Mathematics

12:17 AM

could you possibly take a look at my question? it seems it has received very few attentions and probably is out of interest of community...

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Q: calculate geometry properties using boundary coordinates

Calculation of geometry properties of various shapes are useful specifically in engineering purposes. This is very easily formulated when the shapes of the geometry are rectangular, circle, etc. However, when the shape of becomes a combination of these and becomes more complicated, such calculati...

AMDG

2:56 AM

I thought I read somewhere that you can define ln(x) with exp(x). Is this true and what is the identity (besides the obvious e^y = x)?

Thorgott

For $x>0$, $\ln(x)$ is the unique real number satisfying $e^{\ln(x)}=x$

AMDG

Yeah, but what is that inverse? Can you not define logarithms using exponential functions?

A closed form not using any logarithms

from what I can see based on things such as the infinite series of tanh(x) etc., we don't really need ln(x) to compute ln(x)... we can use exp(x) somehow. It also means we can compute arctan without i using exp(x) for all real values of x--somehow. We'd have to simply adjust the coefficient, not that I'm saying that would be easy :)

Some unknown function f(x) for all real values of x as exp(f(x)*x) maps to arctan(x), therefore.

AMDG

3:29 AM

"infinite power tower"

That sounds so good and so funny at the same time

mathworld.wolfram.com/LambertW-Function.html

manooooh

Hello all! I'm trying to prove that if $(B,+,\cdot)$ is a Boolean Algebra, then for all $x,y\in B$, $x+y=x+y'$ then $x=1$

But I can't prove it

I started with $x+y=x+y'$, then $x+y+y=x+y'+y$, so $x+y=1$, then?? Also, starting from LHS, then $x=x+xy=(x+x)(x+y)=x(x+y')=x+xy'$, then??

jeea

8:17 AM

If two matrices have equal traces and determinants will they necessarily be similar?

ok so if they also have same characteristic polynomial then?

Martin R

9:20 AM

There is a website where you can enter an (approximate) real number, and it returns (algebraic?) expressions equal or close to that number. – I can't remember the name of that site, can someone help me?

Ice Wizard

9:58 AM

https://math.stackexchange.com/a/3840927/811225
How was the expansion written?

Stupid question inc

10:19 AM

is dirichlet theorem hard to prove?

Stupid question inc

10:40 AM

what should I learn before understanding dirichlet theorem's proof ?

Stupid question inc

10:57 AM

why don't you first prove if an+b is odd for a and b is positive integer coprime and n is just positive integer

alright I have a hypothesis that tells math community doesn't really give a sh

feynhat

@jeea No.

Balarka Sen

Sup @feynhat

feynhat

Yo. Tell me something about free group with countably infinite generators.

Balarka Sen

Uhoh

What do you want to know

feynhat

What do its quotients look like?

Balarka Sen

11:05 AM

Yikes dude. You get all countably generated groups.

feynhat

That sucks.

What does a puncture in the surface with infinite genus deform to? I mean surely it is some product of commutators, right? But can it be an infinite product of commutators?

Balarka Sen

@feynhat That's a good question. Remember that the surface of infinite genus (let's say, one-ended infinite genus surface, for the sake of simplicity) deformation retacts onto a wedge of circles.

If you add a puncture to the first copy of $T^2$ in $T^2 \# T^2 \# \cdots$, what happens to it under this 'deformation retraction from infinity'?

It won't be a product of commutators, product of commutators aren't nullhomotopic in the infinite genus surface anymore.

feynhat

Hmm... I can't see what happens to this puncture under the def ret.

Balarka Sen

11:20 AM

Here's a general question. Suppose $M$ is a noncompact manifold which deformation retracts to a subcomplex $X \subset M$ (this is usually called the core of $M$). What is the homotopy type of $M \setminus p$ where $p$ is some point on $M$?

feynhat

@BalarkaSen Does the def ret take the puncture to two meridians in the first copy of T^2?

Balarka Sen

What do you mean by "the puncture is taken to ____"?

feynhat

I mean the image of the loop around the puncture under final stage of def ret.

Balarka Sen

But the image of the deformation retaction of $\Sigma_\infty$ and $\Sigma_\infty \setminus \{p\}$ are not comparable, are they? Their core subcomplexes are different!

If it's a wedge of $2\infty$-many circles for the first case, it's a wedge of $2\infty + 1$-many circles for the second case. Do you see that?

feynhat

Oh yeah. That makes sense. Where exactly is this extra circle being introduced? A meridian of the first T^2 right?

Balarka Sen

11:33 AM

Yup, exactly. Let me draw some picture.

So on the right I have a picture of an apeirogon. It's the standard picture of a $4\infty$-gon in hyperbolic plane, which is drawn so that the vertices lie on a horocycle in $\Bbb H^2$, which is a circle in $\Bbb H^2 \cup \partial \Bbb H^2$ tangential to $\partial \Bbb H^2$ at exactly a point (the point drawn in dirt green)

The horocycle is drawn in red, the apeirogon is drawn in blue. You can quotient the apeirogon by hyperbolic isometries to get $\Sigma_\infty$, paste the edges consecutively like you'd do for a finite $4n$-gon, except for infinitely many quadruplets of edges.

I have marked the domains which quotient to handles on $\Sigma_\infty$ downstairs, so that $\Sigma_\infty = T^2_1 \# T^2_2 \# T^2_3 \# \cdots$

The deformation retraction of $\Sigma_\infty$ to the wedge of $2\infty$-circles is given, at the level of the apeirogon (i.e. before quotienting) by pushing inwards from the dirt green point (the point at infinity), the def ret is shown by the dirt green arrows.

Now if you have a puncture at the orange point, you deformation retract to wedge of $2\infty$-circles, wedge a lollipop. This is because of what happens to deformation retracts if you add a puncture: see the image on the left. The top image is the deformation retraction of the "glass of water" $D^2 \times I$ to $D^2 \times \{0\} \cup \partial D^2 \times I$, by throwing rays from a point above the "glass".

If you have a puncture in the interior of the glass though, you get deformation retracted to a bubble around the puncture, with a stick connecting the bubble to the base of the glass.

feynhat

11:52 AM

Damn.

I was drawing a literal T^2 # T^2 #...

I didn't know this could be realized as a quotient of a $4\infty$gon.

Balarka Sen

Yeah, it's a fun picture. Of course, you can also see it from your picture, with more concentration.

Stupid question inc

what does it means by order of magnitude of prime? Does it mean magnitude of prime number?

size of prime number is $\pi(x)$ but it says the order of magnitude of $pi(x)$ is x/log(x)

so what does it mean?

feynhat

What if it is infinite on both ends? @BalarkaSen

Does it still def ret to wedge of $2\infty$ circles? I suppose, it does.

Balarka Sen

@feynhat Yup. You might find it interesting to draw the corresponding apeirogon

Balarka Sen

12:14 PM

In general if $M$ is a noncompact smooth manifold you can always find a deformation retraction to a codimension $1$ embedded subcomplex. Choose a triangulation (blue), and take a simplex, it's barycenter, and draw a smooth embedded path starting at the barycenter and escaping to infinity (red)

Now "insert a finger" along the red path by drawing a tubular neighborhood of it (orange) and deleting it from $M$. There's a deformation retraction from $M$ onto $M$ with orange tube deleted, and the deformation retraction looks like a finger move.

Finally project outward from the boundary of the orange tube to deformation retract that simplex to its boundary (top right, deformation retraction shown in green).

One has to do this systematically to all simplices to find a def ret of $M$ to a subcomplex. To do this you need to find a red path which connects all the barycenters of all the simplices, and this can be done. Then you insert a finger along this path "from infinity" and do the above procedure.

AMDG

Curious, is there any particular reason why using the function of a circle creates such a good approximation of the circular functions? Even putting them into the circular functions, you get approximations of the hyperbolic functions.

Maybe I should just make an AI to solve this for me...

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$Mathematics$ Mathematics