Mathematics - 2012-06-15 (page 4 of 4)

Benjamin Lim

22:00

@MattN what chapters of AM does it cover?

Eugene

never heard of them

Matt N.

I was enjoying myself greatly though, I think the book is totally awesome. Until suddenly a chapter of extensions and contractions in rings of fractions appeared...

Benjamin Lim

Kiran Kedlaya

Kiran Sridhara Kedlaya (; born July 1974) is an Indian American mathematician. He currently is an Associate Professor of Mathematics at the Massachusetts Institute of Technology. At age 16, Kedlaya won a gold medal at the International Mathematics Olympiad, and would later win a silver and another gold medal. While an undergraduate student at Harvard, he was a three-time Putnam Fellow. A 1996 article by The Harvard Crimson described him as "the best college-age student in math in the United States". Kedlaya was runner-up for the 1996 Morgan Prize, for a paper in which he substantially im...

Matt N.

But I'm done with that too now. whew

Benjamin Lim

@MattN hahahahah

Matt N.

22:00

@BenjaminLim Everything.

Benjamin Lim

oh crap

shit I can't help with primary decomposition

Matt N.

I can help myself, don't worry.

Benjamin Lim

@MattN I have answered a lot of your questions to the best of my knowledge

Matt N.

Yeah, primary decomposition is the next chapter!

@BenjaminLim I know, I upvoted you!

Benjamin Lim

for example that question on how to apply the correspondence between prime ideals in the ring and the localisation

experimentX

22:02

hi everyone again ... how to parametrize ds during line integral??

$ ds $

Matt N.

Yes.

I upvote every single one of your answers.

Benjamin Lim

thanks :D

Matt N.

Just for the record.

And now I should go to bed. Have a lot of primary decomposition and intersection numbers to do tomorrow.

Good night!

Benjamin Lim

nite!

@DavidWheeler how is it going?

Matt N.

Oh yes: and all of your questions, too. (if I see them)

Benjamin Lim

22:05

@robjohn Sorry for the mistake. But I'm very confused with the notation and what are the entries of that matrix

robjohn

@BenjaminLim No need to apologize for a typo. I will look at that question in a bit and see if I can help.

Benjamin Lim

@robjohn Thanks love you XOXO

Eugene

it's surprising how many algebraic and differential geometry concepts coincide.

Jonas Teuwen

@robjohn Hi!!

@robjohn Apparently your advisor did a lot with semigroups!

Eugene

i wonder who's been trolling me lately...

Peter Tamaroff

22:13

@Eugene ???

Eugene

someone has been downvoting me without feedback recently.

i don't mind downvoting i just worry the answers/questions i post are wrong

since it's becoming a trend though i guess i'm less worried.

David Wheeler

@BenjaminLim see the edit?

Eugene

@PeterTamaroff for instance this answer i gave

Peter Tamaroff

@Eugene Well, you can always report that on meta!

Eugene

the guy was downvoting it and retracting his downvote for about 5 minutes.

why would i? if it's not wrong who cares?

lol

Peter Tamaroff

22:18

@Eugene I mean, if it is one aisled case, then let it be. But if it is becoming a trend, we can take action.

Eugene

@PeterTamaroff nah. like you (correctly) observed before, rep is just for fun.

David Wheeler

i am trying to avoid gaining any rep.

Benjamin Lim

@DavidWheeler Thanks, but I'm still wondering why in $Df(h(x))$, we have derivatives with respect to $x$...

David Wheeler

i have only 3 purposes on this site: to learn, to help, and to discuss. the reputation scorecard at best only measures 1 of the 3 with any accuracy.

Peter Tamaroff

@BenjaminLim What's the question?

@DavidWheeler You can always answer in a CW form

Benjamin Lim

22:21

@PeterTamaroff math.stackexchange.com/questions/158575/…

Eugene

@BenjaminLim you really need to read rudin. all the questions you've had so far are in rudin

David Wheeler

because f is a function of k+n variables. we have to take partials with respect to all k+n of them, even if those variables are "re-arranged" by h

Benjamin Lim

hmmmmm

Eugene

this is covered in rudin chapter 9

David Wheeler

so first we compute Df AT h(x), and then compute Dh at x, and multiply. that's what the chain rule does.

Benjamin Lim

22:24

ok

so in the example of $e^{x^2}$

we had $h(x) = x^2$

what does it mean then to compute $Df$ at $h(x)$?

David Wheeler

the single-variable case, doesn't prepare you for the multi-variate case, because a 1x1 matrix is the same as an entry.

in your example, the 'f" is the exponential function.

Benjamin Lim

yes

Eugene

@DavidWheeler indeed! for multivariables you have to "guess" the derivative first then prove that it is the derivative!

David Wheeler

and Df is the 1x1 matrix with the single entry f'(a), whatever point we evaluated at.

Benjamin Lim

so actually I differentiate with respect to $h$ and then only evaluate at $x$ no?

David Wheeler

22:26

the derivative of exp is itself, so our matrix is the 1x1 matrix exp(h(x)) = $e^{h(x)}$

Benjamin Lim

@DavidWheeler yes, but the point is we differentiate with respect to $h$

then evaluate at $x$

and not differentiate with respect to $x$

and then evaluate at $h$

Eugene

@DavidWheeler it was so bizarre when i first learnt it.

David Wheeler

no...we just differentiate "exp", and then stick in h(x) for where we evaluated at.

Benjamin Lim

ah so the $u$ there was actually still $x$ then???

David Wheeler

the form dy/dx = dy/du*du/dx only works out in the single variable case

Benjamin Lim

22:28

it was just a "dummy"

And to think that our useless lecturer just stated the theorem with his mouth...

@DavidWheeler I think I get it

if $F(x) = gf(x)$

then $F'(x_0) = g'(f(x_0))f'(x_0)$

in the $g'$ term

you first evaluate the derivative in the usual style

David Wheeler

it's like the derivative of a function from R^n to R^m is "bigger" than f, unlike where m = n = 1, and they're "the same size"

Benjamin Lim

i.e. the matrix has partials wrt all the variables

David Wheeler

yes

Benjamin Lim

and then evaluate at the point $f(x_0)$

David Wheeler

exactly

Benjamin Lim

22:31

Ok yes I think I misunderstood the result of the chain rule

but in several variables a whole lot of shit gets more confusing

Eugene

@BenjaminLim read rudin for god's sake!

Benjamin Lim

@Eugene FOR GOD'S SAKE I AM LOOKING AT PAGE 214 OF IT NOW WHERE THE CHAIN RULE IS

:P

David Wheeler

yes, that's why it's easier to think of Df(a) as just an mxn matrix whose entries change when a changes.

in other words: a "matrix of functions"

Benjamin Lim

yes

Eugene

if everyone would just read rudin no one would have trouble with multivariable analysis

David Wheeler

22:33

when we have a composite function, we have to multiply 2 matrices together, to get the dimensions right

Benjamin Lim

@Eugene rudin actually deteriorates after chapter 7

@DavidWheeler Yes I was getting a bit messed up with the dimensions

Eugene

@BenjaminLim i highly disagree with this statement.

Benjamin Lim

@DavidWheeler for the second matrix with the identity in it, have I go the dimensions right? I believe so

@Eugene Rudin I learnt is not the right place to go for the first time

especially several variables

David Wheeler

yes, because you were thinking of h(x) as in the "domain" of D(f), and then thinking, how does Df(h(x)) wind up the right size?

Benjamin Lim

I like Munkres' Analysis on manifolds

yes

Eugene

22:36

@BenjaminLim i learnt it from rudin. once you get used to it it's a fantastic book for all occasions.

Benjamin Lim

@Eugene I am not doubting that chapters 2,3,4 are excellent

@DavidWheeler yes

David Wheeler

when i first learned multi-variate calculus, we learned linear algebra first.

Eugene

@anon you have to answer the ridiculously difficult ones so you end up not racing with anybody.

Benjamin Lim

@DavidWheeler You know what's messed up and really ****** up, our lecturer for the analysis course skimmed through this shit really quickly and then went to inverse and implicit function theorems

Peter Tamaroff

Anybody wants to proofread this, just in case anything isn't clear? I'll go grab something to eat! BBL

Eugene

22:38

@DavidWheeler i still found it weird how the derivative for multivariables must be "guessed" first.

Peter Tamaroff

@BenjaminLim F**k him!! LOL

Benjamin Lim

@PeterTamaroff I am writing a formal letter to the maths department after the course about this.

Eugene

@PeterTamaroff i understand none of this. must be all wrong then. =)

Peter Tamaroff

@BenjaminLim Seems fair. Good for you.

David Wheeler

well, like to think of it this way: a derivative is the "best" linear approximation available.

Benjamin Lim

22:39

@DavidWheeler Ah one more thing

robjohn

@JonasTeuwen yes, he did; what have you found?

Benjamin Lim

You know we say that $f$ is differentiable at $a$

if there is a linear map $L$ such that

Eugene

ok somebody IS trolling me!! he did it again!!

Jonas Teuwen

@robjohn Oh, I just saw a lot of semigroup results where they cited Stein.

David Wheeler

a certain limit is 0, yes.

Benjamin Lim

22:40

$\frac{f(a+h) - f(a) - L(a)\cdot h}{|h|} \rightarrow 0$ as $h \rightarrow 0$

Now

robjohn

@JonasTeuwen Ah. I am not surprised.

Benjamin Lim

How do we get from here that $f(a+h) \approx f(a) + L(a)\cdot h$?

What happened to the $|h|$ in the denominator of the fraction before @DavidWheeler?

Jonas Teuwen

@robjohn 8-))). Or Lacey... Or whatever.

@robjohn Tuomas presented a cute generalization of the Carleson theorem (getting the Hunt version for free) for an intermediate space to an UMD space and a Hilbert space.

(Open problem: Is every intermediate space between an UMD space and a Hilbert space an UMD space?)

David Wheeler

for that limit to be 0, means L(a).h approaches f(a+h) - f(a) faster than h does.

Benjamin Lim

@DavidWheeler faster than $h$ does to...?

Eugene

22:43

@robjohn if someone downvotes an answer it should be in his reputation record right?

Jonas Teuwen

@robjohn And Tao Mei presented a $(H_1)^* = \text{BMO}$ result using operator theoretic tools (so only reasoning with semigroups, not kernels)

robjohn

@Eugene They do lose a reputation for it.

David Wheeler

faster than h approaches 0 (we use |h| so we can find a ratio)

Benjamin Lim

@Eugene yes

@DavidWheeler yes.

Jonas Teuwen

So... Now my task is to find out if it would match my operator 8-)).

Benjamin Lim

22:44

However how come the $|h|$ guy just disappeared?

Eugene

@robjohn well now i should see if i can find the troller then

Jonas Teuwen

@robjohn I was getting really ill, so I booked a flight this afternoon to get home (otherwise tomorrow). There was only one seat left today for all flight carriers to Amsterdam apparently! So they charged me an arm and a leg, so to speak 8-)).

Eugene

hmm.. no luck..

Benjamin Lim

@DavidWheeler

robjohn

@JonasTeuwen Too sick to wait until tomorrow? I hope you feel better.

Jonas Teuwen

22:48

@robjohn I didn't want to end up in a Spanish hospital, I do not speak the language, and their knowledge of English is in general quite poor.

David Wheeler

rewrite the limit as $\lim_{h \to 0} \frac{f(a+h) - f(a)}{|h|} = \lim_{h \to 0} \frac{L(a)\cdot h}{|h|}$

robjohn

@Eugene I don't know how you would go about finding someone who has downvoted you.

Benjamin Lim

@DavidWheeler Ah ok

the limit on the right hand side

is approximately just the numerator @DavidWheeler?

Eugene

@robjohn look for some suspects then check their if they lost 1 rep recently

David Wheeler

so for small |h| the LHS = RHS

Jonas Teuwen

22:49

@robjohn data.se?

Eugene

looks like i failed though.

Benjamin Lim

@DavidWheeler thanks

robjohn

@JonasTeuwen I don't think that is updated recently enough.

David Wheeler

it's only a good approximation "near a"

Jonas Teuwen

@robjohn Aha.

So he has to wait a bit 8-).

David Wheeler

22:50

the cool thing is, there is a unique L for which that is true.

which is why we talk of "the" derivative, instead of "a" derivative.

Benjamin Lim

@DavidWheeler yes I found out yesterday about uniqueness :D

Eugene

yeck. it's just getting annoying. downvotes with no explanation

David Wheeler

which justifies the expression the "best" linear approximation

robjohn

@Eugene It's annoying, but there's not much other than asking why that we can do.

David Wheeler

and basically, we leverage information in neighborhoods of "a" by using what we know about the linear map L

Eugene

22:53

@robjohn fair enough. thanks for the advice!

David Wheeler

for example, if L is non-singular, we can find a "local inverse"

Benjamin Lim

yeah inverse function theorem

anon

one could theoretically write a program to track the public "# of downvotes" stat on a select group of user profile pages and keep timestamps of their changes...

David Wheeler

and the derivative of the local inverse HAS to be $L^{-1}$

Benjamin Lim

yes

that is it

David Wheeler

22:55

so once you get past the gobblty-gook of all the partials, it's really just plain ol' linear algebra

the topological structure of R^n lets us do this magic, as long as we "stay local"

Jonas Teuwen

@robjohn These PDE people have weird spaces. Like $L^{\frac32}$...

Eugene

@anon probably not worth the effort.

Jonas Teuwen

So then I think, of so they just don't have the right exponent, or are not able to prove that.

robjohn

@JonasTeuwen The dual to $L^3$ :-)

Jonas Teuwen

Next slide: "And $\frac32$ is optimal".

David Wheeler

22:59

manifolds take this idea, and run with it. we basically turn problems of stuff happening on a surface, to stuff happening in the (euclidean) tangent planes, where everything is much easier.

Jonas Teuwen

@robjohn Yes, or $L^{\frac45}$.

Benjamin Lim

@DavidWheeler You have inspired me

to do a courseon manifolds next sem

robjohn

@JonasTeuwen Ooh, below $1$. That is strange.

Jonas Teuwen

@robjohn Oh, no, it must have been $\frac54$.

Eugene

oh my... adventure time is now an iphone game.

Jonas Teuwen

23:00

@robjohn But for Hardy it is okay 8-).

robjohn

@JonasTeuwen There is a lot more to use with $L^{\frac54}$

@JonasTeuwen yes

Jonas Teuwen

@robjohn We don't know if for our spaces $H^p = L^p$ for $p > 1$ 8-).

There was some brother that had a Hardy space without any maximal function.

Jordan

I hate word problems with ambiguous wording

Jonas Teuwen

So I am like: So how do you verify what kind of functions are in your space? 8-).

"I don't know."

Jordan

"The solution is kept thoroughly mixed and drains from the tank at the same rate." does that mean the solution in is the same as solution out? Meaning y' is 0?

David Wheeler

23:03

@Jordan, you have an example of this?

robjohn

@Jordan To me, that sounds as if the components of the solution are always in the same proportions.

Jordan

Maybe the full context will help "A tank contains 1000L of brine with 15kg of dissolved salt. Pure water enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after t minutes and after 20"

But then that means it never leaves? So I don't understand what I am calculating

robjohn

I would think the solution is draining at 10L/min

David Wheeler

The solution is being diluted by the adding of the water. They mean that you can disregard the fact that in actual practice, the fresher water would be at the top, and the saltier water at the bottom.

Jordan

so y'(t) = 10?

robjohn

23:07

The brine is being replaced at a rate of 1%/min

Jordan

$\frac{dy}{dt}$ = rate in - rate out

But I have no rate out?

David Wheeler

you have 2 things happening simultaneously: one is a mixture (at the current level of concentration) is exiting, another is pure water is entering.

Jordan

It looks like it is 0

at the same rate

robjohn

@Jordan "drains from the tank at the same rate"

Jordan

or no?

Benjamin Lim

23:09

@robjohn several variables, confusing as hell

robjohn

$15e^{-t/100}$

$\frac{\mathrm{d}}{\mathrm{d}t}\text{salt}=\text{flow}\times\frac{\text{salt}}{ \text{volume}}$

Peter Tamaroff

@robjohn Yay!

ablmf

If we have a sigma algebra F, and we know A \cap B \in F, do we have A^c \cap B \in F?

Jordan

what is the e for?

robjohn

@Jordan it is part of the solution to the following differential equation

Jordan

23:18

I didnt follow how it was derived

I actually have no idea how to set up this problem still

robjohn

@Jordan The flow of salt out of the container is the density of salt in the container times the flow out.

@Jordan The flow of salt into the container is the density of salt in the incoming water times the flow in.

Jordan

I am having trouble understanding that

robjohn

suppose there is x kg of salt in the container, then the density of salt is x/1000 kg/L

the flow of salt out is 10L/min times x/1000 kg/L = 1/100 times x kg/min

Jordan

why 10L?

robjohn

@Jordan that was given as the flow

Jordan

23:25

in but not out

robjohn

16 mins ago, by robjohn

@Jordan "drains from the tank at the same rate"

Jordan

I thought that meant the water and salt drains at the same rate as eachother

robjohn

@Jordan The salt is in a solution of 15kg / 1000L

If 10L of solution leaves, then .15 kg of salt leaves.

but that lessens the solution of the salt

Benjamin Lim

@DavidWheeler youtube.com/watch?v=3NPQJKaDLgY

Jordan

I follow that

robjohn

23:33

@Jordan So suppose there are x kg in the 1000 L vat.

the amount flowing out would be x/1000 kg/L times 10 L/min = x/100 kg/min

there is no salt coming in, so we don't need to worry about that.

Thus we get $\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{x}{100}\text{ kg/min}$

Jordan

isnt that negative?

ok, so I need to make a differential equation out of this?

robjohn

That is a differential equation

Jordan

it minutes = t?

I am not sure how to solve this equation I guess

robjohn

$\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{x}{100}$ is the differential equation with time in minutes.

$\frac{\mathrm{d}x}{x}=-\frac{\mathrm{d}t}{100}$

Integrate

$\log(x)=\log(15)-\frac{t}{100}$

30 mins ago, by robjohn

$15e^{-t/100}$

@Jordan better now?

Jordan

log15?

robjohn

23:45

@Jordan to match the initial amount of salt in the vat (when integrating you get a constant of integration)

Jordan

where does that come from in the $\frac{dx}{dt} = -\frac{x}{100}$?

robjohn

@Jordan when integrating $\frac{\mathrm{d}x}{x}=-\frac{\mathrm{d}t}{100}$

Jordan

I get $ln|x| = \frac{-x}{100}$

robjohn

@Jordan what happened to your constant of integration?

Jordan

I forgot it

robjohn

23:48

@Jordan and what would it be to make $x=15$ at $t=0$?

@Jordan and that is wrong... it is $\frac{-t}{100}$, not $\frac{-x}{100}$

Jordan

oh yeah

oh so raise both sides by e?

robjohn

@Jordan and remembering the constant of integration.

Jordan

so $x = e^{\frac{-t}{100}} + K$

robjohn

@Jordan no. where does the constant of integration go?

Jordan

I dont know

robjohn

23:51

The integration gave $\log(x)=\frac{-t}{100}\color{red}{+C}$

Then exponentiation gives $x=e^Ce^{\frac{-t}{100}}$

$x=15e^{-t/100}$

12 mins ago, by robjohn

30 mins ago, by robjohn

$15e^{-t/100}$

The $15$ matches the $15\text{ kg}$ originally (at $t=0$)

M.B.

Is TeX supposed to render here?

(in chat, that is)

robjohn

@MB If you have the bookmark

@MB The easiest place to get it might be here

M.B.

@robjohn Ah, thank you.

robjohn

@MB you're welcome. Enjoy!

Transcript for

$Mathematics$ Mathematics