/mg/ math general: Collatz Edition

Cannot be formed in ZFC

No, I mean what IS it? Like what are its properties? What is its cardinality?

What? How could that be? What about The set of all axioms that are consistent with ZFC

about 8 or 9

>What is the set of all axioms?
It's a category, not a set.

No, those are the axioms of ZFC. I mean all the axiom. And I think that what I am trying to get here is:

What are the properties that an axiom should suffice? What should be in the set of all axioms?

What the fuck is a category man. Tell me I need to know. And tell me about the category of all axioms

I highly doubt that the axiom scheme of comprehesion would allow it, as I cannot think of a set that the set of all axioms could reside in.

What's the mathematical equivalence of forgiveness? Allow for 'time to process'?

But that doesn't make sense. We can talk about natural numbers because there is a set of natural numbers. Where do axioms come from? How can I say "Let A be an axiom". Where does that "A" come from? Where are axioms?

>What the fuck is a category man.
en.wikipedia.org/wiki/Category_(mathematics)

I'm not a "man".

The naturals can be formed in ZFC due to the empty set axiom(look at von neumann construction of naturals). To ask what is an axiom is much more of a philosophical question than a mathematical question.

Are you from the set of all girls (male)?
Also that wikipedia article says nothing about axioms! It is all just a bunch of arrows I am talking about real math not little kid games. Where do axioms come from?

>To ask what is an axiom is much more of a philosophical question than a mathematical question.
That is wrong because in philosophy you can't know nothing.

If Godel proved a theorem that holds with all axioms then that means he at some point had to say "Let A be an axiom". When he uttered those words. where did A come from? If axioms do not belong in a set then how could someone prove something about all axioms?

>14,88

You are referring to Incompleteness Theorem, which is where Godel bitched about 'symbol shunting' instead of actual mathematics.

What does that MEAN?

Are you from the set of all Hands(Left)?

Are you from the set of all Hands(Right)?

Or are you Hand-ed(Left|Right)?

-ed = suffix

No I am not a hand but I am left-handed

11
0!1
0|1
Left|Right
Female|Male
Color|Blue
Hot|Cold
Up|Down

Time series econometrics.

There are like zero online resources for this shit and my textbook isn't helpful at all. Muddling through though

>Is there an upper limit on how long you can keep making moves without either violating the rule against repeating steps or arriving at the base of the mountain?
yes

>When he uttered those words, where did A come from?
A does not exist. It's a made up object which embodies the 'ideal form' of an axiom. For example, when we say something like "Let x be a number", we are not consulting the set of all numbers, and drawing out one of our choice. We are simply considering the properties of the 'ideal form' of a number, which all numbers must be faithful to; indeed, the set of all numbers is exactly those objects which are in alignment with this ideal form.

But if there is an ideal form all axioms take then why can't we just make S the set of all the objects that have those properties?

define "ideal form"

Anti-semitic numbers are illegal user.

What's the formal definition of infinity in abstract algebra/analysis?

There is no formal definition. We call something "infinite" when it has more parts than any counting number, goes on in an endless succession of steps, is a point in an order after all counting numbers, or similar.

>What's the formal definition of infinity in abstract algebra/analysis?
Depends on the context
en.wikipedia.org/wiki/Extended_real_number_line
en.wikipedia.org/wiki/Riemann_sphere

Nu professor said that a set is infinite if every subset of it is a bijection with the entire set.

no. if SOME subset of it is a bijection with the entire set. the empty set will not be in bijection with anything else and it's a subset of every set.

[math]Show \space me \space some \space cool \space stuff \space you \space can \space do \space with \space LATEX[/math]

[math](^\frown \smile ^\frown)[/math]

[math] \space [/math] ▲
▲ ▲

-a = -1 * a
(-a)*1 = -1*a

L:((-a)*1)/1 = -a
R:((-1)*a)/1 = -a

If I'm understanding that correctly, you've essentially got an easy language you can build for this.

It goes:
for n = 1 => infinity: up*(n + 1), down*n

I don't think that solution works, unfortunately, because the next time you repeat the 'down' step, the former down steps are a substring of it.

you never studied the collatz conjecture? lol

at height 2^n, go up 2^(n-1), down 2^(n-1), up 2^n
dunno if i understood the question 2bh

Should clarify you can only move on step up or down per time-point. The problem with your solution is that at height 2^(n-1) you went down 2^(n-1) steps. So the next time you go up and down, you will be repeating those 2^(n-1) downward steps.

you go down 2^(n-1) steps from height (2^n + 2^(n-1)) = 3*(2^(n-1)) down to height 2^n

if some PROPER subset of it is in bijection with the entire set. [math]A\subset A[/math] always

>infinite sets
No such thing.

>being a zenophobe

>a set is infinite if
>if
Yes, "if".

what is 4|n denoting here?

[eqn]
\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}^{\sum\limits_{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}^{\sum\limits_{\sum\limits_{}^{}}^{\sum\limits_{}^{}}}}}}
[/eqn]

4 divides n

I know that this probability function is a multivariate normal distribution in d-space. How do i next proceed in deriving the classifier? Do I have to get an expression from the sum over C1,2 or ?

i like real analysis

[math]{\color{red}{\text{(USER WAS BANNED FOR THIS POST)}}}[/math]

okay here's a problem I couldn't solve for the last hour

Given a1 + a2 + a3 ... + an = k
Where each a and k are positive integers
Describe lcm(a1, a2, a3 ... , an) in terms of k

Given that n people died in a forest, describe their gender in terms of their jobs.

Well the original question is just finding the maximum order of S12
I'm autistically overcomplicating things here because I don't like proof by exhaustion

>Describe lcm(a1, a2, a3 ... , an) in terms of k
Be more specific, it's not clear at all what kind of answer you're expecting

Is it 35?

TURN ON CNN

Is Goldbach Conjecture true?
arxiv.org/pdf/1710.04081.pdf

>We answer the question positively. In fact, we believe to have proved that every even integer 2N≥3×10^6 is the sum of two odd distinct primes. Numerical calculations extend this result for 2N in the range 8−3×10^6. So, a fortiori, it is shown that every even integer 2N>2 is the sum of two primes (Goldbach conjecture). Of course, we would be grateful for comments and objections.

Ugh my brain hurts and I can't write for shit
I meant find the max value of lcm(a1, a2...) for a specific k
No definitely not. disjoint cycles of order 3,4,5 form a order 60 permutation group in S12

The maximum order of S_n, which is
max_{1 =< a1, ..., a_m integers, a1+...+a_m = n} lcm(a1, ..., a_m).

I think there's no nice formula for this, but there's probably asymptotic formulas and bounds for this. Actually I think that Landau did find and prove some asymptotic formula.

>The maximum order of S_n
There's only one order for a group, what is "maximum" supposed to mean?

I've meant order of elements of S_n, sorry.

Then you have the upper bound of n!.

And n is a lower bound. Great, we have something.

>we have something.
Speak for yourself.

What the fuck it actually exists

Yeah, Landau, I knew it.

This doesn't seem to help for n=12 though, the exponential bound gives g(12) =< 314.

I'm studying parsing theory, and found this definition:

A sequence of nodes [math](a_0, a_1, \dots , a_n), \mbox{ } n \geq 0[/math], is a path of length [math]n[/math] from [math]a_0[/math] to [math]a_n[/math] in a graph [math]G[/math] if for all [math]i = 0, \dots, n - 1 \mbox{, } (a_i, a_{i + 1})[/math] is an edge of [math]G[/math].

I would like you to help me: what about when [math]n = 0[/math]? But before, a bit of context: the book I'm reading defines a graph [math]G=(A,R)[/math] where [math]A[/math] is the set of nodes and [math]R[/math] is a relation on [math]A[/math]; the elements of [math]R[/math] are called the edges of [math]G[/math].

If [math]n = 0[/math] then, a path of length zero from [math]a_0[/math] to itself is a sequence whose only element is [math]a_0[/math], then, for all [math]i = 0, -1[/math], [math](a_i, a_{i + 1})[/math] is an edge of [math]G[/math]. But in this particular case, the two posible edges are [math](a_0, a_{-1})[/math] and [math](a_{-1}, a_0)[/math], but none of these is an edge (or an element of [math]R[/math]); actually, [math]a_{-1}[/math] not even exists in our given sequence. Is it possible to have a zero length path given the above definition? Is it vacuously true? why?

a zero length path is just a point, no edges involved

Yes, but why? I mean, that could be deduced from that definition.

>Yes, but why?
because you have a node and there's no edges that don't fit the edge criteria

>n^2

0 is odd?

>Odd numbers larger than 2 are prime;
>3 is prime, 5 is prime, 7 is prime, 9 is left as exercise, 11 is prime, 13 is prime...

How to proof with set identities?
(A − B) − C = A − (B ∪ C)

show elements are the same

>Hartshorne

Threadly reminder to work with physicists.

[math]x \in (A\setminus B)\setminus C \Leftrightarrow (x \in A) \land (x \not\in B) \land (x \not\in C) \Leftrightarrow (x\in A)\land (x \not\in B\cup C) \Leftrightarrow x\in A\setminus (B\cup C)[/math].

Is this correct?

probably
I thought it was \ instead of -

>tfw I'm too dumb

mine is wrong. I'm confused between the set builder notation and set identities

fucking hate snakes

How to do the inner summation? (the one with 4) I will figure out the rest.

that summation doesn't make sense.
are you sure you didn't exchange the two sums?

$sum_{i=1}^j 20-8i$

...

>doesn't know about the linearity of summation

[math]\sum\limits_{a} \sum\limits_{b} f(a, b) = \sum\limits_{b} \sum\limits_{a} f(a, b)[/math]

1) I don't think you know what linearity means
2) That's not true when the b's depend on the a's.
3) It does not make sense when the a's depend on the b's

p. sure that won't work for some sequences if a or b tends to infinity

[eqn]
\sum_{i=1}^{j} \sum_{j=1}^{4} 2(j-i)=\sum_{i=1}^{j} 20-8i
[/eqn]
[eqn]
\sum_{j=1}^{4} \sum_{i=1}^{j} 2(j-i)=\sum_{j=1}^{4}2\left (j\left [\sum_{i=1}^{j}1 \right ]-\sum_{i=1}^{j}i = \right ) =\sum_{j=1}^{4}2\left (j^2-\frac{j(j+1)}{2} \right ) = \sum_{j=1}^4 j^2-j = 20
[/eqn]