Mathematics - 2014-10-26 (page 2 of 5)

@JasperLoy Nah :/

@columbus8myhw Bye.

@Sawarnik Let's hope he does not overdo it so that they get reversed.

Probably they will :/

Then they should reverse the few thousand points that X received.

12:04 PM

How many has he given? I think more than $3$ in one day does it

@Committingtoachallenge 5 a day is safe. Any more is not safe.

Let $V$ be a vector space over $\mathbb{K}$ with a basis $e_i, i \in I$ and $V^*$ be the dual vector space. Consider $T \in V^*$ defined by $T(e_i)=1 \forall i$. Assuming V is finite dimensional, prove the covectors $e^i$ span (I have already shown linear independence) and write T explicitly as a linear combination of the covectors $e^i$.

Oh really? How many have you been getting? $5$?

english.stackexchange.com/questions/204610/… Can you guys take a look at this question on Eng SE?

@JasperLoy I think there has been more for me :/ :(

12:07 PM

$8$ today Jasper? That will definitely be reversed

@Committingtoachallenge You'll get 5 today, chill :D

@Committingtoachallenge Not really, there are 2 upvoting accounts there.

They seem to be pushing everything above $5$ votes?

I have gotten $2$, are they you Saw?

xD

@Committingtoachallenge Yeah.

How do I get a longer equal sign on latex?

12:09 PM

Phew, one more I got :( :O

An equal sign that would get aligned to the \stackrel command

@Sawarnik He is clearly trying to push the boundaries.

@JasperLoy Yeah.

How much rep for upvotes?

$50$?

@Sawarnik I have never seen this happen before, someone using 2 accounts to do this.

12:11 PM

Doesn't an upvote give you 10 rep?

Depends.

On?

I meant how much is required for the privilege, but no, upvotes give different amounts for answers and questions

5 for question, 10 for answer

All I've got so far is the following:

For a vector $v \in V$, define $e^i(v)=v^i$
Then $v = \sum{v^ie_i}$
$T(v) = \sum{v^iT(e_i)} = \sum{T(e_i)e^i(v)}$

If I said it were me would you guys give me heaps of upvotes :)

12:13 PM

What do I need to do from here to show that the covectors span?

A: If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$

0

solving the equation $a+1/a=\sqrt{3}$ you will get $a=\frac{1}{2}(\sqrt{3}+i)$ or $a=\frac{1}{2}(\sqrt{3}+i)$ plugging this in $a^4+1/a^4$ we get $-1$

LoL

There is SO much activity today on MSE, all questions get answered within 3 minutes damn

Not mine :(

Link?

Q: Can someone explain linearisation on nonlinear systems to me?

Took an hour and he didn't stay to clarify

2

I want to find all critical points of the following nonlinear system: $$\def\b{\begin{pmatrix}}\def\e{\end{pmatrix}}$$ $$\b y_1' \\ y_2'\e = \b 5y_2 -15 \\y_2^2 - y_1 ^2\e$$ Then use linearisation to find the type and stability of the critical points. So first of all, finding the critical po...

You have an answer lol

12:18 PM

Yeah but I don't understand it :''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''(

:________________(_

Me neither

He didn't say what $h$ even was

Anyway I will go now, byebye everyone!

Cya @Sawarnik (It isn't me btw, but I do like upvotes!)

Oh, bye @Calculus :)

What is the star for ??

@Calculus Yup :D

@JasperLoy Its never ending...one more.

@Sawarnik He seems more interested in finding the max than giving us points now. Be prepared to have them all reversed soon.

Yes, I know.

12:24 PM

I am glad you didn't retire @Jasper

@Committingtoachallenge Who knows, I might retire from life soon.

What...?

@Committingtoachallenge He never will, so don't worry.

I hope to be reborn in Germany.

What did you say was the last letter of your country, I am pretty sure you said something about that

12:27 PM

@Committingtoachallenge E, I think.

I won't reveal my location anymore. I don't want people spying on me.

@Alizter what's up with all these minor edits?

The world shouldn't be aware of what was discussed on those last 4 messages

2

(remove)

@UserX Yup.

Agreed

12:31 PM

They will never know :P

I once thought my country was very good, but now I think the exact opposite.

Over time, I see more and more sick things happening here.

Every country has its problems, but some are much more sick than others.

Sure, this is a good place for making money, but money is not all there is to life.

Australia is very good other than our 'stop the boats' policy and our prime minister not believing in climate change

@JasperLoy money IS all there is to life.

@UserX I heard you plan to be a medical doctor. Why are you so interested in math then?

AmWhy, lab, Hagen. This pic describes perfectly what they're currently doing at the newest section;

@JasperLoy because everyone has a hobby.

12:39 PM

@UserX If no grad school in the US accepts me, I don't know what I will do.

@JasperLoy you're the most clingy 32 year old I know

@UserX What is the meaning of clingy?

@UserX I am 33, not 32.

@JasperLoy too emotional in a childish-kind way. Do you have a master?

@UserX No, I don't.

@UserX Aha, I like to be clingy then.

@JasperLoy can you apply for PhD if you don't?

Grad school is for PhD right?

12:43 PM

do you like to hang on to things?

@UserX In the US, masters and PhD are done together in grad school. Masters are offered if you drop out of PhD halfway in many places.

@IceBoy Yes, I do. I need to focus on solutions to problems rather than problems though.

focus is good

@JasperLoy what field will you pursue?

@UserX Math.

@UserX They are not minor.

12:44 PM

Lol

I mean what field in maths

@Alizter you capitalized the first letter of a sentence...

@UserX I don't know yet.

@UserX Who else was going to do it?

It was not an old question anyway

Seems minor to me...

If it is on the front page it is not old

@JasperLoy I thought the same too. @UserX Why can't you pursue a math career then?

12:47 PM

@UserX So what if it is minor? I don't get any rep for it. I am doing it to make the question better. Many people don't click when they don't like format/grammar. A capital letter at the start of a question is quite important.

@Alizter Well said. I miss Sarah, lol.

Minor edits are only a problem when you are bumping an old question and you are doing it for rep

@Sawarnik because if I go to medschool I'll be bored to death till I finish uni. If I go to math uni I'll be bored to death after I finish uni

And I plan to live longer than 30

Do the math ;P

What do you mean? :/

@JasperLoy :P

@UserX I am not sure how long I plan to live till. If I don't get well or can't go to grad school, life is not worth living.

12:50 PM

I utterly hate teaching and it's hard to become full-time researcher

If not impossible

Oh.

@JasperLoy can OCD be cured?

Yes, research is tough.

@UserX Depends on your personal definition of OCD and also of cure. For me, the answer is yes.

@JasperLoy medical terms don't have personal definitions.

12:52 PM

Hmm.

if you hate teaching, then you're not learning properly

@UserX Some things in life are not well-defined. I will leave it at that, instead of entering a book-long discussion on OCD.

@IceBoy -_-

@JasperLoy How did you get OCD in the first place? Your father? :O

@IceBoy isn't that a little too subjective?

@Sawarnik It's a long story. But yes, he did a lot of bad things to me.

12:54 PM

Obviously UserX would know more about OCD, afterall he wants to become a doctor. You wouldn't know anything being around it your entire life :P

@UserX ok, in my opinion :)

@Commiting I haven't studied all disorders. But I certainly disagree with the "doctors read from books, they can't understand what it's like" ideology(one JasperLoy has adopted)

@Committingtoachallenge Yup :D :D

@UserX Good, the first step in becoming a good doctor is to know that doctors don't know it all.

Doctors have the best memory powers in the world.

And worst handwriting.

12:57 PM

@Sawarnik that describes me perfectly

@Sawarnik -_-

However doctors can also focus in texts easily, which doesn't describe me

@IceBoy Why?

@UserX :D

@Sawarnik but a counter-example is my mother. Her handwriting is perfect.

@Sawarnik for the same reason you did it to me :P

12:58 PM

@IceBoy ok :P

:D

@UserX Rare exception.

But really, your mother should have done something about your smoking.

:O you smoke?

@Sawarnik like what?

carry

stop you from smoking

1:01 PM

UserX is too cool not to smoke

It's my descision. Why would she do anything other than tell me the dangers?

cancer cancer cancer cancer cancer cancer cancer cancer cancer cancer cancer cancer

2

Noone is smoking to be cool or because "muh problems mah life is hard you dun know nothin"

@Committingtoachallenge Did you put yourself on the map?

People smoke because they got hooked one way or another

1:01 PM

@UserX ;)

carry

:D

@UserX But your decision was not the best for you.

She should have stopped it.

Basically she was a bad parent

nobody can stop anybody from doing what they want

@Committingtoachallenge Kind of...

1:07 PM

@Ice Say that to a north korean ;)

haha :P

@IceBoy not always.

@Committingtoachallenge that's politics

@IceBoy Why? In normal situations...

Parenting $\neq$ forcing descisions/opinions.

@UserX ;)

1:09 PM

@UserX True but parenting = ensure that their child picks up good habits and not bad ones.

@Sawarnik Good habits are also subjective.

My Dad beat me when I smoked a long time ago

I don't smoke

Like smoking is definitely not.

@UserX decision ~~descision~~

@Committingtoachallenge Beating you for smoking is bad parenting.

1:10 PM

@Sawarnik No, that's duplicating your envision of who you think you should have been on your child.

Aren't you becoming a doctor like your mother LOL

@UserX Nah, no way.

Do you agree smoking is good?

@Commiting My mother tries to talk me out of becoming a doctor.

I have not really looked into the articles that claim that smoking is bad for health, maybe I should check them out before thinking so.

and he fell into the trap xD

1:11 PM

@UserX Lol, why? :D

@JasperLoy It is.

@Sawarnik Not every decision in life is a good one.

@Sawarnik because she's tired all the time and hates her specialty.

@Committingtoachallenge Imagine the irony if he became a cancerologist or something... :D

duplicating your envision of who you think you should have been on your child - My mother tries to talk me out of becoming a doctor.

However, I can tell you that I think antidepressants are largely useless.

If $f$ is a integrable on $[a,b]$, and $f(a) = f(b) = 0$, $\int^a_b = 0$, right?

1:12 PM

@UserX I understand.

Antidepressants are, but anti-anxiety is not

@Committingtoachallenge Yeah.

integrable and continuous*

I have taken three courses at a top 10 university in such things

@Sawarnik ONCOLOGIST* lol

1:13 PM

@UserX That's why I said 'or something' :D

who knows all these silly terms.

He took a good three minutes on that, quick google :D

@Committingtoachallenge Which?

@Committingtoachallenge lol

Hey guys make a non-hamiltonian graph with $p$ vertices each with degree $\frac{p-1}2$

@UserX Why don't you pay heed to her advice of not becoming a doctor :P :P

@Sawarnik does cancerologist sound okay to you?

1:15 PM

@UserX not sure

english.stackexchange.com/questions/204610/… I got 3 downvotes here, do you think my answer is wrong?

@Sawarnik because I can decide for myself.

fasthealth.com/dictionary/c/cancerology.php

@JasperLoy The question seems a bit opinion-like.

@JasperLoy that's a subjective question, people might express their dosagreement through downvotes.

1:16 PM

/me patiently waits

@UserX Maths is not that untiring and ... :P

@VibhavPant No

@VibhavPant You'll perhaps be waiting for long...

heh

@Committingtoachallenge why not?

@VibhavPant What if it was a spike going up and then down

1:17 PM

ah yes

That would hold if $a=b$

totally forgot that

@VibhavPant Are you trying for CMI?

@Sawarnik yes

@VibhavPant Nice :)

@Committingtoachallenge I was thinking the same too really :/ :/

1:18 PM

The complete question is if $f(a) = f(b) =0$, and $f(x) = f'(x) + f''(x)$, prove $f(x) = 0$ in $[a,b]$

@Chris'ssis hey !! when did I ask for hint ? =P LOL but ULTRA-NICE solution there !!! (+1) :D

all those things, yes

@r9m Too many adjectives.

@Sawarnik I can count

Alright.

1:21 PM

@Vibhav $f(x)=c_0 e^{-\phi x}+c_1e^{\phi x}$

I tried integrating to get $\int^a_bf(x) = f(a) + f'(a)- f(b) + f'(b) = f'(a) + f'(b)$

@UserX $f(a) \neq f(b) \neq 0$ in your function

That's the general solution to the DE

@r9m hehe, thanks! :-)

@UserX Can you use something less advanced than DEs?

There is nothing advanced

1:23 PM

@UserX but it doesnt satisfy the initial conditions

(for this question)

I didn't insert any initial conditions.

I just solved the DE in case it might help

the only thing I can guess is that the general solution doesnt satisfy the initial conditions given, so the only possibility is $f(x) = 0$

well, it does. it just requires $c_0=c_1=0$

isnt there any other method?

which is a part of the general solution, albeit the most boring case

1:26 PM

@Chris'ssis I did it on my own first :-) hehe !! but your solution looks compact !! (sos put a link to Eric's solution !! nice solution that one ! ... very nice ;) )

What's wrong with this one?

@VibhavPant sure :)

i suppose what you want is something that lets you say

@UserX What is $m$ for $m^2+m -1 =0$

@r9m nice. What tools did you use? (largely speaking)

1:27 PM

@Semiclassical there could be other combinations of constants where $f(a)=f(b)$ holds?

@VibhavPant Won't it be f'(a)-f'(b) :/

@Commiting why?

sure, they just wouldn't both be equal to zero

@Sawarnik yes

I realized that late

"suppose i've got an ODE which, for whatever reason, I don't want to solve generally. is there some easy way to see if it has a solution given dirichlety boundary conditions (vanishes on endpoints of interval)?"

need to be at most second-order ODE to make sense, to be sure

1:29 PM

@Chris'ssis perron transform and functional equation of zeta (I'm learning these stuff in my ANT course now :P .. so can't keep my hands off them :-) ... )

@Commitingtoachallenge did you mean $m^2-m-1=0$?

@r9m Could you use perron transform there? Interesting. I'm not used to that.

Alright, gtg, byes :)

How is Parseval's identity connected to the pythagorean theorem?

And why did two of my answers get downvoted for no reason?

@UserX Oops, maybe some anti-smoker doesn't like you, lol.

1:40 PM

@JasperLoy not a reason to downvote the most analytic answer among the other two.

1:53 PM

@JasperLoy lol :D

@UserX Link?

Just answered a lhf.

@Chris'ssis btw what happened to the $\displaystyle \sum\limits_{n \ge 1} \frac{H_n^{(2)}}{n^4}$ bet ? :P

@r9m That one is done ...

@Chris'ssis elementarily ? :-)

@r9m sure, totally elementarily

1:59 PM

@Chris'ssis hehe :) okay !!

@r9m I combined my work with @robjohn's work ...

@Chris'ssis oh !! robjohn worked on it too ?? where ? :D

@r9m No, he worked on a different problem.

@r9m Let me see how I can find that work ...

@r9m Check this one first

@Chris'ssis oh !! okay ,, I saw the problem in one of RV's answers and he left a note in his post asking if there was an elementary solution to the same identity .. so I did it :)

@r9m all until $(2)$.

@r9m that work must be combined with another work ... (let me find it)

2:04 PM

oh ! very slow internet here .. pics not loading :(

@r9m can't you see it?

@r9m now use the @robjohn's work here chat.stackexchange.com/transcript/36/2014/7/12. Combine the 2 works and have an elementary proof.

@Chris'ssis now I see it .. lemme read :)

@Chris'ssis thankuu !!

@r9m ;)

@r9m it's possibly the first elementary proof in the world to that series.

@Chris'ssis oh ! when did you do it ? :D (you see I need to know if I won or lost the bet :P)

@r9m Wait, that is not my worth only, but I combined my work with @robjohn's work. I couldn't have done it without that work of @robjohn .

@r9m Well, I don't remember the bet, but I got the proof, that is important. :D

2:13 PM

@Chris'ssis eh ! the problem is easy enough ! the bet is more important to me :P (I stalker .. incorrigible :P)

@r9m lol, that problem is easy? :-) You got an elementary proof to it?

@Chris'ssis I did it in an hour after reading RV's post on I&S and pung RV then and there with the information :P

@r9m Did you post it there? Where?

@r9m I did the problem some months ago when @robjohn got that result.

@Chris'ssis no didn't post it .. just mentioned it is linked to your crazy problem aka $\sum\limits_{n\ge 1} (\psi'(n))^4 $ :-)

@r9m lol :D

2:18 PM

@Chris'ssis HAHA .. that was an awesome problem !!

What is different between $A \cap B =\{x\in A : x\in B\}$ and $A \cap B=\{x\in A\; \text{and} \; x\in B\}$

@r9m Glad you liked that. :-)

@robjohn Fantastic !!! :D :D

robjohn

@Chris'ssis Note that here it is shown that $$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{k^2(n+k)^4}=\zeta(3)^2-\frac{4\pi^6}{‌2835}$$ Combine that with $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4} =\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{k^2(n+k)^4}+\zeta(6)$$ and $$\zeta(6)=\frac{\pi^6}{945}$$ and we get $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta(3)^2-\frac{\pi^6}{2835}$$

@robjohn Exactly. :-)

@robjohn I've never ever seen a paper, a post with that series elementarily done.

2:26 PM

Rob Johnson = 'math god' !!!

robjohn

@Chris'ssis Of course, Mathematica can do it... Sum[HarmonicNumber[n, 2]/n^4, {n, 1, Infinity}]

@robjohn Yeah, Mathematica can do it ... I saw ... :-)

robjohn

Gotta take my dog for a walk... BBL

@Chris'ssis when is the Book coming out ?!!! >.< I can't wait !!!!!

bbl .. dinner

user20997

2:47 PM

what is the radical of a maximal ideal of $\mathbb{C}[x_1,...,x_n]$?

cfr.org/interactives/GH_Vaccine_Map/…

3:07 PM

@user20997 Remember that the radical of a prime ideal is itself, and that maximal ideals are prime.

Move that pic around your screen

I can't be the only one seeing an illusion

3:28 PM

@UserX What do you see?

@UserX Looks like a brain scan.

@MatsGranvik Haha, true.

What do you think of my approach to find the zeros of Riemann zeta directly by finding zeros of polynomials? http://mathematica.stackexchange.com/questions/63541/what-is-the-algorithm-behind-mathematicas-reduce-in-this-equation
I got zero votes here on math stackexchange, so that tells me already something.

Can anyone tell me what this symbol is: ∑

$\sum$

3:36 PM

what is this notation called?

it is a sum, called sigma

Can the person who upvoted my posts today and yesterday confess? You can email me if you don't say it here.

ah, summation

@MatsGranvik that's introduced in calculus?

Because it surely is not present in algebra

@JohnMerlino What are you studying?

It's certainly present in parts of algebra :)

3:37 PM

trigonometry, but I did not find that in the trig book. I found that on this site.

Summation is just a notation. It is found everywhere in math where there is addition, lol.

@JohnMerlino I have a blog post on the subject:
http://mobiusfunction.wordpress.com/2013/01/12/the-summation-symbol/

I have a vector space $V$ over a field $\mathbb{K}$ with a basis $e_i, i \in I$. If $V$ is infinite dimensional, how do I prove that T (where $T(e_i)=1$) is not a linear combination of the covectors $e^i$?

All I know is in every linear, quadratic, polynomial, logarithmic equation I have encountered, that symbol was not present

Yes, because there is no need for it.

3:40 PM

You mean $T(e_i)=1$ for all $i$?

@TedShifrin Yes

and $T \in V^*$ where $V^*$ is the dual vector space.

Well, you're only allowed finite linear combinations, so write down the general such.

Hush up, @Mike

LOL :)

mouth zipped

@TedShifrin If V is infinite dimensional, then must I be infinite dimensional?

What was removed?

3:43 PM

$I$ is infinite by definition of dimension.

@TedShifrin Because for the finite case, i've got that $T=\sum_{i \in I}{e^i}$

Yes, $I$ must be infinite. The question is whether it is uncountable ...

Yields the floor to @Mike now :)

I should be working :P

@MikeMiller @TedShifrin How do I know whether I is uncountable or not? And why does I being countable mean there are finitely many linear combinations?

I'm not sure why uncountability was mentioned. It doesn't matter here. I don't understand your second question.

3:48 PM

@MikeMiller I must have misinterpreted the response. Ignore that question then. I'm still confused as to how I can show that T is not a linear combination of the covectors.

@user112495 A linear combination of the covectors looks, in general, like $a_1 e^{i_1} + \dots + a_ne^{i_n}$. Agreed?

Regardless, @user112495, you are allowed to take only finite linear combinations of your basis covectors.

Upper $i$ @Mike

thanks

@MikeMiller Yeah.