Awesome lectures. I watched 263, 364a, and 364b. All lectures had relevant material to my research. However, I find Prof. Boyd best when he is sitting down with a notepad handy. The quick divergence talks he gives provide a lot of information not found in any book. The 364b lectures miss this presentation style. I hope Standford re-shoots 364b in the format of 263 and 364a...
39:32, Professor Boyd said that "the subdifferential of a norm at the origin is the unit ball of the dual norm". But when it comes to 2-norm, 2-norm is differentiable everywhere, so the subdifferential should be a single point. However, the dual norm of 2-norm is still 2 norm, so the unit ball of the dual norm suggests the subdifferential is a set(unit circle) instead. How can this be? Anyone help me with this? Thanks.
You're thinking about the 2-norm squared, which is indeed differentiable everywhere. The same does not hold for the 2-norm, which is the *square root* of the sum of squares. This function when plotted for e.g. R^2 is in fact a circular cone, with a sharp "corner" at 0, and indeed there are an infinite amount of support planes at the origin which globally underestimate the 2-norm, whose unit normal vectors are the subgradients when projected onto R^2. Since the 2-norm of these vectors is at most 1, it is easy to conclude that they are within the unit 2-norm ball.
@@jiaruiwang1609 No problem! (For the sake of completeness, one can also look at the function f(t) = ||tv|| = |t|, where v is a unit vector. Since f is non-diff, so is the 2-norm).
Awesome lectures. I watched 263, 364a, and 364b. All lectures had relevant material to my research. However, I find Prof. Boyd best when he is sitting down with a notepad handy. The quick divergence talks he gives provide a lot of information not found in any book. The 364b lectures miss this presentation style. I hope Standford re-shoots 364b in the format of 263 and 364a...
Thanks for this 'non-causal' presentation. I am sure community really appreciated this effort.
Thank you for the reenactment of this lecture, and the availability of the rest :).
Thanks to Professor Stephen Boyd for his great video.
jump to 16:30
39:32, Professor Boyd said that "the subdifferential of a norm at the origin is the unit ball of the dual norm". But when it comes to 2-norm, 2-norm is differentiable everywhere, so the subdifferential should be a single point. However, the dual norm of 2-norm is still 2 norm, so the unit ball of the dual norm suggests the subdifferential is a set(unit circle) instead. How can this be? Anyone help me with this? Thanks.
You're thinking about the 2-norm squared, which is indeed differentiable everywhere. The same does not hold for the 2-norm, which is the *square root* of the sum of squares. This function when plotted for e.g. R^2 is in fact a circular cone, with a sharp "corner" at 0, and indeed there are an infinite amount of support planes at the origin which globally underestimate the 2-norm, whose unit normal vectors are the subgradients when projected onto R^2. Since the 2-norm of these vectors is at most 1, it is easy to conclude that they are within the unit 2-norm ball.
@@vicktorioalhakim3666 Nice explanation! I did not know that 2-norm is non-differentiable until now😂. Thank you very much!!!
@@jiaruiwang1609 No problem! (For the sake of completeness, one can also look at the function f(t) = ||tv|| = |t|, where v is a unit vector. Since f is non-diff, so is the 2-norm).
So this is only useful for non-differentiable functions ?
How can the textbook be the same, when we went through every chapter in 364A?
Please tell me how to download the Video subtitles, since I am a foreign, only with the subtitles I can catch the class.
might aswell learn english :)
werent you taught english in school ?
@@robinranabhat3125 it is somewhat hard to take lectures smoothly with school-level English
53:19
Lol, you can tell this guy thrill was beyond comprehension to do this dramatic re-enactment of the first day.
At what level of study is this course taught? PhD or Masters?
Both, in graduate programs
Thank you!