Its CRAZY how someone on yt can explain a concept much more efficiently. I am a math major in the first semester and I am rlly struggeling to understand concepts when professors explain it, or its just hatd for me to understand stuff in the lecture, even worse when friends try to explain it to me… as they are trying to confidently teach me, (the themselves haven’t understood it good enough) I then feel very stupid. But I know it mostly depends on their explanation… so thank you!!!
Very nice video, and a very clear and concise explanation! Note though: Starting at 2:50 it says at the top right: "If a given \delta works for any \epsilon we choose, for any points in the domain". This might be a slightly confusing formulation, since it's more like: For a given \epsilon, we can find a \delta that works for any points in the domain. We cannot find a \delta that works for any \epsilon. But, again, you explained it perfectly, and hopefully people who watch the video will get the right idea anyways. Thank you!
This is the best explanation I have seen explaining the difference between continuity and uniform continuity. Unfortunately, the main thing with these problems is how difficult they are to actually prove.
For the uniformly continuous counter-example, it would be nicer if you kept both the epsilon and delta fixed and moved the blue region closer to the y-axis and picked two points on the curve that are in the blue region but are clearly not entirely in the red region.
Awesome Video! Very clear explanation of the use of Delta-Epsilon in the context of uniform continuity, and the counter example added even more clarity
I think there are more people who understand the concept, but can't write a proper epsilon-delta proof than vice versa. But I do think videos like these are helpful.
Great explanation, thank you. But I have a doubt: In function 1/x if we take interval [0.5,1] then we could apply Heine's Theorem because the function is continous in [0.5,1]. But then the function would be uniformly continous within [0.5,1]. However how is it possible if it is not uniformly continous?
Emsie emstraba this function is uniformly continues only in a certain interval but if talk about whole real lime then function is not uniformly continues
Just to double-check -- uniform continuity is informally verified by checking that for a given pair of points (x,y) it is true that |x - y| < delta => |f(x) - f(y)| < epsilon, correct? I spent years thinking it was the other way around (), and couldn't figure out why my verifications never added up.
In the last statement, that the function f(x)=1/x, what gurantees us that indeed, there is no delta we can find in the second points that you've mentioned?
There is something missing here, because f(x) =1/x is uniformly continuous for all x>1.Then how you're gonna fix the epsilon such that the corresponding delta will work for the whole domain.Because we know by video animation that it won't work
But if you choose the delta appearing at 4.27 the definition still holds true: you'll have values within epsilon for all other (x,y) throughout 1/x on the right that point if you look at the graph, for example where you defined your former delta
So i) continuous means one of the points is fixed, and this works when you consider each point in the domain in turn as a fixed point (the maximum delta can be different for each point), ii) uniformally continuous means there is a single maximum value of fixed delta for which point i) is true. So ii) is a stronger condition.
That's correct. Uniform continuity is stronger. (1) Continuity means "for all epsilon, for all x, there exists delta(epsilon, x) such that for all x_0, d(x_0, x) < delta ==> d(f(x_0), f(x)) < epsilon" (2) Uniform continuity means "for all epsilon, there exists delta(epsilon), such that for all x and for all x_0, d(x_0, x) < delta ==> d(f(x_0), f(x)) < epsilon". Notice the difference between the two is that the quantifiers "for all x" and "there exists delta" have been switched. This means that in the definition of continuity, delta is a function of both epsilon and x, whereas in the definition of uniform continuity, delta is only a function of epsilon alone. So, every uniformly continuous function is continuous, because to find delta(epsilon, x), you can just take the delta(epsilon) guaranteed by the definition of uniform continuity.
how about f(x) = x^2, I never understood this kind of math logic. there is no delta for all the points in x^2. each delta makes each epsilon ^2, delta must become smaller the further right you go. i think mathematicians created this epsilon and delta things to mock us. this video has a good visual representstion than others, i probably made one step closer to get to know the delta and epsilon thing.
Whaaaat!!!!??? 5 minutes is all it took for you to clarify a concept I was trying to figure out for months 😂...Excellent!
Its CRAZY how someone on yt can explain a concept much more efficiently. I am a math major in the first semester and I am rlly struggeling to understand concepts when professors explain it, or its just hatd for me to understand stuff in the lecture, even worse when friends try to explain it to me… as they are trying to confidently teach me, (the themselves haven’t understood it good enough) I then feel very stupid.
But I know it mostly depends on their explanation… so thank you!!!
Man the last 20 seconds...cannot thank you enough. God’s work.
That explanation was so clear. thank you very much.
Thank you. This has bothered me for years. The definition is so abstract and features so many moving parts I was never quite sure if I got it.
Great animation and explanation. It is first time, when I could understand uniform continuity geometrically.
I struggled for 1 week trying to understand continuity. Now finally thanks to you i understood!
The graphical explanation cleared up all confusion I had about the definitions. Thank you.
Very nice video, and a very clear and concise explanation! Note though: Starting at 2:50 it says at the top right: "If a given \delta works for any \epsilon we choose, for any points in the domain". This might be a slightly confusing formulation, since it's more like: For a given \epsilon, we can find a \delta that works for any points in the domain. We cannot find a \delta that works for any \epsilon. But, again, you explained it perfectly, and hopefully people who watch the video will get the right idea anyways. Thank you!
Hands down the best explanation on this topic . 💯🕺
This is the best explanation I have seen explaining the difference between continuity and uniform continuity. Unfortunately, the main thing with these problems is how difficult they are to actually prove.
Amazing! Thank you! There aren't enough good videos on Real Analysis :)
Do you know any good ones
@@josuke6869 I know some if u still need: ruclips.net/video/dQw4w9WgXcQ/видео.html
@@IStMl my disappointment is immeasurable and my day is ruined
@@Mryeo5354 come on it was a joke or just to inspire you
@@IStMl Don't know if you're still alive but thank you. I needed those :)
Great examples and visuals. Very concise and no rambling
For the uniformly continuous counter-example, it would be nicer if you kept both the epsilon and delta fixed and moved the blue region closer to the y-axis and picked two points on the curve that are in the blue region but are clearly not entirely in the red region.
Before finishing wathcing this video I didn't believe that this short 5 min video could actually help me but I was soooo wrong. Thank you so much.
This video has makes my understanding better of continuity. Very good video that's makes everyone impressed.
Great explenation! Really helpful for my upcoming calculus exam!
Awesome Video! Very clear explanation of the use of Delta-Epsilon in the context of uniform continuity, and the counter example added even more clarity
🙏🙏🙏🙏😊😊😊 thanks sir for making me visualise through graphical meaning about continuity and uniform continuity.🙋🙋👌👌
You saved me a weeks worth of frustration my friend. Bless you!
Thanks for explaining dont know why lecturers make it so hard to understand
This video was very well-done and very helpful. Thank you for your hard work on this.
Thanks...it made it a lot clearer for me.
This is how to teach real analysis.....
Anyone can solve question but the real task is to understand the hidden geometry..... 😀😀😀👐👐👐👐🙏🙏
I think there are more people who understand the concept, but can't write a proper epsilon-delta proof than vice versa. But I do think videos like these are helpful.
Great explanation! It's a shame you don't upload any more videos!!!
Direct me to your altar. You saved my little U-grad life. Amen. 😎
Thank you. You helped me. Nice explanation and nice visuals. :)
After about 3 hours of nothing, I finally understand, thanks
Thank you so much. you explained it very Nicely in a crisp and concise manner.
Thanks for the video...😍😍😍
That was an amazing explanation. Thank you so much!
Thank you the visual explanation was so clear
Thats the best explanation 👌🙌
Best explanation that I have seen
Very clear when showing it with graphs!
Outstanding
very good visuals and animations
very nice visualisation
Thanks sir you explained in really well
Awesome intuitive explanation. And brief. Thanks!!!
yes exactly🤗😊
Great explanation!!!
Very good video, thank you so much
Thanks a million! 十分感谢！
This is excellent, thank you, sir
Excellent work. Thanks.
Great explanation, thank you. But I have a doubt: In function 1/x if we take interval [0.5,1] then we could apply Heine's Theorem because the function is continous in [0.5,1]. But then the function would be uniformly continous within [0.5,1]. However how is it possible if it is not uniformly continous?
The function is uniformly continuous on [.5,1]. If you extend your interval to (0,1], or (0, anything], that's when you lose uniform continuity.
Emsie emstraba this function is uniformly continues only in a certain interval but if talk about whole real lime then function is not uniformly continues
This one is really good, would be better if there are some example, theorems and application, could be a great lecture.
This is AWESOME! It helps me to understand totally!
Thank you for the video. A nice little refresher!
Just to double-check -- uniform continuity is informally verified by checking that for a given pair of points (x,y) it is true that |x - y| < delta => |f(x) - f(y)| < epsilon, correct? I spent years thinking it was the other way around (), and couldn't figure out why my verifications never added up.
good explanation.thank u so much
The concept is Seriously now understood by me. Thanks for uploading this video. It would be more better if you sound a bit slow. :') Thanks Ya!
Thanks. This video indeed helped me.
Thank you! Very clear explanation
THANKS MAN
I AM FORM TAIWAN
@@js7564 You mean western Taiwan?
This clarifies everything!
thank you very much!! very good explanation
Great explanation!
Great explanation, thanks!
In the last statement, that the function f(x)=1/x, what gurantees us that indeed, there is no delta we can find in the second points that you've mentioned?
There is something missing here, because f(x) =1/x is uniformly continuous for all x>1.Then how you're gonna fix the epsilon such that the corresponding delta will work for the whole domain.Because we know by video animation that it won't work
But if you choose the delta appearing at 4.27 the definition still holds true: you'll have values within epsilon for all other (x,y) throughout 1/x on the right that point if you look at the graph, for example where you defined your former delta
So, can we say that if the slope of a function is bounded below a certain value, then the function is uniformly continuous?
Thanks for such clear explanation :)
i cant thank you enough for this vid....i can only say i am pleased
So im guessing that its true then that no exponential function will be uniformly continuous? Or any line that has a curvature?
Thx alot, makes it very clear
Thank you ❤❤❤
Ur my fucking hero. Thx for saving me many hours for analysis class.
Really helpful, thank you :)
THAANK YOU SO MUCH SIR !!!
oh man this is helpful. thanks dude.
Great Video!
little bit confusion is there ....didnt understand dat which one we hav to choose first epsilon ...delta ??
I THINK FIRSTLY WE NEED TO CHOOSE EPSILON
We choose epsilon first as delta depends on epsilon
You sir, are amazing
The best of all
yeah that was awesome ... amazing explanation
So i) continuous means one of the points is fixed, and this works when you consider each point in the domain in turn as a fixed point (the maximum delta can be different for each point), ii) uniformally continuous means there is a single maximum value of fixed delta for which point i) is true. So ii) is a stronger condition.
That's correct. Uniform continuity is stronger. (1) Continuity means "for all epsilon, for all x, there exists delta(epsilon, x) such that for all x_0, d(x_0, x) < delta ==> d(f(x_0), f(x)) < epsilon" (2) Uniform continuity means "for all epsilon, there exists delta(epsilon), such that for all x and for all x_0, d(x_0, x) < delta ==> d(f(x_0), f(x)) < epsilon". Notice the difference between the two is that the quantifiers "for all x" and "there exists delta" have been switched. This means that in the definition of continuity, delta is a function of both epsilon and x, whereas in the definition of uniform continuity, delta is only a function of epsilon alone. So, every uniformly continuous function is continuous, because to find delta(epsilon, x), you can just take the delta(epsilon) guaranteed by the definition of uniform continuity.
thank you............. so much...........
how come this is in AUTO & VEHICLES category?
Very clear thanks
Woah. Thankful for days!
how about f(x) = x^2, I never understood this kind of math logic. there is no delta for all the points in x^2. each delta makes each epsilon ^2, delta must become smaller the further right you go. i think mathematicians created this epsilon and delta things to mock us.
this video has a good visual representstion than others, i probably made one step closer to get to know the delta and epsilon thing.
I don't get one thing, the definition says: "For every epsilon there exists a delta...", not "there exists a delta such that for every epsilon..."
Diego Mathemagician first you choose an arbitrary epsilon.
in the last exampl the delta will have to tend to 0 when we get closer and closer to the X axis and there is where we have the contradiction
very helpfull thanks :-)
Thank you.
well explained
Great❤
No more videos?
Thanks, I understand
Thanks man
very clear
Thanks!
LIFE SAVER
Thank you so much :)
Thank you
very nice
Where's the rest of the video?
thank you!