Mathematics - 2015-03-29 (page 2 of 3)

11:03 AM

@Chris'ssis Have you shown anyone the integral producing the primes yet?

@JasperLoy It was a misinterpretation of mine. It generates log of primes for a simple reason, there is a sum that cover the log of all natural numbers. Hence the result.

Jasper Loy

@Chris'ssis Oh, OK. No worries. You can still win many prizes, lol.

@Chris'ssis I see. Still, it might be useful.

Chris's sis

@JasperLoy Who knows ...

Incurrence

@DiscipleofBarney Which post is this referring to?

LeGrandDODOM

Great news: you can play a Super Mario 64 lookalike inside your webbrowser here: mario64-erik.u85.net/Web.html

Incurrence

11:13 AM

@LeGrandDODOM Or you can download a ROM for an emulator

Mary Star

Hello!! I am asked to state the two versions of the Knapsack problem and their differences. Could I formulate it as followed??

The Knapsack problem is the following:

There are $n$ items, where the ith item has a benefit of $v_i$ and it has weight $w_i$. We want to pick some items so that we maximize the total benefit while keeping the total weight of $W$.

The difference between the integer and the fractional version of the Knapsack problem is the following:

At the integer version we want to pick each item either fully or we don't pick it. At the fractional version we can take a part of …

LeGrandDODOM

@Incurrence remember this is illegal, plus you don't get HD textures

Incurrence

@LeGrandDODOM True about legality, but the HD textures are possible on a ROM

How are you @Jasper?

@Jasper A class mate likely saw my old username.

Disciple of Barney

@Incurrence You changed your name, have you quite commiting to a challenge? Someone was posting under the name Nilpotency

Jasper Loy

@Incurrence Hi Alex. I am not too good. Every time I try to solve my OCD, new obstacles pop up, but I will keep trying.

Incurrence

11:21 AM

@DiscipleofBarney I haven't quit the challenge, no, just changed the name so my classmates don't know my account

@DiscipleofBarney Not me, I won't be making any new accounts

@JasperLoy How often do you see your doctor?

Jasper Loy

@Incurrence I am thinking of changing my email again. For some reason, I don't quite like it.

@Incurrence Once a month just to get meds. I don't want therapy now, because I want to work through things on my own for now.

Incurrence

@JasperLoy Changing your email address? Why not J...a.....sper.....l...o..y@gmail.com

Jasper Loy

@Incurrence Dots don't matter in gmail.

Incurrence

I troll people with that dot thing, since noone knows it hahaha

Jasper Loy

@Incurrence I know it.

Incurrence

11:22 AM

I know, you showed me it

Disciple of Barney

@Incurrence I am pretty sure every user on math.se is you... except for me

Incurrence

@DiscipleofBarney Hahaha why do you say this xD

Jasper Loy

@Incurrence You have 20 MSE accounts?

Disciple of Barney

@Incurrence There are just so many of you, you might even be Jasper

Incurrence

@JasperLoy No

Jasper Loy

11:24 AM

@DiscipleofBarney We are brothers in spirit, both afflicted with great suffering.

Incurrence

@DiscipleofBarney Did you really think MK and Nilpotency were me?

<- I like this blue icon

Disciple of Barney

@Incurrence I sort of thought MK could have been you, pretty sure Nilpotency is MK though. here is the question. I bet you use the green icon on other accounts to throw me off your trail.

@JasperLoy Spirit Brother, that could be you knew email spirit....brothers@gmail

Incurrence

@DiscipleofBarney So you are tracking me down are you ahahha

@DiscipleofBarney I'll tell you if you get me ahaha

@DiscipleofBarney I may have made one today, but you will never find it(and it is embarrassing so I hope you don't lol)

Disciple of Barney

@Incurrence Was that problem in your assignment?

Incurrence

I wouldn't post a beast question like that on another username, since it isn't trivial

@DiscipleofBarney It looks like it, but it is different

@DiscipleofBarney Different variables and mine doesn't mention the bracket

and mine literally says that it is all $n\times n$ matrices, nothing about invertibility

Disciple of Barney

11:31 AM

@Incurrence The guy uses the same line formatting as MK, pretty sure it is the same guy. It doesn't say anything about invertibility (in the comments the user was confused about what $\mathfrak{gl}$ was though)

Incurrence

I meant comment 2

But the question says $\mathfrak{gl}_n$ is the space of $n\times n$ matrices here

The big giveaway for me is that I put > fields at the start haha

(or am I doing that to throw you off ;)

I am tempted to make more accounts to see if you can find them now LOL

Disciple of Barney

Nil was confusing $GL_n$ and $\mathfrak{gl}_n$ and I guess felt like it was conflicting information.

Haha

Incurrence

How old are you Barney?

Disciple of Barney

22

Incurrence

Ahh okay, for some reason I thought you were much older

Originally I thought you were ~40 for some reason, but you seemed to enthused by my accounts lol

Disciple of Barney

11:36 AM

Why did you think I was 40

Incurrence

@DiscipleofBarney Dunno, you said you were Paul (?)Plummer(?), and I thought you were the Paul with the browny-red icon, who was on here who was 40

Do you know complex analysis well?

Disciple of Barney

Ah, mistaken identity. Sadly I don't, plan to start learning at some point @Incurrence

Incurrence

It's fun, but I suck at it so far

Disciple of Barney

Yah, it looks like fun and has some great theory behind it.

Incurrence

11:49 AM

$\log_e(z)=4i\implies z=\cos(4) + i\sin(4)$ right?

Disciple of Barney

are you mobius, yes

Yes to your question

And you do use lines!

Two, maybe four down because now I think you are lying about the other accounts

Incurrence

@DiscipleofBarney Why do you think I lie about them?

@DiscipleofBarney Those 101% aren't me

Disciple of Barney

Nobody uses lines

Incurrence

You mean the ctrl+r lines?

Disciple of Barney

Except for your 30 accounts,

Disciple of Barney

11:56 AM

^ those ones

Incurrence

That's ctrl+r in the thing, and it is somewhere in the help center

Disciple of Barney

Now if I go backwards in the logs and look at what assingments you have been talking about on what day I can cross reference that with questions on that day. %D

Incurrence

Want some intials for accounts?

D,K,C,U,P,E,A,A,P,S,A,A,C,C,T

Disciple of Barney

No, I think I am done hunting you down and exposing all your accounts

Incurrence

Oh one of those C's is now an I :)

Disciple of Barney

11:59 AM

hah

Incurrence

I'm not Möbius, I just checked his account and he is too pleb :P

Disciple of Barney

Pretty sure your mobius

Incurrence

My English is too good, he is a foreigner haha

Disciple of Barney

see lines and logs

mobius stuff

Incurrence

I have 400 people in my complex analysis class and that is similar to one of the questions

Disciple of Barney

12:03 PM

Oh, not 10

Incurrence

But it is a nice find :)

10 people in my algebra

Well 16

Disciple of Barney

Big difference

Are you candygirl16

Incurrence

Definitely not LMAO

That's not even an account

Disciple of Barney

Oh really

Incurrence

Oh they aren't registered

LMAO the lecturer said at the start of the semester:

"If you go on the internet and just post your problems on math stack exchange under the name candygirl16, you'll get your questions answered and you'll get the assignment done, but you'll LEARN NOTHING"

2

Disciple of Barney

12:06 PM

No one would expect candygirl16 to be Committing to a challange,/Incurrence. Hahahaha

Incurrence

LOL

Disciple of Barney

That is perfect

Incurrence

It's not even registered

All of mine are registered

Disciple of Barney

Or so you say

Incurrence

$z^i=i$

How did you find that one? What is the method that is gaining so many false positives?

The line thing?

It's a funny theory, I'll give you that

Disciple of Barney

12:10 PM

I don't know

Incurrence

I have 29 accounts after 1.7 years, I don't make 4 a day lmao

Disciple of Barney

Sure, I guess you only asked questions under Incurrence today, although I wouldn't be surprised if you were Candygirl16 and mobius and Nilpotency and Makoto K.

Incurrence

;)

Disciple of Barney

Is the complex analysis class basically and engineer class, is that why there are so many. Is you algebra class linear or is it Lie Algebras

Incurrence

Algebra is Abstract-algebra II and it is notoriously hard

Disciple of Barney

12:18 PM

Ah, what are you guys covering

Incurrence

Complex analysis is compulsory and watered down with biology-engineers-finance-other students

Disciple of Barney

(that is what I figured)

Might not be that fun then

They ruin everything

Incurrence

Don't know, he makes it up as he goes

Disciple of Barney

Ah maybe one of those guys that don't give a $f(x)$

Incurrence

We did linear operators <---> matrices, learning the connection and then category theory for 2 days and now group theory

exactly that sort of guy

Disciple of Barney

12:20 PM

Math ed majors were also really bad at my school

Incurrence

ed=education?

Disciple of Barney

Yah

Incurrence

To become highschool teachers?

Disciple of Barney

Yes

That is weird, it wouldn't let me send yah again

Incurrence

Yes, they are painful here, but don't need to do complex analysis at all

Disciple of Barney

12:24 PM

Must be some sort of protection

Incurrence

Yep can't send the same message twice, it joins them

Disciple of Barney

Are you doing homework right now (or should be)

Incurrence

Yep, trying to do some proof of $\tanh^{-1}z=...$

Except I have to go hang out the washing now...(10:30pm)

brb

Disciple of Barney

Okay

Khallil Benyattou

I've got an ansatz question to do with a differential equation! $$ \dfrac{\text{d}^2 x}{\text{d}t^2} + x = 5t\cos(t) + 4\sin(t) \qquad (\star)$$ $(\star)$ is the differential equation I'm trying to solve. I've found the complementary solution (of the homogeneous case of $(\star)$) but for the particular integral, I've tried an ansatz of $x = (at + b)\cos(t) + (ct + d)\sin(t)$ but it doesn't seem to be producing the same general solution as Wolfram. Is there something wrong with my guess for $x$?

Oh ... Is it because $b\cos (t) + d\sin(t)$ already appears in the complementary function that I need to multiply by $t$ for the particular integral?

Incurrence

12:40 PM

Food finally :), haven't eaten since 10h ago

Disciple of Barney

Thats no good, hopefully it is not poisoned

Incurrence

I mean I haven't eaten anything in 10 hours, not that the food is 10h old lol

Jasper Loy

Hi @Incurrence

Incurrence

Feeling any better JL?

Disciple of Barney

I got that, I was just saying I hope its not poisoned

:)

Incurrence

12:41 PM

I do too barney

Do you have other accounts @Dis

Are you another account of mine?

Jasper Loy

@Incurrence I need some help from some people to solve some part of my problems. I hope they will cooperate. If they don't it will be hard for me. So I pray they will help.

Incurrence

What sort of people?

Disciple of Barney

I think I once used an unregistered account, but this is my only one besides that one. I might be just another one of yours though, it is a very deep philosophical problem that your accounts have been considering.

Jasper Loy

@Incurrence Some 'friends'.

Disciple of Barney

Good luck finding that question though

Incurrence

12:46 PM

@JasperLoy Friends because they don't seem real?

Jasper Loy

@Incurrence They are not too understanding in certain aspects.

Incurrence

That is pretty normal, especially when one gets into a trend of thinking in certain ways

This is where you were saying that they think you aren't trying to get better isn't it

Jasper Loy

No, no, I am referring to something else now.

Khallil Benyattou

(Can anyone confirm my suspicions about the DE question I posted about a screen's-height above?) ^_^

Incurrence

Do you know variation of parameters?

Disciple of Barney

12:50 PM

I found the question (I didn't know where it was) here

@Incurrence

Incurrence

@DiscipleofBarney OMG that's one of my accounts, how the heck did you find it???

Disciple of Barney

Haha

Incurrence

That's the third one ever

Disciple of Barney

Oh and it is registered

Jasper Loy

Anyone here uses gmx mail?

Incurrence

12:56 PM

What is Gmx mail?

Jasper Loy

It is a mail service, like gmail etc.

Incurrence

Nope, I have used hotmail, gmail, bigpond mail

Jasper Loy

Do you have a religion?

Incurrence

Nope

Jasper Loy

OK, still, you can pray for me. =)

Incurrence

1:00 PM

I will pray for you :)

Disciple of Barney

I am a disciple of Barney if that helps you

Jasper Loy

I will pray for you too. =)

I don't know who that is

Incurrence

wut

Khallil Benyattou

@DiscipleofBarney Hahahahaha!

Jasper Loy

Who is Barney

Khallil Benyattou

1:01 PM

Please no, @Jasper. Pls ...

y u do dis jasp

OMG

I am going now, bye. Pray that I get my miracle @Incurrence.

Incurrence

I shall

Disciple of Barney

1:02 PM

yesterday, by Kaj Hansen

@DiscipleofBarney http://cdn.slowrobot.com/2232015031586.jpg

Or my profile pic

teadawg1337

Hello everyone

Disciple of Barney

Hello

Who are all these people that star questions but don't vote on them?

I feel people star questions more than vote for them.

teadawg1337

Dunno, the starring system seems somewhat superficial IMO

Disciple of Barney

Seems like a ton of questions have one star and no votes, seems a little backwards but whatever

Saal Hardali

2:00 PM

"We can represent points in $T^2$ as pairs $(e^{i\theta_1},e^{i\theta_2})$."
I understand that this is a parametrization of the torus as a product of circles in $\mathbb{C}^2$. But then i am given a map defined in terms of the parametrization like so:

$$(e^{i\theta_1},e^{i\theta_2}) \mapsto F(e^{i\theta_1},e^{i\theta_2})$$

How am is supposed to check it's smoothness? what are the actual charts on $T^2$?

(the quote is from a book)

user112495

3:12 PM

I've been given the quadratic form $q(v) = x_1x_3 + 2x_1x_2 + 2x_1x_4 + 2x_3x_4$ and I need to compute its rank and signature. I intially used the substitutions $x_1 = y_1 + y_3, x_3 = y_1 - y_3, x_2 = y_2, x_4 = y_4$ and eventually got that I could write the quadratic form as $(y_1+2y_4)^2+(y_1+y_2)^2-y_1^2-4y_4^2-(y_3-y_2)^2$. However, this would suggest that the quadratic form has rank 5 which clearly can't be the case. Am I just generally going about this the wrong way?

user112495

3:47 PM

Is it because I've got some variables in several different squares?

Thomas Andrews

Just finished a lovely math nerd book, "Carry on, Mr. Bowditch." It's for boys, but it is a fun read as an adult.

teadawg1337

@ThomasAndrews I'm intrigued, what's the author's name?

Balarka Sen

Hi @DanielFischer

Daniel Fischer

Hi @BalarkaSen

teadawg1337

Nvm, that was a silly question, I can just use google @ThomasA

Balarka Sen

3:56 PM

@DanielFischer I've decided to revise algebra now that I have done a lot of topology and algebraic topology in the past few months. The long-time goal is to study linear algebra and multivariable calculus before studying cohomology.

such a horrible internet connection.

Daniel Fischer

@BalarkaSen Where I come from, one learns linear algebra before algebra and topology.

Balarka Sen

True, true. But seeing that I didn't do it, I am going to do it the other way.

I have finally met the need to know a little multivariable calc.

user112495

Hmm. I'm still not sure how to do this. Why does what i've done above not work?

What is the best method for calculating the signature of a quadratic form?

Sayan Chattopadhyay

hi@BalarkaSen

evinda

4:11 PM

Hello @DanielFischer
I have to determine if the proposition "Polynomial: good, exponential: bad" is right and justify my answer.
Could I say the following?
Suppose that we have input of size n and we want to run an algorithm to get an information about the input.
Since $O(n^a) \subset O(b^m), a,b,m,n \in \mathbb{N}$, the exponential algorithm will require more time to terminate, and so the proposition is right.

Balarka Sen

@DanielF I enjoyed homology though. Initially it might be a bit hard to grok since there is no way to visualize higher homology groups (or visualize them easily : the only geometric description of higher dimensional cycles I know of involve bordism), but the algebraic tools provide a generic algorithm of some sort for computing homology of arbitrary spaces. Much easier than fundamental groups.

Thomas Andrews

Sorry, I turned off the notifications when mentioned, and meant to turn it on again. @teadawg1337

Set in 1770s-1803, it's about a boy who grows up to re-invent scientific navigation. He wrote a navigation book that was used for 150 years (and is still required in all US Naval vessels.)

David Wheeler

4:53 PM

Sailors were one of the earliest trades to recognize the practical power of math. Like, save your life stuff n shizz.

Mary Star

Could someone help me at the following??

0

Q: Integer - Fractional version of the Knapsack problem

There are two versions of the Knapsack problem, the integer and the fractional one. With what type of programming can these two versions be solved?? Which of the two versions has the optimal substructure property ?? Which of the two versions has the greedy choice property ??

@DanielFischer Do you maybe have an idea??

Karim Mansour

5:18 PM

morning people.

ᴇʏᴇs

Crab people

Karim Mansour

lol

Ben Beri

When I google for circle formula, it gives me $A=πr2$, isnt it $(x - a)^2 + (y - b)^2 = R^2?$

Antonio Vargas

5:34 PM

@BenBeri, the area of a circle with radius $r$ is $\pi r^2$. That's what $A = \pi r^2$ means. The second equation you wrote describes the set of points of some circle.

Karim Mansour

5:46 PM

@DavidWheeler here?

want to ask a question about some corollarly related to sylow theorem

Jasper Loy

6:14 PM

@ᴇʏᴇs Hi

David Wheeler

Hi

Karim Mansour

Hey david I don't understand something in this proof

Let G be a group of order pq, where p and q are primes such that p > q. If p doesn't divide (p - 1), then G is isomorphic to $Z_{pq}$

Jasper Loy

How can p divide p-1?

Karim Mansour

q

not p

Jasper Loy

So edit your question.

Karim Mansour

6:21 PM

Let G be a group of order pq, where p and q are primes such that p > q. If q doesn't divide (p - 1), then G is isomorphic to $Z_{pq}$

David Wheeler

How many sylow $q$-subgroups do you have?

Karim Mansour

The number of sylow q-subgroups must divide |G| = pq, and hence must be one of 1,p,q,or pq.

David Wheeler

You can say more.

Karim Mansour

and any sylow n-subgroups must be of this number 1 + kn for some non-negative integer k.

So for the case of sylow q-subgroups

David Wheeler

I mean, clearly we cannot have $pq$ as the number of sylow $q$-subgroups, nor can we have $q$.

Karim Mansour

6:26 PM

yes if we have $pq$ then we must have that pq = 1 + qk which means that q divides 1

and we can't have q since that would mean we will have q = 1 + qk which is impossible.

David Wheeler

So it's down to $p$, or $1$.

Karim Mansour

yes

David Wheeler

If $1 + kq = p$, then $kq = p-1$, but....

Karim Mansour

yehh

coz q doesn't divide (p - 1)

David Wheeler

so the sylow $q$-subgroup is normal.

Karim Mansour

6:29 PM

so we must have 1 sylow q subgroup hence we have that is normal

but now I want the case for

p

Balarka Sen

so you have a short exact sequence 1 --> Z_q --> G --> Z_p --> 1. can you prove that this splits? what's the section?

Karim Mansour

If we have p then we can't have p or pq by same logic

we can't have q I see because we must have p > q

so that is why

oke understood thank you @DavidWheeler

David Wheeler

Now you have $G = P_2P_3$ where $P_2,P_3 \lhd G$ and $P_2 \cap P_3 = \{e\}$.

Balarka Sen

you have yet to prove that G is isomorphic to Z_pq

Karim Mansour

yeh

G = HK

where |H| = p and |K| = q

Balarka Sen

6:33 PM

if H and K is normal in G = HK and has trivial intersection, then G = H \times K (prove it).

David Wheeler

Now give a counter-example when $q|(p-1)$

@BalarkaSen that's probably already an earlier theorem he's covered.

Karim Mansour

yeh

lets see

Balarka Sen

@DavidWheeler possibly, yes.

David Wheeler

You want to show $hk \mapsto (h,k)$ is a bijective homomorphism.

Karim Mansour

yeh

that is covered in my book

Balarka Sen

6:40 PM

for the counterexample, why not begin with $p = 2, q = 3$?

that has the property that $q|(p-1)$. is $\Bbb Z_6$ the only group of order $2 \cdot 3 = 6$?

Karim Mansour

no

for example S_3

Balarka Sen

true. but what went wrong?

Karim Mansour

coz based on the proof above we can't use the fact

that p doesn't divide (p - 1)

Balarka Sen

sure. but what happens to the sylow subgroups if q divides p - 1?

i want to you think about their normality. the gist of the proof for q not dividing p - 1 was the normality of both of the sylow subgroups

Karim Mansour

ohh

coz then it will not be normal

since it is not the only sylow-q subgroup of G

Balarka Sen

6:45 PM

right! but how many sylow-q subgroups will there be, then?

second sylow theorem says something about it.

think about the easier case of p = 2 and q = 3. numbers help :)

Karim Mansour

hm

If we have P and K are sylow p-subgroups of a group G then there exists x $\in$ G such that P = $x^-1Kx$

Balarka Sen

you have a single 3-sylow subgroup. how many 2-sylow subgroups are there?

Karim Mansour

so

in this case

Balarka Sen

my bad. it's the third sylow theorem.

David Wheeler

A more advanced view is this: if $q|(p-1)$ and the sylow $p$-subgroup is normal, call it $N$, then for every element $h \in H$ where $H$ is any sylow $q$-subgroup, we have $hNh^{-1} = N$, that is: conjugation by elements of $H$ yields an automorphism of $N$

Karim Mansour

6:49 PM

yeh

Balarka Sen

@KarimMansour what does it say about the number of 2-sylow subgroups?

Karim Mansour

1 + pk so we have either 1 + 2 = 3 or 1 + 4 = 5 so it can only have 1

sylow 2

since

I mean 3

since 3 divides |G|

Balarka Sen

it's 1 or 3, isn't it?

because both of them divide 6 and are 1 modulo 2.

David Wheeler

In the case where $N \cong \Bbb Z_p$, the automorphism group is iso. to $\Bbb Z_{p-1}$, and we need the $q|(p-1)$ hypothesis to get a non-trivial conjugation action.

Karim Mansour

yeh

Balarka Sen

6:54 PM

@KarimMansour the 1 case is already been dealt with. what happens when you have 3 2-sylow subgroups?

David Wheeler

Read the comments to my answer here: math.stackexchange.com/q/1211612/23285

Karim Mansour

then our sylow subgroups won't be normal

K is normal in G iff K is the only sylow p-subgroup in G

David Wheeler

What are the Sylow 2-subgroups of $S_3$?

Karim Mansour

can we find them by brute forcing the elements of S3 right ?

I mean finding which elements hav

have order 2

Balarka Sen

@KarimMansour true, but then you have 3 2-sylow subgroups and your group G acts on these 3 sylow subgroups by conjugation by sylow's second theorem...

Karim Mansour

6:58 PM

yeh

Balarka Sen

so the permutation representation gives you a homomorphism G \to S_3, doesn't it?

can you prove that this is an isomorphism?

evinda

Hey @AntonioVargas
How would you justifty that the proposition Polynomial:good, exponential: bad is right?

David Wheeler

Yes, in $S_3$ there is a one-to-one correspondence between elements of order 2, and the sylow 2-subgroups. since $4\not\mid 6$.

Karim Mansour

oh

that is nice

Antonio Vargas

@evinda I saw your proposed reason for it the other day. It's pretty hand-wavy but I think it captures the gist of the idea.

evinda

7:01 PM

@AntonioVargas Should I use the fact that $O(n^a) \subset O(n^m)$?

David Wheeler

More generally, if $p$ is an odd prime, then the dihedral group $D_{2p}$ has how many elements of order 2 (that is, sylow 2-subgroups)?

Balarka Sen

@KarimMansour are you referring to what i was saying or were you talking about david's messages?

Karim Mansour

to both I didn't know one could define an isomorphism between the sylow 2 p subgroups of them

in general

Balarka Sen

? we haven't defined an isomorphism between sylow 2 subgroups

we just proved that there are two groups of order 6, Z_6 and S_3

Karim Mansour

because david was mentioning that we could have one to one corespondence between order 2 elements of S_3 and sylow 2 subgroup of Z_6

Balarka Sen

7:06 PM

oh sure

you mean sylow 2 subgroups of G, not Z_6.

Karim Mansour

yeh

Balarka Sen

that's just the usual permutation representation coming from sylow's second theorem.

Karim Mansour

i see

my book doesn't mention group actions but the way dummit describes those theorem is by group actions.

David Wheeler

So, in general, we have TWO types of groups of order $pq$, Type I (where $q\not\mid p-1$) is cyclic, Type II is more complicated.

Balarka Sen

You can get an explicit presentation for the type II nonetheless without knowing the notion of semidirect products

David Wheeler

7:15 PM

The ones in type II MIGHT be cyclic, but they might not be.

Karim Mansour

oh

we didn't cover semi direct products

David Wheeler

Right, you just have to specify, for a generator $x$ of THE sylow $p$-subgroup, and a generator $y$ of A sylow $q$-subgroup, what $yxy^{-1}$ is.

You know it's going to be some power of $x$, and the choices are limited.

Balarka Sen

Artin takes this approach of just giving the presentation of the groups without explicitly introducing semidirect products anywhere. I found this quite fun, particularly when they introduce the Todd-Coxeter algorithm to check whether some group collapses or not (i.e., when the type II groups are cyclic or not, etc)

It's quite reasonable for small groups, say.

Karim Mansour

I am using hungerford

and dummit

David Wheeler

You're aiming high, which is admirable. I like Herstein, myself, it's easy to read. Fraleigh is alright, but he goes on and on about stuff.

Aluffi, Chapter 0, seems to be the "in-thing" these days.

Balarka Sen

7:27 PM

weird name

columbus8myhw

Assuming that there exists a constant $k$ such that $2^x\ge kx+1$ for all $x$, and without using calculus, I believe you can prove that:
$$k-\frac1n<1-\frac12+\frac13\dotsb\frac1n<k+\frac1n$$

(Of course, calc tells you that $k$ is $\ln2$, but for that we need to define $e$, which is annoying.)

And the premise, that there is such a $k$ with $2^x\ge kx+1$ for all $x$, isn't too hard to believe. Playing with Desmos tells you that it's probably $\approx0.7$.

(I don't remember the details, at the moment, but I think it involves plugging in stuff like $x=\frac1{nk}$ into the equation for various integers $n$. Or maybe it's $x=-\frac1{nk}$. Maybe both. I don't remember exactly.)

Something for you guys to think about.

Mike Miller

7:49 PM

morning

ᴇʏᴇs

Morning @MikeMiller

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$Mathematics$ Mathematics