Mathematics - 2013-04-14 (page 1 of 4)

« first day (984 days earlier) ← previous day next day → last day (4042 days later) »

12:11 AM

If $\displaystyle f^{(n)}=a_{n-1}f^{(n-1)}+\cdots+a_1f'+a_0$ then $$\frac{d}{dt} \begin{pmatrix}f \\ f' \\ \vdots \\ f^{(n-2)} \\ f^{(n-1)}\end{pmatrix}=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \dots & 1 \\ a_0 & a_1 & a_2 & \cdots & a_{n-1} \end{pmatrix} \begin{pmatrix}f \\ f' \\ \vdots \\ f^{(n-2)} \\ f^{(n-1)}\end{pmatrix}.$$ Writing the column vector above as $\vec{f}$ and the constant matrix as $A$, we have $\vec{f}^{(k)}=A^k\vec{f}$, and we also have that $f^{(n+k)}$ is the last component of $\vec{f}^…

@anon What you be talking about?

a comment you made in the preliminaries of the question you linked (and my chat message was in direct response to the message where you linked it, and I wrote a comment on main)

@anon I see.

12:26 AM

@anon do you know about integer partions?

what about em?

@anon Could you derive ramanujans identity, $$\sum_{n=0}^\infty p(5n+4)x^n=5\frac{\prod_{n=1}^\infty (1-x^{5n})^5}{\prod_{n=1}^\infty (1-x^{n})^6}$$ using only the relation $1_{n\equiv b \text{ mod a}}=\frac{1}{a}\sum_{t=1}^ae^{2 \pi i t(n-b)/a}$

or is more in depth knowledge required?

if I expand the product expansion for the partion function, will it factor nicely into the form above, or are other techniques required that is

People studying calculus but insisting on $1+\sqrt x=\sqrt{1+x}$. My reaction exactly.

nvm it is lol

Gustavo Bandeira

@PeterTamaroff xD

They should buy mathematica.

12:37 AM

@GustavoBandeira Mathematica?

Gustavo Bandeira

Show them.

Why?

@GustavoBandeira HAHAHAHA

@Ethan I'll think about it later.

Gustavo Bandeira

@PeterTamaroff You can use the double equal to see if they're equal.

The result will be as W|A.

@anon alright, I didn't want to attempt to expand it anyway with out knowing

12:39 AM

@PeterTamaroff insisting?

also, fun fact: while studying the lesser-known construction of the $p$-adics out of the finite field of $p$ elements using Witt vectors, the identity $\sqrt[p^n]{x+y}=\sqrt[p^n]{x}+\sqrt[p^n]{y}$ in ${\bf F}_p$ is absolutely critical to the explicit computations.

I don't understand

@anon I mean, they insist on writing $\sqrt{x+y}=\sqrt{x}+\sqrt y$.

user image

(or there could be comments from Assad that are deleted I cannot see.)

user image

lol the first one is stupidly obvious now that I think about it, I am still getting used to manipulating weirdly indexed sums

user image

fuuuuuuuuuuuuu

12:49 AM

I don't understand yours

(they are search results)

@anon do you have anything interesting you have written that I could understand?

click on various tags in my profile and see

I semi-intend to finish writing notes on $p$-adics and related topics by the end of mayish

@anon lol I can't really understand any of it

what subject is this? calculus?

12:57 AM

@anon I was trying to understand some convolution identitys involving the divisor function a while ago, but most of it required knowing about modular forms

@anon so I might find your post on them interesting if only I could understand it lol

I found some very messy combinatorial arguments explaining some eisenstein series identitys though

I have only a handful of posts relevant to modular forms

they are a topic I haven't studied.

yes, I know of and a little about them. that second one I did mostly by googling and reading Wikipedia.

@Ethan in this answer I link to a survey of partitions that may be of help to you

@anon Would it be wise to learn some enumerative combinatorics first? I don't really know any combinatoral principles or any of that

@robjohn

@PeterTamaroff yes?

@robjohn I just realized I copied the problem wrongly.

The statement is "..for each particular $x$ there exists some $M$ such that, $$\left| {{f^{(n + k)}}\left( x \right)} \right| \leqslant {2^k}{N^{k + 1}}M$$ for every $k$."

1:07 AM

@PeterTamaroff Ah

@robjohn I edited the question.

Before, it read:

@PeterTamaroff I've been away for awhile, so I haven't had a chance to look at it any more

"... there exists some $M$ such that, $$\left| {{f^{(n + k)}}\left( x \right)} \right| \leqslant {2^k}{N^{k + 1}}M$$ for every $x$."

@PeterTamaroff so it is not uniform

@robjohn Not at all. It is pointwise.

1:09 AM

and $M$ depends on $x$

@robjohn Yes.

@PeterTamaroff how did you get me that book a while back? the one on number theory

@Ethan Which one?

That doesn't tell me much more.

apostals

@Ethan Apostol. You can just google it and download it.

Mariano Suárez-Alvarez

1:47 AM

(there are countries where things are a bit more difficult :-) )

@MarianoSuárez-Alvarez Hehe, true.

@Ethan Don't do anything that can get you into trouble!

@MarianoSuárez-Alvarez Do we define the Riemann Stieltjes integral for non monotonic integrators too?

(In the case $\alpha'$ is RI we get $fd\alpha=f\alpha'dx$, I am wondering about more "exotic" cases)

Mariano Suárez-Alvarez

2:04 AM

only for functions of bounded variation

(and a function of bounded variation is the difference of two monotonic functions; this is called the Jordan decomposition)

@MarianoSuárez-Alvarez $$\sum_{a+b+c...=n}_{0\leq (a,b,c,...)\leq n} 1=\frac{(n-1+k)!}{k!(n-1)!}$$?

Mariano Suárez-Alvarez

I don't understand the notation

if a, b, c sum to n, then they are at most n, and of course their gcd is at most n

All possible tuples (a,b,c,...) with each element $0\leq (a,b,c,..) \leq n$

Mariano Suárez-Alvarez

hm?

such that in each tuple the sum of each element equals n

each tuple is of length n

has n elements*

2:11 AM

@MarianoSuárez-Alvarez Oh, OK. I think Apostol studies functions of bounded variation.

Mariano Suárez-Alvarez

and what does $0\leq(a,b,...)\leq n$ mean?

It should be

$$\sum_{a+b+c..=k}_{0\leq (a,b,c,...) \leq n}1=\frac{(n-1+k)!}{k!(n-1)!}$$ where there are n elements in the tuple

Mariano Suárez-Alvarez

I still don't know what $0\leq(a,b,\dots,)\leq n $ means

every element in the set is between 0 and n

is an integer between 0 and n

Mariano Suárez-Alvarez

that's very weird notation

2:13 AM

each set has n elements

@MarianoSuárez-Alvarez how would I write it typicallY/

Mariano Suárez-Alvarez

you want to count the number of ways of writing $k$ as a sum of $n$ integers, taking into account the ordering of the summands

yes

Mariano Suárez-Alvarez

see: words are very helpful

lol sorry

Is it equal to what I wrote?

$$\frac{1}{k!}n(n+1)(n+2)...(n+(k-1))$$

@Ethan Note the lhs is $${n+k-1\choose k}$$

Mariano Suárez-Alvarez

2:16 AM

the number of ordered partitions of $n$ into exactly $k$ parts is $\binom{n-1}{k-1}$.

@Ethan Have you read of combinations with repetition, Ethan?

is that including zero?

Mariano Suárez-Alvarez

try it for a small number :-)

but you seem to be restricting both th enumber of parts and their size

so I don't know

but google for 'partition of integers'

@MarianoSuárez-Alvarez $$\frac{1}{(1-x)^n}=\sum_{k=0}^\infty \frac{(n-1+k)!}{k!(n-1)!}x^k$$

Mariano Suárez-Alvarez

what am I supposed to do with that?

google for 'binomial series'

I don; t know what you are asking, really

2:21 AM

@PeterTamaroff that was what I wanted, thanks

the n+k-1 chose k

Mariano Suárez-Alvarez

in modern latex one types that as $\binom{m}{k}$ , not $m\choose k$

for reasons too long to explain here

@MarianoSuárez-Alvarez Yeah. I am not a TeX connoisseur.

Mariano Suárez-Alvarez

Ethan, notice that i still don't know what you wanted :-)

@PeterTamaroff I know: that is precisely why I say such things :-)

@MarianoSuárez-Alvarez Yes. Thanks.

@MarianoSuárez-Alvarez :(

2:29 AM

@anon see this.

...the plain TeX commands:\over, \choose, \atop, \above etc.
are not allowed. Essentially the infix syntax of these commands caused a lot of bother. They are replaced by prefix style commands based on the new command \genfrac (see AmS-LaTeX version 1.2 User's Guide [amsldoc.dvi |amsldoc.ps] pages 13-15), in particular: \frac must be used rather than \over and \binom must be used rather than \choose.

@PeterTamaroff here's a cookie... :-)

@JayeshBadwaik NOM NOM NOM.

"In the en they exists as a consistent definition, you cannot be agnostic about it."

@user1 Whu...?

2:34 AM

@PeterTamaroff The post that bumped this: math.stackexchange.com/questions/154/…

@user1 Weird that the word "agnostic" pops up.

@user1 ... no you're not.

The galling thing is that not so useful answer is what made the entire thing now CW... :|

@J.M. Why? =O

it has ... an upvote? :blinks:

@PeterTamaroff math.stackexchange.com/posts/360931/revisions

feck.

2:37 AM

lmao

@J.M. She just broke a bit of the internet.

CW can't be undone, can it?

Wonder how many of us get to do that.

@J.M. You can say "fuck" here, no-one will flag.

@anon Well... mods can undo. Whether they will for this situation is altogether different.

@PeterTamaroff I'd say "fuck", but the Irish pronunciation is just so charming...

@J.M. HAHA

@MarianoSuárez-Alvarez What do you think of Apostol's "Mathematical Analysis"? And what do you think of it compared to Rudin?

@J.M. @user1 If you happen to want to comment of the above, I'd appreciate.

Mariano Suárez-Alvarez

2:41 AM

they are both good books

The last I remember at looking at those two was that I liked them. But that was about a decade ago.

@Ethan "You're", "than".

lol

gibberish is my first language

4

2:55 AM

@PeterTamaroff Yes, Jasper...oops, lost myself, yes, Peter ;-)

@amWhy Hehehe. Someone has to do his work now that he's gone.

Muahaha :-)

@skullpatrol hahaMua

(removed)

Interesting: I see Marvis has gone anonymous

3:03 AM

@amWhy Not his first time, IIRC. He goes anonymous in between name changes.

"IIRC" = " If I remember correctly" ...cool...never made that connection before!

@J.M. You can tell I don't text much...etc...Learning all the time ;-)

@amWhy Methinks it was in active use in newsgroups, long before SMS. :)

@J.M. I have no doubt that's true...but, alas, I haven't really used them... so I haven't acquired "proficiency" in net-speak

@J.M. It is an internet acronym I think, not a text acronym. Actually, I have rarely seen it used as a texting

s'okay.

@JayeshBadwaik yes, as I said, I've been seeing it in newsgroups... man, those were the days. :)

3:11 AM

@JayeshBadwaik Well, you can tell I don't "text"! ;-)

:-)

(FWIW, I'm not much of a texter, either.)

I keep wanting to read "lol" as "lots of laughs" even though I know it is "laughing out loud"!! ;-D

I text occasionally, but it is always in english. :)

That actually works nicely, too.

3:13 AM

I was a texter oncer, when I was a teen.

@J.M. IMHO I can do without... :)

@J.M. How did you get the "signature: "insanity and genius" to appear in your chat profile pop-up?

click the "edit" button by the word "about" in your own chat user profile

It's under about, I think.

yes, what anon said. :)

@anon I must seem like a newbie!

3:18 AM

@amWhy We all start out that way, no worries.

@J.M. To be honest, I've never really participated in "chats" elsewhere... Some can be way-too-fast paced for me to follow...and I'm a little on the shy side :-)

Newbie, newb, noob, or n00b is a slang term for a or , or somebody inexperienced in any profession or activity. Contemporary use can particularly refer to a beginner or new user of computers, often concerning Internet activity, such as online gaming or Linux use. It can have derogatory connotations, but is also often used for descriptive purposes only, without a value judgment. The term's origin is uncertain. Earliest uses probably date to late twentieth century U.S. military jargon, though possible precursor terms are much earlier. Variant forms of the noun include newby and newbe...

Gustavo Bandeira

4:00 AM

@amWhy But are you get shy of showing us your anonymous identity?

Shouldn't anonimity be a remedy for shyness?

@GustavoBandeira Yes, it should be...I mean I'm really likable and I really like people. I'm just extremely sensitive and it doesn't take much for me to feel stupid/ashamed/judged, etc... And I NEVER forget that I am interacting with people, not "anonymous robots" (unless I'm deluding myself!) ;-)

Gustavo Bandeira

@amWhy Yes. They aren't robots, although I'm deeply suspicious of anon being one...

@GustavoBandeira hahahahahaha!

user image

@J.M. lol I saw that image in my head when I added "unless I am deluding myself!"

Gustavo Bandeira

4:06 AM

@amWhy I'm also making an intensive course on how to act when I feel stupid.

@amWhy Whenever you feel like that, don't go away of MSE.

@GustavoBandeira harumph

Gustavo Bandeira

@anon :*

4:29 AM

how would I approach evaluating $$\lim_{x \to 1}\frac{x^{\omega}(1-x^{a})^{\omega}}{a}\sum_{t=1}^a\frac{e^{2\pi i t\omega (n-b)/a}}{(1-xe^{2\pi i t (n-b)/a})^\omega}$$
For arbitrary integers ($\omega$ , $b$ , $a$) with, $1\leq b\leq a$

you made $\omega$ an integer? now you must be trolling.

why am I trolling?

I don't understand

(removed)

Wait I messed up

It should be

(removed)

4:36 AM

in number-theoretic contexts where there are roots of unity, if $\omega$ isn't a phase parameter in some waveish object, it is the third root of unity, not an integer. it's like making $\epsilon$ a very large number, or making $\zeta$ an integer while you're working with dirichlet series, etc.

"if $\omega$ isn't a phase parameter in some waveish object" - e.g. angular frequency

nvm

at any rate, looks like the RHS may pop out of some complex integral residue calculation

Mariano Suárez-Alvarez

what RHS?

is there an equal sign there?

I messed up on it

There shouldn't be n there at all lol

one sec

4:38 AM

the sum. it's on the right-hand side of the expression. whatever.

Mariano Suárez-Alvarez

heh

border line meh

@MarianoSuárez-Alvarez hey mariano. in this blog post, it is suggested that the order type of the exponents in the Teichmuller expansion of $\sqrt{-1}\in\overline{{\bf Q}_2}$ might be $\omega^\omega$. but wouldn't the exponents be a subset of $\bf Q$, and $\omega^\omega$ is uncountable?

$$\lim_{x \to 1}\frac{x^{\omega}(1-x^{a})^{\omega}}{a}\sum_{t=1}^a\frac{e^{2\pi i t(\omega-b)/a}}{(1-xe^{2\pi i t/a})^\omega}$$
For arbitrary integers ($\omega$ , $b$ , $a$) with, $1\leq b\leq a$

(do a ctrl+f for order type, the paragraphs are pretty self-contained)

4:44 AM

@anon The last comment there is so hilariously out of place. Damn bots... :D

Mariano Suárez-Alvarez

w^w is countable

it is just countably many copies of w, one after the other in the shape of w

(notice that the notation is not the same as for cardinals)

($\aleph_0^{\aleph_0}$ is uncountable)

if we write $A=\{1+\tfrac1n:n\geq1\}$, the $A$ has order type $\omega$

and is contained in $[1,2]$

ah, $\omega^\omega=\bigcup \omega^n$ is a union of countable things

Mariano Suárez-Alvarez

hm no

oh

Mariano Suárez-Alvarez

$\omega^\omega$ is the concatenation of countably many copies of $\omega$

Wikipedia surely has the definition of ordinal adition

then of multiplication

and then of exponentiation

they are all iterated concatenaions of ordered sets

4:49 AM

I thought concatenation was ordinal addition?

Gustavo Bandeira

I've learned about inverse images.

like, append $\beta$ to the end of $\alpha$ and you get $\alpha+\beta$

Mariano Suárez-Alvarez

that is what i said :-)

Gustavo Bandeira

I understand the concept, but I can't understand this notation:

$$f^{-1}(H):=\{x\in A:f(x)\in E\}$$

Mariano Suárez-Alvarez

multiplication is repeated concatenation

and exponentation is what it is :-)

4:51 AM

@MarianoSuárez-Alvarez to me, "concatenation of countably many copies of $\omega$" would seem to mean $\underbrace{\omega+\omega+\omega+\cdots}_\omega=\omega^2$

Mariano Suárez-Alvarez

yeah, I skipped a level

although I suppose if you concatenate beyond that it will still be a countable number of copies but a different "shape"

Mariano Suárez-Alvarez

the concatenation of countably many copies of the concatenatin of countable many copies of w sounds too silly

that's why we invented the notation :-)

ah, so this is what it looks like. hard to remember.

also here

Mariano Suárez-Alvarez

«intuition about w^w» seems so misguided

that's like asking for intuition for 4

« first day (984 days earlier) ← previous day next day → last day (4042 days later) »

Transcript for

Apr '1314

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile