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$\max\{x_1,x_2\} = \cases{x_1, \text{if }x_1 > x_2\\x_2, \text{otherwise}}$ You can define like that the maximum of any finitely many elements. When the parameters are an infinite set of values, then it is implied that one of them is maximal (namely that there is a greatest one, unlike the set $\{-\frac{1}{n} | n\in\mathbb{N}\}$ where there is no greatest element)
$\begingroup$ I prefer $\max\{f(x_1,\ldots,f(x_n)\}$ with curly braces and no parentheses. In this instance, the parentheses don't actually help, and the curly braces remind you that the thing whose maximum is sought is a set rather than a tuple. $\endgroup$
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19 de set. de 2017 · Since composition of convex functions is convex, we only need to show max (x, y) is convex. But max (x, y) = x + y 2 + | x − y 2 | and | ⋅ | is obviously convex. A function f: Rn → R is convex if and only if its epigraph epif = {(x, t) ∈ Rn × R ∣ f(x) ≤ t} is a convex set.
Rewriting this, E[Z] ≤ logn t + tσ2 2. Now, set t = √2logn σ to get. E[Z] ≤ σ√2logn. The reason Sivaraman set t = \sqrt {2\log {n}}/\sigma is because that is the point at which the upper bound is at a minimum. You can see this by taking the derivative of the bound with respect to t and setting it to zero.
11 de mai. de 2020 · But let's take x = 2, then (1 - 2) ^ 2 will be (-1) ^2 which is nothing but 1 and according to op's max function, 1 should be returned. But since you gave the condition of x >= 1, we always return 0 even when x is something like 2. I think in comments what Andre Holzner said is correct.
Adding/subtracting gives max (x, y) = 1 2(x + y + | x − y |), min (x, y) = 1 2(x + y − | x − y |) There are 3 separate cases you must cover. You've covered (1). But this is just one possibility. You must also show that the equality holds for a = b. Again, would not be too difficult.
29 de fev. de 2016 · I have a question similar to this one, but am considering sub-Guassian random variables instead of Gaussian.
19. Let X ⊂Rn X ⊂ R n and Y ⊂Rm Y ⊂ R m be compact sets. Consider a continuous function f: X × Y → R f: X × Y → R. Say under which condition we have. minx∈Xmaxy∈Y f(x, y) =maxy∈Y minx∈X f(x, y). min x ∈ X max y ∈ Y f (x, y) = max y ∈ Y min x ∈ X f (x, y). From this we have that maxy∈Y minx∈X f(x, y) ≤minx∈ ...
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