2:17 PM
Hi. I have trouble in quardratics
I think minimum should be $min.(y)=\dfrac{min. Of x^2 -x+1}{max. x^2 + x+1}$
But I am getting wrong x_x
@LeakyNun

@Fawad when one is minimum, the other one doesn't have to be maximum, meaning that your min (y) is unattainable

@LeakyNun so how to get minimum of y?

y=(x^2-x+1)/(x^2+x+1)
(x^2+x+1)y = x^2-x+1
(y-1)x^2 + (2y)x + (y-1) = 0
find the range of y such that the quadratic equation has solution.

Or given x Is real then $\Delta\gt 0$
y(min.)=1/2 :D thanks

why can't delta be 0?

2:25 PM
@LeakyNun it can be.

3:03 PM
@ShaVuklia ooo, waves

@Semiclassical She posted her physics question above, if you can help.

2 hours ago, by Sha Vuklia
I don’t understand one thing about the superposition of waves. My book says the following: Say we have $f_1(t)=A_1\cos\omega_1t+A_2\cos\omega_2t$ and $f_2=0$. Then the infinite wave that travels in the positive $z$-direction is given by: $y(z,t)=A_1\cos(\omega_1t-k_1z)+A_2\cos(\omega_2t-k_2z)$. In the same way, when $f_1(t)=f_2(t)=A\cos\omega t$, then we get $y(z,t)=A\cos(\omega t-kz)+A\cos(\omega t+kz)$.

I can see the pattern here; so we write $\pm k_iz$ in the argument with $\omega_it$, and apparently we then take the sum, with equal amplitude, that travel in the opposite direction. My q

Well, she also said she'd asked a question. so I figured her question might change a bit once she got back here.

I think Doubly Modern would be a good name to juxtapose Semiclassical.

lol
hypermodern

3:15 PM
@Jasper dui lian lol

(which is actually a school of chess play, I think)
which ironically means that hypermodernism in chess is about a hundred years old now...

modern means nothing

non-math question: What's an idiom for someone who perseveres in pursuit of a bad idea?

any time is modern from that time's frame of reference

3:20 PM
yeah, that works pretty well
when it comes to historical stuff, I think modern tends to be associated with the first half of the twentieth century?

@Semiclassical how are people going to call that after a thousand years
change is the only constant

meh.

@LeakyNun yelloha

god dag

I find it interesting enough how and why people use the word now

3:22 PM
godafton =p

@Semiclassical Maybe I would say myopic or short-sighted for that.

Ehm well leaky , I still dont get group actions

jag ater et apple

But am a bit better now
ett *
äter*

lol

3:23 PM
:D
haha

then study examples of group actions

@KasmirKhaan I think to get something one has to think about it for a long time outside chat rooms.

@Jasper been 6 hours on library ><

Is any example given?

but spent only 1 hour on group actions
let me tell you what I got so far

3:24 PM
@KasmirKhaan I see. But seriously, we cannot expect to understand everything quickly. But 6 hours is quite a long time indeed.

g_1(g_2 a )
g_2 a is in the set A

Kasmir, definitions don't make you understand; examples do.

so here g_1 is acting on it

definitions are the abstraction of the properties demonstrated in an example

Yes yes let me just finish what I igot =p
ermm good point ><

3:25 PM
ok

(g_1g_2) is an element of G since G is a group
(g_1g_2) a
is also an element of G acting on an element a in the set A
those are equal , the action ( . ) is not binary
fails the closure , that was not in the book why it is not binary but I figured that one out ><

the action is not binary?

@Jasper here's a terrible mixed metaphor: "Stubborn as a shortsighted mule flogging a dead horse."

it is not

why not?

3:27 PM
well
that is what is written on the book but let me give you what I understood

so what is your point now?

my point on what?

let me give you what I understood

on all this

semi how dare you -.-
well my point is that , the group action still not clear to me

3:29 PM
actually, that was me saying that Leaky should let you finish your point :)

Oh thanks then =p

puts semi on ignore

what I got so far is that , we have a set A
and a group G
elements of the group acting on this set
what is the idea behind that ? I dont knowyet
there are few important facts on this
i) sigma_g is a permutation of A

ping me when you finish

where sigma_g = g.a
@LeakyNun well if you put it that way am done =P

3:31 PM
heh

@LeakyNun can you give me an example ?

is there any example given?

yes but they are not very clear

so what is the example given?

for any nonempty set A the symmetric group S_A acts on by sigma . a = sigma (a) for all sigma in S_A . the assoociated permutation representation is the identity map from S_A to itself

3:34 PM
ugh, more concrete?

there are but long ones wont look pretty if i type them here

alright

do you still have that link of the book ?

are you familiar with D6?

yes
you use D_2n right?

3:35 PM
right

then yes

consider D6 acting on the three vertices

okay but I dont get what "acting " is
like intuivtivly

well
ρ1 acts on 1 to give 2
it carries 1 to 2

okay hmm ( 123) is what we get when we multiply by rho
lets call the rotation "r "
and reflections "s"

3:39 PM
what do you mean (123) is what we get?

since D_6 is isomorphic to S_3

alright

we use S_3 because of easy type set =p

(123) acts on 1 to give 2
r * 1 = 2

okay

3:40 PM
r^2 * 1 = 3

eh, I'd be careful with that notation. (1) in there is not a permutation

ok?

sure, that works.

Hmm yes r^2 will send the elemnt 1 to 3

and r * 2 = 3
@AlessandroCodenotti buongiorno

3:42 PM
Yes so far so good

and r * (r * 1) = 3

Yes

so we have demonstrated associativity
now is it true that ga=a implies g=e?

Hmm dont know

use this example

3:44 PM
one second then :D
r^3 (1) = 1
but r^3 = e so that dont work
Not really sure here because we doing the action on the left only

continue exploring
what do you mean action on the left?

we multiply by g on the left
ga

is there action on the right?

I think yes =p

continue exploring.

3:50 PM
from ga =a , we can get (a'g) a =1

I mean exploring the xample
and there is no action on the right
and stop using what is obvious to you
without justifying it
the rules are given, not a dot more, not a dot less.
1. g(ha) = (gh)a
2. ea = a
a is an element of a set.
a' is hardly meaningful.
D6 has 6 elements
you have only tried 3 and gave up

am still trying
there cannot be such element
ga=a in D_6 at least

show me what you have tried

each product is a permutation of the vertices

and?

3:57 PM
well D_6 wont be a group if we can find other elment than the identity such that ga =a

what is s1 * 1?

and from the definiton we have that 1a =a
what is s ?

reflection

the notation comfuses me now
s1 should be (23) right?

right

3:59 PM
(23)*1 = (23)

hmm?

or what is it am missing
1 is the identity ?

no, 1 is the vertex labelled 1
this is group action

one way to keep it straight would be to write the elements of A differently than the elements of G
e.g. (23)[1]=[1]

@Semiclassical hey!
alright, (123)[1] = [2]; (132)[1] = [3]

4:01 PM
You mean the permutation that changes 2 and 3 and keeps 1 fixed

you could instead do |1> but then you'd get people calling you a physicist :P

it does not do anything to 1

@Semiclassical I mean, hey you gave it away

ah.
oh well.

so this is not a normal product

4:02 PM
????

from what I understood now
it is a bit odd
why would we have an element from S_3 acting on a vertex of A

but you accepted that (123)[1] = [2] and (132)[1] = [3] @_@

Hmm yes I did that but after a bit thinking realizing am doing different thing
hmm let me keep thinking
so we have a set A = {1,2,3}
we doing this magic operation on the vertices and we permuting them
(12) [3] = [3]
and (123)[1] = [2]
what kind of black magic is this -.-
is the operation on the set not beteen groups?

a group action is a function that takes an element of a group and an element of a set and gives an element of said set

@LeakyNun what is a said set?

4:09 PM
said = aforementioned

okay
so this product keeps us in A
hmm if we do ( (1234) (14) ) [4]
we get [2]

right

and if we do first (14) [4]
then on the left by (1234) we get 2 also :D
(1234) (14) = (234)
so the first property work =p
and 1a=a
we need the identity of G to do nothing

now what is (234)[1]?

is keeps it [1]
so there is an elemnt ga =a
such that g is not e ofc
pretty neat

4:14 PM
prove that {g:ga=a} for a fixed a is a subgroup of G

I know there's a name for that subgroup, but I forget what it is.
Stabilizer?

right

stabiliser, centraliser, normaliser, interesting terms.

well is it not empty because 1a=a

I feel like you're going to be extremely hand-wavy in the following steps

4:17 PM
we assumed ga=a and ha=a , we want to show gha =a
what is hand wavy ? ><

relying on intuition
not justifying any step

oh i just wrote the definiton so far

You should include a statement like: Suppose $g,h$ are elements of $\{g\in G:ga=a\}$

which is why I said "following steps"
@Semiclassical that's alright, lol

g (ha) = ga =a
used the associative axiom

4:19 PM
good

we need inverse now :D
1.a =a
g'ga =a
g' (ga) = a
g'a=a
used axiom for associativity and 1a=a
ga=a by defintion

good

:Dd
so we have the closure and inverse and non-empty so it is a subgroup

right

@LeakyNun what got me stuck today was group actions and things related to functions, like preimage and left inverse and stuff like that

4:26 PM
ok

I know they are important when we gonna do homomorphisms so really should get those
Can you help me with those? :) i put couple of notes on them
or we can finish this first =p group actions

what?

I mean, do we continue with group actions now or can I ask about functions
only those 2 topics I did not get from chaper 1 and 2

whatever you like

okay , a map is surjective if it has a right inverse
if we have like x^2
from R^+ to R^+
what is its inverse?

4:30 PM
sqrt(x)

sqrt is what I think of , but
hmm why is that a right inverse?
like what is the definiton of right and left inverse in genral
if we have f : A--> B

If you don't know a definition, look it up.

the book only sais that injective means left invser
inverse*
and surjective has right inverse

4:31 PM
nothing else was mentioned ><
Okay I will :D
okay from what I understood
f: A-->B
g:B-->A
gf (x) = x is the left inverse
and fg(x) = x is the right inverse
is that right?

you wrote the same thing twice.

oups ><

Is $\lim_{x \rightarrow 0} x . \tan{\frac{1}{x}} = 0$

what is the left inverse? @KasmirKhaan
@BAYMAX no

its of the form $0 . \infty$

4:43 PM
@LeakyNun it is the same as in groups =P
gf(x) = x
f has a left invser

g is the left inverse.
gf(x)=x is not the left inverse.

the limit doesnot exist!

oh yes that what i meant to say =p
but it works the same as in groups

@BAYMAX yes
@KasmirKhaan right

exept here its composition

4:50 PM
@BAYMAX it may help to let $z=1/x$ and study the limit instead as $z\to \infty$.

@TobiasKildetoft how tedious can 7 get lol

yes $\lim_{z \rightarrow \infty}\frac{\tan(z)}{z}$
@Semiclassical
I think then i should apply l'Hospital rule