Mathematics

user131753

5:30 AM

@XanderHenderson Did I not make this explicit in my comment "in the academia (in Logic at least) like one obtained from, say, ILLC (here I am specifically talking about this degree)"?

user131753

@XanderHenderson You mean in University of Barcelona, right?

Secret

5:49 AM

[Random]

0<1<c<2

c<1+c<2c<2+c

2c<1+2c<3c=w<2+2c

w<1+w<w+c<2+w

0<1<c<2<3<4<... w

w<w+1<w+2<w+3<w+4<... w2

w+c<1+w+c<w+2c<2+w+c

w+2c<1+w+2c<w2<2+w+2c

---->

1<c<2

1+2c<w<2+2c

1+w+2c<w2<2+w+2c

1+w<w+c<2+w

Philippe .Hansen-Estruch

hey, could someone explain the proof here: math.stackexchange.com/questions/1470427/…?

Alex

6:09 AM

@Philippe.Hansen-Estruch What part of it?

Philippe .Hansen-Estruch

lmao all of it, um I don't understand this statement cd(n,n+1)=1 , it suffices to show that 2Sk is divisible by n and by n+1 separately.

Alex

@Philippe.Hansen-Estruch Well gcd(n,n+1) is the greatest common divisor, so you need some k|n and k|(n+1), n=kq_1, (n+1)=kq_2, so 1=k(q_2-q_1) and hence k is a positive unit of \Bbb Z

and you want to divide S_k by S_1, so you want to divide 2S_k by n and n+1

(which are coprime)

Philippe .Hansen-Estruch

hmm ok I see

OH I SEE, because since s1 is equal to n(n+1)/2

if you take 2Sk

Alex

Indeed

Philippe .Hansen-Estruch

it must divide n and n + 1

Alex

6:14 AM

Yep, checking S_1 divides S_k, may as well check 2S_1 divides 2S_k

and you need n(n+1) to divide 2S_k, which are coprime

Philippe .Hansen-Estruch

So what does he do after he knows that 2Sk is divisible by n+1 and n

Alex

Nothing, if 2S_k is divisible by n(n+1), then this is precisely the statement that S_1 divides S_k

Philippe .Hansen-Estruch

I see, so how would you go about proving that n and n+1 divides 2Sk, still confused on that part

0celo7

hahaha of course people have different conventions for stereographic projection

why even attempt to be consistent, right

skullpatrol

6:32 AM

Who are you laughing at? @0celo7

0celo7

it's called insanity

skullpatrol

icic

Philippe .Hansen-Estruch

where did alex go :(((

Alex

6:47 AM

Oh sorry, needed to get coffee

0celo7

I can see how this result should go, at least

Thankfully

@Semiclassical I know how to compute the Green's function with zero work.

skullpatrol

to make theorems @Alex

Alex

@skullpatrol :')

What do you do these days @skullpatrol?

Are you not a bot anymore?

0celo7

I think skull is an AI

an old lady AI

Alex

They've been around for so many years

Philippe .Hansen-Estruch

6:50 AM

so how would you go about proving that n and n+1 divides 2Sk, still confused on that part

skullpatrol

@Semiclassical I just found out Feynman won the Putnam?

Semiclassical

@0celo7 well, sure. if there's no work required, that means you're talking about Green's function under free boundary conditions and that's easy mate. (I'm awake too late to be confident about whether I'm making a joke or just being nonsensical for its own sake)

0celo7

No ignore that. I think my approach is right but I can't get the right scaling

Basically, maximum principle guarantees me that if I have the right growth at the pole, I'm done

Semiclassical

lol, if you think my statement was anything other than ignorance to begin with then you're out of your gourd as well

tfw you realize the MATLAB issue you've been banging your head against for the last day is not actually something you need to do

0celo7

it is 2Am

so probably

Semiclassical

6:59 AM

I was trying to figure out how to implement the code for plotting the 2D case...but the solution my students needed to provide was for just a 1D slice of the 2D case, so I didn't need to worry about it ugh

0celo7

I need $$\frac{1}{1-\cos\theta}\sim \frac{1}{\theta}$$

but

Semiclassical

labors in ignorance

0celo7

that's not true

at least I don't think it is

Semiclassical

yeah, it's not

blows up quadratically

0celo7

as $\theta\to 0$ of course

Semiclassical

7:00 AM

right

0celo7

@Semiclassical Doesn't it blow up worse?

Does the $\theta^4$ term do anything?

Or does the slowest term dominate

Semiclassical

$\frac{1}{\theta^2+a\theta^4}=\frac{1}{\theta^2}\frac{1}{1+a\theta^2}\sim \frac{1}{\theta^2}$

so yeah, slowest term is what matters

0celo7

ah, right

Hmm, that's definitely the wrong power then

mmm, there's more going on here.

I've got another term in the numerator that does go to zero.

Like a sine!

It's a conspiracy

Secret

(cont.): Ok so the thing works, and thus it is a linear order non archimedian commutative semiring with a notion of finite step blowup implemented into it:

0celo7

Huh.

I think I get $1/\theta$ if I estimate the top correctly

It would be nice to get an exact formula for this but my current method is sufficient for my purposes.

Secret

7:07 AM

Now that the framework is set up, this can be expanded further in Rambles

We can finally have "interfinite" elements, those which lies between the integers but a certain integer multiple k of them will allow them to become infinite

WeavingBird1917

Hi, could someone tell me what it's called when a function contains a derivative or anti-derivative of itself?

Semiclassical

"Contains"?

WeavingBird1917

For example a(x) = b - c, where c is the second anti-derivative of a

Secret

integral equations?

Semiclassical

^

It's a solution of a particular differential equation.

0celo7

7:11 AM

so THAT'S what PDE means

WeavingBird1917

Thanks, now I can search it up.

Semiclassical

or integro-differential equations

if you've got both derivatives and antiderivatives at once

@0celo7 lool

Secret

yeah, those are very common in stochastic systems

0celo7

Schoen and Yau seem to think this is exactly the Green's function. No way, there's got to be some factors

Like, you're gonna be missing some $\pi$'s and shit

Semiclassical

lol

0celo7

7:15 AM

@Semiclassical speaking of $\pi$'s and shit i.gyazo.com/12a28c1fe4d7337a5eaa5dac48644e39.png

Semiclassical

lol

0celo7

alright I've gotta go before I get my second wind and stay up until 5

cheerio

Semiclassical

night

skullpatrol

cya

Secret

sweet dreams

(cont.) actually.... I made a mistake. Now I am suspecting it is the opposite that is true

Proposition: Let $A$ be a Non-Archimedean semiring with a linear order. Then interfinite elements does not exist

Proof:

Suppose there exists some element $c$ such that $kc = w$ and $a < c < b$ for some naturals $a,b,k$ and $w$ is infinite wrt all elements less than itself. Then we have:

$x+y < w$ and $xy < w$ for all $x,y \in A$

Now multiply $a < c < b$ by $k$ and using the property of $c$ to obtain:

$ka < w < kb$

Since $k$ is finite, and hence $xk < w$ for all $x \in A$, the above is a contradiction

Hence $c$ does not exist

skullpatrol

8:10 AM

||||||||

Secret

Interfinite elements is only possible in partial orders, or when the binary operations is not order preserving. In such a scenario, this is equivalent to having an algebraic structure equipped with a n-ary map $A^n \mapsto A$ where the domain is the subset of finite elements but the image is the subset of infinite elements, in addition to its usual operators

Such map is necessary noninjective, and thus cannot be inverted, hence interfinite elements, like their infinite counterparts, have no inverses

The Cult of Infinity:

0<1<2<3<4<5<... Finite

<<<<<<<<... Unbounded

(Updating notation)

$[]_0$

$[0]_0$

$[0,1,2,3,4,5,...,n]_{n}$

$[0,1,2,3,4,5,...]_{\text{Unbounded finite}}$

$[0,1,2,3,4,5,...]_{\omega}$

$[0,1,2,3,4,5,...,\omega]_{\omega+1}$

$[0,1,2,3,4,5,...]_{\aleph_0}$

$[c,2c,3c,...,(n-1)c,w]_n$

$[\omega,c\omega,c^2\omega,c^3\omega,...]_{\omega_1}$

$[0,1,2,3,4,5,...]_{\omega_1}$

$[0,1,2,3,4,5,...]_{\aleph_{\lambda}}$

(Translating to human readable language...)

0. Nothing $[]_0$ (Object does not exist)

2

1. Zero $[0]_0$ (Minimum, identity)

2. Finite $[0,1,2,3,...,n]_n$ (Exhaustible in n steps)

skullpatrol

8:33 AM

Does "Nothing" mean no thing or not a thing?

Secret

it means the absence of any mathematical object

i.e. before numbers are being constructed

skullpatrol

So, undefined.

Secret

pretty much, yeah

skullpatrol

Undefined is not a (defined) thing.

Secret

it is not a mathematical object either

so the definition of the nothing holds fine

Perhaps a better way to say is:

We start with an empty canvas, and then adding things in one by one

skullpatrol

8:37 AM

A contents of the empty set.

Secret

so $[]_0$ is basically the empty set in this WIP framework

skullpatrol

The set with no members

Secret

There could be more than one of these, depending on whether you have equality in the underlying logic framework

skullpatrol

The null set

Secret

3. Interfinite $[c,2c,3c,...,(n-1)c,w]_n$ (An infinite object that is exhausted in n steps, if taken apart in a certain way)

4. Quasifinite $[c,2c,3c,...,(n-1)c,w]_{\text{(infinite)}}$ (Same as interfinite, except for every step it is taken apart, its size remains unchanged)

5. Strictly amorphous $\{0,1,2,3,...,n\}_{(uncountable)}$ (Generally unordered, when being partition the number of times equal to its grade, all those partitions are of size 1)

6. Amorphous $[0,1,2,3,...,n]_n\{0,1,2,3,...\}_{(\text{uncountable})}$ (Cannot be partitioned into two infinite sized partitions)

7. Infinite dedekind finite $[...,0,1,2,3,...]_{\mathscr{D}}$ (Shrink indefinitely when attempted to exhaust it, but is inexhaustible)

8. Tarski finite $[...]_{(\text{infinite})}$ (Every partition is upper and lower bounded)

9. Countable $[0,1,2,3,...]_{\aleph_0}$ (Can in theory be exhausted in steps)

10. Closed upwards $[0,1,2,3,...,\alpha]_{\alpha+1}$ (Has a maximum)

11. Quasicountable (No idea, has to behave like something that is between countable and uncountable)

skullpatrol

9:00 AM

the-cult-of-infinity

Secret

12. First uncountable $[0,1,2,3,...]_{\omega_1}$ (Cannot be exhausted by counting or by any algorithms that runs in steps)

WeavingBird1917

Can anyone solve this differential equation, calculator can't do it: h''(t) =68000 / (3340 - 68 * t) - (3.98576 * 10^14) / (6.3781 * 10^6 + h(t))^2

Secret

that ODE looks quite nonlinear, try moving - (3.98576 * 10^14) / (6.3781 * 10^6 + h(t))^2 to the LHS and made the substitution z(t) = 6.3781 * 10^6 + h(t)

that should make it look like cauchy euler, I think...

Secret

13. Extensive $(\text{No notation})$ (There is an extremely fast growing function such that f(|S|) of it is equal to itself, otherwise for any g that grows slower than f, g(|S|) < |S|)

14. Limiting $[\alpha_1,\alpha_2,\alpha_3,...]_{\lambda}$ (The limit of an increasing hierarchy of infinities)

15. Regular $[0,1,2,3,...]_{\alpha}$ (Uncountable to the point that any algorithm has to run at least $\alpha$ steps to reach it)

16. Inaccessible $[0,1,2,3,...]_{\kappa}$ (No algorithm, no matter how fast it accelerates and self improving, that is slower than an algorithm running $\kappa$ steps, can reach it)

(What is the most abstract way to generalise a powerset anyway...?)

17. Weakly compact $[0,1]_{\kappa}$ (All algorithms, no matter how fast they are, how good they are, and how self improving they are, cannot make any progress in reaching $\kappa$ except oscillate between 0 and 1)

typo: if they are worse than an algorithm that can compute $\kappa$

18. Strongly compact $[0,1]_{\kappa}$ (Same as 17, except even a self improving algorithm that accelerates as fast as $\kappa$ get stuck)

Secret

9:41 AM

19. Berkeley $[0]_{\kappa}$ (Cannot be reached even if the universe is expanding infinitely fast (whatever that means))

20. Reinhardt $[0]_{\kappa}$ (Cannot be reached even if the universe is expanding impossibly fast)

X. Proper classes/containers: $[0,1,2,3,..._{\text{Class}}$ (These are so large that no size can be meaningfully assigned to them, but all sizes can never reach it)

Lim. Absolute infinite $\infty$: (The inconsistent notion that is bigger than anything that is defined here)

#. Transinfinite $..._{\text{Final}}$ (Refers to a size so indescribably large that had this were set theory, the set difference S-S is not empty, but a proper subset with the same size. This is logically inconsistent and thus it cannot exist. Put it in another way, no operations can reduce its size)

Now heading to the opposite direction:

-1. Infinitesimal $\epsilon$ (Arbitrarily close to zero)

So in short:

Nothing: There is no computer

Zero: The computer is not calculating

Finite: The computer terminates after some calculation

Countable: The computer does not terminate, but the calculation is running step by step

Uncountable: The calculation cannot even approximate the answer without knowing the answer

Inaccessible: A self improving AI had no hope of obtaining the answer without knowing it

Compact: The calculation stuck at 0 or 1 and making no progress without the answer

Berkeley: The calculation will not finish even if it goes forever

Reinhardt: The calculation is impossible

Proper classes/containers: The computer explodes as soon it started the calculation

Absolute infinite: This is nonsense

Transinfinite: You are trolling

Infinitesimal: The calculation is running, but it is so slow that it is basically idling

And even briefer summary:

Zero: No calculation is running

Finite: Calculation

Countable: Calculation goes forever, but in steps

Uncountable: Calculation cannot approximate the answer

Inaccessible: Calculation get stuck somewhere far away from the answer

Reinhardt: Calculation is impossible

Or in one line: None, something, step by step, unreachable, get stuck far below, impossible

And that, is the Cult of Infinity

Slereah

10:45 AM

ams.org/journals/proc/1971-027-01/S0002-9939-1971-0267588-7/…

oh boy

More collaring theorems

Silent

11:38 AM

I have this problem, and its solution:

If $\tau = (1,2)(3,4,5)$ determine whether there is a $n$-cycle $\sigma$ ($n\geq 5$) with $\tau = \sigma^k$ for some integer $k$.

And its solution:

But in the second para I can't understand two things: 1. What does $\{1,2,3,4,5\}$ can't be fixed by $\sigma$ and 2. Why wlog, we have to consider only two cases $\sigma(1)=6$ and $\sigma(4)=6$?

Bogey

Hi, question - lets say you have 2 cent and 5 cent coins. You're calculating the number of combinations with which you can use those coins to split 1 USD. Can you use this result to say how many combinations there would be to split 2 USD, without recalculating? Both 1 and 2 USD are a multiple of the coins lowest common multiple, but not sure yet if that helps at all..

Slereah

Isn't that related to the McNugget problem

mathworld.wolfram.com/McNuggetNumber.html

Or the coin problem, as it may be

Coin problem

The coin problem (also referred to as the Frobenius coin problem or Frobenius problem, after the mathematician Ferdinand Frobenius) is a mathematical problem that asks for the largest monetary amount that cannot be obtained using only coins of specified denominations. For example, the largest amount that cannot be obtained using only coins of 3 and 5 units is 7 units. The solution to this problem for a given set of coin denominations is called the Frobenius number of the set. The Frobenius number exists as long as the set of coin denominations has no common divisor greater than 1. There is ...

Bogey

i'm hoping it is :) I'm working on a variation of the coin problem, but my amounts are getting so large that it becomes unsolvable by any algorithms ( i believe )

Vrouvrou

hello

Bogey

problem is essentially - you have a set of available coins, and try to make change for a given amount. You need to count the number of combinations that would allow you to split the used amount of coins (amount = number of coins, not value) evenly into 2 cups

so for example one possible combination of change for 25 cent could be 5 cent, 5cent, 5cent, 10 cent - which works (because 4 total coins = even number), and there are 2 ways to distribute this combination into the cups (cup A: 5x 5 cent, cup B: 1x10, 1x5 .. or cup A: 1x10c, 1x5c and cup B: 2x5cent)

but i need to solve this for fairly large numbers, so i don't see any good way of solving this at the moment - i can't run through all possible combinations (these get WAY to huge), i can't think of a way to cache sub-results (because there are way too many)

but hoping there might be some way that if the target amount is a multiple of the LCM of the coins i can somehow use that fact to simplify the problem

Silent

12:34 PM

@BalarkaSen, will you please look at my question?

Balarka Sen

Depends on the question probably

Is it what you linked above? That seems too much algebra for me right now

Silent

yes that one.

@BalarkaSen never mind.

Alessandro Codenotti

1:12 PM

I'm reading what you wrote in the endspace room @Balarka

Balarka Sen

@Alessandro Nice! Although it's kind of like an infinite rambling

Maneesh Narayanan

3:12 PM

0

Q: prove $\int_{-C}\vec{F}d\vec{R}=-\int_{C} \vec{F}d\vec{R}$

We need to prove $\int_{-C}\vec{F}d\vec{R}=-\int_{C} \vec{F}d\vec{R}$ Usual attempt in Textbook: Suppose $C:\vec{R(t)},a\leq t\leq b$. Then $-C:\vec{R(-t)},-b\leq t\leq -a$. I am able to verify the result. Can I define as $-C:\vec{R(a+b-t)},a\leq t\leq b?$ $\int_{-C}\vec{F}d\vec{R}= \int_{a...

please point my mistake :'(

It is my request.

user193319

3:29 PM

0

Q: Finest Locally Convex Topology on a Group Ring

Let $G$ the free group on $n$ generators, and let $\Bbb{C}G$ denote the complex free group ring (which is the same the group algebra, since $G$ is a discrete group). Endow $\Bbb{C}G$ with the finest locally convex topology. Is much known about such a topology on $\Bbb{C}G$; are there any study re...

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$Mathematics$ Mathematics