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Univalent - Wikipedia
Univalent foundations – a type-based approach to foundation of mathematics; Univalent relation – a binary relation R that satisfies x R y and x R z implies y ... en.wikipedia.org
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Univalent function - Wikipedia
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. en.wikipedia.org
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UNIVALENT Definition & Meaning - Merriam-Webster
1. monovalent sense 1 2. being a chromosomal univalent univalent 2 of 2 noun : a chromosome that lacks a synaptic mate. www.merriam-webster.com
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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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The Origins and Motivations of Univalent Foundations - Ideas
The origins and motivations of univalent foundations: A personal mission to develop computer proof verification to avoid mathematical mistakes. www.ias.edu
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univalent category in nLab
In simplicial type theory, a Rezk complete category, univalent category, or saturated category is a complete Segal type in which all hom-types ... ncatlab.org
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univalent, adj. & n. meanings, etymology and more
univalent is a borrowing from Latin, combined with an English element. Etymons: uni- comb. form, Latin valent-em, valēre. www.oed.com
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UNIVALENT | definition in the Cambridge English Dictionary
UNIVALENT meaning: 1. (of atoms or molecules) having a valency of one: 2. (of chromosomes) existing as a single…. Learn more. dictionary.cambridge.org
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[PDF] The Basic Theory of Univalent Functions - University of Regina
The motivation for why someone should care about univalent functions as a prerequisite for L oewner 's equation and SLE is also currently missing. N onetheless ... uregina.ca
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What does "univalent" mean? - Mathematics Stack Exchange
A bundle is said to be univalent if every other bundle is a pullback of it in at most one way (up to homotopy). math.stackexchange.com
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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 . wikipedia.org
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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 $$g(z)=z+c_{n+1}z^{n+1}+c_{2n+1}z^{2n+1}+c_{3n...
You have $g(z)=zh(z^n)$ and $f(z)=zh(z)^n$ (for the evident $h$). Now if $f(z)=f(w)$ (for some $z,w$), i.e. $zh(z)^n=wh(w)^n$ then $z^{1/n}h(z)=w^{1/n}h(z)$ for _some_ choice of $z^{1/n},w^{1/n}$, i.e. $g(z^{1/n})=g(w^{1/n})$, so $z^{1/n}=w^{1/n}$ and thus $z=w$.
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Univalent foundations
Such types are called "propositions" in univalent foundations. It is claimed by the creators of univalent foundations that the univalent formalization of sets in Martin-Löf type theory is the best currently-available wikipedia.org
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complex analysis (Univalent function ) The Distortion Theorem tells us that if $f$ is a univalent function on $\mathbb{D}:=\\{z:|z|<1\\}$, then $|f'(z)|\leq 12\,|f'(0)|$ for $|z|\leq\frac12$. By iterating this, prove ...
Gah, what a horribly wasteful way to estimate the derivative. You are basically asked to go from $z$ to $w$ in horizontal steps. This is not how one should move about in the half-plane model of hyperbolic plane. Oh well. Begin with the estimate $$|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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