Mathematics - 2019-02-22

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Rithaniel

03:09

You know, I occasionally see people talk about allowing division by zero and seeing if you can somehow make it work. Though, it's always from the perspective of zero as an additive identity and playing around with redefining division. I wonder if you could approach it from another direction. Start out with just division and see if you can redefine addition such that zero remains an additive identity.

(Also, an ellipse field sounds interesting, @AkivaWeinberger Though, wouldn't it be roughly the same structure as a vector field? How would the two differ?)

Jacksoja

03:24

@MatheinBoulomenos can you please send me link to your blogg ?

1 hour later…

BAYMAX

04:44

A: Equation for a smooth step function

The sigmoid function, $S(x) = \frac{1}{1+e^{-x}}$, achieves close to what you need (with appropriate scaling and shifting of the function). Do you need the function to be exactly $\pm 50$ when evaluated at $\pm 10$? If so, a polynomial option would be to use something called the smoothstep func...

I was thinking about the first answer

how to think of the middle function of that piecewise funtion?

$100*(3(0.05x+0.5)^2-2(0.05x+0.5)^3 - 0.5)$

for $-10 < x \leq 10$

I can see that the form $3x^2 - 2x^3$ was used but still how?

to get the expression $0.05x + 0.5$ ?

1 hour later…

Derek Adams

06:18

I've a really silly question, but how in the world do I show that $3(6 + 7t) \equiv 4 \pmod{8}$ simplifies to $5t \equiv 2 \pmod{8}$?

Tobias Kildetoft

06:30

@DerekAdams Just do as you usually would if it was a regular equation and then remember to replace all numbers by their remainder modulo 8

Adan

07:16

hello

Q: How do I calculate fixed amount of monthly payments to pay off a debt with interest?

If my balance owed is 10,000 and my APR is 27%, if I wanted to pay the debt in full in 4 months, what would be my payments per month? here is the website where it takes inputs: https://www.creditkarma.com/calculators/debtrepayment According to the site, my payments would be 2,642 per month. ho...

Derek Adams

@Tobi

@TobiasKildetoft Thank you, very silly oversight on my part

Akiva Weinberger

07:34

@Rithaniel For one thing, you can't put a nonvanishing continuous vector field on a sphere (the hairy ball theorem) but you can put an "ellipse field" on one (just put a unit circle on each point)

Hm. I guess there's no continuous ellipse field on a sphere that avoids circles?

Also, here's something: if you start with the "unit circle everywhere" field on a plane, and then deform the plane (do some continuous bijection $\Bbb R^2\to\Bbb R^2$ to it), that deformation also deforms the ellipse field

so then you can ask which fields on the plane came from deforming the unit-circle field on the plane in that way

(If that deformation is holomorphic then all the circles stay circles)

On the other hand, if you take a unit-circle field from a piece of some curved surface like a sphere, and deform them onto the plane, you'll get different ellipse fields

Like, stereographic projection of a sphere will result in a field with circles that get bigger as you move away from the origin

Different projections from the sphere give you these things

Tissot's indicatrix

In cartography, a Tissot's indicatrix (Tissot indicatrix, Tissot's ellipse, Tissot ellipse, ellipse of distortion) (plural: "Tissot's indicatrices") is a mathematical contrivance presented by French mathematician Nicolas Auguste Tissot in 1859 and 1871 in order to characterize local distortions due to map projection. It is the geometry that results from projecting a circle of infinitesimal radius from a curved geometric model, such as a globe, onto a map. Tissot proved that the resulting diagram is an ellipse whose axes indicate the two principal directions along which scale is maximal and minimal...

Akiva Weinberger

07:56

Oh also SURPRISE this is actually Riemannian geometry kinda

Well not really but they are connected

2 hours later…

MatheinBoulomenos

10:07

@Jacksoja wlou.blog

Frank Science

I need check a very simple fact. Given a ring $R$ and a formal power series $f=\sum_{n\in\mathbb N}a_nT^n\in R[[T]]$. If $a_0\in R^\times$ is invertible, then it follows from invertible function theorem that $f$ has an inverse $g\in R[[T]]$ ($f(g(T))=g(f(T))=T$), correct?

Secret

@Rithaniel hmm..., most literature played with multiplication and the distributive law. Addition seemed to be a place to explore further

Frank Science

10:29

$R$ is not necessarily commutative. For the commutative case, this seems to follow from Lagrange inversion theorem, but I think that the conclusion remains correct.

2 hours later…

MatheinBoulomenos

12:33

@FrankScience I think you mean $a_1 \in R^\times$

and $a_0 = 0$

user193319

12:57

Let $A$ be a Lebesgue measurable set of finite measure, and let $\{f_n : A \to \Bbb{R}\}$ be a sequence of measurable sets converging to $f : A \to \Bbb{R}$ pointwise. Does there exist $B \subseteq A$ such that $m(A \setminus B) = 0$ and $f_n$ converges uniformly to $f$ on $B$? I think I was able to prove this using Egoroff's theorem. I just want to verify that it is in fact true.

MatheinBoulomenos

@user193319 yes, this is correct

user193319

Sweet! Thanks!

MatheinBoulomenos

13:18

@user193319 wait, no, I made a mistake, this is not true. Consider $A=[0,1)$ and $f_n=x^n$, this converges pointwise to the zero function.
But if $B \subset [0,1)$ is of measure zero, then for any $m \in \Bbb N$, there is a point in the interval $[1-1/m,1-1/(m+1)]$ which is not in $B$ (because else $B$ would have positive measure). Using that each $f_n$ is monotonic, we get that $\sup_{x \in [0,1) \setminus B} |f_n(x)| \geq (1-1/m)^n$, letting $m \to \infty$ gives $\sup_{x \in [0,1) \setminus B} |f_n(x)| \geq 1$ which implies that $f_n$ doesn't converge uniformly to $0$ on $[0,1) \setminus B$

user193319

Why are you assuming that $B$ is of measure $0$? I want $A \setminus B$ to be of measure $0$, not $B$.

MatheinBoulomenos

replace $B$ with $A\setminus B$

corrected version: if $B \subset [0,1)$ is a subset such that $[0,1) \setminus B$ is of measure $0$, then for any $m \in \Bbb N$, there is a point in the interval $[1-1/m,1-1/(m+1)]$ which is in $B$ (because else $[0,1) \setminus B$ would have positive measure). Using that each $f_n$ is monotonic, we get that $\sup_{x \in B} |f_n(x)| \geq (1-1/m)^n$, letting $m \to \infty$ gives $\sup_{x \in B} |f_n(x)| \geq 1$ which implies that $f_n$ doesn't converge uniformly to $0$ on $B$

I can guess your proof attempt (since I think I made the same mistake) and I know what goes wrong: using Egoroff's theorem, you can find for each $N$ a set $B_N \subset A$ such that $m(A \setminus B_N) < 1/N$ and $f_n$ converges uniformly to $f$ on $B_N$. Now if you set $B=\bigcup B_N$, then you will have $m(A \setminus B)=0$, but you can't get uniform convergence on $B$ from uniform convergence on each $B_N$

user193319

Hmm...true. Thanks for the counterexample.

Frank Science

14:14

@MatheinBoulomenos Yes, $a_1\in R^\times$. It should have been $f=\sum_{n\in\mathbb Z_{>0}}a_nT^n$.

However, the appearance of $a_0$ does not affect the correctness.

Tug' Tegin

14:51

math.stackexchange.com/questions/174401/…. I cannot understand the second line of Arturo Magidin's answer where it says $\exists i \in I$; can somebody explain what $I$ stands foright?

Noob

Q: Is saying 'This statement is true' a logically valid statement?

I understand how 'This statement is false' is not logically valid, but what about 'This statement is true'? I've always heard self-referential statements are not logically sound, but I can't really give a great explanation for why this one would not be. Anyone have a good argument one way or the ...

logic

can somebody describe the answer in simple words ?

2 hours later…

MatheinBoulomenos

17:00

@FrankScience but composition of formal power series need not be defined if $a_0$ is non-zero

Perturbative

Hey everyone!

Apparently the map $L$ is not an isomorphism of rings, but if it's not an isomorphism of rings what type of isomorphism is it?

Because $H^{\star}(X;R) \otimes_R H^{\star}(F;R)$ and $H^{\star}(E; R)$ are both rings

Oh perhaps $L$ is an $R$-module isomorphism

MatheinBoulomenos

17:16

@Perturbative yes, that's what is meant

Perturbative

17:31

Ah okay thanks @MatheinBoulomenos

1 hour later…

Saeid Nourian

18:40

Hi, is anyone here interested in 3D math graphs? If so, would you like to join this new StackExchange site about Graphing Calculator 3D software? area51.stackexchange.com/proposals/120787/…

user193319

19:01

$Mathematics$ Group Theory

Let's discuss group theory!

2 hours later…

Hermit with Adjoint

20:58

That's the splash screen for category theory software I'm writing

Why do I write code like this? Because I has too many Adjoints.

Ultradark

Is it possible to map the lattice of the natural numbers to the unit square

frogeyedpeas

21:39

what type of map?

oh by lattice you intend to have a structure preserving map

1 hour later…

Jake S

22:59

Hello, is anyone around that could help me figure out why a very simple calculation of Stirling numbers of the second kind seems to be giving me the wrong number?