Mathematics

Meow Mix

1:05 AM

Hi

Eric Stucky

g'morning

Ivan Ivanov

1:34 AM

Hello

Is it possible CFE to have terminals like S->P , P->x

I know it can be written as S->x directly but is this wrong?

Zee

2:07 AM

The smell of rotten dreams in the air...

arctic tern

4:05 AM

Call a 2-subset of {1,...,2n} a "duad." Call a collection of n disjoint duads (i.e. a partition of {1,...,2n} into duads) a "syntheme." Call a partition of (2-subsets of {1,...,2n}) into 2n-1 disjoint synthemes a "synthematic total," or "total" for short. How many totals are there, as a function of n? For n=3, the standard answer is 6, which is related to the construction of the (unique) outer automorphism of S_6.

Semiclassical

Hmm. Now I want to make some pun on 'duad-enom'

SBM

@Semiclassical no; intestinal issue?

Semiclassical

I didn't say it would be a good pun.

SBM

@Semiclassical oh

Automorphism?

Semiclassical

shrug

Abhishekstudent

4:29 AM

I am a bit confused with this limt

$$ \lim_{x\to 0}[\frac{tanx}{x}] $

$$ \lim_{x\to 0}[\frac{tanx}{x}] $$

where [] r5epresents gif

or greatest integer function

I am unable to understand how the answer comes out to be 1

I think that the answer should be 0 and 1

@Semiclassical

@SBM

SBM

@Abhishekstudent

At x=0 tan x ≈ x right?

Abhishekstudent

yes

SBM

So what happens if you divide two things that approach the same value and take limits

Abhishekstudent

1

SBM

You get the limit is one, right?

Abhishekstudent

4:36 AM

yes

BUT, WHY ISNT THAT THE CASE WITH SINX

arctic tern

for x close to 0, tan(x)/x is in between 1 and 2, so [tan(x)/x]=1

Abhishekstudent

when I am doing this by replacing tanx with sinx

arctic tern

for x close to 0, sin(x)/x is in between 0 and 1, so [sin(X)/x]=0

BAYMAX

1

Q: Which of the following sets in $\mathbb R^2 $ have the positive lebesgue measure?

Which of the following sets in $\mathbb R^2 $ have the positive lebesgue measure? For twe sets $A,B \subset \mathbb R^2$,$A+B$={$a+b:a\in A,b\in B $} 1)S={$(x,y):x^2 +y^2=1$}. 2)S={$(x,y):x^2 +y^2<1$}. 3)S={$(x,y):x=y$}+{$(x,y):x=-y$}. 4)S={$(x,y):x=y$}+{$(x,y):x=y$}. Since lebesgue measure...

Abhishekstudent

ok

when, x<0 tanx < x

arctic tern

4:38 AM

and both are negative

so tan(x)<x divided by x (when x<0) gives tan(x)/x>1

I mean, tan(x) and x are odd functions, so tan(x)/x is even.
if you know tan(x)/x>1 for x>0, then you know it for x<0 too.

Abhishekstudent

op

ok

Leaky Nun

4:52 AM

Is $S_\omega$ generated by 2 elements? What about $S_{\omega_1}$?

arctic tern

$S_{\omega}$ is uncountable isn't it?

(bijections $\Bbb N\to\Bbb N$, effectively)

unless it's supposed to be permutations with finite support,
in which case no it's not generated by 2 elements

BAYMAX

5:17 AM

I think in the above question I posted in, the answers both have differet answers for 3 rd question ,

also how the measure of a circle is positive

it can be thought of as a curve (closed curve)of finite length

which must have measure zero

what is the Lebsegue measure of a unit circle

$$ ZERO $$

Leaky Nun

@arctictern no finite support; why not?

oh, simply because it is uncountable...

arctic tern

yes

Leaky Nun

5:33 AM

that's... an interesting proof

heh

it's cheating

hmm

Why isn't a mouse an elephant? You could look at body hair or tail color, but the most obvious distinguishing feature is size.

BAYMAX

Now I remember movie SING and Ratatouille

Balarka Sen

5:45 AM

@arctictern lol

Hallo

Hello

hololulu @Daminark

(Kek @Baymax) How's it going everyone?

BAYMAX

gong rough and smooth for me :)

Daminark

5:56 AM

Smooth, but what about analytic?

SBM

Hope things get better soon @BAYMAX

Daminark

Lol for real though good luck!

BAYMAX

thanks

SBM

When is a function analytic?

Balarka Sen

if there is a convergent power series expansion at every point

SBM

6:01 AM

oh

BAYMAX

or may be differentiable at that point and there exists some neighborhood in which the function is differentiable at all the points in that neighbothood !

Balarka Sen

that's not equivalent to being analytic

you're thinking of complex functions and complex differentiability

Daminark

Baymax, I imagine you're thinking of complex differentiable

Balarka Sen

sniped

Daminark

Wai does this happen to me?

BAYMAX

6:04 AM

yeah daminark

Daminark

I think you need to be careful about real analytic and complex analytic

Also @Balarka met Farb today

He's very fun

yes

(Also very bouncy)

heh

so we cannot tell real analytic at a point in terms of differentiability?

like is there only power series criterion?

Daminark

6:07 AM

Yeah, real differentiability is a very weak condition

You don't know as much as smoothness

SBM

smooth? differentiable infinitely everywhere?

arctic tern

differentiable k times does not imply differentiable k+1 times (any more than continuous implies differentiable), and being differentiable k times for all k (smooth) does not imply analytic (for real functions)

Balarka Sen

real differentiability is very not the same as complex differentiability

the latter, despite similarities, is secretly a PDE condition

BAYMAX

he he Cauchy Riemann equation

Zee

It's just simply the result of the CR equation

SBM

6:09 AM

@BalarkaSen Cauchy Riemann equations

Balarka Sen

The complex differentiability => analytic proof is secretly a bootstrapping proof

Daminark

Sniping left and right

@Balarka that's via the Cauchy integral formula, right?

BAYMAX

hits all secretly with a LAW 80 :)

Balarka Sen

@Daminark Yeah

BAYMAX

google.co.in/…

Who watched Wonder Woman?

Zee

6:12 AM

What is the weakest condition on a complex function for it to be complex differentible?

Daminark

It's tricky to say which is weakest

Zee

Well, continues implies diff in complex functions

Daminark

Continuous? That's not really true

Zee

sure its true

Balarka Sen

lol

Daminark

6:14 AM

Wat?

No

arctic tern

sigh

walked into that

Balarka Sen

#trolled

Daminark

is confuzzled

arctic tern

the weakest condition would be those conditions equivalent to the condition, including the condition itself.

any weaker and it wouldn't imply the condition

#helpful

Daminark

That is a true statement

So we'll say analyticity

:P

Zee

6:18 AM

en.m.wikipedia.org/wiki/Looman–Menchoff_theorem

Daminark

The hell?

Zee

I thought the word "complex" implied CR equation

To rephrase , a CR complex function that is continues is analytic

Looman menchof theorem

Daminark

Oh well if we're assuming it satisfies CR that's another story entirely

I thought you were saying any continuous function period, and was like whoa hold on a second wat?

Balarka Sen

complex function just means f : C --> C

SBM

yes

Zee

6:23 AM

I realize that, I just wasn't aware I need to spell every detail

Balarka Sen

you don't, but you do need to use standard terminology

Daminark

Like, your statement as stated was wrong

Zee

Sure but this is an informal discussion

Anyway, my original question holds

Daminark

Yeah but you need to be clear about what you're saying

Zee

How far can we weaken a CR function to imply smoothness?

Daminark

6:28 AM

Hmm

Zee

You can go as far as distributions but I don't if that's the weakest condition

Alessandro Codenotti

Isn't a smooth complex function also analytic?

SBM

should be

Zee

Ya

Alessandro Codenotti

So you can't weaken CR to imply smoothness because that'd imply analyticity which implies CR or am I missing something?

Zee

6:33 AM

Am asking the question assuming CR holds

You can't weaken CR since being differentible (even once) implies CR

Alessandro Codenotti

I don't understand the question, assuming CR you already have smoothness

Zee

Only if the function is continues , the continuity can be weakened, the question is by how much?

Leaky Nun

@arctictern what about with finite support?

arctic tern

@LeakyNun Given any finite number of permutations with finite support, every element in the subgroup they generate will have support contained in the union of the originals' support, so the subgroup generated must be finite.

Leaky Nun

@BAYMAX 1:0, 2:pi, 3:infty, 4:0

@arctictern thanks

Might I say that the 3 group axioms are too strong?

BAYMAX

6:50 AM

why 4th is zero having only the line $y = x$ but 3rd is infity ?

@LeakyNun

Leaky Nun

For example, I only need existence of left-identity and left-inverse

@BAYMAX 3 is the whole plane

arctic tern

@LeakyNun too strong for what?

BAYMAX

oh

Leaky Nun

in other words, span{(1,1),(1,-1)} = R^2, if you are familiar with linear algebra

@arctictern well, itself.

BAYMAX

yes

4th is zero

Leaky Nun

6:52 AM

yes

BAYMAX

?

why he took the sme set added with itself?

Leaky Nun

yes, 4th is 0.

arctic tern

@LeakyNun logical strength of hypotheses is not the only consideration in a definition - it should also match our intuitive idea of what is being defined.

Leaky Nun

@BAYMAX I would have no idea.

arctic tern

groups distill the abstract properties of symmetry groups and permutation groups

BAYMAX

6:53 AM

but the last set means only the line $y = x$

Leaky Nun

@arctictern sure. Can we make it weaker still?

BAYMAX

?

Leaky Nun

@BAYMAX yes.

arctic tern

@LeakyNun I recall reading you can ensure a binary operation defines a group with a single axiom.

BAYMAX

so measure of the line being zero area so measure is zero

Leaky Nun

6:54 AM

What is the dimension of the trivial vector space?

@arctictern yes, that ab' in G for each a and b in G.

@BAYMAX yes

BAYMAX

Thanks@leaky

Daminark

Well, this is ensuring that a given set is a subgroup, but you would still need to know that there's some identity lurking somewhere, and that inverses exist

arctic tern

@LeakyNun uh, no, that doesn't imply associativity

Daminark

So I'm not sure that this is really saying less

Also yeah associativity :P

Leaky Nun

@arctictern then what is it?

arctic tern

6:56 AM

@LeakyNun you mean what is the algebraic structure with a binary operation having identity and two-sided inverses?

Leaky Nun

@arctictern no, I mean what is the single axiom

arctic tern

trying to find it again

Leaky Nun

Is there any result regarding the solutions of x^2 + y^2 = 1 in a ring, say Z_m?

arctic tern

break it into parts with CRT, lift from Z_p with Hensel maybe

@LeakyNun If a magma M satisfies a=b((((dd)a)c)(((dd)b)c)) for all a,b,c,d then a((bb)b) is a group operation on M.

(And there is a converse, so it is an equivalent characterization of groups.)

Leaky Nun

7:16 AM

@arctictern wat

arctic tern

yeah, exactly

SoumyoB

7:35 AM

soundcloud.com/soumyo-biswas/hot-alkali-metal-beginning-only

^I just made this song

anyone care to tell me if it has any errors?

Samuel Schlesinger

7:55 AM

hello, I'm curious if anybody is aware of any first order statements one can make (say about graphs, but any vocabulary suffices) which is a tautology in all finite models but is not in all infinite models

for instance "every DAG has a sink"

this is clearly true in all finite DAGs but not in infinite ones

user147690

8:22 AM

@SoumyoB Sounds good to me. Everything felt like it fit together well, so I imagine there were no 'errors' (although I'm not entirely sure in regard to that, what I should have been looking for).

Mats Granvik

8:43 AM

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$Mathematics$ Mathematics