Mathematics - 2017-09-07 (page 5 of 6)

Ted Shifrin

18:00

This room is no fun without snipes.

@10Replies: Yeah, but you will.

10 Replies

Ouch.

SteamyRoot

The splitting in partial fractions should be really easy, too.

Daminark

@Steamy welcome to the sniped club

SteamyRoot

;_;

10 Replies

Ok, so how do I know what to substitute for U in a multitechique thing. What is my goal?

Semiclassical

18:01

@SteamyRoot Right. But you definitely want to do that splitting in terms of $u$ not in terms of $x$

Daminark

We have Hagoromo here so it's not that bad

Ted Shifrin

The $x$ sitting in the numerator is an invitation to substitute $u=x^2$, @10Replies, and then you see how nice it gets.

SteamyRoot

Well, doing the substitution is easier

Semiclassical

oh, sure.

SteamyRoot

But because the denominator is $(x^2 -1)(x^2 +1)$

you get something very symmetrical, whether you substitute or not

Semiclassical

18:02

I more mean that if one tried to do parts in terms of $x$ instead of $x^2$

in terms of $x^2$, it's nice.

SteamyRoot

You don't want to split the $x^2 -1$, yeah :P

Semiclassical

in terms of $x$ you'd end up with a term $1/(1+x^2)$, and that's unnecessarily painful.

10 Replies

Hang on a sec. Anyone know how to install latex in Opera?

SteamyRoot

Though it's still quite... doable

Semiclassical

sure

10 Replies

18:03

Yay, now i have latex

SteamyRoot

I'm going to go on an "educated guessing spree" here

and assume that the factor $4$ in the assignment is there to make the final result look nicer

10 Replies

So to use X^2 as U I have to split x^4 into x^2*x^2

right?

Ted Shifrin

@10Replies: Better, $x^4 = (x^2)^2$.

(Same thing, of course.)

SteamyRoot

and therefore the partial fraction decomposition will be $\frac{1}{x+1} + \frac{1}{x-1} - \frac{2x}{x^2 + 1}$

Semiclassical

yeah, I made an error when writing it out

not bad that way either.

Kasmir Khaan

18:13

using that G is abelian, f(xy) = (xy)^-1 = y'x' = x'y' = f(x) f(y) @TedShifrin this is the second implication

Ted Shifrin

Yup.

Kasmir Khaan

I thought about if it is nessasry to do it this way

Ted Shifrin

Often one direction of and if-and-only-if proof is easier than the other.

Kasmir Khaan

f(xy) = f(yx)

Oh yeah =p noticed that

Ted Shifrin

That'll work, too.

Kasmir Khaan

18:15

what I meant to say is, do I need a stronger argument to why y'x' =x'y'

I know we have ableian

Ted Shifrin

If you want an analysis example of that difference in difficulty (you can find this in my videos), look at the equivalence of the $\delta$-$\epsilon$ definition of continuity and the sequential definition.

That is just the definition of abelian ... arbitrary two elements, right?

Kasmir Khaan

the lectures from multivarible?

Ted Shifrin

Yeah. Talking about continuity (probably second lecture on that).

Kasmir Khaan

Yes but I thought about if we did not have automorphism

@TedShifrin I have them saved on my favorit list :D many stuff are in hard detail so did not watch all

Anyway, if we did not have automorphism

f(x) would not be in G

say from G-->H

Ted Shifrin

Yes, so?

Kasmir Khaan

18:17

so assuming G is abelian does not mean H is abelian

Ted Shifrin

If they're isomorphic, one abelian implies the other is.

Prove that.

Kasmir Khaan

Okay , you mean I prove it for me , or it is needed for the exercice?

isomorphic groups are by definition having same structure

Ted Shifrin

for you ... you asked the question.

Kasmir Khaan

Oh yeah yeah =p okay

its 2 line proof =p

f(xy) = f(x)f(y)
f(yx) = f(y) (fx)

but x,y element of G which is abelian

f(x) and f(y) are elements of H

so H is also abelian =p

Ted Shifrin

How do you show arbitrary elements of $H$ commute?

There's something important you have to say.

Kasmir Khaan

18:23

elements in H are in the form of f (-)

Ted Shifrin

because?

Kasmir Khaan

Ehm because the function is from G--> H

Ted Shifrin

Nope.

Kasmir Khaan

I just showed that f(x) f(y) = f(y) f(x)

because f(xy) = f(yx )

we know that x and y commute

because G is abelian

Ted Shifrin

You're not answering my question.

Kasmir Khaan

18:25

How do we know arbatray elements of H commute?

Ted Shifrin

Why is every element of $H$ of the form $f(x)$ for some $x\in G$?

You need to stay focused on the question.

Kasmir Khaan

Okay let me think

it is by definiton right?

the image of f is H

Ted Shifrin

Right. An isomorphism is a surjection (onto).

Kasmir Khaan

=p

Did not understand the question good

Ted Shifrin

But you need to be very careful to ask yourself questions and answer them explicitly in the proof.

Kasmir Khaan

18:28

Yes that is exactly what I want to get good at :D

Ted Shifrin

You need to start your proof by saying, Note that every element of $H$ can be written ...

Kasmir Khaan

Its an important point =p I needed to say that f(x) and f(y) were elements of H

because of the bijection critera

Ted Shifrin

No, that's not the issue.

You need to start with $z,w\in H$ and say $z=f(x)$ and $w=f(y)$ for some $x,y\in G$.

Kasmir Khaan

to prove it for all elements of H

Ahh okay

Ted Shifrin

BTW, I like to use letters that remind me where I am. In other words, I like $x\in X$ and $y\in Y$, etc., just to help keep track.

Kasmir Khaan

18:31

Okay ill start using small g and h =p

Daminark

$\mathbb{Q} \in G$

Semiclassical

don't make this weird, demonark

Ted Shifrin

You could use $g,g'\in G$, $h,h'\in H$, but if you use $'$ to mean inverse, that won't work.

Semiclassical

$g_1,g_2$ etc. would also work

lots of different ways

Kasmir Khaan

Good idea semi :D

Ted Shifrin

18:33

Yeah, but subscripts get tediouuuuus.

Semiclassical

eh, two isn't bad

but this is as much a matter of preferences as anything

Kasmir Khaan

U know what :D ill figure that part out later =p the more important thing now is getting the hang of how to prove things in a rigor way

Semiclassical

if it works, it works

Ted Shifrin

I meant it as a serious suggestion. My students who used letters haphazardly often had more trouble keeping track of where things were.

Semiclassical

(I'm also not all that index-allergic)

Part of what for a rigorous proof is presenting the logic in a clear, transparent way.

Kasmir Khaan

18:35

part b ) G X G --- > G that sends (x,y) to xy .

Semiclassical

So picking your notation to support what you're doing is an important component.

Kasmir Khaan

GXG is defined by (x,y) * (x_1 , y_2) = (x x_1 , y y_1 )

Just posted the question so we are in same boat =p

dont answer it :D

Ill start working on it now

Semiclassical

You're trying to show that's a homomorphism?

Kasmir Khaan

homomorphism => abelian

and abelian => hom

we assume one thing and prove the other

Ted Shifrin

Yup.

Semiclassical

18:38

right.

Kasmir Khaan

okay let me see if i got the notation right

f ( (x,y) (x' ,y' ) = (xx' , yy' ) = xx'yy'

is this correct?

Semiclassical

$f$ is GxG to G

so the output should be what?

Kasmir Khaan

oh yes

one step missing

xx'yy'

Semiclassical

first and third terms make sense

middle one doesn't

Kasmir Khaan

yepp noticed that now ><

f ( (x,y) (x' ,y' ) = xx'yy'

but can we say that (x,y) belong to GXG ?

Semiclassical

18:45

sure. but you can fix the second term that you had originally

are x,y in G?

if so, then that's the definition of (x,y) being a member of GxG

Kasmir Khaan

so we work on 2-tuples

oh come to think of it (x,y) = (x,y) ( 1,1)

by that defintion of the product

once I get the notation straight in my head , i think the problem is easy

f ( (x,y) ( x',y') = f (x,y) f( x',y') = xyx'y'

by homomorphism

f ( (x,y) (x' ,y' ) = f (xx' , yy' ) = xx'yy'

xyx'y' = xx'yy' , multiplication on the right by the inverse of y' and on the left by the inverse of x' , we get yx' = x'y hence abelian

Semiclassical

right. one thing to notice here is that you didn't actually end up caring at all what $x,y'$ were

Kasmir Khaan

What do you mean ? =p

Semiclassical

well, your ultimate conclusion is on x' and y

Kasmir Khaan

I had trouble understanding the operation

Semiclassical

18:59

the only role x,y' play is to be removed at the end using inverses

Kasmir Khaan

Yes

ahh yes =p so this works for any elements

Semiclassical

so we could pick x,y' however we want and the proof will go through the same.

is there a particularly simple pair of elements x,y'?

Kasmir Khaan

Got it :D but have to put this in better way =p

hmm

do you mean numbers?

or a relation between them

Semiclassical

I mean: Given an arbitrary group G, what's the simplest choice of elements x,y' from G?

Kasmir Khaan

if they were inverse of each other

or wait ._.

Semiclassical

19:01

eh, that wouldn't really help. they're sitting at opposite ends of the two products

Kasmir Khaan

let me think a bit

oh

if they were the same element

but that is way to avious that it is abelian

Semiclassical

sure, that's permissible. what's the simplest element they could both be equal to?

Kasmir Khaan

identity

Semiclassical

right. so let's pick x=y'=e

then your proof goes like this. you've got f((e,y))=ey=y, f((x',e))=x'e=x'

what about f((e,y)*(x',e))?

Kasmir Khaan

we get ex' *ye

x'y

Ah I see what is happning now :D

Semiclassical

19:06

right.

Kasmir Khaan

But it is not general if we use the identity right?

specially when proving abelian

Semiclassical

sure it is. you still haven't picked what x',y are

Kasmir Khaan

identity commutes with all so =p

oh okay :D

smart idea

Semiclassical

to wrap this up: in specifying a product of elements in GxG, it seemed like we had to pick 4 elements altogether

but to show abelian we only need to have two elements.

Kasmir Khaan

It is alot easiar and better that way =p

Semiclassical

19:08

now, to clarify: this is how you show that homomorphism implies abelian

to show that abelian implies homomorphism, you do need to pick two elements in GxG

so in that sense you still need 4 elements from G

it's just that, as Ted emphasized, one can simplify the homomorphism -> abelian direction

Kasmir Khaan

One direction is easy and other is less easy =p

Semiclassical

Right.

For that reason, it's important to get the hard direction right.

Kasmir Khaan

the thing that comfused me is GXG

not used on that notation

Semiclassical

yeah

one other way to look at this

suppose you've got (x,y) in GxG

Kasmir Khaan

so the elements of GXG are of the form (x,y) 2 tuples?

Semiclassical

19:11

right.

Kasmir Khaan

OKay

Semiclassical

first element is from G, and second element is also from G

then f(x,y)=xy, but that's also true of f(e,xy). so (x,y) and (e,xy) are mapped to the same element.

Kasmir Khaan

so we have G is a group

and GXG takes one element from G and other from G as a 2 tuple

Okay nice >< will this come in many problems?

Semiclassical

I wouldn't say GxG 'takes from G and G'

GxG isn't an operation

GxG is a set, composed of pairs of elements from G

Kasmir Khaan

Okay =p

Semiclassical

19:13

product stuff comes up a lot, yeah

Kasmir Khaan

We did not do any of this kind in class

Semiclassical

I suspect one variation on this problem would be to consider G1 x G2 where G1,G2 are both subgroups of some group G.

Kasmir Khaan

gotcha =p

Ill write down the other implication now and see how it goes =p

You still here?

Semiclassical

sure

Kasmir Khaan

I might need help , I got a list of exercices to do tonight =p

thanks :D

if am not mistaking this is one line proof @Semiclassical

f( (x,y) (x',y') ) = f (xx',yy' ) = f (x,y) f (x',y' )

Semiclassical

19:26

Which direction are you doing?

Kasmir Khaan

abelian ==> homomorphism

but it is wierd because did not use abelian here

Semiclassical

you're missing details in here, though you can still keep it one-line

Kasmir Khaan

multiplication of real numbers is abelian just that

Semiclassical

who says you're doing real numbers?

Kasmir Khaan

oups ><

true that

Semiclassical

19:28

your first equality is the definition of multiplication in GxG

that's fine, but you should use * to distinguish multiplication in GxG from multiplication in G

so, (x,y)*(x',y') or some such

Kasmir Khaan

f( (x,y) * (x',y') ) = f (xx',yy' ) = xx'yy'

Semiclassical

okay. now, in your second equality you're using the definition of f---good.

Kasmir Khaan

but I need to somehow have f(x,y) f(x',y')

Semiclassical

okay. what's the assumption you haven't used yet?

Kasmir Khaan

abelian

Semiclassical

19:30

right. so if it's abelian, you can swap stuff.

what do you want to swap, given how things worked out earlier?

Kasmir Khaan

thats the thing

from f (xx',yy' )

why cant we say that f (xx',yy' ) = f(x,y) f(x',y')

based on that kind of multiplication

Semiclassical

You need to show that's true.

Kasmir Khaan

I dont see how the abelian part is nessasry here

Semiclassical

I'm not seeing what argument you're making.

Kasmir Khaan

Okay let me try again

Semiclassical

19:32

How are you going from f(xx',yy')=f(x,y)f(x',y') ?

Kasmir Khaan

by reverion what f does

or wait

Semiclassical

okay. but you need to spell that out.

Kasmir Khaan

yeah steps are missing

Semiclassical

that statement is true, to be clear. but you haven't stated an argument.

Kasmir Khaan

let me put it in order

yeah noticed that =p

ehmm this is what I have trouble with

Semiclassical

19:35

Review what assumptions you have. You know that (x,y)*(x',y')=(xx',yy'). You assume that multiplication between elements in G is commutative. And you know that f(x,y)=xy.

That's all.

If you want to get a conclusion, you need to use those assumptions and nothing more.

(well, and the fact that x,x',y,y' are elements of a group G)

Kasmir Khaan

f( (x,y) * (x',y') ) = f (xx',yy' ) = xx'yy' , am thinking of how we can rearrange the order of the elements in this xx'yy'

Semiclassical

right.

to get a sense for that, look at what you want to end up with

you want to end up with f(x,y) f(x',y')

what are these two terms equal to?

Kasmir Khaan

yeah I do it in reverse :D

but grrrr

we have the right order for those

no no no

doing it wrong =p

Semiclassical

f(x,y) =?

f(x',y') = ?

Kasmir Khaan

f(x,y) = xy , f( x',y') = x'y' , so the product is xyx'y'

this is the part I did wrong =p

Semiclassical

19:39

right.

so you start from xx'yy' and want to get xyx'y'

Kasmir Khaan

So i have to say that xx'yy' = xyx'y' ( abelian )

Semiclassical

exactly

Kasmir Khaan

:D

Semiclassical

so, what's the chain of equalities?

Kasmir Khaan

f( (x,y) * (x',y') ) = f (xx',yy' ) = xx'yy' =xyx'y' = f(x,y) f (x',y' )

thus we got homomorphism

Semiclassical

19:41

okay. to make this explicit, what assumption goes with each equality?

Kasmir Khaan

first equality is by definition

Semiclassical

definition of *

Kasmir Khaan

yes =p

f (xx',yy' ) = xx'yy' def of *

Semiclassical

you'd want to say which definition you're using (definition of multiplication * in GxG)

Kasmir Khaan

xx'yy' =xyx'y' ( abelian )

oh okay

Semiclassical

19:42

that's the third equality. what's the second?

Kasmir Khaan

f (xx',yy' ) = xx'yy' this is also by def of * in GXG

Semiclassical

no.

there's no * in there.

Kasmir Khaan

oh yeah

Semiclassical

moreover, you're going from GxG to G.

Kasmir Khaan

that one by the function

Semiclassical

19:43

right. it's the definition of f.

Kasmir Khaan

let me put it all correct now =P

f( (x,y) * (x',y') ) = f (xx',yy' ) definiton of * in GxG
f (xx',yy' ) = xx'yy' by definiton of the function f
xx'yy' =xyx'y' we used abelian

xyx'y' = f(x,y) f (x',y' )

last one is by both def of * in GXG

and def on the function f

Semiclassical

where is * showing up in the last one?

Kasmir Khaan

oh darn it, it is not there

><

I mean to say

Semiclassical

both f(x,y) and f(x',y') are in G, so the multiplication between them is just the one in G. no * in sight.

Kasmir Khaan

we did f in reverse

Semiclassical

19:49

I'd just say that's the definition of f again.

Kasmir Khaan

Okay =p

Semiclassical

if I was to write this out, I'd probably write it like

Kasmir Khaan

next problem is to show that GL_2 ( z /2 ) is isomorphic to S_3

Semiclassical

A = B (reason why)
= C (reason why)
...

Kasmir Khaan

Iam gonna do that on texstudio

Semiclassical

19:51

bah, equal signs should be aligned

yeah, use align environment for that

you can use \tag to get the text conveniently

Kasmir Khaan

ill revist all the detail and show them to you once am done with all =p

deadline next week

Okay ill copy that command =p

I used \text

on the other assigment

Semiclassical

nice thing about \tag is that the text comes out aligned

Kasmir Khaan

Ill try that , thanks:D

Semiclassical

not sure it works on mathjax, but let's see

\begin{align}
A
&= B \tag{text} \\
&=C\tag{text}
\end{align}

yep, there you go

Kasmir Khaan

By GL_2 ( z/2 ) they mean general linear groups of 2x2 matrices over modulus 2 right?

copied :D

Semiclassical

19:54

though I guess maybe \tag is more for labelling equations e.g. eq (1)

up to you

by modulus you mean "arithmetic mod 2"?

(not |z|=2)

Kasmir Khaan

oups

mod 2

Semiclassical

right.

Kasmir Khaan

yes arithmatic mod 2

so 0 and 1 =p

Dont answer it =p

Semiclassical

so 2-by-2 matrices whose elements are 0,1 and where matrix multiplication gives a new element of GL_2(Z/2)

Kasmir Khaan

im gonna think about it then send you =p

Semiclassical

19:55

wasn't planning to, hah

Kasmir Khaan

:D

Semiclassical

I guess where I'd start is trying to figure out what the isomorphism is

Kasmir Khaan

they have to like have same order

Semiclassical

i.e. how you're supposed to associate GL_2(Z/2) with S3

yeah, that's a big hint

it helps that there's just not a lot of elements in GL_2(Z/2)

Kasmir Khaan

yes try to find a function =p

they have to have same number of elements for iso to exist

s_3 has 6

so GL also has 6

Semiclassical

19:57

I forget. Does GL require that all matrices be invertible?

Kasmir Khaan

because we have det is not 0

yes =p

Semiclassical

right

that kills off a number of possibilities

Kasmir Khaan

not 18 ><

Semiclassical

otherwise you'd have 2^4=16 matrices and that's waay too many

Kasmir Khaan

Ehm yeah true

am gonna list all elements to see what function makes sense

Semiclassical

19:59

right.

Kasmir Khaan

damn it :D

told you not to help ?å

=p*

Semiclassical

i said nothing :P

Kasmir Khaan

haha

Semiclassical

but yeah, that's a fair cop

DemCodeLines

20:26

"Assume that price ($) of one pair of keys is given as p. As a result, the number of expected sales is provided by $f(p)=24e^{-3p}$. So expected income from sales is derived from $f(p)=24pe^{-3p}$. In this case, what's the price that maximizes gross income?"

How would I solve this?

Semiclassical

If you want to maximize a smooth function, what branch of math is that?

DemCodeLines

Calculus?

Semiclassical

yep

more precisely, differential calculus

so what should you do if you want to maximize a function in differential calculus?

Transcript for

$Mathematics$ Mathematics