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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
 
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
 
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
 
@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
its about the definition
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
 
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
 
look it up on google
 
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
 

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