай бұрын
  • Using the Darboux definition is more-or-less just applying the squeeze theorem to the Riemann definition.

    @cparks1000000@cparks10000002 ай бұрын
  • At around 12:20, you switch from max/min to sup/inf. This is good to handle the problem you pointed out, but opens a new problem: sup and inf can be infinite, so what if there is a vertical asymptote? For example, if you want to find the integral of log(x) between 0 and 1, how would you find the limit of L(f,P) as ||P|| tends to 0? Since one of the intervals will be sure to contain the asymptote, then for that partition inf[f(xi*)] will always be -infinity, which would make every L(f,P) undefined, no matter how you partition it.

    @3snoW_@3snoW_2 ай бұрын
    • Riemann/Darboux's original notion of integrability was not meant to tackle such issues. Those are resolved using the definition of improper Riemann/Darboux Integrals, which is taking the limit of the integral as a or b approaches such asymptote

      @HaramGuys@HaramGuys2 ай бұрын
  • At 15:20, what tool were you using? Was it Python-based?

    @iksootm@iksootm2 ай бұрын
    • Wolfram Mathematica

      @EpsilonDeltaMain@EpsilonDeltaMain2 ай бұрын
  • Hi EpsilonDelta, appreciate the effort put in to this video. Would you be interested in new video topic suggestions? What would be the best email address to contact you at?

    @elinmalikzade@elinmalikzade2 ай бұрын
  • tbh I think the darboux integral is more intuitive in isolation but the riemann integral is more intuitive in the context of formalizing calculus

    @gabitheancient7664@gabitheancient76642 ай бұрын
  • I'm only appreciating analysis 30 years after doing it in college.

    @audience2@audience22 ай бұрын
  • The failure of the Riemann integral for functions like the one showed that had one constant value for rationals and another for irrationals is a classic and fun one. There’s an interesting pedagogical/thought experiment that points toward more advanced treatment of the integral (Lebesque, eg) by way of “rescuing” the Riemann integral in that function by taking the sampling step of selecting tags in a “statistical” way. So in each sub interval, you _could_ pick a rational. But if you were to choose randomly, you never would, since rationals are infinitesimally rare compared to irrationals. The probability of selecting a rational from a finite real interval is zero. So that tagged partition with all rational tags can be ignored since😊 it basically would never ever be picked by random selection. The probability of picking that from all the possible samples is zero. It’s a handwavy, “physics class math” sort of explanation. And to make it rigorous you’d have to reinvent the same ideas of measure theory anyway. But pedagogically I think it’s an interesting bridge.

    @cyrilanderson@cyrilanderson2 ай бұрын
  • Interesting. I actually learned the Riemann Integral in University at 2001. And I wouldn't say it's useless, if the calculation may contain a limited amount of errors. You see, if you have a sampling (a value recorded for timestamps), you might want some sort of "integral" of that, but that input is not even continuous. What's useful is not the formulation of the Riemann Integral itself, but the idea behind it (take those rectangles and their area up, and you'll have a sum with some error estimation).

    @cfhay@cfhayАй бұрын
  • Newton and Leibniz did _not_ invent the idea to approximate the area by rectangles and refining the approximation by using more and more rectangles. That idea existed centuries, if not decades, before them. E. g. Fermat, Pascal and Cavalieri, among many others, used that idea extensively.

    @bjornfeuerbacher5514@bjornfeuerbacher55142 ай бұрын
    • That is not what they said though, "(...) and the basic idea of partitioning random shapes into rectangles and triangles *has existed since ancient times* , however *the integral as we know today* was invented by Newton and Leibniz, which uses rectangles to approximate the area (...)".

      @LinaWainwright@LinaWainwright2 ай бұрын
    • Also, I remember learning that Cauchy did it before Riemann, but his idea was to have arbitrary partitions but with a fixed tag on the leftmost point of each subinterval. Later, someone proved Cauchy's and Riemann's definition are equivalent, so the tags aren't adding anything of value. Very interesting. All in all, Darboux will usually be the best one for working with proofs.

      @uwukia@uwukia2 ай бұрын
    • @@LinaWainwright Huh? So it says that the idea to approximate the area with rectangles was invented by Newton and Leibniz, just as I said. And as I said, that's wrong, the idea had been used by others before.

      @bjornfeuerbacher5514@bjornfeuerbacher55142 ай бұрын
    • @@bjornfeuerbacher5514 the idea did exist but those people did not apply in the same vein. It's no wonder squaring a circle wa seen as impossible until they went and did it, because of said ideas.

      @StellaEFZ@StellaEFZ2 ай бұрын
    • @@StellaEFZ Actually, what they did was not really "squaring the circle" in the sense this was originally meant.

      @bjornfeuerbacher5514@bjornfeuerbacher55142 ай бұрын
  • Are you using Manim for animations?

    @prashantsharma-mc6hh@prashantsharma-mc6hh2 ай бұрын
  • the lebesgue animation was genius

    @paulwaller3587@paulwaller35872 ай бұрын
  • the video is very good.

    @user-im3lg5sl6c@user-im3lg5sl6c2 ай бұрын
  • 4:22 was it intentional not to put the differential in that integral?

    @giuseppenonna2148@giuseppenonna21482 ай бұрын
    • The notation ∫f(x)dx = ∫f(t)dt = ∫f(u)du clarifies that we are choosing to integrate the function over some specific choice of independent variable. ∫f is 1800s mathematicians saying it doesn't matter what letter we use, it's all the same anyway, so lets be lazy

      @EpsilonDeltaMain@EpsilonDeltaMain2 ай бұрын
    • In analysis, it's sometimes conventional to just write the integral without a "dx", since for single-variable functions, it not only makes no difference, but also can't be rigorously defined. You only get to a rigorous definition with the theory of "differential forms", which comes much later!

      @amritlohia8240@amritlohia82402 ай бұрын
  • 15:17 what's the software that you are using???

    @peterwan7945@peterwan7945Ай бұрын
    • Thats wolfram mathematica

      @HaramGuys@HaramGuysАй бұрын
  • Did the formal definition of the integral not come before that of the derivative? Jacob Bernoulli created the definitions and even the notation, namely the large S standing for sum, by himself in the late 1600s. Then the derivative was later created or formalized by the likes of Euler of Cauchy. I'm not sure if I am entirely correct, but I do believe the integral came first.

    @kyantum@kyantum2 ай бұрын
    • Newton and Leibniz definitely used differentiation as well, since their notation has survived to the present day. (Leibniz used the familiar df/dx, while Newton put a dot over the dependent variable, which I can't get on my phone's keyboard.)

      @tomkerruish2982@tomkerruish29822 ай бұрын
    • The large S was invented by Leibniz, not by Bernoulli. Bernoulli invented the _name_ integral. But both did not really give a _formal_ definition. After all, the whole concept of limits was only rigorously established long after Bernoulli.

      @bjornfeuerbacher5514@bjornfeuerbacher55142 ай бұрын
    • @@tomkerruish2982 Yes, they used it - but they never rigorously _defined_ what this actually means. They kept talking about "infinitesimals" without specifying what that is actually supposed to mean.

      @bjornfeuerbacher5514@bjornfeuerbacher55142 ай бұрын
    • Newton and Leibniz made calculus for physics, but it was not rigorous. Euler, cauchy, riemann, etc. created analysis, which basically formal calculus.

      @icodestuff6241@icodestuff62412 ай бұрын
    • @@icodestuff6241 Well, Euler's approach was not really "formal" and/or "rigorous" in the modern sense - he did his calculations by freely using an "infinitely large" and an "infinitely small" number... But I agree on the others.

      @bjornfeuerbacher5514@bjornfeuerbacher55142 ай бұрын
  • The Apostol text uses the Riemann definition of the integral

    @robshaw2639@robshaw26392 ай бұрын
    • Apostol FTW!

      @tomkerruish2982@tomkerruish29822 ай бұрын
    • @@tomkerruish2982 yes, his two volumes are imo the best freshman calculus course for aspiring math majors

      @robshaw2639@robshaw26392 ай бұрын
    • Even bartle has it

      @AmanKumar-sk3nk@AmanKumar-sk3nk2 ай бұрын
    • Well, many have it. I hated this "YoU NevEr HEarEd BefOre" about every well known thing. I mean even if you once checked Wikipedia it is there

      @sychuan3729@sychuan37292 ай бұрын
  • You say we often tag the rightmost or the leftmost point on the interval, but Darboux integrals, which you claim is most often the taught variation, tags the minimum and maximum of the function on each interval, which is a variation you don't really mention in the intro.

    @MasterHigure@MasterHigure2 ай бұрын
    • The Darboux integral is the one that's most often taught *in analysis courses*. The version where you tag the rightmost/leftmost point is the one that's most often taught *in introductory calculus courses*. Hopefully that clarifies things...

      @amritlohia8240@amritlohia82402 ай бұрын
  • We first learned darboux integral in my uni, then equivalence with riemann and then lebesgue

    @quantumsoul3495@quantumsoul349512 күн бұрын
  • For Riemann, it is not true that you have to know the result in advance. In case of convergence, every single sequence of tagged partitions converges to the same value, so you can just choose one that's convenient to determine the value. Note that you can do that *before* proving convergence because all you do at this step is to say *if* the integral exists, *then* it has to have this value; you then in the second step use that value to prove that the integral actually exists. I'm also not sure that Darboux integration should be considered as a separate type of integral rather than as a method of evaluating Riemann integrals.

    @__christopher__@__christopher__2 ай бұрын
  • Nah, mate, I actually had the Riemann definition in my Calculus 2 course. You are just tripping. The sad thing is that there was no time to go into the other ways of integrating.

    @VeteranVandal@VeteranVandal2 ай бұрын
    • Stated once to be forgotten or actually used the definition in any meaningful ways to prove anything??

      @HaramGuys@HaramGuys2 ай бұрын
    • @@HaramGuys All he did to make the Riemann integral harder is not take a uniform partition like he did in the second example. That simplification has nothing to do with the differing definitions and works perfectly well with the Riemann integral too. He even mentions taking the left or right endpoints which is all he did in the second example for L(f,P) and R(f,P) as x^3 is monotone increasing.

      @abebuckingham8198@abebuckingham81982 ай бұрын
    • @@abebuckingham8198 no, because you need a general proof for the riemann definition. It was okay in the second part because that was explicitly for the upper and lower bounds. But the original riemann definition requires *for all partitions with norm ||P||* . Hence the Darboux definition is better because it lets you simplify the Riemann definition

      @icodestuff6241@icodestuff62412 ай бұрын
    • @@icodestuff6241 You just prove that if it converges for one that it converges for all of them. This is very easy to do in full generality. That's what motivates the U(f,P)-L(f,P) definition in the first place.

      @abebuckingham8198@abebuckingham81982 ай бұрын
    • ​ @abebuckingham8198 Sorry, but your statement is just false. Take, for example, the integral of 1/sqrt(x) from 0 to 1. The function is not integrable, yet the right-riemann sum converges. The proof is as follows. Taking N partitions we have, we get dx = 1/N, and x_n = n * dx = n/N. Thus the right riemann sum is S[ 1/sqrt(x_n) * dx ] = S[ 1/sqrt(n/N) * 1 / N] = S[ 1 / sqrt(N * n)] where S represents summing from n = 1 to n = N. Since N is a constant, we can pull it out, giving 1/sqrt(N) * S[ 1 / sqrt(n)]. S[ 1 / sqrt(n)] is bounded by 1 + 2sqrt(N) which can be proven by integrating 1/sqrt(x) from 1 to N. Thus, 1/sqrt(N) * S[ 1 / sqrt(n)] < ( 1 + 2sqrt(N) ) / sqrt(N) = 2 + 1/sqrtN Therefore, as N goes to infinity, the right-riemann sum converges to a value less than or equal to 2.

      @icodestuff6241@icodestuff62412 ай бұрын
  • Riemann-Stieltjes deserves more fame🤗

    @Bruh-bk6yo@Bruh-bk6yo2 ай бұрын
    • And Henstock-Kurzweil, of course!

      @Bruh-bk6yo@Bruh-bk6yo2 ай бұрын
  • 15:08 first thought: "how do he get complex numbers there?"

    @pawerzonca646@pawerzonca646Ай бұрын
  • the thumb suggests that what is in the curriculum does never suck, which is misleading in the least.

    @16876@168762 ай бұрын
  • As many others I also did learn both definitions in school so the title does not apply to me. I also do not agree with the Darboux integral being overall better - try proving linearity... for Riemann it is a piece of cake.

    @ruze11v19@ruze11v1917 күн бұрын
  • I did learn it!

    @harryschmidt4465@harryschmidt44652 ай бұрын
  • Wtf that's exactly what I was tought. Twice, to be exact: in school and in Uni.

    @atreidesson@atreidesson2 ай бұрын
  • the algebra for x^3 Riemann integrability is no way a cosmic horror, just a little tedious

    @vinesthemonkey@vinesthemonkey2 ай бұрын
  • Idk what you’re talking about I’m doing this in my first year in uni

    @ioannischrysostomou7012@ioannischrysostomou70122 ай бұрын
  • Nah Reimanns was the first definition of integrability I was taught

    @peamutbubber@peamutbubberАй бұрын
  • Okay. Let's make something clear. Sure, int_(a)^(b)(f(x)dx) finds the area under the curve of y=f(x) from a to b, but it also finds the length of the curve y=int(sqrt(f(x)²-1)dx) from a to b

    @seroujghazarian6343@seroujghazarian63432 ай бұрын
  • Weirstrass function still remains integrable nowhere

    @DeathSugar@DeathSugar2 ай бұрын
    • It's integrable since it's continuous. Its not differentiable anywhere

      @HaramGuys@HaramGuys2 ай бұрын
    • @@HaramGuys my bad, you're right.

      @DeathSugar@DeathSugar2 ай бұрын
    • ​@@DeathSugarfinally someone who takes responsibility

      @NamanNahata-zx1xz@NamanNahata-zx1xz2 ай бұрын
    • @@NamanNahata-zx1xz Person online who takes responsibilty caught live on camera!

      @nemanjalazarevic9249@nemanjalazarevic92492 ай бұрын
    • @@NamanNahata-zx1xz"takes responsibility". Why be in the business of proving people wrong in the first place? Can you take responsibility for that? Lol.

      @jaredroussel@jaredroussel2 ай бұрын
  • Many people appraised the video but I kind of disagree. For constructive criticism, I think you should not delve too much into "deeper" topics without explaining concepts or ideas. For example, you have mentioned tat irrational numbers are a dense subset of real numbers at 5:48 . Yes, for someone who knows some higher level math this might be a better clarification but for the general viewer I think this video is intended to, I think it just causes a ton of confusion.

    @enatrage2083@enatrage20832 ай бұрын
  • my brain hurts

    @aflaqueahmad300@aflaqueahmad3002 ай бұрын
  • All you did was use an arbitrary partition for the first example and a uniform partition for the second because as you said, it's easier. It's an artificial complication.

    @abebuckingham8198@abebuckingham81982 ай бұрын
    • Not an artificial complication. Riemann's definition is the way it is because he wanted to make a definition (any partition and any tags) that is as general as possible to be able to tackle all pathological fringe cases. Darboux's improved definition requires you to only provide any reasonable upper bound on the difference between upper and lower integrals, and the regular partition just happen to be a convenient choice for a bound for a specified function like x^3. but definition itself still assumes arbitrary partition It was much later proven that even if you relax the condition of the definition of riemann integrability to sequence of regular partition (but still arbitrary tag) to converge, then it is equivalent to the Riemann's definition. This is a highly nontrivial theorem that takes quite a bit of work to prove, but guess how its proven?? using Darboux's definition

      @HaramGuys@HaramGuys2 ай бұрын
    • @@HaramGuys All he did was apply the squeeze theorem to the Riemann definition. The fact that if it converges for one sequence of refined partitions it converges for all of them is very easy to prove in full generality. It's what motivates the U(f,P)-L(f,P) definition in the first place.

      @abebuckingham8198@abebuckingham81982 ай бұрын
    • @@abebuckingham8198 that just strengthens the argument that riemann's original definitions sucks

      @HaramGuys@HaramGuys2 ай бұрын
  • Wow, i understood nothing from that, i understand why we don't teach this definition of Riemann sum, like bro i understood barely anything from this.

    @AttyPatty3@AttyPatty32 ай бұрын
    • Git gud

      @cparks1000000@cparks10000002 ай бұрын
  • who's "we"?

    @zaero2379@zaero23792 ай бұрын
  • lobotomy analysis

    @FahesHorizon@FahesHorizon2 ай бұрын
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