Mathematics - 2012-08-13 (page 2 of 6)

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Kannappan Sampath

5:00 AM

@BenjaLim Hey!

@KannappanSampath I'm doing fourier analysis :D

Kannappan Sampath

@BenjaLim I know nothing about it. :(

@JonasTeuwen Somehow I'm finding it easier to grasp fourier theory than algebraic topology....

@JonasTeuwen With algebraic topology you have pics and stuff

@BenjaLim Because - up to this point - it is still easier.

but the book is so not rigorous

@JonasTeuwen For me when I read S&S

I usually go straight to the exercises.

5:01 AM

For all chapters?

no

I am only at chapter 3

@JonasTeuwen Ok thanks for checking my work

@JonasTeuwen I'm off to do more exercises :D

@BrianMScott

@BenjaLim Good luck! :-).

@JonasTeuwen thanks.

Close votes please thx.

5:08 AM

@Matt tb is around, Matt. Thought you'd like to know.

5:18 AM

Good morning.

@Matt Hi.

5:33 AM

@JonasTeuwen Quick question

@JonasTeuwen

@BenjaLim Sure.

I have my function $f(x) = x$ if $x$ is rational and $0$ otherwise.

I think such a function is integrable....

But I think I have a problem

I can see that my $L(P,f) =0$ for any partition $P$

But now what about $U(P,f)$?

The problem is how do I determine the sup of $f$ over a subinterval?

@BenjaLim It is not integrable.

Oh, wait... maybe it is.

The usual dirichlet function is not integrable

The sup will take up the $x$ and the inf will take up the $0$.

5:35 AM

wait why will it take $x$?

Because any interval will contain a rational point.

yes but consider the sup of $f$ over $[0,\sqrt{2}]$ say

Yes = $\sqrt{2}$.

Why?

oh yes

ok

But now we have a problem...

How can $U(P,f) = 0$....

It is not.

5:38 AM

so the function is not integrable...

Yeah, I think so.

@PeterTamaroff It is countable...?

It is $0$ except at rational points.

@JonasTeuwen What about

So the Lebesgue integral is $0$.

$$f(x) =\begin{cases} \frac{1}{a} &, x= \frac{a}{b} \in \Bbb{Q} \\ 0, &x \notin \Bbb{Q}\end{cases}$$

Dunno: try. I did not sleep, bro 8-).

5:39 AM

@JonasTeuwen It is not Riemann integrable.

@JonasTeuwen Ok.

@PeterTamaroff Tell Benjamin.

@BenjaLim What are you trying to do now?

@PeterTamaroff Proving that the new function I defined above is riemann integrable

@BenjaLim Did you conclude the former one isn't?

5:41 AM

I did

but this new one above

it's different from the one before.

@BenjaLim Over what are you integrating?

I want to check if it's riemann integrable.

$[0,1]$

@BenjaLim Well, the lower Darboux sum is clearly $0$.

yes I can see that.

Now think about the upper Darboux sum.

5:43 AM

It's the upper sum that I have to check is zero.

@BenjaLim Right.

Well.

@PeterTamaroff I have never actually though about integrals like that in my life.

@BenjaLim "like that"?

yes in terms of riemann sums

@BenjaLim And how else did you think about them then?

5:44 AM

Riemann is about partitions, not these crazy upper and lower sums 8-).

@JonasTeuwen Like this:

@JonasTeuwen Darboux sums are defined on partitions...

@PeterTamaroff Yeah, so?

@JonasTeuwen Like this:

You take the supremum, which does not have to make sense.

For more general spaces (Banach?).

5:45 AM

@JonasTeuwen Then I understand what you mean by "crazy upper and lower sums"

For each $\epsilon >0 $, I choose $n$ large enough so that $\frac{1}{2^n} < \epsilon$

@BenjaLim Sure. Why can you do that?

Archimedean property DUH

Now I paritition my interval into $[0,\frac{1}{2^n}] ,\ldots $ blah

Why is it duh. You ask questions which can be about the same level.

Yes.

@JonasTeuwen Now any rational in such a subinterval must satisfy 1/2^n < \frac{a}{b} < 1/2^{n-1}$

5:47 AM

But apparently you are very good at the fundamentals, but less at computation :-).

@JonasTeuwen What do you mean I ask questions which can be about the same level

@JonasTeuwen huhuhuhuhu?

@BenjaLim He was being mean.

@BenjaLim Why?

Don't worry..

@JonasTeuwen What do you mean I am very good at fundamentals but weak at computation???

5:47 AM

@PeterTamaroff I was not.

@BenjaLim I did not say weak.

Tune down with the question marks.

@JonasTeuwen less at computation rather yes

@JonasTeuwen Why do you think so?

Yes, it is not an insult. If you ask me to check your work... then I can ask questions right?

@JonasTeuwen yes.

I'm just wondering why you say I'm less at computations and very strong at foundations.

@BenjaLim Because most people would rather know how to do the second part (well... kinda) and not know why you can find such $n$.

ok.

@JonasTeuwen Back to the problem.

5:49 AM

It is actually a good thing at this point. So don't worry.

@JonasTeuwen I want to know what the denominators of rationals look like in a subinterval

Why can't $\frac{a}b = 2^{-n}$?

It can.

so?

there is no problem.

Because you write $<$ above.

The point is

ok my mistake

5:50 AM

And what is $f(2^{-n})$?

@CLarue It is 2^{-n}

@BenjaLim No

@JonasTeuwen The point is that the denominator $b$ is always less $\frac{1}{a2^{n-1}}$?

Unless we're on that older function

@BenjaLim It is $1$.

5:51 AM

Anyway, be right back. Need to make a phone call.

@PeterTamaroff no.

$x = \frac{1}{2^n}$

@BenjaLim You wrote the stuff the other way around, maybe? You wrote $\frac{a}{b}\mapsto \frac{1}{a}$

Oh sorry

AH FARRKKKKKKKK

Sorry

Yes it is inverted :D

@BenjaLim So is that the function or not?

@PeterTamaroff It's suppose to be $f(x) = \frac{1}{b}$ if $x =a/b \in \Bbb{Q}$

5:53 AM

aha

@PeterTamaroff So the upper riemann sum is now less than

the length of each subinterval by choice is $ \frac{1}{2^n}$

@CLarue sorry man

@PeterTamaroff that does it :D

@BenjaLim Then, this is correct.

@CLarue Since $a \geq 1$

@CLarue $1/b < 2^{n-1}$

and then the riemann sum is less than $\epsilon$ if we choose $n$ suff. large.

hence is integrable :D

Yes, you already chose your $n$ above, but we can forget that ;)

Done!

@CLarue I have an examples of a function that is integrable, not zero but whose mean squared norm is zero!

6:54 AM

@BenjaLim Easy. $1$ at $1/2$ $0$ otherwise.

7:23 AM

@t.b. Now you made me wonder, just to confirm: when you say "no problems whatsoever": surely, for $L^1$ functions it's not clear whether the Fourier series converges (in any sense) to the function. Right? So if we want nice convergence in $L^2$ norm we can only have that for $f$ in $L^2$.

1 hour later…

8:30 AM

Good morning

Place is pretty empty right now.

@BrianM.Scott Yep - often quiet this time of day

8:46 AM

@OldJohn Considering that it’s 4:45 for me, and at least that early for anyone in the continental U.S., I suppose that it’s not too surprising. And I suppose that it’s a little early for most Western Europe.

@BrianM.Scott Yes, indeed

Just finishing breakfast here ... and trying to decide what to work on today

@OldJohn (I cheat: I’ve been up all night, and I’ll be going to bed in two or three hours.)

@BrianM.Scott Interesting hours you keep!

@OldJohn One of the great virtues of retirement: I can keep the hours that my body likes!

@BrianM.Scott How is retirement? I retired a few weeks ago, and have not stopped smiling since then :)

8:53 AM

@OldJohn I retired a year ago last June, and I’ve finally reached the point at which it feels real, rather than an extended summer vacation. For the most part I’m quite enjoying it. (Some of the associated paperwork has been a bit of a pain.)

@BrianM.Scott I used to worry about the financial aspects of retirement, but now find that I have almost everything I want or need, so that worry has just evaporated

@OldJohn I’m keeping an eye on it, but the pension plan for the Ohio university system is pretty good, so I’m not particularly worried. I’m also not at all extravagant, though I do buy an inordinate number of books.

@BrianM.Scott I have the same problem, in that I seem to buy books at a faster rate than I can read them. Not too surprising I guess when some of them are fairly tough (for me) number theory texts

Hello,

@Steenrod Hi

9:03 AM

anyone here?

@OldJohn My purchases are mostly fiction, and when they’re not, they’re most likely to be historical linguistics. But they still add up in a hurry, and the books still pile up faster than I can read them. What an acquaintance calls my strategic book reserve is large enough to keep me going for quite a while.

@BrianM.Scott I suspect I actually have enough to last me for the rest of my days ... but that doesn't seem to stop me from buying more

@Steenrod No many people here at the moment, I'm afraid

Oh,I shall leave you gentlemen to yourself. :) Have a nice day(rathr good night)I do not want to interrupt people who are so elder to me

@Steenrod Why not? We’re friendly.

@Steenrod No worries about interrupting - we do it all the time :)

9:06 AM

Thank you.Ok,I shall request you to refer to my question.

@OldJohn And you’re younger than I am, too!

2

Q: Complete course of self-study

I am about 16 years old and I have just started studying some college mathematics.I may never manage to get into a proper or good university(I do not trust fate) but I want to really study mathematics. I request people to tell me what topics an undergraduate may/must study and the books that you...

reference-request big-list

@BrianM.Scott Yep :)

@Steenrod I have been looking at that question just now

@Steenrod wondering if you might want to include some complex analysis in the list

@Steenrod I’ve been thinking about that. A lot of the books that I might recommend are old enough that there may be better ones out there now.

@old John sure I got inspired when I saw that a Japanese worker took up math with so much gusto.

(Makoto Kato)

9:09 AM

For topology the common recommendation of the book by Munkres is certainly reasonable. If you like it and want something more advanced, the book by Willard is quite good.

@BrianM.Scott I have found that there is a huge collection of lecture notes and books freely (legally) available online these days - and some of them are excellent

@OldJohn The number keeps growing. I’ve still not seen a set of general topology notes that I really like, though. I’m not fond of the Topology Without Tears that a number of people are using.

I am ready to wait for a day or two at least before I accept an answer.So, @Brian M.Scott may go to sleep and answer me later.

:)

@BrianM.Scott I learned most of my topology from 2 sources - Willard and the old Schaum's outline book with hundreds of exercises

@OldJohn That’s a reasonable combination, I suspect.

9:12 AM

@Steenrod Have you looked at any of the free and legal maths stuff available online?

Not yet.

I bought Herstein.

But I think I will have to downloading books

*resort to

@BrianM.Scott My math career was a bit unconventional - did badly as an undergrad for various reasons, and did a PhD late in life as a part-time external student

@OldJohn One book that I’ve not seen and would like to see is the text by Sheldon Davis; I knew him when he was just starting, and his interests would be perfect for the kind of gen. top. text that I’d like to see.

@Steenrod There are some very good ones - some I have seen for algebra and number theory especially

@Steenrod You can edit a comment by hitting up-arrow, though only for one minute after it’s posted.

9:15 AM

Thanks.

@BrianM.Scott I have heard people muttering about the Mendelson book that Peter is using - I never actually looked at that one, but I do remember using a couple of books by Hu back in the early 70s - but never seen them since

@Steenrod: Whatever you do, I recommend that you try a variety of areas in order to find out what you like best. Don’t feel obliged to stick to the most common ones, either; for instance, if you find that you’ve a taste for set theory, give it a try. (The book by Hrbacek and Jech is very good if you decide that you want a rigorous introduction.)

I see.

@OldJohn I’ve heard of the Mendelson but never seen it; I don’t know Hu at all.

@Steenrod Herstein was my undergraduate text, back in 1966-7, I think. It’s still a very good book, if a bit old-fashioned, and if you can do the hard problems in it, you definitely know what you’re doing.

@BrianM.Scott , the book on set theory is not available for download.Cn you please recommend an alternative one?(I am being able to get 95% of Herstein's exercises till now)

9:24 AM

@BrianM.Scott I will second that - it was (and probably still is) an excellent text

@Steenrod Then you’re doing well. Let me think about the set theory for a little.

It is best if you include it as an answer in that question.

I should be gone now.

Thanks both of you.

@Steenrod OK - bye for now and good luck

9:44 AM

Between 12 - 14h I will hear the test results 8-).

9:54 AM

@JonasTeuwen good luck

@JonasTeuwen ayt?

What test results?

Oh noes, maybe I just missed him now. Do you know about Fourier series, Old John?

I mean, a bit more than what I asked yesterday.

@Matt Hmm - very little - I did some years ago, but have forgotten most of it :(

@Matt sometimes even the act of explaining a problem to someone else helps - even if they don't understand fully - so fire away

Ok, I have a theorem and I'm trying to understand what it's telling me:

user image

It's about the coefficients of the Fourier series of the derivative of $f$.

So one half of the theorem tells me what the coefficients of the $\alpha$-th derivative are.

@Matt yep

But the problem is, what are $n_i$ in the equation of the coefficient?

Oh, it's dawning on me.

10:01 AM

@Matt yes?

@JonasTeuwen Would you believe it today was the first time I proved something was integrable using first principles

Maybe I start out with a Fourier series of $f$ with terms $a_n e^{2 \pi i \sum n_i}$. Then I apply $D^\alpha$.

I have to apply the product rule somehow, $(fg)^\prime = f^\prime g + fg^\prime$.

But what is $(\prod_i f_i)^\prime$? Tumeni symbols.

Because I also have to replace $\prime$ by $D^\alpha$.

@Matt which book is this from?

@OldJohn No book, it's from these very unfortunate lecture notes here.

@Matt thanks

OK - so am I right in thinking that $n = n_1 + n_2 + \dots +n_d$?

10:06 AM

I think so, yes.

OK - so your exponential thing a few lines up could be written as a product of exponentials, maybe?

One wants to apply $D^\alpha = (\partial_{\alpha_1}, \dots , \partial_{\alpha_d})$ to $a_n \prod_i e^{2 \pi i n_i}$, I think.

@OldJohn Yes, that's what I'm saying : ) But computing stuff like this makes my brain go funny.

@Matt similar here - but maybe we can take it a small step at a time :)

Maybe in 2D first?

Say, $f = f(x,y)$ and $D^\alpha = (\partial_x, \partial_y)$.

Then the domain is $\mathbb T^2$.

@Matt excellent idea :)

10:11 AM

And let the $n$-th Fourier term of $f$ be $a_n e^{2 \pi i (n_x + n_y)}$.

Then the second partial derivative would be $D^\alpha = D^2 = ( \partial_x^2 , \partial_x \partial_y, \partial_y^2)$.

@Matt probably (starting to hope @JonasTeuwen reappears!)

Is $D^\alpha f = D^2 f = \partial_x^2 f + \partial_x \partial_y f + \partial_y^2 f$?

@Matt I would think so ... but I am not an expert - I think I only did Fourier in 1-dim :(

Need to see if I can find it on the internets then.

I had Alinhac but had to take it back to the library. I'm sure it says in there somewhere.

@Matt when is the exam?

10:19 AM

On Monday.

The only problem is that I'm still reduced (as always) by the previous oral exam.

I think $D^\alpha f$ for $|\alpha| = 2$ would be $\partial_x \partial_y f $, for example.

Or, $\partial_x^2 f$ and also $\partial_y^2 f$.

Depending on the $\alpha_i$.

@Matt that looks right

So the derivative $D^\alpha = D^2$ of $a_n e^{2 \pi i (n_x x + n_y y)}$ would be of the form $a_n \partial_x^2 e^{2 \pi i (n_x + n_y)}$ for example.

@Matt now I am worried - what is $n_x$?

@OldJohn The $x$-index of the character. We're in 2D so the characters come with numbering $\chi_{n_x n_y}$ as opposed to $\chi_k$ for one dimension.

shouldn't we be doing $D^\alpha$ of $f$ rather than the exponential expression?

10:27 AM

Which would be $a_n e^{2 \pi i n_y y} \partial_x^2 e^{2 \pi i n_x x} = a_n e^{2 \pi i n_y y} 2 \pi i n_x \partial_x e^{2 \pi i n_x x} = a_n e^{2 \pi i n_y y} (2 \pi i n_x)^2 \partial_x e^{2 \pi i n_x x} $.

@OldJohn I assumed that we had a Fourier series of $f$ of the form $f = \sum_k a_k e^{2 \pi i k}$.

@Matt yes ...

@OldJohn So instead of taking the derivative of $f$ and verifying the theorem I was trying to do reverse engineering and taking the derivative of the Fourier series.

@Matt Ah - I see :)

It might not make much sense : (

But the guy who found the theorem must've somehow got to the conclusion he got. And wouldn't that be by taking the derivative of the series?

And afterwards verifying what conditions he needs for that to hold?

@Matt not necessarily, I think

but maybe your approach above is getting there

the coefficients you are getting are starting to look like the ones in the theorem

10:33 AM

@OldJohn I'm satisfied with the first half of the theorem. Now for the second half: do you know what the $\varepsilon_k$ are on the RHS of the inequality?

Good day!

@Matt not offhand - I would need to go back through the notes, I think

@Nimza Hi

@OldJohn Yeah. And not find it : )

@Matt really ? :(

Well unless you're prepared to read every page from the very start to the very end of all of the notes.

I wish this was a book so I could throw it at a wall and subsequently throw it out of the window.

10:36 AM

@Matt that is seriously annoying

@Matt Which book are you using?

None, it's notes.

Okay.

Never mind. I'll have to skip this. So annoying.

See you all later!

Bye

I was reading this thread on r/math where the poster is asking if there are any algebraic structures which have three different operations, addition, multiplication and a third which cannot be written as some combination of the previous two and one of the reply was that you can take any specific mapping from $AxA$ and call it an operation. Is this something non-trivial? Is this somehow related to categories?

One of the guys said it might not always be possible to find the third operations. The condition might be too strict.

10:45 AM

@Matt At doctors office. Back at 15h. I seem to be fine.

Maybe somebody is familiar with linear algebra and can advise me something? scicomp.stackexchange.com/q/3044/1180

@Nimza $S_{m}$ is a matrix?

@JayeshBadwaik no, just numbers

@Nimza $S_{m}$ is a function/polynomial?

11:00 AM

@JayeshBadwaik $S_{m}$ is a generic function. But for me it is sufficient to know it's values only in $(m+1)$ mutually distinct points

@Nimza Why can you just not calculate $f - P_m$ at the $m+1$ points?

@JayeshBadwaik I don't know how to separate $S_m$ from $P_m$ :( I know only the RHS

Okay, so you want to approximate function $f$ by a polynomial of degree $m$ and then, determine the error as a function?

@JayeshBadwaik hm... didn't think about it. Good idea, but what if $S_m$ is a polynomial too or $P_m$ has degree less than $m$?

@Nimza What is the context of the problem?

11:07 AM

@JayeshBadwaik It can't help I think :( reconstruction of Riemann surface

@JonasTeuwen That's great! I assume it means that your tongue thingie has been cured?

@Nimza Somewhat more specific. For example, what is $f$ and where do the factors $S$ and $P$ come out?

@JayeshBadwaik nothing more specific, $S_m$ is a symmetric function for some holomorphic functions, $f$ is a generic function that is measured by some device

So, you are getting data points from a device, and you have to find out the function which represents it?

@JayeshBadwaik roughly speaking yes, I have data from device and I have to find a surface which produced such data, tomography

11:19 AM

Hi folks

@OldJohn hi again)

Anyone ever get annoyed by textbooks with "Basic" in the title - only to find they are not basic at all?

I have :(

@skullpatrol My favourite example is probably Weil's book "Basic Number Theory" (not managed to get past first 2 pages yet) - have you a good example?

@Nimza And you are trying to find a polynomic approximation?

11:30 AM

@OldJohn Not of hand, but as @DavidWheeler would say "one man's ceiling is another man's floor."

@JayeshBadwaik no, I'm looking for any possible way to find $S_m(\xi_k)$

@skullpatrol very true!

@Nimza But then there are so many methods. You can approximate using fourier, polynomial and any other set of functions you can think of. Each of then will give you a different answer.

@JayeshBadwaik say, I approximated $f$ as polynomial. What's next? I think I can have problems if $S_m$ is also polynomial

@Nimza You do not approximate like that. You take a function $S$ and then the other part is the error notation

and then you say for a particular $S$, the corresponding maximum error is this

that's it

now all you have to do is to find the coefficients of polynomial which is now a quiet easy problem.

11:36 AM

@JayeshBadwaik didn't understand... what I have to approximate using $f$ as given data?

You have $f(z)$ right? Now you take a generic polynomial $S_{n}(z) = a_{0} + a_{1} z + ... + a_{n} z^{n}$

ok

Now try to find out co-effecients $a_{n}$ such that $S_{n}(z_{i}) = f(z_{i})$ for all $z_{i}$ you know. Say you know $k$ points from the device. So you would need $n >k$.

YesThis is the approximation function and then, you have to find out the error, which you can specify as the maximum bound.

yes, so the problem is to find "error" $P_n$

Yes

Now, the error $P$ actually can be specified only as a max term. So, the concept is error(z) will be the max value of error at z

11:42 AM

@JayeshBadwaik okay, thanks. But I'm confused a little by the fact that theoretically $S_m(\xi_k)$ can be found explicitly

@Nimza who says so? If you have finite points, how can you construct infinite points with zero error? Unless, there is some structure on $f$, that is $f$ is a specific type of function.

For example, a sine wave,then we just need to know six points to construct it explicity.

@JayeshBadwaik no, I have only to find function values in finite points $\xi_k$, $k=0,\ldots,m$

Okay, so you have a set of values at some points, but those points are not the points you want to find your values at, so you want some kind of interpolation/extrapolation ?

No... I know values of $f$ in any point and I want to know values of $S_m$ in arbitrary $(m+1)$ point. I'll try a little now with CAS

but your $S_{m}$ is hte approximation of $f$ right?

11:48 AM

@JayeshBadwaik don't think so because $P_m$ can take arbitrary big values in general (so it is a bad approximation)

what is $P_{m}$?

? polynomial

@Nimza but how do you find it? is it pre-specified? how do you know it can take arbitrary large values?

@OldJohn My favorite quote on this topic is from Feynman:
Einstein was a giant. His head was in the clouds, but his feet were on the ground. Those of us who are not so tall have to choose!

@JayeshBadwaik roughly speaking, I didn't make any researches with $P_n$, I only found that it arises here (so I have no right to make any assumptions on $P_n$)

11:54 AM

@skullpatrol :-)

@JayeshBadwaik great thanks for help, I'll try something with it

@Nimza welcome :-)

@skullpatrol Are you now a patrol wolf?

Sorry. you are a skull, seems like wolf in small dim

@skullpatrol Nice quote

@OldJohn Thanks :-)

@Nimza Its actually surprising your question hasn't yet been downvoted considering how vague it is. Refine it as soon as possible to make it more clear.

@skullpatrol are you suscribed to r/math?

11:59 AM

@JayeshBadwaik What is r/math?

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