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univalent
univalent, a. and n. Chem. (juːnɪˈveɪlənt, juːˈnɪvələnt) [f. uni- + L. valent-em, pr. pple. of valēre to be worth.] A. adj. 1. Having a valency of one; having the combining power of one atom of hydrogen or other radical. Also, in recent Dicts. (1891–), univalence, univalency.1869 Eng. Mech. 19 Nov. ...
Oxford English Dictionary
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Univalent
Univalent may refer to:
Univalent function – an injective holomorphic function on an open subset of the complex plane
Univalent foundations – a type-based approach to foundation of mathematics
Univalent relation – a binary relation R that satisfies
Valence (chemistry)#univalent – 1-valent.
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Univalent function
In mathematics, in the branch of complex analysis, a holomorphic function on an open subset of the complex plane is called univalent if it is injective Examples
The function is univalent in the open unit disc, as implies that .
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show univalent $g(z)=z+c_{n+1}z^{n+1}+c_{2n+1}z^{2n+1}+c_{3n+1}z^{3n+1}+\cdots$ can be written $g(z)=\sqrt[n]{f(z^n)}$ with $f$ univalent Show that the univalent function $$g(z)=z+c_{n+1}z^{n+1}+c_{2n+1}z^{2n+1}+c_{3n...
You have $g(z)=zh(z^n)$ and $f(z)=zh(z)^n$ (for the evident $h$). Now if $f(z)=f(w)$ (for some $z,w$), i.e. $zh(z)^n=wh(w)^n$ then $z^{1/n}h(z)=w^{1/n}h(z)$ for _some_ choice of $z^{1/n},w^{1/n}$, i.e. $g(z^{1/n})=g(w^{1/n})$, so $z^{1/n}=w^{1/n}$ and thus $z=w$.
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Univalent foundations
Such types are called "propositions" in univalent foundations. It is claimed by the creators of univalent foundations that the univalent formalization of sets in Martin-Löf type theory is the best currently-available
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complex analysis (Univalent function ) The Distortion Theorem tells us that if $f$ is a univalent function on $\mathbb{D}:=\\{z:|z|<1\\}$, then $|f'(z)|\leq 12\,|f'(0)|$ for $|z|\leq\frac12$. By iterating this, prove ...
Gah, what a horribly wasteful way to estimate the derivative. You are basically asked to go from $z$ to $w$ in horizontal steps. This is not how one should move about in the half-plane model of hyperbolic plane. Oh well. Begin with the estimate $$|f'(\zeta)|\le 12 |f'(z)|,\qquad\text{where }\ |z-\ze...
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Univalent function with parameter. Let $a, b, z_0 \in \mathbb{C}$. Find the highest value R, at which the function: $f(z) = z^2 + az + b$ is univalent in the disk: $|z - z_0| < R$ I use the definition of univalent fun...
You are not on the right track. The derivative of $f(z)$ is $f'(z) = 2z+a$, which is zero at the point $z_1 = -a/2$. Hence the highest value of $R$ is at most $R = |z_1-z_0|$. On the other hand it is easy to see that the function is injective on the region $$ U = \\{ z : |z - z_0| < |z_1 - z_0| \\},...
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Univalent Möbius transformation Let $f(z)=\frac{(az+b)}{(cz+d)}$ be a Möbius transformation where $c \neq0$. Through the process of actually computing $f^{-1}$ show that $f$ is a univalent function whose domain-set is...
To give a slightly higher level perspective on things.... Let $\bar{\mathbf{C}} = \mathbf{C} \cup \\{ \infty \\}$ be the Riemann sphere a.k.a. the projective complex numbers. It turns out that a Möbius transformation is an _invertible_ function on $\bar{\mathbf{C}}$, and is the continuous extension ...
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Some confusion about definition of univalent function I was studying about univalent function from this pdf . But in this pdf, it is written that function is univalent if it is injective. But I have read many places t...
The theory of "univalent functions" in complex analysis is all about univalent analytic functions. say "univalent".
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Derivative of an univalent function I know that the derivative of a univalent function cannot be zero. But I am wondering if a function has non zero derivative, then is it univalent? Under what conditions the converse...
Let $f$ analytic in the unit disc and locally univalent ($f'(z) \ne 0$ anywhere); then consider the Schwarzian derivative $S(f)(z)= \frac{d}{dz}({\frac Nehari's theorem says that if $|S(f)(z)| \le 2(1-|z|^2)^{-2}$ for all $|z|<1$, then $f$ is univalent and $2$ is best constant for which the result holds
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Problem related with univalent function I am stuck with the following problem: As wikipedia says : "In mathematics, in the branch of complex analysis, a holomorphic function on an open subset of the complex plane is...
As Daniel Fischer said, the inequality should have been $|a|<1$. I edited the article.
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What is the use of introduction of term univalent function? > What is the use of defining a new term univalent function although there already existed a definition of bijective, one-to-one functions ?
The definition has 3 parts (in bold) to it, so it is more convenient to create a definition instead of writing it out every time:
> A _univalent_ function
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Question about univalent branch Problem: Let the holomorphic function $f$ have a zero of order $m$ at the point $z_0$. Prove that there is an open disk centered at $z_0$ in which there is a univalent branch of $f^{1/m...
By Taylor's theorem $$f(z)=a_m(z-z_0)^m +a_{m+1}(z-z_0)^{m+1}+\cdots$$ in a neighbourhood of $z_0$ where $a_m\ne0$. Then $$f(z)=c^m (z-a_0)^mg(z)$$ for some $c$ where $g$ is holomorphic and $g(z_0)=1$. Then $g$ has a holomorphic $m$-th root near $z_0$. (Compose $g$ with a holomorphic branch of $w\ma...
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Prove that if derivative of a holomorphic function is non-zero, then it is locally univalent Let $f$ be a non-constant entire function such that $f'(z)≠ 0$. Then $f$ is locally univalent.
Now choose $\rho\in(0,r)$ so that $|z|<\rho$ implies $|f(z)|<\delta$ and it follows that $f$ is univalent in $\\{|z|<\rho\\}$.
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$\omega(z)=z+\frac 1 z$ is univalent function? I need to find if > $\omega(z)=z+\frac 1 z$ is **univalent** function **My attempt:** First, I know that: $\omega(z)=z+\frac 1 z$ $=(x+\frac{x}{x^2+y^2})+i(y-\fra...
If _univalent_ means _injective_ , then no: $\omega(\pm i)=0$.
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