Contraction Mapping Theorem & Finding Fixed Points of Functions

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  • เผยแพร่เมื่อ 31 ธ.ค. 2024

ความคิดเห็น • 51

  • @yoyokojo651
    @yoyokojo651 3 ปีที่แล้ว +4

    How interesting, I just learned banachs fixed point theorem for generell metric spaces last week!

  • @BA-ul7rl
    @BA-ul7rl 10 หลายเดือนก่อน

    Thank you! You saved my life! So many other videos just talk theory without any examples.

  • @aliguliyev1866
    @aliguliyev1866 7 หลายเดือนก่อน

    Very clear and direct explanation. I liked this video so much. Thanks a lot.

  • @micomrkaic
    @micomrkaic 3 ปีที่แล้ว +1

    This is very nice. Fixed point theorems are used all over the place in theoretical economics, e.g. in infinite horizon dynamic programming. The difference is that the fixed point in the particular case of dynamic programming is a function, not a number.

  • @FluffleOW
    @FluffleOW 2 ปีที่แล้ว

    Thank you so much!! I'm watching your videos to try and get my head around these concepts for my last ever econ module before I graduate!

  • @242math
    @242math 3 ปีที่แล้ว +1

    this contraction mapping theorem is fascinating, great job prof

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว

      It’s very interesting!

  • @kristimeacham6987
    @kristimeacham6987 2 ปีที่แล้ว +1

    Thank you so much! This made things really clear. I would definitely watch more analysis videos from you!

    • @ProfOmarMath
      @ProfOmarMath  2 ปีที่แล้ว

      Thanks Kristi! Hopefully I can make more 😁

  • @hrs7305
    @hrs7305 3 ปีที่แล้ว +2

    Cool , I have seen few variants of it
    1st- where nth iterate of f is a contraction then f has a unique fixed point
    (By nth iterate of f I mean fofo....f (n time composition)
    I also has some cool applications like
    cos(x) has a unique fixed point in the interval [0,2pi]
    Notice that cos is not a contraction bus cos o cos is a contraction ( sin(cosx)*sinx < 1 )
    2nd - on a compact metric space a distance decreasing map gives you a unique fixed point , this is nice because it is easier to find distance decreasing maps than to find a contraction (contraction is a distance decreasing map)
    distance decreasing map means
    f : X -> X
    d(f x , f y) < d(x , y)

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว

      Definitely like these generalizations

  • @thedoublehelix5661
    @thedoublehelix5661 3 ปีที่แล้ว +1

    Fixed points are really cool! Especially because they can be explained to people without much background in math. By the way, an easier proof for why that equation has a unique solution can be obtained just by differentiating 0.5cos^2x-x and noticing that it is negative on that interval.

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว +1

      I did notice that haha. Should have picked a more subtle situation!

  • @riadsouissi
    @riadsouissi 3 ปีที่แล้ว +6

    I like any video about fixed points. In the same subject, I would love to see a video on the Brouwer fixed point theorem and Borsuk-Ulam theorem
    This said, since you used the derivative of cos(x)^2, I feel it is a bit like a catch 22 since showing that the function x-1/2cos(x)^2 is always increasing (derivate is always positive) and its end points imply a unique fixed point 😅

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว +1

      I actually thought about this haha, good catch!

  • @wesleydeng71
    @wesleydeng71 3 ปีที่แล้ว +3

    Cool! By the way, the actual fixed point is approx. x=0.417715

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว +1

      The theorem gives an iterative way to approximate, nice!

  • @xCorvus7x
    @xCorvus7x 3 ปีที่แล้ว +2

    4:01
    So, all contractions defined like this are continuous.

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว +1

      Actually if we replace C with any constant the function will be continuous

    • @xCorvus7x
      @xCorvus7x 3 ปีที่แล้ว

      @@ProfOmarMath Sorry, I'm not sure I can follow.
      Have you not considered C to be arbitrary?
      If a function has a discontinuity, couldn't you always find an x and a y that are closer to each other than f(x) and f(y) are, simply by approaching the discontinuity from different sides?
      What would a discontinuous function look like that meets this criterion for a special C?

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว +1

      That inequality holds for all x,y and not for a specific choice of x,y so if you let y be the point where the discontinuity occurs then f(y) will be too far from f(x) when x is very close to y

    • @xCorvus7x
      @xCorvus7x 3 ปีที่แล้ว

      @@ProfOmarMath Okay, that's true.
      But in your first response, have you not said that for the contraction f to be continuous, we need C to be arbitrary?
      I thought, you had done so in the video.

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว +1

      For a contraction C has to be at most 1. It happens to be the case that if we look even beyond contraction mappings, any function f that satisfied the condition |f(x)-f(y)|

  • @robertgerbicz
    @robertgerbicz 3 ปีที่แล้ว

    More direct way for the 14:09 place, without using the mean value theorem:
    We know: cos(2*x)=2*cos(x)^2-1
    from this cos(x)^2=(cos(2*x)+1)/2
    so 1/2*(cos(x)^2-cos(y)^2)=(cos(2*x)-cos(2*y))/4=-1/2*sin(x+y)*sin(x-y), where used an addition formula.
    since abs(sin(z))

    • @ProfOmarMath
      @ProfOmarMath  3 ปีที่แล้ว

      Definitely, I think yoav mentioned this earlier

  • @mustafa-damm
    @mustafa-damm 3 ปีที่แล้ว

    Incredibly useful

  • @aashsyed1277
    @aashsyed1277 3 ปีที่แล้ว

    I appreciate you so much.....!!!!!!!!!

  • @matteosevenius313
    @matteosevenius313 ปีที่แล้ว

    triangle what? 7:29

  • @quinnpisani180
    @quinnpisani180 2 ปีที่แล้ว +1

    17 mins later, still confused.

  • @timzhou5971
    @timzhou5971 6 หลายเดือนก่อน

    🤯🤯🤯

  • @aashsyed1277
    @aashsyed1277 3 ปีที่แล้ว +1

    1st comment

  • @aashsyed1277
    @aashsyed1277 3 ปีที่แล้ว +1

    1st like

  • @aashsyed1277
    @aashsyed1277 3 ปีที่แล้ว +1

    first view