Mathematics - 2016-11-19 (page 5 of 10)

DHMO

15:00

@Null it is brilliant.

what do you mean by 40?

Null

40 is the biggest number :D

DHMO

is it a reference to anything?

Null

@DHMO its the biggest number displayed by trees on the earth surface

Ramanujan

Nov 13 at 7:08, by DHMO

physicsists always abuse maths

DHMO

by the definition of $A \subset B$, $\exists z \in E: z \notin A \land z \in B$.
Then, $let X=A\cup\{z\}$.
By some definitions, $X\cap B=X$.
By some definitions, $X\cup A=X$.
$\blacksquare$

@Ramanujan yep. just like $1+2+3+\cdots=-\dfrac1{12}$

@Null nice

@SteamyRoot is this false?

Ramanujan

15:03

@DHMO but I think it was proven by sir Ramanujan

SteamyRoot

Is what false?

DHMO

5 mins ago, by SteamyRoot

@DHMO what are you saying? That any linear and stable summation method sends $1 + 2 + \cdots$ to $-1/12$?

@Ramanujan but people can still abuse it if they don't know what the statement means here

Mahmoud

@Null That is the best video on YouTube :D

user161151

hello people, i am new

DHMO

@user161151 welcome

SteamyRoot

15:04

Yes, that's false.

DHMO

@SteamyRoot why?

Ramanujan

@DHMO but he showed proof for it…was that wrong?or we are missing some thing in mathematics?

Astyx

Re-read it I think

Mahmoud

Welcome @user161151 :D

DHMO

@Ramanujan I didn't say sir Ramanujan is wrong. I said that (other) people would abuse it if they do not understand what that statement means.

Astyx

15:05

Ramanujan was quite wrong on this topic I think

DHMO

lol

Astyx

At least the proof he supposedly gave is

Mahmoud

Now I understand why was Mathematics separated from Physics.

SteamyRoot

@DHMO It was proven that a stable and linear summation method can never assign a finite value to $1+2+3+\cdots$.

DHMO

@SteamyRoot can you show me why it is wrong?

@Mahmoud have you read my solution?

Mahmoud

15:08

Yes, ''by some definitions...'' lol

SteamyRoot

$1+2+3+\cdots = a$. Then $0 + 1+ 2+3 + \cdots = 0+a = a$.

DHMO

@Mahmoud heh, those are the definitions you are supposed to know

the definition of $\subset$

Mahmoud

It's nicely proved tho.

DHMO

thanks

@Mahmoud any more question to share?

SteamyRoot

Now substract second from first term-wise to get $1 + 1 + \cdots = x - x = 0$.

DHMO

15:09

@SteamyRoot never mind then

SteamyRoot

Repeat the same steps (add zero, substract), to obtain $1 + 0 + 0 + \cdots = 0$

user161151

these days, i figured i have a lot of gaps in math.

DHMO

glad to learn

@user161151 for example?

Mahmoud

Lot more.

Wait a sec

user161151

@DHMO i can't solve algorithmic challenges on PE

DHMO

15:10

@user161151 I see. Computational mathematics.

user161151

@DHMO yes.

Null

Tell me your opinion: do you think any logic argument with "...because ..." is likely to be flawed?

SteamyRoot

@Ramanujan Ramnujan didn't "prove" that $ 1 + 2 + 3 + \cdots = -1/12$.

DHMO

@Null why?

Ramanujan

@SteamyRoot then who?

SteamyRoot

15:11

He came up with a summation method to assign finite values to divergent series (of which many exists), but his way was particularly nice

And agrees with the Riemann-Zeta regularization

user161151

@DHMO know any good resources to fill the gaps?

Null

@DHMO I'm just thinking this way. "Because" leads to the need to explain things, but some things can't be explained.(yet)

DHMO

@user161151 sorry, no idea

@Null where did you see that word?

meow-mix

good morning

DHMO

@meow-mix hi

Astyx

15:12

Hi

meow-mix

hello @DHMO and @Astyx

Ramanujan

@meow-mix iam also present

Hi

meow-mix

hi

Mahmoud

@DHMO Prove that $(\forall x \in \mathbb R) : x-\sqrt{x^2+1}\lt 0$ And some more questions.

Null

@DHMO A friend of mine tried to learn about psychology. They got statistics displayed as diagrams. Then he said "ah!... this is so BECAUSE BLA..." And i just cringed, since something makes sense for you doesn't equal that it's the truth.

meow-mix

15:14

the reason we can derive 1 + 2 + ... = -1/12 is because ramanujan assigned a real value to a divergent sum (namely 1-1+1-1+... = 1/2)

DHMO

@Null I see what you mean. You mean that when you say "because", you are relying on intuition, which has been shown many times to be unreliable.

meow-mix

also called Ramanujan Summation

DHMO

@Mahmoud I'll omit $\forall x \in \Bbb R$.

SteamyRoot

@meow-mix actually, Ramanujan's summation goes way further

meow-mix

what do you mean by that?

Null

15:16

@DHMO exactly

DHMO

$x-\sqrt{x^2+1}<0$
$\iff x<\sqrt{x^2+1}$
$\iff x^2<x^2+1$ (needs some more explanation but)
$\iff 0<1$

Null

(im not for banning the word because, just the "intuitive" reasoning)

DHMO

@Null yep. I see what you mean.

SteamyRoot

The sum $1-1+1-1+\cdots$ is $1/2$ by MANY summation methods, like cesaro summation

DHMO

well, you can't deny that intuition is good for something

Mahmoud

15:17

@DHMO I feel like I'm loosing my mind.

SteamyRoot

Abel summation too

DHMO

@Mahmoud where?

Mahmoud

How is this defined $p+p^2+p^3+p^4+...=\frac{1}{1-p}$ ?

meow-mix

the point is

its still a divergent sum

SteamyRoot

But the sum $1+2+3+4 +\cdots$ is neither cesaro nor abel summable

DHMO

15:17

@Mahmoud this is not defined. this is proven.

meow-mix

@Mahmoud only converges in $-1 < p < 1$

DHMO

@Mahmoud let $S=p+p^2+\cdots$. Try to find $pS$.

Mahmoud

@DHMO @meow-mix For all p ?

Null

@DHMO i think it is good to solve very direct problems, like "where do i get food". But it fails the more abstract it gets. (bayes for example..)

meow-mix

no

DHMO

15:18

@Mahmoud only for $-1<p<1$

@Null fair enough

Mahmoud

Hmm.

DHMO

@Mahmoud I thought you are having an exam tomorrow on set theory

Mahmoud

@DHMO Trying to forget about it, it's on Monday :(

Astyx

Tomorrow would be cunning

Null

@DHMO but maybe i'm now biased because you tend to kinda agree with me. also i used the word because haha ;)

SteamyRoot

15:19

I'm not sure trying to forget about an exam will help you pass it though :/

DHMO

@Null that is a valid "because" statement talking about cognitive biases.

@Mahmoud then why are you sharing not-set-theory-related questions?

meow-mix

hey guys, should i ask my parents to buy me artin's algebra?

SteamyRoot

It's definitely not a bad thing to have...

Mahmoud

Because @DHMO $p=-1$ Gives $1-1+1-1+...=\frac 12$

meow-mix

well

that sum diverges

just like any sum $p + p^2 + \dots$ for

Mats Granvik

15:22

$$\lim_{s\to 1} \, \left(\zeta (s)-\frac{1}{s-1}\right)$$

meow-mix

$p \geq 1$ or $p \leq -1$

DHMO

@Mahmoud if $-1<p<1$ then $p$ cannot be $-1$.

Mahmoud

Yes, but it gives the same answer @DHMO :P

DHMO

@Mahmoud abuse of mathematics #314159

Mahmoud

LOL @DHMO

DHMO

15:23

@Mahmoud any set-related question to share?

Danu

@TedShifrin I think I'm making some progress now :)

Mahmoud

@DHMO Okay I will share, a sec please.

DHMO

thanks

Ramanujan

:314159

What does it mean?

DHMO

#314159 means "number 314159"

or "the 314159th one"

meow-mix

15:26

so wait, the polynomial ring $R[x]$ is just a ring of functions from R to R?

(for some ring $R$)

Astyx

No

meow-mix

and you know, the functions are polynomials, obviously

Astyx

formally it's not functions

It's nearly-zero sequences of elements from $R$

meow-mix

err wat

Astyx

Polynomials are much more than just functions

s.harp

15:27

@meow-mix if you consider $R=\mathbb Z_2$ then there are polynomials in $R[x]$ that have the same "function" associated to them but are different

for example $x^2$ is the same as $x$ as a function on $\mathbb Z_2$, but $x^2$ is not $x$ in $\mathbb Z_2[x]$

Astyx

A polynomial is just an expression $\sum_{n=0}^{\infty} a_n X^n$ where $X$ is the unknown and $(a_n)_{n \in \Bbb N}$ is a sequence of selements from $R$ that becomes stationnary to 0 at some point

Sophie

Let $\sqrt{n}=[a_0;\overline{a_1,a_2\dotsb a_n}]$ and $a_i=\max(a_1,a_2\dotsb a_n)$. If $\frac{p}{q}=[a_0;\underbrace{a_1,a_2\dotsb a_n}_{\text{this k times}}\dotsb a_1,a_2\dotsb a_{i-1}]$ then $p^2-nq^2=\pm 1$

meow-mix

why is $x^2$ the same as $0$? $1^2 = 1$?

Sophie

Amended conjecture, I hope it is correct

s.harp

oh my bad, sorry $x^2=x$ in $\mathbb Z_2$ not $0$

Astyx

15:30

I read too fast

My bad

Mahmoud

Prove that : $$\begin{Bmatrix} A \cup B= A \cup C \\ A \cap B \\ \end{Bmatrix}= A \cap C$$ $\implies B=c$ @DHMO

meow-mix

its fine

DHMO

@Mahmoud first-order logic?

you forgot \end{matrix}

meow-mix

so wait, what specifically is a polynomial then?

also, on a polynomial, multiplication is defined by the ring's multiplicatiion

DHMO

also you can put `&` to align:
$\mbox{\begin{matrix} A \cup B & = & A \cup C \\ A \cap B & = & A \cap C \end{matrix}}$
produces
$\begin{matrix} A \cup B & = & A \cup C \\ A \cap B & = & A \cap C \end{matrix}$

meow-mix

15:31

and addition of terms is defined by the ring's addition, right?

Astyx

Yes

Well it's not the same multiplication and addition

But you get the idea

DHMO

@Mahmoud you can uses $\mbox{\begin{cases}}$ instead

meow-mix

why wouldn't it be the same?

DHMO

@Mahmoud I don't understand that notation

s.harp

because the "indeterminates" are not in the ring

the multiplication and addition is constructed from the multiplication/addition in the ring, but it is not the same operation because $x*x$ does not exist in $R$ simply because $x$ is not an element of $R$

meow-mix

15:33

what??

if the indeterminates aren't in the ring

how is multiplication and addition defined?

Astyx

$3x^2 \times 2x$ is not defined in $R$

Convolution of the sequences of the coefficient of your polynomial

Mahmoud

Prove that :$$\begin{Bmatrix} A \cup B & = & A \cup C \\ A \cap B & = & A \cap C \end{Bmatrix}$$ Implies $ B=C$

meow-mix

I'm way confused, sorry. what is "convolution"?

s.harp

$$\sum_k a_k x^k \cdot \sum_n b_n x^n = \sum_k \sum_{n=0}^k a_n b_{n-k} x^k$$

Astyx

^

meow-mix

15:35

no

wait

hmmm

Astyx

$b_{k-n}$

meow-mix

so polynomials are only determined by their coefficients?

two polynomials are the same if and only if they have the same coefficients

Astyx

Yes

Well in some sense

meow-mix

and, multiplication / addition can be applied because all of the coefficients are in $R$?

Astyx

Yes

Null

15:37

only if it's a polynomial ring tho, OR?

Astyx

I can't seem to find a good paper describing the construction of polynomials

(on the web and in english)

Mahmoud

And @DHMO $A,B,C$ are parts of $E$

Null

@Mahmoud was that just an excercise for the chat or do you have problems with sets?

meow-mix

oh, so thats why we denote the set of all $a\sqrt{2} + b$ for rational $a,b$ by $\mathbb{Q}[\sqrt{2}]$?

Astyx

Surely

Mahmoud

15:40

I'm kind of preparing for an exam @Null :I

DHMO

firstly let's use definitions:

$A\cup B=A\cup C$
$\iff\forall x: (x\in A\cup B) \iff (x\in A\cup C)$
$\iff\forall x: (x\in A \lor x \in B) \iff (x\in A \lor x \in C)$
Let $P(x)$ be the proposition $(x\in A \lor x \in B) \iff (x\in A \lor x \in C)$.

$A\cap B=A\cap C$
$\iff\forall x: (x\in A\cap B) \iff (x\in A\cap C)$
$\iff\forall x: (x\in A \land x \in B) \iff (x\in A \land x \in C)$
Let $Q(x)$ be the proposition $(x\in A \land x \in B) \iff (x\in A \land x \in C)$.

Let $R(x)$ be the proposition $x \in B \iff x \in C$.

Null

@Mahmoud to get a feel you can draw yourselfes Venndiagramms. I think the proofs itself are not the hurdle ;)

Sophie

@meow-mix This is field extension notation

Field extension

In abstract algebra, field extensions are the main object of study in field theory. The general idea is to start with a base field and construct in some manner a larger field that contains the base field and satisfies additional properties. For instance, the set Q(√2) = {a + b√2 | a, b ∈ Q} is the smallest extension of Q that includes every real solution to the equation x2 = 2. == Definitions == Let L be a field. A subfield of L is a subset K of L that is closed under the field operations of L and under taking inverses in L. In other words, K is a field with respect to the field operation...

DHMO

@Mahmoud That is what I meant earlier by "first-order logic"

Balarka Sen

@meow-mix You just define it. If $R$ is a ring, $x$ is an indeterminate (literally a symbol, something which doesn't have anything to do with any element of $R$), then define $R[x]$ by declaring elements to be of the form $\sum_{k = 0}^n a_k x^k$, and define multiplication by "expanding out" like you do for actual polynomials with real coefficients, say.

Mahmoud

15:42

@DHMO Yes.

DHMO

@Mahmoud do you have a better approach?

meow-mix

so $R[a]$ for some $a \in R$ is just a set of expressions (whose values are determined by plugging $a$ into a polynomial with coefficients in $R$)?

hi @BalarkaSen

Sophie

@meow-mix you don't need $a$ to be in $R$. For example $\mathbb{R}[i]=\mathbb{C}$

meow-mix

mmmmmhhh

Null

@DHMO i'm not native, is proposition the same like statement?

meow-mix

15:44

so say you have a polynomial

Balarka Sen

@meow-mix $R[a]$ is not in general a thing. But how you defined it makes it the same as $R$.

DHMO

@Null yes, i'm not native either

meow-mix

well yeah, because $a \in R$ @Balarka

Balarka Sen

Correct. For the other inclusion just look at the 0 degree polynomials.

meow-mix

but for $a \notin R$?

Mahmoud

15:45

$(A \cup B) \cup (A \cap B)=(A \cup C) \cup(A \cap C)$ Maybe ? @DHMO

DHMO

@Mahmoud and then?

Null

@Mahmoud that's a tautology

like apple=apple

Balarka Sen

@meow-mix Then you have to specify what $a$ is. For a polynomial ring it's an indeterminate, a symbol, or if you like a transcendental.

DHMO

@Null i know what @Mahmoud means

Mahmoud

Lol @DHMO I made a horrible mistake there.

meow-mix

15:46

yeah but

if $a$ is a value not in $R$, that isn't indeterminate

for example

$\mathbb{R}[i]$

Null

@DHMO ah lol, did not see the edit! @Mahmoud

DHMO

@Mahmoud and then?

@Null I know what he means before the edit

Null

@DHMO nice intuition then...

meow-mix

$i$ isn't in $\mathbb{R}$ but isn't indeterminate either

DHMO

@Null no, it's because we were dealing with a question

Balarka Sen

15:48

@meo-mix This is a subtle point in understanding these stuff. You have to define $i$ first. $\Bbb R[i]$ without defining $i$ does not make sense.

The way to go is to look at the quotient ring $R[X]/(X^2+1)$.

meow-mix

oh boy quotient rings

Mahmoud

I don't know how to make it rigorous @DHMO :(

Null

@DHMO joining sets and adding the intersection is kinda redundant i'd say

DHMO

@Mahmoud I don't even know how you would do it even not rigorously

Balarka Sen

Essentially what you're doing is you're just specifying $X^2 + 1 = 0$. Then your $X$ becomes what we know as $i$.

meow-mix

15:49

@BalarkaSen $(X^2 + 1)$ is an element of $R[X]$ right?

arctic tern

no

Mahmoud

A venn diagram shows it clearly.

Balarka Sen

It's an ideal of $R[X]$.

Null

$x\in(A\cup B)\iff$ x is in A,B or both

Sophie

is Pinter's algebra book good?

DHMO

15:51

@Null yes

$x\in A\cup B \iff (x \in A \lor x \in B)$

@Mahmoud what...

Null

^

heather

what property is it where any number divided by itself equals 1, as long as that number $\neq$ 0?

meow-mix

well

DHMO

@heather don't try to prove properties of real numbers without understanding one of the defintions of the real number

meow-mix

division is multiplication by an inverse i believe

DHMO

15:52

but basically the property of multiplicative inverse

@Mahmoud you mean $(A \cup B)\cap(A\cap B) = B$?

Mahmoud

How can I prove that $LHS$ is $A \cup B$ @DHMO

Null

@meow-mix by the definitions yes!

meow-mix

and since all non-zero elements of $\mathbb{R}$ are units (its a field)

Null

@meow-mix but you can very easily define something like devision for yourself ;)

DHMO

@heather multiplicative inverse theorem: for all non-zero real $x$, there exists $x^{-1}$ such that $x\circ x^{-1} = x^{-1}\circ x=1$

and then the definition of division is $\dfrac ab := a\circ b^{-1}$

therefore $\dfrac xx = x\circ x^{-1} = 1$ (where $x\ne 0$)

heather

15:54

@DHMO, okay, thank you. I couldn't remember the name of it. That makes sense.

DHMO

@Mahmoud I don't get your approach. What would you do after proving LHS is $A \cup B$?

Null

(except 0, but that is obv)

meow-mix

$0$ isn't a unit :D

DHMO

@Null I said non-zero already but I'll make it clear

meow-mix

i know, i was explaining why 0 isnt included

Null

15:55

@DHMO that's why "(...)" ;)

DHMO

heh, because that is the theorem

meow-mix

@BalarkaSen then why isnt $X^2 + 1$ an element of $R[X]$?

Balarka Sen

It is.

Null

I just think it is good to mention the typical traps :D

Balarka Sen

$X^2 + 1$ and $(X^2 + 1)$ are different things

the first is an element, the second is an ideal

Mahmoud

15:56

@DHMO NeverMind.

meow-mix

oh, i just realized... that by parenthesis

DHMO

@meow-mix When you say $R[X]/(X^2+1)$, the $X^2+1$ there means $(X^2+1)R[x]$ or an ideal

meow-mix

you meant the ideal genereated by $X^2 + 1$

Balarka Sen

Yes.

DHMO

@Mahmoud any more question to share?

Semiclassical

15:56

(A) is the ideal generated by the element A

meow-mix

sorry. i thought you were trying to make it look cleaner

Mahmoud

Ok.

Balarka Sen

Ideals are what you quotient your rings by, so...

meow-mix

ok ok

Null

Is (A) a notation that is exclusive to the ideal or is $(-1)\in\mathbb{R}$ valid too? (in the latter i defenitly dont mean that the ideal is part of R)

Semiclassical

15:58

It's a matter of context, to be fair. When I entered in chat and saw A!=(A) I was a bit perplexed until I looked at the entire history

Danu

Ah, I see now @TedShifrin where the well-definedness comes from. Anti-symmetry allows you to take all but one of the $v_1\wedge\dots\wedge v_k\in \bigwedge^k E$ to have arbitrary $L$-component (without modifying the result), so you get a well-defined map from $\ell\wedge [v_2]\wedge \dots \wedge [v_k]$ into $\bigwedge^k E$

Semiclassical

Because in general contexts I'd say (A)=A just as a matter of notation.

Null

@Semiclassical I see...

meow-mix

bad notation

maybe if they used [A] i wouldn't be confused

Null

$A^{I}$ and avoid sets with name I?

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$Mathematics$ Mathematics