Mathematics - 2015-08-29 (page 1 of 2)

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Rudy the Reindeer

12:09 AM

@anon But the question asks for disjoint cycles and transpositions. The answer writes it as transpositions. A correct answer would have written permutations as products of disjoint cycles and transpositions. (The question of writing a permutation as product of transpositions is so easy that it is unlikely that the OP needed help with that especially if he already managed to do disjoint cycles) It seems to me the answer completely ignored half of the question!

12:26 AM

Hi @Ted.

12:42 AM

Slow day in here, @MikeM.

Mhm.

12:58 AM

hi

1 hour later…

2:05 AM

Quite the slow night, indeed

3 hours later…

4:36 AM

Hello yall

Wowowweewow

Nadie están hablando en esta casa

5:11 AM

Posted on the hbar too

How can I calculate the parallel transport numerically?
e.g. if I have two points that are very close together and I know the metric tensor at those points
I'm thinking of something along the lines of https://en.wikipedia.org/wiki/Parallel_transport#Recovering_the_connection_from_the_parallel_transport
But I'm not sure what the gamma means in that equation (Christoffel symbol?)

Soham Chowdhury

5:48 AM

user image

Large trucks frequently run into mathematicians and physicists. It's not new.

6:25 AM

Hello@Balarka

hi

I had pinged you earlier, did you see the message, @Remember?

Well @Balarka I did pay attention . the only thing different there was we were talking about a short exact sequences between any two arbitrary groups and having an abelian group in the middle. But when we were discussing splitting lemma we thought of only abelian groups

No, we were talking about short exact sequence with all the terms abelian groups.

So is there any general idea of the splitting lemma which applies to arbitrary groups which are not necessary abelian . This is my question

Yes.

6:29 AM

What is it?

It's considerably harder, and involves semidirect products. Get used to short exact sequence of abelian groups first.

Okay.

And one question :

You haven't yet proved that $N \to G \to \Bbb Z^n$ splits for any $n$, $N, G$ are abelian.

So I proved that $\{e_1,e_2,...,e_n\}$ generates $\Bbb{Z}^n$

$\{\cdots\}$.

Right.

6:33 AM

Now I have to follow the same procedure, But before that I got to find the identity element of $\BBb{Z}^n$

Now you have to construct a section $s : \Bbb Z^n \to G$.

@Rememberme Identity element of $\Bbb Z^n$ is just $(0, \cdots, 0)$.

Okay so construct a section

Recall what we did for $n = 1$.

Yes thinking on those lines

isn't $G \to \mathbb{Z}^n$ never surjective for finite $G$ and $n > 0$? is this not a short exact sequence?

6:35 AM

I have never said it's finite.

you never said it wasn't

it's clear that it's not finite from the fact that the sequence is exact!

that said, it'd be superfluous to mention it

how's that? it's an unwritten assumption without which your statement doesn't work

I am given that the sequence is exact.

oh I see

6:38 AM

of course, if I was not given that, not even $G$ being infinite would imply exactness :P

it'd be a pointless exercise in that case.

user147690

Is anyone familiar with Cardano's method here?

for solving a cubic?

I guess it's fair enough that you didn't explicitly say the sequence was exact by assumption since this is clearly not the first time you've brought this up, so whatevs. that was my only problem

user147690

Yep, I was just wondering if it finds all roots, or only a single root

it finds all roots, I think. in fact it gives you 6 roots, and you have to cancel the 3 incorrect ones.

user147690

6:42 AM

Hmm I must have missed something then, I obtained the same root in the $\pm$ case and couldn't see any other cases

@SamuelYusim yeah, well, and splitting wouldn't exactly make sense for non-exact sequences

which was another reason I felt I should bring this up

but once again everything is right with the world

heh.

what kind of math have you been thinking about then, @SamuelYusim?

I've learned a bit about character theory over the last couple days and it's pretty awesome

ah, I dunno much about that.

6:47 AM

basically the idea is that in representation theory eigenvalues of matrices come up a lot, and to solve any problem at even the most basic level you'd think you'd need to find the eigenvalues of a large (read: >2) number of large (read: >2x2) matrices, but actually you only need the sums of the eigenvalues to do most stuff, so character theory just builds on properties of the traces of large sets of matrices

that's as best as I can probably explain it without explicitly using the word 'representation'

how does eigenvalue of matrices come up in representation theory? i'm curious.

let me think about what to say without actually defining a representation

@Balarka This is what I thought till now
$s : \Bbb Z^n \to G$ which takes $\{e_1,e_2 ,\cdots , e_n\} \mapsto g$ for some fixed $g \in G$. Now I can see that I have to use the fact that every element can in $\Bbb{Z}^n$ can be written as a linear combination of $e_i$'s. But how should I use that idea

What do you mean by $\{e_1, \cdots, e_n\} \mapsto g$? I can't parse that.

I mean to say that I am defining the map 's' such that it takes the basis vectors to a fixed element g in G

6:52 AM

That doesn't work, sorry.

Oh.

$N \to G \stackrel{f}{\to} \Bbb Z^n$ be your short exact sequence. The section $s : \Bbb Z^n \to G$ must satisfy $f \circ s = \text{id}_{\Bbb Z^n}$.

If you send all your basis vectors to an arbitrary element $g \in G$, then $f \circ s$ takes all the $e_i$ to some fixed element in $\Bbb Z^n$.

You want to map $e_i$ to $e_i$ by $f \circ s$.

@Remember Go back to the discussion we had earlier.

Okay got that. Let me think again

Let's look at the case $n = 1$, as you seem to have forgotten what we did. We have the exact sequence $N \to G \stackrel{f}{\to} \Bbb Z$. How did we define $s : \Bbb Z \to G$?

Well we mapped 1 to an element g and then linearly extended it

6:59 AM

Which element $g$, in particular?

g=s(1)

That's how you are defining $s$!

You can't send $1$ to an arbitrary element $g$.

You have to satisfy $f \circ s = \text{id}_{\Bbb Z}$

Hello, i have a small question please in a metric space a bounded sequence means the existence of closed ball $B'$ such that $(x_n)\subset B'$ ?

Wait I am telling . don't tell

or we can choose any ball ?

7:00 AM

actually screw it, I'll just go all in. A representation of a finite group is, loosely, a particular way of representing the group elements as automorphisms of some complex (or maybe not, I don't know how general it gets) vector space. The main question people want answered in representation theory is: what are all the ways I can do this for a given group $G$?

yeah, that much I know :D

Well, it turns out that with finite groups you can always decompose an arbitrary representation into a combination with repetition of a finite number of distinct irreducible representations, put together in a nice way, i.e., $A$ and $B$ are two different matrices corresponding to $g \in G$ implies $$\begin{bmatrix} A & 0 \\ 0 & B\end{bmatrix}$$ is another one that you can take.

mhm.

Now basically if you know the eigenvalues of the group elements in the irreducible reps, you can figure out what kind of combination of irreducible ones an arbitrary one is. This is where the fact that you only need the sum of the eigenvalues comes in: it just makes the general problem for finite groups a lot easier than anyone would ever expect.

(I wrote that all at once but it was too long so I had to break it up, but yeah)

If f is the map from G to $\Bbb{Z}$ then $g \in f^{-1}(1)$

Is that good @Balarka

7:02 AM

$f^{-1}(1)$ is a set, not an element of $G$.

You want to write $g \in f^{-1}(1)$

That is, $g$ is an element of $G$ such that $f(g) = 1$.

@Remember Now explain why this works.

@SamuelYusim ah. that's very interesting.

Well now if I map 1 to g where g is an element of $f^{-1}(1)$ then when I again map g using f it maps it to 1. so it maps the element 1 again to 1

Precisely! That's it.

So same logic here

@Remember Now get back to the sequence $N \to G \stackrel{f}{\to} \Bbb Z^n$

You have to define the section $s : \Bbb Z^n \to G$.

Map the basis vectors to an element g such that $f(g)= \{e_1,e_2,\cdots, e_n\}$ . Now linearly extend the map

7:07 AM

Ugh. I have already told you about why that doesn't work.

Okay. wait then

Why do you want to map all the basis vectors to the same element of $G$?

It's not nessesary that preimage of all the basis vectors by $f$ contains a common element.

So you mean map each basis vector to different elements in $G$?

It might be different, it might be the same. We just don't know.

Just label them differently to work in full generality.

That is, choose an element $g_1 \in G$ such that $f(g_1) = e_1$. Choose an element $g_2 \in G$ such that $f(g_2) = e_2$... choose an element $g_n \in G$ such that $f(g_n) = e_n$.

So you mean take $e_1 \mapsto g_1, e_2 \mapsto g_2 \cdots e_n\mapsto g_n$. I get it . Now define these the way we did earlier where n=1. Now every element in $\Bbb{Z}^n$ can be written as linear combination of these . So hence we can map every element such that $f\circ g =\text{id}$

7:11 AM

But what are $g_i$'s?

Again, they can't be arbitrary.

There are not arbitrary

Yeah, so how are they defined?

They are defined the way we defined g in the case n=1

Can we do that?

Be explicit. Write it down.

I mean to say that if $f: G \to \Bbb{Z}^n$ and it is defined $f(g_i)=e_i$ for $1\leq i \leq n$

Will this work?

7:14 AM

Huh? Are you sure you haven't made a typo?

I have

Right. So $g_i$ are defined that way.

@Remember But how do you know that such $g_i$'s exist?

We need a property of $f$ here.

it is surjective

Exactly!

OK, now how are you defining $s : \Bbb Z^n \to G$?

off topic - what can a field $F$ with 4 elements look like? If $0,1\in F$ then $1+1,1+1+1\in F$, so how can $F$ be anything other than $Z/4Z$?

7:18 AM

$F$ is field-isomorphic to $(\Bbb Z/4\Bbb Z, +, \times)$.

$s((a_1,a_2,\cdots,a_n))=(a_1s(e_1),a_2s(e_2),\cdots,a_ns(e_n))$ and $s(e_i)=g_i$ for $1\leq i \leq n$ . Will this work ?

Yep.

Well done.

Good. and thanks

However, note that $s$ wouldn't be a homomorphism if $G$ weren't abelian.

Okay.

7:21 AM

Note a second thing : you have defined all your homomorphisms generator-wise so far. It is very similar to what you do with vector spaces, where you define linear maps basis-wise.

This indicates abelian groups might be something close to a vector space.

Indeed, this is how the theory of modules comes to play.

Ahh... The axioms of a vector space have a similarity with abelian groups

nods :)

Soo fascinating. Isn't it?

In a sense, abelian groups are "$\Bbb Z$-vector space". But this doesn't make sense in a literal sense, as $\Bbb Z$ is not a field.

It's a ring, however. Indeed, vector spaces over rings are precisely what we call modules.

where rings act as fields

I guess

7:25 AM

yep

@Balarka Really off topic: If I construct an SES of topological groups what all can we say about it in terms of topology?

I don't really know much about topological groups. I guess fiber bundles are short exact sequences there, but I have no idea.

Oh.

@Balarka Do we need to think about open connected sets to think about line and multiple integrals??

I am asking this because there is this guy in IISc who has asked me to do Multivariable calculus and functional analysis . I am good with functional analysis but I am not getting the motivation for Multivariable calculus @Balarka

7:52 AM

Multivariable calculus is nice.

@Balarka Whats your definition of dimension. Is it that how many parameters you require to define a single point or is it something else?

It's technical. I am trying to define dimension of an affine algebraic variety, btw, not a topological space.

So how do you define dimension of a topological space

Lots of ways to do it

It's in the last chapter of Munkres. Look there.

last chapter.. let me look through it...

Also does a homeomorphism preserve dimension . By that I mean to say a homeomorphism can only be between two such spaces which have same dimension ?

7:57 AM

You have to define dimension before you ask something like that.

I always wanted to do this :P

@BalarkaSen But what according to your definition .. ?

Dimension theory for topological spaces is interesting, although I don't want to bother about it too much.

@Rememberme I have just told you above that I am defining dimension for affine varities, not topological spaces.

Oh. :(

Affine varieties are technical objects.

No no. You will know some definition about dimension in topological spaces . Now according to that definition is what I state true?

8:01 AM

yes, for the definitions I know.

the whole point of dimension theory is to find an easy homeomorphism invariant for topological spaces.

it'd be useless if homeomorphisms didn't preserve a certain notion of dimension, then :P

Ahh...

but I don't care much about it. homotopy and (co)homology are already very good homeomorphism invariants, and can detect non-homeomorphic spaces much better than [whatever notion of dimension you have]

Well I don't know anything about homotopy :P

8:32 AM

@BalarkaSen $F$ has characteristic $2$. so I think it should be the Klein-4 group instead.

yeah, whoops, you are right.

9:01 AM

@RudytheReindeer OP said "I am asked to write this permutation as a product of disjoint cycles and also as a product of transpositions." That's two different parts. The first part was done by the OP - product of disjoint cycles. For the second part, while you may think it "so easy that it is unlikely that the OP needed help," if you go back and read the question you'll see the OP calls the textbook example "unclear" and says twice s/he is "not sure what this means" - they want help.

Even without those pronouncements, I wholeheartedly disagree that a newbie will necessarily know and understand how to express permutations as products of transpositions if they can already find disjoint cycles.

I am still unclear on what it is you actually think is not being answered. The question wants the permutation (a) as a product of disjoint cycles (which OP did), and also (b) as a product of transpositions. How does your interpretation of the question differ? Please be clear.

Do you think it wants us to (a) write the permutation as a product of disjoint cycles and also (c) as a product of disjoint transpositions? (IOW, having the word disjoint apply to both cycles and transpositions.) This can't be valid, since (c) is not even possible in general, nor possible in this specific instance.

Or do you think (d) that the exercise wants OP to express the permutation as a product of things, each of those things either a cycle or a transposition, and all of the cycles are disjoint? That's vacuously accomplished by expressing it as a product of disjoint cycles, so that interpretation of the question is absurd. I really think you should speak up about how your interpretation differs from everyone elses' (including maybe OP's).

Or do you impose upon amWhy the responsibility of explaining how one finds disjoint cycles generally, or in this problem in particular, even though OP already did that without expressing any issue with that? If so, that's rather uncharitable - doing that would be a nice touch, but not necessary if the OP didn't need any help with it. OP already did 1(a) themselves, amWhy helped with 1(b) which the OP needed help with, and amWhy addressed OP's thoughts on part (2) as well. I'd call that >1/2.

2 hours later…

11:28 AM

@anon Consider the ideal $(x, y^2)$ of $k[x, y]$. This is a primary ideal, as the quotient ring $k[x, y]/(x, y^2) \cong k[y]/(y^2)$ is nontrivial and the only zero divisors - $ay$ where $a \in k$ - are nilpotent. Now I want to show that this ideal is not of the form $\wp^n$ where $\wp \subset k[x, y]$ is a prime ideal.

Atiyah-Macdonald proves that $(x, y^2)$ is not $\wp^n$ where $\wp$ is the radical of $(x, y^2)$. How does the general thing follow, then?

oh, nevermind, I get it. assume $(x, y^2) = \wp^n$. Then radical of this ideal is precisely $\wp$. I was being silly.

12:00 PM

hi @AlexClark

user147690

Hey @BalarkaSen

How're you doing?

user147690

Just typing up a question for you, so good :P(just about Sylow theorems)

ok, sure.

user147690

Sorry still typing, it's looking longish, it's about groups of order 56 not being simple

12:09 PM

take your time.

user147690

I want to prove that groups of order $56$ cannot be simple.

So $|G|=56=2^3*7$ and if $G$ is abelian any subgroup will be normal, and there are non-trivial, proper subgroups here, so let's assume that $G$ is non-abelian:

Since $56=2^3*7$, we have $n_2= 1$ or $7$, and $n_7=1,2,4,8$ with $n_2 \equiv 1{\pmod 2}$ and $n_7 \equiv 1{\pmod 7}$.

The latter modulo equivalence eliminates $n_7=2,4$ I believe,

If $n_2=1$ or $n_7=1$ we are done, so we need now only check $n_2=7$ and $n_7=8$

user147690

Is that fine so far?

It's fine. Why don't you work with just the Sylow $7$-subgroup?

user147690

I want to prove that none of them can be simple

See, it's a counting problem. If $n_7 = 1$, then you have a normal subgroup, hence it cannot be simple, you're done.

If $n_7 = 8$, you have to draw some kind of contradiction by counting the elements in your group. Artin tells you how do to it

Recall that intersection of any two groups of prime order is the trivial subgroup.

And you have $8$ of those huge $\Bbb Z_7$'s sitting inside your group $G$, overlapping only at $1$.

Count!

OK, I take back what I said about contradiction. $n_7 = 8$ is actually possible, but it forces $n_2 = 1$ (this is what you have to prove by counting), so you're fine.

user147690

12:19 PM

$8*8=64$ distinct elements of order 2,4,8 or something, let me think

No, the $\Bbb Z_7$'s are overlapping at $\{1\}$! You are overcounting.

user147690

Sorry, why must they intersect trivially?

Sorry, that's not the relevant fact you need.

What you have to prove is that if $N, H \leq G$, $|N|, |H|$ are both prime, then $N \cap H$ is trivial.

Right, so this is a special case of the previous fact I was stating for $|N| \neq |H|$. In this situation, you need the case $|N| = |H|$.

Hint : what's the theorem you use whenever you see something about order of subgroups?

What is the website to render this chatroom LaTeX again

@BalarkaSen Why couldn't the professor have asked me that question yesterday...I know at least this one

@DanielFischer How long till the book is released? :)

user147690

12:27 PM

I think I can prove that fine, so I'll go do that, but I think my main issue is understanding the counting problem

user147690

I'll go have a shower and sleep and work on it tomorrow morning and I'll tell you how I do

user147690

Thanks for that @BalarkaSen

Take that all the Sylow 7-subgroups intersect pairwise trivially as a blackbox, and then count it, @AlexClark, if you want.

OK, let me know when you get this one.

@r9m Err, gonna be a while.

Chris's sis the artist

@r9m Would you give it a try?:-) $$ \int_0^{\pi} \arctan^4\left(\frac{\sin (x)}{2 \sqrt{2}}\right)\csc (x) \, dx$$

3 hours later…

3:26 PM

Morning.

Morning

McDonald's stops serving breakfast at 11:30am here which is in 2 minutes

Morning, @MikeMiller.

@morphic: And now it's three minutes ago...

Yes...

Things in life sometimes change so suddenly.

3:35 PM

Er. I have been told to do Shaferevich after I do primary decomposition (which I am doing now). I am not quite sure if I know enough commutative algebra to get into algebraic geometry just yet.

I am personally more inclined to stick to commutative algebra for a few weeks :S

(I haven't read Noetherian normalization and the proof of Nullstellensatz, for one)

You're to read Shaferevich's "The Socialist Phenomenon"? Not a bad choice by any means but it seems sort of orthogonal to what you're doing.

Hello!!!

lol, no, I have been told to do Shaferevich's "Basic Algebraic Geometry"

I didn't know the book you mentioned existed before today.

You'll probably be fine. The really serious commutative algebra comes in with schemes.

I mean, you can literally change what you're studying if it's too much.

@MikeMiller I am not sure what you mean by that. (I don't think prof wants to get me into schemes just yet. Probably he's trying to go through projective varieties and afterwards GAGA before the totally general thing, from what I gather from his statements)

3:41 PM

I mean that if you don't know enough commutative algebra you can just learn more.

I don't know if I don't know enough commutative algebra, that's the whole point of the question :P Prof's telling me I can pick up things as I go, but I am a bit queasy about that.

Simaltenously, I do want to study algebraic geometry. I just don't want to turn into a bad (and French) algebraic geometer.

I don't know what's unclear about what I'm saying. If it turns out you don't know enough, you can learn more. So what does it matter if you don't know enough?

ah, so you're suggesting to give it a go and see if the commutative algebra I know makes me comfortable with Shaferevich. If not, come back and study more commutative algebra. is that right?

Yes

Seems like a good suggestion. Thanks.

3:46 PM

or just do whatever you want 'cuz you can

I dunwanna know a bunch of theory for the sake of the theory and not being able to any problems. anyway, I'll probably spend a few more weeks on commutative algebra and singular cohomology before jumping into Shaferevich.

4:20 PM

@MikeMiller prof made a suggestion : if I do atiyah-macdonald minus dvr's, dimension theory, etc (exercises included), hatcher ch. 3 (exercises included) and fortster's riemann surface ch. 1,2, then i can study algebraic geometry.

i can sleep in peace, as that will take me quite a few months

4:36 PM

I've forgotten what one uses DVRs for in algebraic geometry.

He's saying minus dimension theory?

4:47 PM

whoops, no, he's saying minus chapter 5 (DVRs, valuations), 9 (dedekind domains) 10 (completions, which I fortunately already am familiar with). i misread that as 11.

Um, @Balarka, you need multivariable analysis to do Forster.

I know, I have been told to do that. :)

I think I'll just ignore you until you're done with that.

Goodnight, @Mike.

@TedShifrin C'mon. I'm just a kid, and I have been told to learn commutative algebra and cohomology. Plus, I have to keep up with my schoolwork.

I will surely do multivariable calculus. But ten things at a time is addling my brain :(

Maybe I'll join Ted's boycott.

The point is that there are some things you literally cannot do without calculus.

Like, for instance, anything with smooth manifolds.

4:51 PM

You're just doing everything back-asswards, and I'm tired of it.

Done

OK, I cannot possibly get any help with you two ignoring me.

I resign. I am opening my copy of your book.

5:13 PM

lol

@TedShifrin If I am not on ignore, I have tried exercise 5. (a) on ch. 2 where you're asking to parameterize the epicycloid. Can you have a look?

Otherwise, I'm going to mail you.

You're not on ignore.

Balarka: Of course I'm happy to take a look. I'm not insisting that you do every exercise. If you want to concentrate a bit more on the stuff that advances you more quickly, I won't fight.

nah, you're right about this. I'm just too jumpy about big, fancy words.

anyway, here's what I tried. Take the circle of radius $b$ centered on the origin.

By "stuff that advances you more quickly," I meant Chapters 3, 5, 6, a bit of 7, and a lot of 8.

5:24 PM

I think I should do some of the exercises in ch. 2 too, it's been a while since I did calculus.

OK, good, I won't object. And there really are some fun exercises in there.

ok, the small circle of radius $b$ is rolling on the the circle of radius $a$. Assume it started from $(0, b)$, and the point that was lying on $(0, b)$ has moved to position $p$, just like your picture.

I'll be happy when you can define a smooth manifold for me, which is just chapter 3... but the rest is good too.

Oh, and the I- function theorems.

To parameterize $p$, I am borrowing your idea of parameterizing the cycloid. Draw a vector between the center of the two circles (pointing outwards the origin), and join the center of the small circle to $p$ (pointing towards $p$)

$\vec{OP}$ (the vector from the origin to $p$) is precisely sum of those two vectors. The first vector has coordinate $[(a + b) \cos(\theta), (a + b) \sin(\theta)]$, so that is taken care of.

No, smooth manifold is chapter 6.

You have to be very careful with angles there, @Balarka. Very.

5:29 PM

The second vector is troublesome. You have to make an angle $\theta'$ between the two vectors. $\theta'$ can just be expressed in terms of $\theta$, as $b\theta = a \theta'$

Oh, to P, yes. Now the next step gets interesting.

This is hard to do without a blackboard.

@Ted: But what he really needs to know to define smooth manifolds is what it means to be $C^\infty$. ;D

That said, apparently the second vector is nothing but $[-a\cos(\theta + \theta'), -a \sin(\theta + \theta')]$, by basic trigonometry.

Yikes orientations.

First, a needs to be b. Second, be very careful.

OK, so I think the parameterization of $OP$ should be $[(a + b)\cos(\theta) - a \cos(\theta + \theta'), (a + b) \sin(\theta) - a \sin(\theta + \theta')]$.

@TedShifrin $b$ is radius of my large circle, $a$ is radius of my small circle.

I've switched your notation for my own convenience, and I presume to your great distress.

5:33 PM

That's backwards from the problem.

How is that convenience?

I find it intuitive that $b$ is larger than $a$ :D

rolls all eight eyes

You're up to eight?

You should see a doctor.

O_o

That last time it was six, I think.

5:35 PM

@Fargle :)

oOo_OoO

2

Hello, @Ted.

haha

You doing ok, Fargle?

Absolutely. Never better. And you?

5:36 PM

Glad to hear that, kiddo. Very glad.

BTW, Balarka, #7 with the tractrix is a very interesting calculus problem.

Yes, I am looking at that right now. It looks nasty. What do you mean by leash pulled taut?

The bad thing about my chosen tags is that they seem to be a common place for physicists to come.

Good question. That means the leash is going to be tangent to the dog's path.

I thought that was your goal, @MikeM.

@TedShifrin hmm, interesting.

I am going to think on it/chew on it while having dinner.

Bon appétit.

Chris's sis the artist

5:45 PM

@robjohn you're pretty silent these days.

1 hour later…

6:48 PM

@MikeMiller oh my, horrors upon horrors :P

it's terrible

which tags specifically?

just geometry tags tend to get a lot of you folks

ah

ugh, the tractrix thing looks pretty nasty.

hmm. I guess one has to use the observation $y'(\theta) = \tan \theta$ somewhere.

1 hour later…

Chris's sis the artist

8:23 PM

Even the quadratic form I created required so much cleverness to be calculated

Anyway.

The cubic version is already a kind of A-bomb, and for the quartic version there is no description to assign.

9:18 PM

@Chris'ssistheartist what kind of integrals are you tinkering with today?

Chris's sis the artist

@Semiclassical then cubic and quartic versions.

as in, higher exponents of arctan?

Chris's sis the artist

@Semiclassical Yeap

yeah, that's pretty horrific

Chris's sis the artist

@Semiclassical That was possible because I created a new specific integration tool.

9:20 PM

from your comments above, it sounds like the quadratic case is tough, the cubic is only just tractable, and the quartic is horrific

Chris's sis the artist

Yeap. :-)

i won't bother asking about the quintic, then :P

Chris's sis the artist

:D

what was the argument of the arctan in there, just to have it in my memory banks?

Chris's sis the artist

@Semiclassical Screenshot to have it entirely

9:22 PM

thanks

Chris's sis the artist

welcome ;)

i'm curious to see if mathematica can even do the quadratic case

Chris's sis the artist

@Semiclassical poor odds :D

it can do the linear version just fine, but that doesn't mean much

Chris's sis the artist

It's a mean integral, you can try it to see.

(or better say, evil integral)

9:28 PM

what's interesting to me is that the even v. odd exponents give integrands with different periodicity ($\pi$ v. $2\pi$)

that suggests to me another generalization: What are the Fourier expansion of those integrands?

Chris's sis the artist

9:41 PM

Just to know I had in mind a different idea.

oh? fyi, my main reason for finding the periodicity interesting is that, in the linear and quadratic cases, the integrand looks almost sinusoidal (which really means that one of the Fourier coefficients dominates the rest)

Chris's sis the artist

Right. Let me know if you make some progress this way.

sure. (i probably won't :/)

Chris's sis the artist

:D

@Semiclassical I like your observations on the problems in general. Looking at a problem from many points of view is a desired thing always.

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