They're whole thing is that axioms should be motivated by what we see in the real world and we don't technically see infinite objects in the real world, and so we shouldn't have axioms that allow for infinite mathematical objects
My understanding for the Dehornoy order is, some symbolic manipulation involved in solving a set-theoretic question appeared coincidentally similar to the braid group relations
Set theorists should actively look for analogies between the infinite cardinal world and finitistic objects
That could be useful in the sense of, "If we could use this approach to prove this finite theorem, then the translation of the approach would prove this transfinite theorem which is known to be false and/or independent of ZFC, thus this approach won't work to prove the finite theorem"
One of the most powerful thing you can do in math is to show that an approach won't work, so you know not to waste time looking there
Like trying to prove something about a manifold, but without using the Hausdorff separation axiom
If it's not true of non-Hausdorff spaces in general, then your approach won't work
While playing with some ideas in graph theory, I stumbled onto this conjecture:
If $G$ is a planar graph and $H$ is a spanning subgraph (a subgraph with the same vertex set), let $F_H$ be the number of faces of $H$, and let $C_H$ be the number of connected components of $H$. Then:
$$\sum...
Yeah, I suppose so. Set theory takes the approach of "strip away all the extra rules and just look at the things underneath it all" whereas Category theory takes the approach of "look at the whole thing, extra rules and all."
fundamental theorem of real analysis: any theorem which seems nice and true (resp. any remotely nice class of functions) has at least one, incredibly contrived, counterexample (resp. nonexample)
If $f:[0,1]\to\Bbb R$ is a real function then what's $\int_0^{-1}f(|x|)dx$
It's $-\int_0^1f(x)dx$
The minus sign is because, even though the values of $f(|x|)$ are the same if you slide $x$ from $0$ to $1$ or from $0$ to $-1$, you're going in the other direction
Similarly, $\int_0^if(|x|)dx$ is $i\int_0^1f(x)dx$
@Thorgott By "upwards" I mean "the path of the integral is moving upwards in the complex plane" (rather than just left or right like in real analysis)
The extra factor of $-1$ in the real analysis case comes from $dx$ being negative. The extra factor of $i$ in the complex analysis case comes from $dx$ being imaginary
Think of $\int_a^b1dx$ for complex numbers $a$ and $b$ (for any path between the two), and the Riemann sum definition of an integral. Justify to yourself that this should be $b-a$
Hint: $\sum(t_{i+1}-t_i)$ (this looks like adding a bunch of vectors head-to-tail in the complex plane)
Why does $\oint x^{-1}dx=2\pi i$ (where the path of the integral is the unit circle)? Roughly: $x^{-1}$ is rotating clockwise and $dx$ is rotating clockwise
so the rotations basically cancel out, so their product is constant, and you basically end up adding $i\cdot|dx|$ for a length of $2\pi$
(The circle in $\oint$ means the path of integration is a loop)
Whoops: $x^{-1}$ is rotating clockwise and $dx$ is rotating counterclockwise
I was about to bring that up. It seems the contribution in the left direction and the right direction are each identically $0$ due to the symmetry around the imaginary axis. On the other hand, the downside contribution happens on the side with negative real part whereas the upside contribution happens on the side with positive real part, so the contributions match up, each equal to $\pi i$, and we get $2\pi i$ in total.
And if you stop partway - say $\int_1^{e^{i\theta}}x^{-1}dx$, along an arc of the unit circle, you get $i\theta$
@Thorgott Think of $x$ as changing over time, so at $t=0$ you have $x=1$ and at $t=2\pi$ you have $x=1$ again but it's gone around the circle
$dx/dt$ is the tangent vector, so it's rotating counterclockwise along with $x$
but what about $1/x$?
We can write $x=e^{it}$, so $1/x=e^{-it}$, which is basically traveling around the unit circle in the other direction
$dx/dt$ at $t=0$ is $i$, and $1/x$ at $t=0$ is $1$. So $(1/x)\frac{dx}{dt}=i\cdot1=i$
In general, $(1/x)\frac{dx}{dt}=e^{-it}(ie^{it})=i$
In fact we've basically just done a change of variables - $\oint x^{-1}dx=\int_0^{2\pi}e^{-it}(ie^{it}dt)=\int_0^{2\pi}i\,dt=2\pi i$
Honestly if you just started learning this stuff, understanding why $\int_a^b1dx$ is $b-a$ (and not, say, the length of the path between $a$ and $b$) is good enough
I mean, for me. If you're taking a test on this you might want to understand a bit more before the test comes around
but like I dunno if I would have ever learned this if I had the threat of a test looming over me
Huh, the parametrization corresponds pretty directly to how the rotations cancel out. This also seems to illustrate how the contour integral can be thought of as the average of function value * tangent at that point over the path times its length (although that interpretation wasn't as intuitive to me). This is neat.
For example, for $n=1$, $\int_0^{2\pi}e^{ix}dx$ is basically the average of all points on the unit circle (times $2\pi$)
Yeah
So now if we look at $\oint x^ndx$, the contour integral around the unit circle
and let $x=e^{it}$
we get $\oint x^ndx=\int_0^{2\pi}e^{int}(ie^{it}dt)$
${}=i\int_0^{2\pi}e^{i(n+1)t}dt$, which is $0$ for $n\ne-1$ and $2\pi i$ for $n=-1$
This is one explanation for why $\oint x^{-1}dx$ is nonzero but all other $\oint x^ndx$ is zero
The other explanation? The fundamental theorem of calculus still holds for contour integration: $\int_a^bf'(x)dx=f(b)-f(a)$
When $n\ne-1$, the function $x^n$ has the antiderivative $\frac1{n+1}x^{n+1}$
and $\oint$ is basically a fancy way of writing $\int_1^1$
so, because we have an antiderivative, the Fundamental Theorem of Calculus assures us that the path doesn't matter, and we end up with $\frac1{n+1}1^{n+1}-\frac1{n+1}1^{n+1}=0$
Why doesn't this work for $x^{-1}$? Because it has no antiderivative on the complex plane (or I suppose $\Bbb C\setminus\{0\}$)
because you can't continuously define $\ln$ over the entire complex plane (even excluding zero)
because if you could, we'd have $0=\ln(1)=\ln(e^{2\pi i})=2\pi i$
the existence of a continuous log branch on a domain is equivalent to the existence of an antiderivative of $x^{-1}$ on the domain. you can use the other direction to construct two continuous log branches that have non-constant difference on their non-empty intersection by choosing two "sausages" that together contain the unit circle and then the non-constant difference is a consequence of the non-vanishing of the $x^{-1}$ integral (tangential, but it's cool)
I guess the intuitive explanation of the vanishing of the integral for $n\neq-1$ in vein of the above would be that the functions clockwise rotation is now faster or slower (depending on whether n is positive or negative) than the counterclockwise rotation of x, so they dont cancel out and since $n$ natural, we get a full multiple of either clockwise or counter-clockwise rotations and those cancel each.
I'm not very familiar with the Dehornoy order but another example where large cardinals are surprisingly related to small objects are Laver tables @Akiva
The $n$-th Laver table is the "multiplication table" of a weird operation on a $2^n$ elements set. If you look at the first row of the $n$-th table it is always periodic, let $f$ be the function mapping $n$ to the length of this period. The only known proof that $f$ diverges (but incredibly slowly) assumes the existence of an $I_3$ cardinal which is an incredibly strong assumption, very close to outright inconsistency
i) Is the following function defined with $\mathrm{x}_{1}>0, \quad \mathrm{x}_{2}>0$ quasiconvex? Is it
also strictly convex? In each case, provide a full explanation.
$$
\mathrm{C}=\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right)=3 \mathrm{x}_{1}^{4}+5 \mathrm{x}_{2}^{2}
$$
I guess for thi...
What is history of polynomials? We study in uni their theory and how to find roots, but what really motivated to find x in $x^2+5x=6$ ? I've read that it is related to finding side of the field with given area. But what about cubics ?
hey guys, I think I remember seeing a multivariate calculus book that had stereograms in its figures. Can someone confirm i'm not making this up, and remind me what its called?
(a stereogram is a picture you stare at cross-eyed to see a 3D picture)
For a linear system $(m=4, n=2)$ made up of random numbers, what is the probability,
roughly, that the system has one solution?
For a linear system with $m=n=2$ made up of random numbers, what is the probability,
roughly, that the system has no solution?
For second question I think probability ...
Let $z \in \mathbb{C}$ find the answer to the equation
$z^2 + iz - \frac{10}{4} = 0$
Using the PQ-Formela one gets $ \frac{-i}{2} \pm \frac{i. (11)^{1/2}}{2} $ Does one stop here and say $z$ = what was just mentioned or does one need to do further calculations?
Try to construct two any two none identical planes that do intersect, you will see that their intersection is a straight. I do not know how to proof this mathematically.
Probably means a system of four equations in two variables (or two equations in four variables?)
The question then becomes "choosing values at random, what is the probability that the system will be linearly dependent and a vector of numbers, also chosen at random, is not a valid solution for this system of equations?"
I just went to the town hall to register myself (which you should do within 2 weeks of moving in) 3 months late, pretended to not be able to speak German to get out of a fine lmfaooo
I know that what taking square roots for reals, we can choose the standard square root in such a way that the square root function is continuous, with respect to the metric.
Why is that not the case over $\mathbb{C}$, with respect the the $\mathbb{R}^2$ metric? I suppose what I'm trying to ask i...
yeah this was the history section of the Wikipedia article but yes I should go to a library. Far out though the sheer genius of all the big names really hits you with this sinking feeling like why tf do even try when they were this incredible hundreds of years ago without computer aide
I mean the reason I want to know is because abstract algebra is killing me as in it doesn't come second nature and requires a shit load of effort and Im just worried that im too old to even make trying to learn it worth while
like it would be good news if I read that he conceived all of that hitting 40 and still living with mother. Its a long shot but who knows maybe home ownership was equally unrealistic then
I'm looking at a mapping of the form $x\mapsto\frac{x}{1-\langle x,y\rangle}$ in $\mathbb{R}^n$, where $y$ is fixed and $\langle\cdot,\cdot\rangle$ denotes the scalar product. Does this have a name and/or geometric interpretation?
I'm trying to think about how this maps the unit ball. The plane gets left fixed. The image is also certainly unbounded in the $z$ direction. Due to the restrictions on $x,y$ when $z$ approaches $(0,0,1)$, I'm imagine something conic-like, but I'm not sure.
Im writing a little exam tommorow... and i have some problems to understand how to derive the Hesse normal form to calculate the distance of a vector to a plane... E: ((x-a) * n) / |n| why do we divide trough |n| ? Its returns the distance from x to the plane... But wouldnt this return the distance from the normalized vector n to the point x instead ?
I'm trying to get a sense of joint distributions of two random variables, where one is discrete with values in $\mathbb{N}$ and the other is continuous with domain $(0,\infty)$ or $[0,\infty)$
This doesn't correspond to taylor series, is what I've learned, just now
The plane is the orange thing... n is the red vector which symbolizes the z axis... our point x is represented by the black vector arrow ( where the green marker points to )... no other values... the green marker represents the distance... :) The question here is why do we devide trough |n| ... wouldnt this return the distance b in the pic ?
And if so... isnt this the distance from our point x towards our normalized vector n instead of the distance from our point x towards the orange plane ( green arrow ) ?
While playing with some ideas in graph theory, I stumbled onto this conjecture:
If $G$ is a planar graph and $H$ is a spanning subgraph (a subgraph with the same vertex set), let $F_H$ be the number of faces of $H$, and let $C_H$ be the number of connected components of $H$. Then:
$$\sum...
