Mathematics - 2013-06-06 (page 1 of 2)

Mechanical snail

00:05

What's a curve that asymptotes to the line $y=-x$ on the left and zero on the right?

Like a squashed rotated hyperbola.

anon

"the" curve?

Mechanical snail

@anon There's a common one I remember seeing somewhere.

Peter Tamaroff

@Mechanicalsnail 1/x-x ? Like this?

anon

the angle made between y=-x (for negative x) and the positive x-axis ray is 3pi/4 radians. define any hyperbola y^2/a^2-x^2/b^2=1 with b/a=tan(3pi/4)/2. then rotate the hyperbola clockwise by pi/8.

Mechanical snail

I remember now; it was the Black-Scholes price of a call option.

anon

00:11

actually I think I made an error, but that's the idea

@Mechanicalsnail yeah, I was just thinking that...

Peter Tamaroff

@anon LOL

Mechanical snail

anon

$$\left(x+\cot\frac{\pi}{8} y\right)^2-\left(x-\tan\frac{\pi}{8}y\right)^2=k^2 $$

for $y,k>0$

Mechanical snail

00:30

@robjohn Nonsense! That triangle has two acute angles. This is a triangle with just right angles.

robjohn

@Mechanicalsnail too bad Goldilocks lives in Flatland...

gekkostate

01:06

Hey guys, I was wondering if someone can help me with parallel lines. Can someone please explain the equidistant property for parallel lines?

anon

can you explain what that is?

Peter Tamaroff

01:24

@anon How the hell does one write math in Wikipedia?

anon

what do you mean?

amWhy

@robjohn Okay, robjohn...back to Goldilocks, are we? :-P

Peter Tamaroff

@anon Well, dollar signs don't work.

anon

mull over the meaning of the phrase "reverse engineering"

Peter Tamaroff

@anon Yeah, it ain't working either.

anon

01:31

le sigh

what article

Peter Tamaroff

Apparently, it uses something like <math> <\math>

anon

indeed

Peter Tamaroff

And sometimes {math | stuff inside }

anon

I'm not familiar with that one

Peter Tamaroff

@anon I just added the definition of net

"More precisely..."

anon

01:33

I must go get ice cream before the place closes.

Peter Tamaroff

@anon I am going to sleep, then.

Good byes.

Tony Stark

Wow, that

there is chat here

neat

what software to draw that man?

anyone?

Ozera

@anon lol. Gimme some

robjohn

02:04

@amWhy No, Mechanical snail brought it up.

AlanH

02:39

@robjohn I'm reading through a proof (closure of a connected set is connected), which they show by contradiction, and it doesn't make sense to me that they say that $cl(S) = A\cup B$, where $A, B$ are subsets of $cl(A)$. How can it be possible that a closed set is made up of two open sets?

@anon or could you help me clarify this confusion? I think robjohn is afk

robjohn

@AlanH are you sure it is not the union of the closures?

anon

(1) Closed sets in general may or may not be the union of two disjoint open sets. Depends on the topology. (2) The proof is presumably supposing by hypothesis that cl(S) is disconnected (and hence, by definition, of that form), in order to yield a contradiction from the premise that S is connected.

robjohn

@AlanH As anon says, two clopen subsets means the set is not connected.

AlanH

@robjohn Let $S$ be connected and let $cl(S)$ be the closure of S. Assume for the sake of contradiction that $cl(S)$ is not connected. Then there is some disjoint non empty open sets $A,B \subset cl(S)$ s.t. $A \cup B = cl(S)$....

anon

@AlanH exactly. what's strange about that?

robjohn

02:43

@AlanH yes, read what anon said first and then what I said after

AlanH

@anon I might be reading it wrong, but it seems like, for example, [0,1] is the union of two open sets that are subsets of [0,1].

anon

do you understand what "proof by contradiction" means?

robjohn

@TonyStark look here

@AlanH but [0,1] is connected

AlanH

@anon face palm

@robjohn sorry... i'm getting confusd with all the negation

anon

Suppose we want to prove A implies B. The idea is to assume A is true and B is false, then arrive at a contradiction, thereby proving A's truth necessitates B's truth.

robjohn

02:45

@AlanH Think of $[0,1]\cup[2,3]$ as a topological space

AlanH

@anon yeah i know what proof by contradiction is. bah.

anon

:)

robjohn

off for a while, bbl

skullpatrol

later

DanZimm

04:03

@Ethan ohai

Ozera

04:14

Can anyone help with: bit.ly/11GfExw ? I'm just not understanding the hint given to me by someone. I can't find a pattern with a n board for tribonacci #'s

DanZimm

@Ozera did you look at Andre's answer?

Ozera

That huge long list? I tried to, but it was hurting my eyes

@DanZimm the only mention of tribonacci #'s is "Let C = the tribonacci constant..." and that doesn't really make sense to me with what hint is given to me by a different user

DanZimm

@Ozera hrm, I was just trying to make sure you saw his answer, not really where to begin to find a real world application of the tribbinacci sequence

Ozera

@DanZimm The hint was: " try finding a combinatorial interpretation of tiling a board of length n using some set of tiles"

I've been trying to do that, but I haven't been getting anywhere

DanZimm

@Ozera have you tried tiling with three different tiles?

Ozera

04:25

@DanZimm What do you mean?

DanZimm

tile a board but instead of with two different tiles, with three different kinds

Ozera

If there is a 2x1 board how do you tile with three different kinds?

Maybe we can use a 3xn board I guess

I really don't see how that helps thoguh

Ozera

04:52

Well, i'm thoroughly stuck now.

Matt N.

05:16

@robjohn Yes, my computer is a laptop. But I should buy a new battery: if I unplug it it dies within 10 minutes. So if I want to sit somewhere else I need to move the plug, too. But I'm too lazy most of the time.

mixedmath

05:39

hello

anyone around?

Mariano Suárez-Alvarez

no one

NONE

Jayesh Badwaik

Zero matter....

anon

05:58

@MarianoSuárez-Alvarez I have a very ill-defined sorta kinda question maybe. This is the best version I have crafted. Do you think it make any sense?

Mariano Suárez-Alvarez

/me looks

makes sense

anon

my motivation is that, since p-adic fields have solvable galois groups, roots of polynomials are always expressible in terms of nested radicals (although there is ambiguity in "which root" is being referred to everywhere a radical appears). however there cannot be a uniform nested radical formula (involving only rational constants) for roots of general nth degree polynomials (n>5), so there may be a "partitioning" lurking underneath of elements according to applicable generic nested radicals

Mariano Suárez-Alvarez

you should probably eexplain why you want the set S there

anon

wlog we could just make S={x,y} for the question as it stands, but if we want to impose further restrictions on S (for example, topological, if the fields L, F are topological) then this is a good template to add things to

Mariano Suárez-Alvarez

also, in the last formual you want a for all m\geq 1 at the start

Iguess?

or somewhere

anon

06:06

it can wlog be >1 because otherwise the polynomial for that specific s_i can be subsumed into the next g_i()'s and the h() at the end.

Mariano Suárez-Alvarez

well, you need to quantify somewhere on n

«$x\sim y$ if there exists an $n\geq1$ and...»

anon

ah, yes

so I guess restrict x,y to having the same degree over F, namely n, where n is fixed

Mariano Suárez-Alvarez

what exactly do you want $x\sim y$ to mean?

in words

anon

I want x~y to mean that there is a fixed "nested radical formula" that works for a certain space of polynomials (in that it characterizes all solutions), of which x,y are both roots of certain polynomials in this space

Mariano Suárez-Alvarez

what do you mean by «a certain space of polynomials»?

anon

06:11

I mean a subset of the space of all polynomials

Mariano Suárez-Alvarez

which one?

anon

$\exists$ one

Mariano Suárez-Alvarez

hm?

does the empty set of polynomial work?

ot the set $\{(T-x)(T-y)\}$

anon

the idea is that no nested radical formula works uniformly for all polynomials, but there might be nested radical formulas that work for certain classes of polynomials, and so we should be able to create a "patchwork" - a family of formula, each nested formula valid for a certain class of polynomials (valid meaning characterizing all the roots), and every polynomial is contained in one of these classes of polynomials (so that the "patchwork" covers all possible polynomials)

Mariano Suárez-Alvarez

that is true

but that does not explain what $x\sim y$ wants to mean :-)

anon

06:15

it means x and y are roots of two polynomials in the same patch

Mariano Suárez-Alvarez

but that is always true, no?

anon

perhaps it'd be better to just define ~ directly on polynomials

Mariano Suárez-Alvarez

ah, forget that

given two polynomials of the same degree, you can say they are related if there is a formula on the coeffcients which gives the roots of both

anon

that looks best

Mariano Suárez-Alvarez

it might be useful to consider only pairs of polynomials with the same monomials

instead of just the same degree

a coarser relation for elements would be: two elements of $\bar K$ are related if the smallest Galois subextensions of $K$ which contains each of them have groups with the same factors in a composition series.

this relation is «one has to take the some number of roots in the same order to construct $x$ and $y$».

more or less

($k$ should contain all roots of unity for this to be close to true, though)

(for polynomials, the analogue would be by looking at the Galois groups of their splitting fields)

Little Child

07:42

da hello.. anyone here ?

Andrew Salmon

Hey

Little Child

Heya !

Andrew Salmon

Not seen you before...

Little Child

new here. Joined a couple of days ago

Andrew Salmon

Ah...welcome, I guess.

Little Child

07:47

Thank you

I came to ask a few quick questions on limits

Andrew Salmon

OK

sure

Little Child

which won't quite suit a full blown question :)

just keeping the site tidy :-) as much as I can

Andrew Salmon

I think it's appropriate to ask these on chat, if they're quick.

Little Child

In limits if $\lim_{x\to0}$ then a fraction of the form $\frac f(g)f(x)$ will always evaluate to either +ve infinity or -ve infinity

it is f(g) by f(x)

Andrew Salmon

Do you mean $\frac{f(g)}{f(x)}$?

Little Child

07:50

Yes !

Andrew Salmon

What is $g$? Is it a constant?

Little Child

Yes !

Andrew Salmon

Not generally. Just let $f(x) = x+1$, and let $g = 1$, for example.

Then your limit evaluates to $2$.

Little Child

Ok so this equals 2

Andrew Salmon

Do you mean $\frac{c}{x}$, perhaps?

Little Child

07:52

err... 1/2 right ?

Yes that would be proper c/x

Andrew Salmon

I'm having a bit of difficulty understanding your question.

Little Child

I was using the video lectures by patrickjmt on YouTube

Andrew Salmon

Are you asking for $\lim_{x \to 0} \frac{c}{x}$?

Little Child

where he begins teaching about limits involving infinity

he says well if you have something like $\frac6x$

and $\lim{x\to0}$

then you get answer as infinity

I was just trying to further my understanding on my own, maybe come up with a more algbraic notation of what he just said

Andrew Salmon

Well, not quite. You have to pay attention to the fact that from the positive direction, $6/x \to +\infty$, but from the negative direction $6/x \to -\infty$.

Thus, $6/x$ does not have a limit at $0$.

Little Child

07:56

It would become undefined if I plug x = 0

Andrew Salmon

However, it is proper to say that $\lim_{x \to 0^+} \frac{6}{x} = +\infty$.

Little Child

limits are all about getting close to a given number but not quite that number right? :)

Andrew Salmon

Well, to take limits, the general idea is that you plug in numbers close to $0$ but not equal to that number.

Exactly.

Little Child

so, in our case x will never equal zero

Andrew Salmon

Right.

Little Child

07:57

so let me just put it all in one line so you can see what is in my head

Andrew Salmon

You don't need to worry about the value of your expression at $0$, only in some neighborhood (some small interval) around $0$.

Little Child

As $\lim_{x\to0}$, $\frac{c}{x}$ will equal $+\infty$ or $-\infty$

Andrew Salmon

No, as $x \to 0$, $c/x$ does not converge.

If something goes to two separate values, then it simply doesn't converge--we don't say that it goes to both values.

Limits are unique. There can only be one.

Little Child

so limit is undefined ?

Andrew Salmon

Well, in this instance we would say it doesn't converge or does not exist rather than undefined.

Little Child

08:02

Sorry, my bad, DNE (Does not exist)

Andrew Salmon

Right.

Little Child

So, in all such cases limit is DNE

anon

except when c=0 the limit exists

"undefined" and "dne" are fairly interchangeable in the context of limits

Little Child

@anon got it cause it will always be zero :)

I am beginning to get the hang of it :-)

Now, I am beginning to understand all what I was taught in high school :-)

@anon where are you from ? :-)

anon

omaha

Little Child

08:08

Ok so from U.S. of A.

skullpatrol

Have you seen this?

Little Child

@skullpatrol I used patrickjmt so no I havent

skullpatrol

:-)

Little Child

Well, I don't have to master limits. I have to get a basic idea

so I can move to derivatives and integration

then to Laplace Transform

On the whole, I don't find skullpatrol "less useful" on chat. He provides comic relief much of the time

@skullpatrol try this

2

Chris Dugale

Is anyone learning algebraic geometry right now?

anon

08:14

I just started this a few days ago, but I'm just doing the category section right now.

(so I'm not sure I really count)

Chris Dugale

No, you count

I just finished an undergraduate degree, and I'm off to grad school in the fall. There's a great course being offered called arithmetic geometry, but it assumes the algebraic curves course, which is being offered in the following semmester.

My ugrad school is very small, and i'm going to Waterloo for grad

feeling kind of underprepared

Little Child

@anon uhm.. what math do I have to learn for image processing ?

anon

probably linear algebra and possibly something fourier/signal analysisy

I've never looked at image processing

Little Child

@anon I find image processing fun but i su*k at math

so basically I have to rely on work of other people

which, at times, is quite limiting

CBenni

how do I prove that the language $L=\{ww\mid w\in\{a,b\}^*\}$ isnt context free >_< I cant find a word for that the pumping lemma helps

Montaigne

09:09

Does someone has an idea why the subset of first order sentences with a finite model are recursively enumerable?

Jayesh Badwaik

09:20

@LittleChild at times? I might say always... :-)

Dominic Michaelis

11:01

schwartz space isn't normed right ?

with the normal definition of the schwartz space and it's family of pseudo norms

Raindrop

11:50

@Ethan @Bageer btw I changed my mind, I'm not interested in the Putnam anymore. I am now doing full-blown effort to become an expert photonics researcher. (I might change my mind about this too)

Dominic Michaelis

12:31

hi

user1

hi

robjohn

what's up?

user1

I was surprised to see how many people looked at an algebraic topology question considering how few algebraic topology questions are asked on main.

robjohn

@user1 I did not look at it. I've never taken algebraic topology

user1

@robjohn Do you ever notice such a phenomenon on questions you do look at?

robjohn

12:45

@user1 I've never really paid attention to the views. I have been surprised by how many votes some questions or answers get.

Little Child

$\lim_{x\to\infty}x^{\frac{1}{x}}$ what will this equate to ?

DNE right ?

user1

@robjohn Yes, I only considered votes (I did not expect to ever have a 12-vote answer on an algebraic topology question considering how few questions fit into that subject).

Little Child

because it comes out to $\infty^{th}$ root of $\infty$ which is undefined right ?

@robjohn any clue ?

robjohn

@user1 the map from $e^{i\theta}\mapsto e^{2i\theta}$ is not homotopic to the identity either, but it is not as easy to show, right? (my understanding is pretty basic)

@LittleChild about what?

Little Child

@robjohn see the limit I posted :)

robjohn

12:51

@LittleChild no, that is 1

Little Child

Why ??

Brian Rushton

@LittleChild you have to take natural logs and use L'Hopitals

robjohn

$$
\lim_{x\to\infty}\frac{\log(x)}{x}=0
$$

Little Child

mmhmm.. ok . patrickjmt says the same. He said it is an indeterminate form. What does that mean ?

user1

@robjohn You are correct that it is not homotopic to the identity. Kahen claims such a thing is easy to show (and if you know the theory of coverings, then I would agree).

robjohn

12:52

@LittleChild it is an indeterminate form, but not indeterminate

Dominic Michaelis

@robjohn isn't that pretty easy? because the mapping from the first over [0,pi) is surely a homoemorphism

but for the second one it is not

Little Child

ok I dont wanna go so deep. I will stop here :)

robjohn

@user1 yes, but not as easy for the layman as the explanation you gave. That is why it got so many votes, I think.

Little Child

I probably need a limits cheat sheet to keep track of these special cases like @robjohn just showeds

user1

@robjohn Do you think laypeople comprise a large portion of the voters in that question?

robjohn

12:55

@LittleChild that one is useful and you can prove it using L'Hopital

@user1 By laypeople, I mean non-algebraic topologists...

Little Child

found a cheat sheet :)

user1

@robjohn In fact, I guessed that, but I am glad you confirmed it. My question still stands.

robjohn

@user1 Not knowing how many algebraic topologists are here, I don't know. However, I was extrapolating from your statement that their questions were rather sparse.

user1

@robjohn :)

Little Child

@robjohn what is a conjugate ?

in fractions ?

robjohn

13:04

@LittleChild in what context? complex analysis?

Little Child

in limits, one way to solve is to use a conjugate . What is it ?

conjugate of a denominator, I guess

robjohn

@LittleChild example? what is the limit?

user1

@DominicMichaelis What's your objection? (You know robjohn and I were talking about maps $S^1\to S^1$, right?)

Little Child

There is no example. I was just asking cause it was bobbling in my head

Ok, here you go. Not sure if a conjugate will come into play

$\lim_{x\to5}\frac{6}{x-5}$

@robjohn There you go. No trigonometric or algebraic identities involved so I guess the only possible way is a conjugate ?

robjohn

@LittleChild not in that one... that is simply $\frac60$, so it is $\infty$

Dominic Michaelis

13:07

@user1 well it is still the same cause the first maps half of cycle which is disconnected when you remove an inner point to the cycle which doesn't have the propertie

Little Child

@robjohn so directly plugging in will give an answer ? no operations needed ?

:)

robjohn

$\lim\limits_{x\to1}\frac{x-1}{\sqrt{x}-1}$

Little Child

plus or minus$\infty$

ok , so waht is conjugate ?

robjohn

@LittleChild multiply the top and bottom by $\sqrt{x}+1$ that would be the conjugate of $\sqrt{x}-1$

user1

@DominicMichaelis How does that mess with them being homotopic?

Little Child

13:12

Those are just two terms in the bottom function. What if like there were 3 or 4 x terms ? root(x), $3x$ , -4

robjohn

$\lim\limits_{x\to1}\frac{x-1}{\sqrt{x}-1}\frac{\sqrt{x}+1}{\sqrt{x}+1}=\lim\limits_{x\to1}\frac{x-1}{x-1}(\sqrt{x}+1)$

Dominic Michaelis

@user1 oh i see missread something

Little Child

answer is 2 then ? :)

robjohn

@LittleChild yes

Little Child

so conjugate = -1 $\times$ denominator.

all the +ve terms become negative and vice versa

Dominic Michaelis

13:16

could someone check if this works

robjohn

@LittleChild In complex numbers, $x-yi$ is the conjugate of $x+yi$, but in number fields, they are often the alternate roots of some defining polynomial. for quadratic polynomials $x-\sqrt{y}$ is the conjugate of $x+\sqrt{y}$

Dominic Michaelis

I mean if the pullback is really the empty set

Little Child

@robjohn you explained that better than my high school teacher

user1

@DominicMichaelis Assuming the given arrows are inclusion maps (monics), then yes.

Just use the fact that $\emptyset$ is an initial object in Set.

Dominic Michaelis

jupp monics

oh right

i am in hausdorff spaces but that doesn't change something

Donkey_2009

13:26

Hup norm.

Little Child

13:44

@robjohn u still here ?

Lord_Farin

15:03

Those who like the Unanswered queue short might like the following (CW) answers: #413042 and #413055.

robjohn

@LittleChild back

@Lord_Farin That is nice

blob

16:37

hello

an urn has 3 white balls and 3 black balls, what are the probability to get 2 white and 1 black with replacement?

I think 3/6 * 3/6 *3/6

robjohn

16:52

@blob It is 3/8. Each choice of black or white is 1/2, and there are three choices that will lead to 2 whites and 1 black

so $3\times\left(\frac12\right)^3$

wwb wbw bww

similarly there is a 3/8 chance of getting 2 black and 1 white. However, there is only a 1/8 chance of getting all black and a 1/8 chance of getting all white. Add them up and you get 3/8+3/8+1/8+1/8=1

Minimus Heximus

17:11

Let $a_n$ denote the maximum number of people in year n (eg. 2013). Finite future will not determine whether $\sum_{n=1}^\infty a_n$ is convergent. So is there any way to prove or disprove its convergence! ? ( What I tried: I tried epsilon delta definition but i failed)

Dominic Michaelis

@MinimusHeximus That is philosophy not math

blob

@robjohn and withouth replacement?

robjohn

@blob that is a bit more complicated. I am in the middle of something. I will go over that if no one else has helped in the mean time.

blob

@robjohn i get with the binomial 9/20

skullpatrol

The son of blob

Lord_Farin

17:30

@blob We can follow a similar approach. We have to select from $6$, then $5$, then $4$ balls, and we have $3$, $2$, and $3$ options (or $3$, $3$, $2$). Therefore, the probability of any particular sequence of drawings is $\frac{3 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4}$. As before, there are $3$ possible drawings, totalling $\frac{9}{20}$, as you obtained.

blob

@Lord_Farin (3*2*3)/(6*5*4) ?

Lord_Farin

@blob That's the same, no? :)

blob

it's 18/120

Lord_Farin

This procedure works with any drawing without replacement; the general formula is $$\binom{m}{a_1, \ldots, a_k}\frac{(b_1)_{a_1}\cdots(b_k)_{a_k}}{(n)_m}$$ for selecting $a_i$ out of $b_i$ balls of colour $i$ for each $i$, with a total of $m$ balls drawn out of $n$ total balls.

@blob Don't forget that we only considered one of bbw, bwb, wbb by this computation.

In that formula the $\binom{m}{a_1,\ldots,a_k}$ is a multinomial coefficient, and $(n)_m$ and similar denote Pochhammer symbols, also known as falling factorials.

blob

@Lord_Farin i am confused

Lord_Farin

17:39

@blob About the specific case or the general formula?

blob

the previus formula is 18/120=3/20 != 9/20

*previous

Lord_Farin

Well no. For each of the possible orders of drawing bbw, bwb, wbb, the probability is $\dfrac{3 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4}$. Since we have three possible orders, the total probability is $3 \dfrac{3 \cdot 3 \cdot 2}{6 \cdot 5 \cdot 4} = \dfrac {9} {20}$.

I'm sorry if I didn't explain that clearly.

The factor $3$ is in fact the "multinomial" coefficient $\binom{3}{1,2}$, which in this case is just a binomial coefficient.

blob

ok thank you

Lord_Farin

@blob Welcome. (I hope it wasn't too intimidating to get the general formula thrown at your feet; if it was, sorry for that. I can't help myself sometimes. :) )

blob

don't worry now it is ok

what are you studying?

@Lord_Farin

Lord_Farin

17:50

@blob I'm involved in the interplay between logic and category theory. Finishing my MSc.

What about you?

blob

i'm studying computer science

bachelor

Lord_Farin

Nice.

blob

so you are studying mathematical logic in depth

Lord_Farin

@blob Indeed. Particularly, employing the branch of logic called model theory to derive results about special kinds of categories.

blob

this field it is very related to computer science, especially for language and compiler theory

looking at wikipedia

Lord_Farin

17:59

@blob Indeed it is! It's closely tied to theoretical computer science.

Type theory, lambda calculus, that sort of thing.

blob

can i add you to friend list, how to do it?

Lord_Farin

@blob I'm not sure if it's possible.

Relevant: meta.stackoverflow.com/questions/886

blob

ok i bookmark your profile :)

Lord_Farin

@blob :)

skullpatrol

"Lord_blob" :)

42 mins ago, by skullpatrol

The son of blob

Lord_Farin

18:09

@skullpatrol How paranoid is it of me to always open YouTube links in private browsing tabs?

blob

one of the best movie ever

skullpatrol

Better safe than sorry :) @Lord_Farin

Lord_Farin

@skullpatrol It's mainly my "recommended videos" feed getting messed up that I worry about; I use YT practically exclusively for music videos.

skullpatrol

@Lord_Farin You could open a "test" account.

Lord_Farin

@skullpatrol Managing two accounts is far above the amount of effort I'm willing to put in.

skullpatrol

18:16

Yeah, I rarely use my dummy account...

...I was going to have one account for each interest.

Lord_Farin

@skullpatrol That's just plain overkill to manage (although I confess to using different browsers for different interests :) ).

skullpatrol

agreed

Peter Tamaroff

18:34

@Lord_Farin I don't even...

Lord_Farin

@Peter What about this one?

Peter Tamaroff

@Lord_Farin Oh, that is forgivable.

Lord_Farin

@PeterTamaroff Perhaps. I got offended by the "correction" and the displayed lack of effort.

Nasser

Was just wondering if there is a name for this PDE?

it is close to one way wave equation, but has extra term a*q(x,t) there on the right.

Peter Tamaroff

@Nasser A "disturbed" wave equation?

skullpatrol

18:50

Peter Tamaroff

@skullpatrol That's kind of a drum?

skullpatrol

@PeterTamaroff A solution of the wave equation in two dimensions with a zero-displacement boundary condition along the entire outer edge.

Vibrations of a circular membrane

The vibrations of an idealized circular drum head—essentially an elastic membrane of uniform thickness attached to a rigid circular frame—are solutions of the wave equation with zero boundary conditions. There exist infinitely many ways in which a drum head can vibrate, depending on the shape of the drum head at some initial time and the rate of change of the shape of the drum head at the initial time. Using separation of variables, it is possible to find a collection of "simple" vibration modes, and it can be proved that any arbitrarily complex vibration of a drum head can be decompos...

robjohn

@blob That's right... I got it looking at it sideways. There is the same chance of getting 2 blacks and 1 white as there are of getting 2 whites and a black. Similarly, there is the same chance of getting 3 blacks as there is 3 whites. 3 whites have a 3/6*2/5*1/4=1/20. that gives 1/10 for all one color. and 9/10 for the other two, each of which must be 9/20.

Cortizol

Can we say that function (I will be more precise) would are some kind of vectors, because they belong vector space (for example) $L^3(X,\mu)$ which is vector space (it is Banach, but leave that is not important now). I am meaning, do we called elements of vector space with "vector", or we use this word "vector" only for vector space $E^n$, Euclid vector space. I mean, this is notation question.

Peter Tamaroff

@Cortizol A vector is, by definition, an element in $V$ where $(V,+,\cdot)$ is a $K$ vector space.

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