Mean Value Theorem Proof

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  • Опубліковано 10 лис 2024

КОМЕНТАРІ • 64

  • @blackpenredpen
    @blackpenredpen 5 років тому +27

    Rolling in the deep!

  • @6612770
    @6612770 5 років тому +25

    Note to Students : This should not be confused with the theorem practised by Car Dealers when they give you a Trade-in price!

  • @jamiewatt6814
    @jamiewatt6814 5 років тому +10

    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

  • @danielescotece7144
    @danielescotece7144 5 років тому +11

    Ah yes Lagrange's theorem, Rolle's big Brother

  • @filippodifranco8225
    @filippodifranco8225 5 років тому +5

    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).

  • @veryveryveryangry
    @veryveryveryangry Рік тому

    Needed a quick reminder about this topic. Fantastic video.

  • @mfonpeter124
    @mfonpeter124 8 місяців тому

    You are an amazing teacher as always. Love your videos 🔥🔥🔥

  • @jessicasanchez134
    @jessicasanchez134 3 роки тому +1

    you have been saving me this semster!!!

  • @CaptchaSamurai
    @CaptchaSamurai 3 роки тому

    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!

  • @mertunsal7335
    @mertunsal7335 5 років тому

    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.

  • @MuPrimeMath
    @MuPrimeMath 5 років тому +4

    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!

  • @ramlaaal
    @ramlaaal 6 місяців тому

    you made it so simple. i literally wanted to roll over the floor.

  • @justinbishop54
    @justinbishop54 2 роки тому

    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!

    • @drpeyam
      @drpeyam  2 роки тому

      Right, the linear function that you’re mentioning is the one in the video

  • @h-67shreeshaa54
    @h-67shreeshaa54 3 роки тому +1

    Thanks..I loved u r explanation 😍😍😍😀🙏

  • @alexanderkalinin988
    @alexanderkalinin988 5 років тому

    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!

  • @marianarose8657
    @marianarose8657 3 роки тому

    this video made my day. thank you!

  • @ghifarazka4981
    @ghifarazka4981 2 роки тому

    Thanks! Love your explanation sir!

  • @remlatzargonix1329
    @remlatzargonix1329 5 років тому +7

    What about the mean value theorem for integrals....I would like to see that proof!

    • @drpeyam
      @drpeyam  5 років тому +2

      Dr. Peyam’s Super Awesome French World Cup Victory Special ua-cam.com/video/tro1H0xbDJo/v-deo.html

  • @stefanosvasileiadis2732
    @stefanosvasileiadis2732 5 років тому +2

    Great as always. Can you upload a video on how to prove rolle's theorem?

    • @drpeyam
      @drpeyam  5 років тому +3

      Sure!

    • @drpeyam
      @drpeyam  5 років тому +7

      There you go: Rolle Theorem Proof ua-cam.com/video/QvcaTltKHVg/v-deo.html

  • @NH-zh8mp
    @NH-zh8mp 5 років тому

    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] )

  • @Karthik-lq4gn
    @Karthik-lq4gn 4 роки тому

    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.

  • @tarikabdelhadibenaouda
    @tarikabdelhadibenaouda 2 роки тому

    thank's for this amazing content .

  • @arbitraryculture4568
    @arbitraryculture4568 4 роки тому

    I love these types of videos :D

  • @sayansarkar2585
    @sayansarkar2585 Рік тому

    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?

  • @FunctionalIntegral
    @FunctionalIntegral 5 років тому

    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.

    • @drpeyam
      @drpeyam  5 років тому

      I don’t know much about those, sadly

    • @drpeyam
      @drpeyam  5 років тому

      Fematika has great videos about it though

    • @FunctionalIntegral
      @FunctionalIntegral 5 років тому

      @@drpeyam Thanks for your reply. I don't really like Fematika. I like your nice and refreshing style of teaching.

  • @weimeuret7689
    @weimeuret7689 4 роки тому +1

    f(my problem)= Solution to my problem = Search (key words of my problem + peyam)

  • @ujwalsmanhas1093
    @ujwalsmanhas1093 2 роки тому

    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) .

    • @drpeyam
      @drpeyam  2 роки тому

      But what you say is not an actual average, just a physical one, you’re confusing it with 1/2 (f(a) + f(b))

    • @ujwalsmanhas1093
      @ujwalsmanhas1093 2 роки тому

      @@drpeyam true .

  • @datsmydab-minecraft-and-mo5666
    @datsmydab-minecraft-and-mo5666 4 роки тому

    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?

  • @baguette9669
    @baguette9669 Рік тому

    thank you so much

  • @Neubulae
    @Neubulae 5 років тому +2

    Rolling… with two tangents each on one side and a secant in the middle… Now I am thinking about sandwich rolls.

    • @ChristopherEvenstar
      @ChristopherEvenstar 5 років тому +1

      Remember that time f(a) equaled f(b)? Oh wow, that was great. I almost died.

  • @clarawessels9605
    @clarawessels9605 4 роки тому

    Thank you!

  • @alvideor
    @alvideor 2 роки тому

    Doesn't setting g (x) to this special function result in loss of generality?

    • @drpeyam
      @drpeyam  2 роки тому

      Nope

    • @SimsHacks
      @SimsHacks Рік тому

      for any f, you can create the g. so it works for all f

  • @oliverletterer3854
    @oliverletterer3854 5 років тому

    Ana 1 practice exercise?

  • @GhostyOcean
    @GhostyOcean 4 роки тому

    Could you use the intermediate value theorem to prove this?

    • @GhostyOcean
      @GhostyOcean 4 роки тому

      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.

    • @GhostyOcean
      @GhostyOcean 4 роки тому

      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…

    • @drpeyam
      @drpeyam  4 роки тому

      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

    • @GhostyOcean
      @GhostyOcean 4 роки тому

      @@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.

  • @handrez.
    @handrez. Рік тому +1

    damn

  • @郑枨玚
    @郑枨玚 4 роки тому

    ☺☺ blessed

  • @dgrandlapinblanc
    @dgrandlapinblanc 5 років тому

    Thanks. En français on parle de théorème des accroissements finis. For the final with the different repetitions on parle de bêtisier (bêtises, erreurs, stupidités...)

  • @ahmadkalaoun3473
    @ahmadkalaoun3473 5 років тому +3

    Yaay first view :)

  • @MathIguess
    @MathIguess 5 років тому +1

    Who disliked this and why? :O

    • @drpeyam
      @drpeyam  5 років тому +4

      I don’t know, there’s someone who keeps disliking all my videos 😞

    • @MathIguess
      @MathIguess 5 років тому +2

      @@drpeyam well I like your videos a lot :D and so do most people :D so whatever xD

  • @trexbattle
    @trexbattle 5 років тому

    Okayyy u mite say wats goin on jus wait for it we’ll see wats going on

  • @munsif2320
    @munsif2320 3 роки тому

    Hi peyan mean by Tamil😅😂

  • @megalul4141
    @megalul4141 5 років тому

    haHAA WeirdChamp 👉🚪