Rolle’s Theorem Proof
Вставка
- Опубліковано 20 вер 2024
- In this video, I prove Rolle’s theorem, which says that if f(a) = f(b), then there is a point c between a and b such that f’(c) = 0. This theorem is quintessential in proving the mean-value theorem in Calculus. Along the way I prove Fermat’s theorem, which says that if f has a maximum/minimum at a point c, then f’(c) = 0
Dr Peyam proves that English is a tonal language.
This is trivial
-This post was made by Lagrange's GANG
this is one of those things that feels so obvious it doesnt seem like it should need to be proved
I think the point is to prove that the definitions we gave for continuity and the derivative necessarily have the expected properties. If we used a definition that gave a derivative at 0 for |x|, for example, it wouldn't be true.
@@iabervon but the definition for derivative came from the geometric/graphical interpretation from smooth continuous functions specifically, so i cant imagine how it wouldnt have had this property
@@nathanisbored Just because something feels right doesn't mean it's true. Building rigorous proofs for those intuitive ideas is the basis of maths after all.
What is clear and intuitive to you might not be so intuitive for others .... That's why we have proofs .... This is the nicest thing about math ....
@@moawyahabdulrahman8782 except that proofs are often not clear and intuitive either. they're logically consistent, but rarely try to appeal to intuition
This is amazing, as I have a differential calculus final later today, and we learned about Rolle's Theorem, but I always wondered the proof for it.
Adding to this, I should point out that Fermat's Theorem says that If f(x) has a local minimum at x=a, then f'(a)=0 or f'(a) doesn't exist. However, Rolle's Theorem has an additional stipulation, that f(x) must be differentiable on (a,b), which covers the "doesn't exist" portion of Fermat's Theorem.
Does this work, if f(a) is not equal to M or m? What f(a) is a medium value?
2:31 And this is what mathematicians use everyday to stay fit.
Hahaha
Just want you to know your videos are helping an Indian college student. a lot. wish my college professors were like you. THANKS AGAIN, STAY SAFE.
I just love you man. Here in Germany you have such a good repute!
The proof really helped me out understanding the subtle concepts. You’re the best. But I still have two problems bothering me regarding rather minor issues first, @ 9:27 where you starts mentioning about inequality for the numerator f(c+h)-f(c)=0, why is there equal sign? As far as I know, shouldn’t it be h>0 according to the definition of limit as h is never zero because h is approaching 0 but never zero itself? I really appreciate your superb proof and I’ll find it truly grateful if you could answer my question. Have a great day!❤
2:48
Hi Dr. Peyam :)
Btw great video
This seems super obvious when looking at a graph, but if u were doing or reading some function with no immediate access to it's graphical representation, this theorem would prove useful in that regard cause it'll help you intuit what the problem is saying
Isn’t it just a special case of the theorem that there exists a c in R with f‘(c)=(f(b)-f(a))/(b-a)
Yes but you need Rolle’s to prove the theorem that you mentioned
@@drpeyam you can use the MVT to prove Rolle's Theorem and vv. Thus they are equivalent. A logical fact worth mentioning.
well Dr. Peyam
wanna tell u about this discovery
mitchel rolle was quit curious about intermediate velocity of a particle tracing a undistorted path usually if someone ask then we gonna say oh yess its avg. speed was |X2 -X1 |/ t2 - t1 ok but what ab8 its velocity ,speed while covering its journey then he had gone through basics of algebra and connect with the defination of velocity v(t) = X'(t) , t €(t1 , t2)
0< v(t) < |x2 - x1| / a (t2 -t1)
0< a
can you help me.
We consider (F) to be a numerical function.
- When the derivative of (F) is greater than or equal to 0, we say that the function (F) is increasing
- When the derivative of (F) is greater than 0 & not equal to 0, we say that the function (F) is strictly increasing
- When the derivative of (F) is less than or equal to 0, we say that the function (F) is decreasing
-When the derivative of (F) is less than 0 & not equal 0 , we say that the function (F) strictly decreasing
Our teacher said that there are exceptional cases where the derivative of (F) can be equal 0 at the ending points. than we say that the function is strictly increasing (or, strictly decreasing )
I really didn't get a good idea about this. So I want an example of this exceptional case ... I hope that you will make a video to explain it accurately ... Thank you and may God protect you, Dr. Peyam
This is great. Like the ham sandwich theorem. I prefer a ham sandwich on a rolle with butter, mayonnaise, and mustard, with a big stack of ham.
Yum, now I’m hungry! 😋
Here is my proof (though not rigorous, I have never heard of Rolle's theorem before) :
There are two cases for the continuous function f where f(a) = f(b) and we'll proceed by showing the two cases are equivalent,
CASE 1: f'(a) is positive if f'(b) is negative or vice versa
CASE 2: f'(a) and f'(b) have same sign i.e, they are both positive or both negative. In this case it is quite obvious that there must exist some x = d where d < b such that f(a) = f(d) and f'(d) is negative if f'(a) is positive or vice versa. This is simply case 1.
Consider the points a,b again. It is possible to choose a and b such that they are arbitrarily close to each other and f(a) = f(b), i.e f(a) = f(b) and a - b can be made arbitrarily small and so if a - b ---> 0 then f'(a) - f'(b) ---> 0 (because, from a, if we go to the right the slope increases but from b if we go left the slope decreases) but if a and b can be made arbitrarily close to each then there must a exist a point c to which a and b are approaching to (since f is continuous), where f'(c) = f'(a) - f'(b) = 0 (where, c = a = b) and that concludes the proof. I am not going to watch the video until I get a reply...
Am I right? You hearted my comment...Woo!
I'd like to see more of these proofs videos .
Why is there a max AND a min on [a,b] when there will be a max OR a min?
The second statement is consistent with the first, but since "max AND min" is stronger and also true he went with that. Keep in mind [a,b] is a closed interval so an extremum might be at an endpoint.
I hate math but I love to watch Dr Peyam's math videos
I’m Japanese and I wanna make Japanese subtitles if there English one
But I don’t know whether I do it...
Just enabled translations, so maybe you can try it out now
"Not to be confused with Fermat's last theorem"
Instructions unclear, finished proof using Fermat's little theorem 🤔
Somewhere down this path, the world stops. And I end up back where I was.
So deep, haha!
Is that a flat or round world? ;)
@@madhuragrawal5685 I don't think it matters.
Does Extreme Value theorem have to be proven as well?
LUB Property→Monotone bounded convergence→Bolzano-Weierstrass→EVT→IVT→Rolle's Theorem →MVT, Cauchy's MVT→Integral MVT
Yes, EVT is needed for Rolle
@@duckymomo7935 lub? Could you expand that please?
@@madhuragrawal5685 It seems that the order provided is necessary to prove each theorem. So the Least Upper Bound property must be proven first
@@ThePharphis yeah, I just didn't know what lub stood for. I'll look into it on my own I guess, now that I know what it's called
@@madhuragrawal5685 I didn't either, but it came up on a google search!
hey dr peyam! Im a 2nd year student and in my linear algebra course im trying to show that the infinite dimensional vector space R^infinity = {(x_1,x_2,...) | x_i in R} has a basis which is not countable. Could you do something on these infinite dimensional vector spaces? i find them fascinating. thank you and i love your videos :)
Not true though, the basis is countable, check out my video on infinite dimensional space for that
@@drpeyam I have just had a look at the video. Are you claiming that {e_1,e_2,...,e_n} is a basis for all n implies that {e_1,e_2,...} is a basis for R-infinity? because that's simply incorrect. consider (1,2,3,4,.....), no finite linear combination of e_i can represent this vector in R-infinity(there's always trailing zeros). If not that then i'm unsure of how the basic can be countably infinite? Furthermore what would you suggest a basis for R-infinity is?
wwebadgerse It depends on how you define a basis, but the standard definition is the one of a maximally linearly independent set, and the one I mentioned is maximally linearly independent (ie adding any other vector to it makes it linearly dependent)
Can a Maximum and a Minimum both exist within (a,b) , and if so would it have any consequences?
yes, they can, but the main part of is to prove that derivative in max or min is zero
Brilliant
If we got an inflexion point as a maximum, we couldnt derivate on that point so the conditions arent suffisant
Theorem only applies to differentiable functions. Otherwise it makes no sense.
ich bin vollkommen von der Rolle
You can't hide from me!
I mean its like the most obvious thing ever, but I guess if you ask the question "Why is it so obvious?" well here is the simplest answer I can give - The function is continuous. The first derivative of a is negative if b is positive or vice versa. If we go from a to b, the first derivative (slope of tangent) goes from positive to negative or vice versa. In both cases we have to cross a point where the slope is 0 (because the function is continuous) if we go from negative to positive or vice versa. That's it...I think xkcd made a joke about this theorem once...
Did you consider the trivial case where f is constant on [a,b]. then for all x in the interval f(x)=f(a)=f(b). But the derivative of a constant function is 0. Thus pick any c in the open interval and the derivative will be zero.
Dr. Peyam...gracias por este y todos sus videos.
Video is 13:37 long ELITE 😆
i would've liked but i dont wanna disturb the 666 likes atm.
this guy always starts the video by saying "thanks for watching" and I'm like "wtf, are you kidding me? THANK YOU for making this video"
OMG I love your accent ♥️🤣
me too. do you know where it's from?
My name is Min but you didn't say hi :(
Hi!!!
nicee
u bring me to past time, calculus basic and calculus I. Im so cringe lol
The Comic Sans is terrible, but the cool video length makes up for it. 🤘
Hahaha
Brilliant