I discovered your channel when I was in Sixth Form (16-18 year olds in the UK) about 5 years ago. At the time, I had only just started getting into maths and I thought the crazy integrals you did were really funny and interesting (I remember the cube root of tan(x) in particular). I showed some of the teachers the DI-table method you use for integration by parts, and one was so impressed she said she would use it too. I went on to do maths at uni and I've just found out I will graduate with a First. Even though I ended up not doing much like what you do on your channel (I preferred algebra, in particular ring theory, Lie algebras and algebraic geometry), you undoubtedly played a big role in my early interest in maths. Thank you for inspiring me.
@Ehud Kotegaro I assume you meant to make this a separate comment lol. But yes, I was also thinking that the result immediately follows from the theorem that a function is continuous at a point if and only if its limit at the point is just its value at the point. On the other hand, you could think of this as proving 1/x is continuous at 2.
@@finlay5789 I assume you're doing maths at uni? If so, good luck, but be aware that maths at uni isn't about calculating limits and integrals like most of this channel (this is geared towards Americans), you're learning theorems and proofs. This particular video however is a good example of what you might do in Analysis.
Lol i remember my teacher giving me LESS points on an integral i solved with the DI method on a test because he'd never heard of that method. The function was something like x^3*sinx and i just couldnt be bothered solving that with standard IBP
If you wanna know how crazy calculus gets, goes even further. I know we only did limits here, but if you do differentiation and integration, all of it fits under the topic of differential forms. Basically, if you want to differentiate and integrate over arbitrary smooth manifolds (that are not R^n), then you just need to introduce the concept of a tangent space, the ideas of a smooth manifold (charts, atlases, partitions of unity), and some concepts from abstract algebra like exterior derivatives, wedge products. This defines differentiation and integration both intrinsically and extrinsically. And now you can now obtain the fundamental theorem of calculus in any dimension using something called Stokes’ Theorem. Then beyond this, you can study the algebra behind the exterior derivative, wedge product, and the types of algebraic structures that admit these (exterior algebras). Then you can go even further from that and learn about how category theory unifies a lot of these types of algebras that admit similar types of structure. Mathematics is one deep rabbit hole (:
Then you can exclude the law of excluded middles and get sets of numbers in [0, 0] that are not {0} and also have that every function over domain R be continues and infinitely differentiable. Seriously.
@@xinpingdonohoe3978 It's CollegeBoard's grading system for AP exams, which are standardized exams for high schoolers to prepare for college. I got a 5 on AP Calc BC as well
Good thing I will be a legal adult by the time I take my AP Calc BC exam. I can use the epsilon-delta theorem for limits without having to worry about committing fraudulent behavior!
I did my calculus classes long time ago. I wasn’t a bad student nor a genius, but I also liked math in general. I’m over 30 now and just realized how calculus works de facto. Man I just wish I could redo my classes with the insights you give us. 14 years ago I would be totally addicted to your channel hahahaha.
What is the difference between Newton and Leibniz calculus? Newton's calculus is about functions. Leibniz's calculus is about relations defined by constraints. In Newton's calculus, there is (what would now be called) a limit built into every operation. In Leibniz's calculus, the limit is a separate operation.
It’s hard to see why it might seem necessary for this example because it’s continuous at that particular point. But how do you show lim x->2 (x²-2x)/(x-2)=2. You probably recognize the common and will cancel it out. Doing a proof like this is the reason WHY we’re allowed to certain things like this.
I think the point is, how do you really know that a function is continuous? I mean, we can all envision what y = 1/x looks like on a graph, but we know that from graphing a whole bunch of points and connecting the dots. How do you really know that there aren't any discontinuities, other than the one at x=0 of course? Can you prove it? Sure you can say "I don't see why it WOULDN'T be continuous", but that's not really a proof. I get the feeling that, 99.9% of the time, these proofs are an abstract consideration; usually you and I can tell whether a function is going to be continuous just by looking at it. Even so, mathematical rigor is its own reward.
@@kingbeauregard thank you Joe and kingbeauregard. I now understand why we need to go through all these difficult proofs but it is necessary if we want to be rigid. And now I can see the beauty of this proof now. Thank you.
It’s less a ‘proof’ and more the actual definition of a limit. Now, we have further proofs which say if a function is of a certain form (e.g. for this one, rational functions where the denominator is not zero at this location) then the function is continuous. And we know that continuous functions have limits that are relatively easy to evaluate.
Thanks so much for posting today. Down with the flu or maybe even COVID, and feel horrible right now. But your video made me forget my sickness for a little bit🖤♥️🖤♥️
@@blackpenredpen Using calculus, I predict that the virus is dying off at a rate of 0.67% every hour. Using this, I predict that I will be fully recovered by [100/0.67 hours, too tired to do calculations] from now! (Totally a joke, just trying to hang in there)
Ive had to do similar proofs like this in my proofs lesson, but the problems were much simpler than this. Still, I find real analysis terrifying. (The problem was proving that for all epsilon > 0, if 0
Everyone has Real Analysis PTSD. Although I would say back when I took Complex analysis, that course truly killed me, but strangely, abstract algebra, number theory, and linear analysis was an easier pill to swallow xD.
@@xghoulxx I can only speak for me and my classmates, but at least in Germany, real analysis 1 and 2 are probably the easiest classes in the early years at university. Abstract algebra is a little harder, others even worse..
I remember using that result for some questions without ever proving it while wondering if I was allowed to or not. Though the proof is so trivial that it was indeed allowed to be used.
Here is why we need it. You and I have seen the graph of y = 1/x, so we know what it looks like. But how do we really know it's continuous? How do we know that it doesn't have a discontinuity at x=3.2493 for example? Epsilon-delta is a way to prove that a function really is continuous. Calculus requires that we work with continuous functions, so it is necessary to prove that functions are continuous. (Fortunately, most functions are.) The goal of epsilon-delta is to show that, as f(x) gets closer and closer to "1/a", x must also get closer and closer to "a". Think of the graph of y = 1/x, and imagine a rectangle whose center is at (a, 1/a). Make this rectangle tall enough that the function never touches the rectangle's top or bottom edges. This rectangle has a height of 2*epsilon, and a width of 2*delta. Now, can you shrink this rectangle smaller and smaller (so it becomes less tall and also less wide), all the way to zero, and the function still never touches the top or bottom edges? If you can construct rectangles that do that, then it means that the limit really does approach 1/a. So, the epsilon-delta method is really about defining the proportions of the rectangles: if you can mathematically prove that you can create rectangles that operate as described, then the function must be continuous. That means establishing a relationship between epsilon and delta. It may be that the relationship works only when x is near a; that is fine, since limits are about a function's behavior near a specific point.
It doesn't have to be a nightmare. IMHO there are two parts to think about: the concept, and the technique. The concept: Suppose you are trying to prove the limit of f(x) = L at x=a. So, imagine a rectangle centered at (a, L) that has proportions such that f(x) never touches the top or bottom edges of the rectangle. Now, can you shrink that rectangle down to nothing, such that the function never touches the top or bottom edges? If you can mathematically prove that such a rectangle exists, then the limit must exist too. "delta" is all about the width of the rectangle, and "epsilon" is all about the height of the rectangle. Sooooo, all of this math is about figuring out whether such a rectangle exists, and if you pick a given epsilon, what size does delta have to be? The technique: You start with | f(x) - f(a) | < epsilon, and you want to wrestle with it until you get to the form |x-a| < (some function of epsilon). That |x-a| will become our delta. So you have to do a lot of algebra, and you can use one special trick: you can say that, if we limit our x values to a small distance from a, then within that range, the function will never cross the line |x-a|*(some constant that you determine with some side math). At that point, you've switched over to determining your epsilon against that line rather than the original function, but that's fine: since that line has the limit you want, so will the original function.
I recall my first week or so of high school calculus being taught this "adult" nonsense before we really got to the good stuff like derivatives. I say "nonsense" because basically it was teaching how to prove the obvious, which could have been done a different way, especially with the invention of the calculator. (Also, even test makers didn't waste time by having it on the AP exam, which by the way I scored a 5 on which allowed me to place out of Calc I and II in college.) In this example, you would have one student input 1.99 into the equation 1/x where he gets .5025 on his calculator, and see if another student can come up with a closer number to 2 (such as 1.999) on his calculator and see if his answer is closer to .5 than .5025 (which of course it is, .50025), and then ask if anyone can come up with an x value, but not 2, which results in an answer even closer to .5. The point of the exercise is that when you are dealing with limits, someone can always come up with a number closer to the limit than someone else. So, for example, one student posits a 1. with a billion 9's behind the point, another comes along and counters with a 1. with a trillion 9s behind the point. All that could be done in less than one day in class. Maybe for those students interested in pursuing a degree in math, which thrives on proofs, the teacher could provide a few homework exercises for them to prove the obvious just for their enjoyment.
Please don't call it "English" Alphabet... it's Latin, ok? Most other European languages use it, too. I think only Greek and some languages that use Kyrillic don't use it.
Slightly disappointing for me since I would never discover the trick of choosing a delta that is the min of a two element set that includes 1(or some other convenient constant) Delta is a function of Epsilon as your limit proof showed. Thanks for the mild criticism of the term arbitrary in conjunction with phrase “arbitrary epsilon” Epsilon > 0 is all that is necessary 😀
We should talk about the arbitrary values for a second. When you do epsilon-delta proofs, you're going to need to rework | f(x) - f(a) | into a form that is something like |x - a|*(some expression without any x's in it). You're usually going to reach a point where you can't get rid of any lingering x's through sheer algebra, and | f(x) - f(a) | has turned into |x-a|*g(x). That's when you cheat. Since we're really only concerned with values in the vicinity of x=a, we can restrict ourselves to a region that is as arbitrarily small as we choose. Then we can do some math and determine that, in that region, g(x) has a maximum value "M", so we can swap out g(x) and instead work with |x-a|*M. Why are we allowed to do this cheat? Well, it's a squeeze proof. If we are saying that |x - a|*M is bigger than |x - a|*g(x) over the entire region in question, and |x - a|*M has a limit at (a, L) (and of course it does, it's a straight line), then it follows that |x - a|*g(x) must likewise have a limit at (a, L). So then, why a minimum of "1" specifically? Primarily for mathematical ease, but also, it doesn't make us trip over that discontinuity at x=0. A minimum of "1" is a fine choice if we're concerned with x=2. It's a terrible choice if we're concerned with x=0.5.
can anyone do this? They give three prime numbers, p, q, r. Solve that ³√p, ³√q, ³√r are not three characters (not necessarily consecutive characters) of any arithmetic suite.
I came on and was ready to see some crazy looking adult calculus, then I realized I did this, proof last semester, I also realized that I am in fact and adult.😄🕊️
no we do not have to prove it using the epsilon delta definition as the definition already proves the existence of limits. this is a finite limit, so by the proof of that we already have the proof in place and we can simply turn the mathematical handle.
Youre using a circular argument. What prevents anyone from saying that the left limit is 1/3 and not 1/2? It may seem trivial to you, but he showed the more rigorous way of doing it because later on if you continue down the math route, you will need to do epsilon-delta proofs a lot
@@nestorv7627 computer programmer so I am only interested in turning the mathematical handle. I am not saying that the epsilon delta proof is invalid or unnecessary. I am saying that the limit of finite series is already been proven. It is a valid argument to use already existing proofs without having to keep proving the existing proofs. Prove that 1+1=2. You see we do not always prove everything in a proof.
Well, we just need to say that 1/x is continuous in the interval [1;3], which means by definition that for a in [1;3]: lim x->a of f(x) = f(a), from there, lim x->2 of 1/x = 1/2 !
In our university they are currently teaching real analysis and I have had problems with Limits and continuity. Really great to find this video, it helped me a lot! Thanks BPRP
So all you have to do is prove that x-a is smaller than something and f(x)-L is also smaller than something?? How does that prove anything. Where does the epsilon delta definition come from
The epsilon delta definition can seem obtuse since he didn't really explain what epsilon and delta actually were in this video. I'm pretty sure he has other videos on it, but let me try to explain: The reason we use absolute values is because both delta and epsilon are _distances_ from some point. Delta is a distance in the input space (usually the x-axis), and epsilon is a distance in the output space (usually the y-axis). So when we say that |x - a| < d, what that means is we can pick a point x in the input space which is within a distance d of the point a. Then, putting this point into the function we want that f(x) will be within the distance e of the point L in the output space. And make sure this holds for any epsilon! So, what this means is if we are given some small area around a point in the output space, we should be able to pick a small area in the input space which gets mapped entirely within the area in the output space if the function is continuous. If we can't do that, the function is discontinuous and our limit won't necessarily converge.
This seems unnescessrily complicated both in definition and in proof. The way I was taught the definition of limit didnt have the delta part it just had that there exist a such that |f(a)-L|
I am confused, what about the f(x)-L bit why wasn't that explicitly mentioned in the proof? Why does simply substituting epsilon for delta work so neatly when they are addressing two related butt different inequalities?
I am not sure at all that this was what you asked for - it’s probably not. “For any ε there exists a δ such that…” means that δ is a function of ε, at least in this context, not sure if this is always the case. |f(x)-L| < ε and 0 < |x-a| < δ(ε) the proof is complete once δ is found. At least in this case, probably in most, it is easiest to find some other function h(•), such that |f(x)-L| < h(δ(ε)) in this case he found h(x) = x/2 |1/x-1/2| < δ(ε)/2 and 0 < |x-2| < δ(ε) Now you can just notice ( maybe ) that if δ(ε) = ε, nothing breaks, and the proof would thus be complete.
There is a theorem that is quite easy to prove which states that power functions are continuous in every point of their domain. If a function is continuous in a certain point by the definition of continuity of a function the limit can be evaluated by evaluating the function in that point. Who knows the proof of this theorem and uses simple substitution should than be considered as part of the “adult calculus” set.
Oooh, Ima take a stab at it. If we're trying to prove the limit of x^n at x=a, then: | x^n - a^n | < epsilon | (x - a) * (x^(n - 1) + x^(n - 2)*a + x^(n - 3)*a^2 ... + a^(n-1)) | < epsilon |x - a| * | x^(n - 1) + x^(n - 2)*a + x^(n - 3)*a^2 ... + a^(n-1) | < epsilon So from there, set an arbitrary limit on delta, figure out what the maximum value of the second absolute value is over that set of x-values (let's call it "M"), and we're left with |x - a| * M < epsilon.
@@kingbeauregard Well done! But instead to take the maximum it is better to take a majorant of that set, because to apply Weierstrass theorem f needs to be continuous in the closed interval [a-h;a+h], where h is a real number. The proof would than be circular. You can also prove it, considering Lim |x^(n+h)-x^n| for h->0. It becomes than Lim|x^n||x^h-1|=0 because x^h-1∼(x-1)h for h->0.
BPRP’s proof is like the Calc 1, week 1 homework assignment. Your version is like what the tutor tells the student to turn in because it’s more efficient and fits the topic though the student may not directly understand it, haha.
@@stephenbeck7222 After doing a LOT of thinking about epsilon-delta (and I'm not saying it's high-quality thought, just the best I'm capable of), I feel like the practical approach to arriving at a delta is, start with | f(x) - f(a) |, find some way to peel off an |x - a| term, and remove any x's from whatever's left. It just so happens it's real easy to pull an |x - a| out of | x^n - a^n |. I'll take it!
