Mathematics - 2017-09-08 (page 1 of 8)

Dodsy

00:00

haha :)

10 Replies

Where does the two come from? du?

Semiclassical

you started off with $\displaystyle \int \frac{4x}{1+x^4}\,dx$, right?

10 Replies

Yes

Semiclassical

so if you substitute $u=x^2$, what's $du$?

10 Replies

2x

Semiclassical

00:02

that's $\frac{du}{dx}$

$du$ would be $2x\,dx$

10 Replies

where does dx come into play?

Semiclassical

well, dx is a differential

so du had better be one as well

so du = 2x dx is at least consistent. du=2x wouldn't be.

then $\displaystyle \int \frac{2}{1+(x^2)^2}2x\,dx=\int \frac{2}{1+u^2}\,du$

Kasmir Khaan

@Semiclassical to prove that GL_2( Z/2) is isomorphic to S_3
how do I proceed after finding a correspondence between each element from GL to S_3 ?

I found a 1-1 map but how do I continuelike rigoursly ?

Semiclassical

well, you need to prove it preserves the group operation

Kasmir Khaan

okay hmm

should I do that for all elements?

that does not sound like a good proof =p

Semiclassical

00:06

that's what I'm wondering myself.

I think one can get away with only checking some fraction of them, but

I'm not entirely certain.

Kasmir Khaan

okay thanks =p

Semiclassical

it seems like there should be a smarter way.

10 Replies

Why is the two in the first one? The origional was 4x/(x^4+1)

Kasmir Khaan

@anon is it true that if we have an injective homomorphism , it is automaticly isomorphism ? when the order of both groups are the equal ?

Semiclassical

because $du/dx=2x$ contains a factor of 2.

anon

00:08

@KasmirKhaan a one-to-one map between finite sets of the same size must also be onto

Kasmir Khaan

@anon I want to learn how to make good proofs , dont have alot of experience

anon

communication skills

also basic logic, and specialized jargon

Semiclassical

I'm not going to continue to repeat this point. if you're not understanding how u-subs work, read that section.

10 Replies

Ok, so now what do I do once I have 2/(1+u^2) du ?

Kasmir Khaan

@anon okay after I found the correspondance between all 6 elements , how to argue that it is an isomorphism ?

@anon jargon ? =p

Semiclassical

00:09

@10Replies expand by partial fractions

oh, wait

anon

@KasmirKhaan try to figure it out

Semiclassical

something's wrong

10 Replies

You can't factor x^2+1 right?

Semiclassical

was the denominator always 1+x^4? i somehow thought it was 1-x^4 originally.

10 Replies

Me too. I wrote down x^4-1 in my notebook, but, i must have missread it.

Semiclassical

00:11

6 hours ago, by 10 Replies

How does one go about integrating something like 4x/(x^4-1). I'm not sure how to start?

that version is easier because the denominator factorizes

Kasmir Khaan

@anon I need to show that they both obey the same group structure, so f (xy) = f(x) f(y) thus homomorphic , but the map i found is not like f(x) = e^x or something, I constructed it from the finite elements, you see what I mean ?

10 Replies

That would explain some of my confusion.

Semiclassical

this one is trickier as a result.

10 Replies

Would it be easier to do trig substitutions instead then? Or can I still use partial fraction?

Kasmir Khaan

@anon If I do them all case by case , it would not be a good proof imo, because if the size were larger , it would take forever ><

Semiclassical

00:12

can you confirm which version it is?

10 Replies

x^4 + 1

Semiclassical

that's a good question. i'm going to cheat

drat.

Kasmir Khaan

@anon I dont want an answer , I need to figure it out bymyself, just a hint or something on how to think right =p

Dodsy

drat is a good exclamation.

Semiclassical

okay, the answer comes out in terms of arctan.

10 Replies

00:13

anon

@KasmirKhaan how many maps have you ever shown are a group homomorphism? stop avoiding doing the work and just do it. you'll learn more advanced modes of thinking later.

Semiclassical

...ow.

Kasmir Khaan

@anon okay thanks =p ill keep trying

Semiclassical

kinda funny that the first term there is the easier to integrate despite looking worse.

but yeah, you definitely want to do a trig sub.

10 Replies

Yeah, not only did they give me a ridiculously hard problem... it's also combined with a deceivingly easy problem. I already got the first one, hence why I didn't ask about it to begin with.

Semiclassical

00:15

in fact, $\int \frac{du}{1+u^2}$ basically is an inverse trig function

if you've got access to even a small table of integrals, you'll probably see that one

I guess the hint I'd give is that you want to choose $u$ so that $1+u^2$ becomes simpler

if you do $u=\cos\theta$, though, then that becomes $1+\cos^2\theta$ which isn't nice.

$u=\sin \theta$ is similarly bad.

so you'll want either $u=\sec\theta$ or $u=\tan \theta$.

one of those will work, the other won't

so pick one and see what happens.

Dodsy

night semi, ttyl

Semiclassical

night

10 Replies

I'm not sure how to do trig substitution. I googled it, and I don't see any square roots in this problem

Semiclassical

doesn't mean you can't do it. the same principle holds

10 Replies

What exactly am I trying to substitute?

Semiclassical

00:25

well, i gave you two options

either $u=\sec\theta$, or $u=\tan\theta$

10 Replies

Is this a different U than in intgral ( du/(1+u^2))?

Semiclassical

wait.

i'm wrong. sec(theta) isn't an option to try.

u=tan(theta) is the only one you'd want to attempt

10 Replies

I don't know how to go about attempting, thats what I'm saying.

Semiclassical

it's the same u.

so here's the logic

10 Replies

1 + tan^2 = sec^2?

Semiclassical

00:27

right. that's why this substitution will help.

so if u=tan(theta), then what's du?

10 Replies

but doesn't U = x^2?

Semiclassical

sure.

10 Replies

x^2 doesn't = 1+ tan^2.... right?

Semiclassical

you could do $x^2=\tan\theta$ directly.

no, but nowhere have I suggested that it should be

Kasmir Khaan

@anon I came to conclusion that I have to write down all the product f(xy) = f(x) f(y),nothing else makes sense to show that the structure is obeyed

Semiclassical

00:29

the point is that if x^2=tan(theta), then 1+x^4=1+tan(theta)^2=sec(theta)^2.

so at this point we treat u=tan(theta) with u being understood as a function of theta. what's du/d(theta)?

10 Replies

sec^2(theta)

Semiclassical

right. so du=sec^2(theta)d(theta)

but as you noted earlier, 1+u^2=1+tan^2(theta)=sec^2(theta)

so du/(1+u^2)=d(theta)

hence the integral becomes $2\int \frac{du}{1+u^2}=2\int d\theta$

and that last integral is easy if you don't trick yourself into thinking its hard :)

skullpatrol

...and don't forget the + C

10 Replies

Semiclassical

denied

skullpatrol

00:37

drat

10 Replies

The last integral is easy? Is it just theta?

Semiclassical

yuuup (and the 2)

and now you just need to back substitute to get back to x

you've got $u=x^2$ and $u=\tan\theta$, so $2\theta=...?$

10 Replies

2arctan(u)

Semiclassical

sure, and u=x^2

so that's

10 Replies

2 arctan(x^2)

Semiclassical

00:41

ding

10 Replies

Do I bring the 4 back in, or are we done with that?

Semiclassical

well, let's check to see by differentiating

if that's our antiderivative, we should be able to differentiate w/r/t x and get back 4x/(1+x^4)

best way to do that is to use the chain rule with u=x^2 and remember how differentiating arctan works

...which, not going to lie, seems painful.

10 Replies

I should not have waited 2 years between calc 1 and 2, I don't remember any of the derivative or integrals of trig functions or log and ln and whatever.

Semiclassical

the thing is: no, you don't bring the four back in. the 4 has been there the entire time; we split the initial numerator up as 2*2x

the 2x went with the differential (since du/dx=2x) and the other 2 stuck around after

so the answer is complete.

10 Replies

yay

skullpatrol

00:45

Why the two year gap?

10 Replies

Think the gap is about 1.5

Took calc I in 9th grade, then I took linear algebra in 10th and then 11th I have no more math at my highschool, so I walk over to the university nearby for calc II

skullpatrol

Hmm...

10 Replies

I took AP calc, which didn't help because I learned just enough calc to pass a test, but, not enough for me to remember a significant portion of it.

I got a 5 on AB and 4 on BC though, so it worked temporarily lol.

skullpatrol

...what textbook are you using?

10 Replies

essential calculus: early trancendentals

by James Stuart

Only 15 problems left :/ I did 7 in far more time than I would like to admit

Will you guys hate me if I choose to never take another math class after calc II? I'm ready to be done with this whole math thing

skullpatrol

01:04

::puts 10 replies on ignore::

::changes username to 9.999... requests::

:-D

Kasmir Khaan

hi chat

Who can help me proving isomorphism of groups?

Semiclassical

01:30

@KasmirKhaan I'm not confident this will work, but

there's only two groups of order 6. one of them is nonabelian and isomorphic to S3, the other is abelian and isomorphic to Z/6.

Kasmir Khaan

I moved on to another exercice, Z/7 * isomorphic to Z/6

Semiclassical

ah. that should be a lot easier.

Kasmir Khaan

@Semiclassical I dont think we are allowed to use such ideas because we did not prove them =p

let me tell you what I got so far

Semiclassical

you'd have to prove more than what I just said, yeah.

Kasmir Khaan

Z/7* = <3> = <5>

Z/n is cyclic for all n

Z/6 is generated by {1,5}

Semiclassical

01:33

1) prove that GL2(Z/2) is a group of order 6. 2) prove that there are only two groups of order 6. 3) prove that it's not Z/6. 4) conclude that it's S3.

Kasmir Khaan

Its smart but we are not allowed to use something we did not go thru =p

so I will make a table of both groups

Semiclassical

yeah, i figured.

Kasmir Khaan

and show that structure is the same =p

it is the same as in the second example

Semiclassical

you may be able to make a few shortcuts.

Kasmir Khaan

I think he picked small groups for that reason

Semiclassical

01:35

for instance, $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ definitely acts as the identity under matrix multiplication

Kasmir Khaan

so we do them by incpection =p

Yes I got all the products to work , without using any fancy stuff i dont think I can do better than that

Semiclassical

If we label that as $f(e)$, then we definitely have $f(e)f(a)=I_2 f(a)=f(a)=f(ea)$

so that establishes the $f(e)$ row of the operation table, and similarly for the column

Kasmir Khaan

How to make table on texstudio?

Semiclassical

but this only does so much.

there's a few ways, I think.

none of which I remember too well

Kasmir Khaan

haha =p

Semiclassical

01:38

\begin{array}{|c|c|} \hline \\ e&a \\ \hline \\ a& e \\ \hline \end{array}

Kasmir Khaan

gave error

Semiclassical

well. that would be right except that it's too tall :/

tbh you're best off googling it

Kasmir Khaan

I think Ill just hand write them if it will be too much to write ><

he does not like that but

btw semi

on the second example

if a map 3 to 1 , 5 to 5 , 2 to 2

the first 2 are generators and the order of 2 is 3

2^3 = 8 = 1 mod 7

and 2+2+2 =6 =0 mod 6

1 to 0

idenity maps allways to identity

Semiclassical

one way to describe that simply

write each element 1 through 6 as 3^x

and then map 3^x -> x

Kasmir Khaan

Oh nice :D

it will be also easiar to prove hom, surjective and injective that way

Semiclassical

01:45

this preserves the multiplication structure in (Z/7)*, since 3^x 3^y = 3^(x+y)\mapsto x+y=(x+y)

and it's also easy enough to establish the bijection: 3^0 = 1, 3^1=3, 3^2=9=2, etc.

Kasmir Khaan

Yepp it works perfectly :D

Semiclassical

this is a map from (Z/7)* to Z/6, to be clear, not the other way around

Kasmir Khaan

Yes we used that Z/7* is generated by 3

Semiclassical

right.

this same idea should work for any (Z/p)^* with p an odd prime, in fact. one just needs to find a generator.

and one can then show an isomorphism with Z/(p-1).

Kasmir Khaan

because both are cyclic right ?

Semiclassical

01:49

right.

if it were (Z/n)^* with n composite then it'd be trickier

Kasmir Khaan

this stuff is intresting but a bit tricky =p

Semiclassical

I forget if there's an easy way to identify a generator of (Z/p)*, come to think of it.

Kasmir Khaan

i think by fermat theorem

a^p-1 = 1 mod(p)

or wait that cant be right ><

Semiclassical

it's a^(p-1)=1 mod p

but it doesn't really help here. 2^6=64=1 mod 7, but 2 isn't a generator.

Kasmir Khaan

yes >< I think one just have to find a generator by incepection

then once that found , the others will follow

<a^k> = < a^gcd(n,k) >

Semiclassical

01:54

What I want to say is that if p is prime then (p-1)/2 generates the (Z/p)^*

Kasmir Khaan

where a is a generator and n is the order of the group

Semiclassical

but i'm not at all certain that's true.

Kasmir Khaan

=p

Semiclassical

google time

Kasmir Khaan

haha

Semiclassical

01:57

right, this generator is known as a primitive root mod p

Kasmir Khaan

oh thanks :D

hmm am having trouble proving hom here

Semiclassical

even with primes it doesn't seem like there's a simple pattern to the primitive root. neat

Kasmir Khaan

Yeah when we did fermat in other classes , some numbers gives wrong answer

they are not prime and they "pass " as primes

Semiclassical

well, suppose you've got two elements of (Z/7)*. since 3 generates this group, these two elements must be of the form 3^x and 3^y where x,y are between 0 and 6.

what's the product of these two numbers in (Z/7)*?

Kasmir Khaan

3^(x+y)

But do we do f : 3^x --> x

or f :x -->3^x

its comfusing having 3^x first

Semiclassical

02:01

either works.

if you find x->3^x clearer, go ahead

in that case you've got a map from Z/6 to (Z/7)*

Kasmir Khaan

lets do f: 3^x --> x

yes I know it should work both ways =p

Semiclassical

okay, so (Z/7)* to Z/6.

Kasmir Khaan

f(xy) = 3^xy right?

that is not what we want

Semiclassical

indeed not.

let's go the other direction, just to save some energy

Kasmir Khaan

okay ><

f:x ---> 3^x

grrr

same thing I get

Semiclassical

02:03

the group operation in Z/6 is addition, and we want to prove that f preserves this operation

Kasmir Khaan

okay

Semiclassical

so what does that mean here? i.e. what does it mean for f to preserve addition in Z/6?

Kasmir Khaan

f(xy) = f(x) + f(y)

the first product mod 7

Semiclassical

careful.

Kasmir Khaan

the second sum is mod6

oh

got it backwards

Semiclassical

02:05

right. the way you wrote it would work for 3^x |-> x.

Kasmir Khaan

f(x+y) = f(x) * f(y)

Semiclassical

right. that's what we want for x->3^x.

but what's f(x)*f(y), based on what we decided about f?

Kasmir Khaan

f(x+y) = 3^(x+y) = 3^x 3^y = f(x)* f(y)

Semiclassical

right.

the only thing one should be a bit careful about, I think, is that 3^(6)=1

so it's legitimate to think of 3^(x+y)=3^x 3^y with x+y taken mod 6.

Kasmir Khaan

it should be 1

oh

Semiclassical

02:07

right.

one might want to be careful with the writing at that point.

okay, back another time

Kasmir Khaan

okay thanks for help ! @Semiclassical

J. M.

02:26

It can be a bit annoying when relevant unupvoted comments get buried under petty upvoted ones.

Oh well.

Typhon

02:39

indiedb.com/games/block-builder

10 Replies

concept is cool... graphics are hideous

Faust

03:10

morning

yhey typhon

@TedShifrin second day was harder graph theory is throwing me a curve ball also i didnt like my first geometry class hope it gets better also i got 1 analysis question for you next time i find you ^^

Daminark

What are you doing in graph theory?

Also I can try to help with analysis

Faust

wierd words

There exists a set $X \subset V (G) - \{u,v\} $
such that some vertex in X is adjacent to
both u and v.

what does that say?\

anon

there is a set of vertices (not including u,v) which has a vertex x adjacent to both u and v

Faust

ok

the analysis question is harder i will type it up after a 15 min nap or something

i wanna prove something in abetter way than i fugred out

Ted Shifrin

03:56

hi @anon — you doing anything interesting at school this term?

Kasmir Khaan

Hello Ted :D

Ted Shifrin

Don't you ever sleep, @Kasmir?

Kasmir Khaan

haha

its 6 am here been up all night

Ted Shifrin

that's not a good way to start the semester

Kasmir Khaan

but it is ok because we dont have lectures this week

Semiclassical

03:58

I guess that's a "no."

Ted Shifrin

I think Balarka has singlehandedly ruined the sleep cycles of half the room

Kasmir Khaan

I know but drank lots of coffe

Transcript for

$Mathematics$ Mathematics