Proving Uniform Continuity using ε-δ

[@ishaanvatus3536]

When I saw ε-δ in the thumbnail, the distance between my finger and the screen approached zero from above

[@stewartmoore5158]

"Neeeeeeeerd" - Homer Simpson

[@megauser8512]

LOL!

[@thedudethatneveruploads2617]

The limit of my finger as it approached the screen approached zero

[@neilgerace355]

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

[@mohammedal-haddad2652]

That's one good mathematician's joke.

[@leftistadvocate9718]

"you have to play around with the expressions like you would play around with your girlfriend at home" you earned yourself a subscribe

[@PapaFlammy69]

:^)

[@davidbrisbane7206]

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

[@HolyMith]

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.

[@gdsfish3214]

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.

[@vukstojiljkovic7181]

Oh my god. You made me laugh :D

[@RC32Smiths01]

Dang, now not that sounds like really strong analysis and intricate proof right here. Appreciate the information and enlightening indeed!

[@RC32Smiths01]

@@PapaFlammy69 Ahh it's great to tune in man!

[@janus3042]

5:08 Papa Flammy: "Your Girlfirend at Home" Me: "What Girlfriend?"

[@bilyanaconsulova405]

Came only to hear "I just kicked a child", job done.

[@djalalmaster1018]

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

[@northernberger]

You could've posted this before my exams on this shit, not two days after

[@V-for-Vendetta01]

Yeah this happened to me twice, once on the Basel problem and once related to radius of convergence ;-;

[@davidbrisbane7206]

By the Mean Value Theorem, there exists a point c∈(y,x) such that |(cosy - cosx)/(y - x)|

[@epeseferma2173]

Alternatively,f'(x)=1-sinx => 0

[@diaz6874]

0:09 Mathematicians kick children just for fun. xD

[@HaiNguyen-cz2bj]

I had to double check to make sure I clicked on the right video. This was brilliant

[@k.c.sunshine1934]

It's so sexy to hear a competent mathematician speak/write in complete mathematical sentences.

[@thedarksword3495]

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 🤦‍♂️

[@riccardoorlando2262]

Yeah, maybe calculate the fundamental group of the torus with Seifert-Van Kampen? Should lend itself to pretty pictures

[@Skylitzz_]

There's no doubt in my mind that you would be able to explain this much better than my analysis professor in your sleep.

[@walterodimm]

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!

[@donaastor]

so what is a continuous function then?

[@jonkeuviuhc1641]

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! :)))

[@alexandersanchez9138]

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.

[@alexandersanchez9138]

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.

[@Vaaaaadim]

I'd be happy to see more videos on Real Analysis stuff

[@yuanmingluo2455]

I think it is a kind of circular reasoning when you use integral to prove continuity.

[@lordmetzgermeister]

Came here for the meme. Nice. Nice meme. (check auto-generated english subtitles at 0:10)

[@tahafakhech7712]

I love it papa, I wish you could do more epsilon proofs videos.

[@Hyebze]

So good !!

[@PapaFlammy69]

:)

[@69ms]

Can you do a video computing the limit of a function using just the definition of limits?

[@69ms]

Flammable Maths ty

[@0511270511]

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)

[@shashwat1330]

@@0511270511 limit will exist when x is irrational, but not when it is rational.

[@Fru1tyy]

Yo Papa I was wondering , would you ever make a video evaluating eta’(1) ?

[@jperez7893]

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

[@caldera99]

Is the 218218 code a reference to a high quality cultured manga or is it some joke i missed?

[@djordjesankovic1007]

I hate maths but I watch your videos anyway. I love watching things that I dont understand at all

[@lachlanpfeiffer8199]

This is the first step to converting you to a math lover :)

[@djordjesankovic1007]

@@lachlanpfeiffer8199 Yes but I will become math lover in another life 😁

[@sergioh5515]

My favorite vid as of rn

[@sumittete2804]

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 ?

[@b.blokzijl1189]

Please do more epsilon delta proofs! You explain them amazingly ❤💪💪

[@Ferolii]

nice way to delimit abs(cosx-cosy). Last year when i say this problem didnt even think of it

[@JustinsRealmMC]

Now let’s do some hardcore epsilon-delta proofs

[@LordOfNoobstown]

integarals

[@someperson9052]

You have such beautiful handwriting

[@proxyfeint]

More of these!!

[@samas69420]

10:15 why 2δ? is it because the thing in abs is the sum of two quantities both less than δ?

[@martyalden]

Oh my god, I have an exam of this today, hahaha what a coincidence

[@hussainsajwani8192]

Instead of using the integral here, would saying that |cos(x)-cos(y)|

[@captain_ali_01]

subscribed within the first 20 seconds of the video.

[@plasmacrab_7473]

Could we also say |cos(x)-cos(y)|

[@ChrisChoi123]

huh that code 218218 looks oddly similar to the "nuclear code", if you know what I mean

[@samuelwilkin5]

Captions at 0:10 Good morning fellow mathematicians I just kicked a child

[@Anthony-db7ou]

I’m here for the child kicking

[@derpfish4385]

0:08 Subtitles did a great job here 😄

[@loganwall2943]

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?

[@loganwall2943]

Flammable Maths unfortunately, it’s the same delivery date on amazon, I probably shouldn’t have waited this long to order it

[@SoapFX]

Sehr nice Papa. Vor allem, weil ich im ersten Semester bin und die Analysis mich auffrisst 8)

[@egillandersson1780]

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 ?

[@egillandersson1780]

@@PapaFlammy69 Ok ! Thanks

[@egillandersson1780]

@@PapaFlammy69 I was wrong with this notion of uniform continuity 😟.

[@IshaaqNewton]

1:54 😵😵😵 I was Watching some Mysterious Arts of Pyramid on the Chalkboard at the first time.

[@Suanhossien]

Wow bro thanks a million. Could you do measure theory proofs as well?

[@QuiescentPilot]

Can you try doing another one of these, but in a higher-dimensional function? I never really figured out how tbh

[@brunoberganholidias5790]

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.

[@paulg444]

He floats across the stage like a mathematical Fred Astair !

[@YitzharVered]

Damn that continuity be uniform! 😩😩

[@listentome5583]

What college course would help introduce the topic(s) in this video?

[@cosenza987]

the Mathematical Mathematics Memes group is having fun with this

[@cosenza987]

@@PapaFlammy69 damn what a madlad

[@Ennar]

Nice idea to show that |cos x - cos y|

[@andrewhaar2815]

Without loss of generality, though, you can say x

[@Ennar]

@@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

[@angelmendez-rivera351]

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.

[@Ennar]

@@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.

[@angelmendez-rivera351]

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.

[@matron9936]

One bijective boi

[@johubify]

Morning maths is the best

[@anim8623]

follow the damn train CJ

[@Ferolii]

It would be nice if you do some topology exercises like homomorphisms or homotopy

[@hoodedR]

Finally...

[@hoodedR]

@@PapaFlammy69 😊😊

[@SlipperyTeeth]

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

[@alexandersanchez9138]

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.

[@SlipperyTeeth]

@@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"

[@alexandersanchez9138]

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.

[@donghaefishy41]

Hello hope you realise you’re famous on Mathematical Mathematics memes on facebook now :D

[@ruroruro]

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.

[@angelmendez-rivera351]

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.

[@angelmendez-rivera351]

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.

[@ruroruro]

@@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"?

[@SlipperyTeeth]

@@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"

[@ruroruro]

@@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.

[@samueljele]

Good old times

[@sachatostevin6435]

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 :)

[@sachatostevin6435]

@@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

[@xaxuser5033]

delta=epsilon-2 also works

[@nnniv]

Papa flammy i have my math exam on monday. wish me good luck :3

[@lukehibbs6723]

MORE EPSILON DELTA PROOFS

[@honolululuke158]

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???

[@honolululuke158]

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

[@prikroymenya2794]

Flammy why did you post this 2 days after my Real Analysis final exam :(

[@Bartleby388]

>mfw he explains the triangle inequality without apples and oranges

[@liosittler]

no to demonstrate you give a epsilon and find the delta!!

[@eva-jd2zg]

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*

[@MathIguess]

Top notch memes

[@ferdydemier6830]

More analysis stuff please!

[@ferdydemier6830]

Also, couldn't you just say the difference between the two cosines is 2 at max?

[@aadityabhetuwal5990]

All you had to do was follow the damn Greta , CJ.

[@paulg444]

you could use mean value theorem to bound that integral too.

[@francesca1734]

I only wanted to say that you are both clever and handsome Love from Italy

[@francesca1734]

I have found someone who wrote the same things ahah

[@Ricocossa1]

Hey, this is supposed to be a fun channel... lol

[@kwirny]

Spielerei :D

[@tanvec]

5:45 I never knew that abs(-x) = abs(x) factorial. Kappa

[@shaqramento3238]

Now do it without an integharal🙂🙃🙂🙃🙂🙃🙂🙃🙂🙃🙂

[@UrasSomer]

If you reply I will eat a tangerine with its peel

[@biswaranjan17]

I'm watching this video with 2x speed :D

[@tellemfr]

moar analysis videos

[@josephtarantin1931]

Chess is the best sport

[@meinegute4261]

Let us spiel around a little bit! Wenn du umziehst dann werd doch mein Analysis Tutor, haben gerade genau das in der Vorlesung

[@meinegute4261]

@@PapaFlammy69 Technische Uni Hamburg, erstes Semester Technomathe

[@Vincentsgm]

when u are a climate activist but u take the train xdé

[@Luiigii52]

not gonna lie, this would have helped me a lot a year ago. Either way, great job.

[@jiaming5269]

😋😋😋

[@RealOmegamonX]

FIRST OIAAAAAHAUHEEHEE

[@mariochavez3834]

si\ (t)

[@cirnobyl9158]

5 kicked children disliked this video.

[@doria_bolognese]

'How dare you' 🤭🤫