Proving Uniform Continuity using ε-δ
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- Опубліковано 19 сер 2024
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Today we are going to proof uniform continuity of the funktion cos(x)+x. For this we make use of the abstract epsilon delta definition and use some spicy integral approximations to deal with everything. Enjoy! =D
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When I saw ε-δ in the thumbnail, the distance between my finger and the screen approached zero from above
"Neeeeeeeerd" - Homer Simpson
LOL!
The limit of my finger as it approached the screen approached zero
4:32 "It just makes sense" If Fermat had just written that in the margin, Andrew Wiles would have saved himself a ton of work
That's one good mathematician's joke.
"you have to play around with the expressions like you would play around with your girlfriend at home" you earned yourself a subscribe
:^)
Better ... you have to play around like you would with your best friend's girl friend at home when your girl friend is not looking
Ah, a good ol epsilon delta proof. Just like the neanderthals had to do it. Luckily the ancient egyptians figured out that these proofs can be simplified for differientiable functions if the derivative is bounded.
Oh my god. You made me laugh :D
Epsilon-delta proofs always confused the hell out of me, but you explained that you need to cast some spells to get delta in terms of epsilon and then you're all good. Better than any of my lecturers.
5:08
Papa Flammy: "Your Girlfirend at Home"
Me: "What Girlfriend?"
Dang, now not that sounds like really strong analysis and intricate proof right here. Appreciate the information and enlightening indeed!
@@PapaFlammy69 Ahh it's great to tune in man!
Came only to hear "I just kicked a child", job done.
Nice proof💚✨
U can also use The mean value theorem in order to evaluate |cos x -cos y|
Simply, the theorem says that there exists c in [y,x] such that
f'(c)(x-y)=f(x)-f(y)
If we consider our f is cosx, so we get
-sin(c)(x-y)=cos x- cos y
Hence |cos x-cos y|=|sin(c)(x-y)|≤|x-y| since |sin(c)|≤1
You could've posted this before my exams on this shit, not two days after
Yeah this happened to me twice, once on the Basel problem and once related to radius of convergence ;-;
0:09
Mathematicians kick children just for fun. xD
Alternatively,f'(x)=1-sinx => 0
I think it is a kind of circular reasoning when you use integral to prove continuity.
It's so sexy to hear a competent mathematician speak/write in complete mathematical sentences.
By the Mean Value Theorem, there exists a point c∈(y,x) such that
|(cosy - cosx)/(y - x)|
When I saw the sigma and delta I thought this was a chemistry video from another channel about sigma bonds. btw papa can you do a video about TOPOLOGY next
Edit: realized this is an eplison 🤦♂️
Yeah, maybe calculate the fundamental group of the torus with Seifert-Van Kampen? Should lend itself to pretty pictures
Came here for the meme. Nice. Nice meme.
(check auto-generated english subtitles at 0:10)
Basicaly this guy proved that if f and g are continuos then h is continuos , where h(x)=g(x)+f(x) and that x and cos(x) are continuos ( two elementry functions). All preaty basic.
Though the proof for cos x is nice, that's not enogh, that's not Papa-Flammy-Enough!!!!
So I chalange you Papa Flammy! Proove something harder! Prove the Reimann Hypotesis! :)))
I hate maths but I watch your videos anyway. I love watching things that I dont understand at all
This is the first step to converting you to a math lover :)
@@lachlanpfeiffer8199 Yes but I will become math lover in another life 😁
Make another epsilon delta for functions that blow up at small values of an independent variable
Oh my god, I have an exam of this today, hahaha what a coincidence
Can you do a video computing the limit of a function using just the definition of limits?
Flammable Maths ty
Can you also do ones where you show that no limit exists, using the definition of a limit, say for
F(x)= x, where x is rational
-x, where x is irrational
or even F(x)=sin(1/x) because I just wanted to see other ways to do these ones. (Reference Michael Spivak mcalculus' Chapter 15 Q20)
@@0511270511 limit will exist when x is irrational, but not when it is rational.
I had to double check to make sure I clicked on the right video. This was brilliant
Basically, what you proved to handle the cos case is that functions with bounded derivatives are uniformly continuous. That Lemma actually proves the whole statement (and many others) in a single stroke.
So good !!
:)
Morning maths is the best
My favorite vid as of rn
Also, randomly, here’s a vastly superior definition of the derivative because it generalizes to topological vector spaces seamlessly (from the usual setting of Banach spaces):
Define f’(a) = u(a) when u(x) is a function, continuous at a, so that f(x) = f(a) + u(x)(x-a) on some neighborhood of a. This definition generalizes nicely to multivariable functions by letting u(x) be a matrix. Then, the chain rule for the Fréchet derivative pops straight out of direct verification. I can’t take full credit for this convention; it was my professor, Jim Morrow, who brought it to my attention. You should make a video about it so that cool bois and grills can start using the *true* derivative definition.
i love you papa, for the next assignment we have to show that every continuous func. is also e-d-continous. so this for sure does help a lot!
so what is a continuous function then?
I’m here for the child kicking
There's no doubt in my mind that you would be able to explain this much better than my analysis professor in your sleep.
I'd be happy to see more videos on Real Analysis stuff
Now let’s do some hardcore epsilon-delta proofs
0:08 Subtitles did a great job here 😄
integarals
It’s very nice to proof like this. But i more or less people have an intuitive sense that this is uniformly continuous. What is more interesting is to prove a function by this method and find out by the proof that our assumptions about the uniform continuity is false! That has bigger implications in the greater scheme of things. Can you please show such an example and end up proving that it is not continuous and not uniform? Does some equations of a tangent function be an example of this? And can you prove it over the complex domain as a generalization?
I also watched your video on e^x + cos(x). How many roots does this have. All of it’s roots
You have such beautiful handwriting
subscribed within the first 20 seconds of the video.
Yo Papa I was wondering , would you ever make a video evaluating eta’(1) ?
More of these!!
If a function is uniformly continuous on a closed interval, could we refine the definition of uniform continuity by replacing the condition |x-y| < δ and |f(x) - f(y)| < ε with |x-y| ≤ δ implying |f(x) - f(y)| ≤ ε ?
Captions at 0:10
Good morning fellow mathematicians I just kicked a child
Finally...
@@PapaFlammy69 😊😊
One bijective boi
nice way to delimit abs(cosx-cosy). Last year when i say this problem didnt even think of it
Sehr nice Papa. Vor allem, weil ich im ersten Semester bin und die Analysis mich auffrisst 8)
Nice idea to show that |cos x - cos y|
Without loss of generality, though, you can say x
@@andrewhaar2815 you are right that absolute value is increasing for nonnegative reals, but again, without paying attention to bounds, the RHS integral is negative. You could say wlog x
EnNarr91 I don't see how your last comment proves anything. cos is an even function, so without loss of generality, you can assume 0 < x < y. That alone fixes the problem.
@@angelmendez-rivera351 funny how two comments are made presumably to argue against me and instead making a fix in otherwise correct proof. Just like my original comment.
EnNarr91 Your comment is just flawed as the proof itself. And the proposal we explained to criticize your comment is actually what fixes the proof in the first place. Jens' proof is invalid only because he failed to state the premise that 0 < x < y causes no loss in generality. But one can also understand that given the missing premises, if we take his premises as true, then the missing premise is a hidden implicational consequent of the premises he did present. I see no problem with that from a formal logic perspective. All it means is that his conclusion is an implication as opposed to a singular consequent.
follow the damn train CJ
1:54
😵😵😵
I was Watching some Mysterious Arts of Pyramid on the Chalkboard at the first time.
huh that code 218218 looks oddly similar to the "nuclear code", if you know what I mean
I love it papa, I wish you could do more epsilon proofs videos.
Is the 218218 code a reference to a high quality cultured manga or is it some joke i missed?
Could we also say |cos(x)-cos(y)|
Please do more epsilon delta proofs! You explain them amazingly ❤💪💪
10:15 why 2δ? is it because the thing in abs is the sum of two quantities both less than δ?
delta=epsilon-2 also works
Hello hope you realise you’re famous on Mathematical Mathematics memes on facebook now :D
Instead of using the integral here, would saying that |cos(x)-cos(y)|
He floats across the stage like a mathematical Fred Astair !
the Mathematical Mathematics Memes group is having fun with this
@@PapaFlammy69 damn what a madlad
Papa flammy i have my math exam on monday. wish me good luck :3
MORE EPSILON DELTA PROOFS
5:32
So you used that cos(x) is differentiable to show that x+cos(x) is continuous
Bold move.
Might as well of just used that x is differentiable to save you the step of the triangle inequality
Also, you didn't have to provide the disclaimer for the negative sign missing, because you chose the bounds of the definite integral correctly
Phoenix Fire differentiable doesn’t imply uniformly continuous on non-compact domains. For instance, f(x)=x^2 actually isn’t uniformly continuous on its entire domain, but is on any compact set.
@@alexandersanchez9138
6:45 He used that the derivative was bounded
If a function is differentiable and its derivative is bounded, then it is uniformly continuous
Essentially, if you take out the triangle inequality step, his proof is just "x+cos(x) is differentiable and bounded, so it is uniformly continuous"
Phoenix Fire you’re totally right. It seemed like a bunch of people were saying that that there was a circularity issue, and so I just commented to address that.
What college course would help introduce the topic(s) in this video?
Wow bro thanks a million. Could you do measure theory proofs as well?
>mfw he explains the triangle inequality without apples and oranges
Damn that continuity be uniform! 😩😩
no to demonstrate you give a epsilon and find the delta!!
Thank you, Papa !
I'm not comfortable with these epsilon / delta proof. This one is very clear.
But I have a question : when you write cos(x)-cos(y) as a integral, don't you consider implicitly the continuity of sin(x) without proof ?
@@PapaFlammy69 Ok ! Thanks
@@PapaFlammy69 I was wrong with this notion of uniform continuity 😟.
Do integrals rely on continuity in order to exist/work? I'm not in college, so I haven't taken these classes yet, but if the Integral relies on continuity to work, wouldn't this video be a circular argument? Genuinely curious.
I wanted to buy one of your merry Christmas hoodies, but it won’t arrive until January, so now what am I supposed to do?
Flammable Maths unfortunately, it’s the same delivery date on amazon, I probably shouldn’t have waited this long to order it
Can you try doing another one of these, but in a higher-dimensional function? I never really figured out how tbh
you could use mean value theorem to bound that integral too.
It would be nice if you do some topology exercises like homomorphisms or homotopy
のtフィ絵dtあｇしん。あっっっっっっｈ
Edit. Rip. That was supposed to say not first ahhhhhh but I hadn't switched my phone keyboard. Good morning, bruhs. Papa, thank you for the maths cheer. *Watches ad in deep suspense waiting for video to start*
I find your proof a little problematic.
As far as I am aware, to formally take the integral/derivative of sin/cos you need to (at least) show that they are continuous (possibly piecewise). But if you know, that they are continuous, then you don't need the integral trick, since it's trivial to demonstrate, that a sum of 2 continuous functions (identity and cos) is continuous.
Edit: yeah, I know, that uniform continuity is different from regular continuity. If you want to prove the uniform continuity, consider the following: sin(x) is a periodic, everywhere differentiable, continuous function. Therefore, it's uniformly continuous.
RuRo Not actually. If you use the definition of the derivative, the only thing you need to do to prove the derivative exists is to show the limits of sin(x)/x and (1 - cos(x))/x exist as x -> 0. I do not think these require proving continuity on sin(x) and cos(x) at x = 0.
RuRo Also, even if this was a requirement, we know sin(x) and cos(x) are everywhere continuous. There is no need to prove a well-known nameless elementary theorem to prove this. Continuity everywhere, after all, does not assume uniform continuity.
@@PapaFlammy69 Sure, continuity doesn't imply uniform continuity, I never said it did. But I don't see, how that is relevant? The sum of 2 uniformly continuous functions is uniformly continuous.
@Angel Mendez-Rivera
Continuity is *defined* as the existence of the 2-sided limit as the argument approaches some point. So yes, continuity **is** required for differentiability. If the function is discontinuous at some point, then it doesn't have a derivative at that point.
It seems to me, that you are asserting, that "sin(x) is continuous, differentiable and its derivative is cos(x)" is a more basic fact than "sin(x) is uniformly continuous"?
@@ruroruro
x^2 is continuous, differentiable, and its derivative is 2x, but x^2 isn't uniformly continuous on R
The problem I have with Flammable Maths proof is that he used that cos(x) has a derivative (-sin(x)) that is bounded (by -1 and 1) which does imply uniform continuity
Essentially, the triangle inequality step wasn't necessary because he could've treated x along with cos(x) (derivative being 1-sin(x) which is bounded by 0 and 2), and then his proof amounts to just stating "the function is differentiable and its derivative is bounded, so it is uniformly continuous"
@@SlipperyTeeth I am not claiming, that uniform continuity follows from continuity, I just find it weird, that for some reason both the differentiability and continuity of sin(x) are "basic" claims, that don't need any proof, while the uniform continuity of sin(x) is not.
I mean, it's a bounded, periodic, continuous function without any singularities, special points etc. In my opinion, either you accept that sin(x) is a "well behaved" function and you get to assert that it's infinitely differentiable, uniformly continuous and smooth OR you want to be rigorous about your proofs and then *nothing* about sin(x) should be known apart from your definition of choice and you have to prove every statement about it.
IMHO, claiming, that uniform continuity is somehow a much more complicated property than continuity and differentiability is special pleading.
Good old times
I only wanted to say that you are both clever and handsome
Love from Italy
I have found someone who wrote the same things ahah
Flammy why did you post this 2 days after my Real Analysis final exam :(
Top notch memes
moar analysis videos
Can't you just argue for cosx that derivative is always bounded between - 1 and 1, therefore the cosy-cosx/(y-x) is also in that boundaries???
Which in fact is the definition of the differential quotient, thats absolute value is smaller than the derivative at its maximum value over some interval
Chess is the best sport
hey Papa, my hero, (you've probably been asked this a bajillion times already) can you please make a video on some of those weird zeta values that make people think of weird summations to infinite series' like the -1/12 thing n stuff?
i'll totes share your channel everywhere (even more) if you do that :)
@@PapaFlammy69 legend, thanx! i'll go through them now :)
I think i'm old enough now that you can teach me the birds and the bees of -1/12
Spielerei :D
I'm watching this video with 2x speed :D
More analysis stuff please!
Also, couldn't you just say the difference between the two cosines is 2 at max?
Now do it without an integharal🙂🙃🙂🙃🙂🙃🙂🙃🙂🙃🙂
Hey, this is supposed to be a fun channel... lol
All you had to do was follow the damn Greta , CJ.
5:45 I never knew that abs(-x) = abs(x) factorial. Kappa
If you reply I will eat a tangerine with its peel
Let us spiel around a little bit!
Wenn du umziehst dann werd doch mein Analysis Tutor, haben gerade genau das in der Vorlesung
@@PapaFlammy69 Technische Uni Hamburg, erstes Semester Technomathe
not gonna lie, this would have helped me a lot a year ago. Either way, great job.
si\ (t)
FIRST OIAAAAAHAUHEEHEE
when u are a climate activist but u take the train xdé
😋😋😋
'How dare you' 🤭🤫