Mathematics - 2017-01-09 (page 3 of 7)

Balarka Sen

7:01 AM

$a^4 + a^2b^2 + b^4 = a^4 + 2a^2b^2 + b^4 - a^2b^2 = (a^2 + b^2)^2 - (ab)^2$

Similar for the second.

Ramanujan

For second we should add and subtract √2ab ?

Balarka Sen

No, 2a^2b^2

Or rather, (√2ab)^2

TheGreatDuck

-1

Q: Are these alternate axioms of differential algebra (on real number functions) consistent?

In another question I asked, I asked about using a particular axiom to define the derivative so that I could negate it in general. However, I realized that there are two alternative systems in particular that I find interesting*, and that the core alternative truths that describe them as I desire...

I think I made this question a little better than it was

feel free to see the edits i made to be more precise

did I improve the question by adding the links and stuff? Is it crystal clear now?

Ramanujan

Rhs is (a^2)^2 - (ab√2 + b^2)^2 = a^4 - 2a^2b^2 - 2√2 ab - b^4 @balarka

TheGreatDuck

wait...

you two still discussing that a^4-b^4=0 thing?

Ramanujan

7:10 AM

@TheGreatDuck no

TheGreatDuck

oh ok then

carry on

also

turns out my boat is buggy

cannot actually fly it

XD

i didn't realize till I went to fly it, lol

Ramanujan

Lol

Balarka Sen

@Ramanujan You messed up. RHS is not that.

TheGreatDuck

@Ramanujan gotta look very carefully at it. I assume that the condition for it to be flying requires the player to be in a weird position or something.

:p

Idk what's more annoying. The bugs or that I have to put an entire mod into one line of code. I really should get around to making a converter for the file but I'm lazy.

sans-level lazy if you catch my drift

(when it comes to this anyway. Probably cause I'd rather get stuff done than worry 'bout making it easier. XD)

Ramanujan

@TheGreatDuck did you showed you work to friends?

Balarka Sen

7:18 AM

That's all wrong. $(a^2)^2 - (\sqrt{2}ab + b^2)^2 = (a^2 - \sqrt{2}ab - b^2)(a^2 + \sqrt{2}ab + b^2)$.

Not what's up in the first line.

Ramanujan

?

@BalarkaSen typo?

Balarka Sen

I don't understand your question. $a^2 - (\sqrt{2}ab + b^2) = a^2 - \sqrt{2}ab - b^2$.

No it's not a typo, I'm pointing out what you did wrong.

Ramanujan

Alright,i got that

So we can't write rhs in a^2 -b^2 form?

Balarka Sen

Yes you can. $(a^2 + b^2)^2 - (\sqrt{2}ab)^2$

TheGreatDuck

@Ramanujan yes. A couple. They made a joke about the game being "warcraft".

now they refer to it as "warcraft"

XD

it's funny

@Ramanujan oh yeah. To point out another part that's funny. There's a scene where mario is called into court not for killing a zombie (it's already dead) but for murdering it by chopping it's head off.... WITH A SHOTGUN!

Ramanujan

7:28 AM

Mario?

TheGreatDuck

XD

yup

Ramanujan

That game ends?

TheGreatDuck

of course

all games have endings

but that's not the end

the end if when you literally fight chuck norris

because it was made in 2009

and had to go there

I think fighting the moon in space is a bit more epic.

Ramanujan

What? Where chuck Norris come?

TheGreatDuck

a 2d animated chuck norris that literally serves as a cutscene trigger is cheezy as... the moon.

Ramanujan

7:30 AM

If game was made in these years,that I suppose john skeet would be there in game :P

TheGreatDuck

@Ramanujan basically, they (the villains) wanted an unbeatable weapon to kill everyone. So they made a clone of chuck norris because they used google and looked up "unbeatable weapon, roundhouse kick, and some other stuff"

and read the site "chuck norris facts" for weeks

trust me, there are parts in the plot that sound totally stupid

and yet the overall plot makes total sense in hindsight

to some degree the characters can just be... a cross between idiots and flat out random.

though to be fair.. in 2009 chuck norris was definitely the one to beat

that was the meme

I'd probably just play it yourself when you have time

Ramanujan

It is pc game?

TheGreatDuck

@Ramanujan gamejolt.com/games/neen-ten-doo-nightmare/185160

yes

windows based

but I hear it works on linux for no apparent reason as well

shrugs

Ramanujan

I don't have pc :P I only have one Android device and that to running on KitKat :D

TheGreatDuck

oh

well then i dont recommend playing it

watch playthroughs

vinny was pretty funny back in the day

Ramanujan

7:35 AM

Hmm

TheGreatDuck

depending on your opinion of his language (cursing)

monotonetim finished it in one video though

so he's the most complete to watch

vinny lost some of his last video

so you don't see the ending

anyway

tis late

Ramanujan

You ended that game?

TheGreatDuck

i didn't make it

but i played it

all of it

euclid

@BalarkaSen hi

Ramanujan

No,i mean in playing how far you have gone

TheGreatDuck

7:37 AM

actually have known about it since 2011

i finished the whole game

Balarka Sen

Hi

euclid

@BalarkaSen i saw you are interesting in number theory.its mean you work in analytic number theory also?

Liad

Hi. i got a small question in probability.
If the expiriment is to flip a fair coin n times, and X is the number of double "H" , is it correct that X distribution is Bin(n,1/4) ?

Balarka Sen

@euclid I am interested in number theory but I do not know anything about it

euclid

look like me.thanks:)

Arrow

7:50 AM

Suppose $\alpha:A\to B$ is a fiber bundle. Suppose for every pair of points $b_1,b_2\in B$ there's an isomorphism $\phi_{b_2b_1}:\alpha^{-1} \left\{ b_1 \right\} \cong \alpha^{-1} \left\{ b_2 \right\} $ and that these isomorphisms satisfy the cocycle condition $\phi_{b_3 b_2}\circ \phi_{b_2 b_1}=\phi_{b_3 b_1}$. Why does it follow that $\alpha$ is a trivial bundle?

Balarka Sen

@Arrow Fix a $F = \alpha^{-1}(b)$ for some specific $b$ in the base. You're probably supposed to see that the bundle-morphism $A \to B\times F$ given by sending each fiber $\alpha^{-1}(x)$ to $\{x\} \times \alpha^{-1}(b)$ by $\phi_{bx}$ is an isomorphism.

Ramanujan

8:22 AM

What is it doing? How is it doing??

:(

@BalarkaSen hi

rob

@Ramanujan Example 1, line 0: left polynomial, starting with a^4, multiplied by right polynomial

line 1: coefficients of left polynomial, sorted by powers of a

Arrow

@BalarkaSen sorry if this is silly, but I don't understand how the cocycle condition shows isomorphy to the trivial bundle

rob

line 2: coefficients of right polynomial

line 3: line 1 * 1, which is first time in line 2

line 4: line 1 * -2, which is next nonzero term in line 2, and offset

@Ramanujan The offset groups the products by powers of a.

line 6: coefficients of left polynomial * right polynomial

Is that helpful?

Ramanujan

8:39 AM

Order is by taking coefficient of a then b?

rob

@Ramanujan The polynomials have a special shape: each term has $a^n b^m$ with $n+m=\text{constant}$

left polynomial $n+m=4$, right $n+m=3$

Ramanujan

Ok

I didn't get 3rd line

Pissed off layman

Welcome to the math chatroom @JohnRennie and rob :-)

rob

@Ramanujan my "line 3"?

under the first horizontal line?

Ramanujan

Yes,under horizontal line

John Rennie

8:43 AM

@Pissedofflayman I'm just spectating. What I know about maths you could write on a flea's bum (in a large font).

rob

Those are the coefficients of the left polynomial multiplied by $a^3$

Pissed off layman

:D

rob

line 4: coefficients of left polynomial multiplied by $-2ab^2$

Ramanujan

Why are we multiplying?

rob

@Ramanujan Your image seems to show a method for multiplying two polynomials ... ?

line 0: (left polynomial, order $a^4$) multiplied by (right polynomial, order $a^3$)

Ramanujan

8:46 AM

Ok,each coefficient is multiplied by coefficient of right side polynomial

rob

@Ramanujan Yep. It's a bookkeeping trick for the distributive property of addition and multiplication.

Ramanujan

In fourth line , why we left some space there?

(Second line after horizontal line)

rob

line 4: coefficients of left polynomial multiplied by $-2ab^2$

First coefficient multiplies a term like $a^5 b^2$

In line 3, first coefficient multiplies $a^7$

so line 4 is shifted over two spaces from line 3, so the $a^5$ terms are aligned vertically

(If right polynomial had a nonzero $a^2$ term, there'd be another line shifted by only one space)

Ramanujan

Ok

Vrouvrou

Please, i have this two conditions $$\limsup_{|t|\rightarrow0}\frac{B(t)}{\Phi(t)}<+\infty,~\text{and}~ \limsup_{|t|\rightarrow+\infty}\frac{B(t)}{H(t)}=0$$ what is the good interpretation : $$B(t)\leq c_{\varepsilon}\Phi(t)+\varepsilon H(t)$$ or $$B(t)\leq c_{\varepsilon}\Phi(t)+\varepsilon H(t)+\beta $$ ?

Lu Bu

8:54 AM

Why doesn't my mathjax render in real time?

rob

@Ramanujan Make sense now?

Ramanujan

9:18 AM

@rob that's pretty much how i used to multiply variables

rob

@Ramanujan Okay. Glad to help.

Ramanujan

But instead taking only coefficients I use to take variables too

Thanks

sittian

hi

Ramanujan

@sittian hi

sittian

i have two little doubt regarding set theory

phi is subset of every set ?

whether it is empty set ?

tell

rob

9:21 AM

@Ramanujan Yes, I usually carry the variables around. But the resemblance to the base-10 multiplication that I learned in elementary school (and long division, too, it looks like) suggests that maybe I could save some ink by this "detached coefficients" trick.

sittian

it may b {{},}

?

Ramanujan

@sittian I am not familiar with set theory,somebody other will help you :)

sittian

don't put ur name ramanujan then

user402003

@sittian that doesn't matter here

it doesn't matter sittian

Ramanujan

@sittian yes, phi is a subset of every set

Vrouvrou

9:40 AM

can someone answer me ?

Secret

The first few pages of Maclane's category theory book suggest I am doing some kind of category theory stuff here, where axioms (known as equations in model theory or identities in universal algebra) and operations are both objects and morphisms, and theorems, lemmas, propositions, corollaries etc. are functors of the category of axiomatic systems

LeGrandDODOM

@DanielFischer have you had a look at this question ?

Secret

Therefore under this framework, theorems are morphisms that take in equations or a system of equations and spit out equations

and a a theorem will behave like a bijective map for some given pair of input and output equations when the property "iff" holds

DHMO

9:58 AM

@Ramanujan can you prove that A subset B and B subset C implies A subset C?

LeGrandDODOM

crackpots everywhere

Secret

Note the diagram commutes

Ramanujan

10:17 AM

google.co.in/…

DHMO

@Secret could you explain?

Ramanujan

@DHMO let c subset of A and A subset of B

Then C is subset of B

DHMO

@Ramanujan thank you for restating my question

@Fargle b8

Secret

(I cannot say I am really good at category theory yet because I have only read a few pages of Maclane, thus it is possible someone may point out I am saying nonsense) Basically, category theory is a genralisation of maps and sets to mathematical objects. In category theory there are arrows called mophisms, which are generalisation of maps from objects to objects. An object can be anything ,from mathematical structures to logical statements. For a given type of object, it and the morphisms form a category. A functor is like a map that takes categories as inputs and spit out categories

actually a small mistake here: the diagram actually does not commute.

DHMO

@Secret what would be the problem if I replace all the $\subset$ in your diagram with $\not\subset$?

Secret

10:27 AM

well short answer is that the proof will fail, because now $x \in A$ will be mapped by $A \not\subset B$ to many possible things (e.g. $x \in C$, $x \not\in B$, $x\in A$). I have not read whether mophisms must map one object t one object, thus we might be leaving cateogry theory altogether

DHMO

What is a morphism?

Secret

Roughly speaking, it's a map obeying composition law and send objects to objects in a category (have not read much in detail yet)

DHMO

And what on earth is a category?

Secret

e.g. homomorphism is a type of morphism that preserve the properties of the object

e.g. group homomorphism preserve the group structure

A category is a pair made of objects and morphisms

kinda a very extreme generalisation of ordered pairs I guess...

DHMO

Interesting.

Secret

10:32 AM

Category theory is very powerful when it comes to organising patterns found in many mathematical structures. I am still really a noob on it though

one of its application in abstract algebra is solving the universal property of a given algebraic structure, that is given the following diagram:

Where M is some algebraic strucrture (e.g. a group), A is its group extension and B is some subgroup. Given that you know the group homomorphism h and f, one aspect of the universal property is whether g is unique such that the diagram commutes

(because it would be quite messy if the mapping from structures to structures depends on the order you apply the morphisms)

also typo: H should be h

It is however said that one should not start reading category theory unless one has enough knwoleedge on abstract algebra, which is why I postponed reading maclane for now, even though so far the stuff is quite stragithforward

(but really I only have just read one page back on that day when the topic is raised, so it might be actually much more difficult than I thought)

Therefore, those diagrams that I am using now in my powerpoint are really just convenient shorthands of analysing some of the proofs I am working on until later

Yassin Rany

10:49 AM

hey guys, hello!

someone here have some experience with compactness in topology?

Alessandro Codenotti

Maybe, depends on your question

Yassin Rany

about finite instersections of compact subspaces

Alessandro Codenotti

Just ask

Yassin Rany

in almost all proofs using the definition of compactness as the topological space in which every open cover have a finite subcover

there is a step that i can understand and in my standpoint is not true

Let C be an open cover of U1∪U2.

Then C is an open cover of both U1 and U2.

this is the part that i am having trouble

i think C would not be an open cover of both

what do you guys think? my argument is that C may contain open sets that are not in the relative topology on U1 or in U2, so C would not be made only of open sets

Akiva Weinberger

11:09 AM

@YassinRany Look at the link on the top-right to enable ChatJax

There's two definitions of an open cover of $C$. One is that it's a family of sets $U_i$ such that every set is open in $C$ (and thus a subset of $C$) and such that the union is $C$.

On the other hand, sometimes our set $C$ is a subspace of some other set, call it $X$. So $C\subseteq X$

Then the other definition is that it's a family of sets $U_i$ such that every set is open in $X$, and such that the union contains $C$

To get from the second to the first, replace each $U_i$ with $U_i\cap X$ (which are open in $C$ due to the subspace topology).

Oh, sorry, the notation I just used was slightly different from that of your question — I was using $C$ as a topological space and $\{U_i\}$ as an open cover rather than the other way around

In any case, the definition of compact works no matter which version of "open cover" you use

Yassin Rany

11:26 AM

please do not worry with notation

i think i understood

thank you very much

if i may ask you know a book reference about this?

Ramanujan

Yassin Rany

is because i was using the definition that (in your notation now) a subset C of the topological space X is compact if it is compact as a topological space in its own right, with the topology being the interection of C with all open sets of X

@AkivaWeinberger "))

Ramanujan

I think x = 1/3(√5x^2 + 4d^2)

Balarka Sen

Yikes

N3buchadnezzar

Does anyone remember where the proof for $\int_0^{2\pi} \log |1-e^{ix}|\,\mathrm{d}x = 0$ is? I remember seeing it on this page a while back, and can not find it again.

Balarka Sen

11:37 AM

@N3buchadnezzar Hmm.

N3buchadnezzar

Want to avoid invoking the residue theorem, and so forth

Balarka Sen

This sounds like mean value property, actually.

N3buchadnezzar

Yeah, problem is I am using that integral to prove the mean value property

Balarka Sen

Oh dear.

N3buchadnezzar

Or rather Jensen Formula, but that's like the same thing

Balarka Sen

11:39 AM

Huh? No, Jensen is not the same as mean value property. The latter is easier to prove directly.

N3buchadnezzar

Yeah, trying to follow the steps of Alfhors / Rudin atm.

Balarka Sen

I can give you a quick proof with mean value property: note that $f(z) = 1 - z$ vanishes nowhere in a small closed disk around $0$ (w/ radius $< 1$ say). Then you can lift $f$ to the cover $\exp : \Bbb C \to \Bbb C^*$, aka, there is a holomorphic $g$ such that $f(z) = \exp(g(z))$. So $\log |f(z)| = \exp(\Re(g))$. The rest is MVP.

(log|f(0)| = log(1) = 0)

@N3buchadnezzar Rudin proves mean value property using the Jensen's formula? Strange.

Actually there's a subtlety in what I did. Using what I wrote I think I only get $\int_0^{2\pi} \log|1 - \varepsilon e^{i\theta}| d\theta = 0$ for any $\varepsilon < 1$. Because I am not using the whole of unit disk as the domain for $f$.

So I guess one has to take limit as $\varepsilon \to 0$.

I don't like that at all; the integrand does not seem uniform under that limit. But I give up. Also I meant $\epsilon \to 1$.

Akiva Weinberger

11:59 AM

@YassinRany Yeah. They're equivalent

Balarka Sen

Hi, @AkivaW

Jester Tran

hi, does anyone here trade or invest in financial assets?

Null

12:23 PM

can someone clarify please what "$\epsilon'$" is here? math.stackexchange.com/a/1133823/346682 and wether all after that is only "$\epsilon$"

mick

How to type a cyberoot ? Not \cuberoot ??

Null

@mick \sqrt[3]{x}

Yassin Rany

12:54 PM

@AkivaWeinberger thank you again! ))

s.harp

@Null Wikipedia

Aresloom

short question guys is it true, that if <x,y> = 0 with some scalarproduct that <x,y> = 0 with every other scalarproduct too? So orthogonality of vectors depends on the scalar product used?

Alessandro Codenotti

I feel that the answer to a mathematical question is always much harder than the question itself to understand

Null

@s.harp hi, i just think there's a typo. But I don't know. certainly $\epsilon<\epsilon$ makes no sense

s.harp

1:09 PM

@Null there is a typo, he means $0<\epsilon'<\min(a^{1/n},\epsilon)$ not $0<\epsilon<\min...$

Null

@s.harp ah ok, that was my assumption, but I didn't want to edit it on my assumption ;)

Secret

1:23 PM

Is it ever possible to have a poset of some kind such that x < x?

DHMO

@Secret no

Alessandro Codenotti

No @Aresloom, use $\begin{pmatrix} 2 & 1\\1 & 3\end{pmatrix}$ as your inner product, $(1,0)$ and $(1,-1/2)$ will be orthogonal with this inner product but not with the standard one (assuming I didn't do a calculation mistake along the way)

Aresloom

@AlessandroCodenotti thank you very much, you were almost right $(-1/2,1)$ instead of $(1,-1/2)$

euclid

@arctictern hi

Alessandro Codenotti

Woops, I must have mixed them up

Secret

1:34 PM

or one can get an inner product with (1 2) instead of (2 1) as the 1st column

Alessandro Codenotti

That doesn't work, you need a positive definite symmetric bilinear form

Secret

oops (playing with too much relativity cause me to forget the positive and symmetric requirement of inner products)

Ramanujan

In a cube,if we are filling some liquid, then what is increasing with time? 1) total surface area (or) , 2) lateral surface area.

s.harp

@Secret What meaning are you giving to $<$? Usually one uses it for a strict partial order, that is an irreflexive transitive relation. Irreflexivity is the impossibility of $x<x$ for any $x$. Read for example Wiki page

Ramanujan

?

Secret

1:46 PM

Yeah, sorry for the mistake, my mind have been wandering around wildly recently, what I have in mind might be a generalisation of partial ordering, by nuking the reflexive property. I don't even know if such even makes sense

s.harp

@Ramanujan look at the following picture:
http://www.beginnersschool.com/wp-content/uploads/2015/05/conebnw.jpg
what happens to the surface area when you fill it with water?

Balarka Sen

Hi @Alessandro

Ramanujan

@s.harp increase , but I want to know for cube will TSA increases or LSA?

MartianCactus

a square is inscribed in the circle, find the ratios of the areas of the square and circle

Mike Miller

mornin

Balarka Sen

1:51 PM

Hi

MartianCactus

please help me

s.harp

@Ramanujan perhaps the picture was not ideal, imagine a very fat cone, with a wide opening angle

Secret

still procrastinating on chat while I should be proofing stuff. But my brain got so locked up right now due to the hot weather

Mike Miller

it's raining here :(

Ramanujan

@s.harp I want in case of cube not ~~cone~~

s.harp

1:53 PM

ah, i thought for a general shape, well then both the total and the lateral surface area are increasing...

Alessandro Codenotti

Hi @Balarka!

Ramanujan

@s.harp and if i want to know change in surface area with respect to time (differentiating) what formula should I take?

MartianCactus

a square is inscribed in the circle, find the ratios of the areas of the square and circle PLEASE HELP ME

Balarka Sen

What's new, @Alessandro?

Alessandro Codenotti

I got an answerto my topology question from a while ago, but I'll need some time to understand it

s.harp

1:55 PM

@Ramanujan if you are a filling a cube that is lying on a side, how does the height change over time

Secret

square circle problem: hint: how is the radius of the circle relate to the side ofthe square, then work from there

Ramanujan

@s.harp as lateral surface area changes

s.harp

What is the explicit dependence of the height of the water as a function of time

Martin Svanberg

So I'm looking at this question: Y ~ X² where X ~ U[-1, 1]. Find the CDF of Y.

My solution is to look at CDF_Y(t) = P(X² < t) = P(-√t < X < √t) = (1 - P(X < -√t))P(X < √t) = (1 - CDF_X(-√t))(CDF_X(√t)) although this does not give me the right answer, and I can't figure out why my approach is wrong.

MartianCactus

oh ok!

Martin Svanberg

1:59 PM

I've already seen the answer; I just want to know why I'm wrong.

MartianCactus

we take the diameter

and then make it equal to the diagnonal of the square

DHMO

Exercise: Prove that the derivative of x^n with respective to x is nx^(n-1), where n is any real number and x is a strictly positive real number.

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