Mathematics - 2016-10-08 (page 1 of 5)

WDUK

00:08

when simplifying a fraction with a surd in the denominator, is it generally easier to rationalise the denominator first?

or should i do that last

arctic tern

@WDUK if your first step isn't multiplying top and bottom by the bottom's conjugate, what other step are you considering?

WDUK

simplifying just the numerator first

since it has some brackets i want to clear

its pretty ugly in general as ive not done much with powers that are fractions

arctic tern

you need to multiply things up top and multiply things down below. since those two steps are independent of each other, it makes no difference which you do first

Danu

This is some of the coolest stuff I've seen in a while.

Dat 90 MB picture is worth downloading.

WDUK

looks bit like fractals

Danu

00:17

Exactly

The article goes into that.

WDUK

00:46

is (5x)^2 = 5^2 + x^2 or 5^2 * x^2

im assuming + because otherwise it would be ^4

arctic tern

@WDUK what does blah^2 mean?

WDUK

power of 2

arctic tern

which means?

WDUK

squared

arctic tern

which means?

0celo7

00:47

power of two

WDUK

multipled by itself

arctic tern

there we go

WDUK

i know that lol

arctic tern

so (5x)^2 means 5x times 5x

WDUK

i think you misread

5^2 + x^2
or
5^2 * x^2

arctic tern

00:48

(5x)^2=(5x)(5x)=(5*5)(x*x)=5^2 x^2

@WDUK no, I haven't misread anything. I'm teaching you how to think.

WDUK

okay so its multiple

arctic tern

going back to your "otherwise it would be ^4" comment, it would be (what exactly)^4 ?

there is a^2 a^2 = a^(2+2) = a^4, because both are just a*a*a*a

but this is not a^2 a^2, it is more like a^2 b^2, because the 5 and x in 5^2 x^2 are not the same

WDUK

ah okay

that explains my mistake in my calculation then

Forever Mozart

01:14

too much caffeine I'm about to jump through the roof

WDUK

im about to flip my table from frustration

Forever Mozart

lol

I know the feeling

WDUK

well damn i got it wrong again

0celo7

01:45

@ForeverMozart halp

I'm bad at topology

Forever Mozart

what is the problem

0celo7

$X$ is a metric space and $\{K_i\}$ is a countable number of compact sets such that $K_{i+1}\supset K_i$ and $\bigcup K_i=X$, then I conjecture any compact $C\subset X$ is contained in some $K_j$.

Forever Mozart

It is not true

0celo7

Yeah?

Forever Mozart

oh wait they are increasing...

oh it is true

do you see why?

@0celo7

0celo7

01:53

No, that's why I'm asking...

I tried a contrapositive proof

A direct proof

Forever Mozart

Suppose for each $n$ there exists $x_n\in C\setminus K_n$.

Then $(x_n)$ has a subsequence that converges to a point in $C$.

Say $p$.

There exists $N$ such that $p\in K_N$.

0celo7

@ForeverMozart Yes, by compactness.

@ForeverMozart Ah.

Wait, no, why is that an issue?

Forever Mozart

oh a little more

you can make each $x_n$ miss an open set around $K_n$.

and then you will get a contradiction

0celo7

uhhh

Forever Mozart

you have to be more careful about construction

what you need is that for each $n$ there is an open set around $K_n$ that contains only finitely many members of the sequence $(x_n)$.

but I think you can do this recursively

0celo7

02:02

Jesus

@ForeverMozart Is there not something easier that uses metricity directly?

Forever Mozart

the contradiction will be that $p\in K_N$ but has a neighborhood containing only finitely many members of the subsequence

0celo7

I have a $\sigma$-compact metric space, and I can prove something on each set in the compact covering, but I need it to work on any compact subset

Forever Mozart

oh I have a beautiful proof

0celo7

Not nets please :P

Forever Mozart

no

Ok,

you can assume that there is no $n$ such that $K_n=X$.

0celo7

02:08

How come?

Forever Mozart

cause otherwise $C\subseteq K_n$

0celo7

Oh, right.

Still going for a contradiction?

Forever Mozart

no

this is direct

0celo7

Hmm, ok.

Forever Mozart

So you can assume that $K_n\subsetneq K_{n+1}$ for each $n$

0celo7

02:09

Ok.

Forever Mozart

(there will be a subsequence where this is true)

right?

Ok for each $n$ pick $x_n\in K_{n+1}\setminus K_n$.

0celo7

@ForeverMozart Yes I agree.

Especially given the way I'm constructing the $K_n$s.

@ForeverMozart Ok.

Forever Mozart

Now define a function

$f(x_n)=n$

0celo7

@ForeverMozart Continuous somehow?

Forever Mozart

wait I think its the same problem

let's see

ok here's my final try

0celo7

02:22

@ForeverMozart Does it help if the space is separable?

Forever Mozart

I don't think so

0celo7

countable dense subset separable

Forever Mozart

pick $x_n\in K_{n+1}\setminus K_n$ such that there is an open set $V$ with $x_n\in V\subseteq \overline V\subseteq X\setminus K_n$.

0celo7

Ok

Forever Mozart

omg

Suppose for each $n$ there exists $x_n\in C\setminus K_n$.

0celo7

02:30

yes

Forever Mozart

lol

0celo7

dude

@ForeverMozart Is it really easy?

Forever Mozart

Is this true:

For each $n$ there is an open set $U$ containing $K_n$ such that $C\setminus U$ is infinite?

0celo7

No, $X$ could be finite.

Er, well

I guess it's not finite

Forever Mozart

WAIT

0celo7

02:38

If it's finite it's compact (maybe), and I'm assuming $X$ noncompact

@ForeverMozart I'm waiting...

Forever Mozart

Well

pick $x_0\in C\setminus K_0$.

There exists $n_0$ such that $x_0\in K_{n_0}$

pick $x_1\in C\setminus K_{n_0}$.

Enjoys Math

0

Q: The smallest grammar of a formal language string represents certain finite group presentations.

Let one possible smallest grammar of $S = aaaaaa $ be: $S \to BB \\ B \to aaa $ Symbolically transform the rules of the smallest grammar like so: $$ S \to 1 \\ B \to b \\ a \to a $$ To yield presentation equations: $$ 1 = b^2 \\ b = a^3 $$ Intuitively I think that $G = \{1, a, b, a^2 b, a^2,...

Forever Mozart

Let $U_0$ be open with $K_0\subseteq U_0$ and $x_0\notin U_0$

pick $x_2\in C\setminus K_{n_1}$

Enjoys Math

02:53

group theory applied to smallest grammar unsolved problem in CS

@BalarkaSen

Forever Mozart

@0celo7 I have solution but it is not easy to write

0celo7

Jesus

Forever Mozart

it requires a recursive construction

0celo7

I need this for an "easy exercise"

Are you sure separability does not help?

Forever Mozart

it is easy

if you see it

0celo7

02:54

This doesn't seem like it should be hard

Forever Mozart

just hard to write

Enjoys Math

that's what she said

Forever Mozart

yeah I got the picture

let me continue from above

Enjoys Math

I'm not convinced

j/k

:D

Forever Mozart

OMG

@0celo7 it is not true

LOL

DUH

0celo7

03:01

dude

Forever Mozart

LOL

HAHAHAHAHAHAHA

0celo7

:(

Forever Mozart

Let $X=\{0\}\cup\{1/n:n\in\mathbb N\}$.

Let $K_n=\{0\}\cup\{1/k:k\leq n\}$.

$X$ is compact

but it is not contained in any of the $K$'s

0celo7

what's the topology?

subspace from $\Bbb R$?

Forever Mozart

subspace of $\mathbb R$

0celo7

03:04

is that space separable

Forever Mozart

of course

0celo7

right, it's countable.

duh

Well holy crap my homework just got wayyyy harder

Forever Mozart

why?

0celo7

@ForeverMozart I'm trying to prove Arzela-Ascoli on $\sigma$-compact metric spaces.

So I have a countable covering by compacts $K_n$

Forever Mozart

hmm

yes

0celo7

03:05

I can get a convergent subsequence on each $K_n$

and I can get a subsequence that converges uniformly on any $K_n$ by a diagonal argument

But I'm to show that I can make the convergence uniform on ANY compact subset of $X$

So, I figured any compact subset of $X$ should be contained in one of the $K_n$s

that would give me the necessary convergence

@ForeverMozart What if $X$ is noncompact

can you still get a counterexample?

Forever Mozart

yes

0celo7

$\{0\}\cup\{1/n\mid n\in\Bbb N\}\cup\{n\mid n\in\Bbb N\}$?

Forever Mozart

yes

0celo7

Then take $K_n=\{1/k\mid k\le n\}\cup\{k\mid k\le n\}$

Alright I'll write up the argument as much as I can, then send him (my advisor) this with your counterexample

@ForeverMozart should I credit you?

Forever Mozart

no its just the simple convergent sequence example

0celo7

03:14

k

thanks

Forever Mozart

that's the first thing I thought of, but I didn't trust it initially

0celo7

I thought of this too :/

I ruled it out for some reason while eating noodles

0celo7

03:53

@ForeverMozart Ok, the proof can be saved.

But ONCE AGAIN

the hint he gave was wrong

I'm getting kinda pissed at him

Forever Mozart

he told you to prove that thing?

@0

@0celo7

0celo7

@ForeverMozart No, he told me to prove AA on $\sigma$-compact metric by proving it on the compact cover, the using a diagonal to find a subsequence that works everywhere

but one can use separability directly and that works

(I think)

I'll write up the details tomorrow

Forever Mozart

maybe he meant an cover of compact neighborhoods

0celo7

@ForeverMozart what am I covering?

Forever Mozart

i dont know

0celo7

04:04

I have two days until this is due, I have to sleep on it

Forever Mozart

i am over-caffeinated

my brain is sort of fried

user228700

Hi :-) Do u know the formula for finding the sum of squares of natural numbers starting from a number that is not 1?

Forever Mozart

its something like $n(2n+1)(n+1)/6$

if $n=1$ you get 1*3*2/6=1

if n=2 you get 2*5*3/6=5=1+4

if n=3 you get 3*7*4/6=14=1+4+9

so I think thats it

0celo7

@Kaumudi yes

get the formula going from 1 to your number

then from 1 to k

subtract

user228700

@0celo7 Ah, yes! OK, thanks!

0celo7

04:40

@ForeverMozart you there?

Your counterexample is wrong

The Kn do not even cover

{0} isn't contained in any of them, so now could you cover?

Oh I'm stupid

Ignore.

user228700

I've got a small question regarding statistics; I've been given the number of goals and the corresponding number of matches for a team A. The variance of another team B is given. To find which team is more consistent, am I to compare their variances? How does this work..?

Ralf17

08:18

Hello!

I have a question:

How can I show that it is impossible to add two mixtures of candy worth $4 per pound and $5 per pound to obtain a final mixture worth $6 per pound?

Danu

write out the price of a general mixture

with a% of the 4 per pound and b% of the 5 per pound

Secret

09:29

Catholic semigroup

In mathematics, a catholic semigroup is a semigroup in which no two distinct elements have the same set of inverses. The terminology was introduced by B. M. Schein in a paper published in 1979. Every catholic semigroup either is a regular semigroup or has precisely one element that is not regular. The semigroup of all partial transformations of a set is a catholic semigroup. It follows that every semigroup is embeddable in a catholic semigroup. But the full transformation semigroup on a set is not catholic unless the set is a singleton set. Regular catholic semigroups are both left and righ...

Why are they giving rather arbitrary names to subtypes of algebraic structures...?

Balarka Sen

@Adeek Did you figure out your linear algebra problem?

user228700

09:47

Hello :-)

Secret

Special classes of semigroups

In mathematics, a semigroup is a nonempty set together with an associative binary operation. A special class of semigroups is a class of semigroups satisfying additional properties or conditions. Thus the class of commutative semigroups consists of all those semigroups in which the binary operation satisfies the commutativity property that ab = ba for all elements a and b in the semigroup. The class of finite semigroups consists of those semigroups for which the underlying set has finite cardinality. Members of the class of Brandt semigroups are required to satisfy not just one condition but a...

Too many stuff to copy, I'm just going to make a footnote isntead

Moving on to monoids

user228700

I've a small doubt regarding statistics: when is the best situation to use the mean, median and mode to represent the central tendency of a set of data? I know that if the set has extreme values, it is meaningless to report the mean of the data as the central tendency. What are the situations in which mode and median must be used, then?

Secret

Generally, median is the best because it is not sensitive to extreme values like mean nor asymmetric distributions which there may be more than one mode

median cannot be skewed very badly unless half of the data is filled with extreme values

user228700

Riight, OK...

user228700

I was reading up on this and came across this:

user228700

09:56

"You will notice, however, that the mean is not often one of the actual values that you have observed in your data set. However, one of its important properties is that it minimises error in the prediction of any one value in your data set. That is, it is the value that produces the lowest amount of error from all other values in the data set."

user228700

What do they mean by "It minimises error in the prediction of any one value in the data set"?

Balarka Sen

I like mean for that reason too.

The mean-square deviation around mean (aka standard deviation) is the least.

user228700

Oh, okay...

Ramanujan

0

Q: Equation for Distance of the Straight line from the Origin

By reduction of the equation $ax + by + c = 0 $ of a straight line to the normal form , we get $$\left(\frac{-a}{\sqrt{a^2 + b^2}}\right)x + \left(\frac{-b}{\sqrt{a^2 + b^2}}\right)y = \frac{c}{\sqrt{a^2 + b^2}}$$ And, $$p= \frac{∣c∣}{\sqrt{a^2 + b^2}}$$ And my textbook says that $p$ is the dist...

geometry

user228700

What is the definition of the term "outliers" when speaking about data wrt statistics..?

user228700

09:59

(I've looked but I haven't found anything that makes sense yet :/)

Secret

There are many definitions, one of these that is commonly used in analytical chemistry is a data point that lie outside of 3 standard deviations from the mean

Balarka Sen

I don't think there's a rigorous definition. Look at the data {1, 1.5, 2, 2.3, 3, 156}

156 is an outlier.

user228700

OK, I will accept that there is no rigorous definition...

Secret

In practice, what is and is not an outlier depend on the interpretation of the data wrt to some hypothesis

and interpretations are not always objective...

Balarka Sen

But @Secret's definition sounds like a manifestation of the 68-95-99.7 rule.

user228700

10:02

@Secret When u say "lies outside of 3 standard deviations from the mean", do u mean $\mu+3$?

Balarka Sen

@Kaumudi No, $\mu \pm 3\sigma$.

$\sigma$ is the standard deviation of the data.

user228700

Oh, okay...

user228700

@BalarkaSen Googling this rule...Ah, I see.

Secret

More sophisticated ways to detect outliers include student t tests and other statistical checks

these are commonly used in analytical chemistry

user228700

Oh, do u practise analytical chemistry?

Danu

10:05

Hi @Balarka

Balarka Sen

Hi @Danu.

Danu

I just read the "Hard Lefschetz theorem"

Secret

Well, I have done a course on that in my 2nd and 3rd year undergrad, as well a month of part time experience in a food safety company

Danu

But I thought the previous "Lefschetz theorem on (1,1)-clases" was harder :p

Balarka Sen

@Danu cool, eh?

user228700

10:06

@Secret I see. OK, thanks, guys :-)

Danu

@BalarkaSen I don't know...

I'm not sure how to use it

Do you know how to use it?

Balarka Sen

The weaker one says that cohomology and homotopy of a smooth complex variety is determined by a hyperplane section of it. That can be used to compute homology groups of varieties!

That's interesting in itself, to me.

Danu

That's not at all what it says to me :P

Balarka Sen

Odd, because that's how I know the weak Lefschetz theorem stated.

Danu

What is the "weak" one?

The one on (1,1)-classes?

Balarka Sen

10:13

The inclusion of a hyperplane section of an n-dimensional variety inside the variety is a homology isomorphism upto dimension n/2 IIRC.

I don't know about (1, 1)-classes

Danu

That's not at all the phrasing of any theorem I'm seeing in here :P

Balarka Sen

Weird.

Danu

Just check out 3.3.13... Perhaps I'm just not realizing what it "means"?

Balarka Sen

What's $L^{n-k}$?

Danu

The Lefschetz operator (wedging with $\omega$, the Kahler form) applied $n-k$ times.

Secret

10:18

Monoid

In abstract algebra, a branch of mathematics, a monoid is an algebraic structure with a single associative binary operation and an identity element. Monoids are studied in semigroup theory as they are semigroups with identity. Monoids occur in several branches of mathematics; for instance, they can be regarded as categories with a single object. Thus, they capture the idea of function composition within a set. Monoids are also commonly used in computer science, both in its foundational aspects and in practical programming. The set of strings built from a given set of characters is a free monoid...

How is that fit the idea of a generator if S does not really reproduce M after applying the monoid binary operator?

Balarka Sen

@Danu I haven't seen this version before. But it still sounds interesting: it gives obstructions for smooth manifolds of even dimension being complex algebraic varieties.

That's one of the central questions out there: how restricted is the topology of complex algebraic varieties?

Danu

Why does it give that?

Note that this whole story is only for compact Kahler manifolds, but okay

Balarka Sen

Not all smooth manifolds satisfy $H^k \cong H^{2n-k}$ for all $k \leq n$...

Danu

okay

Balarka Sen

Philosophically, I guess it says topology (or at least cohomology) of complex varieties is determined by it's topology in half it's dimension.

@Danu Here's an interesting problem, since we're discussing this. What are the possible fundamental groups of smooth complex algebraic varieties?

It's known that no free groups come up as such, I believe.

Not even $\Bbb Z$.

Danu

10:28

My supervisor has some papers on this.

Balarka Sen

May I know whom your supervisor is?

Danu

Dieter Kotschick

arxiv.org/find/math/1/au:+Kotschick_D/0/1/0/all/0/1

He has some papers on Kahler groups.

Balarka Sen

Are Kahler groups precisely the complex algebraic variety groups?

Danu

I'd assume Kahler groups are fundamental groups of Kahler manifolds.

But I don't know anything more :P

Balarka Sen

Yes, that's true.

But I don't know if Kahler manifolds are complex varieties, and vice versa.

Danu

10:31

Complex varieties is more general than complex manifolds, right.

They are allowed non-smooth points

Balarka Sen

Assume smoothness.

Danu

So smooth complex varieties are AFAIK the same thing as complex manifolds

Kahler is of course an additional condition

Balarka Sen

Yeah

Danu

I don't know if that restricts the fundamental group---I assume it does.

Balarka Sen

I am forgetting who's theorem the smooth complex variety <=> complex manifold thing is.

Chow's theorem?

Yeah, google verifies.

Danu

10:35

Ehh... I don't know. In my book, it follows straight from the definition of a complex variety...

Balarka Sen

How do you prove that every complex manifold is a complex variety???

I think you're thinking of the other direction.

Which is of course obvious.

Danu

So in fact Huybrechts only defines subvarieties

Balarka Sen

Sure, how do you prove every complex manifold appears as a complex subvariety of CP^n for some n?

Danu

I don't think that's true---isn't that the definition of projective?

Balarka Sen

It is true, that's what Chow's theorem says.

Every complex submanifold of CP^n is a subvariety.

Danu

10:38

So... how do you get noncompact things? as open subsets?

Balarka Sen

Oh, I implicitly assumed compactness.

Danu

lol

:D

Balarka Sen

:P

Secret

"A subset S of M is said to be a generator of M if M is the smallest set containing S that is closed under the monoid operation, or equivalently M is the result of applying the finitary closure operator to S"

@danu Is $S$ closed under the monoid operation of $M$, or only $M$ is and applying the monoid operations on $S$ gives $M$?

Danu

Anyways, I still don't know what you're on about exactly---in Huybrechts, he talks a lot about projective complex manifolds, which are the ones that embed into (a closed subset of) $\Bbb P^n$.

If all compact manifolds were projective I'm pretty sure he'd have said so.

Balarka Sen

10:42

That should be just Whitney, @Danu.

Ramanujan

Hi nerds

Danu

@BalarkaSen What should be?

Balarka Sen

Embed into some $\Bbb C^n$, projectivize, and you have an embedding into $\Bbb P^n$.

Danu

You certainly can't embed stuff holomorphically into $\Bbb C^n$, can you?

You know how holomorphic maps are constant...

Balarka Sen

Oh, holomorphically, nah.

Danu

10:43

lol

But that's the whole point of complex geometry

Balarka Sen

But why should that matter?

Danu

So when I talk about projective

I mean biholomorphic to a subset of $\Bbb P^n$

Balarka Sen

I am confused how that's relevant to what I am saying, about these being algebraic varieties.

Danu

1. I don't know what an "algebraic variety" is to you. 2. I know what an analytic subvariety is to me---it's something locally defined by the zero set of finitely many holomorphic functions. 3. It follows from that definition that smooth analytic subvarieties are complex submanifolds, immediately.

Balarka Sen

I was going to pick on your for that.

Danu

10:46

:D

Now, from my definitions

Balarka Sen

By algebraic variety, I mean a subset of P^n cut out by a bunch of homogeneous polynomials.

Danu

Obviously all Kahler things are complex varieties

@BalarkaSen So you mean something you cut out of $\Bbb C^n$ by hom. poly's, then projectivize?

Balarka Sen

3 I agree with and is obvious. I am saying the converse should be true.

@Danu Sure, if you want.

Danu

@BalarkaSen That is also a trivial statement in my definition :P But I guess it doesn't coincide with "algebraic variety"

Balarka Sen

3 is trivial, nobody's bothering you with that.

Danu

10:49

@BalarkaSen Ok. I know that under the name "complete intersection"

No, the converse of 3---that a complex submanifold is exactly a smooth analytic subvariety

Balarka Sen

How is that trivial?

Sorry, that's not the converse.

Danu

It's jsut the definition of analytic subvariety for me

Okay

Balarka Sen

The converse is algebraic.

I never said analytic.

Danu

I know

And I kept on repeating that I didn't use your definition :P

Balarka Sen

Huh?

Danu

10:50

2 mins ago, by Danu

@BalarkaSen That is also a trivial statement in my definition :P But I guess it doesn't coincide with "algebraic variety"

Balarka Sen

Your definition of "algebraic subvariety" is "analytic subvariety"? double huh?

Danu

Never mind

Balarka Sen

@Danu Also, I am a bit skeptic now that this is equivalent to what I said. Projectivizing may result in more zeroes, I think I have seen that before.

Danu

@BalarkaSen OK

Could be

Balarka Sen

Lots of subtleties :(

Danu

10:55

That's why you stop thinking about varieties and stay thinking about manifolds :D

Balarka Sen

Here

Equation of projective closure need not be the homogenized polynomials which cut out the affine variety.

@Danu Fair enough.

Danu

I have a very naive question

Why does one often do things over finite fields in algebraic geometry?

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