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$\begingroup$ 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. $\endgroup$Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

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.Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

21 de ago. de 2011 · M (x) is a function. Taking the maximal number amongst the parameters.

**max**{x1, x2} = {x1, if x1> x2 x2, 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 {− 1 n ...Defining, Z = [

**max**i Xi] By Jensen's inequality, where the last equality follows from the definition of the Gaussian moment generating function (a bound for sub-Gaussian random variables also follows by this same argument). Rewriting this, E[Z] ≤ logn t + tσ2 2. Now, set t = √2logn σ to get. E[Z] ≤ σ√2logn. Share.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.6 de mai. de 2019 · The Tracy-Widom distribution gives the limiting distribution of the largest eigenvalue of a random matrix (in the $\beta$-Hermite ensemble, where $\beta$ is 1,2 or 4).

I need to prove the ∞ -norm of a matrix, which is here. ‖A‖∞ =

**max**x ≠ 0 {**max**1 ≤ i ≤ n | (Ax)i |**max**1 ≤ i ≤ n | xi |}. I can't find the the proof anywhere and would appreciate if someone could explain every step of the proof. Thank you! matrices. solution-verification. proof-explanation.