For the example at the end concerning infinitely many discontinuity of the derivative, visualize a function that resembles the behavior of the near-zero part of x^2sin(1/x) at integer points. Copy-and-pasting a part of x^2sin(1/x), I realized after making the video, does not quite work because it is an odd function and there will be problem of whether the function will be differentiable at the boundary between two equal parts (even if we change the interval we copy-and-paste). However, the intuition still stands.
yes, and specifically it is possible to let these intervals "(-2,2)" be smaller and smaller!!. Ex copy it onto [1/2,1], and [1/3,1/2), and [1/4,1/3) and so on. Then the derivative has *infinitely* many discontinous points on a *bounded* interval (0,1]!
At 5:50 ....acc to the definition of derivative we are getting a finite value of derivative of the function at x=0 but by the usual method we are not getting so as limx→0(cos 1/x ) isnt defined .....why such thing is happening????
Please tell very important here if going through first derivative principle f'(0+)=0 but differentiating the function gives that f'(0+) is osscilating,, can we say that first principle dont give us derivative at f(0+)
The cool thing is that if you change the x^2 to x^3, then it is not a counterexample. I’m still trying to find an intuitive reason for this difference.
If you look at the graph of the counterexample, you can see that the derivative has magnitude 1 every time it crosses 0 (except at 0), and opposite signs each time. The function stays continuous by going less far each time as you approach 0, but this doesn't help the derivative. With x^3, the closer crossings are also flatter, so the derivative can be continuous.
Yep. I found it helpful a while back to consider the chain of functions, starting with f1(x)=sin(1/x) like he does in the video, and as discussed f1(x) does not have a limit as x approaches 0. Then you have f2(x) = x*sin(1/x), which does have a limit at x = 0 (from here we can make the piecewise function with f(0)=0 to make it continuous) but does not have a derivative. Next is f3(x) = x^2*sin(1/x) which has a first derivative that itself is not continuous. And continuing, f4(x) = x^3*sin(1/x), a function with a discontinuous 2nd derivative. And we could keep going on.
From the definition of the derivative (like he did in the video), the derivative at x=0 would actually be f'(0)=lim h->0 ( f(0+h)-f(0) )/ (0+h-0)= lim h->0 (h-0)/h=1
Yoav Shati Your function is no different from f(x)=x, therefore it's obvious that the derivative is 1 everywhere, as shown more formally in other comments.
That's a common mistake in peace wise functions, when you have a single point defined like f(a)=b people assume that the function is constant or a horizontal line at that point (a,b). I made this mistake a few times and it really feels satisfying when you find out why what you did is wrong and understand the subject more.
Some functions can be continuous, differentiable and have continuous derivative everywhere, whereas other functions (like this) don't. You can't prove a certain property is true/false *sometimes*, only that it's ALWAYS true/false. For the *sometimes*, all you can find are examples where it's true and examples where it's false.
It is interesting that the derivative function is only discontinuos at one point and bounded and hence the derivatice function is riemann integrable for every interval [a, b], and in particular if you integrate the derivative over [0,x] you will just get f(x). But if you modify the function a little bit this doesnt have to happen: g(x) =x^2 sin(1/x^2) at x nonzero and g(0)=0 is differentiable function whose derivative is not riemann integrable over any [a, b] that contains 0
For the example at the end concerning infinitely many discontinuity of the derivative, visualize a function that resembles the behavior of the near-zero part of x^2sin(1/x) at integer points. Copy-and-pasting a part of x^2sin(1/x), I realized after making the video, does not quite work because it is an odd function and there will be problem of whether the function will be differentiable at the boundary between two equal parts (even if we change the interval we copy-and-paste). However, the intuition still stands.
yes, and specifically it is possible to let these intervals "(-2,2)" be smaller and smaller!!. Ex copy it onto [1/2,1], and [1/3,1/2), and [1/4,1/3) and so on. Then the derivative has *infinitely* many discontinous points on a *bounded* interval (0,1]!
How does this cutting-pasting concept work to show that f' is discts. at infinitely many points on the real line? I didn't get it at all. :(
Please make a video on (x - 3)(x - 1)|(x - 3)(x - 1)|
Hands down the best Video of reql Analysis on RUclips. LetssolveMathProblems is the master of Mathematics.
This is the standard example in the calculus class I took
At 5:50 ....acc to the definition of derivative we are getting a finite value of derivative of the function at x=0 but by the usual method we are not getting so as limx→0(cos 1/x ) isnt defined .....why such thing is happening????
How will the function change if instead x²sin(1/x) it will be x²sin(1/x²)?
Please tell very important here if going through first derivative principle f'(0+)=0 but differentiating the function gives that f'(0+) is osscilating,, can we say that first principle dont give us derivative at f(0+)
The cool thing is that if you change the x^2 to x^3, then it is not a counterexample. I’m still trying to find an intuitive reason for this difference.
If you look at the graph of the counterexample, you can see that the derivative has magnitude 1 every time it crosses 0 (except at 0), and opposite signs each time. The function stays continuous by going less far each time as you approach 0, but this doesn't help the derivative. With x^3, the closer crossings are also flatter, so the derivative can be continuous.
Yep. I found it helpful a while back to consider the chain of functions, starting with f1(x)=sin(1/x) like he does in the video, and as discussed f1(x) does not have a limit as x approaches 0. Then you have f2(x) = x*sin(1/x), which does have a limit at x = 0 (from here we can make the piecewise function with f(0)=0 to make it continuous) but does not have a derivative. Next is f3(x) = x^2*sin(1/x) which has a first derivative that itself is not continuous. And continuing, f4(x) = x^3*sin(1/x), a function with a discontinuous 2nd derivative. And we could keep going on.
Cool try it.
What about square root of x into sin(1/x)?? Please help me out.
Nice video. Loved it!
Couldn't you use {x at x=/=0 and 0 at x=0} as a counterexample?
The derivative is 1 for all x but 0, and it's 0 at x=0
From the definition of the derivative (like he did in the video), the derivative at x=0 would actually be f'(0)=lim h->0 ( f(0+h)-f(0) )/ (0+h-0)= lim h->0 (h-0)/h=1
We cannot because the derivative is 1 (NOT 0) at x = 0 as well. Note that f'(0) = lim as x->0 of (f(x)-f(0))/(x-0) = lim as x->0 of x/x = 1.
Yoav Shati Your function is no different from f(x)=x, therefore it's obvious that the derivative is 1 everywhere, as shown more formally in other comments.
That's a common mistake in peace wise functions, when you have a single point defined like f(a)=b people assume that the function is constant or a horizontal line at that point (a,b).
I made this mistake a few times and it really feels satisfying when you find out why what you did is wrong and understand the subject more.
tell pl me why are you looking at limx->0 , but nio limx->0- and x->0+ ? lim(sin1/x) x->0+ = 0
f(x) = { x sin 1/x, x is not equal to zero. { 0, x=0
Is this function continuous every where or continuous only at x=0 plz tell me
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Thank you so so so so much!! This video was so helpful!!!!
I'm ok that you showed an example, but next time could you prove it by using theorems? Thanks :)
Some functions can be continuous, differentiable and have continuous derivative everywhere, whereas other functions (like this) don't. You can't prove a certain property is true/false *sometimes*, only that it's ALWAYS true/false. For the *sometimes*, all you can find are examples where it's true and examples where it's false.
Using a counterexample to prove that something cannot be true is completely rigorous.
Nice, keep doing great work! 👍
Very good explanation, thanks 👍🏻
Wow i learnt so much from this video!!Amazing
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It is interesting that the derivative function is only discontinuos at one point and bounded and hence the derivatice function is riemann integrable for every interval [a, b], and in particular if you integrate the derivative over [0,x] you will just get f(x). But if you modify the function a little bit this doesnt have to happen: g(x) =x^2 sin(1/x^2) at x nonzero and g(0)=0 is differentiable function whose derivative is not riemann integrable over any [a, b] that contains 0
That is a cool insight I never considered. Real analysis is full of surprises. =)
You cant have “jump” incontinuity for a derivative. It will be continious or the limit wont exist
It is exactly what I want😍
Show that the function x^2+sin x is continuous in all real values of x
Show that if f and g are continuous then so is f+g. Then show that x^2 and sinx are continuous
Thank you!
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Sounds like Ronny Cheng
Yo
Easier example is modx lol
herroo i tanka iu for sorving dis probremu