Mathematics - 2012-11-09 (page 3 of 5)

mick

17:00

on the other hand , GarbageCollector collect a garbage integral hahaha no offense

robjohn

@BrianM.Scott: with my recent spotty attendance, I've not seen you recently. How are things?

mick

@Charlie arrgh , just like throwing notes away .. . garbage ... throw away ...

robjohn

@GarbageCollector That would be really big...

Garbage Collector

@mick how about this one $$\int a^{\textrm{d}x}$$? :-)

J. M.

@GarbageCollector I'm not quite sure where you're getting these notions, but: look up the multiplicative calculus (product calculus).

robjohn

17:02

@GarbageCollector That would be $\int1+x\log(a)$

Brian M. Scott

@robjohn I’ve been keeping pretty busy here; not much has been going on otherwise, though it’s very nice that the damned robo-calls have finally ended.

mick

@Charlie feel free to look at my last 2 questions !
besides did you get the joke yet ??

J. M.

@BrianM.Scott "robo-calls"?

robjohn

@BrianM.Scott post-election?

Charlie

@mick no

Brian M. Scott

17:03

@robjohn Yep.

robjohn

@J.M. automated election ads

mick

@Charlie did you look at my 2 last questions ?

Brian M. Scott

@J.M. Automated, taped calls in aid of one candidate or another in the recent U.S. elections.

J. M.

@robjohn Oh, so that's what they're called... phone spam, then. :)

Charlie

@mick no mick, not yet...

robjohn

17:04

@J.M. precisely

mick

@jdoe do you still mind ? :)

Charlie

No!!

J. M.

@robjohn Here I thought all you guys had to contend with were telemarketers...

robjohn

@J.M. these are essentially automated telemarketers for politicians.

mick

@Charlie ???

Brian M. Scott

17:06

@J.M. We can opt out of those via a national don’t-call list, though violations do occur. Unfortunately, charities and political ads seem to be exempt.

Charlie

@mick another charles

robjohn

@BrianM.Scott yeah, we are supposed to be on the national don't call list, but we still get an awful lot of calls.

mick

@Charlie your emotional over nothing again :)

J. M.

@Charlie It's not like "Charles", "Chuck", or "Charlie" are quite uncommon... :)

mick

CHUCK NORRIS

:p

J. M.

17:07

@BrianM.Scott Nasty...

Charlie

@mick it's not emotional, it1s just a little joke

jdoe

why don't phone companies stop spam callers

Brian M. Scott

@robjohn My solution is simple. Virtually no one should be calling me, so in general I simply don’t pick up. And I don’t have an answering machine.

mick

i will probably get flagged for chuck norris jokes soon

J. M.

@jdoe Apparently, not in their best (economic) interest (yet)...

mick

17:08

@J.M. agreed

robjohn

@BrianM.Scott Same here. If I don't recognize the caller ID, I don't answer. If it is a valid call, they will leave a voice mail.

Brian M. Scott

@J.M. As long as one isn’t a proper Charlie.

mick

we live in the age of smart phones and stupid people !

Jayesh Badwaik

@jdoe they get paid!!

@BrianM.Scott In my android phone, I have an app named "SPC"

I use it to block those specific numbers/ number matching

Charlie

@JayeshBadwaik i thought every phone had this...

Brian M. Scott

17:10

@robjohn I have neither caller ID nor voice mail, so they’re simply out of luck. The one or two people who do have license to call me know how.

J. M.

@mick Well, the "smart phones" do an excellent job of spreading stupid stuff faster...

Jayesh Badwaik

@Charlie every phone? I do not think so. Android has free apps.

mick

@J.M. thats why they are smart phones , they do things smarter than stupid people can , unfortunately stupid ppl control them

J. M.

"I love the computer. It allows me to make more dumb mistakes efficiently..."

mick

@J.M. explain ?

Charlie

17:12

@JayeshBadwaik some old phones of mine had it... "black list" it was called

J. M.

@mick ...a joke, that one.

mick

i asked 2 recent questions ...

Jayesh Badwaik

@Charlie yup, but it was very specific, in here, I can block calls having a specific string in their phone numbers, or the calls without any phone numbers etc etc

Charlie

@JayeshBadwaik hmmm okay

robjohn

"The computer allows you to make mistakes faster than any other invention, with the possible exception of handguns and tequila." — Mitch Ratcliffe.

Jayesh Badwaik

17:13

So, basically, I block all numbers from Noida, where I have no one I know, and it is the oriign of all such calls

Charlie

If we call to my phone company they block those calls too, @jayesh

mick

@robjohn tila tequila ;)

Jayesh Badwaik

@Charlie well, my phone company used to spam me with offers :P :P

@robjohn making mistake fasters allow you to improve faster

mick

Does anyone here know a detailed proof of apery's constant ?

in particular 1) the identity 2) the recursion 3) how that implies irrationality ?

jdoe

17:27

@mick, what do you mean by "the identity"?

mick

@jdoe an alternative sum expression for zeta(3) used in apery's proof.

jdoe

@mick, he gets a pair of something like 5th order recurrences whose ratio approximates zeta(3) faster than possible if it were rational (rationals repel each other)

kush

would it be a true statement if I said almost all real numbers are not integers?

mick

@jdoe that is 2) and 3) yes

@kush i think so

jdoe

@kush, do you have a definition of "almost all"?

kush

17:29

@mick I am trying to understand the concept of almost all and it seems it means all except finitely many

mick

because the integers have smaller cardinality

@kush that depends on context

jdoe

@kush, that's not true

almost all means everything except a set of measure zero

mick

@kush usually means something else

kush

@jdoe ah because integers are not finite

mick

@kush what jdoe said

kush

17:31

@jdoe I'll go and read up on measure now

Thank you

mick

@kush seems a good idea

jdoe

@kush, you can understand "measure zero" without any measure theory, though

it's especially simple

kush

@jdoe wikipedia says "the measure of the empty set is required to be 0"

jdoe

@kush, yeah but you can define a "set of measure 0" without doing measures and stuff, directly just with sets and epsilons

kush

and that if A is a subset of B, the measure of A is smaller than the measure of B.

@jdoe I think I am lost now :(

jdoe

17:35

@kush, measure theory takes a lot of work to set up

@kush, it is possible to define what a "measure 0" set is directly, so that you don't have to do that hard work

kush

@jdoe how would you define measure zero?

jdoe

E has measure zero iff for all epsilon > 0 you can cover it in intervals whose lengths sum <= epsilon

so we can cover Z in the union of intervals [i-epsilon/(2i^2),i+epsilon/(2i^2)] (i ranging over all integers), this has total length epsilon

this proves that Z has measure zero, so almost all reals are irrational

made a mistake it needs to be /4i^2

mick

i want a new badge :)

Garbage Collector

17:53

Excuse me. Do moderators get paid in any StackExchange site?

kush

@jdoe wow that's very nice

@GarbageCollector The answer as far as I know is no.

3

Q: Are moderators paid to moderate?

If not why would they devote so much time editing questions and so forth? What do they get in return for these services?

discussion

Ed Gorcenski

@GarbageCollector Not the community elected ones, as far as I know.

But there are SE employees who are paid, I believe.

2

Garbage Collector

@EdGorcenski OK. Thanks.

@kush Thanks for the link. Useful!

kush

@GarbageCollector you're welcome :)

robjohn

@kush you are correct

jdoe

18:05

I need some advice on succeeding

Ed Gorcenski

Do the opposite of failing!

jdoe

well basically I have a lot of notes to catch up on and I'm not sure how

I wish i knew what to do

Ed Gorcenski

Take one thing at a time.

Ignore the pile of stuff to catch up on.

jdoe

ok

Ed Gorcenski

And just start somewhere. Don't try to take on the whole pile; in fact, completely ignore the fact that you have a whole pile.

It will intimidate, frustrate, and inhibit you.

jdoe

18:11

that's a good idea!

Ed Gorcenski

Just choose the most sensible thing, and do that first.

And when you're done, repeat.

Take warm showers, drink lots of fluids, and sleep.

jdoe

Iv'e got an algebra question

if I have the inclusion K -> L from a field extension L/K, and elements l in L,k in K is also in lL then how do we get K/(k) -> L/(l)

that doesn't even make sense. I fixed it

Charlie

@jdoe LIsten to Ed! He's got awesome advices!

jdoe

yeah I agree

I can write k = l l' for some l' in L

this isn't true without additional assumptions

Charlie

18:32

Good night @JayeshBadwaik, sleep tight!

mick

yoyo im back

the hippest mick on mse :)

Charlie

hey

mick

@Charlie hi

i posted a new question just now

I hope it is clear

Charlie

you make too many questions

mick

i do not believe i can make to many questions

but feel free to balance with an answer :)

I do not believe in too many questions

unless they are copies of eachother

and im serious now , no smilies

Charlie

18:43

why are you getting emotional?

mick

im not emotional , just making my point

imho it is worth defending

Charlie

@mick keep calm

mick

why do you think i make to many questions , assuming you meant it

I am calm , but more indian music and i will go crazy !

Charlie

@mick why "more"?

jdoe

lol

mick

18:46

►

the only cool indian song :)

Charlie

@mick it's cool, but not the only one

mick

►

kitt sofort her komen !

there are knight rider apps on iphone :)

jdoe

i might just be too tired to do more math tonight

Ed Gorcenski

An hour of sleep is worth 2 of studying... as long as you use your waking hours well.

mick

@jdoe hug :)

an hour of chat is worth ...

2

jdoe

18:51

thanks

hahaha

Nimza

Good evening!

Charlie

@nimza!!!

Nimza

hello @Charlie :)

Charlie

wassup?

Ed Gorcenski

An hour of chat is worth negative three hours of work ;)

jdoe

18:54

lol

Charlie

If we go to asylum do they let us take paper and pencil?

Nimza

hello @robjohn, do you know, is it possible to find explicitely function $f$ if we know $(f)^{(2)}$, $(f^2)^{(3)}$, ..., $(f^{m})^{(m+1)}$ up to some $m$ in some open domain?

@Charlie nothing interesting, how are you?

mick

►

Charlie

@Nimza fine, fine

Nimza

@Charlie that's super!)

Charlie

18:56

@Nimza just working on an assigment, those things

Nimza

@Charlie m :)

mick

@Nimza hi

i asked 3 new questions today

robjohn

@Nimza you know those over the whole domain?

mick

going to watch the news guys
brb

Nimza

@robjohn aha! they are analytical (holomorphic) here)

jdoe

19:01

I don't know why but I suddenly got so tired I can barely stay up

Nimza

@robjohn It can be reduced to nonlinear system of algebraic equations, but I don't like it because I think that there are no way to find it's solution explicitely

Charlie

@jdoe go to sleep!what time is it? 12h30 a.m?

jdoe

it's 19:00

Nimza

hello @mick!

jdoe

I wan to go to bed at 22:00 I think

but that's so far away

Ed Gorcenski

19:04

@jdoe Sleep when you're tired.

jdoe

really??

Charlie

what?????

jdoe

I could try it, I don't know what would happen

Ed Gorcenski

Seriously, if your work is your priority in life right now, as you have intimated, then do what you must to get it done. If you're tired now, sleep.

jdoe

yeah it is my #1 priority

Ed Gorcenski

19:05

What would happen is you would get rested, probably wake up at an unusual hour, feeling unusually good. Chances are you'd then do something productive.

jdoe

thank you

robjohn

@Nimza It looks as if $m=3$ might work

Nimza

@robjohn hm? Do you have some idea?

@robjohn heh, in fact the problem is to find only two complex numbers! first and seconds Taylor coefficients of $f$

robjohn

Knowing $f''$, we know $f'''$ and $f''''$ as well. Then
$$
(f^2)'''=(2ff')''=(2f'f'+2ff'')'=6f'f''+2ff'''
$$
$$
(f^3)''''=(3ff')'''=(3f'f'+3ff'')''=(9f'f''+3ff''')'=9f''f''+12f'f'''+3ff''''
$$

From there, we solve the 2 linear equations in 2 unknowns to get $f$ and $f'$

Nimza

@robjohn wow! yes, that works! Great thanks.

@robjohn And I think we can generalize this to $(f_1+\ldots+f_k)^{(2)}$, ...,$(f_1^m+\ldots+f_k^m)^{(m+1)}$. So nice and beautiful!

robjohn

19:26

@Nimza for a possibly greater $m$

Nimza

@robjohn aha. now I'm thinking about it

Nimza

19:40

@robjohn oh, I think that in this case we will receive nonlinear system :( because we have to add equations $f_1^{(s)}+\ldots + f_k^{(s)} = g^{(s)}$ and $f_{i}^{(s)}$ are not known coefficients in this case

robjohn

@Nimza hmm... that seems a problem.

Dan Brumleve

downvoted everyone else here for being wrong math.stackexchange.com/a/233768/1284

am i being overly pedantic?

(or wrong myself for that matter? my answer is 1 out of 5 and those are poor odds)

everyone else seems to be saying that an upper bound on the error term will tell you exactly how many digits of the partial sum are correct. but i don't think that's true.

robjohn

@DanBrumleve Two answers, at least, were correct. I hate to say that that means yours looks wrong.

Brian M. Scott

@DanBrumleve You are, and the other answers are in any case not wrong. At worst they fail to answer the letter of the question.

Dan Brumleve

@Brian i guess none of them are wrong outside of the context of the question

but i meant with everything taken together

Brian M. Scott

19:46

@DanBrumleve They correctly bound the error. And I think that just about everyone will realize that isn’t quite the same as identifying the number of correct digits before rounding.

Dan Brumleve

having an error bound seems necessary to actually compute the first n digits. but we still don't know on that basis exactly how many terms as a function of n that we'll have to add up.

Brian M. Scott

@DanBrumleve And we all know that, so what’s the point?

Dan Brumleve

so i guess a most complete answer would include a discussion of the error bound as well.

Brian M. Scott

@DanBrumleve Do you really think that André’s answer didn’t answer the question as the OP intended it to be understood? If so, how do you account for the fact that the answer was accepted?

Dan Brumleve

No, apparently it was good enough.

mick

19:56

actually i think the answers were quit good ...

Dan Brumleve

well in my defense i've observed that real-numbers-as-infinite-digit-sequences vs. real-numbers-as-cauchy-sequences is a topic that can be confusing to many, sometimes because the language we use is more specific than our intentions... so on that basis i hope my pedantry is of some service.

here is a recent thread that exemplifies what i mean: math.stackexchange.com/questions/217893/…

err not my answer specifically but the whole discussion

Charlie

@mick now, this time, could you help me with a question?

help me to think...

Ed Gorcenski

20:22

##frustrated grumble##

This software does not define angles using the right hand rule

And angles are taken clockwise from the vertical axis.

Dan Brumleve

@robjohn i didn't downvote you yet this time.

robjohn

@DanBrumleve yet, nice :-)

Dan Brumleve

i'm chilling out on that. still i wish you would mention the digit issue.

nope only the other 4. it was someone else.

mick

@Charlie im sorry i did not hear the ping ....
Did you look at my recent questions ?
I cannot promise you I can help you , I ask more questions then i give good answers ...

im not that good

and i write i with a small i way to many times

Charlie

hmm..thanks, i'll try to figure it out myself. thanks

mick

20:29

how is it possible i did not hear the ping ? i had troubles logging in too.
Maybe Im afraid to leave now :p

@Charlie good that is more mysterious :)

im gonna listen some more to knight rider :)

Charlie

@mick it has to do with injection

mick

Now i wonder if I should talk about math here or just post the next question :)

@Charlie if you graduated you should have no trouble with that ? maybe open the book again. either how , thinking of an example function usually does the trick ...

@Charlie did you post the question ?

is there a badge for reputation over 9000 ? like supersaiyan math ? :)

@GarbageCollector hi

Garbage Collector

@mick Hi mick,

mick

did i post to much garbage ? :)

maybe you can smell it

Garbage Collector

Is there still any zero related discussion?

Charlie

20:35

Let $F:\mathbb {R^2}\rightarrow \mathbb {R^2}$ the transformation
$$u=x; v=xy$$
show that if $x\neq 0$ so there is a neighborhood of $(x,y)$ that is mapped to a neighborhood of $(x,xy)$ and it is an injection.
The thing is , I think i should use the inverse theorem in this case.Is that right?

@mick i won't ask your help again

:)

Garbage Collector

@mick I will collect them.

Ed Gorcenski

@Charlie I think you should not need that theorem

Charlie

@EdGorcenski no? are you sure?

mick

@Charlie i bet this is homework , smells like it ...

Ed Gorcenski

By contradiction, assume F(x_1,y_1) = (u,v) = F(x_2,y_2) for (x_1,y_1)\neq (x_2,y_2)

Charlie

20:38

@EdGorcenski i'll think bit more...

Ed Gorcenski

From the left, you have u = x_1, v = x_1y_1

mick

I think Ed is correct

Ed Gorcenski

From the right, you have u = x_2, v = x_2 y_2

This means that x_1 = x_2, and there is a similar equality that can be stated for the v term.

Do you see?

mick

I do :p

anon

observe (x,xy)=(a,ab) iff (x=a and (y=b or x=0=a)). for every point (x,y) with x nonzero, there is a neighborhood which has no vanishing x-coordinate. so the map restricted to this nhbd is injective, as two points map to the same iff their coordinates are equal i.e. they are the same point.

Charlie

20:40

@EdGorcenski yup

mick

@Charlie sorry I was too slow :)

Ed Gorcenski

And do you see why we need the condition that $x \neq 0$?

Charlie

@EdGorcenski hmm yes

mick

zero will be proud

@JohanLarsson hi !

Ed Gorcenski

A good gut check when doing textbook/homework problems is to see if you consume all of the hypotheses floating around in the statement ;)

Johan Larsson

20:42

hej

anon

@JayeshBadwaik no, that's just with sets (and is actually true, in contrast to what I referenced; there are in fact counterexamples to the claim on groups)

mick

@EdGorcenski so true !

Charlie

@EdGorcenski Thanks, Eddie!

mick

darn , girls dont need me :p

Charlie

no

@anon thanks Anon!

anon

20:44

$\rm Charlie \in\{women\};Charlie\ne\{women\}$

mick

@Charlie thanks :)

Charlie

@anon why you say that?

anon

mick said "girls don't need me" in response to your interacting with Ed rather than him, it seemed

Charlie

@anon hahaha

Brian M. Scott

@anon But $\text{Charlie}\in\{\text{women}\}\leftrightarrow\text{ Charlie}=\text{women}$!

anon

20:48

that didn't go as planned

Charlie

@anon what was your plan?

anon

fix: replace $\rm \{w\}$ with $\rm w$

Brian M. Scott

@anon Much better! :-)

Charlie

{Charlie} is a singleton

anon

{Charlie}, rather. Let's not make the reverse mistake.

Brian M. Scott

20:51

@Charlie No, Charlie is a point. $\{\text{Charlie}\}$ is a singleton!

Charlie

@BrianM.Scott :D

mick

Im a class apart

Argon

@Charlie Chaplin

Charlie

:)

mick

he is not a she ?

Charlie

20:54

@mick me?

mick

yes ?

Charlie

who knows...

Ed Gorcenski

This exam needs to post already... or to at least not post until after I get back from seeing the new Bond movie.

Brian M. Scott

Who nose? Knot aye.

mick

I asked 3 questions today , Im so proud :)

Charlie

20:56

@mick How many times do I have to say I'm a girl?

Argon

Where is Jasper? It is Friday!!!

mick

@Charlie But charlie chaplin ?

charlie chaplin is not a girl

Charlie

@mick Charlie $\neq$ Charlie Chaplin

Argon

:)

mick

but argon said so :)

@Charlie

so I wrote : he is not a she

Argon

20:57

@mick She likes Chaplin.

That's it.

Charlie

^thanks

mick

aha but does Chaplin like her ?

Garbage Collector

Who stay in Germany?

Charlie

german people

Argon

XD

mick

20:58

its full of germans there :/

Charlie

@mick yes

Garbage Collector

Seriously, I want to know more about German.

Germany

mick

why on earth do you ??? xD

Garbage Collector

I am in Asia.

mick

they started 2 world wars.
You should know that.

Johan Larsson

20:59

I go to south Germany for work now and then, one thing that is really good is the food

Transcript for

$Mathematics$ Mathematics