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.
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 😅
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.
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 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?
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
@@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.
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)|
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))
How interesting, I just learned banachs fixed point theorem for generell metric spaces last week!
The timing!!
Thank you! You saved my life! So many other videos just talk theory without any examples.
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!
Thank you!
Very clear and direct explanation. I liked this video so much. Thanks a lot.
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.
Definitely!
this contraction mapping theorem is fascinating, great job prof
It’s very interesting!
Thank you so much! This made things really clear. I would definitely watch more analysis videos from you!
Thanks Kristi! Hopefully I can make more 😁
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 😅
I actually thought about this haha, good catch!
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.
I did notice that haha. Should have picked a more subtle situation!
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)
Definitely like these generalizations
4:01
So, all contractions defined like this are continuous.
Actually if we replace C with any constant the function will be continuous
@@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?
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
@@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.
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)|
Incredibly useful
Very much
triangle what? 7:29
Cool! By the way, the actual fixed point is approx. x=0.417715
The theorem gives an iterative way to approximate, nice!
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))
Definitely, I think yoav mentioned this earlier
I appreciate you so much.....!!!!!!!!!
You too aash
@@ProfOmarMath THANKS
@@ProfOmarMath YOU ARE WELCOME
🤯🤯🤯
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