When Dr Casey writes the Bolzano-Weierstrass theorem you can feel the extra pulse of the chalk hitting the board, which gives you a sense of awe you can't find in the books. There should be a warning about the camara moves at the beginning, some individuals may find them disturbing, even dangerous.
This theorem is proven when a convergent subsequence is given. Such a subsequence consists of Xn which is sup{Xk:k>=n}. We have proved this subsequence is monotone and bounded, thus it's convergent by our previous theorem.
1:02:40 Alternatively instead of defining an n_0 is one can define the sequence of indices for the lower and upper bounding sequences as n'_k = {1; k = 1 {n_(k-1)+1; k>1 That way you simply have a_(n'_k) - 1/k < x_(n_k)
17:55 There's a problem with the proof here. x_n < sqrt(2/[n-1]) only holds for n > 1. For n = 1 the right hand side is undefined. You would have to replace the right hand side with something like a_n = {2; n 1
Hey self-learners, be aware of a minor typo in the example of x_n = (-1)^{n} @ 40:11 It should be k >= n in both sup and inf. Not n >= k.
Real one. Thanks
When Dr Casey writes the Bolzano-Weierstrass theorem you can feel the extra pulse of the chalk hitting the board, which gives you a sense of awe you can't find in the books. There should be a warning about the camara moves at the beginning, some individuals may find them disturbing, even dangerous.
horrible camera work
its not that bad, ive seen a lot worse on youtube
51:19 it is not straightforward to me why the BW theorem is correct given the previous theorem.
This theorem is proven when a convergent subsequence is given. Such a subsequence consists of Xn which is sup{Xk:k>=n}. We have proved this subsequence is monotone and bounded, thus it's convergent by our previous theorem.
damn good explanation
Bravo!
this is nice
damn cool~~
1:02:40 Alternatively instead of defining an n_0 is one can define the sequence of indices for the lower and upper bounding sequences as
n'_k =
{1; k = 1
{n_(k-1)+1; k>1
That way you simply have
a_(n'_k) - 1/k < x_(n_k)
56:00
cool
Hi, I would like a sandwich with x
17:55 There's a problem with the proof here. x_n < sqrt(2/[n-1]) only holds for n > 1. For n = 1 the right hand side is undefined. You would have to replace the right hand side with something like
a_n =
{2; n 1