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14:17
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?
14:25
@LeakyNun it can be.
15:03
@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
15:15
@Jasper dui lian lol
How about bimodern, 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
@Semiclassical flogging the dead horse
15:20
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
@LeakyNun yelloha
god dag
I find it interesting enough how and why people use the word now
15:22
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*
15:23
: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
its about the definition
let me tell you what I got so far
15:24
@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 ><
15:25
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?
15:27
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
15:29
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
15:31
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
15:34
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?
15:35
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"
15:39
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
15:40
r^2 * 1 = 3
eh, I'd be careful with that notation. (1) in there is not a permutation
sure, that works.
Hmm yes r^2 will send the elemnt 1 to 3
and r * 2 = 3
@AlessandroCodenotti buongiorno
15:42
Yes so far so good
and r * (r * 1) = 3
so we have demonstrated associativity
now is it true that ga=a implies g=e?
Hmm dont know
use this example
15:44
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.
15:50
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
15:57
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
15:59
(23)*1 = (23)
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]
16:01
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
16:02
????
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?
16:09
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
16:14
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
16:17
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
16:19
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
: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
16:26
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?
16:30
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
look it up on google
16:31
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 $0 . \infty$
16:43
@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
16:50
@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

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