Winding numbers and domain coloring

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Recommends edm · 2018-03 · Mathematics Stack Exchange · on “What are the higher-dimensional generalisation of intermediate value theorem and bisection method?”

I watched this video by 3Blue1Brown on how to find zeroes of a continuous map from $\Bbb R^2$ to $\Bbb R^2$: [this video] He generalises intermediate value theorem on $\Bbb R$ to the plane to become "winding number not zero implies there is a zero of the function", and bisection method to bisecting rectangles, using the fact that winding numbers of loops add together nicely. I think this generalisation can be done to any higher dimensions. Here, I try to formulate the generalisation: Suppose $U\subseteq\Bbb R^{n+1}$ is open, convex and nonempty. Let $f:U\to\Bbb R^{n+1}$ be continuous. Suppose $S\subseteq U$ is a set that is homeomorphic to the $n$-sphere and for all $x\in S$, $f(x)\not=0$. We res

Recommends Roy Sht · 2020-06 · Mathematics Stack Exchange · on “Intuition for a Proof of the Fundamental Theorem of Algebra (According to 3Blue1Brown)”

I have recently watched this [this video] video by $3$ Blue $1$ Brown, regarding winding numbers and the fundamental theorem of algebra. I am trying to formalize the idea he shows in the video (starting at 21:26). Specifically, I am trying to connect his proof to a one I've seen in complex analysis classes: Theorem. Fundamental theorem of algebra. Let $P(z)=z^d+a_1 z^{d-1}+a_2z^{d-2}+\ldots+a_{d-1}z+a_d\in \Bbb{C}[x]$ be a monic complex polynomial of degree $d$ . So the sum of all orders of $P$ 's zeroes is exactly $d$ . Proof. For a function $f$ , denote the sum of orders of all its zeros as $N_f$ . Let $Q(z)=z^d$ . It is obvious that $N_Q=d$ . Subtracting $Q$ from $P$ gives a polynomial of degr

Recommends Arthur · 2020-10 · Mathematics Stack Exchange · on “Bisection method complex roots”

There is a winding number variation on the bisection method. Here [this video] is 3blue1brown's video on the subject. I don't know to what degree it's actually usable in practice, though. (This winding number idea is the core of my favourite proof of the fundamental theorem of arithmetic.)

Recommends K.defaoite · 2020-11 · Mathematics Stack Exchange · on “Algorithm to find roots of a scalar field”

3Blue1Brown has a beautiful visualization [this video] of an ingenious root finding method for two variable equations.

Recommends Accelerator · 2023-11 · Mathematics Stack Exchange · on “Do colour space models form vector spaces?”

I guess another type of color modeling is that in $\mathbb{C}$, a pretty way to illustrate 2D functions is by associating each point in the output space with a color. Specifically, you can use hues to correspond to the direction away from $(0,0)$, and you can use shading to correspond to the distance away from $(0,0)$. It's typically called "domain coloring", and 3Blue1Brown talks about it in a video here [this video] .

Cites Felix · 2018-04 · Mathematics Stack Exchange · on “Color hues from rotation around origin”

What I'm trying to reproduce is a set of functions that convert a vector to RGB values so that the hue goes full circle around the origin. After seeing a 3Blue1Brown video [this video] about graphically solving 2D-equations which have 2D outputs, like complex functions, I wanted to replicate the algorithm in Matlab. [LINK] The brightening of the colors radiating from the origin is somewhat trivial, but it's the colors I'm interested in. I decided to fiddle around in MS Paint to see what kinds of values fully saturated RGB colors have. For a red color (brightness half the way) the values are obviously $[255,0,0]$. Similarly for green and blue. Yellow, greenblue (sorry not English :D) and pink have

Cites D.R. · 2023-05 · Mathematics Stack Exchange · on “Which series expansion does the Complex Argument $\arg(z)\in\mathbb{C}\cup\{\hat\infty\}$ where $z \in \mathbb{C}$ have?”

You could expand the Taylor series for $\ln(z)$ at other points (e.g. $z = \pm i$ ) to extend the values of $z$ for which the series converges. Because the argument of $z$ is the same as that of $z/|z|$ (on the unit circle), these 3 patches of $\ln(z)$ will cover about 3/4 of the unit circle. One can extend further, by considering 2 different Taylor series of $\ln(z)$ at $z=-1$ ; one using $\ln(−1)=−i\pi$ and one using $\ln(−1)=i\pi$ as the constant term of the series (the higher derivative terms will be the same). In this way we get full coverage of the unit circle. Further Remarks: the above was mostly a copy-paste of my previous comment, but since you are so kind to reward the bounty to m