/mg/ - mathematics general

[Anonymous No.15235870]

Previously >>15213505

Cirno is the strongest edition.
Talk math.

[Anonymous No.15236125]

He showed up with a hole in a proof that AC -> well ordering theorem.
The parts we have are
let X be a nonempty set
assume AC
let c : P(X) \ {} -> X be a choice function
now apply transfinite recursion to functions F with domain an ordinal alpha and codomain X such that F(beta) = c(X \ F[beta])
and the question is: how do you make this proof work at limit ordinals?
it does go through, but there is a little extra that's missing here...

[Anonymous No.15236129]

>>15236125
should be in here

[Anonymous No.15236163]

Why are degenerate chains/simplexes useful in algebraic topology? Those are simplices for which the last coordinate doesn't make a difference. I don't understand why considering them helps us to define the homology.

[Anonymous No.15236171]

>>15235870

Could anyone explain the statement I've heard tossed around, that "in constructive mathematics, every function from the reals to itself is continuous"?
Or am I misunderstanding something?

[Anonymous No.15236174]

>>15236163
Are you looking at simplicial homology of a simplicial complex? Or singular homology of a general topological space?

[Anonymous No.15236177]

>>15236174
Singular homology of a general topological space.

[Anonymous No.15236190]

>>15236177
I don't have a good all-encompassing answer, but as an example, you use degenerate 2-simplices in order to show, for example, that a concatenation of two (concatenatable) paths/edges/1-simplices is homologous to their sum.

[Anonymous No.15236199]

>>15236177
I don't have a good all-encompassing answer, but as an example, you use degenerate 2-simplices in order to show, for instance, that a concatenation of two (concatenatable) paths/edges/1-simplices is homologous to their sum (as a chain). (You can show this as an exercise.)

[Anonymous No.15236200]

>>15236190
Nothing is missing.

[Anonymous No.15236220]

>>15236199
Why do some definitions of singular homology seem not to use degenerate simplices? E.g. https://en.wikipedia.org/wiki/Singular_homology

[Anonymous No.15236230]

>>15236220
Could you point out where that gives an example of a definition of singular homology that doesn't use degenerate simplices?

I don't understand how that would work really. For example, you wouldn't even be able to discuss homology of the one-point space in such a framework. (Since any positive-dimension singular simplex in the one-point space is degenerate.)

[Anonymous No.15236235]

>>15236171
https://mathoverflow.net/questions/164694/are-all-functions-in-bishops-constructive-mathematics-continuous

PS: don't bother with constructive mathematics, it's inferior in everyway to classical. Unless it's explicitly impossible (some topos shit), excluded middle is ALWAYS based and true.
>t.guy who wasted embarrasing amounts of time on philosophical bullshit

[Anonymous No.15236249]

>>15236235
Lmfao, thank you for the pro tip anon. Based asf

[Anonymous No.15236266]

>>15236230
So the wikipedia definition is wrong?

[Anonymous No.15236268]

>>15236266
What?

[Anonymous No.15236275]

>>15236268
>>15236230
Hatcher also doesn't seem to mention degenerate simplices.

[Anonymous No.15236279]

>>15236268
The wikipedia definition doesn't mention degenerate simplices.

[Anonymous No.15236283]

>>15236275
Why should he? I'm not following what you're trying to say.

[Anonymous No.15236286]

>>15236283
These examples seem to show that talking about degenerate simplices is unnecessary in the definition of singular homology.

[Anonymous No.15236290]

>>15236286
Sure. It's also unnecessary to mention that 2 is an even number whenever we use it, but it is.

[Anonymous No.15236297]

>>15236290
Why do mathematicians complicate the definitions and waste time by quotienting out by degenerate simplices?

[Anonymous No.15236298]

>>15236286
Sorry, I just realized you were asking if it is useful to consider degenerate simplices. I thought you were asking why we don't discard them. The answer is, we need them.

[Anonymous No.15236305]

>>15236297
They quotient out boundaries, not degenerate simplices, you midwit. Learn 2 read the fucking definition.

[Anonymous No.15236311]

>>15236305
Peter May quotients out by both. So does Serre in his thesis. Also Fomenko if I'm recalling correctly.

[Anonymous No.15236334]

>>15236311
Here he is taking the chain subcomplex (of the standard singular chain complex) spanned by the nondegenerate chains.
He is arguing that this is equivalent, in each index, to quotienting by the degenerate simplices.
Note, however, that the degenerate simplices do not form a chain sub-complex, unless I'm mistaken.

The "quotient" I'm referring to, which I thought you were referring to, is the quotienting of the cycles by boundaries to get the homology. That is still what Peter May is doing here: he is using the homology of the "nondegenerate" singular chain complex.

What's your question exactly?

[Anonymous No.15236356]

>>15236311
Here Peter May is using B the chain complex spanned by the nondegenerate simplices, as a subcomplex of C the standard chain complex generated by "all" singular simplices.

He mentions that each B_n is the quotient of C_n by an appropriate subgroup, namely D_n the span of the degenerate simplices. Note however that the D_n do not form a chain subcomplex, unless I'm mistaken.

My assumption is there's some theorem stating B and C have isomorphic homology (or are homotopy-equivalent chain complexes), since the latter is what gives the "standard" definition of singular homology.

[Anonymous No.15236385]

>>15235870
The lean proof assistant is FUCKING bullshit and I hate it.

[Anonymous No.15236386]

>>15236311
Actually, what I said in >>15236199 is wrong, my mistake.

You can show the statement I mentioned without using degenerate simplices. (You can take the "flattened" 2-simplex involved in that exercise, and if it's degenerate, then vary it a little to make it non-degenerate, but still "flattened".)

[Anonymous No.15236409]

>>15236356
Why would he complicate the definition by quotienting them out?

[Anonymous No.15236424]

>>15236409
Maybe because (chains built from) nondegenerate simplices are easier to work with for some purposes , would be my guess , but honestly idk.

[Anonymous No.15236605]

[math]\begin{aligned} E\left[\hat{\theta}_{MLE}\right]=E[Z] & =\int_\theta^{\infty} z n\left(1-2\left(1-e^{-2(z-\theta)}\right)\right)^{n-1} 2 e^{-2(z-\theta)} d z \\ & =\int_\theta^{\infty} z n\left[1-F_X(z)\right]^{n-1} F'_X(z) d z\end{aligned}[/math]
Anyone see the trick here? Where [math] F_{X} [/math] is denoted to mean the CDF of random variable [math] X [/math], and [math] F'_X [/math] is its derivative i.e. the PDF.

[Anonymous No.15236741]

>>15235870
Is the ice maid smart?

[Anonymous No.15236747]

Hello, if [math]A[/math] is a set and [math]\sim[/math] an equivalence relation on [math]A[/math], then the map [math]\pi : A \to A{/ \sim}[/math] that sends an element to it's corresponding equivalence class is often called something like the "canonical map".
After learning about natural transformations in Category Theory I now wonder: Is this map called canonical because there is a natural transformation hiding there (which would justify the name) or is the usage of canonical here not meant like that and it's just an informal way to refer to it?

[Anonymous No.15236765]

How do graphing calculators work versus things like Wolfram Alpha or Desmos?

I plotted a function y = -arctan(x/10million)-arctan(x/500million)

My super old TI graphing calculator plots the function and I can zoom to the y-value in question and find the correct solution. When I try to plot this on desmos, it displays the line y=0.

When I ask Wolfram to solve for x at the y-value in question, it says "no solutions exist."

I don't understand how modern tools can't do this, but a super old calculator does it just fine.
Is it because the magnitude of the numbers is so large?

[Anonymous No.15236770]

>>15236765
What's the y-value in question?

[Anonymous No.15236801]

>>15236747
I think you could could set something up as follows:
- Let C be the category with objects pairs (A,E) where E is an equivalence relation on the set A , and morphisms (A,E)-->(A',E') are functions A-->A' preserving the relations E,E' respectively
- Let D be the category of sets
- We have a functor F : C-->D sending (A,E) to E ,
and a functor G : C-->D sending (A,E) to A/E .
- Then, the canonical morphism A --> A/E gives a natural transformation F --> G .

[Anonymous No.15236807]

>>15236801
> F : C-->D sending (A,E) to E
Typo : F should send (A,E) to A , not E

[Anonymous No.15236824]

>>15236770
-120

[Anonymous No.15236844]

>>15236824
And there's the issue.
WA (and probably Desmos) are reading your input as radians, not degrees.

[Anonymous No.15236868]

>>15236844
yeah but on desmos, i just did y= and then put it in, so there was no left side.

[Anonymous No.15236879]

>>15236801 >>15236807
Thank you so much for the fast response anon.
I'll work through the details tomorrow but I think I understand this.

Out of curiosity, is this particular result that the canonical projection map gives rise to a natural transformation common knowledge (possibly as a standard example in textbooks) or did you come up with it on the spot? I did google around before asking here but didn't find much. Hence why I almost assumed to get a negative answer (i.e. that the usage of canonical here has nothing to do with natural transformations)

Having a bit of an existential crisis right now because I've seen so many things called canonical or natural in my past classes and I just assumed those were informal terms. So there is a chance most (if not all) of those examples are actually natural in a precise sense (using the the language of category theory)? Kinda wish someone at least told me that lol

[Anonymous No.15236884]

>>15236868
well, in that case it might be the magnitude thing. I don't actually use Desmos, so I'm not aware of the specifics
but WA will give you an actual answer if you specify you mean degrees

[Anonymous No.15236897]

Prove that the set of integers which have an odd number of factors is the set of square numbers.

[Anonymous No.15236909]

>>15236897
number of factors = product of (exponents in factorization+1)
number of factors = even if any exponent is odd
if all exponents in factorization are even then
integer = (factorization with halved exponents) squared
therefore all odd factor count are squares
n/a=b -> n/b=a
factors come in pairs and factors are even
unless a=b then n=ab=a2 then n is a square
therefore all squares are odd factor count

[Anonymous No.15237022]

>>15236879
Not sure honestly, it may be an exercise somewhere.
I'm no algebraist, but reading MacLane's CWM was a great help to me in understanding category theory for my quals lol.

[Anonymous No.15237024]

Is there a finite non-Abelian p-group, which is not an internal semidirect product of a pair of nontrivial subgroups?

[Anonymous No.15237052]

>>15235870
Sorry if this is a sqt question
Is basic mathematics by serge lang a good review before doing calculus again? I don't really need a mathematician's understanding. Thus is mostly for fun and to maybe help with a career in bioscience/programming/data science down the line
But I would like to develop an intuitive understanding of math again before I go into calc again (does it cover everything I need before calculus)?

[Anonymous No.15237084]

youve been working on a problem for a long time but to finish it you need to either
>do a bunch of heroically difficult calculus
>or build an expensive computer with a shit ton of ram to do it linearly
but you can neither get the motivation nor cash to do either even though you know you could retire if you do

[Anonymous No.15237128]

Do you love Japan?

[Anonymous No.15237540]

>>15236741
her power is maximum.

[Anonymous No.15237544]

Let [math] S [/math] be the set of closed intervals in [math]\mathbb{R}[/math] with the inclusion order relation. Show that for any two elements [math] A,B \in S [/math] there exists [math] sup \{A,B\} \in S [/math] where supremum is the element of [math] S [/math] for which [math] A \subset sup \{A,B\} [/math], [math] B \subset sup \{A,B\} [/math] and for any other upper bound [math] M [/math] of [math] \{A,B\} [/math] we have that [math] sup \{A,B\}\subset M [/math].
I believe that for [math] A=[x_{1},y_{1}][/math] and [math] B=[x_{2},y_{2}] [/math] I have [math] sup \{A,B\}=[min(x_{1},x_{2}),max(y_{1},y_{2})] [/math], I'm just having a hard time showing that this is the least upper bound under inclusion. Any hints?

[Anonymous No.15238035]

>>15237544
Where are you stuck?

[Anonymous No.15238067]

>>15238035
I think I got it, for any upper bound [math]M=[x_{3},y_{3}] \in S[/math] of [math]\{A,B\}[/math]. We have that [math]A \subset M[/math] and [math]B \subset M[/math], so [math]A\cup B \subset M[/math], in particular the smallest element [math]min(x_{1},x_{2})[/math] and the largest element [math]max(y_{1},y_{2})[/math] of the union must be elements of [math]M[/math]. Thus [math]x_{3}\leq min(x_{1},x_{2})\leq max(y_{1},y_{2}) \leq y_{3}[/math] and so [math][min(x_{1},x_{2}),max(y_{1},y_{2})]\subset M[/math].

[Anonymous No.15238191]

Math is beautiful

[Anonymous No.15238196]

A Russian bot walks into /mg/ math general and immediately corrects himself,
>"My deepest apologies sir, I didn't even see you!" he says.
>"That's quite fine, young bot," the general replies, "I have quite a lot of work to do organizing the front for the next expeditionary advance! Run along, now."
>The young Russian bot continues onto his destination, toy airplane in hand, making "zoom" and "rat-tat-tat" and "kra-cow kra-cow" noises.
>On each wing the identification of the aircraft is painted, "36" on the left wing and "36" on right.
>On the tail, only the starboard displays the model of the aircraft, "8"
>"Off to bomb the West!" the young bot announces confidently.
>The general chuckles to himself.

[Anonymous No.15238218]

>>15238191
>>15238191
What is [math] 1 - 1 + 1 - 1 + \dots ? [/math]
It seems impossible to add up all the terms, however consider what happens when we multiply the series by [math] -1 [/math]
[eqn] (-1) ( 1 -1 + 1 - 1 + \dots ) = -1 + 1 - 1 + \dots [/eqn]
This is just the series subtracted by [math] 1 [/math]. Thus, the series has a value of [math] 1/2 [/math]. Wow math is le heckin beautiful!

[Anonymous No.15238249]

Formalization would be a death sentence for topology.
None of the topologists care about being rigorous.

[Anonymous No.15238256]

>>15238218
>What is 1−1+1−1+…?
nothing, it is ill defined. It does diverge indefinitely.

[Anonymous No.15238274]

>>15235870
Just here to briefly explain to pseudos the rigorous definition of the delta distribution and its fourier transform because I find it really interesting as it highlights nicely the mathematical thought process:
The delta distribution is a linear, continuous functional and as such maps functions to values. It is rigorously defined as
\delta(f) = f(0) for some function f, where the domain of \delta initially is the space of all smooth functions with compact support. Now this is straight forward and obviously well-defined.
Now, to define its fourier transform (and related fourier transform), one moves from the aforementioned function space to the space of all Schwartz functions, i.e. functions for which any derivative decays faster to zero than any monom for x -> +- infinity. (this implies smoothness of the functions of course)
For this function space, one can show that the fourier transform always exists and is bijective, i.e. we never leave this space of functions if we compute the fourier transform. This is not true for the initial function space considered.
Now, to define the fourier transform of \delta, one observes that for any L1 function, the identity
\int_{-\infty}^{\infty}\hat{f} gdx = \int_{-\infty}^{\infty}f \hat{g}dx
where \hat{f} denotes the respective fourier transform, i.e. we can move around the fourier transform from one function to the other without changing the value of this integral.
Adopting this fact to the delta distribution, which is justified by so called delta sequences of functions which converge to the delta function in the limit. For any delta sequence, this identity also holds and as such it has to hold for the delta distribution as well.
So we get the following definition:
\hat{\delta}(f) := \delta(\hat{f}).
Then we get the well known fact
\hat{\delta}(f) = \delta(\hat{f}) = \hat{f}(0) = \int_{-\infty}^{\infty}f(x)e^{-j0x}dx = \int 1f(x)dx = <1,f> and it yields \hat{\delta} = 1 in the distribution sense.

[Anonymous No.15238308]

>>15238249
https://www.info.ucl.ac.be/~pvr/decon.html
> The broader movement that goes under the label "postmodernism" generalizes
>this principle from writing to all forms of human activity, though you have to
>be careful about applying this label, since a standard postmodernist tactic for
>ducking criticism is to try to stir up metaphysical confusion by questioning
>the very idea of labels and categories. "Deconstruction" is based on a
>specialization of the principle, in which a work is interpreted as a statement
>about itself, using a literary version of the same cheap trick that Kurt Gðdel
>used to try to frighten mathematicians back in the thirties.
> Deconstruction, in particular, is a fairly formulaic process that hardly
>merits the commotion that it has generated. However, like hack writers or
>television producers, academics will use a formula if it does the job and they
>are not held to any higher standard (though perhaps Derrida can legitimately
>claim some credit for originality in inventing the formula in the first place).

[Anonymous No.15238358]

magic magic magic

[Anonymous No.15238541]

>>15238358
Your exponent is n some times, when it should be i.


Anyway, here's some more cuties:

[math] \dfrac{1}{1-(z+w)} = \sum_{k,m \geq 0} \binom{k+m}{k}\, w^m\, z^i [/math]

[math] \dfrac{z^s}{(1-z)^{s+1}} = \sum_{k=s}^\infty {k\choose s} z^{k} [/math]

[math] \dfrac{1}{(1-z)^s} = \sum_{k,m \geq 0} \left[\begin{matrix} k \\ m \end{matrix} \right] \dfrac{1}{k!} s^m z^k [/math]

[Anonymous No.15238553]

>>15238218
There's also some cuties tying to that

[math] \sum_{k=0}^\infty z^k = \dfrac{1}{1-z} [/math]

[math]\sum_{k=0}^\infty (-1)^k \cdot (1 - \epsilon)^k = \dfrac{1}{2-\epsilon} [/math]

[math]\lim_{\epsilon\to 0}\sum_{k=0}^\infty (-1)^k \cdot (1 - \epsilon)^k = \dfrac{1}{2} [/math]

Make of that what you will.

[Anonymous No.15238569]

>>15238274
The delta distribution is a linear, continuous functional and as such maps functions to values. It is rigorously defined as [math]\delta(f) = f(0)[/math] for some function [math]f[/math], where the domain of \delta initially is the space of all smooth functions with compact support. Now this is straight forward and obviously well-defined.
Now, to define its fourier transform (and related fourier transform), one moves from the aforementioned function space to the space of all (((((Schwartz))))) functions, i.e. functions for which any derivative decays faster to zero than any monom for [math]x\rightarrow\infty[/math]. (this implies smoothness of the functions of course)
For this function space, one can show that the fourier transform always exists and is bijective, i.e. we never leave this space of functions if we compute the fourier transform. This is not true for the initial function space considered.
Now, to define the fourier transform of \delta, one observes that for any L1 function, the identity
[math]\int_{-\infty}^{\infty}\hat{f} gdx = \int_{-\infty}^{\infty}f \hat{g}dx [/math]
where [math]\hat{f}[/math] denotes the respective fourier transform, i.e. we can move around the fourier transform from one function to the other without changing the value of this integral.
Adopting this fact to the delta distribution, which is justified by so called delta sequences of functions which converge to the delta function in the limit. For any delta sequence, this identity also holds and as such it has to hold for the delta distribution as well.
So we get the following definition:
[math]\hat{\delta}(f) := \delta(\hat{f})[/math].
Then we get the well known fact
[math]\hat{\delta}(f) = \delta(\hat{f}) = \hat{f}(0) = \int_{-\infty}^{\infty}f(x)e^{-j0x}dx = \int 1f(x)dx = <1,f> [/math] and it yields [math]\hat{\delta} = 1[/math] in the distribution sense.

[Anonymous No.15238653]

The eigenfunctions of the Laplacian with Dirichlet boundary conditions on the interval [math][0,\pi][/math] are given by [math]\sin (n x)[/math], [math]n \in \mathbb{N}[/math]. The integrals [math]\int_0^\pi \sin (n x) \, \mathrm{d} x [/math] vanish like [math]n^{-1}[/math]. Is there a general result that for any (sufficiently regular?) bounded domain [math]U \subset \mathbb{R}^d[/math] the integrals of the eigenfunctions of any selfadjoint extension of the [math]d[/math]-dimensional Laplacian vansh like [math]O (n^{-d})[/math] as [math]n \to \infty[/math]?

[Anonymous No.15238670]

naive question maybe but why do most abstract algebra classes start out with (and to focus on) group theory and tend to skip completely through semigroups and monoids?
are semigroups and monoids too simple or due to their simple nature not as nice as groups? or is it just that they don't pop up as often as groups do in mathematics?

[Anonymous No.15238717]

>>15238670
They don't pop up as often and their theory is boring. There isn't a classification of finite simple monoids.

[Anonymous No.15238719]

>>15238670
Genreating quotient spaces for semigroups is less intuitive than in groups because it's done by congruence relations rather than normal subgroups. Your average shitter can't understand quitient spaces in groups or rings so this harder formulation would be even worse.
A course on semigroups/monoids would be incomplete without talking about Green's relations which would likely be too advanced for entry level stuff. It's not entry level but I will say the free semigroup is easier than the free group
There is no point in talking about semigroups unless you talk about this which would take up a lot of time in an algebra course, but that's not to say semigroups aren't studied, there's a reasonably large intersection between semigroup theory and theoretical computer science.

[Anonymous No.15238744]

>>15238670
In contrast to what's been said above, I rather think it's because in the 100 years from groups in Galois work, to the use of Frobenius Lie's ideas in physics, it was just important for some other work. Groups and ideals are in turn parts of rings, which was a hot topic in works of Hilbert, Noether and the like.
Monoids and semigroups are certainly not boring, and they have rich special cases. Chaining together functions, especially algebra values functions, gives loads of such structures, and representations.

[Anonymous No.15238993]

mfw I completely missed the second page on my midterm because I somehow unknowingly flipped two pages at once every time I thought I was flipping just the first page

[Anonymous No.15239037]

>>15238993
Post exams I am always paranoid I have done this, and the paranoia doesn't fully die down until I get the grade or exam back.

[Anonymous No.15239067]

Explain the completeness theorem to me WITHOUT using compactness.

[Anonymous No.15239194]

>>15239067
Given a theory T if p is provable from T, then p holds in all models of T. Soundness, which is the other direction, also holds.

[Anonymous No.15239196]

>>15238569
thank you. How do you do that here

[Anonymous No.15239222]

>>15239194
that is exactly backwards

[Anonymous No.15239253]

>>15239194
Prove it faggot
NO COMPACTNESS

[Anonymous No.15239290]

I hate heckin matherino

In calc 2 we derived a nice equation for the tangent line to a parametric equation:
x = f(t)
y = g(t)
slope at t: g'(t)/f'(t)

now in calc 3 we have a vector r(t)=<f(t), g(t)> and the tangent is given by r'(t)=<f'(t), g'(t)>

[Anonymous No.15239313]

>>15235870
sauco?

[Anonymous No.15239358]

>>15239196
use [nath] and put your latex here without the $$ [/nath] but with math instead of nath

[Anonymous No.15239393]

>>15236125
Zero. There are no busses in gensokyo

[Anonymous No.15239593]

Hi!
I have found new problems that look cool. Hopefully they're cool. I think they might be easy but I'm not sure, haven't done em yet. I'll try them at the library today when I go there. I appreciate anyone replying whether they've solved it or not. Good luck to anyone who decides to attempt!

[Anonymous No.15239687]

>>15239693
>{8:1,24:1,32:1,36:2}

[Anonymous No.15239751]

so I took a linear algebra class, I took a linear optimization class, and still don't truly understand matrices enough that I may apply them. How can I change this? Do I need to go grind on kahn academy?

[Anonymous No.15239788]

>be me
>academics exam is incoming
>hey maybe chans knows maths better than me, i can get sme advice
>open sci
>now i feel like a caveman

[Anonymous No.15239797]

I can't do algebra...

[Anonymous No.15239986]

What is: [eqn] \lim_{x \to 1} \lim_{n \to \infty} x^{2n}? [/eqn]

[Anonymous No.15240143]

>>15239593
Say for a) there should be cycles of length 2 or 4.
We may have:
- one cycle of length 4
- one cycle of length 4 of and a transposition
- i transpositions where i=0,1,2,3.

There are 6 choose 4 choices of numbers for a 4-cycle which can be arranged into 3! different 4-cycles, in total 15*6=90 4-cycles

1 transposition can be chosen in 6 choose 2 ways.
2 transposition can be chosen in (6 choose 2)*(4 choose 2)/2 ways.
etc

Am I feeding a bot?

[Anonymous No.15240168]

>>15239067
The proofs of completeness in first order or propostional logic don't use compactness and their statements are dstinct.
>Complete
A logical system is complete if everything that is true about that system has a proof
>Compact
A logical system is compact if some set of sentences [math]\Sigma[math] has no model (valuation that makes all sentences true in prop. logic) then there is some finite subset of [math]\Sigma[math] that has no model (or prop. equivalent).
The way compactness is proved goes along these lines: we need a proof of a contracidtion from our infinite set but proofs are finite so we only used finitely many of our sentences

[Anonymous No.15240247]

>>15240143
Dude why would I be a bot??
Do you just not like the way I speak? That makes me a bot now? I'm not human enough for your liking? Tell me.
>aM i fEeDiNg A bOt??
I don't know. You tell me
I'm not even gonna check your answer. I don't care. Check it yourself, you don't want a bot checking your precious answer.

[Anonymous No.15240542]

>>15236879
Yes, it means they have some (often the simplest, if there is a choice of multiple) universal property. Watch Borcherds' video on limits/colimits on youtube

[Anonymous No.15240543]

>>15240168

I'm asking to explain it because I don't 100% understand the proof

[Anonymous No.15240546]

>>15239290
Okay? These two are not contradictory

[Anonymous No.15240562]

>>15240543
Look at the proof of completeness for propositional logic first; the proof in the case of first order logic follows the same structure with a few extra steps to ensure it satisfies all the requirements of first order logic like dealing with =. Some of the details in the first order case are quite tedious to unravel so I'd say to just look at a sketch proof

[Anonymous No.15240757]

any books you recommend on representation theory? I'm an undergrad cuck so don't go so hard

[Anonymous No.15240817]

What does [math] \theta_{i+1} = \theta_{i} /\sim_{l,r_i}[/math] mean in this context? Since each [math]\theta_i[/math] is a partition of [math] X [/math] set difference wouldn't make sense.

[Anonymous No.15240860]

>>15240817
Think I got it

[Anonymous No.15240952]

>>15240817
It's the factor/quotient space

[Anonymous No.15241016]

I'd like to request some help from you guys. I'm tutoring a probability class and one of the students came to me with a question I couldn't solve. say we have n urns. we choose an urn at random and put a ball in it. we repeat this process until one urn gets 2 balls. what is the probability of using exactly p balls?

[Anonymous No.15241020]

What are some applications of regulators to infinite series?

[Anonymous No.15241028]

>>15241016
you should really be removed from tutoring a probability class immediately if you can't solve that

[Anonymous No.15241029]

>>15241028
it kind of landed on my lap. is it really that easy? could you give me a hint at least?

[Anonymous No.15241034]

sin(x)+sin(pi*x) is aperiodic. How would I go about finding some f where sin(x)+sin(pi*x)+sin(f) is periodic?

[Anonymous No.15241044]

>>15241016
Sounds like a cute question. Some thoughts:

The chance that the "game" terminates in the first round is 0, since no urn has a ball.
Given the game is still going (that's a conditional probability, not an overall one) the chance that the "game" terminates in the second round is 1/n, since 1 out of the n urns has a ball in it from round one.
Given the game is still going , the chance that the "game" terminates in the third round is 2/n, since 2 out of the n urns has a ball in it from the previous two rounds.

Now the game is surely finite, since after the n'th round, if the game is still going, the n+1'th round must terminate (there'd be a ball in each urn.)
The changes of terminating in the k'th round, call it L_k, has L_1=0 and
[math] \sum_{k=2}^{n+1} L_k = 1 [/math]

For some p, you're asking for L_p.
It might not be necessary to find L_k, but it's surely sufficient.
But also L_p should be 1 minus the chance of terminating before p mins the chance of terminating any time after.

With what I've said I'm sure we can plug all the data together.
If nobody takes over from here, I'll think it over myself (I think it's just some multiplications of 1-foo) in a few hours

[Anonymous No.15241049]

>>15241044
To elaborate, I think it's a chain of the form:
Losing before p equals
r_2 + (1-r_2) * r_3 + (1-((1-r_2) * r_3)) * r_4 + ...
where r_k = k/n

But I might be overcomplicating it

[Anonymous No.15241081]

>>15241049
>>15241044
thanks, anon. I'll try to think of something

[Anonymous No.15241089]

>>15241016
[eqn] n (p-1) \frac{1}{n^2} \left( \frac{n-1}{n} \right)^{p-2} [/eqn]

[Anonymous No.15241103]

>>15240952
Thank you.

[Anonymous No.15241114]

i love classification results so much, bros. here's a basic, but neat one i learned just now: in [math]\mathbb Q_p[/math] the only roots of unity are the [math]p-1[/math] many [math]\mu_{p-1}=\{x\in\mathbb Q_p\mid x^{p-1}=1\}[/math] for [math]p[/math] odd and [math]\{\pm1\}[/math] for [math]p=2[/math].

[Anonymous No.15241121]

>>15241081
I'm >>15241044 and put together the script that I think represents your situation (double check it)
I also tried the formula of that anon who had posted something (but not deleted). That one looks off a bit for larger p


n k rate (anons rate)
2 2 0.49991 0.5
2 3 0.50059 0.5
3 2 0.3319 0.3333333333333333
3 3 0.44441 0.4444444444444444
3 4 0.22196 0.44444444444444453
4 2 0.24806 0.25
4 3 0.3712 0.375
4 4 0.27977 0.421875
4 5 0.09418 0.421875
5 2 0.19922 0.2
5 3 0.31772 0.32
5 4 0.28786 0.38400000000000006
5 5 0.15239 0.4096000000000001
5 6 0.03818 0.4096000000000001
6 2 0.16573 0.16666666666666666
6 3 0.27723 0.2777777777777778
6 4 0.27875 0.34722222222222227
6 5 0.18467 0.3858024691358026
6 6 0.07654 0.40187757201646096
6 7 0.0163 0.401877572016461
7 2 0.1432 0.14285714285714285
7 3 0.24496 0.24489795918367346
7 4 0.26408 0.3148688046647231
7 5 0.2002 0.3598500624739693
7 6 0.10883 0.3855536383649671
7 7 0.03609 0.39656945660396614
7 8 0.0062 0.3965694566039662
8 2 0.12515 0.125
8 3 0.21975 0.21875
8 4 0.24655 0.287109375
8 5 0.20646 0.3349609375
8 6 0.12803 0.366363525390625
8 7 0.05873 0.38468170166015625
8 8 0.01697 0.39269590377807617
8 9 0.00244 0.39269590377807617
9 2 0.10924 0.1111111111111111
9 3 0.19732 0.19753086419753085
9 4 0.22989 0.2633744855967078
9 5 0.2033 0.312147538484987
9 6 0.14274 0.3468305983166522
9 7 0.07496 0.369952638204429
9 8 0.03054 0.38365458776755595
9 9 0.00722 0.3897443431289457
9 10 0.00097 0.3897443431289457
[Finished in 21.748s]

[Anonymous No.15241132]

>>15241121
in fact, if I didn't type it out wrongly, it looks like anons proposal (4th column) wasn't even normalized

[Anonymous No.15241156]

>>15241121
>>15241121
Wait, just realized how to compute it analytically:
You got a chain of states from
2 to n+1, with losing state n+1,
the transition probability to go from k to n+1 is (k-1)/n,
and the transition probability to go from k to k+1 is the remaining 1-(k-1)/n.
Now write down the transition matrix P and solve PL=L, with L the terminating vector.
Might still be overblown but pretty confident that will work analytically
afk gl

[Anonymous No.15241201]

>>15241156
Sorry, I'm mistaken, that vector is not your solutions, just another property of your system

[Anonymous No.15241204]

>>15241016
[eqn]\frac{(n)_{p-1}(p-1)}{n^p}, \quad \quad (x)_k = x(x-1)\cdots (x-k+1)[/eqn]
4chan really is a place for brainlets...

[Anonymous No.15241212]

>>15241204
Yes, that looks like it makes sense - the chance of even getting to p (a product of not losing, I suppose that's the falling product) times the chance of losing from p.
(I had computed the chance of terminating given you're in p, but that's almost already part of the setup)

[Anonymous No.15241216]

>>15241204
I really has become a safe space for college idiots lately

[Anonymous No.15241217]

I used to think that constructivists were a bunch of weirdos and I still do but I get where they're coming from now: a lot of the nonconstructive proofs are not very satisfying. When you prove BPI by saying, "oh just use compactness", you feel cheated. For me, it's a similar feeling to when you use a group/semigroup presenation to assert existence of something: yes it exists but you know almost nothing about it.

[Anonymous No.15241248]

>>15241204
>>15241212
that does make sense. thank you both so much for your help, anons

[Anonymous No.15241279]

>>15241212
There are n_p-1 ways of choosing the first p-1 boxes. The last one has to be among the first p-1, so multiply it by that. Then multiply by probability of that combination which is n^-p.

[Anonymous No.15241372]

the heckin chain rule is too much for me bros

[Anonymous No.15241438]

Hello,

I'm 29 and I love mathematics, but unfortunately math doesn't love me back.

If you have to relearn math from scratch where would you start? Does anyone have the picture or anything I can breadcrumb?

[Anonymous No.15241444]

>>15241438
You're 29 and love mathematics... but you need to start from scratch? What happened?

[Anonymous No.15241448]

>>15241438
I'm somwhat curious as to why you would like maths if you aren't particularly good at it.
Regardless, learning all of maths is quite a tall order since there are many different branches. What are you particularly interested in?

[Anonymous No.15241451]

>>15241444
>What happened?
I'm ashamed to say: hard pornography use, heavy alcohol dependency, and other embarrassing things.

I am not a good man, but maybe I can be.

Please help me.

[Anonymous No.15241457]

>>15241451
>Please help me.
I'm sorry bro but I am not wise enough to know what would be ideal in your situation.

If I was to be in your situation (without actually knowing if that's the right path to take) I would start with Basic Mathematics (I have heard a lot of people recommend it), don't be fooled by the title, it's not for children (perhaps it is, but for really smart children).
You want to study as many maths as possible. If something intruiges you, search for resources.
I would suggest books but If you don't want to study a specific book I guess you can try online lectures? The most important thing is to realise that you trully want to learn and that doing exercises and problems are the true way of learning. Do not do exercise for the sake of it, do them because you want to learn something.

Don't take my words too seriously though.

>I am not a good man, but maybe I can be.
Are you implying that learning math (and other similar things you'll do I guess) is going to make you a good man? That's not the case.

[Anonymous No.15241458]

>>15241448
>I'm somwhat curious as to why you would like maths if you aren't particularly good at it.
I'm sorry, I don't have the words within me right now to explain it's appeal to me. One day, when I immerse myself completely, I may tell you but as I am right now I may only romanticize, confuse, and mislead. I hope that day come soon.

>What are you particularly interested in?
From an onlooker's perspective, from the ground up. It may take decades for me, but I assure you - I will honor it.

[Anonymous No.15241464]

>>15241217
This is the useful form of constructivism, and it's probably what most people believe without codifying it anyway; constructivism as a general philosophy that constructive proofs are better than non-constructive ones is a very good belief to hold. Constructivism as foundational autism trying to build an alternative mathematics according to your anal-retentive tastes is not

[Anonymous No.15241466]

>>15241457
>I'm sorry bro but I am not wise enough to know what would be ideal in your situation.
It is alright, I am grateful for what you have provided and I intend to put it to use immediately.

>Are you implying that learning math (and other similar things you'll do I guess) is going to make you a good man? That's not the case.
It is certainly a noble start. Basic Mathematics.

It may not be much, but you have given one more person hope in your life.

[Anonymous No.15241520]

>>15241217
It's a broad term. People like Wildberger and people like the HoTT crowd could well both be called constructivists, but probably don't know of each other at all.
>>15241464
I don't think there's all too many people who advocate classical mathematics breaking constructive math - and then so it's really just an appeal to keep more track of what principles you use.
On the one hand, there is Russian constructivism on the one hand, which with modern computability theory can be said to sit within the classical theory. I think it's fair (and actually fun) to keep track of when you compute something computable (recursive) and be aware of when you leave this - this programmability. Something like this is baked into type theorists work.
Somewhat relatedly, either more conservatively or orthogonal to that, there's finitists. I think you can let them do what they enjoy and don't engage when you go beyond it.
Then there's Browerian analysis, which is finitist in a sense, but uses classical math breaking axioms. "If you can evaluate a higher order function, F(f) in N, then the fact that your were able to evaluate means you inspected only finite amount of data from f." I think this not out of bound conceptually, and just shows how classical math, where it doesn't hold, is build in a loosey goosey hopeful manner.
But nobody after WWII was really interested into such a hardcore intuitionism, so who do you even insult if you reject it.
tl;dr I don't think it's too much to ask to keep track when you leave computable functions. I think people are just afraid of seeing [math]\neg\neg[/math] in their formulas. But they say something.

[Anonymous No.15241548]

>>15241438
first to dispel some myths:
math is not innately virtuous - however, learning new things is good for you and if you're going to learn, it might as well be math.
math is not difficult nor does it require you to devote your life to it - you /need/ to believe in yourself and your ability to learn. In particular, stop self-identifying with being a bad person that received some divine gift in the form of an opportunity to study math.

as for reccs
the other anon already mentioned basic mathematics
i'll just add khan academy onto that, because having worked video examples is pretty valuable

finally, if you're studying math for the sole reason of it being some sort of redemption to put virtue into your life, then drop it, you won't get what you want from math nor will you get far in math. However if you do like math for the sake of math itself, you will get a lot of fun and satisfaction from it.

[Anonymous No.15241565]

>>15241458
From the ground up is an unsual way of wanting to do things and if you try to persure that then it may be too much for you. Think of maths like a building built into the side of a hill with its entrance part way up: you enter on the first or second floor, where you have an intuitive understanding of what's going on, maybe you've learnt some calculus or algebra before, if you want then you go to the higher floors and learn the more advanced techniques and ways of thinking about certain areas (matric spaces may be on one floor and topology might be on the floor above). If you choose then you can go to the lower floors where you learn about the foundations of maths, analysis is the foundation of calculus but it is not necessary to know if you only want to do calculus; logic and set theory sit below them at the base of the building but, again, they aren't 100% necessary for everything.
Saying you want to start from the very bottom is understandable but maths doesn't work like that, it's better to pick on topic of particular interest and read about that.

[Anonymous No.15241633]

>>15241548
Thank you. I do appreciate this.

>if you do like math for the sake of math itself, you will get a lot of fun and satisfaction from it.
Perhaps what I am hanging on to is some counterfeit which is so difficult to let go. I will keep this message with me.

>>15241565
I understand where you are coming from - strands seem to emanate from a common trunk which may or may not cross in-between interests and may leave it out entirely.

>it's better to pick on topic of particular interest and read about that.
If you do not mind me asking, what interests you?

[Anonymous No.15241642]

>>15241633
I quite like abstract algbra and topology but am less interested in calculus and statistics, but if I'm honest I like them because they pose interesting problems and maybe not for their own sake.
If you want to learn abstract algebra then I'm certain there's a few online letures about the topic

[Anonymous No.15241942]

What do the witnesses of a Big O estimation actually mean?

[Anonymous No.15241970]

>>15241942
that you had sex

I wouldn't bet on it happening

[Anonymous No.15242396]

>>15241942
>witnesses of a Big O estimation
If you mean the constant term, it's the intercept of the log-log plot of computational cost against problem size.
If you mean the threshold "for sufficiently large N" beyond which the cost is bounded by your big-O formula, then any other larger (finite) number would work just as well, so it doesn't really mean much (though the constructivists ITT might beg to differ).

[Anonymous No.15242412]

>>15236235
Your reference to the law of excluded middle is making me want to read my copy of From Frege to Godel now.

[Anonymous No.15243072]

>>15239986
Nice latex friend. Let’s throw out the initial lim and ask, for each [math]x \neq 1[/math]: What is [math]\lim_{n\rightarrow \infty}c^{2n}[/math]? Does it approach a value as x approaches 1 from the left? Does it approach a value as x approaches 1 from the right?

[Anonymous No.15243107]

>>15243072
It is [math] (0,1, \infty) ? [/math]

[Anonymous No.15243114]

lads I got a PhD in mathematics, starting a solid senior-level job later this year.

however, I still do not have horny qt3.14 Asian girls approaching me. can someone clarify what else I need to do to accomplish this.

[Anonymous No.15243683]

Writing my master's thesis, and I have a specific field with some funky arithmetic defined. The fact that the arithmetic respects the field axioms is not extremely trivial (kinda is albeitever), but can be checked quite fast if you whip out some pen and paper. Since I'm in CS I don't want to spend too much time on the mathematics, so I don't want to show how the arithmetic respects the field axioms (unless I really have to). How do I express this in the paper?
"That the following bla bla respects the field axioms is ...."??????? WHAT DO I WRITE!!!!!!

[Anonymous No.15243703]

>>15243683
I don't think you can go wrong is just stating the provable fact. If you want you can even add a hint for the harder part of the proof, if any.
That said, it's not clear what you mean by "an arithmetic". Do you deal with an algebra over Q or what is the structure? I don't think "arithmetic" is used for some structure too often, in the most fields

[Anonymous No.15243762]

>>15236171
The important thing to recognize is that in constructive mathematics, a function is some which, for a particular input, gives a particular output.

For example, let 1={0} and let CH denote the continuum hypothesis. Adopt the full Separation axiom, for convenience.
Consider the subset
[math]f\subset \{0\}\times \{0,1\}[/math]
defined by
[math]f = \{x \in\{0,1\} \mid x=0 \land \text{CH}\}[/math]
Via case analysis, we can prove
[math](\text{CH})\lor \neg \text{CH})\to \big((f=x\mapsto 1)\lor (f=x\mapsto 0)\big)[/math]
Not assuming excluded middle and not assuming or rejecting CH, this subset does not define a function.
The moral here is that, generally, constructively should restrict your idea of how big a function space has to be. E.g. if you assume all functions are programmable, then there's not more functions from N to N than there are indices for Turing machines (not more than countable).
So keep in mind that "function" is not apriori as broad as in the classical case.

Now akin to a binary Cauchy sequence, you can think of most common real numbers as being given via a countable infinite sequence of 0's and 1's, or propositions, i.e. a predicate P(n) over N.
For simplicity, consider a predicate that is decidable, meaning 'P(a) or not P(a)' holds for all 'a'. Say you know each value is computable (that is, P is captured by an actual function from N to {0,1}). Say you evaluate the first 10000 terms and they are all 0 and you can't find a counter case. Here, you might still not be able to formally proof that all values of the predicate/function are 0.
You know
forall n. [P(n) or not P(n)]
but not
[forall n. P(n)] or not [forall n. P(n)]
This issue means that given a real number 0, you can't generally know if it's 0 or not.

This together makes it so that in some systems, it's hard to formulate a "function" which is discontinous and actually provably a function. In turn, you can consistently postulate that they don't exist at all.

[Anonymous No.15243768]

Forgot the tuple bracket. I.e. I should have written

[math] f = \{(0, v) \} [/math]
with
[math] v = \{x\in \{0,1\} \mid x=0 \land \text{CH}\}[/math]

[Anonymous No.15243776]

The important thing to recognize is that in constructive mathematics, a function is some which, for a particular input, gives a particular output.

For example, let 1={0} and let CH denote the continuum hypothesis. Adopt the full Separation axiom, for convenience.
Consider the subset
f⊂{0}×{0,1}

defined by
f:={(0, v)}
v:={x∈{0,1}∣x=0∧CH}

Via case analysis, we can prove
(CH)∨¬CH)((f=x1)∨(f=x0))

Not assuming excluded middle and not assuming or rejecting CH, this subset does not define a function.
The moral here is that, generally, constructively should restrict your idea of how big a function space has to be. E.g. if you assume all functions are programmable, then there's not more functions from N to N than there are indices for Turing machines (not more than countable).
So keep in mind that "function" is not apriori as broad as in the classical case. You of function spaces as potentiall sparse

Now akin to a binary Cauchy sequence, you can think of most common real numbers as being given via a countable infinite sequence of 0's and 1's, or propositions, i.e. a predicate P(n) over N.
For simplicity, consider a predicate that is decidable, meaning 'P(a) or not P(a)' holds for all 'a'. Say you know each value is computable (that is, P is captured by an actual function from N to {0,1}). Say you evaluate the first 10000 terms and they are all 0 and you can't find a counter case. Here, you might still not be able to formally proof that all values of the predicate/function are 0.
You know
forall n. [P(n) or not P(n)]
but not e.g.
[∃n. P(n)] or not [∃n. P(n)]
This issue means that given a real number x, you can't generally know if x=0 or not.

This together makes it so that in some systems, it's hard to formulate a "function" which is discontinuous and actually provably a function. In turn, you can consistently postulate that they don't exist at all.

[Anonymous No.15243778]

>>15236171
>>15243776
Also your statement is not automatically true at all, it's just possible to assume this as an axiom

[Anonymous No.15243810]

>>15243683
Checking field axioms is always pretty easy and would ammount to maybe half a page at most.
>(F,+) is an abelian group because ...
>(F\{0},*) is an abelian group because ...
>Distributivity holds because ...
That's it

[Anonymous No.15243901]

>>15243776
how is v in {0,1}

[Anonymous No.15243964]

>>15243901
[math]Sx:=x\cup \{x\}[/math]
[math]0:=\{\}[/math]

[math]1:=S0=\{0\}=\{\{\}\}[/math]
For a proposition [math]P[/math] and any set [math]S[/math], define
[math]w:=\{x\in \{0\}\cup S\mid x=0\land P\}[/math].
If [math]P[/math] is provable, then
[math]w=\{x\in \{0\}\cup S\mid x=0\}=\{0\}=1[/math]
If [math]\neg P[/math] is provable, then
[math]w=\{\}=0[/math]

It's probably cleaner if I write, without splitting v from f,
f := {p in {0}×{0,1} | p_left={x∈{0}∣x∈CH}}

[Anonymous No.15243973]

>>15243964
i thought this was in the context of constructive mathematics where we can't do a case analysis on P (or CH in the previous post)

[Anonymous No.15243974]

>>15243901
Sx:=x∪{x}
0:={}
1:=S0={0}={{}}

For a proposition P and any set S, define
w:={x∈{0}∪S ∣ x=0∧P}
If P is provable, then
w={x∈{0}∪S ∣ x=0}={0}=1
If ¬P is provable, then
w={}=0

It's probably cleaner if I write, without splitting v from f,
f := {p in {0}×{0,1} | p_left={x∈{0} ∣ CH}}
(The variant with v expresses the same idea but here it's properly defined as a subset of {0}×{0,1})

[Anonymous No.15243978]

>>15243973
Yeah I probably wrote it down sloppily. You want to define the object f to make the case it's not provably a function, but would be if you adopt LEM.
To make that case you don't reall need to label v as an element of anything

[Anonymous No.15243984]

My v was only a would-be bit, I don't need to make f a subset upfront to my the case that the function space is allowed to lack some elements that are classically functions

[Anonymous No.15244014]

>>15243776
>words words words
Are construtivists the leftists of mathematics?

[Anonymous No.15244018]

>>15244014
they're the vegans of mathematics

[Anonymous No.15244027]

>>15244014
It's just math. For everything that's compatible with classical mathematics, you can optionally say the same thing by carrying a computability predicate around with you, and instead of "for all functions holds" say "for all computable functions hold". A predicate akin to
https://en.wikipedia.org/wiki/Kleene%27s_T_predicate
Nothing expressed here is non-classical, in that light, it's just a matter of what you take "for all" to mean.

It probably speaks for the foundational nature of the notion(s) of computability that a syntactic/formal proof calculus (just one principle away form the classical one) can be taken to speak of those kinds of objects (the computable ones).

[Anonymous No.15244057]

>>15244018
Although the constructivists can still taste meat.

If [math]C[/math] are your constructive axiom and [math]N[math] are some non-constructive ones, and the classical mathematician can prove the theorem [math]A[/math], i.e.
[math]C\cup N\vdash A[/math]
then there's a nice (constructive, in a trivial sense) proof that [math]N\to A[/math], i.e.
[math]C\vdash N\to A[/math]

It's just more bookkeeping and harder proofs. But not less content.

[Anonymous No.15244063]

>>15244018
Although the constructivists can still taste meat.

If [math]C[/math] are your constructive axiom and [math]N[/math] are some non-constructive ones, and the classical mathematician can prove the theorem [math]A[/math], i.e.
[math]C\cup N\, \vdash\, A[/math]
then there's a nice (constructive, in a trivial sense) proof that [math]N\to A[/math], i.e.
[math]C\, \vdash \, N\to A[/math]

[Anonymous No.15244119]

Best book on number theory?

[Anonymous No.15244123]

Hello /mg/.
I'm working on proving Theorem 53 in pic rel. So I'm trying to show that given Y and ~, (Q,q) satisfies (i)-(iii) in Def. 5 iff it satisfies (1) and (2) in Theorem 53.
I managed to prove the forward direction, i.e. (i)-(iii) implies (1),(2). The converse is giving me trouble though.
Specifically, how do I go about showing (ii) from the assumptions (1) and (2). I think my problem is that I'm not sure how to use (2) since I don't know choice of (Z, k : Y -> Z) would get me anywhere.
Does anyone just have a hint at how to start this proof please

[Anonymous No.15244125]

>>15244119
Manin number theory

[Anonymous No.15244127]

>Constructive mathematics doesn't even have a full proof theory

LMAO! And people are unironically constructivists

[Anonymous No.15244137]

>>15244125
Any particular reason you reccomended this one or is it plain and simple good?

[Anonymous No.15244171]

>>15244119
Jeff's Number Theory.

[Anonymous No.15244252]

I have an idea: "bubble popping anti-math" or "Sebastian vs. the Nothing"
It is as follows. The principle is observing the "set theory ghetto" and
the particular habits of its inhabitants as well as the particular
structure of some proofs in set theory, such as some proofs that various
statements are equivalent to AC: the well ordering theory, the trichotomy
of cardinalities of sets, Zorn's lemma, and so on.
The method is as follows: first declare certain tools to be "native" to
set theory and "essentially unused" outside set theory. Then take the
implications of those tools inside set theory, such as AC<->WOT as axioms
and delete the tools used to prove them such as
- transfinite recursion
- transfinite induction
- axiom of regularity (AR)
and add much weaker axioms as replacements, such as
>for all x, x is not a member of x, in symbols
>(forall (x) (</- x x)) [1]
>∀x x ∉ x
to compensate

[Anonymous No.15244256]

So inside your subject area, you only have relatively weak tools to prove
facts, not the entire power of set theory
if new facts of set theory are required when extending the subject area,
then those facts can be established inside set theory as consequences of
ZFC, then taken as new axioms for a new version of formal axiomatic
mathematics corresponding to new theorems in your subject area
In fact, this question generally, "How has your subject area of math made
use of facts of pure set theory lately?" could be used as the basis for
a mathematical newsletter on the topic of new developments in a subject
area you study
The objective is classification and converting subject areas of math into
"formal silos" that each rely on taking various statements among
- AC1<->AC2 where AC1 and AC2 are two equivalents of AC in ZF
- a statement of "pure set theory" relying on AR
as axioms, in other words specializing ZFC to various subjects of math by
removing axioms that give information about the "internal structure" of
sets and forcing the theorems to rely on axioms that obey a discipline of
data abstraction specialized, specific, and particualr to that subject
[1] see "EYE" document at https://imgur.com/a/Gzp2Ile

[Anonymous No.15244264]

>>15244252
But why, exactly?

Firstly, I think separation (class intersection, basically) proves essentially all the induction needed in normal math.

Secondly, also why add the implications back you mention back? The example you gave will be provable in very weak systems, it doesn't need something like AC or WOT to prove AC<->WOT. You probably only need the existence of powersets and separation.

I think, anyway, that this is all well understood. Consider e.g.
https://en.wikipedia.org/wiki/Reverse_mathematics#The_base_system_RCA0

[Anonymous No.15244265]

>>15244252
>>15244256
So the whole point of this is to use these new axioms specific to various subject areas of math as a "Dewey Decimal System for Math"
Essentially a math search engine, but something along the lines of LexisNexis, pre-internet in vision and scope, able to handle queries such as
>Give me the subject areas of math a that depend on AC1<->AC2, AC3<->AC4, and show me where they are used
https://www.youtube.com/watch?v=SkypZuY6ZvA

[Anonymous No.15244268]

>>15244264
see
>>15244265

[Anonymous No.15244278]

this board is at times indistinguishable from /x/

[Anonymous No.15244291]

>>15244264
There's another reason: making good on the claim in Kunen:
>Virtually all results in the branches of mathematics based on set theory hold even in the absence of regularity [1]
The thinking is to put this into effect by looking at "cheap knock offs" of AR such as ∀x x ∉ x designed to keep set theory in set theory, essentially making it formally impossible to prove various results of set theory in a subject that isn't set theory
So set theory (and hence the rest of math) can't formally "contaminate" any given subject
https://www.youtube.com/watch?v=ZyhrYis509A
[1] https://en.wikipedia.org/wiki/Axiom_of_regularity

[Anonymous No.15244301]

>>15244278
the music on /x/ is way better
https://mp3.hardnrg.com/MadamZu-March_2004.mp3

[Anonymous No.15244337]

>>15239687
#_nabrnastar.py_Na[Na,*]
#_The_monoid_semiring_and_semimodule_over_the_natural_numbers_of_finite_linear
#_combinations_of_natural_numbers_under_multiplication_with_natural
#_coefficients
#
class_ScalarError(TypeError):
____pass
class_Elt(object):
____def___init__(self,d):
________if_isinstance(d,int):
____________d_=_{d:1}
________self.d_=_d
____def___add__(self,other):
________ks_=_set()
________ks.update(self.d.keys())
________ks.update(other.d.keys())
________nd_=_{}
________for_k_in_ks:
____________nd[k]_=_0
____________if_k_in_self.d:
________________nd[k]_+=_self.d[k]
____________if_k_in_other.d:
________________nd[k]_+=_other.d[k]
________return_Elt(nd)
____def___mul__(self,other):
________nd_=_{}
________for_k1_in_self.d:
____________for_k2_in_other.d:
________________k_=_k1*k2
________________if_k_not_in_nd:
____________________nd[k]_=_0
________________nd[k]_+=_self.d[k1]_*_other.d[k2]
________return_Elt(nd)
____def___rmul__(self,other):
________if_isinstance(other,int):
____________nd_=_{}
____________for_k_in_self.d:
________________nd[k]_=_other_*_self.d[k]
____________return_Elt(nd)
________raise_ScalarError(other)
____def___repr__(self):
________return_'<'_+_repr(self.d)[1:-1]_+_'>'

[Anonymous No.15244343]

>>15244265
https://youtu.be/gzbDdgWiaS0

So you want to drive the theories out of "Cantors paradise"?
As I pointed to above, there's a strong case to be made that second order arithmetic is already very strong compared to the math people do. There's just some outlier in academic math, those which need vast categories for homological algebra or whatnot.
So you'd basically say "arithmetic is captured by this theory" and from then on you're hard pressed to find anything left to classify which isn't modeled within this theory.

>>15244291
It's absolutely true that regularity is not really used much anywhere - to the extent that removing it doesn't give you a theory that's very new. As I said, second order arithmetic induction is already strong. Epsilon induction isn't used for much.

[Anonymous No.15244352]

(Mind you, Peano arithmetic, which has no sets at all, is already bi-interpretable with the hereditarily finite sets and their membership, meaning ZF-Inf+not Inf)

[Anonymous No.15244366]

>uses ZFC-Inf+!inf to study theoretical CS

[Anonymous No.15244377]

>>15244366
I can't tell if that's a judgement or an appeal

[Anonymous No.15244400]

>>15244377
>are you listening to the club drugs
https://mp3.hardnrg.com/MadamZu-March_2004.mp3
P
>O
R
>N
S
>T
A
>R

[Anonymous No.15244419]

>>15244400
I have a small house worth of crypto in a browser extension, I'm not gonna click random files

[Anonymous No.15244427]

>>15244419
>bicycle day digits

[Anonymous No.15244721]

>>15235870
Where can I get the pdf for this book? It's not on libgen or z-lib. Does anyone in here have it by chance?

[Anonymous No.15244756]

Does anyone have any idea what he means by "TOW"? I assume it's a function, googling didn't help

[Anonymous No.15245001]

>>15244721
What part do you need? Half the chapters are papers available on arxiv. Just copy from the table of contents into google.
Like this one:
https://arxiv.org/abs/2002.00879
>>15244756
Tower function 2^2^2^2^2^2^2^2^2^2

[Anonymous No.15245108]

>>15244125
What's the best book with exercises?

[Anonymous No.15245427]

>>15244427
Explain. Is this something people would know?

[Anonymous No.15245718]

>>15238191
define what it means to add up infinitely many terms and suddenly it isn't so heckin beautiful anymore

[Anonymous No.15245726]

>>15238553
What you make of it is that if by definition
[eqn]\sum_{k=0}^\infty z^k := \frac{1}{1-z}[/eqn]
for all z then surely it would be equal to 1/2, but then it wouldn't be the case that
[eqn]\sum_{k= 0}^\infty z^k = 1+z+z^2+z^3....[/eqn]

[Anonymous No.15245863]

>>15245718
you have to say this out loud every time the professor defines anything
>define the singular homology and suddenly it isn't so heckin beautiful anymore
>define the splitting field and suddenly it isn't so heckin beautiful anymore
>define the minimal polynomial and suddenly it isn't so heckin beautiful anymore
>define the one point compactification and suddenly it isn't so heckin beautiful anymore

[Anonymous No.15245871]

>>15245108
Niven Zuckerman Montgomery 5/e

[Anonymous No.15245907]

>>15245726
I'm not sure if you're making a proposition for a theory or just a claim.

I don't know of any theory that takes the formula \sum z^k = 1/(1-z) as definition. It's a result in analysis, valid for z in a certain domain. My limit result involving 1/2 is also provable in analysis (indeed, I gave the proof). In the theory of analysis, there's no value z so that the left hand side is 1/2 whole the right hand side is simultanously defined.
You can set up theories of infinite sums (not in the sense using the metric over the reals for their convergence) to make things work natively (pic related), although that's somewhat exotic math.

[Anonymous No.15246275]

>>15245001
I don't need it now, I was just checking but thanks. If I end up buying the pdf, I'll try to remember to come back here to upload it.

[Anonymous No.15246310]

>>15244337
cool stone

[Anonymous No.15246318]

>>15246275
Give me your preferred anonymous file sharing method and I'll give you it

[Anonymous No.15246399]

>>15235870
Does the plant fuck her?

[Anonymous No.15246524]

>>15246310
Let Na denote the natural numbers.
For Na[Na,*], indeed any monoid semiring Na[C] where C is an abelian monoid, there is a semilinear map taking natural values, the *height* of an element, in symbols | p | where p <- Na[C] defined by | e[c] | = 1 for all c <- C
> | ? | : Na[C] -> Na
Since { e[c] : c <- C } generates Na[C] as a C-semimodule over Na, this extends to a unique Na-semilinear map on Na[C].
For any given naturals n,k, there are only finitely many elements p <- Na[C] such that p <- < e[i] : 0 <= i <= k > and | p | = n
This can be used to prove that Na[Na,*] is a countable set
Note: Na[[Na,*]], formal linear combinations of elements in Na,* with natural coefficients, is a completely different animal, and in general we even need to add a condition to an abelian monoid C to prove that multiplication in Na[[C]] exists, namely
> (*) for all c <- C, the set { (a,b) : c = ab } is finite
however, we are in luck for C = Na,+ and C = Na,* as both satisfy (*)
>Exercise. Let C be an abelian monoid satisfying (*) above, and let E = Sum[n=0]^oo e[n] <- Na[[C]]. Calculate E^n explicitly for n = 1,2,3,... where C = Na,+ and C = Na,*

[Anonymous No.15246543]

>>15246524
errata: formal power series, not formal linear combination

[Anonymous No.15246580]

>>15246524
What is e[c] in the 2nd line?

[Anonymous No.15246588]

Posting here since the stupid questions thread reached the bump limit:

Given a closed/open convex subset C of the plane, with C not containing the origin, does there exist a straight line (extending infinitely in all directions) through the origin which is disjoint from C ?

[ An anon in the sqt already gave a counterexample if "closed/open" is dropped in the above; I'll leave it as an exercise ]

[Anonymous No.15246598]

>>15246588
>in all directions
Typo: "in either direction", not "in all directions"

[Anonymous No.15246602]

>>15246524
errata: penultimate line should read C = Na \ {0},*
it's worth pointing out that if we adjoin oo to the naturals with
>oo + n = n + oo = oo
>oo * n = n * oo = oo if n >= 1
>oo * 0 = 0 * oo = 0
and call this object Na^oo then we can once again handle abelian monoids C that do not satisfy (*) when constructing the semiring of formal power series in C with coefficients in Na^oo, Na^oo[[C]]
HOWEVER
the main problem with this (abuse of) notation is that Na^oo does not refer to a semiring with no additional structure...it refers to a semiring plus our way of using oo as a special symbol to handle the result of attempting to add up infinitely many nonzero terms in a product in Na[[C]] where C does not satisfy (*)
Spivak does this
https://archive.org/details/spivak-m.-calculus-2008/page/57/mode/2up?view=theater

[Anonymous No.15246625]

>>15246580
elements of Na[C] are finite formal linear combinations of elements in C...if C is the natural numbers, then replace c with e[c] to avoid confusion, so { e[c] : c <- C } is a basis for Na[C] as a semimodule over the natural numbers

[Anonymous No.15246653]

>>15246625
Oh so the e[c] are the standard basis elements for the free module

[Anonymous No.15246678]

>>15246653
yes
compare polynomial ring in Hoffman and Kunze
https://archive.org/details/LinearAlgebraHoffmanAndKunze/Linear%20Algebra%20-%20Hoffman%20and%20Kunze/page/n125/mode/2up?view=theater
this will take you down the traditional Galois theory and adjoining solutions to polynomial equations to fields route

[Anonymous No.15246689]

>>15246678
What is (*) in your original question?

Can you upload a picture of the exercise or smth because I can't parse your writing

[Anonymous No.15246798]

>>15241942
>witnesses of a Big O
We call a number a witness for something when it can be “plugged in” to replace an existential quantifier. In the case of Big O, the statement begins with “there exist a lower bound M and a constant factor c…” so a witness could mean finding an actual value for M, for c, or for both

[Anonymous No.15246835]

How do inductive definitions work?

in my discrete structures class the lecturer defined [math]\leq[/math] on [math]\mathbb{N}[/math] like this:
(i) [math]0\leq n[/math] for every [math]n\in\mathbb{N}[/math]
(ii) If [math]m\leq n[/math] then [math]s(m)\leq s(n)[/math] where [math]s[/math] is the successor operation on [math]\mathbb{N}[/math].

Why am I allowed to do define a relation like this? Usually in class we define relations explicitly so I expected something like [math]m\leq n[/math] iff (...)

Also isn't there more than one relation that satisfies (i) and (ii)? I'm assuming we want the one that relates elements only if they were put into a relationship by (i) or (ii) but how do I even knoq such a thing exists (and how do I maybe formulate this more precisely?)

[Anonymous No.15246901]

>>15244721
>>15246275
https://www.dropbox.com/s/lgs083g8uk3emyz/harmonic%20analysis.pdf?dl=0
I might take this down at some point.

[Anonymous No.15246923]

>>15246835
Have you covered proofs by induction?

[Anonymous No.15246925]

>>15246835
The lecturer probably also said that the trichotomy holds which would rule out relations like the full relation.
Many constructions in logic and set theory are inductive like this, the philosophy is that everything can be built from the empty set (or 0) - a notion called Zermelo heirarchy

[Anonymous No.15246945]

>>15236897
I'm too lazy to write it out but the gist is that, if a number has an odd number factors, the all of the primes in it's prime factorization must be raised to an even power (trust me bro). Therefore, there exists a natural square root (just take the same prime factorization, but divide every power by 2)

[Anonymous No.15246944]

>>15246318
Nice, I don't really have one but anonfiles looks alright, you might have to send a screenshot/qr code of the link though because 4chan flags it as spam

[Anonymous No.15246957]

>>15246901
>>15246944
Lol, the page just refreshed as I replied and I see this, anyway, thank you very much anon, much appreciated.

[Anonymous No.15247012]

>>15246835
Most likely what your lecturer was doing is, <= is a relation symbol that is explicitly included in your presentation of Peano Arithmetic (or whatever). <= could be anything consistent with these axioms (if you've seen models this may make more sense).
One can use the induction axiom (plus some other axioms to show the statement
for all m for all n (m <= n and not m>=n and not m=n) or (m >= n and not m<=n and not m=n) or (m=n)
which is the trichotomy, as expected

In terms of "more than one relation", one can show that the addition structure determines <= in the sense of:
for all m for all n: m<=n equivalent (exists a: a+m=n).

But there's not really "one possible structure" in interesting math ever, which you'll see when you study first order compactness, lowenheim-skolem, etc
The above should be interpreted as saying if I know how my model of PA adds numbers then I know how it interprets <= for free.

[Anonymous No.15247634]

>>15235870
I've been self-studying Category theory for the last few weeks with Lane's book. What kind of schizo ultra-turboaustim is needed to come up with these things? It's honestly impressive how it makes connections between seemingly unrelated things

[Anonymous No.15247760]

>>15246689
see
>>15246524
antepenultimate line
>I can't parse your writing
>Can you upload a picture
I'm not quite sure where to direct you for clarification...almost certainly the group ring section of Dummit and Foote
https://archive.org/details/XS8YSXQYXY9QJXSQ/214005482-Dummit-and-Foote-Abstract-Algebra/page/n245/mode/2up?view=theater

[Anonymous No.15248167]

>>15246835
>I’m assuming we want the one that relates elements only if they were put into a relationship by (i) or (ii)
This is what he means, BUT you need a little more than what you have written here. For example you cannot deduce that 0 <= s(0) from just i and ii.
>but how do I even knoq such a thing exists (and how do I maybe formulate this more precisely?)
In addition to the “bottom-up” definition you have provided (start with 0 <=0 look at whether 0 <= 1; add 2 into the mix; etc) there is also a “top-down” version that works as follows:
>consider ALL binary relations xRy on [math]\mathbb N[/math] which satisfy i, ii, and whatever missing iii
>there DO exist some such relations, because the “always yes” relation works
>define [math]\leq[/math] as the unique smallest such relation
This lets you define [math]\leq[/math] using power sets in a framework like ZFC. Some people prefer it this way but some do not, in particular you have to assume power sets exist, and have to prove the “unique least” relation actually exists!

[Anonymous No.15248169]

>>15248167
>For example you cannot deduce that 0 <= s(0) from just i and ii.
Disregard that I suck cocks. Was thinking that (i) just said 0 <= 0 lol

[Anonymous No.15248209]

What an amazingly well structured textbook. An actual pleasure to read.

[Anonymous No.15248951]

>Truly a
>Minefield of a
>Text

>Thought my
>Heart would
>Leap out of my
>Chest!

[Anonymous No.15249018]

is infinitely large napkin still worth reading despite being incomplete?
should i just read textbooks on the individual subjects?

[Anonymous No.15249061]

>>15249018
I'd see it as a motivation to make your own kind of record like that. Not sure if it's good to learn from for anything, but you can use it to double-check formal definitions

[Anonymous No.15249147]

why do most order theory texts study partial orders and not the more universal notion of preorders? especially considering the latter is much more in line with all that category theory stuff

[Anonymous No.15249285]

>>15249147
I'm not going to answer your question, just because...however, I am going to point out the following
Prop. Let C be a group embeddable monoid. The binary relation a <- bC (a.k.a. "b is a prefix of a") is always a partial preorder; it is a partial order iff C^x = { a : there exists b such that ab = ba = 1 } is the trivial subgroup of C.
Definition. Let C be a group embeddable monoid. It is *positive-definite* if ab = 1 implies a = b = 1 for all a,b ∈ C.
Taking one more step, we can look at a group embeddable monoid C and then look at submonoids D <= C such that D ∩ C^x = {1} and C = < D U C^x >, and call such a submonoid a *spire*
Example. Let Re be the real numbers and Re[+] the non-negative reals. Then C = Re[+] x Re, the closed right half-plane, is a group embeddable monoid, in fact it is embeddable in an abelian group, namely Re^2 under addition. We have C^x = {0} x Re.
So what do we find. First of all, it looks like AC is going to give us
>let E <= C be a submonoid such that E ∩ C^x = {1}. Then E can be extended to a spire submonoid of C
and it looks like this is going to be true for general group embeddable monoids C...???
Next, we can consider the topological properties of C = Re[+] x Re and try to prove that the closed spire submonoids are homeomorphic to the projective space of 1-dimensional half-spaces (half-lines or rays) when C is considered as a semivector space over the semifield Re[+]
this is just a hunch...

[Anonymous No.15249301]

The zig-zag lemma still isn't intuitive to me

[Anonymous No.15249381]

>>15249285
meds NOW

[Anonymous No.15249434]

>>15249381
https://www.youtube.com/watch?v=XHRLQOGLxR8
https://soundcloud.com/user-114329832/sets/howls-moving-castle-ost
https://mp3.hardnrg.com/MadamZu-March_2004.mp3
https://www.youtube.com/playlist?list=PLWNPmSJMzoRR8eNal31PK_fIAaHbZILdw
https://vimeo.com/683157880
https://vimeo.com/240041217

[Anonymous No.15249790]

Apparently for any two positive integers x and y, the quotient arctan(x)/arctan(y) is always transcendental, but my prof can't remember where they saw the proof or what it relied on.
I want to find the proof but google is useless. Is there any way of finding the proof that is better than just gradually reading a bunch of transcendental number theory books?

[Anonymous No.15250307]

>>15239593
Since there hasn't been any replies since the last one, I'll give a hint!
It has to do with cycles! Cycle length! Now anyone should be able to solve this problem!

[Anonymous No.15250373]

HE POSTED
https://www.youtube.com/watch?v=cw2ZhHWnGXE

[Anonymous No.15250393]

>>15249147
>why do most order theory texts study partial orders and not the more universal notion of preorders?
This is a silly question. Partial orders and preorders are exactly the same thing up to a quotient, and partial orders are easier to draw.

[Anonymous No.15250407]

>>15239593
a) 415
b) 720

[Anonymous No.15250619]

At what point are you first nutting?
>ring
>with unity
>commutative
>integral domain
>Noetherian
>integrally closed
>Dedekind
>GCD
>UFD
>PID
>Euclidean
>field
>characteristic zero
>algebraically closed
>C

[Anonymous No.15250688]

>>15250407
Anon I'm afraid both of those are wrong!
I'm not sure how you got them but for b) you seem to have gotten 6!, which would imply that all permutations of 6 give the identity permutation when you take the 6th power of it. I don't think this sounds quite right. If you tell me how you got your answers I could try to help you find the real ones. Or if you don't want that, I can just post the solution from the book.

Regardless I respect your attempt. I appreciate the time and effort you've given it and I hope we can guide you to the correct solution.

[Anonymous No.15250854]

>>15249301
Help me out

[Anonymous No.15250876]

>>15250688
a) 301
b) 396

[Anonymous No.15250926]

>>15250876
Interesting. Now b) is correct however a) remains wrong. I suspect a simple arithmetic mistake since you got b) correct and they're basically the same problem solved in the same way. Either way nice job on b)! How did you get it?

[Anonymous No.15251005]

>>15244337
>>> e = Elt({1:1,2:1,3:1,5:1,7:1,11:1})
>>> sorted((e*e).d.items())
[(1, 1), (2, 2), (3, 2), (4, 1), (5, 2), (6, 2), (7, 2), (9, 1), (10, 2), (11, 2), (14, 2), (15, 2), (21, 2), (22, 2), (25, 1), (33, 2), (35, 2), (49, 1), (55, 2), (77, 2), (121, 1)]
>>> sorted((e*e*e).d.items())
[(1, 1), (2, 3), (3, 3), (4, 3), (5, 3), (6, 6), (7, 3), (8, 1), (9, 3), (10, 6), (11, 3), (12, 3), (14, 6), (15, 6), (18, 3), (20, 3), (21, 6), (22, 6), ..., (1331, 1)]
>>> sorted((e*e*e*e).d.items())
[(1, 1), (2, 4), (3, 4), (4, 6), (5, 4), (6, 12), (7, 4), (8, 4), (9, 6), (10, 12), (11, 4), (12, 12), (14, 12), (15, 12), (16, 1), (18, 12), (20, 12), (21, 12), (22, 12), (24, 4), (25, 6), (27, 4), (28, 12), (30, 24), (33, 12), (35, 12), (36, 6), (40, 4), (42, 24), (44, 12), (45, 12), (49, 6), (50, 12), (54, 4), (55, 12), (56, 4), (60, 12), ..., (14641, 1)]

[Anonymous No.15251085]

>>15250926
give me a hint. 350. is that closer?

[Anonymous No.15251145]

>>15251085
That is farther! You need to go down.
>give me a hint
I don't know how to! I don't really know how you got the answers you already did so I'm afraid of accidentally revealing too much. How did you get b)? Normally the way I solved it, it's almost identical to a) so if you got one, you get the other too unless you make a calculation error. Divisors of 4 are 4,2 and 1. I'll say that.

Good luck anon! I believe in you! We'll get the correct answer very soon!

[Anonymous No.15251258]

>>15250619
commutative already does it for me. But my most beloved property is semisimple (not the gay radical version).

[Anonymous No.15251514]

So like uhhhhhhhhhhhhhhh aren't extension fields and polynomial rings the same thing? At least the elements in such structures are the same (?) and the operations are the same

[Anonymous No.15251618]

>>15251514
There's [math] |2^{2^{\aleph_0}}| [/math] extensions of [math]\mathbb Q[/math] that are subfields of [math]\mathbb R[/math].

[Anonymous No.15251639]

>>15251618
shiet mane this algebra stuff got me FUCKED UP fr fr
I was working with extension fields and everything was fine. Multiplication? Just use an irreducible polynomial. Nice. Suddenly I'm not supposed to be working with extension fields, but polynomial rings. Then someone says quotient rings??????? crazy world out here

[Anonymous No.15251680]

>>15251639
shiet mane

But actually, thinking about it twice, I'm not sure if the set theorists couldn't make "uncountable base polynomials" work somehow, so that you could also frame all those field extensions as this sort of quotient rig.
I'm really not an expert here, wait for more senior answers.

[Anonymous No.15251693]

>>15251005
>>> sorted((e+e*e+e*e*e+e*e*e*e).d.items())
[(2, 1), (3, 1), (4, 1), (5, 1), (6, 2), (7, 1), (8, 1), (9, 1), (10, 2), (11, 1), (12, 3), (13, 1), (14, 2), (15, 2), (16, 1), (17, 1), (18, 3), (20, 3), (21, 2), (22, 2), (24, 4), (25, 1), (26, 2), (27, 1), (28, 3), (30, 6), (33, 2), (34, 2), (35, 2), (36, 6), (39, 2), (40, 4), (42, 6), (44, 3), (45, 3), (49, 1), (50, 3), (51, 2), (52, 3), (54, 4), (55, 2), (56, 4), (60, 12), (63, 3), (65, 2), (66, 6), (68, 3), (70, 6), (75, 3), (77, 2), (78, 6), (81, 1), (84, 12), (85, 2), (88, 4), (90, 12), (91, 2), (98, 3), (99, 3), (100, 6), (102, 6), (104, 4), (105, 6), (110, 6), (117, 3), (119, 2), (121, 1), (125, 1), (126, 12), (130, 6), (132, 12), (135, 4), (136, 4), (140, 12), (143, 2), (147, 3), (150, 12), (153, 3), (154, 6), (156, 12), (165, 6), (169, 1), (170, 6), (175, 3), (182, 6), (187, 2), (189, 4), (195, 6), (196, 6), (198, 12), (204, 12), (210, 24), (220, 12), (221, 2), (225, 6), (231, 6), (234, 12), (238, 6), (242, 3), (245, 3), (250, 4), (255, 6), (260, 12), (273, 6), (275, 3), (286, 6), (289, 1), (294, 12), (297, 4), (306, 12), (308, 12), (315, 12), (325, 3), (330, 24), (338, 3), (340, 12), (343, 1), (350, 12), (351, 4), (357, 6), (363, 3), (364, 12), (374, 6), (375, 4), (385, 6), (390, 24), (425, 3), (429, 6), (441, 6), (442, 6), (455, 6), (459, 4), (462, 24), (476, 12), (484, 6), (490, 12), (495, 12), (507, 3), (510, 24), (525, 12), (539, 3), (546, 24), (550, 12), (561, 6), (572, 12), (578, 3), (585, 12), (595, 6), (605, 3), (625, 1), ..., (34969, 6), (37349, 4), (41327, 12), (48841, 6), (54043, 4), (63869, 4), (83521, 1)]
>>> e
<2: 1, 3: 1, 5: 1, 7: 1, 11: 1, 13: 1, 17: 1>

[Anonymous No.15252319]

Is there a specific term for increasing the number of dimensions?

[Anonymous No.15252341]

>>15235870
What if I just ignore sampling theory and observe every individual in the population instead. Will this strategy scale? I only ask because employer doesn't want to miss anything.

[Anonymous No.15252402]

>>15252341
Sampling is used to predict information about the whole population when you have some limited information about it. If you observe every individual in the population, then you have nothing else to do. You have all the information you need.

[Anonymous No.15252546]

>>15251693
Note that Na[Na,*] is isomorphic, as semirings and semimodules over the naturals, to Na[x[1],x[2],...] and Na[[Na,*]] is isomorphic to Na[[x[1],x[2],...]
here x[i] is identified with e[p] where p is the i'th prime
We can pull the same "arithmetic expression of the unique factorization of naturals" that we see in pic related using this isomorphism
sum[n=1]^oo e[n] = (1 + e[2] + e[4] + e[8] + ...)(1 + e[3] + e[9] + e[27] + ...)(1 + e[5] + e[25] + ...)...
= 1/(1-e[2]) 1/(1-e[3]) 1/(1-e[5]) ...
HOWEVER
The underlying "mathematical machinery" required to establish this is mind-bogglingly trivial when compared to the theory of complex functions: namely we can define the product of arbitrary elements of Na[[Na,*]] simply by appealing to the fact that every solution to X = a * b in natural numbers a,b where X is a natural number has finitely many solutions
The theory of analytic continuation of complex functions is not required

[Anonymous No.15252555]

>>15252341
>Will this strategy scale?
No, because it's impossible to observe every individual in the population, be they people of a country, a petri dish of bacteria, the flow of atoms in a pipe, etc.

[Anonymous No.15252572]

>>15252546
Sorry,
The fact that the algebraic structures of polynomial rings and monoid rings are isomorphic is important
The fact that you can take the field of fractions to get an identity similar to the Riemann zeta function prime factorization is separate

[Anonymous No.15252717]

>>15251085
Anon did you get it? Are you working on it? It'd be amazing if you got it right. I'd praise you so much and thank you too. I think you're very capable of getting the first part correct since you got the second one right. I believe in you! I support you! You can do it! So do it!

[Anonymous No.15252807]

>>15250926
(a) I “solved” by adding.
(b) I solved by subtracting from 720.

This failure to re-use existing work may be responsible for one being right and the other being wrongz

[Anonymous No.15252814]

>>15252319
> Is there a specific term for increasing the number of dimensions?
Like, you have a 2d plane and you want to place it inside a 3d space? I don’t know a general term but for that I would say something like “embed into” if the space already exists and I’m describing the way it gets placed in, or something like “add a vector u normal to V and let U be the new 3-dimensional space generated by V and u” otherwise
>>15252717
I hope that’s a didgeridoo that 2hu is smoking

[Anonymous No.15252873]

Where can I find magazines, papers or journals on mathematics for free?

[Anonymous No.15253002]

>>15252807
I see. Well I'll just say that the answer to a) is 256. If you want to, try solving again see if you get the correct answer. Solving by subtracting? Weird. One easy way to do it just to add the number the of permutations who consists of cycles of length 4,2 or 1. Same for b) where it's just 1,2,3,6.
>>15252814
Anon, it's not a touhou, it's from blue archive which is something I've only recently found about. Also it's a swiss roll. What else could it be anon? What popped into your head? Tell me.
>didgeridoo
Wow that's a big joint!
>>15252873
Ahhh, scihub?

[Anonymous No.15253126]

Didn't do well on my calc test. I have all As right now and I aced the last one so it's an upset for me. I got really angry, it took a lot not to make a scene in class. I hit myself on the head a lot on my way home.

I think the solution is to study the areas I slipped up on more. Those were questions I'd only seen once or twice. There was one question where I understood the mechanics of the question, I had done questions like this a million times, I looked it over several times and found no errors, but I still got a nonsense answer (with a 0 on the bottom of a fraction). Idk, very frustrating. I'll be studying more to make up for it. I could've studied more for this test but, as I said, I felt very confident. I know this material.

My head hurts. I need to stop self-harming. Gonna have spaghetti soon.

[Anonymous No.15253151]

What are some counterexamples in algebra?

[Anonymous No.15254577]

What are the Geometrical implications of a change of variable in an integral?
[math]\int F(x) dx = \int f(t) \frac{dx}{dt}dt[/math]
The [math]\frac{dx}{dt}[/math] factor tells us to change the width of our infinitely small boxes from dt to dx!

[Anonymous No.15254585]

>>15235870
I'm 18 + 8 years and want to learn calculus in an autodidactic manner. Where should I start? I remember nothing about algebra and pre-calculus simply trying from recall but I'm sure I would start recognizing stuff once I started to read over the material again. pls no bulli.

[Anonymous No.15254635]

>>15254585
Openstax

[Anonymous No.15254637]

>>15253126
You're a college freshman? Sophomore?
You need to learn to accept losing marks and getting 50% on an exam. It's not high school anymore, where you get grades for free and easy 100%. (even if you ARE still in high school, this still applies)
As a freshman, I had to take a chemistry lab. At the end of the semester, my professor gave me my grade all giddy and asked "are you happy?", and it was an 85%. I wasn't happy because I was used to getting high grades all the time for 12 years. But my grade was one of the highest in the lab.
I later learned to be happy.

I ate up many C's and C+'s my final year in college, and even got a D+ in my sophomore year. Still got a 3.3/4.0 GPA.
I'm in grad school now and last semester I got like 60% on a final despite doing well all semester. I was happy because the final grade was an A (we do up to A+).

[Anonymous No.15254667]

Bro suppose you got an integral [math]\int f(x) dx[/math] and now you like bro what if we make the substitution [math]x=2t[/math] to get
[math]\int f(2t) 2 dt[/math] now really your moving half as quickly down the number line but multiplying all your little areas by 2... shit wild bro fr fr

[Anonymous No.15254904]

>>15235870
>>15254897 /mg/bros help

[Anonymous No.15254953]

4 color theorem..if each vertice only has two other vertices then there will always be one color that can alternate any point on the plane. Why do we need a computer proof for this ?

[Anonymous No.15254991]

>>15252546
half the point of the zeta function is being able to plug in shit
trivial formal power series abstraction is not all there is to it

[Anonymous No.15254992]

>>15251514
every ring is a quotient of Z[xi: i in I] for some sufficiently large index i. just have a variable per element and quotient all relations.

[Anonymous No.15255046]

>>15253126
Exactly what that other anon said. I've taken tests where scores higher than 50% are considered "good" since hardly anyone achieved it.

[Anonymous No.15255077]

>>15254904
Depends on your goal.

[Anonymous No.15255388]

>spectral basis eigen triangular echelon form transverse mapping seidel Jacobi wronskian Markov leontief Cramer adjoins nullspace subspace spanning rank isomorphic preimage kernel tableau

Filtered

[Anonymous No.15255537]

>>15251514
No.
Take the ring Q[x]. Not a field, so not a field extension of anything.
What's the inverse of x? It isn't there.

[Anonymous No.15255671]

>>15253126
>you thought "if i could just get better at this, i wouldn't hate myself so much"
>this is not the case, in fact the converse is true
>"if i wouldn't hate myself so much, i could just get better as this"
>true skill building can only happen in an atmosphere of profound self-compassion and gentleness
remember that

[Anonymous No.15255754]

Are high school math teachers cringe?

[Anonymous No.15255760]

>>15255754
most don't even know basic math, so yeah

[Anonymous No.15255789]

anyone know if there are pdf files of Paul Taylor's "Practical Foundations of Mathematics" that aren't completely fucked in terms of formatting (check libgen if you want to know what I mean)
I would rather not buy the book completely blind. Also; any opinions on that book in general?

[Anonymous No.15255826]

>>15255754
>Are high school math teachers cringer?

[Anonymous No.15255836]

My TA sucks ass
>Continue the sequence

>[math]A_{1}(1), A_{1}(-1), A_{2}(68)[/math]

[Anonymous No.15255874]

>>15255789
I have a 5.3 MB djvu version, "Paul Taylor-Practical Foundations of Mathematics.djvu"
If the record is straight, then I found it in September 2019 and this suggest it's still online somewhere.

It's a fantastic text. This is assuming you know what it's about. Depending on what you're actually interested in, I might have more suggestions.

[Anonymous No.15255935]

>>15255874
I'll see if I can find it.
And I'll gladly take more suggestions.
Couple things that peeked my interest about the book: A thorough treatment of (co)induction and (co)recursion via (co)algebras; some gentle domain theory and the general constructive flavor of the text

[Anonymous No.15255956]

>>15255935
Okay.
Texts by Bart Jacobs are generally good - if sometimes hard. He released a fat pdf on coalgebras a few years ago.
If type theory interests you, what's really good is Type Theory & Functional Programming by Thomson

https://www.cs.ru.nl/B.Jacobs/PAPERS/index.html

https://www.cs.kent.ac.uk/people/staff/sjt/TTFP/ttfp.pdf

[Anonymous No.15256092]

I am 24 years old, finished a highly prestigious school for maths undergrad and currently work in finance. Is it realistic for me to find a high IQ math wife (female)? She doesn't have to be pretty, just smart and like math. I am highly socially awkward.
Do you think the demand for these girls is too high for me to have any chance of success? Where should I look?

[Anonymous No.15256108]

>>15256092
>I am highly socially awkward.
>Do you think the demand for these girls is too high
The demand for the girls would be fine on its own. It's the demand for highly socially awkward people that's very low.

Imagine you find her, and no one else is interested in her. Would you still have your chance, if you can't hold an interesting conversation? How do you imagine her falling in love, if you're not socially engaging?
You need to address the main issue before anything else.

[Anonymous No.15256189]

>>15255956
I'm not a programmer at all (much less a functional programmer), so I'm not sure if the TTFP book is for me haha
The Jacobs text looks very intriguing though, thanks for the link.
Is the text easier than his "Categorical Logic and Type Theory"? I remember checking out that text out of pure curiousity and realizing during the very first chapter already that my mathematical maturity is not enough to read that book yet.

[Anonymous No.15256194]

>>15256108
So you're saying I'm destined to be an incel for life?

[Anonymous No.15256205]

>>15256189
His Categorical Logic and Type Theory is harder, but that book is also more of encyclopedeic nature, approaching many connections and frameworks, and also covers fields which are sort of out of fashion. The coalgebra stuff is a more particular topic, and recent interest. PS I only read both Taylors book and Jacobs tomb pretty much 10 years ago, myself being not as far as I'm now (and I'm not far anyhow, I'm a physicist) and I didn't finish the second.
I'd push back on the idea that the TTFP is for programmers, it's kinda not. It's more about logic and types. But it's also why I asked what you're really interested in.

[Anonymous No.15256298]

>>15256194
Nothing that crude, I'm just saying it lowers your odds. I'd think of it as probabilities, not absolutes.
It's possible to train against social awkwardness. You might never be the hyper-charismatic businessman who can sell water to a fish, but you can improve your skills with practice, like almost anything else.

[Anonymous No.15257292]

Trying to learn calculus.
Going off from general highschool knowledge,
Should I go straight for calculus?
Is algebra needed before that to actually understand?
Maybe geometry?
Would a solid base of (a thorougly learned) something be necessary?
Thanks for any reply.

[Anonymous No.15257311]

Let [math] f(z) [/math] be a complex entire function with [math] f(\mathbb{C}) = \mathbb{C} [/math] , i.e. [math] f [/math] is surjective onto the whole complex plane.

Then do we have [math] f(\infty) = \infty [/math] ? More precisely: does [math] f [/math] extend to a continuous function from the Riemann sphere to itself, sending complex infinity to itself?

[Anonymous No.15257323]

Does this mean that the maximum distance between any two prime numbers is less than 70 million? Forgive me if I sound dumb but I do not have the mathematical rigor and precision to interpret this on my own, I am an undergrad majoring in not math.

[Anonymous No.15257330]

>>15257292
Basic calculus requires an understanding of elementary algebra, and while it's less essential, learning the basics of trigonometry (sine, cosine, their related functions, the unit circle, etc.) is a good idea.
Geometry outside of trigonometry will not prove particularly useful for the basics, and neither will linear algebra (although the latter is vital if you want to go beyond the basics).
Concept-wise, you'll find it beneficial to have at least a rudimentary understanding of things like sums and limits before you go in.
>>15257323
It's not "there is no gap of this size", but rather "even though prime numbers become more sparse the higher you go, you never reach a point where there are no more gaps smaller than this".

[Anonymous No.15257393]

>>15257323
>Does this mean that the maximum distance between any two prime numbers is less than 70 million?
A simple counterexample: consider the family of sequences [math]F_n = \left( n! + 2, n! + 3, \dots, n! + n \right)[/math]
For [math]n \geq 2[/math], the sequence [math]F_n[/math] comprises a sequence of n-1 consecutive composite numbers, so in particular [math]F_{7 \cdot 10^7 + 1}[/math], [math]F_{7 \cdot 10^7 + 2}[/math], [math]F_{7 \cdot 10^7 + 3}[/math], ... is an infinite sequence of gaps exceeding 70 million.

[Anonymous No.15257682]

>monitor this thread loosely for months
>not a single other plooter exists
I walk a lonely road
The only one that I have ever known
Don't know where it goes
But it's home to me, and I walk alone
I walk this empty street
On the Boulevard of Broken Dreams
Where the city sleeps
And I'm the only one, and I walk alone

[Anonymous No.15257767]

>>15255754
I don't think so! Mine were all very lovely people who were at least somewhat enthusiastic about math and teaching it. Classes are usually filled with kids who don't actually want to be there however so they have to teach to an audience who does not care, which I assume must be tiring and disheartening. I think their math knowledge varies widely though. This can't be the case for all, but some I had were genuinely not all that knowledgeable about the field they're supposed to be teaching. As an example, one of them insisted to me that 22/7 was irrational because it was pi. I don't remember how the conversation ended.
>>15256194
He said you needed to address the main issue. To you that means you're destined to be an incel?
>>15257682
Anon I don't know what a plooter means but maybe they're just silently watching this thread too.

[Anonymous No.15257882]

Is there a good application of homology to graph theory?

[Anonymous No.15257988]

>>15257882
same stuff, Anon

[Anonymous No.15258005]

>>15256194
oh, well, don't do that
they'll chop you up and turn you into incellation
cheap way to make incellation that way
don't be an incel
Onions incellation is made of incels

[Anonymous No.15258016]

>>15255754
some are good. most are very retarded and bad. take your kids to a russian school of math program

[Anonymous No.15258119]

>>15257882
Cycle space
https://en.wikipedia.org/wiki/Cycle_space?useskin=vector#Topology

[Anonymous No.15258218]

how was this transition done?

[Anonymous No.15258227]

>>15258218
pic wasnt attached

[Anonymous No.15258251]

why dont you try it and find out

[Anonymous No.15258335]

What are gradients
[eqn]
1=x^2+y^2,
f_x=2x,
f_y=2y
[/eqn]
the gradient is supposedly orthogonal to the curve? Why?

[Anonymous No.15258346]

>>15258335
is like derivative, but fancy
meaning direction where things move most
when you descend them, AGI at bottom sometimes

[Anonymous No.15258387]

>>15258227
See the exponentials in the parentheses? They're of the form (x-y).
Multiply the top and bottom of the fraction by (x+y), then simplify. Convert each exponential to the usual complex number form, a+bi, then simplify.

[Anonymous No.15258414]

>>15238218
*yawn* it's just the average.
Connect the dots on the values.
If you start with the negative value you get the opposite
why is this a big deal again my niggers?

[Anonymous No.15258482]

>>15238218
>>15238256
Retards.
Convergence in R is not the only way to define infinite sums.

Math is a free country; it's free from retards.

[Anonymous No.15258635]

>>15258414
[math] \sum_i x_i = \frac 1n \sum_i x_i [/math] apparently.

[Anonymous No.15258782]

>>15258387
thanks. i solved it differently, but was looking for more ways because it took me way too long. your way is better and what i was looking for, thanks.

[Anonymous No.15258873]

>>15258782
No problem.

[Anonymous No.15259067]

>>15257767
>I don't know what a plooter means but maybe they're just silently watching this thread too
its when you try to work on something but end up just graphing or plotting functions in a sort of artistic way in hopes of gaining the attention of someone who can help you focus. like a portfolio or body of work that doesnt mean much in and of itself

[Anonymous No.15259107]

Conic algebra is a bridge from group theory, ring theory, field theory, linear
algebra, module theory, and so on to computation.

A *conical* monoid M is one such that the multiplication satisfies the *conic*
property,
> a a' = 1 implies a = a' = 1 for all a,a' <- M,
and a *conic section* of a group is a conical submonoid.

A *conical* semiring S is one such that the abelian monoid (S,+) is conical,
and a *conic section* of a ring is a conical subsemiring.

Examples of conical semirings: Na[x], Na[x,y], Na[[x]].

A *conical* semimodule T over a semiring S is one such that the abelian monoid
(T,+) is conical, and a *conic section* of a module M over a ring R is a pair
(S,U) such that S is a subsemiring of R and U is a conical subsemimodule of M,
where M is considered as a semimodule over S.

A *conical* semifield J is one such that the abelian monoid (J,+) is conical,
and a *conic section* of a field is a conical subsemifield.

A *conical* semivector space L over a semifield J is one such that the abelian
monoid (L,+) is conical, and a *conic section* of a vector space V over a field
K is a pair (J,L) such that J is a subsemifield of K and L is a conical
subsemivector space of V, where V is considered as a semivector space over J.

[Anonymous No.15259141]

Graphic artist,. Who needs help explaining why you can't just go from RGB to CMYK,. Can someone help me with the numbers?

RGB swatches are determined by 3 scales,. Red, Green and Blue,. Each of which can be set from 0 to 255 in any combination.

How do I math the exact number of possible RGB combinations?

Likewise CMYK has 4 scales Cyan Magenta, Yellow and Black,. Each of which only goes 0 up to 100 in any combination.
What's the math for the exact number of CMYK values?

[Anonymous No.15259161]

>>15259141
https://www.rapidtables.com/convert/color/rgb-to-cmyk.html

[Anonymous No.15259166]

>>15259141
>Graphic artist,. Who needs help explaining why you can't just go from RGB to CMYK,. Can someone help me with the numbers?
>
>RGB swatches are determined by 3 scales,. Red, Green and Blue,. Each of which can be set from 0 to 255 in any combination.
>
>How do I math the exact number of possible RGB combinations?
>
>Likewise CMYK has 4 scales Cyan Magenta, Yellow and Black,. Each of which only goes 0 up to 100 in any combination.
>What's the math for the exact number of CMYK values?
>>15259161
Hooo boy are your clients going to be upset when the nice vibrant purple on their screen prints out like sludge.

[Anonymous No.15259171]

>>15258335
f(x, y) = x^2 + y^2 - 1

df = 2x dx + 2y dy

In polar coordinates x=r·cos(p), y=r·sin(p), this is

f(r, p) = r^2 - 1

df = 2r dr

i.e. the functions has equal value at equal distances r. The gradient is radial.

(You actually wrote down the curve f(x,y)=0, not a function to take the derivative of.)

[Anonymous No.15259174]

>>15259166
does kludge sound like sludge

[Anonymous No.15259208]

>>15259174
kloooge

[Anonymous No.15259352]

>>15259208
does grudge sound like fudge

[Anonymous No.15259394]

>>15253151
Good question, some stuff I found
https://ncatlab.org/nlab/show/counterexamples+in+algebra
https://mathoverflow.net/questions/29006/counterexamples-in-algebra
https://www.mathcounterexamples.net/category/algebra/
Not enough to fill a whole book, I suppose. Such is the nature of the subject

[Anonymous No.15259683]

>>15259107
conic algebra -> convex algebra
convex monoid
convex binary operator
convex section

[Anonymous No.15259762]

Why did Apostol decide to take such a convoluted approach to defining the Integral in his Calculus book? He starts by talking about the greek method of exhaustion. Introduces Area and Measurable functions as axioms (lol) then has to go through great pains to first define step functions (he never mentions the phrase Darboux or Riemann) and then just assumes that all sorts of step functions for the upper and lower integral can be found (muh exhaustion) and defines the integral by the Archimedean property between the upper and lower integrals (seems a bit circular, but whatever).

Now he has to generalize the integral which is just pretty much a repeat of his step function section, but he has to spend a ton of time showing how to calculate polynomials. The part where he has to integrate the Sine function is even more amusing, because he has to dump a ton of identities that come out of nowhere, but based on about 4 properties Sin/Cosine have, and at this point he still appeals to the geometric definition of Sine (with a promise of the analytic to come). This takes him ages... but eventually he can integrate Trig.

Why does he do this? Why not start with convergence, continuity, power series (define trig functions here analytically), differentiation and primitives, and then integrals? Oh look, now you can define area formally too at this point. He has to cover all of these things anyways, so why make it more difficult for himself? What's the appeal? Just to get engineers integrating early? Is Apostol good, or considered a meme? I already know calculus, but going through the first few chapters reminded me of the install gentoo meme. Is it a troll?

[Anonymous No.15260151]

Hello.
In Cantor's theorem that [math]|A| < |\mathcal{P}(A)|[/math], most arguments for the "strict part" of the inequality show that there is no surjection to the power set by the usual diagonal argument. But what we really want is to show that there is no injection out of the power set.
Now, classically this is no problem as [math](\star)[/math] whenever there is an injection [math]X\rightarrowtail Y[/math] and [math]X[/math] is inhabited then there is a surjection [math]Y\twoheadrightarrow X[/math].
My questions:
1. Does the classical theorem [math](\star)[/math] fail without excluded middle (which is used in the classical proof to construct the surjection)? Is the statement maybe even equivalent to EM?
2. What if we restrict [math](\star)[/math] to only talk about the case we care about here, namely where [math]X=\mathcal{P}(A),Y=A[/math]. That is, can we constructively show that if there is an injection [math]\mathcal{P}(A)\rightarrowtail A[/math] then there is a surjection [math]A\twoheadrightarrow \mathcal{P}(A)[/math].
3. Forgetting about surjections altogether; is there a (constructive) proof that directly shows that there is no injection out of the power set (via some kind of diagonal argument)?

[Anonymous No.15260177]

>>15257330
Thanks anon.
I'm the first anon you replied to.

[Anonymous No.15260488]

new

>>15260487