3 Integrals You Won't See in Calculus (And the 2 You Will)
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- čas přidán 20. 06. 2024
- In Calculus, we usually learn the Riemann integral, or sometimes the Darboux integral in disguise. But there are many problems these integrals can't solve! Like if we want to integrate a function which is discontinuous everywhere, or if we want to integrate with respect to a random process. Let's explore 5 different integrals, starting with the 2 you might see in Calculus, and then 3 more advanced integrals that are often only seen in graduate school -- the Riemann-Stieltjes integral, Lebesgue integral, and Itô integral!
00:00 Introduction
00:32 Level 1 -- Riemann Integral
01:58 Level 2 -- Darboux Integral
04:00 Level 3 -- Riemann-Stieltjes
07:02 Level 4 -- Lebesgue Integral
09:57 Level 5 -- Itô Integral
On sets of measure zero, always bet on Lebesgue
Bear witness to his overwhelming integrability!
another day, another Dr. Sean banger
T
Dr Sean from iCarly!
Darboux is usually lumped in as a case of Riemann, which is a somewhat generalized concept after all. If you know f is continuous, then you can just take left endpoints and equal width intervals, as far as the limit is concerned, but it has to be proved that the choices don't matter.
As far as arbitrary partitions go, insisting on equal spacing could make it annoying to prove that you can split up integrals from 1 to e and then e to pi, for example. By the time you justify the validity of uniform steps, you have already considered arbitrary partitions as the max width goes to zero, and may as well have allowed them from the start.
It's a lot of fine print but if f is continuous, uniform intervals and left endpoints go brrrr.
How to explain integral calculus in 12 minutes... You nailed it perfectly!
1 ☑️
2 ☑️
3 what 😱
Re: peeking into the future, that’s not entirely true. The Stratonovich Integral (midpoint rule) is also an adapted process (meaning it can’t see the future) and even results in the standard chain rule when taken in differential form. The real reason people use an Ito integral is that it is a martingale. Admittedly, this is a bit technical.
Just a small correction. Ito integrals are martingales under the additional (square integrability) condition, but in general, stochastic integrals with respect to martingales (e.g. Brownian motion) are only local martingales rather than true martingales. But like you said, it is a bit technical.
I didn't know about the Itô integral, so I learned something new today. Nice!
I'm so interested in Itô integrals now, Dr. Sean is the GOAT
You are a fantastic teacher. I am new to calculus, and have really wondered why we would need more than a "general integral." This video not only answered the question, but also justified what the hell derivatives are actually measuring (and why we bother taking them). Thank you.
Woah, I am astounded by how easy to understand you made the concepts of the more complicated integrals!
And I am astounded how arbitrarily complicated it can be presented by some teachers. 😋
Brilliant video!👍 Exactly the right amount of depth for me 🤗
Thanks for putting effort into the production and having great audio and video, and double thanks for not using negatively biased graphics. 🙏
Thanks so much! I'm glad you liked it
Great video, but at 9:55 it's more correct to say that we get an irrational number *almost* every time (i.e. with probability one, but it's still technically possible to get a rational number).
A probability of 1 means quite literally every time
@@DistortedV12 not necessarily, if you have infinite options. if you choose a random integer, the probability of not getting 7 is 1, but you can still get 7
nah, i don't think that's right.
Right! Thanks for your comment. I was thinking of sampling X_1, X_2, X_3, ... and meant that we'll see an irrational number each of these times, even if we keep sampling forever (countably infinitely many times). The union of countably many probability 0 events still has probability 0. But if we have uncountably many uniform random variables, then we may indeed get some rational numbers! We'll get irrational numbers almost every time.
@@DrSeanGroathouse Unfortunately I still think that's an incorrect interpretation of the probability here. Even if you only sample one point, it's still possible to get a rational number, for the simple reason that rational elements exist in the set [0, 1]. It would have probability 0, like you say, but probability 0 doesn't equate with impossibility.
Another way to think about it: we agree that the probability of choosing any individual number from [0, 1] is 0. If probability 0 implied impossible, then this would mean that it would be impossible to select any element in the set!
Love the visualizations always makes it easier.
Thank you. Very nice to see this info being simply explained. In the Ito integral, it's interesting how 2nd order terms are important (in a standard deviation sense) because of the nature of the random process, whereas in the other types of integration presented the 2nd order terms are considered insignificant (zero). Would have been nice (a luxury) to include the Generalized Riemann Integral (uses a different type of partitioning).
Hi Dr Sean! The video was very insighftul and easy to comprehend. Thank you very much for your work. I am looking forward to lean more maths in an MBA program than in my undergrad in finance. I was wondering about what measury theory is and how brownian is relevant to stohastic analysis. Thanks again, looking forward for more videos in the future :)
If I'm not mistaken, by "random process" you mean a stochastic process, i.e. a random variable with an index number (often, though not always, interpreted as time) attached?
In the case you showed, the index number has an interpretable direction so talking about it as time is meaningful and some authors would call the process causal.
But what happens if we give up on the causality assumption (as some do in time series analysis, though that is in discrete time) and let the "future" (i.e. events with a high index number) affect "today"? Is the Ito integral still valid?
Amazing video!
Expectation values in quantum mechanics makes so much more sense now
Incredibly enough, I knew all of them. I need to learn #4 for real tho.
In Spanish we have an expression for this kind of visual enjoy, esto es cine chavales
Thanks for sharing 😁
Oye tío, ¿Tortilla de patatas con o sin cebolla?
It's interesting to think the first insane idea I'd have for measuring the curve (butt ton of rectangles) ends up being pretty close to what smart folks do.
Life is so strange and interesting.
gl with your channel , very nice video :)
Thanks so much!
That's what I study. Super interesting. Hope I get some work after my degree
Maybe could have elaborated more why in measure theory we only consider countable sums, not bigger sums than that. I think a beginner watching the video would have wondered: then why isn't the measure of [0,1] 0 since the sum of the measures of all its elements is 0.
TIL our teacher at uni actualy taught us the Darboux integral
Grear video man
Great work
Thanks! I'm glad you liked it
You deserve way more subscribers
So true.
I’d just like to say on thing regarding Itô integrals: one of the reasons it’s mathematically tricky to work with is that you’re integrating wrt a fractal function (Brownian motion is a fractal) meaning that you can’t really make your partitions infinitely thin in the same way as Riemann integrals. This results in some interesting rules regarding differentials
Isn't the Îto integral simply a lebesgue integral where you integrate with respect to a probability measure instead of the lebesgue measure ? It would be more rigourous that way
Would you also be able to cover the Stratonovich integral? (that can be equivalent to the Ito with some manipulation and more useful in physical processes)
Thanks for the idea! I added it to my list for future videos
Excellent video, but I would like to point out that at 6:42 the distribution function should be right-continuous as it is a property of all cdfs
You lost me bro in this one , the context was all over the place since you cramed such a giant topic in under 12 mins😅😅 to learn from this video I need to learn a 72 hours integration class separately 😂😂
shouldn't the CDF be right-continuous? (@6:40)
Need for videos on it0 integral
Define a process for generating a random number that takes less than infinite time which can demonstrate only irrational numbers.
Also, can you go into further depth on the higher level integrals?
use a process that generates only rational numbers then add pi
@@gdclemo I guess rationals and irrationsals are of the same cardinality then
@@fightocondria Nope. You didn't say it should generate ALL the irrationals, but that it should only generate irrationals.
I really have trouble understanding the Itô integral. Is this meant for a fixed small omega, so the value is not well defined, but things like Expected Value and Varianz are?
What isn't said in this video is that you can integrate any "measureable function" (ie a function for which the inverse image is "smooth" enough) with respect to any measure. Normal integrals like Riemann's are just integrals with respect to the lebesgue measure, but there are many more measures one can use. For example, if you integrate a function with respect to the counting measure, you get a series. Probability measures can also be used, and as a random variable is just a measurable function, you can integrate it with respect to the probability measure, and it gives you its expected value : E(X)=∫X(𝝎)dP(𝝎).
Can we get a video on elementary and non elementary functions? :3
Thanks for the idea! I added it to my list for future videos
7:01 "Let us look into an integral that is often only seen in grad school"
Undergrad physics students having to learn it to actually do the maths of Quantum Mechanics (I know we can do without it, but still it is needed to do it correctly)
it's also in any pure maths undergrad, in europe anyways
@@oke5403 I don't think pure math students were forced to learn it in my country (Brazil), but I know your pain
In Germany, the Lebesgue integral was first year stuff. From my perspective, the jump from Lebesgue to Ito was quite harsh. While the first three were all part of the first semester course for me, the last one I encountered just in my Phd studies.
@@patriziosommatinese1820 Damn, Lebesgue at the first year would be a little too much for me, at least depending on how rigorous was required. Itou integral I'll have to learn now in my masters degree because I'm working with a stochastic model
Here in Italy, it's second year stuff for mathematicians and, in less depth, for physicists.
i understood approximately 1.4 of these integrals
At least you are rational about it. 🤭
I'm incredibly confused by the explanation for the Lebesgue integral. Even if there are countably-infinite rational numbers and uncountably-infinite many irrationals (eg. 0.1010010001..., 0.2010010001..., 0.1020010001...) over [0,1], how can we have P(rational)=0?
Is it based on lim(P(rational)) or something bc of (countable-inf➗uncountable-inf)?
In any case, how can an "irrational every time" result be valid when there's a nonzero probability you choose, say, 0.5 if you're pulling any real number over [0,1]? (because, obviously, [0,1] contains 0.5)
Probability being zero does not mean something is impossible in non discrete cases. Imagine you have a uniform distibution. If you imagine probability of one point be bigger than 0, lets say some x > 0, then, because disribution is uniform, you would have probability of any other point also be x, but there are infinitely many points, so total probability wont add up to 1. If you are interested check measure theory courses
The example in the video is the Dirichlet function, there are many videos that go into it. But the idea is that since in the real number line there are uncountably infinite numbers between 0 and 1, but only countable infinite rational numbers, the density of rational numbers between 0-1 is infinitely smaller (the so called measure μ), and they are overwhelmed by the amount of irrational numbers. The total Lebesque integral will be: 1*μ(rational) + 0*μ(irrational) = 1*0 + 0*1 = 0
@@dubvascl5840 Hmm okay. That's counterintuitive but I suppose I've got a reading list now. (Since I realize I didn't mention it before), I've only been up through a calc 2 course, does measure theory rely on much/anything beyond about that level?
@@TheTinyDiamond I dont know system of courses in US, but it should not. I my university i had Lebesgue integral at the end of 2 semester
The Lebesgue integral is constructed by extending a measure from the intervals to many other subsets of the real numbers (although not all, which is precisely why measure theory exists). This measure gives the interval (a,b) measure b-a.
If you fix a point x, you can see that {x} is a subset of (x-epsilon,x+epsilon) for any epsilon larger than zero. But the measure of this interval is 2*epsilon. So the measure of {x} has to be less than 2* epsilon for any epsilon larger than zero - so it must be zero.
The measure of a countably infinite set is just the sum of the measures of its points, which in this case is just zero. So the measure of the rationals is zero. But since there are uncountably many irrationals, you cannot apply the same logic.
Then what did Newton cook about integrals?
Newton and Leibniz got ideas about change and little bits down, but did not establish a formal rigorous foundation.
I am confused about because one p=0 sum will be 0, wouldn’t it be 0 for each irrational number too so then 0 either is chosen
I image it comes from irrationals being uncountably infinite.
9:45 ngl I'm kind of dissappointed the integral is equals 0. Great explanation by the way 🙌
Really? Our prof taught us Riemann-Stieltjes Integration in our first year of college.
Hmm that's weird, I am an undergrad and have seen up to the Lebesgue Integral. The Itô Integral should be the one seen in graduate school.
3:43 this is true only for some well behaved functions. Because we can easily take Riemann integrable functions take have NO infimum or supremum around a point or some points.
that is inprope integral
@@leolrg2783 hummm. It doesn't matter. What I am saying is about the equivalence between Riemann and Darboux integrals. I am pretty sure this equivalence only works for functions that have sup in any interval of their domain, while Riemann integral will work for some functions that don't.
@@samueldeandrade8535 No. A function is rieman integrable then it must be bounded, so that sup exists.
@@leolrg2783 not true. Riemann integrable functions need to be bounded almost everywhere. For example, the function
F(x) := 1/x, if x is rational,
F(x) := 0, otherwise,
is Riemann integrable, but NOT bounded.
@@leolrg2783 what you are saying is valid IF the function is continuous. If the function is NOT continuous, it may be valid or not.
Hello
is Monte Carlo integration mentioned?
so basically rectangles
hi
You have to gatekeep ito!!!!
Itô for the masses? Blasphemy! 😆
Stupid question. How is the probability that we pick a rational number 0 when 1/2, for example, is rational and is an element of [0,1]. 9:45
Same with 1/3. Is it that these rational numbers are so greatly outnumbered that the possibility of picking one is negligible. Still, in precise math, I would still expect it to be non-zero. If someone could explain, I would be grateful.
@@michaelodekunle6591 The probability of an event being 0 does not mean the event is impossible. That's the "tl;dr" of it. An event can be possible and have probability 0 of occurring, as you've pointed out, when the sample space is infinite.
I think it has to do with the different countabilites, but not sure.
you forgot the Kurzweil Henstock integral
Pretty sure it's "Darboo", not "Darboh". At least if we're assuming the name is French.
They are always either French or German in mathematics, especially high level maths. LOL
@@accaciagame1706 maybe 100 years ago. But in modern mathematics you have people like Nash, Turing, Conway and Feynman
@methatis3013 If we exclude New Zealand which has a tiny population, France still has the highest Fields medal per capita and the second net highest winners after the US. The French are unusually good mathematicians even to this day. Nash wasn't that good a mathematician. He just came across a very useful but simple Game theory concept. No where in the league of Newton, Poincarre, Euler etc.
@@methatis3013 nobody's saying English-speaking people are bad at maths. Only bad at even _trying_ to pronounce foreign names correctly. Not that the French are any better. Maybe it's a type of ignorance born of their colonial history, dunno.
Unpopular Opinion: The idea that you can "pick" a number that requires an infinite number of steps (digits) to describe seems dubious to me. Not having infinite precision always gives you some uncertainty about any representation of an irrational value. An infinite process that is used to calculate digits, run for an infinite amount of time, is not the same thing as a point on the number line.
Gee I wonder what 1/3 means
Realistically, there is uncertainty in everything. Exact values only exist as mathematical concepts. Tell me, do you think it's possible to cut a 2x4 to exactly 3.5 feet?
@@InternetCrusader-rb7lsexactly, some numbers only have infinite decimals because of our number system,
A wild Axiom Of Choice appeared! 😄
I'm surprised that Henstock-Kurzweil integral is not here :)
en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral
ito calculus is used in financial mathematics to find solutions to Stochastic differential equations.
In Spanish we have an expression for this kind of visual enjoy, esto es cine chavales
😂
!La magia de este cine es *integral* ! 🤭