3 ms·

You are really super slow on the uptake, reading comprehension. Again, once again, over again, yet again, one more time, this time just for you, my writing is

• sctb 9y ago
  You're being incredibly rude to other community members. Since this is obviously not OK, would you please stop?
• yorwba 9y ago
  The problem with your posts is not that they are hard to understand, it's that they are not nice to read. For one, you come off as arrogant, as if you were the only one who has any idea what they are talking about. You present a solution to help disadvantaged students learn better, and I agree that your solution is fine. But it will only work for students that have the will, time and ability to help themselves, and I think that is only a tiny minority of all students, and even tinier when you only consider economically disadvantaged ones. That your first post didn't mention any of the factors that could make your solution unworkable probably contributes to the downvotes, because you come across as somewhat naive, and not in a cute way. I don't think it deserved to be flagged, but it wasn't a good post either. Your second post is much too repetitive. Repetition is very memorable (you got "Get the book. Read the book. Do the exercises." stuck in my head quite well), but if you don't enrich it with convincing arguments, it just becomes tiresome to read. Your third post is actually quite reasonable (but still really long, not sure if you could have made it shorter). You clarify that your solution is only for talented students and you mention a bunch of things that would be good for such students to know. Your fourth post is really bad. You directly attack someone who was trying to help you with stylistic advice and tell them that they didn't understand your post, deny that there is any problem and proceed to repeat yourself. That you got called out by a moderator should give you a hint how far off the mark your writing is. That said, consider me nerd-sniped! I quite enjoyed doing your exercises, although I had to look up some definitions here and there. Unfortunately, they don't fit in a single comment, so I have to see how many self-replies I am allowed.
  • yorwba 9y ago
    Example: For positive integer n and the set of real number R , show that, on any finite intersection of closed half spaces C in R^n, any linear function that is bounded above on C achieves its least upper bound on C. If C is empty (i.e. the closed half spaces do not overlap), the claim is false, since every real number is an upper bound of every function on C, therefore the least upper bound does not exist. However, for non-empty C and a linear function f bounded above on C, any x in C induces a lower bound l for the upper bounds, which guarantees the existence of a least upper bound u. The point where u is achieved can be determined by an iterative procedure involving a set of half-spaces H (initially H_0 is empty), a point x in C (initially x_0 is chosen arbitrarily) and a search direction d (initially d_0 is the gradient of f). The following invariant is maintained: for any boundary of a half-space in H holds that x is on the boundary and d is parallel to it, and for any x' in C, f(x') > f(x) implies that the dot product d (x' - x) is positive. This holds for the inital conditions, since H_0 is empty and d_0 is the gradient of f, which means that d_0 x' = f(x') > f(x) = d_0 x is equivalent to d_0 (x' - x) > 0. If d ever becomes zero, the invariant implies that there are no points x' in C with f(x') > f(x), which means that f(x) = u. Otherwise, the value of x_{n+1} is chosen from the linear subspace x_n + t d_n, where t in [0, infinity). As this is a closed, convex set, its intersection with C (which is an intersection of closed, convex sets) is itself closed and convex. Since f keeps increasing linearly with t, the upper bound u on f immediately induces an upper bound on t. Summarily, this constrains t to an interval [0, s]. Then x_{n+1} = x_n + s d_n. This step maintains the invariants, since x moves parallel to all boundaries of sub-spaces in H_n and x_n was contained therein, thus x_{n+1} is also contained in all these boundaries. Because the search direction doesn't change, its invariant is maintained as well. At this point, there must be a half-space h that "cut off" the line at x_{n+1}, which means that x was not moving in parallel to it. Setting adding h to H_n keeps x_{n+1} contained within all the boundaries, but makes d_n no longer parallel to them. By also setting d_{n+1} to the component of d_n that is parallel to the boundary of h, this can be reinstated. However, now it must be shown that f(x') > f(x) implies d_{n+1} (x' - x) > 0. From the previous step, it holds that f(x') > f(x) implies 0 < d_n (x' - x) = (d_n - d_{n+1} + d_{n+1}) (x' - x) = (d_n - d_{n+1}) (x' - x) + d_{n+1} (x' - x) where d_n - d_{n+1} is perpendicular to the boundary of h and away from it. For d_{n+1} (x' - x) to be non-positive with the overall result staying positive, (d_n - d_{n+1}) (x' - x) would have to be positive. But this means that x' must be a non-zero distance outside h, and is therefore not in C. The only remaining possibility is that the invariant is maintained. The above algorithm must terminate in a finite number of steps, since there is only a finite number of half-spaces that can be added to H. Because the algorithm can only terminate by finding an x in C where f(x) = u, this proves that the least upper bound is achieved.