Well explained, sir! Kindly explain what you mean on 2:30 by the statement "q in (0,1) implies that the RHS should be greater." Common sense suggests that it should be smaller since it q is a fraction of 1.
This is a nice video, but I believe the statement of the theorem as given here is slightly wrong. Your quantifiers in the definition of a contraction mapping are the wrong way around. It should be ‘there exists a q such that for each x, y’ rather than ‘for each x, y there exists a q’. This may seem inconsequential, but it is very important to the theorem that there is one ‘uniform’ constant q that works for all points in X under the map T. If not, we can potentially have a sequence of points in X such that values of d(Tx, Ty) come arbitrarily close to d(x, y), and this violates what is necessary to prove the Banach Fixed Point Theorem.
@Diego Marra The problem is that such a ‘biggest q’ doesn’t necessarily exist. There exist subsets of [0,1) that have no greatest element in the standard ordering (indeed the set itself has this property).
Frr ko k Jjuuu. J BB ñjkk.n😊
Well Explained. The Canada map example was really helpful.
I really appreciate the map example!
youtube.com/@user-iq5ve6bf8q
Nice explanation Sir.Your this video is useful to me to understand the concept of Contraction mapping.
Well explained, sir! Kindly explain what you mean on 2:30 by the statement "q in (0,1) implies that the RHS should be greater." Common sense suggests that it should be smaller since it q is a fraction of 1.
Wow. So clearly explained. Thank you!
お、おう 大きい地図に小さい地図を重ねると???ってなっちゃうけど、 小さい地図をびろーんと伸ばすとどこか一点は不動っぽいのはわかる・・・かな🤔😀👍
Fantastic explanation!
T:X-R
This is a nice video, but I believe the statement of the theorem as given here is slightly wrong. Your quantifiers in the definition of a contraction mapping are the wrong way around. It should be ‘there exists a q such that for each x, y’ rather than ‘for each x, y there exists a q’. This may seem inconsequential, but it is very important to the theorem that there is one ‘uniform’ constant q that works for all points in X under the map T. If not, we can potentially have a sequence of points in X such that values of d(Tx, Ty) come arbitrarily close to d(x, y), and this violates what is necessary to prove the Banach Fixed Point Theorem.
@Diego Marra The problem is that such a ‘biggest q’ doesn’t necessarily exist. There exist subsets of [0,1) that have no greatest element in the standard ordering (indeed the set itself has this property).
@Diego Marra Very good thing to question though!
good catch!
Best explanation I have ever seen. Keep it up ❤
really helpful
great job boys! <3
To gret
this is a great explanation, helped me with Picard-Lindelof proof
ruclips.net/video/EoImJHpsocQ/видео.html
love that, very easy to understand, thx!
very helpful! thx
very helpful