Mathematics - 2018-11-11 (page 2 of 3)

Rithaniel

05:16

Well, they're essentially the same thing.

Secret

@Nick How to enchant an object: $$\tan ^{-1} \theta$$

Looks like mathematicians are magicians after all

Still a more interesting question can be formulated: Does zorn's lemma apply in the partial order of absurdities. guarentee that there exists a maximal absurdity given any proposition

For our purpose, we can define an absurdity to be a rewriting of an expression under a rewriting system that can be obtained if some of its rules were relaxed (e.g. $\sin$ becomes $s\cdot i \cdot n$). The degree of absurdity is then given by:

$$|\text{absurd}| = S$$

where $S$ is a set containing the rules to be relaxed

Semiclassical

@KemonoChen the former, since it's C at zero

Secret

We knew that human stupidity has no known limits, this suggest there exists propositions $P$ such that $|absurdity|$ has no upper bound hence violating Zorn's Lemma

Akiva Weinberger

@KemonoChen Not much of a reason to prefer one to the other but probably $\tan^2\theta$ 'cause, in my mind $\cos$, $\sin$ and $\tan$ come before $\sec$, $\csc$ and $\cot$

Kemono Chen

Thanks to all of you.

Akiva Weinberger

05:30

That said, if $\theta=\operatorname{arcsec} u$ or some such, probably the second one so it would simplify better

Kaustabha Ray

06:16

What is the number of solutions of the equations $x_1 + x_2 + ... + x_k = n$ where each $x_i$ is an integer \ge 2?

When we are concerned with the number of solutions, we can find it by using the Combination Formula allowing repition $\binom{n+k-1}{n}$. How to account for the fact each integer is $\ge 2$?

Kaustabha Ray

06:30

Would it work by adding up the $k$ 2's and then finding a solution to the equation summing up to $n-2k$?

Akiva Weinberger

07:28

@KaustabhaRay Sounds like a plan

Thoughts on Goldbach: If we could find a sequence of objects $a_n$ (not necessarily numbers) such that there existed two functions $f$ and $g$ where $f(a_n)$ is a decreasing sequence of nonnegative integers and $g(a_n)$ is a Goldbach sequence, then we'd know that the Goldbach sequence terminates

Maybe $a_n\in\Bbb Z^k$ would be a good choice for some $k$

and $a_n$ defined recursively

Secret

What we can get from the sum of two primes other than if one of the primes is 2, then the sum must be odd?

Akiva Weinberger

If neither is 2, then the sum must be even

Oh crap

I meant Collatz

Secret

Speaking about Collatz conjecture, I had an MSE inspired from it. And one major issue is that there seemed to be no easy way to predict how many times one need to divide by 2 depending on what number you started with:

1

A: A strange seemly pattern from some even numbers. Is there a known theorem that can explain this pattern?

You are looking at numbers of the form $2n$. Looking at your enumeration, the most obvious thing is that every second number is divisible by $2$ only once. That this happens is clear, as these are the numbers where $n$ from above is odd. The second pattern you note is that every $3+4k$th term i...

Written in terms of product of prime powers, there seemed to be no pattern on how to predict which numbers are products of exactly two prime powers

A rough thought I had about that is, if on average you divide by 2 not enough times, then as the recursion continues, the sequence can grow without bound

Other things I noticed include: Since n has to be odd in order to be mapped by 3n+1, the result of 3n+1 must be even. Thus the open subquestion is whether 3n+1 will always result in an even number that can be divided by 2 enough times such that it basically cancels out the magnification produced by 3n, and thus ensure the sequence will converge

Trying to map out the orbits of odd numbers n under 3n+1 proved to be way too messy to follow however

Another obstacle is that the map $x+1$ for any integer $x$ has an unpredictable effect on its prime factorisation

196

Q: What is the importance of the Collatz conjecture?

I have been fascinated by the Collatz problem since I first heard about it in high school. Take any natural number $n$. If $n$ is even, divide it by $2$ to get $n / 2$, if $n$ is odd multiply it by $3$ and add $1$ to obtain $3n + 1$. Repeat the process indefinitely. The conjecture is that no ...

o wow, so the obstacle I mentioned is equivalent to the collatz conjecture (and its generalisation)

hmm... is there a reason that we expect the pattern of prime factorisation to be computable at all?

(not that individual prime factorisation of a given number is computable, that is possible if hard as guarenteed by the fundemental theorem of arithmetic, but whether there is some overarching formula that govern which prime powers appear for which integer)

Liad

07:54

@LeakyNun hi

Akiva Weinberger

Also, if you divide by 2 by accident when it's odd, it will never be an integer again no matter what you do

Liad

@AkivaWeinberger hi!

Akiva Weinberger

@Liad Hi

Secret

But how can you divide an odd number by 2 by accident, the collatz map is defined so that for any odd integer, it is mapped with 3n+1?

Unknown MathMan

I am not getting why the properties those are need to be satisfied by isomorphism preserves group structures. Can anyone give an intuitive explanation behind defining an isomorphism?

Secret

08:08

to preserve group structures, the identity must map to the identity, inverses must map to inverses and product of two terms must map to the corresponding product in the other set

Liad

@AlessandroCodenotti hi!

Alessandro Codenotti

08:29

Hi

Icche Guri

Hello , I am looking for solution of this question :

0

Q: How many many steps needed for n*n matrix multiplication?

I have got a strange question in previous year question and that is , if an algorithm needs 21 steps for a 7*7 matrix multiplication then how many steps would it need for n*n matrix multiplication ? I have tried to do 7*7 matrix multiplication and calculated how many multiplications done . Then...

algorithm matrix multiplication calculation problem-steps-recorder

Can you please help me to solve this question ?

Liad

im trying to show that $Y= \{ (t,t^2,\dots,t^n) : t\in K \}$ is a closed subset. so an element of $Y$ is of the form $\sum_{i=1}^n \alpha_i t^i$ where $\alpha_i \in K$. how can i find a polynomial s.t $Y$ is the of zeros of this polynomial? (i thought maybe showing $I(Y)$ is radical, but it didn't work as well.. ) @AlessandroCodenotti

Alessandro Codenotti

Where do the $\alpha_i$ come from?

Liad

K

Alessandro Codenotti

No I mean why are they here?

Liad

08:37

if $y\in Y$

then it is in some ideal of the form $(t,t^2,\dots ,t^n)$

Alessandro Codenotti

$Y\subseteq K^n$, right? It is not an ideal

Liad

yes

each closed subset is intersection of zeros of polynomials, and i just cant find polynomials that Y would be their zeros

Alessandro Codenotti

Start with $\{(t,t^2)\}\subseteq K^2$

Liad

ok

Leaky Nun

@Liad $(t, t^2, \cdots, t^n)$ is not an ideal

it's the element of $K^n$ whose $i$-th coordinate is $t^i$

Liad

08:42

isn't it the ideal generated by those n-elements?

Leaky Nun

I just told you.

Liad

but why do you think that?

Leaky Nun

because if $Y$ is a set of ideals then it doesn't make sense to say "closed subset"

and also $(t, t^2, \cdots, t^n) = (t)$ as ideals so it would be redundant

Liad

Ah. ok.

ok so for $n=2$ we can take $f(x,y) = x^2 -y$.

Leaky Nun

bingo

Alessandro Codenotti

08:48

Good, do $n=3$ now

Also for a generic $n$ how many polynomials do you expect to need?

Liad

$f(x,y,z) = x^3-z^3 + x^2 - y$.

Leaky Nun

nope.

Alessandro Codenotti

Decide how many polynomials are needed first

Adam

so why is this one getting downvoted? math.stackexchange.com/questions/2988963/…

Liad

hmm ok

maybe $(x^3-z)^2 + (x^2-y)^2$ ?

Alessandro Codenotti

08:58

I'm not going to say it a fourth time, decide how many polynomials you need, then find them

Liad

im trying to see how to find it

maybe $n!$ ?@AlessandroCodenotti

Alessandro Codenotti

Intuitively inside $K^n$ what's the dimension of the zero set of a single polynomial?

Liad

hm, n-1 ?

Alessandro Codenotti

Yes, so if you want something of dimension $n-2$ how many polynomials will be needed?

(you can think about linear algebra for intuition, a single linear equation defines an hyperplane, if you want a codimension 2 subspace you intersect two hyperplanes)

Liad

@AlessandroCodenotti why n-2?

Leaky Nun

09:09

because you have a map $\Bbb A^1 \to Y$...

and also $Y \to \Bbb A^1$...

Alessandro Codenotti

What does the dimension of $\{(t,t^2,t^3\}$ look like? Is it a curve or a surface?

Leaky Nun

they're isomorphic...

Mary Star

Hello!!

Let $k$ be a field of characteristic $p$ and let $K=k(x,y)$ be the field of rational functions in two variables and let $F=k(x^p, y^p)$.
How could we show that the extension $K/F$ is purely inseparable?

Leaky Nun

@MaryStar apply the definition.

(and remember freshman's dream)

Liad

@AlessandroCodenotti curve.

Mary Star

09:11

@LeakyNun Do we have to find the minimal polynomial for an arbitrary element of K ?

Alessandro Codenotti

So dimension $n-2$ in the $n=3$ case. If a polynomial gives something of dimension $n-1$ how many do you need to get something of dimension $n-2$?

mercio

are you talking about $n=3$ only or about a generic $n$

Leaky Nun

@MaryStar tell me the definition.

Daminark

Henlo

Leaky Nun

@mercio 3 first, because he gets stuck

hi @Daminark

Liad

09:18

@AlessandroCodenotti you getting at something like $f(x,y,z) = x^3-z$ and $g(x,y,z) = x^2-y$ and then take the intersection of their zeros set?

@Daminark hi!

Mary Star

@LeakyNun The extension K/F is purely inseparable if for every element in K\F, the minimal polynomial over F is not a separable polynomial, i.e. it has not distinct roots, or not?

Leaky Nun

hmm

and equivalently?

Daminark

How's everything going?

Alessandro Codenotti

@Liad good, so now you should also know how to do a generic $n$

Liad

yes.

Alessandro Codenotti

09:21

Also note that all that stuff about dimension was very intuitive but can be formalised in a way that agrees with intuition. Doing so it's not straightforward though

Mary Star

@LeakyNun Could you give me a hint?

Liad

@AlessandroCodenotti cool.

mercio

@Liad what about $h(x,y,z) = y^3-z^2$ ?

Liad

what about it?

Leaky Nun

@MaryStar look up equivalent definitions online

(and prove that they're equivalent)

Alessandro Codenotti

09:25

@Liad Its zero set also contains $(t,t^2,t^3)$

mercio

is it in $I(Y)$ and is it in the ideal generated by $f$ and $g$

Liad

ok but why we are looking at this polynomial?

Alessandro Codenotti

Let $A=\{(t,t^2,t^3)\}$. What Mercio is saying is that you have $A\subseteq V(f,g)$ but do you have the other inclusion? Why is $h$ (or other polynomials) not needed?

Liad

$V(f,g) = Z(f) \cap Z(g)$ ?

Alessandro Codenotti

Yes

Liad

09:30

so we showed $A = V(f,g)$

mercio

I was rather asking about $I(A)$

Liad

ok let me see what is that.

by def. its all the polynomials that vanishes on A.

mercio

yes, so is $h(x,y,z) = y^3-z^2$ in $I(A)$ ?

Liad

yes

mercio

(though if you only want to show that $A$ is closed you don't need to find exactly $I(A)$)

Liad

09:33

i already showed A is closed. but we can find I(A) also .

mercio

and is $h$ in the ideal generated by $f$ and $g$ ?

Liad

yes i think so

mercio

oh right

I might have done something wrong lol

I have to recheck

Liad

are we trying to find generators of $I(A)$ ?

mercio

I want to be sure that $I(A) = (f,g)$ yes

Mary Star

09:37

In wiki, there are the following equivalent definitions:
1) K is purely inseparable over F
2) For each element $ \alpha \in K$, there exists $n\geq 0$ such that $\alpha ^{{p^{n}}}\in F$.
3) Each element of K has minimal polynomial over F of the form $X^{{p^{n}}}-a$ for some integer $n\geq 0$ and some element $a\in F$.

Do we use here the second one? @LeakyNun

mercio

yeah $h$ is in $I(A)$

sorry

Leaky Nun

@MaryStar right

after you proved that they are equivalent

Liad

that's fine. so you want to say that $I(A) = (h)$ ?

mercio

oh no $I(A)$ can't be $(h)$

Liad

so we are aiming at $I(A) = (f,g)$ ?

mercio

09:39

yes

Liad

just trying to see where you going ^^

ok

mercio

(I was mistakenly thinking $h$ was missing)

Mary Star

Ok! So, we have that $x,y\in K$ and $x^p, y^p\in F$. Each $\alpha\in K$ is of the form $\alpha=\sum a_{ij} x^iy^j$ or not? @LeakyNun

Liad

@mercio ok so we have $(f,g) \subset I(A)$ why the other inclusion also works?

mercio

well maybe you can show that $I(A)$ is generated by all of the $x^ay^bz^c - x^{a+2b+3c}$ and then check that those are all in $(f,g)$

Leaky Nun

09:57

@MaryStar sure

Mary Star

10:08

Then we have that $\alpha^{p^n}=\left (\sum_{i,j} a_{ij} x^iy^j\right )^{p^n}=\left [\left (\sum_{i,j} a_{ij} x^iy^j\right )^{p}\right ]^{p^{n-1}}=\left [\sum_{i,j} a_{ij}^p x^{ip}y^{jp}\right ]^{p^{n-1}}=\ldots =\sum_{i,j} a_{ij}^{p^n} x^{ip^n}y^{jp^n}\in F$, right? @LeakyNun

Leaky Nun

looks like you're confusing between forall and there exists

Mary Star

Ah do we maybe consider $n=1$, i.e. :
$\alpha^{p}=\left (\sum_{i,j} a_{ij} x^iy^j\right )^{p}=\sum_{i,j} a_{ij}^p x^{ip}y^{jp}\in F$ ? @LeakyNun

Leaky Nun

correct

Mary Star

Ok!! Thanks a lot!! :-) @LeakyNun

Gandalf

10:23

Hello

I have three people who contributed 50,100,21 usd

Now i have 300 usd that i want to divide between the three people according to their contribution

I want a single formula i can use to divide the 300

mercio

300/3

Gandalf

Divide according to their contribution

My method is first to get the % their contribution represents then work out their share from there

But it would be nice to have a single formula

mercio

that sounds good

how do you get their % contribution ?

Gandalf

50/300 * 100

for instance

mercio

that doesn't sound right at all

Gandalf

10:30

why

mercio

so according to you the % represented by the first person's contribution is 50/300 * 100 ?

so that's 16.6666 % ?

Gandalf

Yes

mercio

then for the second person it is 100/300 * 100 ?

Gandalf

yes

mercio

66.6666... % ?

Gandalf

10:32

haha

wrong formula

that wont get the actual %

mercio

and then the last one would have been 7%

Gandalf

yeah, that will not add up

mercio

yeah

Gandalf

In my head i though the percentages would add up

mercio

you computed them incorrectly

for example why did you involve 300 ?

you are looking at the percentages from the original 50, 100, 21 contribution

Gandalf

10:37

I am calculating some sort of dividend

We have 300 usd as profit

mercio

you should give them all to me

Gandalf

My investors contributed 50,100 and 21 usd each

@mercio haha, not likely

the sec will be on my neck

mercio

shouldn't you be able to compute the respective percentages just by looking at 50, 100 and 21 ?

Gandalf

Now i must divide the profit according to the contribution

@mercio 50 gives 16%

mercio

how did you come up with that ?

what if you have to give out a billion dollar

Gandalf

10:41

The % method is not working out correctly

mercio

is the first guy's share suddenly going to plummet to 0.000005% in that scenario ?

Gandalf

@mercio Yeah it shall

I wonder if there is a formula for calculating dividends in such a case

mercio

yes there is

it involves understanding what "%age of their contributions" means and giving the right formula for it

Gandalf

Its happening in 60 seconds across the board

mercio

do you agree that the %age of their contribution is the same in the 300 case and in the 1000000000 case ?

Gandalf

10:43

yes

mercio

so why are you using 300 or 1000000000 in your formula for computing it

Gandalf

i will never get to a billion

mercio

I am trying to explain how your formula of % = initial contribution / amount to pay out * 100 makes no sense

Gandalf

I agree my formula makes no sense

mercio

so what should you be dividing with ?

so that it sums up to 100 ?

Gandalf

10:46

No idea

mercio

it would be like solving 50/x * 100 + 100/x * 100 + 21/x * 100 = 100

Gandalf

gives 56% at best

mercio

uh what ?

Gandalf

in that formula, it adds up to 56%

if x = 300

mercio

yeah because x = 300 is not the solution

Gandalf

10:50

i dont get that

what would be x

mercio

it would be the number you have to replace 300 with in order to get the correct formula

from your original incorrect formula

Gandalf

300 usd is all i have

Akiva Weinberger

300*50/(50+100+21)

300*100/(50+100+21)

300*21/(50+100+21)

$87.75, $175.50, and $36.75

approximately

(I rounded to the nearest multiples of 25c)

Gandalf

makes sense

thanks

Akiva Weinberger

Imagine you're dividing the 300 into 50+100+21 pieces

You give the first person 50 of those pieces, the second person 100 of those pieces, and the last person 21 of those pieces

That way the proportions work out

and there are 50+100+21 pieces total, so each piece is worth 300/(50+100+21)

Gandalf

10:58

Thanks for the explanation

Akiva Weinberger

No prob

Will Hunting

Good to have a wizard here. If only you could give me a miracle.

Gandalf

Maybe in the evening,right now my wizardly is not fully on

Aleksa

Hey could someone check out my question? math.stackexchange.com/questions/2993694/…

NaCl

11:44

Hi guys! How do the coefficients of a quadric change if I rotate it?

Leaky Nun

11:54

they become rotated as well

so $8$ becomes $\infty$

$i$ becomes $\cdot -$

NaCl

nice troll

appreciate it

Leaky Nun

$T$ becomes $\vdash$

thanks

GodotMisogi

12:42

For proving ℤ[(1 + √-11)/2] is a Euclidean domain, how do you choose the correct point such that the norm condition is satisfied? I can't really visualise it for some reason

Unknown MathMan

12:59

It seems magical when I observe that isomorphism preserves abelianness and cyclic property. But, what was the motivations behind defining isomorphisom (or homomorphism , as definition of homorphism playing the important role) so that we have those property preserved?

Unknown MathMan

13:20

Here is a problem from group homomorphism: If $\mathbb{Z}/10\mathbb{Z}$ to itself has a homomorphisms, b onto homomorphisms and c one-one homomorphisms, find a,b and c.

MatheinBoulomenos

@GodotMisogi I would prove it like this:
Let $\alpha=\frac{1+\sqrt{-11}}{2}$, it suffices to show that for every $z \in \Bbb{Q}(\alpha)$, we can find a $z' \in \Bbb{Z}[\alpha]$ such that $|z-z'|^2 <1$.
Let $z =a+b\alpha \in \Bbb{Q}(\alpha)$
Now we have to do some casework.
Note that if $z'=a'+b\alpha$, then $|z-z'|^2=(a-a')^2+(a-a')(b-b')+3(b-b')^2$
The first case uses the fact that this less than $|a-a'|^2+|a-a'||b-b'|+3|b-b'|^2$. We can always take $a' \in \Bbb Z$ such that $|a-a'|\leq \frac{1}{2}$, if we do that and we put $|b-b'|=w$, then we want the condition $\frac{1}{4}+\frac{w}{2}+3…

Unknown MathMan

Ofcourse the operation is addition. I think $\phi(\bar{1})$ decides the whole map.

MatheinBoulomenos

@UnknownMathMan yeah $\overline{1}$ is a generator, so you're right about that

GodotMisogi

@MatheinBoulomenos Thanks!

Unknown MathMan

@MatheinBoulomenos and if I send $\bar{1}$ to any generator, we will have a onto(and hence one-one, as both has same cardinality) homomorphism. As there are 4 generators, we can have 4 onto homomorphism, right?

Alessandro Codenotti

13:27

Hi @Mathei

MatheinBoulomenos

@UnknownMathMan correct

Hi @Alessandro

Gaurang Tandon

Hey all, I think I discovered a contradiction to the statement that: Show that number of solutions satisfying $x^5=e$ is a multiple of 4. Consider $U(4) = \{1,3\}$. Then the only elements satisfying $x^5=e$ are 1 and 3, which is two in number. Clearly this aint a multiple of 4

GodotMisogi

@MatheinBoulomenos |a - a'| ≤ 1/2, |b - b'| ≤ 1/3 should also work, right?

Gaurang Tandon

is my contradiction correct, or am i missing something?

Unknown MathMan

@MatheinBoulomenos What we can say about a and c then?

Alessandro Codenotti

13:30

I have an algebraic geometry doubt. I know that in general a sheaf isn't completely determined by its stalks, but if $(X,\mathcal O_X)$ is a prevariety then for all open $U\subseteq X$ we have $\mathcal O_X(U)=\bigcap\limits_{x\in U}\mathcal O_{X,x}$ (where the intersection is done inside $k(X)$ so that it makes sense), so it seems to me that in this case the sheaf is actually determined by the stalks

GodotMisogi

On a completely unrelated note, how would you show that for a radical ideal I in a Noetherian ring with a minimal primary decomposition, the primary components are prime?

MatheinBoulomenos

@GodotMisogi yeah, then you do the second case with -1/2≤b-b≤-1/3 and get an upper bound of $\frac{3}{4}+\frac{1}{2}(\frac{1}{2}-\frac{1}{3})=\frac{5}{6}$, which works

I was going for the optimal bound for the first case

@UnknownMathMan your observation about injective=surjective due to finiteness shows that b=c

@GaurangTandon in $U(4)$, you have $3^5=3$

Gaurang Tandon

@MatheinBoulomenos oops, i am sorry for the oversight... :(

Unknown MathMan

@MatheinBoulomenos I am not sure if "not onto means not one-one" holds. By checking case by case it seems b=c, but is there any other way?

MatheinBoulomenos

@UnknownMathMan for a map from finite set to itself, injective, surjective and bijective are all equivalent

Unknown MathMan

13:36

Ok got it, it is the contrapositive statement of "if one-one then onto", which is true as cardinality is same.

@MatheinBoulomenos Yeah

MatheinBoulomenos

@Alessandro well, you still need the inclusions $\mathcal{O}_{X,x} \to k(X)$, otherwise I agree

Unknown MathMan

@MatheinBoulomenos Is there a way to get a rather than checking case by case?

Alessandro Codenotti

@MatheinBoulomenos Yeah, I wrote $\mathcal O_{X,x}$ but actually meant the image of the canonical inclusion in $k(X)$, that was a bit sloppy

MatheinBoulomenos

@UnknownMathMan yeah, if you can convince yourself that you can just send $\overline{1}$ to any element in $\Bbb{Z}/10\Bbb Z$

Unknown MathMan

I think we can't send it to $\bar{0}$

Ok we can, so there are 10.

Thanks :)

MatheinBoulomenos

13:43

@GodotMisogi Let $I$ be a radical ideal in a Noetherian ring $R$, then the primary decomposition of $I$ in $R$ corresponds to the primary composition of $(0)$ in $R/I$, so we may assume wlog that $I=0$ and $R$ is reduced, but if $R$ is reduced, then $0$ is the nilradical which equals the intersection of all (finitely many) minimal prime ideals, since prime ideals are primary, that's a primary decomposition

to see that it is minimal note that the minimal associated primes of $R$ (as an $R$-module) are exactly the minimal primes of $R$, since in general for a f.g. module $M$ over a Noetherian ring $R$, one has that minimal primes in the support of $M$ correspond to minimal prime associated to $M$

@UnknownMathMan np

Liad

im trying to define a functor on $set^3$. i want it to be $F(A,B,C) = A^{B\times C}$

now on a morphism $(A,B,C) \to (D,E,F)$ , i want $A^{B x C} \to D^{ExF}$

given $f: B\times C \to A$ i want to send it to

a function $E\times F \to D$

problem is i need to take preimage of elemennt $(a,b) \in E\times F$ and it could be empty , any ideas how to fix it? @MatheinBoulomenos

MatheinBoulomenos

the natural way to make this into a functor is covariant in $A$, but contravariant in $B$ and $C$

Liad

what covariant means?

MatheinBoulomenos

I don't see any obvious way to make this into a functor that is covariant in all three variables

Alessandro Codenotti

How is the long dash used as a placeholder for functors done in latex?

Leaky Nun

13:54

@AlessandroCodenotti $- \otimes N$ "- \otimes N"

you'd be surprised

Alessandro Codenotti

oh lol

MatheinBoulomenos

@Liad covariant are the usual functors you know, so for a morphism $A \to B$, you get a morphism $F(A) \to F(B)$, contravariant functors reverse the direction, so for $A \to B$, you get a morphism $F(B) \to F(A)$

Alessandro Codenotti

I was sure that would mess up the spacing since - is supposed to be a binary operator

Liad

so here i need to define a functor on $3-set ^{op} $ ?

Leaky Nun

@AlessandroCodenotti $3-5$

MatheinBoulomenos

13:56

@Liad a functor on $\mathbf{Set} \times \mathbf{Set}^{op} \times \mathbf{Set}^{op}$

so what I'm saying is that if you want to make $F(A,B,C)=A^{B \times C}$ into a functor, then the right thing to do is to associate a tripel of morphisms $(A \to D, E \to B, F \to C)$, to a morphism $A^{B \times C} \to D^{E \times F}$. Can you do that?

Liad

yes

just to see if i got you right, $(A,B,C) \to (D,E,F)$ iff $A\to D , E\to B, F\to C$ ?@MatheinBoulomenos

MatheinBoulomenos

yeah, that's how morphisms in $\mathbf{Set} \times \mathbf{Set}^{op} \times \mathbf{Set}^{op}$ work

Liad

that's nice.

Leaky Nun

@MatheinBoulomenos triple

MatheinBoulomenos

okay

Alessandro Codenotti

14:09

@LeakyNun My last set theory pset has an exercise saying principle filter and it annoys me

Leaky Nun

@AlessandroCodenotti you should complain to the department head

MatheinBoulomenos

@Alessandro complain to the ~~principle~~ principal

GodotMisogi

@MatheinBoulomenos I haven't really studied modules, so is there any way of showing this without referring to that?

MatheinBoulomenos

@GodotMisogi what's your definition of a minimal primary decomposition, I think you can get that the intersection of all minimal prime ideals is a minimal primary decomposition of $(0)$ in a reduced Noetherian ring much easier, without mentioning associated primes

Leaky Nun

14:24

@MatheinBoulomenos how do you guys know so much stuff

@UnknownMathMan any sentence you can write using group operations is preserved under isomorphism

abelian is "$\forall x \forall y, xy = yx$"

cyclic is $\exists g, \forall h, \exists n, h = g^n$

i.e. the property can be expressed using group operations

MatheinBoulomenos

@LeakyNun idk

Liad

@MatheinBoulomenos is this the same way a functor $(A^B)^C$ would work?

MatheinBoulomenos

@Liad yes, actually $(A^B)^C$ and $A^{B \times C}$ are naturally isomorphic!

Liad

that's what i want to prove :)

with this functors

Leaky Nun

@MatheinBoulomenos I just proved that the integral closure is idempotent

GodotMisogi

14:33

@MatheinBoulomenos Primary is the usual definition. Minimal in the sense that ∩_{i ≠ j} Qⱼ \notin Qᵢ for any i and rad Qᵢ ≠ rad Qⱼ

Where I = ∩ᵢ Qᵢ is the minimal primary decomposition of proper ideal I

MatheinBoulomenos

@GodotMisogi you can prove that pretty easily for the intersection of all minimal prime ideals

@LeakyNun fun fact: if $A$ is an intergral domain, then the intregral closure of $A$ in its fraction field $K$ is equal to the intersection of all valuation rings $R$ with $A \subset R \subset K$

Leaky Nun

heh...

interesting...

ok I proved one direction now

Secret

Pondering about irrationality again:

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

The issue is that there seemed to be no easy way to bound this function unless $x=0,1$

The reason why Fourier's irrationality proof of $e$ works is because $e^1$ has a tight way to bound the resulting infinite series

thus ensuring the tail to show the expected behaviour of summing to less than 1 regardless of how big the denominator is being magnified

Stone–Weierstrass theorem

In mathematical analysis, the Weierstrass approximation theorem states that every continuous function defined on a closed interval [a, b] can be uniformly approximated as closely as desired by a polynomial function. Because polynomials are among the simplest functions, and because computers can directly evaluate polynomials, this theorem has both practical and theoretical relevance, especially in polynomial interpolation. The original version of this result was established by Karl Weierstrass in 1885 using the Weierstrass transform. Marshall H. Stone considerably generalized the theorem (Stone...

Will Hunting

16:05

Stone's theorem is hard because stones are hard.

Semiclassical

sounds legit

S. Crim

Hi everyone, is there anyone here with sufficient knowlegde on martingales/applied probability?

PabloZ392

16:19

Hello :)

Mary Star

17:14

We consider the $\mathbb{F}_2$-vector space $(2^M, +, \cap)$, where $M$ is non-empty set and $+ : 2^M\times 2^M \rightarrow 2^M: (X,Y)\mapsto (X\cup Y)\setminus (X\cap Y)$.

I want to show that $(2^M, +, \cap )$ for $\mathbb{K}=\{\emptyset , M\}$ satisfies the axioms of associativity.

$(X+Y)+Z=[(X\cup Y)\setminus (X\cap Y)]+Z=([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z)$

Is this correct so far?

RockDock

18:05

Prove that A5 has no normal subgroups N ≠ e , A5?

0

Q: Prove that $A_5$ has no normal subgroups $N$ $≠$ $e$ , $A_5$?

$A_n$ is subset of $S_n$ consisting of all even permutations. I have read the permutation groups theory but I don't know how to solve this problem (I mean how to proceed ). I want little simple explanation compared to that given at Alternative proofs that $A_5$ is simple Also it seems my questio...

abstract-algebra normal-subgroups

please just tell how to approach this problem

user21820

← 3 messages moved from CRUDE

RockDock

-1

Q: Prove that $A_5$ has no normal subgroups $N$ $≠$ $e$ , $A_5$?

$A_n$ is subset of $S_n$ consisting of all even permutations. I have read the permutation groups theory but I don't know how to solve this problem (I mean how to proceed ). I want little simple explanation compared to that given at Alternative proofs that $A_5$ is simple Also it seems my questio...

abstract-algebra normal-subgroups

Mike Miller

@user21820 we don't want that either

RockDock

@MikeMiller Sir please help me to solve this problem

No idea how to approach

user21820

@MikeMiller Lol sorry I didn't realize RockDock already posted it once here.

RockDock

18:10

@user21820 Please help me sir

Just receiving down votes and no suggestions .

user21820

@RockDock How sure are you that I am a "sir"? Also, you should be very specific about which lines of the answers to the duplicate question you do not understand. I'm sure lots of people here can help you if you explain what exactly you don't get.

165

Q: How to ask a good question.

How to ask a good question. This thread has advice on the following aspects of writing a good question on this site. Each item in this list links to an answer below about that specific aspect of question writing. Provide context, $\,$ include the source and motivation for your question, $\,$ a...

support faq etiquette asking-questions context

Anyway, I got to go.

RockDock

@user21820 wait

Sir because of your reputation

Just tell me how to approach this problem

Oskar Tegby

How does one best do a presentation in mathematics? Do I just use PowerPoint, or is there something better available?

Mike Miller

On the chalkboard

Pig

latex, beamer

RockDock

18:14

@OskarTegby Please help me .tell how do i proceed ?

-1

Q: Prove that $A_5$ has no normal subgroups $N$ $≠$ $e$ , $A_5$?

$A_n$ is subset of $S_n$ consisting of all even permutations. I have read the permutation groups theory but I don't know how to solve this problem (I mean how to proceed ). I want little simple explanation compared to that given at Alternative proofs that $A_5$ is simple Also it seems my questio...

abstract-algebra normal-subgroups

Oskar Tegby

43

Q: Alternative proofs that $A_5$ is simple

What different ways are there to prove that the group $A_5$ is simple? I've collected these so far: By directly working with the cycles: page 483 of http://www.math.uiowa.edu/~goodman/algebrabook.dir/algebrabook.html Because it has order 60 and two distinct 5-Sylow groups: https://crypto.stanf...

group-theory finite-groups big-list alternative-proof simple-groups

@MikeMiller: That will take way too long to do a thesis presentation.

RockDock

Thanks @OskarTegby

Zee

Wtf

RockDock

Atleast you helped me . SIR @user21820 and SIR @MikeMiller did not

Oskar Tegby

Apparently, "Keynote" in macOS has support for LaTeX code.

Mike Miller

18:23

@OskarTegby I've presented my thesis a few times, and I disagree. ;)

The term is "beamer" for a sort of powerpoint that lets you use TeX.

But I really strongly advise against slide talks. Most of the time the speakers rush through things too quickly and the audience learns nothing.

The chalk talks force you to slow down, and decide what is really crucial to your talk.

user21820

@RockDock What a joke. Still calling me "sir". And all Oskar did was to post the exact same link that was planted right on the top of your question.

And stop pinging me for no good reason.

3

Oskar Tegby

@MikeMiller: Thanks for your input. I'll definitely consider it.

Mike Miller

@OskarTegby How long is your talk?

Oskar Tegby

I think it's like 45 minutes.

Mike Miller

Yeah, I think chalk talk is generally better in that case. (If it was one of those 15-20 minute speed talks, everything I said goes out the window. But those suck.)

Oskar Tegby

18:35

@user21820: Haha, yeah.

Mike Miller

Maybe talk to your advisor, if you have one.

Ultradark

Hi @amWhy

Oskar Tegby

He suggested me to make slides. I'll see how it goes and if I can learn something from it.

Ultradark

I'm going to post a question here. If you have time to peruse it that would be great

Mike Miller

They have better insight into your project and presentation than I do.

Oskar Tegby

18:37

Yeah, but I'm still happy to have a second opinion for the evaluation of the result.

Mike Miller

No matter what, give at least one practice ahead of time to gauge time. It would help if you have an audience member at the practice who would stand to learn something from your talk - that means they will be able to say if you're going to fast etc.

Oskar Tegby

Haha! I'll try to force some friend to do that.

Ultradark

Question: Intersecting the two-dimensional structure $K_S$ with the vertical line $x=1/2$ yields what kind of local distribution? Does it yield a non random and non periodic distribution? I'm looking for the exact distribution.

$K_S$ is a mathematical structure composed of four elements. $S$ is some ordered, increasing, positive set.

$K_S$ is defined as follows:

$K_S=\{\phi_S(x),1-\phi_S(x),M_S(x),1-M_S(x)\}.$

Expanding the four elements contained in $K_S,$ we get:

$\phi_S(x)=\{e^{s_1A_1},e^{s_2A_1},...\};$ $A_1=1/\ln(x).$

user21820

← 3 messages moved from CRUDE

Transcript for

$Mathematics$ Mathematics