In this video, I give a proof of the mean-value theorem in calculus, by reducing it to a special case of Rolle’s theorem. Featured at the end are also some bloopers, enjoy!
I like how you brought up the physics example because I think it makes the idea quite intuitive for me. If a car goes from A to B in some time t then the average speed can be calculated and this theorem says that at some point between A and B the car must be travelling at the average speed because the car is either travelling at the average speed constantly on the interval [A,B] or at some speed below or above the average at different points on the interval making the average somewhere between the lower and higher speed. At least that is my physics student understanding of this theorem haha
Nice theorem, I always knew it as Lagrange's Theorem. When, long long time ago, I studied it, it was easily proved using Cauchy's Theorem (using g(x) = x), that is instead proved using Rolle's Theorem (that is the deep reason why critical numbers are found setting the first derivative to zero). By the way I recently found a collection of Analysis 1 final exam tests given in my university in Engineering, Physics and Mathematic between 1977 and 1992 (I've done one of these in 1989, yes I'm so old...). Inside the collection there's the following cute test on Lagrange Theorem. Using Lagrange's Theorem proof that Ln(2+sin x) is uniformly continuous in R).
It is really interesting and wonderful thing about math. You can approach different ways and both of them leading you the same place. Like Bolzano Theorem is a special case for IVT, while IVT can be proven using Bolzano, of course one needs to prove Bolzano via set theory and stuff. Also IVT can be proven using the concept connectedness which is really the shortest proof of IVT I have ever seen.
I’ve seen a version that lets you skip a few lines by simply letting g(x) be f(x) minus the linear function that not only has that specific slope but also passes through points (a,f(a)) and (b,f(b)). Then f(x) and our special linear function share values a and b so the difference is zero for both, and rolles automatically applies!
Replacing x by (x - a), in the g(x) makes proof easier to calculate. g(x) = f(x) - [(f(b) - f(a)) / (b - a)] · (x - a) Great video as always! ε> I am super thankful to dr πm for not only clear expositions, but also for liveliness!
Hey, i have similar math expression as f(x+y) = f(x)*f(y) - to find all f,g :R → :R g(f(x) - y) = f(g(y)) + x. Its more difficult to solve it because it have function composition. For you, I think it will be really interesting to solve it!
I guess one can use intermediate value theorem as well . if f is continuously differentiable in (a,b) then there should exist some f' in between f(b) and f(a) so should exist the average value f(b)-f(a)/(b-a) for some c in (a,b) .
When initially you take the statement g(x) = f(x) - ((f(b)-f(a))/(b-a))x, aren't you taking a condition other than the initial conditions mentioned in the statement of the theorem? By this I mean that when you get the final result, you are getting in on the basis of the conditions of the theorem statement PLUS the statement g(x) = f(x) - ((f(b) - f(a))/(b-a))x , which is something we are NOT trying to prove (we are trying to get the result on basis of the conditions in the theorem-statement alone) I know I must be wrong, but wrong exactly where?
I always imagine it this way. If f is a differential and continuous function between a and b and if m is the average change then I would set m to zero, i.e tilt the graph so the line joining a and b becomes parallel with the x axis (i.e, slope or m = 0). Now we can use rolle's theorem to show that there exists a point c between a and b whose slope (or m) is zero. Tilt the graph back to its original form and we have a point c at which the tangent is parallel to the line joining a and b.
This is Larrange theorem in calculus The Larrange theorem in algebra states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G ( |G| = |H|.[G:H] )
Really very nice and clean proof of MVT. Hopefully at some point in future you can upload some videos related to differential geometry and topology. Physicists would appreciate it very much :D.
For example, f(x)=x*sin(x) for x in [0,2.028…], f'(0)=f'(2.028…)=0 but f(0)=0 and f(2.028…)= 1.819… So the IVT only guarantees f'(c)=0 for some c in (0,2.028…) when we need f'(c) to be 2.028…/1.819…
I mean I don’t see how IVT would help, but maybe there might be a way? Also IVT requires f to be continuous at the endpoints, so if you want to do IVT on f’ you have to have that f’ is continuous at a and b as well
@@drpeyam yeah, I noticed this after a little bit of thought. I was thinking that it could work for class C1 functions who's derivative is constantly increasing/decreasing.
Thanks. En français on parle de théorème des accroissements finis. For the final with the different repetitions on parle de bêtisier (bêtises, erreurs, stupidités...)
Rolling in the deep!
Astronaut in the ocean reference
Note to Students : This should not be confused with the theorem practised by Car Dealers when they give you a Trade-in price!
I like how you brought up the physics example because I think it makes the idea quite intuitive for me. If a car goes from A to B in some time t then the average speed can be calculated and this theorem says that at some point between A and B the car must be travelling at the average speed because the car is either travelling at the average speed constantly on the interval [A,B] or at some speed below or above the average at different points on the interval making the average somewhere between the lower and higher speed. At least that is my physics student understanding of this theorem haha
Ah yes Lagrange's theorem, Rolle's big Brother
5:56 That sounds similar to how the pythagorean theorem is a special case of the law of cosines, but we use it to prove the law of cosines!
Needed a quick reminder about this topic. Fantastic video.
Nice theorem, I always knew it as Lagrange's Theorem. When, long long time ago, I studied it, it was easily proved using Cauchy's Theorem (using g(x) = x), that is instead proved using Rolle's Theorem (that is the deep reason why critical numbers are found setting the first derivative to zero).
By the way I recently found a collection of Analysis 1 final exam tests given in my university in Engineering, Physics and Mathematic between 1977 and 1992 (I've done one of these in 1989, yes I'm so old...). Inside the collection there's the following cute test on Lagrange Theorem.
Using Lagrange's Theorem proof that Ln(2+sin x) is uniformly continuous in R).
you have been saving me this semster!!!
You are an amazing teacher as always. Love your videos 🔥🔥🔥
It is really interesting and wonderful thing about math. You can approach different ways and both of them leading you the same place. Like Bolzano Theorem is a special case for IVT, while IVT can be proven using Bolzano, of course one needs to prove Bolzano via set theory and stuff. Also IVT can be proven using the concept connectedness which is really the shortest proof of IVT I have ever seen.
Thanks..I loved u r explanation 😍😍😍😀🙏
What about the mean value theorem for integrals....I would like to see that proof!
Dr. Peyam’s Super Awesome French World Cup Victory Special ua-cam.com/video/tro1H0xbDJo/v-deo.html
I’ve seen a version that lets you skip a few lines by simply letting g(x) be f(x) minus the linear function that not only has that specific slope but also passes through points (a,f(a)) and (b,f(b)). Then f(x) and our special linear function share values a and b so the difference is zero for both, and rolles automatically applies!
Right, the linear function that you’re mentioning is the one in the video
Great as always. Can you upload a video on how to prove rolle's theorem?
Sure!
There you go: Rolle Theorem Proof ua-cam.com/video/QvcaTltKHVg/v-deo.html
Replacing x by (x - a), in the g(x) makes proof easier to calculate.
g(x) = f(x) - [(f(b) - f(a)) / (b - a)] · (x - a)
Great video as always! ε>
I am super thankful to dr πm for not only clear expositions, but also for liveliness!
Hey, i have similar math expression as f(x+y) = f(x)*f(y) - to find all f,g :R → :R g(f(x) - y) = f(g(y)) + x.
Its more difficult to solve it because it have function composition. For you, I think it will be really interesting to solve it!
I guess one can use intermediate value theorem as well . if f is continuously differentiable in (a,b) then there should exist some f' in between f(b) and f(a) so should exist the average value f(b)-f(a)/(b-a) for some c in (a,b) .
But what you say is not an actual average, just a physical one, you’re confusing it with 1/2 (f(a) + f(b))
@@drpeyam true .
Thanks! Love your explanation sir!
this video made my day. thank you!
When initially you take the statement g(x) = f(x) - ((f(b)-f(a))/(b-a))x, aren't you taking a condition other than the initial conditions mentioned in the statement of the theorem? By this I mean that when you get the final result, you are getting in on the basis of the conditions of the theorem statement PLUS the statement g(x) = f(x) - ((f(b) - f(a))/(b-a))x , which is something we are NOT trying to prove (we are trying to get the result on basis of the conditions in the theorem-statement alone)
I know I must be wrong, but wrong exactly where?
you made it so simple. i literally wanted to roll over the floor.
I always imagine it this way. If f is a differential and continuous function between a and b and if m is the average change then I would set m to zero, i.e tilt the graph so the line joining a and b becomes parallel with the x axis (i.e, slope or m = 0). Now we can use rolle's theorem to show that there exists a point c between a and b whose slope (or m) is zero. Tilt the graph back to its original form and we have a point c at which the tangent is parallel to the line joining a and b.
yeah same thought
Does this work for any other function other than g that meets those conditions? why does using the function g cover all the cases if so?
This is Larrange theorem in calculus
The Larrange theorem in algebra states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G ( |G| = |H|.[G:H] )
*Lagrange
thank's for this amazing content .
I love these types of videos :D
Rolling… with two tangents each on one side and a secant in the middle… Now I am thinking about sandwich rolls.
Remember that time f(a) equaled f(b)? Oh wow, that was great. I almost died.
Doesn't setting g (x) to this special function result in loss of generality?
Nope
for any f, you can create the g. so it works for all f
f(my problem)= Solution to my problem = Search (key words of my problem + peyam)
Really very nice and clean proof of MVT. Hopefully at some point in future you can upload some videos related to differential geometry and topology. Physicists would appreciate it very much :D.
I don’t know much about those, sadly
Fematika has great videos about it though
@@drpeyam Thanks for your reply. I don't really like Fematika. I like your nice and refreshing style of teaching.
thank you so much
Could you use the intermediate value theorem to prove this?
If there were a function f such that f'(a)=f'b) but f(a)≠f(b), then you can't use the IVT on f' because it only guarantees f'(a) exists.
For example, f(x)=x*sin(x) for x in [0,2.028…], f'(0)=f'(2.028…)=0 but f(0)=0 and f(2.028…)= 1.819…
So the IVT only guarantees f'(c)=0 for some c in (0,2.028…) when we need f'(c) to be 2.028…/1.819…
I mean I don’t see how IVT would help, but maybe there might be a way? Also IVT requires f to be continuous at the endpoints, so if you want to do IVT on f’ you have to have that f’ is continuous at a and b as well
@@drpeyam yeah, I noticed this after a little bit of thought. I was thinking that it could work for class C1 functions who's derivative is constantly increasing/decreasing.
Ana 1 practice exercise?
Thank you!
☺☺ blessed
damn
Thanks. En français on parle de théorème des accroissements finis. For the final with the different repetitions on parle de bêtisier (bêtises, erreurs, stupidités...)
Daset dorost dash ajab clipee.... Yeki vase complex analysis va maximal analytic continuation Doros kon age mishe .. namely obtaining a Riemann surface via analytic continuation
Yaay first view :)
Who disliked this and why? :O
I don’t know, there’s someone who keeps disliking all my videos 😞
@@drpeyam well I like your videos a lot :D and so do most people :D so whatever xD
Okayyy u mite say wats goin on jus wait for it we’ll see wats going on
Hi peyan mean by Tamil😅😂
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