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Question: Title: $x^x(\ln(x)+1)=b$ Body: I want to solve for $x$ and have tried solving it myself and used many algebraic calculators including wolfram alpha, and I could not get any answer from anything. If you can solve it please show me how, and if it is impossible please explain why to me. Th...

2 hours later…
7:35 PM
A new tag was created by Rodrigo de Azevedo, including a tag-excerpt saying: "The theory that develops differential calculus for functions that are not differentiable in the usual sense."
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How to compute the Clarke subdifferential of the function $h(x,y)=x^2+y^2$?

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I've been reading an article on Clarke critical values of subanalytic Lipschitz functions. There I've come across the following definition(s) of subdifferential: $$f: U \to \mathbb{R}^n, \ \ \ \ \emptyset \neq U \subset \mathbb{R}^n$$ $f$ is locally Lipshitz continuous Let $$x \in U:$$ The Fre...

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In the book [1], Clarke et al. define the generalized gradient for a Lipschitz function $f:\mathbb{R}^n\to\mathbb{R}$ as follows. 8.1. Theorem (Generalized Gradient Formula). Let $x\in\mathbb{R}^n$, and let $f:\mathbb{R}^n\to\mathbb{R}$ be Lipschitz near $x$. Let $\Omega$ be any subset of zer...

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According to various websites, for some function $f:X\to R$ we can define a map $$D(x,v):= \lim_{y\to x} \sup_{t\searrow0} \frac{f(y+tv)-f(y)}{t}$$ and the Clarke Subdifferential (Def 1) is $$\partial f(x) := \{v \in X^* : D(x,v) \geq v\}.$$ The Clarke Subdifferential generalizes the gra...

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The Clarke subdifferential of $f(x)=|x|$ at $x=0$ is a set $[-1, 1]$, and $\{sign(x)\}$ otherwise, just like the ordinary subdifferential because of convex. Now for g(x)= \begin{cases} |x|/2-x^2/4, \text { if }|x|\leq 1,\\ 1/4, \text { otherwise}, \end{cases} \end{equa...

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To give some context: I am aware of the uses of Convex Analysis (and its applications in Convex Optimization), I have been studying (for a while) the developments of Nonsmooth Analysis (and its applications in Nonsmooth Optimization) as traced by Frank Clarke. While all of this work on subdiffe...