The Mean Value Theorem and Fixed Points

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  • Опубліковано 17 гру 2024

КОМЕНТАРІ • 40

  • @JBaker452
    @JBaker452 6 років тому +16

    The bare mathematics are not very exciting, but it is easy to dramatize the effects with a surrounding story. Like, a gallant knight can view from one corner of the forest (0,0) to the other corner (1,1) in search of his princess, who must pass from the left to the right.
    For sure he will spot his princess.

  • @corasnyder5881
    @corasnyder5881 4 роки тому +2

    This video is so clear and helpful. I highly recommend this video.

  • @soyeonkim1435
    @soyeonkim1435 4 роки тому +1

    Thank you for your explanation! I was reading Martin Davis' where the Brouwer's fixed point theorem was mentioned briefly! And your video helped me a lot!!

  • @VerSalieri
    @VerSalieri 6 років тому +3

    Consider a cup of water, put a handful of rice into the cup.... let X denote the space of positions of each grain of rice in the handful of rice in the water (not sugar or salt... those dissolve). Let f be the continuous mapping/rearrangement of rice in the water (in non mathematical terms, stir it in a continuous manner), so f(x) is the new position of the grain after stirring. ..by the fixed point theorem (since X is closed and bounded and hence compact), there should be at least one grain of rice that remains in its place. The tricky part now is to find that fixed grain of rice.....to fully stir sugar in a cup of tea, we should try to counter this fixed point theorem....

  • @yuvalpaz3752
    @yuvalpaz3752 6 років тому +1

    Ah, I remember this question, this is basically the first question one have when studying the intermediate and mean value theorems.
    My favorite exercise of the mean value theorems is:
    Let f be continuous on [a,b] and differentiable on (a,b), with f(a)=f(b)=q and f(k)=q+|a-b|/n for some k∈(a,b). Prove that there exist c∈(a,b) such that |f′(c)|>n.
    ---------------------
    Well, first time I had this there was added detail: assume that f' is integrable, this is easier like this, especially for new in the area.
    -------------------

  • @jkp910
    @jkp910 2 роки тому

    I like this guy's energy.

  • @andoniades
    @andoniades 3 роки тому

    Working on this problem now.
    My head just exploded, but the vid was good non-the-less.
    Thanks

  • @danielcohen227
    @danielcohen227 6 років тому

    you are the best teacher i have seen yet

    • @drpeyam
      @drpeyam  6 років тому

      Thank you! :)

  • @NAMEhzj
    @NAMEhzj 6 років тому +5

    While you're at it, you could also make videos about the Banach- and Brower Fixed-Point Theorems ;)
    Love your videos!

    • @drpeyam
      @drpeyam  6 років тому +1

      Hehehe, tempting :P

  • @stevepa3416
    @stevepa3416 6 років тому +1

    hello dr.payam. I have a request if you don't mind. 1st I'm interested in things concerning whether or not functions will have elementary antiderivativess. in essence lets say I have function fx. what conditions must be met in the function for it to be able to have an anti derivative expressable in elementary functions, and then from this if we are presented some nasty function, we can do tests to see if at Least an elementary antiderivative exists. and second. lets say I have some function, on a closed interval a to b, can we create a list of upper limit lower limit such that those limits, the integral of a function is equal to A. a being any arbitrary real number. like lets say 3 and 7 or. Sqrt2e to 5. let me know if you understand the requests. thank you very much if you see this

    • @drpeyam
      @drpeyam  6 років тому

      For your first question, unfortunately there is no such condition, because very complicated functions could have antiderivatives but simple functions could not. The only think we can say for sure is that polynomials have antiderivatives, for everything else we don’t know.
      For example, (cos(x))^2 has one but (cos(x^2)) doesn’t have one
      Or 1/ln(x) doesn’t have one, and neither does cos(sin(x)), so you can’t just take reciprocals or compositions and hope to have an antiderivative
      For your second question, I’m not 100% sure what you mean

  • @douglasmagowan4918
    @douglasmagowan4918 6 років тому +1

    This proof is missing some key critera in the set up of the proposition. Particularly, f is continous on [0,1] and differentiable everywhere on (0,1).

    • @drpeyam
      @drpeyam  6 років тому

      Yep, by writing f’ not equal to 1, I’m assuming f is differentiable everywhere

  • @TheMauror22
    @TheMauror22 6 років тому +1

    Great vídeo! You should do the proof of the existence and uniqueness theorem for ODE's!!!

    • @drpeyam
      @drpeyam  6 років тому +1

      I love that theorem! It’s on my to-do list :)

    • @TheMauror22
      @TheMauror22 6 років тому

      Dr. Peyam's Show yeaaah! I'm really looking forward to seeing you proving that!!!!!!

  • @draganm1006
    @draganm1006 6 років тому

    Maybe do a video on the Sharkovskii's theorem?

  • @estuardodiaz2720
    @estuardodiaz2720 6 років тому

    love your channel!! please keep doing videos with proofs and theorems hahahah

  • @jorgeeduardopereztasso6134
    @jorgeeduardopereztasso6134 6 років тому

    numeric methods are so cooool!

  • @maxweinstein1537
    @maxweinstein1537 6 років тому

    Don't we also have to assume that f is continuous? First of all, if we allow this, we can easily find functions that climb as high as they can and then reappear beneath the line y=x, so that they will never intersect the line. Also, g's continuity implies f's continuity and the converse (if we take the limit as x goes to some point a of g(x)=f(x)-x, then g is continuous iff g(a)=f(a)-a and the previous equation, which gives you that g is continuous iff f is continuous), which is vital to be able to apply the intermediate value thm.

    • @drpeyam
      @drpeyam  6 років тому

      If f is differentiable, it’s continuous

  • @WoWSchockadin
    @WoWSchockadin 6 років тому +1

    How can any student possibly not be excited when getting teached by you? At least on earth, that's a contradiction ;-)

  • @rizkyagungshahputra215
    @rizkyagungshahputra215 6 років тому +1

    why f'(x)≠1?

  • @jcho1719
    @jcho1719 6 років тому

    I dun like the way you write 'f' but yr video is good 😂

  • @kpremkumar9523
    @kpremkumar9523 4 роки тому +1

    In the begining you promise exactly one fixed point and after you show that at least one fixed point. For. f'(x) not equal to zero. ,why

  • @pitreason
    @pitreason 6 років тому

    But you didn't say that function f must be continious.

    • @drpeyam
      @drpeyam  6 років тому +3

      If it’s differentiable, it’s continuous

    • @pitreason
      @pitreason 6 років тому

      Dr. Peyam's Show ok, i didn’t get that it is differentiable

  • @alanhiguera3484
    @alanhiguera3484 6 років тому +1

    Peyamilluminati flat earth confirmed?

  • @welliguessillbehereforawhi3963
    @welliguessillbehereforawhi3963 5 років тому +1

    lol u really excited

  • @stydras3380
    @stydras3380 6 років тому +1

    WHAT DO YOU MEAN?!

  • @helloitsme7553
    @helloitsme7553 6 років тому

    Ouch a fixed point in time may not be changed

  • @MrQwefty
    @MrQwefty 4 роки тому

    This video isn't very nice. It's so mean!