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4

Take $\psi(t)=(1+t^2)^{\frac 1 2}$, then $\psi''(t)=(1+t^2)^{-\frac 3 2}\geq 0$, $\psi$ is convex on $\mathbb{R}$. By Taylor-lagrange : $$\psi(t)=\psi(s)+(t-s)\psi'(s) +\frac 1 2 (t-s)^2 \psi''(\xi) \geq \psi(s)+(t-s)\psi'(s)~~ \forall t,s \in \mathbb{R}$$ So, for all $t \in [0,1]$ : $$(1+f'(t)^2)^{\frac 1 2} \geq (1+F'(t)^2)^{\frac 1 2} + (f'(t)-F'(t)) ...


4

Hint: If $xy \geq 0$ then $x^2+y^2 \geq 2xy \geq xy$. If $xy <0$ then $x^2+y^2>xy$. Therefore, $f(x,y)=xy.$


3

Your equation can be rearranged to $$ \cos\chi = \sqrt{f}\cos\eta - \sin\eta. \tag{1} $$ Let $\phi$ be an angle (unique up to an added integer multiple of $2\pi$) satisfying $$ \cos\phi = \sqrt{\frac{f}{f + 1}},\qquad \sin\phi = \frac{1}{\sqrt{f + 1}}. $$ By the sum formula for cosine, (1) becomes $$ \cos\chi = \sqrt{f + 1}(\cos\phi \cos\eta - \sin\phi ...


2

This is indeed true but nontrivial, it uses some advanced complex analysis and I am unaware of any purely topological proofs. First of all, the assumption $\partial A= \partial B$ and connectivity of $A$ and $B$ imply that $A$ is simply connected. (The same holds for $B'= B\cup \{\infty\}$.) Now, since $K\ne \emptyset$, $A$ is a proper open simply connected ...


2

Here is a solution (see graphics below), $\gamma([a,b])$ is the concatenation of three (or better said four) circles: The circuit begins in $B$, describes twice the circle with center $C$, then describes the circle with center $D$, all of them with a positive orientation, then, arriving back in $B$, describes the circle with center $A$ in the negative ...


2

$$ f:t \mapsto \frac{t}{1+t^2} $$ is an "almost injective" function: $f(x)=f(y)$ iff $xy=1$. In such a case, however, $$ g(x)=\frac{x}{1+x^4}\neq \frac{y}{1+y^4} = g(y)$$ unless $x=y=1$.


2

If $f: X \to Y$ is Lipschitz with Lipschitz constant $k$, i.e. $d(f(x),f(y)) \le k d(x,y)$, then for any $r$ we have $\mathcal H^r(f(X)) \le k^r \mathcal H^r(X)$, where $\mathcal H^r$ is $r$-dimensional Hausdorff measure. Since any continuously differentiable function on an interval is locally Lipschitz, it follows that the image of a $C^1$ curve has ...


1

There is in fact a standard method for such problems, relying on the definition of injectivity. Here it is: let us assume that $$\gamma(s)=\gamma(t)$$ with the objective to deduce that necessarily $s=t$. or, in an equivalent way, that there doesn't exist a couple $(s,t)$ such that $$\gamma(s)=\gamma(t) \ \ \ \text{with} \ s \neq t \ \ \ \ (1)$$ Assume ...


1

Yes, this batman equation is real. For context, the batman curve is a piecewise curve in the shape of the logo of the Batman superhero originally posted on reddit.com on Jul. 28, 2011. It can written as two functions, one for the upper part and the other for the lower part, as $$f(x) = (h-l) \,H \,(x+1)+(r-h) \, H\,(x-1)+(l-w)\,H\,(x+3)+(w-r)\,H\,(x-3)+w$$ ...


1

Suppose $\gamma$ is a simple, closed, piecewise smooth curve in $\mathbb{C}$ defined on $[0,1]$. Choose $t_0$ where $|\gamma(t)|$ achieves its maximum. Then the curve does not intersect the halfplane $H$ which is tangent to the circle of radius $R=|\gamma(t_0)|$ at $\gamma(t_0)$, except at $\gamma(t_0)$. Then there is a ray through $\gamma(t_0)$ in the ...


1

Look, for example, at pp. 85-91 of Guillemin and Pollack for a version of this valid for smooth hypersurfaces in $\Bbb R^n$.


1

Using Andrew's answer, I could define the following parametrization, which appears to be working great : \begin{align} \big(\, \chi(s), \; \eta(s) \big) = \big(\arccos{s}, \quad \arccos{\frac{s}{\sqrt{1 + f}}} - \arccos{\sqrt{\frac{f}{1 + f}}} \big), \end{align} where $-1 < s < 1$ (or $s = \cos{\chi}$). But I still suspect this could be improved (i.e. ...



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