数学代写|微积分代写Calculus代写|MTH2010

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• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|微积分代写Calculus代写|Multivalued Functions and Riemann Surfaces

Just as for real functions, a complex function $f^{-1}$ is the inverse function of a complex function $f$ when
$f^{-1}(f(z))=z$ for all $z$ in the domain of $f$
and
$w=f\left(f^{-1}(w)\right)$ for all $w$ in the range of $f$
The inverse is under composition (and is not the multiplicative inverse). Inverse functions are important because their application solves $f(w)=z$ for $w$. In other words, they provide the algebraic formula to solve important equations. For example, $f(z)=$ $z^{2}$ has inverse $f^{-1}(z)=\sqrt{z}$, which we can properly define. It solves $w^{2}=z$ for $w$, since (applying the inverse function to both sides) it means $w=\sqrt{z}$. The square root function is certainly not the only inverse function of interest. Likewise, $f^{-1}(z)=\sqrt[3]{z}$ solves $z=w^{3}$. Similarly, $f^{-1}(z)=z^{5 / 4}$ solves $z=w^{4 / 5}$. In the same way, we would like to properly define the logarithm function $f^{-1}(z)=\log z$ to solve $z=\mathrm{e}^{w}$. Each of these inverse function definitions follow the same structural pathway, and this section generally describes how that works.

The square root function provides a good place to start. The Extension Theorem’s demand that the square root function is an extension of its real-valued counterpart implies $\sqrt{z}=\sqrt{x}$ for real $z=x \geq 0$, but how is $f^{-1}(z)=\sqrt{z}$ defined for other complex values? It turns out, as we see below, that the polar representation of the domain value $z$ (and then the straightforward analysis of how the algebra of the inverse function should work) produces the correct definition of $f^{-1}(z)$.

数学代写|微积分代写Calculus代写|Riemann Surfaces

Since a multivalued function takes on more than one output value $f(z)$ at a given $z$, the branches’ output values $f_{1}(z), f_{2}(z)$, etc. will be range sets in the complex plane $\mathbb{C}$. They will each have the same domain. Bernhard Riemann had the clever idea of “gluing” the domain sets together, one on top of the other, to produce a single domain embedded in three dimensions.

The effect is to produce a single “sheet” in 3-space (with the complex plane as its two horizontal dimensions) that makes up the “total domain” of the multivalued function. Such a surface is called a Riemann surface, and the individual pieces of the sheet that correspond to each branch of the function is called a branch of the multivalued function’s domain. There would then be, for example, infinitely many domain sheet pieces above a given value of $z$ when $f$ has an infinite number of branches there. Furthermore, Riemann realized it could be possible to “glue” together the domain pieces to produce two very satisfying results. First, mathematicians can consider the function as defined on the Riemann surface’s domain, and then it is not multivalued. Instead, it maps each domain point on the Riemann surface to exactly one range value in $\mathbb{C}$. Second, this resulting function $f(z)$ could be considered as continuous over its Riemann surface, whereas the multivalued function isn’t continuous across any branch cut. In other words, when complex points $z$ and $w$ are close together in the Riemann surface, the output values $f(z)$ and $f(w)$ can be close together in the range space. In short, the inherently pleasant analytic properties of the function can exist when considering the Riemann surface as the domain, whereas the branch cuts destroy the multivalued function’s property of continuity. Nice!

Here’s the key to doing this successfully: Examine the range output for each branch. “Cut” each branch’s domain (in $\mathbb{C}$ ) along a slice that corresponds to where the branch’s range output has a boundary. This is called a branch cut. Each domain is a flat two-dimensional region. For each one, “pull” one side of the branch cut up into 3-space and the other side down. Then “glue” together the domain pieces of those branches so that the output points glue together, too, forming a continuous map on the Riemann surface domain.

数学代写|微积分代写Calculus代写|Multivalued Functions and Riemann Surfaces

$f^{-1}(f(z))=z$ 对所有人 $z$ 在领域 $f$

$w=f\left(f^{-1}(w)\right)$ 对所有人 $w$ 在范围内 $f$

$f^{-1}(z)=\sqrt[3]{z}$ 解决 $z=w^{3}$. 相似地， $f^{-1}(z)=z^{5 / 4}$ 解决 $z=w^{4 / 5}$. 同样，我们想正确定义对数函数
$f^{-1}(z)=\log z$ 解决 $z=\mathrm{e}^{w}$. 这些反函数定义中的每一个都遵循相同的结构路径，本节一般描述其工作原理。

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MATLAB代写

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