Mathematics - 2012-12-18 (page 1 of 4)

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Argon

...?

Charlie

?

Gustavo Bandeira

?

Argon

?

Gustavo Bandeira

12:35 AM

.?

Argon

.?.

Gustavo Bandeira

$\Huge\text{?}$

Charlie

ok it's getting weird

Argon

Hahaha

Gustavo Bandeira

I have a secret.

The secret is: I have another secret.

Charlie

12:38 AM

@Argon Oh! i forgot to tell you!

Argon

@Charlie Oui?

Charlie

@Argon in a very nice park here there was a giant menorah

Argon

@Charlie Hahaha :) I have a little one.

Charlie

@Argon hehe i don't have any :(

Biologists think they are biochemists,
Biochemists think they are Physical Chemists,
Physical Chemists think they are Physicists,
Physicists think they are Gods,
And God thinks he is a Mathematician.

Argon

@Charlie What uni do you go to?

Charlie

12:41 AM

@Argon to the university with the name of my state :P

Argon

@Charlie ok

Charlie

@Argon you know my state?

Argon

@Charlie ..no :)

Gustavo Bandeira

Acre

O Acre é uma das 27 unidades federativas do Brasil. Está situado no sudoeste da região Norte e tem como limites os estados do Amazonas a norte, Rondônia a leste, a Bolívia a sudeste e o Peru ao sul e oeste. Ocupa uma área de 152.581,4 km², sendo pouco menor que a Tunísia. Sua superfície equivale a menos de 2% do total do país. É um dos e foi o último a ser efetivamente povoado. Sua capital é a cidade de Rio Branco.

Charlie

@GustavoBandeira haaaahahhahaahahaaaahahahaha

Argon

12:49 AM

I've been to en.wikipedia.org/wiki/Acre,_Israel !

Charlie

@Argon WOW! nice!

Jonas Teuwen

@JayeshBadwaik Thanks!

Link

Hey how do I find $\frac{d}{dx} sec^2 x$. I know how to use the chain rule

user19161

Halfway to capping today...

Link

I don;'t get why the answer is is what it is.

Argon

12:55 AM

@Link What do you have so far?

Link

I know what the answer is

I don't get how to do it

at all

blank

user19161

@Link That is the square of the reciprocal of cosine.

Argon

@Link $2\sec x \frac{d}{dx}\sec x$

Link

I know that the chain rule is $f'(g(x))g'(x)$, @JasperLoy, I know that... I don't get why the chain rule works here

huh?

user19161

First, can you differentiate sec x ?

Link

12:56 AM

what?

sec x tan x

Argon

@Link $f(x)=x^2$

$g(x) = \sec x$

Link

okay wait, why?

Argon

$f(g(x)) = \sec^2 x$

Because you know what $f'(x)$ and $g'(x)$ are

Link

...

user19161

OK, so to differentiate $\sec^2 x$, that is simply $2\sec x\times \sec x\tan x$.

Link

12:57 AM

ARGH.

so frustrating

why???

why is there 2 sec x?

Argon

@Link Why to what? The assignment of the functions?

Link

no

user19161

Think of $u=\sec x$.

Link

why is there a 2 sec x?

doesn't help

I never learned it in terms of U

Argon

@Link That is $f'(g(x))$

Link

12:58 AM

:p

Argon

Got it? :)

Link

Nope..

Argon

hehe okay

We consider $\sec^2 x$

And note that this is $(\sec x)^2$, i.e. two functions nested

Jonas Teuwen

I have to write a document about self-reflection for my work 8-(.

Argon

@JonasTeuwen Bleh, I hate self reflection!

Jonas Teuwen

1:01 AM

Reflection is a common wave phenomenon in physics. In particular specular reflection (i.e., self-reflection) is the mirror-like reflection of light from a surface. But that is not what this document is about.

Charlie

@JonasTeuwen self reflection?

Link

Okay

Wait, I think i get it!

Argon

YAY!

Link

$(sec\;x)^2$ makes more sense, applying the chain rule I get: $2(sec\;x)*(sec\;x\;*\;tan\;x)$

so in the end: $2sec^2 x tan x$

right?

Argon

YAY!

Link

1:04 AM

The sad thing is I know integration and stuff, but I have trouble with a stupid trig function.

user19161

You don't need to do too much integration as a medical doctor @link!

Argon

$$\frac{d}{dx}\sec^2 x=\frac{1}{2\pi i} \oint_\gamma \frac{\sec^2 z}{(z-a)^{2}}\, dz$$ where $\gamma$ is a simple closed contour (oriented counterclockwise) around $z$.

user19161

I think I will change into another gravatar...

Argon

@JasperLoy A darker blue square?

Jonas Teuwen

I am sure I can bake a cubical pie.

Good night guys.

Argon

1:17 AM

Good night!

Link

Hey, why is the derivitive of sin 2x = 2 cos 2x?

Argon

@Link Try chain rule again!

Link

wait, why???

there is no power here?

Argon

@Link So? There are nested functions!

Link

Ugh. I keep forgetting the chain rule

confusing it with the power rule...

I get it now.

Argon

1:21 AM

Ok

good

robjohn

@argon: sorry, I have been busy today. Take a look at this for preliminaries. If you understand this, I can get the rest pretty quickly. dl.dropbox.com/u/78279253/Argon_Q.pdf

Argon

@robjohn WOW, thanks!

Charlie

Hey @jas...

Argon

1:41 AM

@robjohn I'm not sure if I get the line "Since the path in starts at ..."

robjohn

@Argon The path in (1) starts at $\frac12-i\epsilon$ and ends at $z$

Argon

@robjohn So $z$ is more general?

robjohn

$x$ is a real number... $z$ and $w$ are complex

Argon

Right

Ok

@robjohn Also, I don't get "Therefore, along the top..."

Khromonkey

Do any of you play chess?

Argon

1:56 AM

@robjohn Other then understanding what it means by "along the top/bottom/real line," I get everything :)

Nur

2:09 AM

I never learnt chess (which means I miss out on all the cool combinatoric problems that involve it). :[

robjohn

2:28 AM

@Argon The function has a branch cut along $[0,1]$ so the value on the top side is different (by $2\pi i$) from the value on the bottom side

Argon

@robjohn Is this always the case?

Also, how do you integrate over a closed contour around the real axis when there is a branch cut there?

(Sorry about the questions)

Argon

2:45 AM

Sweet: $$\int_0^\infty \frac{dx}{x^n+1} = \frac{1}{n} \int_0^\infty \frac{ x^{1/n-1}}{1+x} = \frac{1}{n} B\left(\frac{1}{n},\frac{n-1}{n}\right)$$

user19161

@Argon Yes, this one is simply "blue".

Argon

@JasperLoy Nice new blue avatar!

user19161

Hey @amwhy I have changed from "steelblue" to "blue"!

Nur

@Argon which is $\displaystyle \frac{\pi}{n \sin(\frac{\pi}{n})}$

amWhy

@JasperLoy We're a closer match! :)

Nur

2:50 AM

That's by the reflection formula!

user19161

@amWhy We are in the same equivalence class!

amWhy

@JasperLoy Indeed!

Argon

@Nur Awesome, ya!

user19161

@amWhy Christmas is coming, I hope you get something hot then...

Argon

$$\int_0^\infty \frac{dx}{x^n+1} = \frac{\pi}{n \sin \left(\frac{\pi}{n}\right)}$$

amWhy

2:52 AM

@JasperLoy I'll make a point of it!

user19161

@amWhy Yes. Also, still waiting for you know what, but take your time!

amWhy

@JasperLoy I'm so far behind in emails! I will do, this week!

@JasperLoy It's NOT you...it's EVERYONE...so rest assured, I'm not avoiding you!

user19161

@amWhy ...

amWhy

I've got four MUST DO emails this week...you're one of them!;-)

user19161

I have got no emails to do. I have nothing to say right now.

Nur

2:56 AM

@Argon Yes. I like that integral. Pretty useful too. There's a solution that uses complex analysis, though -- but I never learnt contour integration! :[

Argon

@Nur I have done the proof that $$\int_0^\infty \frac{x^{p-1}}{x+1}\, dx = \frac{\pi}{\sin (\pi p)}$$ with contour integration, actually

Nur

@Argon How did you arrive the beta then though?

Oh, you mean you have done that one as well!

user19161

@anon Did you say you lost much rep lately due to removed users? Well, I gave you a little present just now to make up for it! Have a great week!

Argon

@Nur I used en.wikipedia.org/wiki/Beta_function#Properties

Nur

@Argon well if you're interested here is the calculation: thestudentroom.co.uk/…

Argon

3:03 AM

@Nur Oh, this is what I did.

Pretty cool though!

Nur

It's indeed. Where did you learn contour integration from?

Argon

Good night all!

@Nur Informally self taught

:/

Nur

That's nice. Well, good night! :]

user19161

@Argon Night bro!

Ethan

jasper

user19161

3:14 AM

@Ethan Yes, you need to ping me to get my attention.

Ethan

could you look at my proof:math.stackexchange.com/questions/261159/…

i think theres somthing wrong with it

but I cant find the error

plz

lol

user19161

@Ethan Hover your mouse over the line and click on the right arrow to reply, or use @ plus the first three non-space characters like @eth for example.

Ethan

@JasperLoy could u

look

at my proof

please

user19161

@Ethan I'm not good in this, I think you should get someone else, like maybe @anon to look at it.

Ethan

its just algebra

the first part is

most of it

id put it up as a bounty

user19161

3:18 AM

@Ethan Nah, it's too hard for me. =)

Ethan

lol

ur just lazy

user19161

I am serious, I have forgotten 99 per cent of the math I learnt.

user19161

I am only using 1 per cent of my powers now...

Ethan

lol i forget everything too

do u remember any geometry

lool

i dont

user19161

Hmm, I think I remember vaguely something called Pythagoras theorem, LOL.

user19161

3:21 AM

@ethan I think you definitely have talent for mathematics. With the right training, you will make a great mathematician.

Ethan

im going to end up working at a convience store

like wall mart

i dont have good grades

i have ok grades

user19161

Nah, I think you might even win a Fields medal. I am serious.

Ethan

you dont even know me

lol

user19161

I have seen what you have written.

user19161

And I use "might" which just means possibility!

Ethan

3:23 AM

where are u from

currious

user19161

Me, well, that is a secret for now... I'll tell you more in future...

Ethan

lol

if I replaced an older question with my newer question so I could post it as a bounty, would the be abuse

would i get in trouble

that*

would i

if i just clicked edit

and re wrote the whole thing

rofl

user19161

Wait let me answer a question on the site...

user19161

3:44 AM

Ah asker asked another question in the comments so I had to edit my answer and then edit his question, sigh...

Ethan

lol jasper

look

math.stackexchange.com/posts/256444/revisions, "added 4243 characters, chaged title, started bounty",

I edited an old post of mine

slightly

changed*

user19161

@ethan I answered your easy question.

robjohn

@Argon That is the branch cut for $\log\left(\frac{z}{1-z}\right)$

user19161

@robjohn I need 90 more to cap!

Ethan

I know

I was just covering loose ends

robjohn

3:53 AM

@JasperLoy your mother must be so proud...

user19161

@Ethan Would you like to upvote me for my efforts? Feel free to ask for clarification. I think I provided the best answer there...

robjohn

@JasperLoy I was offline most of yesterday. I was able to make 25 points to offset part of the 50 I lost due to a user removed.

@MikeSpivey Hey there!

Mike Spivey

@robjohn Hey, Rob!

Ethan

dont screw me over with the notation line,

im sure it could be re worded in terms of limits and functions of another varriable to make sense

lool

user19161

@Ethan I know where all your doubts lie ethan, once you study analysis they will be gone.

user19161

4:02 AM

@robjohn Yes, she is proud that I am a very special person. =)

Mike Spivey

I managed to earn the "Outspoken" badge today. I suppose it has something to do with the three starred comments of mine on the sidebar over there. :)

user19161

@MikeSpivey Indeed.

user19161

From math.stackexchange.com/questions/261157/… and math.stackexchange.com/questions/261189/… we see that the asker doesn't know WTF is going on.

user19161

To think I spent so much time answering him only to feel slighted by him now...

user19161

4:18 AM

Oh fine, unregistered user who doesn't care, just here to get answers to HW problems I guess...

Jayesh Badwaik

@JasperLoy There? email...

user19161

@JayeshBadwaik Yes, received. Everyone is so secretive these days!

Jayesh Badwaik

@JasperLoy Secretive, declarative. I am sad here. :-( :-(

user19161

@amWhy Well, after you email me, I will share some secrets with you... Good night.

user19161

@JayeshBadwaik Hmm, let me think of what to reply first...

Jayesh Badwaik

4:32 AM

@JasperLoy okay.

user19161

Over and out!

user19161

Also, I need 70 to cap today, so please feel free to upvote more of my posts everyone yeah? Bye!

robjohn

@MikeSpivey "Posted 10 messages in chat that were starred by 10 different users. "

Mike Spivey

@robjohn Yes, I guess the three from today put me over ten.

robjohn

@MikeSpivey Congratulations!

Mike Spivey

4:39 AM

@robjohn Thanks, although I'm not really sure congratulations are in order. :)

It may be more of a sign of procrastinating on my grading. :)

robjohn

@MikeSpivey another badge. It doesn't happen every day

@JayeshBadwaik I really have no idea.

Jayesh Badwaik

ahh, no problem.

robjohn

@JayeshBadwaik Is it still out of 800?

Jayesh Badwaik

@robjohn no, no, its out of 900. the general gre math section is out of 800.

robjohn

@JayeshBadwaik That must be what I took then. I don't think I took 2 math GREs

Jayesh Badwaik

4:51 AM

@robjohn hmm, yup I guess, in your time, general gre was quite different I think. Thanks anyway.

robjohn

@JayeshBadwaik Sorry I couldn't be more help. It's been a while.

Jayesh Badwaik

@robjohn yeah, I understand. No problems. :-)

Sanchez

5:05 AM

@JaveshBadwalk, I just got curious what your question was. Are you takling about the GRE?

Jayesh Badwaik

@Sanchez yes, the subject GRE. My score was 70%. I was thinking whether it is somewhat decent score.

Sanchez

Ah I see.

Are you applying to math graduate school?

Jayesh Badwaik

I am an engineering graduate, so not really thinking of Ph.D., but probably thinking of masters first.

Ph.D. I will apply to after masters.

Sanchez

I see.

Jayesh Badwaik

BTW, I want to apply to masters in mathematics, not engineering. And my eventual aim is to do a Ph.D..

Sanchez

5:16 AM

Sorry, as I am not familiar with the admission process either, I can't help on this.

Jayesh Badwaik

no problems. :-)

Peter Sheldrick

7:56 AM

um analysis?... math.stackexchange.com/questions/244853/…

algebra? as if...

Chris's sister

8:09 AM

Hi all

libjup

8:41 AM

Hi everybody :-)

Basic question: How do I plot the graph z := 500 - x - y ??

Peter Sheldrick

in octave [x,y] = meshgrid(-5:1/n:+5); surf(x,y,500-x-y)

it's a plane

libjup

Weird... What did I do wrong? it's a linear optimzation question: x<=400 and y<=600 that's no problem to draw... But then there is z left which is 300-x-y
How can I draw this linear function??

Other information given 40x+60x+20x<=10000 and 50x+60x+45z <= 15000

* 40x + 60y + 20z of course

My idea was to replace z somehow

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$Mathematics$ Mathematics