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3 votes

Bounding $\int_1^{+\infty}(x-\lfloor x\rfloor-\frac12)f(x)\,\mathrm dx$ for convex and decreasing $f$

$\def\d{\mathrm{d}}\def\ge{\geqslant}\def\le{\leqslant}\def\paren#1{\left(#1\right)}\def\peq{\mathrel{\phantom{=}}}$Note: This solution is by a friend of mine. Lower bound Since $f$ is decreasing, ...
Ѕᴀᴀᴅ's user avatar
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3 votes
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Is $C = \{ x \in \mathbb{R}^n \:|\: \textbf{1}^T(x)_- \leq \frac{1}{2}\textbf{1}^T(x)_+ \}$ a convex set?

Using $$ (x)_+ = \frac 12 (|x| + x) \, , \, (x)_- = \frac 12 (|x| - x) $$ we have for $x = (x_1, \ldots, x_n)$: $$ \begin{align} \phi(x) &:= \textbf{1}^T(x)_- - \frac{1}{2}\textbf{1}^T(x)_+ \\ &...
Martin R's user avatar
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2 votes

Convexity of $\ln(a^\top x) - \sum_{i=1}^n \ln(x_i) $

Well, we can compute the Hessian. Let $f_a$ denote the map with given $a$. Then $$ \partial_i f_a = \frac{a_i}{a^\top x} - \frac{1}{x_i}$$ and $$\partial_i\partial_j f_a = -\frac{a_i a_j}{(a^\top x)^2}...
stochasticboy321's user avatar
2 votes
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Finding a bound for optimal solution of a quadratic optimization problem

Note that $$ (A+\rho I)^{-1}=\rho^{-1}\big(\rho^{-1}A+I)^{-1} $$ and $$ \lim_{\rho\rightarrow\infty}\big(\rho^{-1}A+I)^{-1}=I\ . $$ Using these two observations, we have \begin{align} x_\rho&=\\ \...
lonza leggiera's user avatar
2 votes

Finding a bound for optimal solution of a quadratic optimization problem

let $x_\rho$ be the solution to $$\min_{x\in \mathbb R^n} f(x)+\rho\|x-y\|^2$$ for some function $f$ which is bounded below by $b$ (and assume that a unique solution exists). Let $\varepsilon>0$ ...
Zoe Allen's user avatar
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2 votes

Finding a bound for optimal solution of a quadratic optimization problem

Some thoughts. Conjecture 1. Let $A$ be a given $n\times n$ symmetric positive semi-definite matrix. Let $\mu$ be a given $n\times 1$ vector. Let $e$ be the $n\times 1$ vector of ones. Let $\rho > ...
River Li's user avatar
  • 39.4k
1 vote

Is $C = \{ x \in \mathbb{R}^n \:|\: \textbf{1}^T(x)_- \leq \frac{1}{2}\textbf{1}^T(x)_+ \}$ a convex set?

Let $n$ be a positive integer and let $C$ be the set $$C = \{x \in \mathbb{R}^n; {\bf{1}}^{\top}x_+ \ge 2\cdot({\bf{1}}^{\top}x_-) \}.$$ It [the set $C$ that is] is indeed convex. We first note the ...
Mike's user avatar
  • 21k
1 vote

Bounding $\int_1^{+\infty}(x-\lfloor x\rfloor-\frac12)f(x)\,\mathrm dx$ for convex and decreasing $f$

An interesting variational problem. I find it more convenient to integrate from 0, so let me restate the problem as follows: Let $B_1(X)=X-\frac12$ be the first Bernoulli polynomial and $\beta_1(x)=...
H. H. Rugh's user avatar
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