Mathematics - 2016-12-20 (page 3 of 6)

Kaj Hansen

07:01

@usukidoll, probably the GRE scores. If they really wanted them and didn't get them, it'd be quite difficult to convince them otherwise regardless of anything else. It's not saying anything about you as a person or a student. It's just some unfortunate circumstances.

Balarka Sen

Yeah, I see where you're coming from. But would upvoting every answer stop the ill-formatted low-quality homework questions from coming? I think the community in total wants more homework questions, which is why they are coming up more and more. I also think the community-moderated review - while a great idea - is falling apart.

usukidoll

yeah

again pin this on my undergrad advisor

Kaj Hansen

Absolutely not @BalarkaSen, but it might increase the ratio of good questions to bad

Balarka Sen

I have been serving through questions for two nights in MSE and I see just so damn many ill-formatted questions.

And half of them gets answered. Certainly half of them do not get closed at least.

usukidoll

there should be a How to Master the GRE books

650 is the minimum score

900 is max

Kaj Hansen

07:04

There are @usukidoll

usukidoll

I know but I freak out at some of the questions

I'm like let's do this one second and the next. WTF I'm out!

Kaj Hansen

I definitely couldn't answer all the questions on my GRE

usukidoll

what was your score?

Kaj Hansen

Don't remember honestly. It wasn't very good though. I will need to take it again if I apply to grad school. However, if you're answering at least 70% of the questions, you're doing pretty well.

Balarka Sen

@KajHansen I suppose.

usukidoll

07:06

;_;

Kaj Hansen

@BalarkaSen, that's def. the case if you're answering accurately. I think getting 70% of the questions correct puts you in like the 80th %tile ?

usukidoll

gre is 650 so that's 48%

praxis is 160 or 2/3rds

Kaj Hansen

Do not take my word for that

usukidoll

I already found the gre percentiles from reddit lel

Balarka Sen

I think you pinged the wrong person

usukidoll

07:08

haha

Kaj Hansen

I definitely know a number of people who scored >80%tile. Literally everyone I know says they skip questions here and there. The test covers more material than "normal" people can absorb in all of undergrad.

Calc / multivariable calc / linear algebra is like 65% of it though. Definitely have a good grasp on those three

Balarka Sen

I have to decide what I want to do now

Kaj Hansen

Get a gold badge

Balarka Sen

too hard

Kaj Hansen

lol

Jacksoja

07:15

hello chat

usukidoll

spicy italian footlong

Jacksoja

i got a question about cauchy integral theorem

anyone keen to help me ?

Balarka Sen

just ask; don't ask to ask

Jacksoja

ok

usukidoll

k. what is 1+1+ sandwich?

Jacksoja

07:17

okay, if i have 2 points or more where my function is not analytic

what do i do ?

if i just have 1 point i can deal with it , can cauchy integral formula be generelised?

Balarka Sen

I can't parse that question. What do you want to do?

Jacksoja

ill type the function

Balarka Sen

Oh, I see. Nevermind, the second line cleared it up.

Jacksoja

line integral of z^3 / ( (z-3 ) ( z^2 +1 )

and my region is the circle with radius 2

so i and -i are the problem

Kaj Hansen

Residue theorem or Cauchy integral formula. My answer here might help? math.stackexchange.com/questions/1299127/…

Balarka Sen

07:19

Write it down as (blah)/(z - 3) + (blah)/(z^2 + 1)

Then use Cauchy on each term individually.

Kaj Hansen

Partial fraction decomp might work too yeah

Jacksoja

thank you @BalarkaSen @KajHansen

i will look at that and come back to you

kaj

is that allowed to split the contour ?

Balarka Sen

Sure, why not? As long as the deformation doesn't "go through" the singularities, the result will be the same.

This is exactly what the homotopy-preservation of contour integrals say.

Kaj Hansen

You can always split the contour and make two separate integrals as long as the two resulting contours "sum" to the original

Jacksoja

thanks very clever

we did not do residue yet

Kaj Hansen

07:24

This is analogous to $\displaystyle \int_a^b f(x) \ dx = \int_a^y f(x) \ dx + \int_y^b f(x) \ dx$

For $a < y < b$

Jacksoja

but they told us in the text this is from residue calculation

yes as long as c1 and c2 make the whole contour c

but for the theory

shouldint the region be closed

if we have a circle and we cut it into 2 pieces

we will have 2 open pieces no ?

Balarka Sen

Note that we cut the region taking care of the orientation of the edges obtained after cutting. They are oppositely oriented, so that the sum of the two edges cancel out.

Jacksoja

oh yes yes this is like greens theorem

thanks guys :D

Balarka Sen

I am not sure where Green's theorem comes up though.

Jacksoja

i meant the idea of orientation of curves and such

Kaj Hansen

07:33

Yeah, also analogous to $\displaystyle \int_a^b f(x) \ dx + \int_b^a f(x) \ dx = 0$. @Jacksoja

For "familiar" real integration

Jacksoja

yes yes :)

kaj what do you study ?

Balarka Sen

to hit the point home, endpoint of $[a, y]$ cancels out with the starting point of $[y, b]$ because they are oppositely oriented

Kaj Hansen

"0" cancels out with "0" and becomes "0" :P

Balarka Sen

yup

I don't know if you know this but what's neat is that all these manipulations with contours is essentially doing homology theory

Kaj Hansen

http://math.stackexchange.com/questions/2065647/which-of-the-following-functions-are-uniformly-continuous.

UPBOAT!!!!

I knew about as much as that sentence

Not much further

Balarka Sen

07:40

upboated

Kaj Hansen

Has something to do with all the 1-forms on the interior of the contour being exact or something?

Balarka Sen

aka voted to close

Kaj Hansen

lolol

It's been 5 years since I last thought about that.

Ali Caglayan

hey @Kaj

Balarka Sen

@KajHansen The 1-forms story is the dual of the contour story, yes.

It's a cohomology theory

Kaj Hansen

07:42

Hey there. I haven't gotten around to the matrix multiplication thing yet, but I haven't forgotten about it. Just been preoccupied with other stuff the past couple days

Ali Caglayan

Yes, there is no rush!

Kaj Hansen

Faster algorithms for doing stuff has been ripe with downright beauty in my experience

I mentioned that square-and-multiply for fast exponentiation....that was one of my favorite results from undergrad due to the shear simplicity + cleverness + utility

lol, question put on hold a mere 5 minutes after it being asked

Ali Caglayan

yeah, its good as the idea is picking up some momentum

So I am reading that discrete fourier transform is basically a representation of a group algebra where the group is $S^1$

Secret

businessinsider.com/…

Reminds of hopf filbrations

Ali Caglayan

@KajHansen have you seen the algorithm for evaluating polynomials? It relies on fast polynomial multiplication

Balarka Sen

07:54

@AliCaglayan I forget what DFT is. Fourier series is representing the function algebra on S^1(of smooth functions on S^1, I suppose) inside the Hilbert space of L^2-sequences, yes?

Or something like that

Ali Caglayan

@BalarkaSen Yes

Kaj Hansen

Yeah, something to do with fourier transform as well?

Ali Caglayan

Yes convolution is the same as multiplication

So represent polynomial vectors in circle group algebra, convoulate then inverse FFT you have your multiplication

So thats something like 3n log n multplications which is O(n log n)

Balarka Sen

@KajHansen I think Fourier transform sends the complex-valued function algebra on R to the Hilbert space of l^2-functions but don't quote me on this

I have forgotten these stuff

Kaj Hansen

haha, I never knew it

Balarka Sen

07:58

I know one or two of the classical things, none of these abstract stuff

usukidoll

how do I get a hat?

Balarka Sen

more to upboat

Ali Caglayan

Yeah so I think DFT is a homomorphism $\Bbb C[S^1] \to \oplus_{i=1}^h \Bbb C^{d_i^2}$ for some integers $d_i$

or some nonsense like taht

@KajHansen I think your power algorithm can be bounded by $\log n + \frac{\log n}{\log \log n} + o\left(\frac{\log n}{\log \log n}\right)$

TheGreatDuck

08:21

Vote on meta a bunch to get a vader het like me

Kaj Hansen

like me

Ali Caglayan

have you seen rogue one?

TheGreatDuck

Yeah

Ali Caglayan

what about you @Kaj

TheGreatDuck

I reluctantly saw it

wasnt my pick

Ali Caglayan

08:23

I quite liked it lol

TheGreatDuck

Its star wars, so... meh

Kaj Hansen

I haven't yet, but I want to

TheGreatDuck

They did a good job though

ill give them that

Kaj Hansen

I've seen all the other films a ton, and I played a wide variety of SW video games excessively as a kid :P

usukidoll

anyone know websites with praxis tests?

practice versions

TheGreatDuck

08:25

...

on what?

Ali Caglayan

@KajHansen I think it is definitely worth seeing in that case

usukidoll

the math content knowledge one... 5161

TheGreatDuck

I dont even know what that is

usukidoll

gotta wonder why there's so many gre practice tests to download but so few based on the praxis x.x

TheGreatDuck

What subject in math?

usukidoll

08:27

...

TheGreatDuck

What?

why wont you give a straight answer?

Ali Caglayan

Is praxis a type of gre test?

usukidoll

no

TheGreatDuck

He wants a math practice test

so what subject?

usukidoll

praxis is for teacher license if you're gonna teach secondary education

I'm a she for crying out loud -_-

TheGreatDuck

08:29

And i knew that how exactly?

Ali Caglayan

Well I am out, Take care @kaj

TheGreatDuck

I better sleep

usukidoll

throws a rock

TheGreatDuck

its already 2:30 AM

Ali Caglayan

its 8:30 am

TheGreatDuck

08:31

*2:30

usukidoll

gre is for graduate school
praxis is for teacher licensing
there we go ;)

Kaj Hansen

lol, get good sleep @AliCaglayan

Ali Caglayan

lol @KajHansen have a good one

usukidoll

finally I found a good book that doesn't mix up stuff

TheGreatDuck

Praxis: The anylitical approach

Remember when we all did that?

XD

usukidoll

08:37

ugh some questions I swear smh

DHMO

08:48

hi chat

Kaj Hansen

Hey there

usukidoll

09:10

hi

Brody

Suppose Euclid were roaming the square grid of a city's streets and decide to have the path of his stroll determined by a turn rule over positive integers: arriving at an intersection, turn left if prime; turn right if not prime.

He starts at 2, and continues his way through the square lattice going left or right by the primality of the succeeding integer. What does his path look like up to integer 50? 100? Any patterns?

M'vy

Probably not. Or else we could predict prime number distribution

Secret

09:30

Puzzle: \/ | | = |, move one stick to make the equation correct

Priblem is, I cannot see any way to introduce a + or - that can get the roman numerals to add up

and making =\= sounds too trivial to be the solution

DHMO

@Secret source?

ccorn

Interesting distance geometry question. Have to think about that.

Secret

this issue's NewScientist, it's one of the puzzles, which is taken from the book: 101 bets you will always win

DHMO

newscientist.com/article/…

Puzzle 1

Secret

yeah. The typical solutions of these match stick puzzles is to either exploit roman numerals or make an operator out of it. But here there seems little space to do so

DHMO

09:37

I agree

usukidoll

facedesk

Secret

We have a total of 7 matches in the problem:

I V - I I I = I (needs 10 matches) ---X
I I I - I I = I (needs 9 matches) ---X
I I - I = I (Need 7 matches, but issue there are two slanted matches and you need to have 1 in the starting problem to do that else it will seems quite cheaty)

V I =/= I (need 7 matches, but is this a bit too cheating?)

so basically, the only two solutions I can came up with sounds too cheating to be true

DHMO

V I I = 7 (needs 8 matches)
V N = I (needs 8 matches)

Brody

@Secret this technically works

One can also form $\geqslant$ for a similar equation

Well, this isn't strictly an equation i.e. equality, so this probably breaks rules

DHMO

09:56

I \ I I = I is not true
I / I I = I is not true

s.harp

$\forall \mathrm{I}=\mathrm{I}$

DHMO

wrong syntax

Secret

I don't think that proposition even have a truth value because it is incomplete

DHMO

V / I = I is not true

V - I = I is not true

Brody

$\forall \text{l}\quad\text{l}=\text{l}$ (requires 8 matches)

DHMO

09:59

wrong syntax

Brody

and it doesn't have a separator thing, notation notation

DHMO

you need to have a colon

Brody

yeppers

compensating with a wide space doesn't quite work :P

DHMO

I - II = I is true if you flip the monitor by 180 degrees

Secret

10:18

It is also true without flipping the monitor by 180 degrees if the roman numerals are elements of the group {I,-I}

Adeek

hey @KajHansen

Kaj Hansen

Hey

Adeek

can you help me proof read my elliptic curves project ?

can I send it to you ?

because i have to submit it by wenesday.

Kaj Hansen

Maybe. I don't know much about ECs

[email protected]

Adeek

just the english also my project is pretty self contained.

it explains everything

Okay I will send it to you right no

now *

I sent it @KajHansen

Kaj Hansen

10:30

See it

Pretty pretty Tex

Capitalize title?

Adeek

Yeah

I will capitalize the title

Kaj Hansen

Pohlig-Hellman attack

Adeek

@KajHansen Elliptic curve and Elliptic curve Cryptography

does that sound good ?

okay

Kaj Hansen

"Elliptic Curves and Cryptographic Applications / Applications to Cryptography"?

Adeek

yeah that sounds good

I will change it to "Elliptic Curves and Cryptographic Applications"

Kaj Hansen

10:34

ok

Adeek

@KajHansen I will go to sleep I will hear from you tomorrow ?

Kaj Hansen

Yeah, I'll be on tomorrow. I'll look over it

Adeek

thanks a lot

This is also self contained on ECC so you will find it very nice.

Kaj Hansen

Sounds great; talk to you soon

Adeek

awesome :)

usukidoll

10:50

:/

ughhhhhhhhhhhhhhh

Dennis

I can't get my head around this answer math.stackexchange.com/a/1978182/400692 but I have no rep to comment on it xD.

I want to ask: What exactly is the random variable "X | X>1". How is it defined?

Because as a random variable it has to map e.g. the dice shows 5, as it is in the set of possible outcomes. But what should it map onto?

DHMO

11:06

2 hours ago, by Brody

Suppose Euclid were roaming the square grid of a city's streets and decide to have the path of his stroll determined by a turn rule over positive integers: arriving at an intersection, turn left if prime; turn right if not prime.

2 hours ago, by Brody

He starts at 2, and continues his way through the square lattice going left or right by the primality of the succeeding integer. What does his path look like up to integer 50? 100? Any patterns?

@Brody This is my program used (the language is Scratch)

This is the pattern up to 50:

This is the pattern up to 100:

Dennis

Seems quite nice to me :)

towc

11:27

is there a technical term for the z when representing 2 as \sz.s(s(z)) in lambda calculus? Void parameter?

DHMO

@towc how is it void?

towc

when representing numbers, z is just there to... be?

the semantic purpose of z is to be available and do nothing, as far as number representations are concerned

Akiva Weinberger

12:19

@towc Is "dummy variable" maybe what you're looking for?

Mohamed Ziad

12:46

i have a question, how come that functions with finite discontinuities are integrable at the dicontinuity?

DHMO

because you can cut the graph right at that discontinuity and obtain two trapeziums, one on its left and one on its right

Akiva Weinberger

@MohamedZiad Say $f$ has a discontinuity at $c$. Then:$$\int_a^bf(x)dx=\int_a^{c-\epsilon}f(x)dx+\int_{c-\epsilon}^{c+\epsilon}f(‌x)dx+\int_{c+\epsilon}^bf(x)dx$$

The middle thing on the right can be made as small as you want

Mohamed Ziad

well, okay it makes sense but i can't find a mathematical proof, is there even one? and i also want to know why this notion doesn't contradict with the fact that f(x) is non differentiable at a jump-discontinuity point ?

SteamyRoot

For every partition of the interval you integrate on, you'll have an error around the discontinuity. But as your partition becomes finer, the error will be smaller.

So the error vanishes in the limit.

There is the Riemann-Lebesgue Theorem, not to be confused with the Riemann-Lebesgue Lemma (which is about Fourier stuff)

Balarka Sen

@MohamedZiad What Akiva wrote is a mathematical proof.

SteamyRoot

12:56

It says that any bounded function on a compact interval that is "almost everywhere continuous", is Riemann integrable.

Mohamed Ziad

i was replying on DHMO

Balarka Sen

Oops

DHMO

@MohamedZiad because you cannot find the slope right at the point of discontinuity; but the area right under the point of discontinuity is (obviously) zero

Mohamed Ziad

okay, i see. thanks all

Nagendra

Excuse me! Can anyone please helpe me solve this problem? I can't understand the question at all. Assume that the moon takes exactly 30 days to complete the cycle and also assume that it rises in the east exactly at 6.48 p.m on the first day. Then at what time will it rise on the 4th day?

Null

13:28

smbc-comics.com/comic/2011-05-05

Null

13:43

smbc-comics.com/comic/2011-05-23

Perturbative

Hey everyone, any hints on how to prove the set of algebraic real numbers (over the rationals) is countable?

SteamyRoot

Countable unions of countable unions of finite sets

Null

over the rationals is probably all we are interested in, just saying

Akiva Weinberger

@Perturbative Countably many polynomials (prove!) and finitely many roots per polynomial

(If we're in ZF it matters that the roots can be ordered from least to greatest for every polynomial but in ZFC we don't care)

@Perturbative To show countably many polynomials: Countably many polynomials of degree $n$ (prove!), countably many possible degrees

Perturbative

13:59

@AkivaWeinberger, If we let $p(x)$ denote a polynomial with rational co-efficients, then there are $\mathbb{Q}^n$ possible polynomials correct?

Sophie

$\int_0^\infty \frac{|\sin(x)|}{x}dx=^? \infty$

Perturbative

@AkivaWeinberger, for any polynomial of degree $n$

Sophie

how do I get the question mark to be on top of the equals sign?

SteamyRoot

@Sophie \overset

@Perturbative More or less, since degree $n$ means the first coefficiënt is nonzero

So technically there are $\mathbb{Q}_0 \times \mathbb{Q}^{n-1}$ polynomials of degree $n$.

But there are $\mathbb{Q}^n$ polynomials of degree at most $n$

Perturbative

If so we have $\mathbb{Q^n}$ possible polynomials, each with at most $n$ possible roots, leading to $n \cdot \mathbb{Q^n}$ possible algebraic numbers. We can show that there is a injection then $f : n \mathbb{Q^n} \to \mathbb{Q}$, hence the set of algebraic real numbers must be countable (since $\mathbb{Q} is countable)

@AkivaWeinberger, would that constitute a proof?

@SteamyRoot $\mathbb{Q_0}$ is just $\mathbb{Q} - \{0\}$ correct?

SteamyRoot

14:06

Yup

Balarka Sen

Hi @Alessandro

Alessandro

Call $\mathbb{Q}_n[x]$ the set of polynomials of degree at most $n$ over $\mathbb Q$, it's clear that $\mathbb{Q}=\bigcup_n \mathbb{Q}_n[x]$ so $|\mathbb{Q}|\le\sum_n |\mathbb{Q}_n[x]|\le |\mathbb N||\mathbb Q|=|\mathbb Q|$

Hi @balarka!

If that renders properly it'll be a miracle since I'm on my phone and can't check

Krijn

@Alessandro It does

Perturbative

@Alessandro, Yep it renders properly

@Alessandro, I don't see how $\mathbb{Q} = \bigcup_n \mathbb{Q}_n[x]$, though, i.e. I don't see how the union of a set of polynomials (with rational co-efficients) would equal the set of all rationals. Are you taking the union of only the co-efficients in those polynomials?

Alessandro

That was supposed to be $\mathbb{Q}[x]$ sorry, the set of all polynomials over $\mathbb Q$

Here too: $|\mathbb{Q}[x]|\le\sum_n |\mathbb{Q}_n[x]|\le |\mathbb N||\mathbb Q|=|\mathbb Q|$

SteamyRoot

14:20

If you want everything condensed alot, you could write it as $$\bigcup_{n \in \mathbb{N}} \bigcup_{\vec{q} \in \mathbb{Q}^n} \left\{x \in \mathbb{R} \mid \sum_{i=0}^n q_ix^i = 0\right\}$$

Ah, drat. Now to find the mistake.

Alessandro

That's because $|\mathbb{Q}_n[x]|=|\mathbb{Q}|$

Perturbative

@Alessandro, No problem, so the union of the set of all polynomials over $\mathbb{Q}$ equals $\mathbb{Q}$?

Sophie

Baghdad sounds like "bag that!"

Balarka Sen

what's new

Alessandro

No, their cardinalities are the same @Perturbative

Perturbative

14:27

@Alessandro, Ah okay, phew

Alessandro

Not much @Balarka. I'm slightly confused by GP's definition of a smooth manifold, but I'll ask you about that when I have both my computer and a copy of the book later!

Balarka Sen

Sure thing.

It defines a smooth manifold as a subset of $\Bbb R^n$, not as abstract topological manifolds with extra structure.

Alessandro

Aha! That solves my doubt already! Because they define smooth manifolds as those locally diffeomorphic to open sets of $\mathbb R^n$ but my doubt was how do you make sense of the "diffeomorphism" if the domain is a subspace of a manifold and not of $\mathbb R^n$?

The other definitions I've seen use transition maps between local charts, which makes sense

Balarka Sen

@Alessandro Yeah :)

In the abstract sense, a smooth manifold is just a topological manifold whose transition functions are $C^\infty$.

Alessandro

That seems more natural. But I have a very short proof of Whitney embedding theorem with GP's definition :P

Balarka Sen

14:40

I am skeptic. Whitney embedding theorem does not say a manifold embeds in some Euclidean space, but R^(2n+1) (in fact R^(2n)) where n is the dimension of the manifold.

Alessandro

Fair enough, it was just an attempt at a bad joke

Balarka Sen

Heh, I figured :) I just pointed out that Whitney is still nontrivial using this definition.

The reason working inside R^n is easier is because you can define the tangent space - the crucial construction in smooth manifold topology which does NOT have a good analogue in topological manifolds - with relatively less work.

Alessandro

Well they do prove it in the book if I remember correctly

Balarka Sen

Yep.

i have 8 hats now :d

Alessandro

Do you also have enough heads to wear them all?

Balarka Sen

14:49

If I don't try to wear all of them simultaneously, why not?

i really hope this guy accepts my answer, in which case i might get another one

SteamyRoot

When did Valve take over the Stack Exchange network anyway?

Evinda

15:32

Hello
Does anyone one you have the book S. V. IAblonskii. Introduction to discrete mathematics online?

Ramanujan

:p

@BalarkaSen,how you earned that secret hat?

Balarka Sen

15:50

@Ramanujan which one

SteamyRoot

I imagine the one you're using at the moment

Balarka Sen

It's described here.

s.harp

Is there a way to see via partial integration that $B(x+1,y)=\frac x{x+y} B(x,y)$ where $B$ is the beta function, which can be defined by $B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$

Sophie

use the relationship with the gamma function

I think that can be proven by substitution then parts

r9m

@user91500 hi! long time no see :)

SoumyoB

16:09

hello everyone

can someone tell me what does the triangle pointing left mean in the first statement in the following link:

Semidirect product

In mathematics, specifically in group theory, the concept of a semidirect product is a generalization of a direct product. There are two closely related concepts of semidirect product: an inner semidirect product is a particular way in which a group can be constructed from two subgroups, one of which is a normal subgroup, while an outer semidirect product is a Cartesian product as a set, but with a particular multiplication operation. As with direct products, there is a natural equivalence between inner and outer semidirect products, and both are commonly referred to simply as semidirect products...

"Given a group $G$ with identity element $e$, a subgroup $H$, and a normal subgroup $N\lhd G$..."

Astyx

@SoumyoB It means N is a normal subgroup of G

SoumyoB

@Astyx isn't that redundant though?

Astyx

Might be

It doesn't bug me though

SoumyoB

yes I see now

thanks

Astyx

My pleasure

SoumyoB

16:33

one more problem- if $N \lhd H$ is a normal subgroup of $H$ then can every automorphism, $\phi$ on $N$ be represented in the form $\phi(n) = hnh^{-1}$ for every $n\in N$ and for some $h \in H$?

SteamyRoot

Hmmmm

Suppose $N$ is the center of a group $H$, then it's characteristic and hence normal.

But since it's in the center, that would mean $\phi(n) = n$ for all $n \in N$.

Actually, I'm making this too difficult. Any group is a normal subgroup of itself.

Your question then reduces to "is every automorphism inner?", which is definitely not the case.

But even if $N$ has to be a proper normal subgroup of $H$, it would not be true.

SoumyoB

shit... our prof never taught these things in the Algebra course

@SteamyRoot hold on lemme check out all the definitions of the terms you have used that I'm not acquainted with

SteamyRoot

Characteristic group means "invariant under automorphisms"

And an inner automorphism $i_g$ on $G$ is of the form $h \mapsto ghg^{-1}$.

SoumyoB

ohhh

yes I see your point

so it is a known fact that every automorphism need not be inner right?

SteamyRoot

Well, consider for example the group of integers $(\mathbb{Z},+)$

SoumyoB

16:42

roger

SteamyRoot

It only has two automorphisms, the identity and $x \mapsto -x$.

But since the group is Abelian, the only inner automorphism is the identity

SoumyoB

I see

but how did you know that these are the only 2 automorphisms?

SteamyRoot

I'd recommend reading the answers on this question for that: math.stackexchange.com/questions/156179/…

SoumyoB

oh wait I think I know the answer

thanks buddy

@SteamyRoot a bit off-topic but could you just take a guess about which year of undergraduate I am in assuming that I'm a math major?

SteamyRoot

Hmmm... That would probably really depend on the country (and university)...

SoumyoB

16:48

I'm assuming you're in the US?

SteamyRoot

Nope, Belgium.

SoumyoB

well since you would know best about your own country, assume I'm from Belgium too

in an average university there

SteamyRoot

For my university, I'd guess you're probably in your second semester of second year, or first semester of third year.

(Note: we have 3 years of bachelor)

SoumyoB

shit

I'm actually in my fifth year

SteamyRoot

Well, I am in a very algebra-centered university.

SoumyoB

16:50

god damn our university has done a very shitty job of teaching its undergraduates

Brody

@DHMO Looks the same as what I've drawn out. The use of software was nice. Don't have much time now, so I might discuss it more later

SoumyoB

I'm applying to various European universities right now and I really hope if I get into one I'm not heavily outmatched

@SteamyRoot how good would you say your university is? As in by any chance is it like one of the best ones in maths in your country or something like that?

SteamyRoot

Well, my country barely has any universities at all

I'm at KU Leuven

SoumyoB

https://www.kuleuven.be/english/news/2015/times-higher-education-world-university-rankings
"KU Leuven ranks 35th, which makes it the highest-ranked university in Belgium and convincingly ranks it among the European top."

I'm applying there too then

SteamyRoot

Good luck :D

SoumyoB

17:00

not that I'm very hopeful

my GPA is hardly 3.0 out of 4

@SteamyRoot is combinatorics well-researched over there?

SteamyRoot

Uhhh... We have a rather large statistics department, that's about all I can say :/

Mithlesh Upadhyay

Someone voting down my this post, after when I did mark linked post as duplicate. What is bad with this? http://math.stackexchange.com/questions/1668618/solving-recurrence-relation

I think they voted down on answers too :(.

SoumyoB

well then I'll check out the maths department myself, thank you for the help @SteamyRoot

SteamyRoot

wis.kuleuven.be/english/general-information

Bonus points if you find me, though it's very unlikely

SoumyoB

@SteamyRoot are you testing my stalking skills? ( ͡° ͜ʖ ͡°)

Mithlesh Upadhyay

17:10

2

Q: The language accepted by the nondeterministic pushdown automaton is___.

The language accepted by the nondeterministic pushdown automaton $M= (\{q_0, q_1, q_2\}, \{a, b\}, \{a, b, z\}, δ, q_0, z, \{q_2\})$ with transitions $$δ (q_0, a, z) = \{ (q_1 a), (q_2 λ)\},$$ $$δ (q_1, b, a) = \{ (q_1, b)\},$$ $$δ (q_1, b, b) =\{ (q_1 b)\},$$ $$δ (q_1, a, b) = \{ (q_2, λ)\}$$ i...

Agawa001

17:50

hey hey @r9m

Alessandro

hi @ted

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$Mathematics$ Mathematics