Pantelis Sopasakis
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 Apr 4 comment What is the merit function? A merit function in the context of optimisation theory is something different. Mar 18 comment integration of $\ln \ln x$ The integral of $\ln\ln x$ is $x\ln\ln x - \mathrm{li}(x)$, where $\mathrm{li}$ is the logarithmic integral function Mar 18 comment Distribution of squared multivariate normal random variable Take a look at mathematix.wordpress.com/2016/03/10/… The chi-squared and the Wishart distribution are applicable depending on whether $\Sigma=I$ or $\Sigma$ is a general matrix and whether $\mu=0$. Derivations, nevertheless, can be cumbersome. Mar 18 comment Can every possible curve be expressed mathematically The answer is yes; trivially let $C$ be a curve. The question should rather be, can every curve be described as $F(x)=0$ where $F$ is a composition of a finite number of elementary functions (or, if you wish, plus a few more, like the Gamma function, etc). Then, the answer is no. Mar 17 awarded Yearling Mar 17 comment If $A:=[0,1]$ and $B:=(-1,2)$ the function $f:A\mapsto B$ is continuous and $f(0)=1$ and $f(1)=0$.Is $f(A)\subset A$ true or false, proove it? No, it's not true. Try to draw a continuous graph of a function $f$ in the $[0,1]\times (-1,2)$ with anchored points at $(0,1)$ and $(1,0)$. Mar 17 revised $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? Fixed typo in figure Mar 17 answered $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? Mar 16 comment $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? @LeonardoFranciscoCavenaghi I'm trying... Mar 16 comment $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? I mean that the fact that $\phi$ is well-defined, smooth and bijective between the two spaces, doesn't make it a diffeomorphism. It's inverse has to be differentiable as well. Mar 16 comment $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? @LeonardoFranciscoCavenaghi I don't see how this post answers the question. One needs to show that $\phi$ is a diffeomorphism, which means that $\phi^{-1}$ must be differentiable. Mar 16 comment $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? I believe that the answer should be along the lines of the inverse function theorem. Mar 16 comment $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? $\mathcal{B}_\epsilon$ is an open ball. Since $\phi$ is defined for $x\in\mathcal{B}_\epsilon(0)$, i.e., for $\|x\|<\epsilon$, we have $\frac{\|x\|\pi}{2\epsilon} \in [0, \pi/2),$ so the domain of $\phi$ is $\mathcal{B}_\epsilon\setminus \{0\}$. Mar 16 comment $\phi(x) := \tan({\frac{\|x\|\pi}{2\epsilon}})\frac{x}{\|x\|}$ is defined on the open ball $B_{\epsilon}(0) \subset \mathbb{R}^n$? What do you mean that $0\notin B_\epsilon(0)$. Isn't $B_\epsilon(0)$ the open ball of radius $\epsilon$ centered at $0$? What is more, you need to show that $\phi^{-1}$ is also differentiable. Finally, $\phi$ is a differomorphism between $B_\epsilon(0)\setminus 0$ and $\mathbb{R}^n\setminus \{0\}$, right? Mar 15 revised Is distance function defined on a convex set is always convex? added illusration of how a non-convex distance looks like Mar 15 revised Is distance function defined on a convex set is always convex? added 505 characters in body Mar 15 revised Is distance function defined on a convex set is always convex? added 828 characters in body Mar 15 comment Is distance function defined on a convex set is always convex? Oh, OK, I will update my answer. Mar 15 comment CDF expected value when Y=X^2 I edited your question and replaced unicode characters for the integral (it should be \int_a^b) and the symbol $\leq$ (should be \leq). Could you please check everything is fine? I believe you should go through the question once again and make it clearer. Please, take a look at math.stackexchange.com/help/notation Mar 15 revised CDF expected value when Y=X^2 I tried to format the question using TeX and to make it more understandable