Mathematics - 2012-04-20 (page 3 of 3)

Dan Brumleve

10:00

good night! light will come here soon i think

x4d33746153706c306974

@DanBrumleve I meant electricity

anon

A k-combination is a subset of k items out of a collection.

x4d33746153706c306974

$user image$

This coefficient is from which binomial theorem?

anon

There's only one binomial theorem I'm aware of, so it wouldn't make sense to ask which one.

x4d33746153706c306974

@anon hmm I thought there are two , one for natural numbers and other for integers , isn't it?

aah, natural one derives from integers one I guess

Dan Brumleve

10:07

well the binomial theorem concerns the coefficients of the expanded product (x+y)^n but here you have a practical application of the binomial formula and i think needn't be concerned about the meaning of the theorem itself.

anon

The binomial theorem is typically understood for naturals n and k, but it generalizes to even complex numbers I think.

x4d33746153706c306974

@DanBrumleve right , thanks a lot :)

Dan Brumleve

... unless there is some rewarding way to rephrase combinations of letters in numeric terms?

x4d33746153706c306974

you guyz are great mathematician and I'm also trying to make my maths good , so that I become good at algorithms :)

Dan Brumleve

good way to do it

for algorithms i recommend try to wrap your brain around quickselect if u haven't already en.wikipedia.org/wiki/Selection_algorithm

(something i found recently that bucked intution)

x4d33746153706c306974

12:56

@DanBrumleve can you help with this question : stackoverflow.com/questions/10240483/…

Well I thought to use set or map but space overhead was too much

Daniil

13:54

Hi!

Rob

Hi!

Matt N.

14:19

@JM Ayt?

I have a question regarding tags.

J. M.

@MattN How may I serve?

Matt N.

This one has created a tag "denseness". I'm quite sure this used to exist and "we" decided to delete it.

What do you think?

J. M.

Yes, it sounds rather too specific...

Matt N.

And he's done it again here.

Can I flag these for deletion? Or what can I do?

I suggested a tag-wiki in which I wrote that it should be deleted. Someone will see that and delete it (or disagree and leave it).

J. M.

You can remove the tags yourself. The system should purge them in a day or two...

Matt N.

14:28

Ah, from the question! Right. : )

I can't make sense of this comment.

@JM Done! : )

Yay : ) And I got accepted : )

That's quite some profile picture they have there. (not the puppy)

J. M.

@MattN It scares me that I can't quite tell if it's a lady's or a guy's midriff...

Matt N.

@JM Looks female to me...

J. M.

I hope you're right...

Matt N.

14:52

The underwear is a give-away, no?

J. M.

...I've seen strange stuff... :o

(i.e. I have been suitably traumatized.)

Matt N.

: D

J. M.

On another note: once upon a time there was this game where you take a look at pictures of cleavage and then guess if it was from the chest or the buttocks. That certainly made me paranoid about pictures... :)

Matt N.

I didn't know there was such a game : )

J. M.

Of course, for maximum effect, they always use guy buttocks...

Matt N.

15:05

: D

You really made that chocolate tea pot : )

J. M.

@MattN It wasn't too hard, and you gave me the idea...

Matt N.

: )

Ilya

15:21

home, sweet home

Matt N.

Ello : )

Ilya

olle!

J. M.

@Ilya That would be Russia, no?

(Also, hi.)

Ilya

@JM hi. I live here more than for one year now. And here it's sweet. In Russian it's not

Matt N.

: D

J. M.

15:25

Heh.

Ilya

@MattN don't imagine candy shops, I didn't mean that - although there is Leonidas shop just below my window here

J. M.

Mmm, Leonidas...

Matt N.

Never heard of it.

J. M.

Good chocolates. Very costly, though.

Ilya

@MattN since you're not in China, google can help you ;)

Matt N.

15:30

: )

bbl

Ilya

bye

@JM: how are you?

J. M.

@Ilya Only slightly better than the last time... :)

Eric Gregor

16:00

i am considering the map $\Phi(x,y)=x^3-y^2$, with Jacobian $\begin{pmatrix}3x^2\\ -2y\end{pmatrix}$. I want to show that $\Phi^{-1}(0)$ is an injective immersion (i think it is anyway). is the point to show that $3x^2=-2y$ is an injective immersion of $\mathbb{R}$ into $\mathbb{R}^2$? will that imply that this is an immersed submanifold?

that is @all

rather i think i want to show that $y^2=x^3$ is an injective immersion

ok, so i think i just want to show that $y\mapsto (y^2,y^{3/2})$ is an immersion (it's obviously injective)

so i compute the Jacobian of this map and get $\begin{pmatrix}2y & \frac32 y^{1/2}\end{pmatrix}$

sorry, the map should be $y\mapsto (y^2,y^3)$

with the Jacobian being $\begin{pmatrix}2y & 3y^2\end{pmatrix}$

this reminds me of localizing in algebra

Ekuurh

16:18

guys, i was wondering something: if phi(n) divides n-1, does that mean phi(n)=n-1?
i can easily prove that it means n is square free, and a Carmichael number, but that's it :)

Eric Gregor

this is a famous curve, i think. i've just never considered it as a manifold

Ekuurh

anyone in here? :)

Eric Gregor

it means n is prime, i think, which is the same thing you are saying

Ekuurh

yea, that's why i'm trying to prove :)

and checked it in a computer for up to a million

so it's probably true. can't seem to prove it though

Eric Gregor

prove that you can't have n be the product of two distinct primes

and try to generalize the argument

Ekuurh

16:24

hmmm. that's probably true. 1 sec

done

it can't be the product of two distinct primes, because
n=pq =>
phi(n) = (p-1)(q-1) = pq-p-q+1 divides pq-1

then pq-p-q+1 divides p+q-2

and the right side is bigger than the left for p,q>2

even p,q>=2

wait no

oh

(p-1)(q-1) divides (p-1)+(q-1)
and the right side is biggger than the left if p,q>2

not sure what about p=2, i'll see what i can do

okay, that implies q-1 divides 1 then q=2, and that's impossible because q does not equal p

eric, are you still there?

Eric Gregor

i'm kind of busy, but lemme see

i'm confused by your argument, could you say it again with a fresh start?

Ekuurh

okay, thanks anyway :)

Eric Gregor

the two prime case should be easy

Ekuurh

okay

i say that for every natural number N, it holds that if phi(N) divides N-1, then N is prime.

i've been able to prove N is square free, and not the product of two distinct primes.

yea, the two prime case is done

and i believe even three primes shouldn't be too hard. (same tricks i did last time, probably more case checking)

Eric Gregor

there should be a more general induction argument you can make

Ekuurh

16:34

square-freenes is easy:
if p^2 divides n, then p divides phi(n). but phi(n) divides n-1, so p divides n-1. but p divides n, so p divides 1. contradiction.

i'm going, thanks eric

Eric Gregor

i'm here

i'm writing something for you

so let $n=p\cdot m$, where $m$ is coprime to $p$. suppose we have shown that $\phi(m)$ does not divide $m-1$

Ekuurh

message me if you have any progress :) good luck with your stuff

Eric Gregor

consider that case now and you should be able to do it

this is just an inductive argument

that's all i have to say, you should be able to finish now

Ekuurh

hmm

i don't think it'll do

wait, what's M?

never mind, thanks for the help, good luck :)

Eric Gregor

fixed. that's my best advice under pressure, i have work to do. i think you can make it work

user979616

18:24

hi all I have a question Ive been asked to solve. But I have no idea where to begin. THe question is y'=(y+e^x)/(x+e^y)

Matt N.

18:37

I just posted an answer tagged differential-equations. I'm not sure I should've done that -- I don't really know much about them.
Does [this](http://math.stackexchange.com/questions/134482/solving-a-differential-equation-involving-y-and-its-exponential) work?

Kannappan Sampath

But, you have not yet separated the variables no?

There is an $x \mathrm{d}y$ there... @Matt

Matt N.

Yes.

N3buchadnezzar

Heya =)

Matt N.

Can't I do $\int x dy = xy + K$?

Kannappan Sampath

Hmm, no, I think. Why is a function of $x$ right?

Matt N.

18:43

That was just an idea.

Might have to think of something better.

N3buchadnezzar

Do any of you two know anything about ellipses?

Matt N.

Nope.

N3buchadnezzar

Okai!

Kannappan Sampath

I think I learnt about them in High school. So, ask.

@MattN the why is $y$ there. Sorry about that...

Matt N.

Already got a comment with a word that I don't know. Oh my. : )

@KannappanSampath Sure : )

N3buchadnezzar

18:45

I need to show that $x = \sqrt{3} \cos t \ , \ y = \sin t \qquad t \in [0,2\pi]$ desribes a ellipse.

Kannappan Sampath

Do you know the standard form of the equation of an ellipse?

N3buchadnezzar

My first guess was to show that it is on the form $(x/a)^2 + (y/b)^2 = 1$. But this seems very simple.

Kannappan Sampath

^Yes, exactly.

That is the way to go. It is simple, so what?

N3buchadnezzar

So without further ado I can just say that $a=\sqrt{3}$ and $b=1$ ?

Kannappan Sampath

Yes.

N3buchadnezzar

18:48

I do not quite know why I can say that though.

I see that it works obviously.

Kannappan Sampath

@Matt f$\rm{d}y=g\rm{d}x$ is exact if $\partial f/\partial x=\partial g/ \partial y$...

Do you know the parametric equation of an ellipse...? @N3buchadnezzar

Matt N.

@KannappanSampath Ooh! Cool! Thank you!

N3buchadnezzar

It is $x = a \cos t$ and $y = b \sin t$

for $t \in [0,2\pi]$

Kannappan Sampath

So, yes.

N3buchadnezzar

But this is what I need to prove, using this would be rather meaningless?

Matt N.

18:52

Deleted it. Shouldn't mess with differential equations.

Kannappan Sampath

Hey, here is what you do: Show that what you have is an ellipse by saying it's equation looks like one. Then, if you had an ellipse, its parametric equation looks like what you have. Since, you have what you have (!), you must ge t what you want. :D @N3buchadnezzar

N3buchadnezzar

Then I will never need to prove anything again :D

Kannappan Sampath

@MattN look up for solving exact equation, it is quite an easy thing.

N3buchadnezzar

I will just say that there is a theorem saying it is true, and since it is true. My proof is complete.

Matt N.

heads out to look up solving exact equation

Kannappan Sampath

18:54

@N3buchadnezzar I mean it is obvious to me. But, I don't understand what you're doing....

Matt N.

Now feeling stupid again. Haven't felt stupid for at least a week.

Kannappan Sampath

Why?^ (If I may ask. :))

Hi @ymar

ymar

Hi guys!

Matt N.

@KannappanSampath This. Of course you may ask : )

@ymar Hi ymar : )

ymar

I think I hit the jackpot with this question. No correct answer and I've already learned a lot.

Matt N.

19:07

I saw and upvoted : )

ymar

Thanks. Qiaochu's answers and the comments are dripping knowledge. :) I love it.

Ilya

@ymar that drives you to the Dark Side, young padawan

ymar

@Ilya I'll have to see those movies some time to understand what people are saying. :)

Matt N.

: )

Kannappan Sampath

Me too. I am very weak when it comes to understanding such humour/ abuses.

Ilya

19:15

Star Wars is easier to catch then Star Trek:
1. only 6 episodes
2. there is N. Portman

ymar

She's a beauty, isn't she.

Matt N.

I'd only count 3 episodes...

heads out to look up Natalie Portman

N3buchadnezzar

@KannappanSampath Mind me asking another question about the ellipse?

Kannappan Sampath

np. :) You may ask.

N3buchadnezzar

Its about integrals :p

Kannappan Sampath

19:20

I don't know if I can answer but will try.

N3buchadnezzar

Show that the arclength of the ellipse can be expressed as $$ L = 2 \int_0^\pi \sqrt{2 + \cos u} \, \mathrm{d}u $$

For the love of god, I can not show this. And I believe it is wrong.

Kannappan Sampath

There is a formula involving the parametric derivatives right?

N3buchadnezzar

Or I am not strong enough in my kung fu trigonometric juggling abilities.

Ilya

@ymar sort of

N3buchadnezzar

$$L = \int_{t=a}^{t=b} \sqrt{ \left( x'(t) \right)^2 + \left( y'(t)\right)^2\,} \,\mathrm{d}t $$

Matt N.

19:23

@Ilya Not my type...

ymar

She's great to look at in movies. But I wouldn't bother to look up her naked photos.

Matt N.

Heh : )

Ilya

@ymar nonono, she's not that kind of girl - that kind of cuteness which somehow contradicts with being sexy

N3buchadnezzar

I get $$ L = 2 \int_0^\pi \sqrt{2 + \cos 2u\,}\,\mathrm{d}u$$

ymar

@Ilya Exactly.

Ilya

19:27

@ymar: in Black Swan erotic scene Mila Kunis was leading, Portman was acting

ymar

That was actually pretty sexy, although I didn't like the movie very much.

Kannappan Sampath

@N3buchadnezzar Then you probably are right...I should work the details...

N3buchadnezzar

@KannappanSampath 2sec

Ilya

@ymar me too. I wouldn't compare it with Requiem for example. B.S. was also kind of expressive, but there is no tragedy. Or it does not touch me. I even would say that Jared was playing much more natural than Natalie

ymar

I didn't see that. What is it?

Kannappan Sampath

19:29

Can some one tell me how to write that something is happening $k$ times in underbrace construct?

Ilya

Requiem for a Dream

Requiem for a Dream is a 2000 drama film directed by Darren Aronofsky and starring Ellen Burstyn, Jared Leto, Marlon Wayans and Jennifer Connelly. The film is based on the novel of the same name by Hubert Selby, Jr., with whom Aronofsky wrote the screenplay. Burstyn was nominated for an Academy Award for Best Actress for her performance. The film was screened out of competition at the 2000 Cannes Film Festival. The film depicts different forms of addiction, leading to the characters’ imprisonment in a world of delusion and reckless desperation that is subsequently overtaken and devastat...

N3buchadnezzar

It seems like
$$ \int_0^\pi \sqrt{2 + \cos 2u\,} \, \mathrm{d}u = \int_0^\pi \sqrt{2 + \cos u\,} \, \mathrm{d}u$$

@KannappanSampath Any idea how to show that?

Kannappan Sampath

@N3buchadnezzar Hmm, no? Put $2u=t$ in the integral on left...

Ilya

@ymar: you didn't see that movie? come on, forget about Star Wars for a while. This is a really great one

ymar

Oh, I 've heard of it. I'll watch it tonight. I finally have money so I can get wasted and watch a movie.

N3buchadnezzar

19:32

@KannappanSampath Then the limits changes, mmm. Will look into it

Seems strange :p

ymar

@KannappanSampath I ate some Indian food today for the second time in my life. Does all Indian food have loads of ginger in it?

Kannappan Sampath

@ymar No, not all. Seems strange to me.

ymar

Oh, good. I hate ginger...

Ilya

@ymar maybe you were ehhh faked, and that wasn't an Indian food? ^_^

ymar

It was an Indian restaurant. And there was an Indian guy in there. :)

N3buchadnezzar

19:36

@ymar We all hate gingers :p

Rob

encrypted-tbn3.google.com/…

ymar

BRB

Ilya

@N3buchadnezzar Cartman?

N3buchadnezzar

@Ilya Yeah and this

Jonas Teuwen

@Ilya Hi!

Kannappan Sampath

19:45

@N3buchadnezzar :)

Ilya

@JonasTeuwen Hi! how are you

Jonas Teuwen

@Ilya Okay, you?

Ilya

okay

drinking glenlivet in small doses :)

and decided finally to update my facebook page, although photos are not of the best quality - made with the phone

Matt N.

@Ilya Ah, never mind, I can't view his friends.

Ilya

@MattN: but you know my surname. I shared photos for everybody

Matt N.

19:52

@Ilya Check FB : )

@Ilya Found them : )

Ilya

@MattN that's you???

Matt N.

@Ilya Yes? Why? It's a made up name.

Ilya

but homecity...

Matt N.

Don't like private details of me on the internetz.

: )

Ilya

but why Russian city?

Matt N.

19:55

@Ilya You troll : D Do you have any pictures of Beijing apart from the motorway? : D

@Ilya If someone wants to stalk me this seemed like a good place to send them to.

Ilya

@MattN wait a minute, I should figure out how to add them

@MattN have you heard about Sobolev spaces?

Matt N.

@Ilya Yes. Don't remind me. I have an exam about them, Futile Attempts, and today I realised that I can't even do a simple exercise the teddy gave me. :,(

Now in panic mode.

Jonas Teuwen

@Ilya Sobolev spaces!!! :-).

Ilya

@MattN Sobolev (and many other strong Russian mathematicians) lived and worked in NS, there is a very strong math department

Matt N.

@Ilya Well, apart from the fact that it's in Russia I'd probably like it there. I hear it's cold. : )

Ilya

20:01

@MattN the amplitude is very huge. -30..+45

Matt N.

@Ilya Eww, 45. I'd never survive that : /

ymar

But you'd be happy in -30? :)

Ilya

@MattN btw, the also photos I've made from the plane also refer to NS area

Matt N.

@ymar Yes : )

ymar

It's never been -30 in Warsaw in my lifetime I think, but -20 is awful. And I like cold weather.

Matt N.

20:03

@Ilya Can't see much though : )

@ymar I've had -25 here once, in a tiny ski resort and it was lovely. Very sunny, plenty of snow and we went cross-country skiing : )

In retrospect I'm probably too lazy to do cross-country skiing.

Ilya

@MattN heh, that's different when you have a continental climate and a wind

Matt N.

@Ilya We have continental climate here. NS does as well, no?

Ilya

@MattN haha, you have continental climate in ZH?

ymar

Sure, it's good for skiing, but not opening the window in your house. And for going out to the shop or anywhere. Also, public transportation fails completely.

Matt N.

@Ilya Of course. It's in the middle of the European continent, how more continental can it get?

@ymar True but they already fail if there is zero degrees and one inch of snow. : D

Ilya

20:08

@MattN middle, ha.

ymar

It gets much more continental in Russia. :) Switzerland is pretty close to the ocean compared to Moscow or Siberia.

Matt N.

Ok. You win : )

ymar

The center of gravity of Europe is somewhere in Poland or Lithuania I think.

Matt N.

What's the centre of gravity of Europe?

Ilya

@MattN nonsense. It's just important for Polish people to be in the center (@ymar :-p)

ymar

20:11

map

Matt N.

Stockholm is probably nicer to live.

ymar

The centre of gravity is the point you pin a needle into, turn it all around, fix the needle to the table, and Europe doesn't fall down. :)

Matt N.

Equally cold, or possibly colder, especially in summer.

@ymar I didn't think you meant it literally : D

ymar

Not really, no. But you can do it with an appropriately scaled model. :)

"After a re-estimation of the boundaries of the continent of Europe in 1989, Jean-George Affholder, a scientist at the Institut Géographique National determined that the Geographic Centre of Europe is located at 54°54′N 25°19′E. The method used for calculating this point was that of the centre of gravity of the geometrical figure of Europe. This point is located in Lithuania, specifically 26 kilometres (16 mi) north of its capital city, Vilnius, near the village of Purnuškės."

Matt N.

Heh : )

I'm going to sleep. Terrible day. : /

Good night.

ymar

20:24

Good night!

Ilya

@MattN: good night

Kannappan Sampath

@ymar Linear Algebra....is killing me.

ymar

@KannappanSampath That's not very nice of her.

I can try to help.

Kannappan Sampath

The heck of me. I fail to understand what a gen. Eigen space is...

ymar

How about this: math.stackexchange.com/questions/42350/…

Is it helpful?

Kannappan Sampath

20:31

But, Arturo's answer does not define what this is?

See, we defined gen. eigen space to contain all vectors $v$ that anihilate some power of $T-\alpha I$.

ymar

Isn't that exactly what Arturo says?

Kannappan Sampath

But, some define it to be $\ker (T-\alpha I)^d$ where $d$ is the power of $t-\alpha$ in $\chi(t)$

ymar

I'm unfamiliar with the notation. What's $\chi(t)$?

Kannappan Sampath

The characteristic polynomial!

ymar

Ah, right.

OK, so the problem is to prove the equivalence of these definitions?

Kannappan Sampath

20:36

Yes.

Firstly, we have... $\ker T \subseteq \ker T^2 \subseteq \ker T^3 \subseteq \dots$

ymar

And the same goes for $T-\alpha I$

Kannappan Sampath

Yes. But, then, the thing is to prove, a vector annihilated by $d+k$th exponent is also annihilated by $d$ for $k \in \Bbb N$.

ymar

Would it be enough to prove that $(T-\alpha I)^{d+1}=0$?

No that's nonsense, wait.

Kannappan Sampath

Yes, that's what I was wondering. :P

We need to prove $(T-\alpha I)^d \equiv 0$ for all $v$ annihilated by some power of $T-\alpha I$.

Oh!!!!!!!!!!!!!!!!!!!!!!!!!Is this obvious?!!!

I think we need to argue with Jordan blocks...

Ah! right. So, $T-\alpha I$ restricted on our gen. eigen space is an upper triangular matrix.

(Have I lost you?)

ymar

No, I'm here, trying to collect my thoughts. Go on, it'll click at some point.

Kannappan Sampath

20:46

For me, this room is always an Aha! moment. I get some beautiful thoughts when I type something up...

anon

maybe this will help too

Kannappan Sampath

@anon Nice link. Thank you. :)

anon

the key part is that the primary decomp thm indicates a gen eigenspace has dimension equal to the algebraic multiplicity, I think

Kannappan Sampath

Yes. You're right.

OK, So, the index of nilpotence of $T-\alpha I$ restricted on our eigen space is what we'd want to know.

So, this is atmost $d$, the size of the block.

But, as anon points out, if there are less than $d$ linearly independent eigen vectors, say, then there are more blocks of smaller size.

So, each block gets annihilated by its size. The whole operator is nilpotent with index equal to the maximum of the size of the smaller blocks. Right?

anon

Since I don't like blocks, how bout this: restricting $T$ to $V_{(\alpha)}$ creates a new linear map and $\alpha$ is the only eigenvalue, which means the characteristic polynomial must be $(\lambda-\alpha)^d$ where $d$ is the dimension of the gen eigenspace which equals the algebraic multiplicity on the larger space $V$. Then invoke Cayley-Hamilton on $T$ on the subspace.

Kannappan Sampath

20:56

[...]where $d$ is the dimension of the gen. eigen space [....] why is that the case?

anon

by the link I gave, which I found only minutes ago and have not yet read :P

(also, woke up with a headache, need pills)

oh wait

no, the degree of a char polynomial is always the dimension of the space the linear map acts on

sorry, I thought you were asking about the "equals algebraic multiplicity on $V$" part

Kannappan Sampath

@anon Hah!! Right.

ymar

(If anyone is waiting for me to make a comment, then I'm afraid my brain's too numb. It doesn't seem to be clicking for me.)

Kannappan Sampath

Curse me. I should have known this.

@anon Agreed with that.

anon

cool, I learned something today

Kannappan Sampath

21:01

@anon you may need to sleep longer. I have seen you were postponing sleep for some time (from transcript.) :P

anon

I'm going to go pill up and take a shower. bbl.

Kannappan Sampath

See you later.

ymar

See you.

anon

I got 10 hrs, it's just I sleep during the day now so light shines in my room and I'm kind of in a zombie half-asleep state :P

Kannappan Sampath

:)

So, @ymar, the definitions are equivalent then. :)

ymar

21:04

Well, I know they are, because I learned it a couple of years ago. I even understood the proof then. But now I don't.

Kannappan Sampath

OK. I need to think about this too. Why don't we move to CA and discuss this ?

ymar

OK.

t.b.

21:41

call for votes

Jonas Teuwen

After all these years, I have not yet found a good way to study math books :-).

I need advice.

t.b.

@JonasTeuwen I'm here for a few minutes, but not much longer...

Jonas Teuwen

@tb Sure. How do you read a math book? 8-).

(Then I'll try the same)

t.b.

21:56

It is hard to say in general. Some books can be read as easily as a novel, others require a lot of work per page, so my approach is obviously different in those two cases... Also, it is a function of what the book is to me: do I read it for general culture or is it part of what is supposed to be my specialty.

Jonas Teuwen

The later usually for me.

Usually I do something like this: I start reading.

After a few days I notice that I know nothing yet.

So then I take a piece of paper...

So I might just as well skip the first step, that might help.

Kannappan Sampath

@tb Worried that you're not seen often!

Chris

@JonasTeuwen: What I do is studying from a lot sources..from a book, the internet (notes, asking questions) and usually have a notebook where I keep the things I consider important..

As for the books, in example, my book in analysis I didn't help me at all, but my book at analysis II (Thomas) is really helpful!!

Kannappan Sampath

Hi and welcome btw. @tb

Jonas Teuwen

That is one crappy book if you ask me, @Chris.

Folland is more to my liking.

Actually now I bought abstract harmonic analysis by Folland. Let's see how that goes!

t.b.

22:01

@JonasTeuwen Well, having a sheet of paper next to the book while reading is of course helpful... I don't scribble very much but I do take a pause at every definition and try to get an idea what it is actually about. I need to place the things in my own world and with the few examples I actually understand.

Kannappan Sampath

@Chris This just does not work well when you do Graduate studies. You're interested in what only 10% world is interested in and they are hard to find.

Jonas Teuwen

I think I actually register nothing when I don't have a sheet of paper :-). It somewhat forces me to actually think about it.

@KannappanSampath Only 10%?

I think that is way too much 8-).

t.b.

@KannappanSampath I'll have to cut down my participation again, I'll be here more or less every day, but more on main than on chat.

Jonas Teuwen

@tb Sure, that's probably a good thing :-). Will miss you here!

Kannappan Sampath

I'll miss you too @tb. No linear algenra help. No nice examples </3

Chris

22:03

We didn't had Folland in our available choices..

@KannappanSampath: So what do you recommend for graduate studies?

Jonas Teuwen

Hmm, usually there are no books available right?

t.b.

Thanks guys, I'll show up, but I shouldn't spend as much time here as I did recently.

Kannappan Sampath

@JonasTeuwen I know but for argument sake, that looks like a good enough number.

Jonas Teuwen

So, in the previous thing I did there were like two books, Villani and Ambrosio et al.

And then many papers.

Kannappan Sampath

@Chris I am not a graduate student as yet.

Jonas Teuwen

22:05

@tb I can imagine that studying harmonic analysis takes a lot of time which you can't afford to spend here, right? 8-)).

Chris

@JonasTeuwen: No, we are given some choices by our professors..

Sorry I meant undergraduate..that's what I am..(studying computer science)

t.b.

@JonasTeuwen Sure, do write up things. Also, do more than just copy the book. I always found it more helpful to phrase things on my own and sketch the main ideas of proofs. Working the proofs through with a specific example is always very illuminating.

Jonas Teuwen

Lately I've tried to actually try to figure out what made the guy/woman actually to write it this way.

Or how they've thought up this.

t.b.

Writing things forces you to slow down and thus you internalize them more easily. The more independently of the book you do that the more the internalization effect.

Jonas Teuwen

But then suddenly it takes much more time to read through the material and I have so much to read...

Okay. I'll try better.

Chris

22:07

There's always too much to read...My approach is at least what you study, be sure that you have understood it fully

t.b.

@JonasTeuwen but there's only so much you can digest... Maybe do several approaches: read ahead an hour or so without spending very much time on the details, just skimming the proofs, then go back and work through the things more slowly. I don't write into my books, but maybe marking parts that surprised you at the first cursory reading might be helpful.

Jonas Teuwen

Yes, I'd like to mark in books, but math books are so expensive that it is quite a shame...

But then again, you have them to learn from, not to be pretty.

t.b.

Use a very soft pencil.

Jonas Teuwen

Yeah... But for normal books I'd use a yellow marker :-). A pencil might be a good compromise.

t.b.

Then there's also the PostIt® option...

Jonas Teuwen

22:11

But anyway, you would only spend a few minutes here. Thanks for the advice!

t.b.

No problem, but I should really go and do some typing again... See you folks, have a nice week-end!

Jonas Teuwen

See you!

Chris

Bye @tb

robjohn

23:25

sometimes, I hate work. I've spent three straight days and just get gripes. I will be glad to get back to MSE.

2

Kannappan Sampath

Hi @robjohn

robjohn

This is much nicer. I haven't really been here in days, and I got 75 rep today :-)

@KannappanSampath howdy. How goes it?

Kannappan Sampath

Good enough.

I am into a lot of thinking about Linear Algebra these days.

So, will you spend more time here now? are things fine?

robjohn

a bit, sorry, I was afk there

@KannappanSampath what's up?

ping me, I have to switch windows to get back to work.

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