Contraction Mapping Theorem & Finding Fixed Points of Functions
Рет қаралды 14,715
3 жыл бұрынThis video looks at an intriguing equation as an excuse to introduct the Contraction Mapping Theorem, a fascinating theorem that lets us find fixed points to functions.
#FixedPointTheorems #ContractionMappingTheorem #FixedPoints
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@aliguliyev1866 2 ай бұрын
Very clear and direct explanation. I liked this video so much. Thanks a lot.
@yoyokojo651 3 жыл бұрын
How interesting, I just learned banachs fixed point theorem for generell metric spaces last week!
@ProfOmarMath 3 жыл бұрын
The timing!!
@BA-ul7rl 5 ай бұрын
Thank you! You saved my life! So many other videos just talk theory without any examples.
@kristimeacham6987 2 жыл бұрын
Thank you so much! This made things really clear. I would definitely watch more analysis videos from you!
@ProfOmarMath 2 жыл бұрын
Thanks Kristi! Hopefully I can make more 😁
@242math 3 жыл бұрын
this contraction mapping theorem is fascinating, great job prof
@ProfOmarMath 3 жыл бұрын
It’s very interesting!
@FluffleOW 2 жыл бұрын
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!
@ProfOmarMath 2 жыл бұрын
Thank you!
@riadsouissi 3 жыл бұрын
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 3 жыл бұрын
I actually thought about this haha, good catch!
@micomrkaic 3 жыл бұрын
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.
@ProfOmarMath 3 жыл бұрын
Definitely!
@hrs7305 3 жыл бұрын
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 3 жыл бұрын
Definitely like these generalizations
@thedoublehelix5661 3 жыл бұрын
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 3 жыл бұрын
I did notice that haha. Should have picked a more subtle situation!
@wesleydeng71 3 жыл бұрын
Cool! By the way, the actual fixed point is approx. x=0.417715
@ProfOmarMath 3 жыл бұрын
The theorem gives an iterative way to approximate, nice!
@mustafa-damm 3 жыл бұрын
Incredibly useful
@ProfOmarMath 3 жыл бұрын
Very much
@aashsyed1277 3 жыл бұрын
I appreciate you so much.....!!!!!!!!!
@ProfOmarMath 3 жыл бұрын
You too aash
@aashsyed1277 3 жыл бұрын
@@ProfOmarMath THANKS
@aashsyed1277 3 жыл бұрын
@@ProfOmarMath YOU ARE WELCOME
@timzhou5971 Ай бұрын
🤯🤯🤯
@matteosevenius313 Жыл бұрын
triangle what? 7:29
@robertgerbicz 3 жыл бұрын
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 3 жыл бұрын
Definitely, I think yoav mentioned this earlier
@aashsyed1277 3 жыл бұрын
1st comment
@aashsyed1277 3 жыл бұрын
1st like
@aashsyed1277 3 жыл бұрын
first view
@xCorvus7x 3 жыл бұрын
4:01 So, all contractions defined like this are continuous.
@ProfOmarMath 3 жыл бұрын
Actually if we replace C with any constant the function will be continuous
@xCorvus7x 3 жыл бұрын
@@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 3 жыл бұрын
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 3 жыл бұрын
@@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 3 жыл бұрын
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)|
@quinnpisani180 Жыл бұрын
17 mins later, still confused.
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