OK, give us an epsilon delta proof that the product of two arbitrary continuous functions is continuous. Is that easy like Sunday morning? Don’t think so 😀
This one makes much more sense. I really like that you care for your subscribers and improve yourself if you see something new. Kudos to you and congratulations to your students!❤️
If you choose delta to be whatever you like, then i think you will prove every function to be continuous. Prove me wrong. Prove that 1/X^2 is not continuous, with the help of this delta epsilon method.
@@anuragguptamr.i.i.t.2329 The function is not defined for x=0. Since the value f(0) doesn't exist, it's not equal to anything, in particular it's not equal to the limit
@@pablop8166 that's why i want it to be proven by the epsilon delta method. If discontinuities of a function cannot be proven by this method and only the continuities can be proven, then this method would become useless.
@@anuragguptamr.i.i.t.2329 Remember that the proof starts with f(a) equals to the limit of f(a) as x approaches to a. But since f(0) does not exist, you can conclude right away that the limit can't be equal to it. You can't use the epsilon-delta definition because you already proved that the initial statement is false
@@pablop8166 i have studied about limits, derivatives, several advanced mathematics topics and the basic definition of derivatives with the help of limits. But, i did never study about this epsilon- delta method during my school, college and post graduation. Bprp steve is the first ever source for me, to get to know about this method. But, now after reading your replies, now i know, why they didnt ever include this method into my curriculum. 🤣
Imagine being his student and giving your own solution to his questions. He's sure to give you an A+ and not mark you incorrect (given that you did it right).
Can you do videos that teach us about math proving methods? (example : direct method or maybe by induction etc...) Please show us the beauty of proofing!!
So as a math PhD student, I have been spending so much time studying for a probability and statistics exam, that when I went to sleep I was thinking to myself, “wait, did I really forget the definition of continuity”, so I looked up prove x^2 is continuous, and this restored my faith in myself lol. Always gotta love epsilon delta
I usually watch this channel once or several times a day to see what you are up to. I enjoy your effortless execution as you masterfully destroy one math problem after another. Don't get me wrong, im no math major, I've failed calculus before, but I still enjoy trying the odd math problem from my university days, seeing how much of the calculus part of my brain has deteriorated in my retail environment. Some days I'm feeling mighty confident in my abilities, neither days I am quickly humbled. You are indeed a genius. But that doesn't stop me from failing with grace
Correct. In a typical calculus class, your 3 sided equation is used to verify continuity of piecewise functions after it is determined that the individual pieces are continuous on their own domains, normally by just asserting they are ‘nice’ (e.g. polynomials or other elementary functions). But without knowing what a ‘nice’ function, you can’t calculate your one-sided limits except by following the method of the video.
Much easier proof: A continuous function is one such that f^(-1)(U) is open for every open set U. Taking preimages commutes with taking unions, so we only need to show this is the case for intervals (a,b). Take 0
Sounds like the definition of continuity from a topology course. It seems as though you have “abstracted the difficulty out of the problem” However, the notion of open set for real numbers depends on arbitrarily small real numbers, so I don’t think you you dodged the bullet 😀. Just my opinion. Thanks for reminding me of the good old days in topology
Can you also give a delta without using the min? So just one delta edit: i found one, you have to use quadratic formula. then its delta= sqrt( |a|^2+epsilon )-|a| in total this comes to delta (delta + 2|a|) = (sqrt( |a|^2+epsilon )-|a|) (sqrt( |a|^2+epsilon )-|a|+2|a|) =(sqrt( |a|^2+epsilon )-|a|) (sqrt( |a|^2+epsilon )+|a|) =|a|^2+epsilon-|a|^2 =epsilon with 3rd binomial formula
g(x) = x is continuous (taking delta = epsilon shows this in definition). The product of two continuous functions is continuous. Hence f(x) = x*x = x^2 is continuous.
@@angelmendez-rivera351 That's true but the proof that the product of two continuous functions is also continuous isn't any longer than the proof in the video.
@@Bodyknock It definitely is longer. I wrote a comment in the main comments section in which I did the full proof. Compare that proof to the following: f : (a, b) -> R, f(x) = x^2. If |x - c| < δ, then |x^2 - c^2| = |(x - c)·(x + c)| = |x - c|·|x + c| = |x - c|·|(x - c) + 2·c| =< |x - c|·[|x - c| + |2·c|] = |x - c|^2 + 2·|c|·|x - c| < δ^2 + 2·|c|·δ. Therefore, for every a < c < b and every ε > 0, there exists some δ > 0 such that if 0 < |x - c| < δ, then |x^2 - c^2| < ε, and such δ > 0 is given by δ = sqrt(ε + c^2) - |c| for every a < c < b and every ε > 0. Therefore, f is continuous.
This reminds me of Russell's Principia Mathemathica proving that 1+1=2 with logical axioms and dozens of pages proving it and then there goes Gödel to say that every axiomatic system is incomplete. Bruh moment.
Too bad you didn't choose to spend enough time in the very beginning to clearly and slowly explain the core idea behind the epsilon delta stuff, maybe visually. This was the most important thing in my opinion, and I missed it.
Thanks for ur input. I just wanted to show how to do the proof for this. If I also introduced the definition than the video would be too long. Plus that deserves a video by itself 😃
I was thinking something like this too. I think you then need to prove 2hx -> 0 and also h^2 -> 0 when h -> 0, but this isn't hard. Definitely easier than the way it was done in the video I think but maybe I'm also missing something idk.
i see why the minimum is used, but it's possible without it. the only difference is that we get δ(δ+2|a|)=ε, which is a quadratic equation so we choose δ equal to the positive solution, and that solution is sqrt(|a|^2+ε)-|a| to me, that seems easier than using a minimum, but i guess that's subjective
Also, you can indeed prove a function is discontinuous using the definition. However, be careful. For example, the function f : R\{0} -> R\{0}, f(x) = 1/x is not actually discontinuous anywhere. On the other hand, the function g : R -> R, f(x) = 1/x if 0 < |x|, f(0) = 0 is discontinuous somewhere: namely, at 0. This is an important distinction. The domain of f is not connected, but f is continuous everywhere in the domain. So you cannot demand for f to be proven discontinuous with the definition, because it is not discontinuous. You can, however, demand for g to be proven discontinuous at 0, since 0 actually is in the domain of g. You can indeed use the definition to do this. How so? Because |x - 0| = |x| < δ with δ > 0, and |1/x - f(0)| = |1/x| = 1/|x| < ε, with ε > 0, and 1/|x| < ε is equivalent to 0 < 1/ε < |x| for every ε > 0. |x| < δ does not imply, for every δ > 0, that 1/ε < |x|, because 1/ε < |x| and |x| < δ implies 1/ε < δ, which is equivalent to δ = 1/ε + γ for some γ > 0. Therefore, |x| < δ = 1/ε + γ implies 1/(1/ε + γ) = ε/(1 + ε·γ) < 1/|x| not 1/|x| < ε.
Yes. For example, consider the function f: R -> R, f(x) = 1/x if 0 < |x|, f(0) = 0. This function is not continuous at 0. In other words, for some ε > 0, it can be proven that for every δ > 0, |x - 0| < δ and |1/x - f(0)| = ε or ε < |1/x - f(0)|. How? Because |x - 0| = |x|, |1/x - f(0)| = |1/x|, and ε = |1/x| or ε < |1/x| implies |x| = 1/ε or |x| < 1/ε, which always happens given that δ = 1/ε.
@@angelmendez-rivera351 proof not clear to me. First, epsilon should be independent of x (unless u are defining epsilon as a function of a fixed x = constant = c). Second, NO choice of delta should work so any delta also must be independent of any choice of epsilon. There obviously is a proof of the fact that your f(x) is discontinuous at 0 but I don’t think this is it. You must show there exists an epsilon such that for all delta there exists an x such that the absolute value of x is less than delta and the absolute value of f(x) is greater than or equal to epsilon. Instead of pursuing that method of “dis-proof” however, how about instead simply proving (easily) that the right hand limit of f(x) as x approaches 0 is infinity (which obviously does not equal f(0) = 0)?
@@jessewolf6806 *First, epsilon should be independent of x* And it is independent of x in my proof. *Second, NO choice of delta should work so any delta also must independent of any choice of epsilon.* That is entirely equivalent to what I wrote. The definition of continuity is of the form "Universal Quantifier on ε (Existential Quantifier on δ (Proposition))." Therefore, the definition of discontinuity is "Negation (Universal Quantifier on ε (Existential Quantifier on δ (Proposition)))." But "Negation (Universal Quantifier on ε (Existential Quantifier on δ (Proposition)))" is truth-table equivalent to "Existential Quantifier on ε (Negation (Existential Quantifier on δ (Proposition)))" which is truth-table equivalent to "Existential Quantifier on ε (Universal Quantifier on δ (Negation (Proposition))," which is precisely the syntax I used in the proof. *You must show there exists an epsilon such that for all delta there exists an x such that the absolute value of x is less than delta and the absolute value of f(x) us greater than or equal to epsion.* This is exactly what I did, symbolically. Perhaps the fact that I did it symbolically made it unclear to you, but this is 100000% what I did. *Instead of pursuing that method of "disproof" however, how about simply proving (easily) that the right hand limit of f(x) as x approaches 0 is infinity* OP wants us to use the limit definition of continuity to prove that a function is discontinuous at a point. So I did. You are begging the question by telling me to *not* answer the OP's "challenge." Look, I do apologize if the language I used in my proof was somehow unclear, but you are just imposing random interpretations onto my proof instead of asking for clarifications, which is what any person *should* be doing, and then saying "see, your proof is bad because of this interpretation I gave it, so how about you don't do what OP asked and try proving discontinuity in a different way." I am not sure how am I supposed to take that and simply accept it.
Let me rephrase what I stated in my first reply. There exists some ε > 0 such that, for every δ > 0, |x - 0| < δ & the disjunct ε = |1/x - f(0)| or ε < |1/x - f(0)|. This is because |x - 0| = |x| < δ and |1/x - f(0)| = 1/|x|. |x| < δ is equivalent to 1/δ < 1/|x|. So there exists some ε = 1/δ such that for every δ > |x| > 0, ε > 0 and ε < 1/|x| is satisfied, which is what needed to be satisfied.
You would need to prove differentiability, and then prove differentiability implies continuity, and both proofs require the same definition used to prove continuity in the first place.
For the record, some people are saying that you can instead just use the fact that x |-> x^2 : R -> R+ is differentiable everywhere, to then conclude it is continuous everywhere. The problem with this is that you would need to prove the function is differentiable, and prove that differentiability implies continuity, and both proofs require using the definition of a limit, which is the same definition you use to prove continuity directly instead. As such, proving continuity directly is actually just literally simpler. If you want to take the differentiability for granted, then you may as well also take continuity for granted. We cannot be cherry-picking about which statements to prove and which to not prove.
My proof would have been similar, except that, instead of saying |x + a| < 1 + 2·|a|, I would have said |x + a| < δ + 2·|a|, which is actually true for any δ > 0, and so I would not have "pre-chosen" δ before hand. The idea is that |x - a| < δ is given, and the triangle inequality implies |x + a| = |(x - a) + 2·a| = < |x - a| + |2·a| = |x - a| + 2·|a| < δ + 2·|a|. The reason I find this to be even simpler and more intuitive is that no further assumptions are being used. With |x - a| < δ and |x + a| < δ + 2·|a|, the consequence is that |x - a|·|x + a| = |x^2 - a^2| < δ·(δ + 2·|a|). Thus the δ to then "choose" would be the largest solution to the equation δ·(δ + 2·|a|) = ε, which always exists for every ε > 0 and always satisfies δ > 0 for every such ε. This does become obvious if you simply solve for δ explicitly: by completing the square, you can prove that δ = sqrt(ε + a^2) - |a|, which for every ε > 0, satisfies δ > 0. I also like this because it avoids using the minimum function.
I feel like we need to see some videos where epsilon-delta makes things easier. Like, proving f(x) = x^2 / [(x^2 + 1)(x^2 + 2)(x^2 + 3)] is continuous. That looks like a nightmarish function, but thanks to the magic of the epsilon-delta process, that denominator can be replaced by the number 6, so all that complexity reduces to proving that x^2 / 6 is continuous.
How does the concept of continuity deal with transcendental numbers? For example, wouldn't the function be discontinuous at the square root of e since e cannot be a solution to y = x^2?
"How does the concept of continuity deal with transcendental numbers?" The same as any other numbers. If a small change in input implies a small change in output then the function is continuous. rational/algebraic/transcendental has nothing to do with it.
An integral is always continous (easy proof using the Riemann definition of an integral) x^2 = int(from 0;to x;2x) QED If you're talking about x^2 being differentiable, then: If a function is continuous, then it's integral is differentiable (definition) 2 is continuous (by definition, I guess?) 2x is differentiable, x^2 is differentiable
Constant is continuous since f(x)-f(a) = 0 < epsilon 😀 so any delta will suffice I think there is some fairly heavy lifting going on trying to prove x^2 is continuous based on integrating 2x. Sure it’s known to be true but proving it all the way back to “first principles” would challenge me I know. Still an interesting other approach to problem. I think I finally get that epsilon delta proof!
@@MyOneFiftiethOfADollar continous, mind. Any integral of a function that is defined on an intrrval has to br continous on that interval. Since the rieman sum is f(x)dx and dx->0 there cannot be a gap in the function.
@@JoQeZzZ are you intimating that blackpenredpen should have said integrability implies continuity and skipped the epsilon delta proof? Your claim is certainly intuitively true but does not constitute a rigorous proof.
I've seen about 200 different explanations of the epsilon delta limit definition over the years, along with hundreds of applications of it. And I *still* don't understand how to actually prove anything with it. It always seems to come down to some sort of magic somewhere that isn't intuitively obvious. And it isn't the same magic every time, either. I'm pretty sure it has to do with how little attention there was to algebra on inequalities back in various levels of education. Take that and mix in the mind-meltiness that limits and the like can get into and it's a perfect storm of confusion, I think.
I think the reason for this is that often proofs are presented in a way that uses further assumptions and are not purely algebraic, although I do think not being adept at absolute value inequalities can also make it harder to understand the proofs. The key aspect of the proof is to understand that, given that |x - a| < δ, you conclude that |x + a| =< |x - a| + |2·a| using the triangle inequality, which is a property inherent to the absolute value, then concluding that |x - a| + |2·a| = |x - a| + 2·|a| < δ + 2·|a|. Ince you are able to make this conclusion on your own, the rest of the proof is, as they call it, "trivial."
Why do we need delta do be min{1, E/(1+2|a|)} and not just E/(1+2|a|) ? Is the min{1 part just there so that we can say that |x-a| < 1 , since delta can now be no greater than 1?
I think epsilon delta is so daunting because the concept isn't clear. I'm pretty sure this is the explanation that would click with me, and might click with other students. Suppose you have a function f, and you want to prove the limit of the function equals L at x=a. So, imagine drawing a rectangle centered on the point (a, L) tall enough that the function never touches the top or bottom of the rectangle. Now, can you make smaller and smaller rectangles -- smaller height and also smaller width -- such that the function never touches the top or bottom of the rectangles? If you can do that, then that means you have a limit. So, the job is to prove the existence of a simple relationship between the height and width, that lets you show that, for any rectangle height, it's possible to set a rectangle width such that the function never touches the top or bottom of the rectangle. And, as the height goes to zero, so does the width. The trick is actually doing the math, which usually involves proving that there is a simpler function that is always further away from the limit than the original function is. Then, if you can prove that the simpler function can be contained in rectangles as described above, it follows that the original function will likewise be contained in those rectangles. You may have to do a few iterations of this, so that eventually you're dealing with a function pretty far removed from the original one. If you can come up with a straight line function, that is the simplest situation. Remember, the point is that, in any rectangle you draw around (a, L), you can be confident that the function never touches the top or bottom edges of the rectangle. A straight line doesn't have any wiggles that could threaten to touch the top or bottom edge, no matter how small your rectangle is. So it's safe to say that, if you can get to the point of a straight line function, you're pretty much done. Pretty often, you will want/need to set a maximum width for the rectangles, so that, below that maximum width, you can introduce a straight line that is always further away from the limit than the original function is. That way, in the "linear" region, you have your straight line function. So the height of our rectangles is 2*epsilon: that's a distance of epsilon above the limit, and a distance of epsilon below the limit. And the width of our rectangles is 2* delta: that's a distance of delta to the left of the point and a distance of delta to the right of the point. If it is possible to mathematically prove a delta that is defined in terms of epsilon, and delta goes to 0 as epsilon goes to 0, that proves our limit. All of that is just the concept. Once you understand that, I think, the math will make more sense.
The definition of differentiability still requires showing a limit exists, a much more difficult limit in general than the one done here. If you don’t want to do an epsilon-delta proof for differentiability then you might as well skip it for continuity too.
Appreciate your honesty and humor regarding the difficulty of this proof! Never understood why delta was assumed to be min of (1, 1/(2abs(a)+1). Triangle inequality did the trick and thank you again for modeling the difficulty faced by people who would be trying to discover that choice of delta the first time they attempted it. Also nothing sacrosanct about 1. Could be any positive constant.
If you choose delta to be whatever you like, then i think you will prove every function to be continuous. Prove me wrong. Prove that 1/X^2 is not continuous, with the help of this delta epsilon method.
Delta can’t be whatever you like. Because if you pick a delta first, then I will pick a smaller epsilon that breaks your delta. Order matters - you have to start with an arbitrary epsilon then choose a delta based on that epsilon that makes the inequality work. It’s the whole point of the definition
Also, I already explained to you elsewhere, that while f : R -> R+, f(x) = 1/x^2 if 0 < |x|, f(0) = 0, is not continuous at 0, which can indeed be proven from the definition, and I already showed you this proof, the function f' : R\{0} -> R++, f(x) = 1/x^2 is continuous everywhere in its domain. So there is nothing to disprove using the definition.
Honestly I feel like this is more confusing when you choose the value of 1, why not just write that since |x-a| < delta then |x+a| < delta + 2|a| instead of having 1 + 2|a| ?
Because the delta is defined with values that you supose you have, not with variables or arbitrary values, in this case, we have like a constant the a, and the epsilon, because the hypothesis gives you the a and the epsilon
Why not just use the limit laws? lim_(x -> a) x = a, so lim_(x -> a) x * lim_(x -> a) x = a * a = a^2, but because lim_(x -> a) x * lim_(x -> a) x = lim_(x -> a) x * x = lim_(x -> a) x^2, it holds that lim_(x -> a) x^2 = a^2.
can't you just prove it by showing that the derivative is defined at every point? if the function isn't continuous it would have a vertical asymptote and the derivative at that point wouldn't be defined.
You'd have to then first prove derivative existing implies continuous and that x^2 has a derivative. Which is actually simple but not simpler than "product of continuous functions is continuous."
I love math Anyway, is there anyone who is a Foreign worker/works abroad here who can answer this questionnaire? I need this for online class, thanks ❤️❤️ - How long you have stayed abroad? - What are the purposes of your stay there? - What are your most unforgettable or memorable experiences there? - How will you compare your country with other countries in terms of culture, religion, economy, politics, law, social practices, etc? - Do you want to go back abroad or to other countries in the future? Why or why not?
When math teacher struggles with this proof: th-cam.com/video/TS5snj73S1g/w-d-xo.html
Q: Prove that f(x) = x² is continuous.
A: Come on, it is obviously continuous!
“Of course”
🤣🤣🤣
I've seen a math professor who would argue that it is not continuous and that there are gaps in the function, so it's not obvious to everyone.
@@pronounjow Oh God
Lemma 1: f(x) is a polynomial. Lemma 2: all polynomials are continuous over the reals. QED.
Proof of the lemmas are left to the reader.
This is easy: 1) f(x) = x is continuous 2) The product of continuous functions is continuous
@@spicca4601 That's trivial, choose epsilon = delta and you're done ;-)
Technically you also need a proof for 2). Which isn't hard, but need to be shown as well
Now proove x is continuous 🤷😅
Yesss!!
OK, give us an epsilon delta proof that the product of two arbitrary continuous functions is continuous.
Is that easy like Sunday morning? Don’t think so 😀
This one makes much more sense. I really like that you care for your subscribers and improve yourself if you see something new.
Kudos to you and congratulations to your students!❤️
I am happy to hear this. Thank you.
The new part starts at 5:14. Thanks to Victor for a much better way!
If you choose delta to be whatever you like, then i think you will prove every function to be continuous.
Prove me wrong.
Prove that 1/X^2 is not continuous, with the help of this delta epsilon method.
@@anuragguptamr.i.i.t.2329 The function is not defined for x=0. Since the value f(0) doesn't exist, it's not equal to anything, in particular it's not equal to the limit
@@pablop8166 that's why i want it to be proven by the epsilon delta method. If discontinuities of a function cannot be proven by this method and only the continuities can be proven, then this method would become useless.
@@anuragguptamr.i.i.t.2329 Remember that the proof starts with f(a) equals to the limit of f(a) as x approaches to a. But since f(0) does not exist, you can conclude right away that the limit can't be equal to it. You can't use the epsilon-delta definition because you already proved that the initial statement is false
@@pablop8166 i have studied about limits, derivatives, several advanced mathematics topics and the basic definition of derivatives with the help of limits.
But, i did never study about this epsilon- delta method during my school, college and post graduation.
Bprp steve is the first ever source for me, to get to know about this method.
But, now after reading your replies, now i know, why they didnt ever include this method into my curriculum. 🤣
Imagine being his student and giving your own solution to his questions. He's sure to give you an A+ and not mark you incorrect (given that you did it right).
I can’t give A+ but I will give the student a t-shirt. Like the winner of my calc 2 video project 😃
@@blackpenredpen I'll take that shirt over an A+ any day of the week.
@@JamalAhmadMalik Yeah he can't give an A+ for giving an alternative solution. It's still a part of the same question
And there’s literally no A+ in the grading system where I teach. I know it’s a bummer.
@@blackpenredpen
If I’m right, you teach in a university in Los Angeles, right?
Behind every successful mathematician is a Pokemon stuffy.
Counter example:
In this case the Pokemon Stuffy is in front of the successful mathematician
I really appreciate your educational videos on mathematics man, and I'm sure a lot of other students do too. You explain well.
Great job on Dr Tom Crawford's channel yesterday. That was crazy to watch.
Can you do videos that teach us about math proving methods?
(example : direct method or maybe by induction etc...)
Please show us the beauty of proofing!!
So as a math PhD student, I have been spending so much time studying for a probability and statistics exam, that when I went to sleep I was thinking to myself, “wait, did I really forget the definition of continuity”, so I looked up prove x^2 is continuous, and this restored my faith in myself lol. Always gotta love epsilon delta
I usually watch this channel once or several times a day to see what you are up to. I enjoy your effortless execution as you masterfully destroy one math problem after another. Don't get me wrong, im no math major, I've failed calculus before, but I still enjoy trying the odd math problem from my university days, seeing how much of the calculus part of my brain has deteriorated in my retail environment. Some days I'm feeling mighty confident in my abilities, neither days I am quickly humbled. You are indeed a genius. But that doesn't stop me from failing with grace
Thank you! I am happy to hear that you enjoy my content. 😊
I appreciate the passion for teaching 🙏 The education system needs teachers like you now more than ever.
If f(X) is continuous function at a point like take 'a' then these must follow this argument:
lim f(x). lim f(x)
x-> a^- = x->a^+ = f(a)
Correct. In a typical calculus class, your 3 sided equation is used to verify continuity of piecewise functions after it is determined that the individual pieces are continuous on their own domains, normally by just asserting they are ‘nice’ (e.g. polynomials or other elementary functions). But without knowing what a ‘nice’ function, you can’t calculate your one-sided limits except by following the method of the video.
You actually need the converse tô prove what you want. If both limits are equal, then the limit exists and is equal to the one you have found
Wasn’t this uploaded a few hours ago?
The other one has been unlisted.
There was probably a little error in the vid 😊
Yea. The new part is at around 5:12
Much easier proof: A continuous function is one such that f^(-1)(U) is open for every open set U. Taking preimages commutes with taking unions, so we only need to show this is the case for intervals (a,b).
Take 0
Sounds like the definition of continuity from a topology course. It seems as though you have “abstracted the difficulty out of the problem” However, the notion of open set for real numbers depends on arbitrarily small real numbers, so I don’t think you you dodged the bullet 😀. Just my opinion. Thanks for reminding me of the good old days in topology
Can you also give a delta without using the min? So just one delta
edit: i found one, you have to use quadratic formula. then its delta= sqrt( |a|^2+epsilon )-|a|
in total this comes to
delta (delta + 2|a|)
= (sqrt( |a|^2+epsilon )-|a|) (sqrt( |a|^2+epsilon )-|a|+2|a|)
=(sqrt( |a|^2+epsilon )-|a|) (sqrt( |a|^2+epsilon )+|a|)
=|a|^2+epsilon-|a|^2
=epsilon
with 3rd binomial formula
ohhh that's exactly what I just did. I also wonder if this is possible or not.
This is the coolest video on TH-cam
Oooh that's why it was unlisted
g(x) = x is continuous (taking delta = epsilon shows this in definition). The product of two continuous functions is continuous. Hence f(x) = x*x = x^2 is continuous.
You need to prove the pointwise product of two continuous functions is also continuous.
@@angelmendez-rivera351 That's true but the proof that the product of two continuous functions is also continuous isn't any longer than the proof in the video.
@@Bodyknock It definitely is longer. I wrote a comment in the main comments section in which I did the full proof. Compare that proof to the following:
f : (a, b) -> R, f(x) = x^2. If |x - c| < δ, then |x^2 - c^2| = |(x - c)·(x + c)| = |x - c|·|x + c| = |x - c|·|(x - c) + 2·c| =< |x - c|·[|x - c| + |2·c|] = |x - c|^2 + 2·|c|·|x - c| < δ^2 + 2·|c|·δ. Therefore, for every a < c < b and every ε > 0, there exists some δ > 0 such that if 0 < |x - c| < δ, then |x^2 - c^2| < ε, and such δ > 0 is given by δ = sqrt(ε + c^2) - |c| for every a < c < b and every ε > 0. Therefore, f is continuous.
@@angelmendez-rivera351 So long that you managed to write it in a youtube comment. Sort of contradicting yourself there....
Love these analysis videos. Do you think you're going to do more linear algebra or abstract algebra?
It depends if I can find good topics or not.
Beware , Pikachus can be really dangerous ,it could give you a really powerful thunderbolt lol .
Victor will protect us 😃
@@blackpenredpen Nah it couldn't last a week
I love your pikachu! That should be an element of the channel carrying everything while solving math from the death star to pikachu!!!
This reminds me of Russell's Principia Mathemathica proving that 1+1=2 with logical axioms and dozens of pages proving it and then there goes Gödel to say that every axiomatic system is incomplete.
Bruh moment.
Too bad you didn't choose to spend enough time in the very beginning to clearly and slowly explain the core idea behind the epsilon delta stuff, maybe visually. This was the most important thing in my opinion, and I missed it.
Thanks for ur input. I just wanted to show how to do the proof for this. If I also introduced the definition than the video would be too long. Plus that deserves a video by itself 😃
The proof? Trivial and left as an exercise to the reader. Wait, what do you mean I can’t say that in an exam?
f is differentiable. hence f is continuous. QED
You need to prove f is differentiable, and prove that differentiability implies continuity.
Couldn't you just do the limit of (x+h)^2 as h -> 0 ?
I was thinking something like this too. I think you then need to prove 2hx -> 0 and also h^2 -> 0 when h -> 0, but this isn't hard. Definitely easier than the way it was done in the video I think but maybe I'm also missing something idk.
i see why the minimum is used, but it's possible without it. the only difference is that we get δ(δ+2|a|)=ε, which is a quadratic equation so we choose δ equal to the positive solution, and that solution is sqrt(|a|^2+ε)-|a|
to me, that seems easier than using a minimum, but i guess that's subjective
When you're doing the proof right but you still included an OFC ... of course.
I mean it is actually very obvious!
LHL = 0
RHL = 0
f(0) = 0
so continuous....?
Also, you can indeed prove a function is discontinuous using the definition. However, be careful. For example, the function f : R\{0} -> R\{0}, f(x) = 1/x is not actually discontinuous anywhere. On the other hand, the function g : R -> R, f(x) = 1/x if 0 < |x|, f(0) = 0 is discontinuous somewhere: namely, at 0. This is an important distinction. The domain of f is not connected, but f is continuous everywhere in the domain. So you cannot demand for f to be proven discontinuous with the definition, because it is not discontinuous. You can, however, demand for g to be proven discontinuous at 0, since 0 actually is in the domain of g. You can indeed use the definition to do this. How so? Because |x - 0| = |x| < δ with δ > 0, and |1/x - f(0)| = |1/x| = 1/|x| < ε, with ε > 0, and 1/|x| < ε is equivalent to 0 < 1/ε < |x| for every ε > 0. |x| < δ does not imply, for every δ > 0, that 1/ε < |x|, because 1/ε < |x| and |x| < δ implies 1/ε < δ, which is equivalent to δ = 1/ε + γ for some γ > 0. Therefore, |x| < δ = 1/ε + γ implies 1/(1/ε + γ) = ε/(1 + ε·γ) < 1/|x| not 1/|x| < ε.
Can you show a case where we use the epsilon delta formula to prove a function is not continuous?
Yes. For example, consider the function f: R -> R, f(x) = 1/x if 0 < |x|, f(0) = 0. This function is not continuous at 0. In other words, for some ε > 0, it can be proven that for every δ > 0, |x - 0| < δ and |1/x - f(0)| = ε or ε < |1/x - f(0)|. How? Because |x - 0| = |x|, |1/x - f(0)| = |1/x|, and ε = |1/x| or ε < |1/x| implies |x| = 1/ε or |x| < 1/ε, which always happens given that δ = 1/ε.
@@angelmendez-rivera351 proof not clear to me. First, epsilon should be independent of x (unless u are defining epsilon as a function of a fixed x = constant = c). Second, NO choice of delta should work so any delta also must be independent of any choice of epsilon. There obviously is a proof of the fact that your f(x) is discontinuous at 0 but I don’t think this is it. You must show there exists an epsilon such that for all delta there exists an x such that the absolute value of x is less than delta and the absolute value of f(x) is greater than or equal to epsilon. Instead of pursuing that method of “dis-proof” however, how about instead simply proving (easily) that the right hand limit of f(x) as x approaches 0 is infinity (which obviously does not equal f(0) = 0)?
@@jessewolf6806 *First, epsilon should be independent of x*
And it is independent of x in my proof.
*Second, NO choice of delta should work so any delta also must independent of any choice of epsilon.*
That is entirely equivalent to what I wrote. The definition of continuity is of the form "Universal Quantifier on ε (Existential Quantifier on δ (Proposition))." Therefore, the definition of discontinuity is "Negation (Universal Quantifier on ε (Existential Quantifier on δ (Proposition)))." But "Negation (Universal Quantifier on ε (Existential Quantifier on δ (Proposition)))" is truth-table equivalent to "Existential Quantifier on ε (Negation (Existential Quantifier on δ (Proposition)))" which is truth-table equivalent to "Existential Quantifier on ε (Universal Quantifier on δ (Negation (Proposition))," which is precisely the syntax I used in the proof.
*You must show there exists an epsilon such that for all delta there exists an x such that the absolute value of x is less than delta and the absolute value of f(x) us greater than or equal to epsion.*
This is exactly what I did, symbolically. Perhaps the fact that I did it symbolically made it unclear to you, but this is 100000% what I did.
*Instead of pursuing that method of "disproof" however, how about simply proving (easily) that the right hand limit of f(x) as x approaches 0 is infinity*
OP wants us to use the limit definition of continuity to prove that a function is discontinuous at a point. So I did. You are begging the question by telling me to *not* answer the OP's "challenge."
Look, I do apologize if the language I used in my proof was somehow unclear, but you are just imposing random interpretations onto my proof instead of asking for clarifications, which is what any person *should* be doing, and then saying "see, your proof is bad because of this interpretation I gave it, so how about you don't do what OP asked and try proving discontinuity in a different way." I am not sure how am I supposed to take that and simply accept it.
Let me rephrase what I stated in my first reply.
There exists some ε > 0 such that, for every δ > 0, |x - 0| < δ & the disjunct ε = |1/x - f(0)| or ε < |1/x - f(0)|. This is because |x - 0| = |x| < δ and |1/x - f(0)| = 1/|x|. |x| < δ is equivalent to 1/δ < 1/|x|. So there exists some ε = 1/δ such that for every δ > |x| > 0, ε > 0 and ε < 1/|x| is satisfied, which is what needed to be satisfied.
Can't you just show that g(x) = x is cont., since f(x) = g(g(x))?
cant we say that bc f(x) is differentiable for all values (-inf, inf), it is therefore continuous by default?
And how do you check differentiability?
You would need to prove differentiability, and then prove differentiability implies continuity, and both proofs require the same definition used to prove continuity in the first place.
Q: Prove f(x)=x^2 is continuous for all real x
A: The proof is trivial and left as an exercise for the grader
Great video :) congrats on 100mil total views!
Thank you!!!
I still don't really get why we chose 1 as one of the values for δ.
I don't get it. Why can the triangle inequality theorem be applied?
Q: Prove f(x) = x^2 is continuous
Other maths branch : Obviously it is continuous
Real Analysis : Here is the proof
Pretty much
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For the record, some people are saying that you can instead just use the fact that x |-> x^2 : R -> R+ is differentiable everywhere, to then conclude it is continuous everywhere. The problem with this is that you would need to prove the function is differentiable, and prove that differentiability implies continuity, and both proofs require using the definition of a limit, which is the same definition you use to prove continuity directly instead. As such, proving continuity directly is actually just literally simpler. If you want to take the differentiability for granted, then you may as well also take continuity for granted. We cannot be cherry-picking about which statements to prove and which to not prove.
I feel like Nintendo is gonna start charging this guy rent any day now
Beautiful!
This is sooo satisfying to watch now, after seeing the a=-1/2 clip
Yay!
may i know how to prove this function is on uniformly continuos on the domain?
Where did that 2a come from in the triangle part
do the defimition of continuity at a point a involve 0
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Hahaha 沒錯
My proof would have been similar, except that, instead of saying |x + a| < 1 + 2·|a|, I would have said |x + a| < δ + 2·|a|, which is actually true for any δ > 0, and so I would not have "pre-chosen" δ before hand. The idea is that |x - a| < δ is given, and the triangle inequality implies |x + a| = |(x - a) + 2·a| = < |x - a| + |2·a| = |x - a| + 2·|a| < δ + 2·|a|. The reason I find this to be even simpler and more intuitive is that no further assumptions are being used. With |x - a| < δ and |x + a| < δ + 2·|a|, the consequence is that |x - a|·|x + a| = |x^2 - a^2| < δ·(δ + 2·|a|). Thus the δ to then "choose" would be the largest solution to the equation δ·(δ + 2·|a|) = ε, which always exists for every ε > 0 and always satisfies δ > 0 for every such ε. This does become obvious if you simply solve for δ explicitly: by completing the square, you can prove that δ = sqrt(ε + a^2) - |a|, which for every ε > 0, satisfies δ > 0. I also like this because it avoids using the minimum function.
There is no need for choosing the larguest, cos any that make It less than the given epsilon is enough.
Awesome! I wish I can be a genius in mathematics as you guys are😥
Geniuses aren't made, they are born geniuses
There's no way geniuses are born.
For being a genius, you just have to practice and learn the right way.
@@levilikesnothavinganycreat4971 You can't be serious, especially when talking about math
@@levilikesnothavinganycreat4971 Do you really think Ramanujan was such a genius because he 'practiced and learned the right way'?
@@tonylee1667 I hope some of my ancestors were geniuses. But if it was true, I would be a genius by now 🤣🤣 someone give us his DNA please 🧬
Is pikachu continuous?
Hi Blackpenredpen!
I feel like we need to see some videos where epsilon-delta makes things easier. Like, proving f(x) = x^2 / [(x^2 + 1)(x^2 + 2)(x^2 + 3)] is continuous. That looks like a nightmarish function, but thanks to the magic of the epsilon-delta process, that denominator can be replaced by the number 6, so all that complexity reduces to proving that x^2 / 6 is continuous.
How does the concept of continuity deal with transcendental numbers? For example, wouldn't the function be discontinuous at the square root of e since e cannot be a solution to y = x^2?
"How does the concept of continuity deal with transcendental numbers?"
The same as any other numbers. If a small change in input implies a small change in output then the function is continuous. rational/algebraic/transcendental has nothing to do with it.
An integral is always continous (easy proof using the Riemann definition of an integral)
x^2 = int(from 0;to x;2x)
QED
If you're talking about x^2 being differentiable, then:
If a function is continuous, then it's integral is differentiable (definition)
2 is continuous (by definition, I guess?)
2x is differentiable, x^2 is differentiable
Constant is continuous since f(x)-f(a) = 0 < epsilon 😀 so any delta will suffice
I think there is some fairly heavy lifting going on trying to prove x^2 is continuous based on integrating 2x. Sure it’s known to be true but proving it all the way back to “first principles” would challenge me I know. Still an interesting other approach to problem. I think I finally get that epsilon delta proof!
@@MyOneFiftiethOfADollar continous, mind. Any integral of a function that is defined on an intrrval has to br continous on that interval. Since the rieman sum is f(x)dx and dx->0 there cannot be a gap in the function.
@@JoQeZzZ are you intimating that blackpenredpen should have said integrability implies continuity and skipped the epsilon delta proof? Your claim is certainly intuitively true but does not constitute a rigorous proof.
Almost easier to show that f is differentiable which then implies continuity 😂
Yeah, the differentiable implies continuity is easier than choosing delta to be min of two values, but I think I finally get that now.
Of course we can put "of course" on the board. We might not get the best grades, but life goes on.
I've seen about 200 different explanations of the epsilon delta limit definition over the years, along with hundreds of applications of it. And I *still* don't understand how to actually prove anything with it. It always seems to come down to some sort of magic somewhere that isn't intuitively obvious. And it isn't the same magic every time, either. I'm pretty sure it has to do with how little attention there was to algebra on inequalities back in various levels of education. Take that and mix in the mind-meltiness that limits and the like can get into and it's a perfect storm of confusion, I think.
With any epsilon you can always find a delta such that for all x, 0
I think the reason for this is that often proofs are presented in a way that uses further assumptions and are not purely algebraic, although I do think not being adept at absolute value inequalities can also make it harder to understand the proofs. The key aspect of the proof is to understand that, given that |x - a| < δ, you conclude that |x + a| =< |x - a| + |2·a| using the triangle inequality, which is a property inherent to the absolute value, then concluding that |x - a| + |2·a| = |x - a| + 2·|a| < δ + 2·|a|. Ince you are able to make this conclusion on your own, the rest of the proof is, as they call it, "trivial."
it is eassy to prove .
FOR INDIANS
Holy your bread is so long O-O. I've been watching a splattering of your videos never thought you can grow that
Q: prove that X² is continuous:
A: Every polynomial function is continuous
How come delta= epsilon /1+2mod(a)
do you know how to solve integral -pi/2 to +pi/2 cos^2x/1+3^x
I have a great challenge for you: what
(x^(-x))^(x^(-x))^(x^(-x))^... (infinite tetration) approaches to? Why?🤔😏
Just use the tangent and the convexity? Both can easily established from basic algebra. Together they give you a funnel forcing continuity.
Why do we need delta do be min{1, E/(1+2|a|)} and not just E/(1+2|a|) ? Is the min{1 part just there so that we can say that |x-a| < 1 , since delta can now be no greater than 1?
Of course it is! QED
"quite easily done"
I think epsilon delta is so daunting because the concept isn't clear. I'm pretty sure this is the explanation that would click with me, and might click with other students.
Suppose you have a function f, and you want to prove the limit of the function equals L at x=a. So, imagine drawing a rectangle centered on the point (a, L) tall enough that the function never touches the top or bottom of the rectangle. Now, can you make smaller and smaller rectangles -- smaller height and also smaller width -- such that the function never touches the top or bottom of the rectangles? If you can do that, then that means you have a limit. So, the job is to prove the existence of a simple relationship between the height and width, that lets you show that, for any rectangle height, it's possible to set a rectangle width such that the function never touches the top or bottom of the rectangle. And, as the height goes to zero, so does the width.
The trick is actually doing the math, which usually involves proving that there is a simpler function that is always further away from the limit than the original function is. Then, if you can prove that the simpler function can be contained in rectangles as described above, it follows that the original function will likewise be contained in those rectangles. You may have to do a few iterations of this, so that eventually you're dealing with a function pretty far removed from the original one.
If you can come up with a straight line function, that is the simplest situation. Remember, the point is that, in any rectangle you draw around (a, L), you can be confident that the function never touches the top or bottom edges of the rectangle. A straight line doesn't have any wiggles that could threaten to touch the top or bottom edge, no matter how small your rectangle is. So it's safe to say that, if you can get to the point of a straight line function, you're pretty much done.
Pretty often, you will want/need to set a maximum width for the rectangles, so that, below that maximum width, you can introduce a straight line that is always further away from the limit than the original function is. That way, in the "linear" region, you have your straight line function.
So the height of our rectangles is 2*epsilon: that's a distance of epsilon above the limit, and a distance of epsilon below the limit. And the width of our rectangles is 2* delta: that's a distance of delta to the left of the point and a distance of delta to the right of the point. If it is possible to mathematically prove a delta that is defined in terms of epsilon, and delta goes to 0 as epsilon goes to 0, that proves our limit.
All of that is just the concept. Once you understand that, I think, the math will make more sense.
Pleas show how can we prove discontinuty of function by delta epsilon
"And always choose 1. Why 1? Because 1 is eazzyy." 😂😂
So, conclusion is that
Solving the Riemann hypothesis is not hard.
イプシロンデルタ 始めて見たけど、これ何の役に立つのかようわからん
定義の厳密性だけなのかな･･･
あとyoutubeで受験問題解説してる人のはさみうち法ってこれやん
これ知ってて高校生に偉そうなこと言うのあかんなあ～って気分😂
Your beard is looking horrible ...... You should have to shave it soon as possible
Thanks for a good video .DrRahul Rohtak Haryana India
When did u decided to do analysis ? It's so funny
Bc this is a good topic. “Easy statement but hard to proof”
I would have just shown that it is differentiable, which is much simpler, and if it is differentiable then it has to be continuous
The definition of differentiability still requires showing a limit exists, a much more difficult limit in general than the one done here. If you don’t want to do an epsilon-delta proof for differentiability then you might as well skip it for continuity too.
@@stephenbeck7222 I was thinking about the limit of the growth rate, but yeah otherwise it is indeed useless 😂
x to the power of twoooooooooooooooooooooooooooooooooooooooooooooooooooooooooo...
I've taken 2 years of various calc classes (calc 1 through diff eq) and not once have a had a teacher proove this so rigorously, thank you so much!!
Prof: obvious, do as homework.
But what if there was a hole in the function? The limit might exist but the function is not continuous.
Why not check the limit of its left hand side and right hand side is equal check?
Did he use an old photo for the thumbnail?
Pls prove that parallel lines converge at infinity
Appreciate your honesty and humor regarding the difficulty of this proof! Never understood why delta was assumed to be min of (1, 1/(2abs(a)+1).
Triangle inequality did the trick and thank you again for modeling the difficulty faced by people who would be trying to discover that choice of delta the first time they attempted it. Also nothing sacrosanct about 1. Could be any positive constant.
If you choose delta to be whatever you like, then i think you will prove every function to be continuous.
Prove me wrong.
Prove that 1/X^2 is not continuous, with the help of this delta epsilon method.
Delta can’t be whatever you like. Because if you pick a delta first, then I will pick a smaller epsilon that breaks your delta. Order matters - you have to start with an arbitrary epsilon then choose a delta based on that epsilon that makes the inequality work. It’s the whole point of the definition
@@stephenbeck7222 then i might choose delta in such a way that i could also prove 1/X^2 to be continuous!?!?!?!?
@@anuragguptamr.i.i.t.2329 You will never find any δ > 0 such that 1/x^2 is continuous at 0 for every ε > 0, so no, you are wrong.
Also, I already explained to you elsewhere, that while f : R -> R+, f(x) = 1/x^2 if 0 < |x|, f(0) = 0, is not continuous at 0, which can indeed be proven from the definition, and I already showed you this proof, the function f' : R\{0} -> R++, f(x) = 1/x^2 is continuous everywhere in its domain. So there is nothing to disprove using the definition.
Please make more videos on bprp asmr
Proof: This proof is so obvious that it does not have to be proven
Honestly I feel like this is more confusing when you choose the value of 1, why not just write that since |x-a| < delta then |x+a| < delta + 2|a| instead of having 1 + 2|a| ?
Because the delta is defined with values that you supose you have, not with variables or arbitrary values, in this case, we have like a constant the a, and the epsilon, because the hypothesis gives you the a and the epsilon
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is your pikachu taking calc2 ?
What is the 1 for?
woah where did you get the pikachu mic
Take a Pikachu and then insert a mic :)
Loved ur microchu
Why not just use the limit laws?
lim_(x -> a) x = a, so lim_(x -> a) x * lim_(x -> a) x = a * a = a^2, but because lim_(x -> a) x * lim_(x -> a) x = lim_(x -> a) x * x = lim_(x -> a) x^2, it holds that lim_(x -> a) x^2 = a^2.
Let c be arbitrary any real no.
Now ,if we prove that at c the f(x)=x^2 is continuous it means it's continuous for all real no.
This is what he did in the video
can't you just prove it by showing that the derivative is defined at every point?
if the function isn't continuous it would have a vertical asymptote and the derivative at that point wouldn't be defined.
You'd have to then first prove derivative existing implies continuous and that x^2 has a derivative. Which is actually simple but not simpler than "product of continuous functions is continuous."
Sehr interessant und sehr informativ
Very interesting and much informations
I love math
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