i love this channel! Who says summer has to be enjoyed away from school? Not me! Math is the best!
@blackpenredpen6 жыл бұрын
Nole Cuber what an awesome comment! Thank you!!
@l3igl2eaper6 жыл бұрын
I'm taking Calculus 4 and Diff EQ this summer. I can't get enough math! hahaha
@asriel5222 жыл бұрын
"That nerd"
@jonathangrey63542 жыл бұрын
@@asriel522 this is indeed very correct
@napv79226 жыл бұрын
I like the fact that it makes so much sense when you are also thinking on the derivative as a linear aproximation, you get the aproximation precisely in the same way, and basically when there is only one posible outcome for that aproximation, you have differetiability.
@slowfreq6 жыл бұрын
You might like to know about the Weierstrass Function, a function which is continuous _everywhere_ but differentiable _nowhere._
@blackpenredpen6 жыл бұрын
armin yes! I studied that back in my grad school life. : )
@gregoriousmaths2664 жыл бұрын
He said that the converse WASNT true
@francescocostanzo82252 жыл бұрын
@@gregoriousmaths266 and he gave a real nice example
@aeoexe6 жыл бұрын
6:38 Actually you are using differentiable at this point right now, because limit of product is equal to product of limit if both limit exists. For example lim sinx/x cannot be decomposed to lim sinx lim 1/x, just to make it rigorous.
@ZoneEEEEEEEEEEEE6 жыл бұрын
aeoexe Since it's given that f is differentiable at a, it's totally fine to split the limit because lim (f(x)-f(a))/(x-a) exists
@aeoexe6 жыл бұрын
That is what i am talking about, because he said limit of product is equal to product of limit at 6:38
@blackpenredpen6 жыл бұрын
aeoexe I should have also said "because we know both limits exit" right there too. But I did mention f'(a) is a "nice number" since we had differentiability
@jennaoneill77184 жыл бұрын
Thank you so much! Your explanation was so clear and concise, and your enthusiasm made it so enjoyable to watch! Thanks for being my quarantine lecturer!
@pyromen3216 жыл бұрын
I'm loving the high energy thumbnails!
@kyoxx12316 жыл бұрын
There exist an alternative definition of differentiable function: Let f:[a,b]→R, then f is differentiable in Xo if exist a function g:[a,b]→R continuous in Xo so that: f(x) = f(Xo) + (x-Xo) g(x), and g(Xo)= f'(xo). By that definition f is continuous in Xo because it's a sum of continuous functions: f(Xo) and Xo are constant numbers, x is continuous in every x and g(x) is continuous by definition. #YAY
@valqora2 ай бұрын
Beautifully done. You should make a college calculus course as you are excellent at going explaining your steps
@rohitg15296 жыл бұрын
The limit of a sum is not always the sum of the limit right? For example lim n->inf of ln(n)-harmonic sum till (1/n) converges. If you split up the sum into 2 limits, you get infinity-infinity, which is undefined. I think that you need to place the restriction that f(a) is defined. P.S. I know that if f(a) is not defined then the function is definitely not continuous at a, but still, you cannot split limits in general. Peace
@offthepathworks91715 ай бұрын
Thank you, best video on YT on this proof!
@albertodelaraza44756 жыл бұрын
The limit of a sum, is the sum of the limits. The same is true for the limit of products. My understanding is that this is true only if the functions inside the limits are continuous. (No?) But, we are trying to prove continuity in the first place. Can we prove continuity by assuming continuity first? It is clear that I am missing something important. Can you please clarify. I need this little clarification because I haven't studied calculus for almost thirty years and I don't remember much of it. (Love your channel, BTW. It brings back much of the love for math I had as a young person. So, thanks and keep up the good work.)
@stephenbeck72226 жыл бұрын
Alberto DeLaRaza you are close. The limit of a sum equals the sum of the limits, on the condition that the inside limits merely exist. Remember you can have the limit existing without being continuous. Also, as another comment pointed out, “the limit existing” means it must be an actual number and not infinity, which is why you can’t say lim sin x / x = lim sin x * lim 1 / x (as x -> 0).
@albertodelaraza44756 жыл бұрын
Ah, I see my error now! Thank you for the clarification!
@hugocoolens6 жыл бұрын
"The limit of a sum equals the sum of the limits": shouldn't we all so add the condition "for a sum consisting of a finite number of terms"?
@JoshuaHillerup6 жыл бұрын
I suddenly find myself very curious about how to find the area of a fractal, as in the integral of a continuous function that is not differentiable anywhere.
@Theraot6 жыл бұрын
For what I read, Daniell integral should do. With that said, I do not claim to understand it.
@JoshuaHillerup6 жыл бұрын
Alfonso J. Ramos yeah, I know it's possible, I just don't understand how to find it.
@farissaadat44376 жыл бұрын
If I'm not mistaken, the exact area of something that is much more complicated like the mandelbrot set is an open problem. I think the area of something simpler like a kock snowflake can be written as an infinite series quite simply, then some work can be done to find the value.
6 жыл бұрын
Awesome proof, it finally made sense to me :) I know abs(x) at x=0 is the counter-example but can you also show in general that "if continuous then differentable" is a false statement or it doesn't lead to any conclusions?
@alegian79346 жыл бұрын
This proof was taught to me in school. So I won't watch till the end, but I still liked and commented :) beautiful proof
@Metalhammer19936 жыл бұрын
man this shows why i never got proofs. whenever i read them i tend to forget what i wanted to show in the first place or what the proof is supposed to show^^
@pierreabbat61576 жыл бұрын
Can you explain why f'=0 implies f is constant if the domain is the reals, but not if the domain is the p-adic numbers?
@pixelpix17286 жыл бұрын
The way I'd prove this is by saying: "If you take the derivative of a function in a point *a*, you will have the slope the line tangent to the function in that point, but you can only have the slope of line tangent the function at that point if, and only if, the point *a* exists in the function (if it is continuous at a), since you can't have the slope of nothing" For example we have the famous function sin(x)/x, we can clearly see it's not conitnuous at 0, so that point doesn't exist in the function. It's derivative is (xcos(x)-sin(x))/x^2 , and you can see that the slope of the tangent at 0 is 0/0. sin(x)/x it's not differentiable at 0 because the point 0 doesn't exist in that function! :3 Is my proof right? QwQ
@hanztimbreza62176 жыл бұрын
sapato :3 Although that is a nice graphical way to think about it, I prefer the more rigorous (and conceptually simpler) proof presented in the video.
@pco2466 жыл бұрын
Is it rigorous to say that the product of a derivative and an infinitesimal quantity is always 0?
@meerable2 жыл бұрын
hm.. |0| = 0. But for f(x)=|x| df/dx at 0 don’t exist?
@francescocostanzo82252 жыл бұрын
Does this count as a real analysis question?
@Ready4Music6 жыл бұрын
Clickbait: Table → Content. 😁
@09stir6 жыл бұрын
Can you help me with the following integral? Integration of x^2 * √(1+4x^2) dx It seems to me you can use integration by parts or something like trig substitution. However i get different results when doing so
@harisrasheed47026 жыл бұрын
integration of 1/x^2 + 3x ?
@timperry69486 жыл бұрын
Great video. Thanks again.
@willful7596 жыл бұрын
Great video as always
@teacherjn20245 жыл бұрын
Hello sir can I have a favor? do you lecture notes for Differentiability in R^2?
@ugursoydan81872 жыл бұрын
thank ypu so much
@WhiteboardMaths6 жыл бұрын
Oh man this came up in my analysis exam two months ago :0
@user-md2rv6cf4j6 жыл бұрын
You are amaizing thank you
@geekjokes84586 жыл бұрын
What was that ending? Vihart is taking over your channel?
@LUKAS-bb4jc2 жыл бұрын
Wait at the end it looks like the tangent formula
@nicholaslau31946 жыл бұрын
Isn't a differentiable function by definition continuous?
@jay_sensz6 жыл бұрын
Yes, and this is the proof.
@helloitsme75536 жыл бұрын
Not by definition, the definition is that when you get closer to a certain x-value , the y-value you are approaching is the same as the value at the specific point of x. If that's true for all x then it's continuous. A differentiable function is a function for which all x-values the derivative is defined
@Geo25rey6 жыл бұрын
When you record a snail you find in your backyard
@user-zb6od8nt4u6 жыл бұрын
Can you solve this please? X^lg(X-2)=1000
@NoNameAtAll26 жыл бұрын
Тарас Заблоцький Log as lg (base 10) or as ln (base e)?
@user-zb6od8nt4u6 жыл бұрын
NoName lg base 10 ,beacouse in spain we use log to say base 10
@alexdarcovich93496 жыл бұрын
What happened to the video where you found the limit of infinite square roots of 0? #yay
@blackpenredpen6 жыл бұрын
Alex Darcovich I will remake it
@qntnfeynman3602 жыл бұрын
Nice
@KeyMan1376 жыл бұрын
Will you discuss how the converse is not true and mention the Weierstrass function? en.wikipedia.org/wiki/Weierstrass_function
@AndDiracisHisProphet6 жыл бұрын
"I don't know what a number is..." Are you sure you are the right person to make math videos?
@AndDiracisHisProphet3 жыл бұрын
@@christinabae2625 Thanks, anti-joke-chicken
@AndDiracisHisProphet3 жыл бұрын
@@christinabae2625 no problem. this comment is so old i don't even remember what i commented. But BPRP and I have some (friendly) jabs at each other from time to time. Ok, to be honest it is me that does all the jabbing, but whatever
@matheus_rml6 жыл бұрын
is there any other example of continuos functions that are not differentiable at some ponit?
@juberto_93836 жыл бұрын
Matheus Ramalho For example f(x)=x^(1/3) is continuous but not differentiable at 0.
@uchihamadara60246 жыл бұрын
abs(x) is continuous but not differentiable at 0
@aeoexe6 жыл бұрын
I am here to blow your mind. en.wikipedia.org/wiki/Weierstrass_function
@stephenbeck72226 жыл бұрын
Here's a nice little set of functions. f1(x) = sin(1/x). f2(x) = x*sin(1/x). f3(x) = x^2*sin(1/x). Assume for each that the function is actually piece-wise defined where f(0) = 0 in order to "plug" the hole at 0. f1(x) is not continuous at x=0 because the function oscillates between 1 and -1. f2(x) is continuous but not differentiable at x=0 because the tangent line slopes also oscillate. f3(x) is differentiable at x=0 and every other point, however the derivative is not continuous, i.e. it is not second-differentiable.
@himanshumallick22696 жыл бұрын
Lab glasses??
@JamalAhmadMalik6 жыл бұрын
#yay!❤
@alejrandom65923 жыл бұрын
I want a tshirt that says d'able
@anandunambiar10322 жыл бұрын
Sir, I don't get the proof..you just proved lim x → a f(x)=f(a)....where is the differentiability here?.. we are asked to prove if a f is differentiabie it is continous
@jascarkamasonskaja73146 жыл бұрын
8:44 it’s me on math classes))))))
@helloitsme75536 жыл бұрын
Does continuity also imply differentability?
@japotillor6 жыл бұрын
HelloItsMe no
@geometrydashbayve50046 жыл бұрын
As explained in the video, a counterexample is f(x)=|x|. This is a continuous function, yet is not differentiable at x=0.
@nejlaakyuz40256 жыл бұрын
Weistrass function,
@user-vm6qx2tu3j6 жыл бұрын
#yayyyyy
@15schaa6 жыл бұрын
You should've put another minute or two of that snail to get to 10 minutes. Do you not watch PewDiePie?