@N3buchadnezzar You want to compute integrals?

I have some.

@JonasTeuwen Only fun ones =)

They are very fun.

=)

Please compute $$\int \sin^\nu \alpha \cos^\mu \alpha J_l(\xi \sin \alpha) \, \textrm{d}\alpha.$$

What is $J_1$ ?

That is the BesselJ function of order $l$. Very common 8-).

Ive actually run into that one a few times belive it or not :p

My first hunch would be to use parts, then use the gamma function.

mind=explode

You will fail miserably if you do that @N3buchadnezzar.

@anon Hmm...?

@N3buchadnezzar If you do by parts you will reduce the order of the powers of $\sin$ and $\cos$ but those will be expressed in previous powers...

don't mind me I'm just being off-topic

@anon This is the place to be off-topic!

@JonasTeuwen My hunch tells me this integral is not elementary, and I guess you should have included the range of $\xi$

:P

That's $\xi$.

Are you trolling me?

It is a positive real number, but that shouldn't really matter what the sign is.

I keep forgetting my greek :/ and I am too lazy to render math.

Hmm, but how is that possible? You're at the university right, then you should have had ~6 years of Greek?

@JonasTeuwen Would another way to solve it use another definition of the bessel function and integrate ?

@N3buchadnezzar I hope so.

But then you get a series of $2F1$, and that doesn't really make things better, does it? 8-).

I was thinking of the integral definition

Den use fubini to switch integration order :p

And obtain an integral which you are unable to compute?

So you made of one integral two integrals.

That doesn't sound like an improvement 8-).

I think the series definition is more the way to go, but I'm not sure.

$$J_1(\alpha) = \frac{1}{\pi} \int_0^\pi \cos(\tau - x \sin \tau) \mathrm{d}\tau $$

You still obtain something horrible.

That's an $l$ not $1$.

Yes, try to compute the remaining integral.

(If you are able to, I'll buy you a beer 8-))

Well, whenever I try to integrate something I often just try a variety of things ;)

Mmm, that's not the right way to go for these integrals I suppose. Unless you have some magical feeling which thing works best.

Making one integrals into two, then switching the limits often works. Aswell as differentiating under the integral sign. Integrating under the integral sign etc.

Yes, I am aware of those methods.

However, it will not work here, I'm quite sure as the argument of your $\cos$ is horrible.

Doing trigonometric substitutions will make it worse.

At least, that's my experience 8-).

Also, it is not the derivative of something simple.

Yes I know, all approaches failed !

Well, I have a solution, but it is a series of $_2F_1$ functions.

:/

8-).

Elliptic ?

Hypergeometric.

And hypergeometric functions are already quite general.

You take those when everything else fails.

So, I suspect that a series of those functions cannot be simplified in an easy manner, so another approach is needed.

$$\int_0^\infty \frac{\log \tau}{\tau^a(1 + \tau)} \mathrm{d}\tau $$

Yes, that one is easy.

Most integrals are not so nice.

I am still having problems picking contours

@JonasTeuwen Hard integrals are calculated nummerically, or defined through series :p

What is $a$ because it is easy in the sense that it diverges.

$a \in (0,1)$

Divergent.

ops

@JonasTeuwen :/ Fixed!

That's better 8-).

Tag inverse-semigroups has been created and used here. I don't think it's a good idea, but I don't want to decide on my own. Who should I ask?

@anon Haha!

Nice. Paul Garrett expanded on his answer in response to a comment of mine even though I'd already accepted someone else's answer.

How about people stop down voting Chung's questions.

@MattN Why should they? They give reasonable reasons for their downvotes.

@ymar Because so many downvotes aren't justified anymore.

I dunno, I think they're justified until the reasons for them have disappeared.

If someone offends you do you punch them in the face once or do you keep beating them until they're dead?

Latter or neither.

In fact, judging by the lynch on most dictators I would say abuse the corpse as well...

But that's not punching in the face. Downvotes are a good way of telling the guy that something is wrong with his questions.

I see nothing wrong with his questions. This one doesn't show any attempts and gets 3 up votes. Huh?

I hate diagram chasing. I hate everyone who ever made me chase a diagram and I hate them even more for the fact they made me chase a diagram!!

I'm an idiot. I just spent 1.5 hour typing an answer which even the asker may not read (judging by how many answers he accepted) and I haven't done my homework for tomorrow.

Noooo

@ymar Welcome to the club, buddy. :-D

What's your homework for tomorrow?

Four exercises in field theory.

I have some idea of how to solve three of them, but I still need to actually do that.

So I think I'll say goodbye now. :)

Goodbye now!

I was right!

See you!

I'm still sitting in my office... T_T

It's 23:00 already.

Heh. I thought you're one hour ahead.

You switched to DST, we only switch on Saturday.

Heh. I didn't know the idiotic switches took place on different dates.

@MattN Oh yeah, especially in Israel. The dates for the switched are actually decided by the Jewish calender instead of global norms for that. Those jerks at the government decided to finally allow Israel to be closer to the norm, but it will take effect only next year I think.

@ymar weren't you going to do some homework?

@MattN Yeah, I was, but I remembered than part of what I need to to that is somewhere here.

Heyo captain jack

Still is confused on how to pick my contours

noone? :/

@Templar I don't think that has a name, but you can formally define it with factorials.

Night folks.

@MattN G`night'

@anon

k<n

do you mean k<m?

wait

looks like $\displaystyle \frac{k!(n+m-k-1)!}{(n-1)!m!}$

no its wrong

i mean my expression

i meant something like

$\prod_{i=n}^{i=k}\frac{i}{i+m)}$

and n > k and m>0

maybe they can be >= but idk

try double dollar signs

$$\prod_{i=n}^{i=k}\frac{i}{(i+m)}$$

it never worked for me in chat so idk if it shows for you

And you can edit previous posts by using the up arrow.

It shows up, but you forgot an parenthesis.

you mean $$\frac{p}{q}\mapsto \frac{p(p-1)\cdots 2\cdot 1}{q(q-1)\cdots (q-(p-1))}=\prod_{i=0}^{p-1}\frac{p-i}{q-i}=\frac{p!(q-p)!}{q!}=\binom{q}{p}^{-1}$$ for $p\le q$, $(p,q)=1$?

Templar, can you see the LaTeX parsed in your browser?

I can't see it in chat, but when I post questions I can

and I think no, but im not sure what you written

@Templar: Please see here.

You need to drag the javascript link into your bookmarks bar, then click it when you're on your chat tab, for the latex here to parse.

yeah it works, thanks

@anon Your formula gives me an error when p<q, eg p=5 and q=10

@N3bu: did you see underneath the formula, where I said $(p,q)=1$?

@anon I did right before you responded :$

And you forgot the gcd part because of laziness, or is it not needed?

It is standard in number theory to abridge $\gcd(a,b)$ to just $(a,b)$.

p=1 q=2 = error

if you're getting an error for that then the error is your implementation methinks :P

@anon Indeed

Curse you maple!

hides

@anon

yes

Is this so far correct?

you and your $\varphi$'s...

lemme check

(I tried in the end write it less backward-3-looking...but it become almost e)

it's correct so far

$\vec{x}=R(\cos\varphi\sin\theta,\sin\varphi\sin\theta,\cos\theta)$

Varying $\theta$ rotates a point about the origin, which does not change its distance from the origin.

(or rotates it about an axis through the origin, can't remember. same answer: radius doesn't change.)

...and the same is for $\phi$ (other angle), also a rotation?

@hhh see upload.wikimedia.org/wikipedia/commons/7/70/…

@anon $\partial_\phi R = \partial_\theta R = 0$?

yes

$\partial_R \phi =\partial_R \theta$?

=0.

yes

yes

@anon so do you know anything about this? I think your last response was about my previous statement which was incorrect $$\prod_{i=n}^{i=k}\frac{i}{(i+m)}, n>k, m>0$$

why are you writing the larger number on the bottom of the \prod?

because it goes from larger to lower

but it doesn't matter

or it does

@anon: $\partial_\phi \theta=\partial_\theta \phi=$?

if i were writing wihtout that product symbol i would start from higher number :D

@hhh: =0. They're all independent.

i can just post original question, from which i am getting something like that plus something more

@anon math.stackexchange.com/questions/124414/…

@anon: dot-product?

two terms cancel out, zero -multipliers there.

@hhh: Oh, I guess there was something I missed in your image. Divergence is not the same as gradient, so to compute $\nabla\cdot F$ you have to put a dot in between the spherical expression for $\nabla$ and F...

Can one use complex numbers and eulers identity to calculate

$$ \sin \left( \frac{5}{12}\pi \right) = \sin \left( \frac{\pi}{4} + \frac{\pi}{6} \right)$$ ?

I know that one can use the sum of two angles, but I would like to try another approach.

@anon: does this fix it?

Yes.

Hi.

@anon Is anon an abbreviation for "anonymous"? If that's the case, you probably will not reply to my question if I ask you where you are from, right? 8-).

It's known here I am from Nebraska in the US, and anon is indeed an abbreviation for anonymous. (Back on 4chan, where everybody has the handle "Anonymous" unless otherwise specified, everyone refers to each other as "anons.")

Unless you are found trying to harm kittens, then you are no longer anonymous, and everyone will soon know who you are.

Of course I am not anonymous, I merely borrowed the nomenclature for an alias.

I am pseudonymous.

@robjohn Definition of trolling: obnoxious, attention seeking behavior that may use incorrect or upsetting statements.

@anon Oh! So you have mastered the art of trolling? You should teach Skullpatrol, he does need some lessons.

:-0

Heh. On 4chan, trolling has a more narrow definition (at least last time I was a regular): your entire discourse must be an intentional practical joke on other participants.

@anon $\hat e _{\theta}=\partial_\theta e_r$ and $\hat e _{\phi}=\partial_\phi e_r$ ?

@anon Getting duckrolled for instance

This is as opposed to someone who is merely oblivious / obnoxious / boring / etc. by nature.

Related:

@hhh If your definitions of $\hat{e}_\theta$ and $\hat{e}_\phi$ are as such, then sure.

@anon not this time, I have $r, \phi, \theta$ in the basis (I do not know how to get from $e_r$ to $e_\phi$ or $e_\theta$: i.stack.imgur.com/Xi3xD.jpg

what are your definitions of the $\hat{e}$'s in spherical coordinates then?

@N3buchadnezzar What is your definition of trolling?

@Skullpatrol Your sole existence

@N3buchadnezzar Hmm... how abstract.

@anon this is my explicit premise, I have done no other explicity premise.

As stated in this version: i.stack.imgur.com/Xi3xD.jpg

well, what do $e_R, e_\theta, e_\phi$ stand for, first of all? if you're using these quantities at all you have to have them standing for something.

re

The definitions of the $e$ terms (in spherical coordinates) should exist in your text prior to this problem. Either $\hat{e}$'s do not need to appear in your problem, or they have also been defined for you in your text. I haven't done calculus in spherical coordinates for probably 4 years and have no memory of what these things mean.

Is this $\frac{\bar x}{|r|} = \hat e _r$ or $\hat e_r = \frac{\bar x}{|\partial_r \bar x |}$?

Good night! Ding ding ding.

@anon I accepted your answer on the posted factorization question, thanks for putting in so much work into your detailed response.

@JonasTeuwen Sleep well.

Good night @Skullpatrol

@JonasTeuwen Good night and hopefully your wrist feels better the next day.

My book does not cover the 3D scenario, but well I can guess what the unit vectors are -- just make sure the dot products between them are 0, right?

I guess

...and the length must be 1 ofc...

...and they must be orthogonal! So the cross-product between two returns the third one.

Can anyone help explain how x^2 > 5 turns into -sqrt(5) < x < sqrt(5) ... I see that it makes sense, but attacking it algebraically I get x^2 > \pm sqrt(5) ... how does the -sqrt(5) term get to be less than x? Compare it to say x^2 = 5... both the + and - sqrt(5) stay on the RHS

don't forget to pay attention to signs, hhh

*sorry to interrupt :D

@stariz77 add dollar signs around formulae

@stariz: If you see that it makes sense, then what you're looking for is a formal argument, not an explanation.

@anon ?

@hhh: $a\times b =c \iff b\times a = -c$

Um, you can't go from $x^2>5$ to $x^2>\pm\sqrt{5}$.

I don't get it. I just don't get it. What is the fucking point in having me copy out a proof from a book if it doesn't stick in my head?

Why are they doing that to me, over and over and over again?

@Asaf: To make someone feel like they're helping you?

And why oh why does it always have to be some diagram chase?!

Who is this someone?! If I can find him I will break his kneecaps.

Helping me... a real help would be to cut me off the hook on this theorem and letting me focus on my thesis instead.

Look at me... 1:20 am and I'm sitting in my office, probably the only person in the building, writing this diagram for the third time after deleting it and starting the proof over...

@anon I mean $x > \pm sqrt(5)$ ... I don't understand why following that algebraic procedure has both positive and negative on the same side. Or is the negative term being multiplied by (-1) implicitly and therefore the inequality is flipped for that term?

@stariz77: The reason there's a $\pm$ on the right side is because you put it there and have thus confused yourself, not because it makes any sense or follows from any "algebraic procedure" logically valid for inequalities. What you need to do is split $x^2>5$ into two cases: (a) $x^2>5$ and $x>0$; (b) $x^2>5$ and $x\le0$. The first becomes $x>5$, the second becomes $x<-5$, and the original is equivalent to "(a) or (b)."

@Templar Very nice way to split it up into two cases.

@stariz77 Do you recognize the difference of two squares?

@Skullpatrol yes... thanks for that guys, all cleared up!

Np