Mathematics - 2012-04-27 (page 1 of 2)

12:48 AM

Suppose $B=\int_A r^{2} \rho_0 d m$ and I need to calculate the moment of inertia for a pyramid that is cut by the x-y-planes -- getting you a piece with 4 surfaces. Now I need to calculate the inertia along Z -axis. I have so far with homogenous density:

$$B=\rho_0\int_0^{3a}\int_0^{-y+3a}\int_?^? r^2 dz dx dy$$

but I am locked with the inner borders, ideas?

Subproblem on page 964 of 8ex. here.

The vertices of the thing are $(3a,0,0,)$, $(0,3a,0)$, $(0,0,3a)$ and $(0,0,0)$ where $a \in\mathbb R$.

Z -axis is along the direction against which I need to integrate the moment of inertia. Because moment of inertia is a cross product -- the Z -term becomes zero but but -- this cannot be zero here...ideas what am I thinking wrong?

The distance is 3a from origin to the point $(0,0,3a)$ along which I need to integrate.

Actually, Heureka! It must be just 3a! But but if I think this as cross-product -- my head gets confused -- let $r$ be the vector orthogonal to the $z$ axis pointing to the points in the unit.

No!

I am thinking too complicated -- it needs $x,y$ coordinates then, yes it is 3a with the inner integral and the integral becomes:

$$B=\rho_0\int_0^{3a}\int_0^{-y+3a}\int_0^{3a} r^2 dz dx dy$$, ditto sorry noise.

robjohn

@leo: I have added an answer to your question. The answer uses the suggestions I made earlier.

ymar

Hi. Could someone please take a look at this answer of mine? I'm afraid it might be incomprehensible...

robjohn

1:04 AM

@ymar sorry, I have to take my dog for a walk. If no one else has taken a look by the time I get back, I will take a gander.

ymar

@robjohn Thanks! If I'm not here by that time could you leave a comment?

hhh

$\int_{A}r^{2}dm$ in moment of inertia, what is the $r^{2}$ let say in the case of cube?

anon

1:35 AM

$r^2$ is $x^2+y^2+z^2$, no matter what $A$ is...

hhh

@anon Thanks!

anon

@ymar: Your answer makes sense to me, it's just that restricting attention to lattices turns out not to be enough generality to truly answer the question (seeing as how probabilistic independence doesn't fit into it, I think..)

ymar

@anon I agree. I just have no idea how to answer the question in full generality. Do you think I should have made it clearer?

anon

Dunno, it was clear enough to me.

ymar

OK thanks! And have you noticed any mistakes? I'm not very good at lattice theory so I could have easily said something stupid...

anon

1:46 AM

I didn't work it out fully because the concept was clear enough, but I didn't see anything obviously wrong.

ymar

That's reassuring, thanks! :)

anon

(Not that I'm "good at lattice theory" ..)

hhh

@anon but then $\int_0^{3a} x^{2}+y^2+z^2 dz$ is not $3a$!

$[z^{3}/3]^{3a}_0=(3a)^{3}/3=9a^3$

anon

So?

hhh

@anon nothing, just clarified my misunderstanding -- hopefully correct now.

$$B=\rho_0\int_0^{3a}\int_0^{-y+3a}\int_0^{3a} r^2 dz dx dy$$

where the last one is inside

Eric Gregor

2:54 AM

@MarianoSuárezAlvarez, i'm trying to understand a part of Lee's Smooth Manifolds. what's the connection between symmetric tensors on a vector space $V$ and covectors? in the text he says "If $\omega$ and $\nu$ are covectors then $\omega\nu=\frac12(\omega\otimes\nu+\nu\otimes\omega)$."

i'm not asking for a proof, per se. but i don't understand why this is under a proposition about "Properties of the Symmetric product"

i mean, i can see that the rhs is the symmetric tensor of $\omega\otimes\nu$

leo

hey, there!

Eric Gregor

but i think i'm just not getting what the meaning of the lhs is. i don't understand why the product of functionals should be a tensor

hi

anon

3:29 AM

@Eric: A pointwise product of functionals is no longer linear, no? We need to conceive of the product as taking two linear maps to a bilinear map on the product of two copies of the original space, hence the product of functionals lies in the tensor product... or something like that.

Eric Gregor

i'll think about that @anon, thanks

robjohn

3:46 AM

@leo: thanks for the acceptance :-)

leo

@robjohn Thanks for you answer!

you help me a lot

robjohn

@leo your answer was well thought out. (+1)

@leo You might want to do the t<0 case, too. It is a bit different

leo

in LaTeX one can use \author{} for type the author's name on the title. I there a way to use that string in some other place? I'm looking for something like \theauthor

@robjohn Thanks :-). I want to improve it by adding some images to show what's happen

robjohn

@leo I was thinking of adding the flow analysis to my answer.

leo

it would be great

I cant edit comments :-|

I suppose is because I'm using Chrome

robjohn

4:02 AM

@leo even within 2 minutes you can't edit them?

anon

everyone should be able to edit their own comments within a 5 minute window. the browser should not matter.

J. M.

@Jonas: i.stack.imgur.com/H91Vk.jpg

robjohn

@leo: which comments, the ones on questions or the ones on chat?

@JM hey there. Nice to see you back.

leo

@robjohn no

J. M.

@robjohn and hi to you also. :)

leo

4:09 AM

There isn't the usual dow-arrow to the left of the comment, so I can't edit it

robjohn

@leo try typing the up arrow after clicking in the text input area

@leo the arrow only appears when you are hovering over the line in question

clicking the mouse in the text input area is just to make sure the focus is in that area

leo

4:32 AM

test(1)

the first way work

leo

4:58 AM

good night !

see you

Jeff

anyone know how to search the math room transcript for a word (like @jeff)?

J. M.

@Jeff There's a search box at the upper right corner.

Jeff

hehe :D. that worked. i found what i needed.

@robjohn are you here? (your icon is gray)

robjohn

@Jeff I'm here working on an answer

Jeff

@robjohn got a moment for a very quick, simple follow up to earlier discussion about the mapping of (z-i)/(z+i)?

robjohn

5:12 AM

@Jeff sure

Jeff

@robjohn you said that if $z$ is real, then $(z-i)/(z+i)$'s absolute value is $1$. I want to say that therefore, any real $z$ maps to " the edge of the unit circle". Is there a more mathematical, or precise way, to say "the edge of the unit circle"? (see, easy question :D )

robjohn

actually the circle is the boundary (edge) of the unit disk. So the real line is mapped to the unit circle.

J. M.

"Is there a more mathematical, or precise way, to say "the edge of the unit circle"?" - "the set of all points $(x,y)$ such that $x^2+y^2=1$"

robjohn

@JM or all complex $z$ so that $|z|=1$

J. M.

See? Different ways to say the same thing!

anon

5:17 AM

circle is already $|z|=1$, it's the disk (open or closed) that's the filled-in one

the circle is the boundary of the disk

J. M.

In the case of the disk, you use $<$ or $\leq$ instead of $=$, depending on whether you want to include the boundary or not.

Jeff

@robjohn ok. IOW if I say "maps the real axis to the unit circle" that means what i want it to mean ("maps the real axis to the border of the unit circle"). cool. thx (again!)

robjohn

@Jeff border of the unit disk

J. M.

Remember: a circle is one-dimensional; a disk is two-dimensional.

Jeff

@anon @robjohn oh. the key difference is unit circle vs. unit disk. or, the unit circle is the "border" of the unit disk. if that's that your point, then i'm good to go.

@JM ty, also.

J. M.

5:28 AM

The most overloaded word in mathematics...

anon

the topological term is "boundary." The unit circle is $\{|z|=1\}$, the open unit disk is $\{|z|<1\}$, and the closed unit disk is $\{|z|\le1\}$, as JM says. the boundary of the closed unit disk is the unit circle, and the open disk has no boundary.

@JM: regular, dual

J. M.

@anon They're overloaded too, but not as much as "normal" I think.

Eric Gregor

in jim's answer here: math.stackexchange.com/questions/97239/… what's the appropriate way to generalize?

anon

true

Eric Gregor

i've been trying to find this for a half an hour

it seems like you should be able to associate to symmetric tensors in the basis a symmetric polynomial in some number of variables and do something with that, but i haven't quite got it

anon

5:32 AM

@EricGregor Me and Mariano were discussing this at some point. Consider summing $T(X_1,\cdots,X_n)$ over all permutations of $\{1,\cdots,n\}$ and then dividing by $n!$.

Eric Gregor

that's the symmetrization

anon

Wait, that's what's in the answer, nvm

err, question

Eric Gregor

that's just the definition of Sym(T)

anon

I should read before I comment :/

Eric Gregor

no worries

i was thinking of defining a notion of a height of a tensor in the basis of the symmetric tensors based on the number of basis elements in the $k$-vector and showing that if $S$ and $T$ disagree on a $k$-vector of a given height they must disagree on one of lower height, and get some kind of contradiction after assuming minimality. but this kind of idea might be the product of weariness

and i couldn't get it to work out anyway

where height is defined by me to be the number of basis elements of $V$ appearing in a $k$-vector

Matt N.

6:01 AM

Morning.

Sleepy x_x

J. M.

6:13 AM

The title raised my hopes too high...

Kannappan Sampath

!!!!

Hi @JM !

J. M.

@KannappanSampath Hi.

Kannappan Sampath

@Ilya The ping for you. Would like to tell you an answer to the CDF problem.

Eric Gregor

this is a dumb question but if you have the wedge product $a\wedge b\wedge a$, is this zero?

Kannappan Sampath

> Have top users received emails about the swag?

Eric Gregor

6:17 AM

i think that's true

(talking to myself still(as usual))

anon

@Eric: The wedge product is associative and anticommutative, so that does evaluate to zero

also $a\wedge a=0$ and $0\wedge a=0$

J. M.

@KannappanSampath Yes.

Kannappan Sampath

Nice. :-)

Just curious, after all.

anon

Whoah, awesome. Thanks Kanna! I need to check my email more.

And meta.

J. M.

Now, I wait for six weeks...

Jeff

6:28 AM

@JM pretty disappointing, huh? :D

anon

What if one of us were more than one of the top users >:)

Kannappan Sampath

@anon You're welcome. :)

Jeff

@anon multiple personality disorder?

Kannappan Sampath

@anon Then, they would send one of the packages to me. :)

(For instance, I was hoping Arturo would get 5 packages for being super awesome and he would give me one of them. :))

Hi @AlexanderAmenta.

anon

I'll fax you the sticker logo :)

Alexander Amenta

6:32 AM

hi @KannappanSampath etc

Eric Gregor

test

anon

@JM Is my account name just my user ID number?

robjohn

@anon I gave robjohn.

I figure if that is not right, they will send another email.

anon

two user names can be the same though. I guess there aren't any other "anons" with 8075 rep, but at one point there were certainly two "robjohn"'s ...

robjohn

@anon I know, and I worried a bit about that.

However, are there a lot of duplicates on the first two pages?

anon

6:36 AM

I'll just ask on meta.

robjohn

and they know who it's from by the form key in the URL we used to fill in the form

Alexander Amenta

so i was reading this question earlier (math.stackexchange.com/questions/137539/…), and it got me thinking

robjohn

@AlexanderAmenta that's what they're supposed to do :-)

Alexander Amenta

(and i don't know whether i should post it as a question, so i'll just throw it out there in chat)

say you have a rectangle, written as a finite almost disjoint union of squares

is there always a way of splitting this union into two rectangles?

i feel like there should be a counterexample but i can't think of one

Eric Gregor

finite almost disjoint??

Alexander Amenta

6:41 AM

the edges can intersect

Eric Gregor

like you partition a rectangle into squares and you are asking if you can create two rectangles out of this partition?

Alexander Amenta

yep

Nimza

Good morning! Did you hear about pseudolines?

Eric Gregor

i think this might be possible @AlexanderAmenta, but i don't know

Alexander Amenta

it would imply the result in the question i linked

@Nimza pseudolines?

Eric Gregor

6:45 AM

how can one show that the volume of the parallelipiped spanned by $(v_i)_1^n$ is $|\det(v_1,\dots,v_n)|$?

Nimza

@AlexanderAmenta yes, each family of curves that satisfy some specific properties is so-called (for example, it must provide a "double fibration")

Eric Gregor

i guess we consider the transformation of the volume of the unit cube

Alexander Amenta

@EricGregor so how do you prove that the determinant gives the volume of the transformed unit cube?

Eric Gregor

that's the question, alex

Alexander Amenta

indeed

(i forgot how to do it)

anon

6:58 AM

one could view how elementary column operations geometrically affect the parallelepiped

that's a possible start

Eric Gregor

well the transformation is definitely a multilinear functional if that's what you're getting at

anon

Suppose the $v_i$ are linearly independent and let $e_i$ be an orthonormal basis for the space. Note that $v_i^\perp$ is generated by the set of all $v_j,j\ne i$. Show that the coset $v_i+v_i^\perp$ intersects $\langle e_i\rangle$. In this way we can transform the parallelepiped into a hyperrectangle with the same volume using column operations$^\dagger$. Now factor the scalar lengths of the sides of this hyperrectangle out of the det and we have a determinant associated to the unit cube.

$^\dagger$ these column operations corresponding to vector addition in $v_i^\perp$ when focusing on the $v_i$ component

Eric Gregor

what is $v_i^\perp$?

anon

the orthogonal complement of $\langle v_i\rangle$.

Eric Gregor

ohh

anon

7:09 AM

this is the n-dimensional generalization of the 2D case, as I picture it

Eric Gregor

but then what do you mean by $v_i+v_i^\perp$ as a coset?

anon

$v_i^\perp$ is a vector subspace, so we view the vspace as an additive group and this is a subgroup. Basically this is $$v_i+v_i^\perp=\{v_i+x:x\cdot v_i=0\}$$

Eric Gregor

ok, i see

anon

When we focus on the $i$-th component in $\det(v_1~\cdots~v_n)$, we can add the appropriate element of $v_i^\perp$ (which leaves the det invariant) to the $i$th component to make it a multiple of $e_i$ in the $i$th component. Do this for each component..

Eric Gregor

what if $v_i$ is $e_i$?

anon

7:12 AM

then we're already done :)

Eric Gregor

err, a multiple of it

do you get my question?

anon

if $v_i$ is a multiple of $e_i$ then we end up with $0\in v_i^\perp$

the whole point of invoking $v_i^\perp$ is so that we can change the $i$-th component to a multiple of $e_i$ without affecting the value of the determinant, nor the volume of the parallelipiped

on the algebraic side we end up with $\det(\lambda_1 e_1~\cdots~\lambda_ne_n)$ and on the geometric side we have a hyperrectangle with the same volume as the original parallelepiped with side lengths $\lambda_1,\lambda_2,\cdots,\lambda_n$

Eric Gregor

so your procedure is to take $v_i$ and send it to a multiple of one of the basis elements

anon

yep

Eric Gregor

and choosing whatever multiple will preserve the volume

anon

7:21 AM

the multiple of $e_i$ it ends up being is determined by the values of $v_i$ specifically and is not something we can choose

blah

Eric Gregor

sure, that's trivial

we can assume we're dealing with genuine $n$ dimensional parallelipipes

@anon what do you mean by "factor the scalar lengths out of the determinant"?

you are assuming that the volume is given by a determinant to begin with

oh, maybe i understand

anon

By multilinearity $\det(\lambda_1e_1~\cdots~\lambda_ne_n)=\lambda_1\cdots\lambda_n\det(e_1~\cdots~‌e_n)$

Eric Gregor

you are making a hyper rectangle and then using linearity

anon

yes

Eric Gregor

ok, i believe this works. i wonder if there is a more elegant way using tensors

Dan Brumleve

7:42 AM

hi i finally exceeded 3k reps and earned the right to vote to close or reopen

is it possible to see a list of recently closed questions?

anon

@Dan: Do you see "Tools" or "Review" link at the very top of the MSE page? Go to this and click the "Tools" tab and there will be a recently closed section.

Dan Brumleve

i have "review" but i don't see "tools" or "recently closed" anywhere

J. M.

"Tools" is 10k+...

@DanBrumleve Can you access this?

Dan Brumleve

This page requires more privileges

J. M.

Yep, 10k+...

Dan Brumleve

7:47 AM

hmm well if i have the right to reopen i ought to be able to see what has been closed

"recently closed" would fit in well with the other subsections in "review"

anon

Dan Brumleve

squinting but i guess i don't want to reopen any of those

J. M.

Anything worth reopening will usually have an associated plea on meta anyway...

skullpatrol

Would that make it a "meta plea"? :D

Jeff

8:04 AM

[smh]: "$1+\frac 12 + \frac 14 + \frac 18 + \cdots + \frac 1{2^n} + \cdots = \Sigma_{n=1}^{\infty}$ which, as we saw from the previous chapter is equal to $1$. [smh]

does anyone else in here get as PO'd at typos as me?

oops, i should have typed $n$ from ZERO to infinity (I made a typo!)

skullpatrol

@Jeff Are you PO'd at yourself?

Jeff

@skullpatrol no. i did not publish that comment with a professional editor and publishing house and i did not print, on the cover, that this book has proven strategies to help you get better scores.

@skullpatrol and, most of all, i corrected myself. the book i'm reading has no errata list anywhere (that I can find) on the internet

and i work for the publisher

N3buchadnezzar

8:29 AM

Hey

Man I wanted to see if there were any in chat who could check if my calculations for a problem was correct or not =(

Guess I will pop in later.

David Wallace

@N3buchadnezzar Speak!

N3buchadnezzar

@DavidWallace I need to find the area of the cylinder $x^2+z^2=5x$ inside the sphere $x^2 + y^2 + z^2 = 25$

Using spherical coordinates I obtained $5000\pi/3$ is that correct?

David Wallace

Umm, give me a few minutes. I haven't done one of these for a few years. Or better still, tell me in more detail how you did it, and I'll tell you if you've made any mistakes.

N3buchadnezzar

Kay

$$ A=\int_0^{2\pi} \int_0^{\pi} \int_0^5 \left( x^2 + z^2 - 5x \right) \, r^2 \, \sin^2 \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\phi =\frac{5000\pi}{3}$$

where $ x = r \cos \phi \, \sin \theta$ and $z = r \cos \theta $

I know my integration is correct (I let maple handle that). What I do not know is wheter this approach is correct or if my integral limits are correct. Although they seem to be when checking against finding the area of a sphere.

David Wallace

8:46 AM

How did you deal with the terms in $\sin^4\theta$ and $\sin^3\theta$?

Oh, OK, I didn't see your comment underneath.. I was busy trying to check your integration.

The integral limits are fine. I just need to think about how the cylinder looks, to make sure that you're getting the inside, not the outside (if you see what I mean).

N3buchadnezzar

Yeah, I was a tad unsure about that too.

David Wallace

I don't think you're going about it the right way. I think you need to turn the equation for the cylinder into an equation for r in terms of theta and phi.

N3buchadnezzar

@DavidWallace Hmmm, I tried that. But I could not figure out my limits.

David Wallace

So you got $\Huge r=\frac{5 \cos\phi\sin\theta}{\cos^2\phi\sin^2\theta + \cos^2\theta}$ I presume?

N3buchadnezzar

\Huge

Hahaha ^^

David Wallace

8:56 AM

Does $$ and $$ work?

Ah, that's what I was looking for.

N3buchadnezzar

I think \Large looks fine, \Huge is a bit overkill =)

David Wallace

So did you get the same expression for $r$ that I did?

N3buchadnezzar

@DavidWallace Uhm, I am a tad rusty in this aswell. Helping out a friend of mine

I presume you used r^2 = x^2 + y^2, with $x = r \cos \theta$ and $y = r \sin \theta$ ?

David Wallace

No. $z=r\cos\theta$ and $x=r\cos\phi\sin\theta$.

$y$ is irrelevant.

N3buchadnezzar

Ah right, then I see.

Now I get the same as you =)

David Wallace

9:06 AM

OK, so you can put in the $r^2\sin^2\theta$ and do the triple integral. Presumably you'll use Maple again?

N3buchadnezzar

Yeah

Hmmm, the limits are the same I pressume.

David Wallace

Yes, the limits are right. They come from the sphere.

N3buchadnezzar

9:23 AM

Seems that equation crippled maple!

David Wallace

Damn; well if Maple can't do it, I don't have any hope of doing it.

Kannappan Sampath

10:55 AM

Hi folks!

anon

hey

Kannappan Sampath

@anon FYI I updated that file there...Thank you so much.

anon

you're welcome!

Kannappan Sampath

And, how do you like the material in general, the designs?

anon

Eh?

Kannappan Sampath

10:58 AM

I meant to ask if you liked the block designs, $(v,b,r,k, \lambda)$ designs.

anon

I only read about 3 or 4 pages.

Or are you asking if I like designs, as a mathematical topic?

Kannappan Sampath

Yes. :)

anon

Somewhere between meh and interesting. I liked the idea of general incidence structures more than narrowing to block designs.

Kannappan Sampath

meh="I don't care" in my dictionary, does it coincide with the rest of the world, firstly?

anon

I need to figure out if the elementary symmetric polynomial $e_n$ can always be written as a linear combination of $n$th powers of other symmetric polynomials...

that meaning is universal for meh in English as far as I'm aware

Kannappan Sampath

11:02 AM

@anon Thank you for confirming that.

anon

err, not symmetric, just other polynomials

aha! maybe I can use $$\left(\sum_{i\in I}x_i\right)^n-\sum_{i\in I}x_i^n$$ as a basis, for $$I\subseteq\{1,2,\cdots,n\}.$$

Oh, and we're working in $k[x_1,\cdots,x_n]$. I guess we'll assume characteristic zero for simplicity. Denoting the above polynomials $s_I$, we can write $e_2=s_{12}/2$ and $e_3=\big(s_{123}-(s_{12}+s_{23}+s_{32})\big)/6$. Not sure if I should work out $n=4$ or just move on to trying for a general formulation now..

Looks like if we split up the $s_I$ into homogeneous parts, each part can itself be rewritten using other $s_I$'s..

smells like induction

Benjamin Lim

12:08 PM

@KannappanSampath

Benjamin Lim

12:41 PM

so

let $x$ be a zero divisor

we look at the principal ideal generated by $(x)$

then this principal ideal is completely disjoint from $S$

$(x) \cap S = \emptyset$

So by Krull's Lemma, we can find a prime ideal $P$ such that

$P \supset (x)$

Kannappan Sampath

Yes, because $S$ was all non-zero divisors..

Benjamin Lim

and furthermore $P \cap S = \emptyset$

therefore

for every $x$ a zero divisor

we can find a prime ideal $p_x$ such that $x \in p_x$

is contained in the union

$\bigcup_{x \in \text{set of zero divisors}} p_x$

so therefore

you know that each $p_x$ is disjoint from $S$

therefore you also have containment $\bigcup p_x \subset \text{set of all zero divisors}$

the set of all zero divisors

$S$ saturated $\iff$ $A - S$ is a union of prime ideals

Definition:

we say that a multiplicative set $S$

is said to be saturated if

$xy \in S \iff x \in S \hspace{2mm} \text{and} \hspace{2mm} y \in S$

Kannappan Sampath

$xy \in S \implies x,y \in S$

Kannappan Sampath

12:59 PM

Bye folks. I might be logged in while I really am not here. Please do not fret if I don't reply that quickly.

@BenjaminLim Sorry that I could not be here for a while. I was away from the laptop while my skype was still logged in. Hope you're not angry with me for that. I did not have much time to do the problems you had asked me to solve. I am sorry again. And, thank you for calling me to find out that..

Matt N.

1:32 PM

whew Now let's see what to do with the remains of today.

Ilya

1:56 PM

@Jonas: are you here?

J. M.

"Relax and enjoy the math!" - wise words to live by. :)

robjohn

3:13 PM

@JM whose words are they?

N3buchadnezzar

@robjohn Someone who never took quantum physics.

J. M.

@robjohn Mariano's.

@N3buchadnezzar If you're not enjoying it, why do it? ;)

robjohn

@JM +1 for Mariano :-)

N3buchadnezzar

@JM Mandatory !

J. M.

@N3buchadnezzar Oh, a class? Too bad...

robjohn

3:16 PM

I added some diagrams to the ODE flow answer.

N3buchadnezzar

@JM You have taught Calculus 2 and 3 before right ? I seem to vagely remember it

J. M.

@N3buchadnezzar Once upon a younger time, yes.

(Actually, no numbers where I taught; just "differential", "integral", and "multivariate".)

N3buchadnezzar

I asked a short question two or three hours ago in chat? Blushes.

J. M.

@N3buchadnezzar which?

N3buchadnezzar

Here it is called multivariate, but I get the impression refering to it as Calculus 3 is more common.

Here

robjohn

3:21 PM

@JM was that before numeral systems? ;-)

J. M.

Oh dear. :)

@robjohn Apparently so.

robjohn

@JM I remember it well :-)

J. M.

@N3buchadnezzar The quick tip is that you can exploit the symmetry of your problem...

N3buchadnezzar

"Now class I will leave this second order ODE on the board, while I go kill this pesky dinosaur."

robjohn

@N3buchadnezzar They probably still teach flint arrowhead making.

N3buchadnezzar

3:29 PM

@robjohn Hopefully

Remember children, if all else fails. Use your club on your classmate and take his stonetable.

Eric Gregor

4:58 PM

hey @MarianoSuárezAlvarez, are you around?

Matt N.

5:29 PM

@robjohn or @JonasTeuwen, ayt?

robjohn

@MattN ?

Matt N.

@robjohn It's ok, I wanted to confirm a retag of a question with one of you but I've already retagged.

robjohn

@MattN okay

skullpatrol

What's up?

J. M.

Bah, it's getting harder and harder to search for dupes...

Jonas Teuwen

6:03 PM

@MattN 8-)).

robjohn

Hmm... I just ran across this upload on Scribd.

@JM more things to search through?

J. M.

@robjohn Yes, and searches for \binom{n}{k} don't work for \binom{k}{n}... :)

robjohn

@JM can you use *? (wildcards)

J. M.

They work in Google, but the results I get are no better than browsing through the tag...

@Jonas: for you

Jonas Teuwen

6:22 PM

@JM :D.

I love monkeys.

skullpatrol

6:59 PM

@JonasTeuwen: Some more for you

Matt N.

7:16 PM

I think Asaf likes pointless discussions.

skullpatrol

7:27 PM

Sounds like Asaf is lonely. @MattN

Matt N.

NARQ alarm.

That was fast! It's been done already.

Jordan

7:54 PM

@Kannappan I don't wan't to be a Kann because life would be too easy, all that money would make me lazy

Jordan

9:07 PM

what a worthless problem, why would I ever need to find the area under the curve of a function that includes no numbers but an a instead

ymar

Let $M$ be an upper-triangular $\aleph_0\times\aleph_0$-matrix over a field $k$ with no non-zero coefficients on the diagonal. Do the columns of $M$ necessarily span the direct sum of $\aleph_0$ copies of $k$?

robjohn

@Jonas @anon: would you review the main entry for binomial-coefficients to see if you agree?

Darn, I thought the excerpt would be used when looking at the tag, and the main entry would be used for learn more...

Nasty thing about being >20K is that you can't get 2 points here and there for any kind of edit, even wiki info.

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$Mathematics$ Mathematics