Mathematics - 2017-01-09 (page 2 of 7)

4:00 AM

But $u$ isn't continuous everywhere

look, in the differential equations class they were either referred to as the solution to an initial value problem or a general solution to the differential equation depending on the problem

Mike Miller

Are you looking for an answer or a debate?

TheGreatDuck

let me put it simple terms then

GFauxPas

me?

Mike Miller

Yeah. What assumptions do you have on $u$?

GFauxPas

4:01 AM

$u$ is the heaviside step function

Mike Miller

Let me think for a minute. The solution should probably still be continuous.

TheGreatDuck

@MikeMiller neither. @GFauxPas asked me what I was doing in some question and I told him and tried to explain why I was doing it but instead of just accepting it he fights on things irrelevant to my point.

GFauxPas

$\chi(\{x > 0\})= u(x)$

TheGreatDuck

@MikeMiller the differential equations professor last semester flat out point blank said the solutions to all differential equations are continuous where defined.

if the diverge or fail to exist that's a completely different issue

GFauxPas

Well sorry that I think some things are relevant that you don't

Im not trying to start any fights

TheGreatDuck

4:03 AM

and he said the sorts of problems we would be doing (such as the one I was showing @GFauxPas) had continuous solutions.

Mike Miller

@TheGreatDuck Sure. For the level of things you're talking about, that's the right perspective. On the other hand, GFP is interested in more general situations, and I want to tell him that solutions in these situations are still continuous.

@GFauxPas Your tone does sound rather confontational.

GFauxPas

I do that sometimes without realizing it, I apologize

TheGreatDuck

@MikeMiller no... he thinks that y'' + 2y' + y = u(x) does not have a continuous solution.

GFauxPas

It's something I try to work on

TheGreatDuck

He thinks it has a distribution as a solution.

Mike Miller

4:04 AM

@TheGreatDuck That's a more general perspective - that of 'dstirubtional solutions' to differential equations. Sometimes those distributions are actually represented by continuous functions.

TheGreatDuck

@GFauxPas it's okay. It's just my point wasn't "I can solve these guys by using the alternate system and converting back". It's "I can do it easier than with the laplace transform".

@MikeMiller ah. Haven't learned that (yet).

Mike Miller

I want to say that any distributional solution to the equation is indeed represented by a continuous function.

GFauxPas

Interesting

Mike Miller

I know :) The stuff you're trying to learn now is best thought about the way you're thinking about it, instead of in terms of distributions.

GFauxPas

he just wants to expand the notion of "constant" to "piecewise constant"

which, if it's something he finds interesting, is worth exploring. why not? I never explored that idea

I dont know what results it gives

TheGreatDuck

4:06 AM

well it's an alternate system

Balarka Sen

Hi @Ted

TheGreatDuck

it's more of a method to solving

Ted Shifrin

Hi @Balarka

Secret

model of a projective plane

►

GFauxPas

listen, exploration is great, I encourage you to try out new ideas.

TheGreatDuck

4:07 AM

@GFauxPas considering that was one the professor did in class cause it was easy to think of and we could remember for reference involved 3 or 4 terms in a partial fraction decomposition and finding inverse laplace transforms I would argue that solving in the alternate system (which I showed is as trivial as if h were a constant) and finding a continuous solution is if not easier, less difficult then dealing with the laplace transform tables. :-)

Mike Miller

OK, yes, solutions to your ODE when $u$ is piecewise constant (or piecewise continuous, with discrete set of discontinuitities) are indeed automatically continuous.

Secret

Why when people make projective plane animations, they love to rotate it so tat the midrib is in the way (thus resulting in the projected 3D figure to look like a roman surface)?

GFauxPas

I'll buy it, Mike seems to know what he's talking about

Mike Miller

Proof by intimidation?

GFauxPas

argumentum ad baculum

TheGreatDuck

4:09 AM

@MikeMiller and what I do is solve as if piecewise constants were constants and then use that fact to find a continuous solution. Basically, the arbitrary constants in that system are piecewise constans which provides the missing parameters that must be picked such that the solution is continuous.

Does that make sense?

Ted Shifrin

One of my favorite phrases, @MikeM

TheGreatDuck

lol

Secret

For example, this is the angle of the projective plane that give a closed cross cap projection

GFauxPas

TheGreatDuck just make sure your solutions make sense if you have to bring them back to the system where discontinuities mean not differentiable there

Secret

Balarka Sen

4:10 AM

The cross cap is not easy to see once you're capped.

GFauxPas

Ted TheGreatDuck wants to make a system where piecewise constant functions have zero derivative everywhere

Secret

I am interested in seeing what that double line region look like when we rotate this thing around in 4D. however, the animator does THIS

GFauxPas

I'm not sure how I feel abou tit

TheGreatDuck

@GFauxPas like I said: weak solution.

Mike Miller

Actually, you only need $u$ to be locally $L^2$.

GFauxPas

4:10 AM

whatever a weak solution is

Mike Miller

@TheGreatDuck I dunno! I'd have to think, and I'm all thunk out today.

Secret

TheGreatDuck

basically it means it's not differentiable everywhere. It's the best solution that can be made on that range.

like how the integral of floor is a weak solution to the equation y' = floor(x)

Secret

that's the wrong 4D rotation to expose the details of that double line (well at least now I know why the roman surface is a immersion of the projective plane)
Youtube link: https://www.youtube.com/watch?v=yUerROXAEtw

Mike Miller

@TheGreatDuck That's true - but one should usually think of the derivative of the Heaviside function as the Dirac delta.

TheGreatDuck

4:13 AM

@MikeMiller well in that system than the weak solution is the solution from what I understand.

it's just that one can choose whether to reject or accept the dirac delta from what @GFauxPas was kind of saying.

GFauxPas

I didn't say it very well

TheGreatDuck

^^^

GFauxPas

but anyway the dirac delta can be created if you want to create such a thing

just make sure the definition makes sense and does what you want it to do

TheGreatDuck

that's the thing though

Balarka Sen

I just can't seem to get myself to do some new stuff. I have been flipping the pages of Hirsch and thinking about random problem for decades

TheGreatDuck

4:15 AM

isn't the dirac delta function an actual proven thing?

it's not something tacked on as an afterthought

GFauxPas

no. you can prove it does things you want it to do

but you have to figure out what it means first

TheGreatDuck

but then by that reasoning how can we say it is the derivative of heavyside?

GFauxPas

well let's see

TheGreatDuck

the derivative of heavyside is the limit definition

GFauxPas

let's say we agree that $\delta(x) = 0$ for $x \ne 0$

TheGreatDuck

4:16 AM

so in some sense the dirac delta should be that limit

GFauxPas

what's the derivative of heaviside at $x \ne 0$?

TheGreatDuck

@GFauxPas it wouldn't be dirac otherwise

GFauxPas

okay

TheGreatDuck

idk. It says "infinite" in the definition given to me.

I would imagine that means infinite slope

GFauxPas

at $x \ne 0$, whats the derivative of heaviside?

TheGreatDuck

4:16 AM

a jump

Mike Miller

@TheGreatDuck Let's think about it differently. What's $\int_{-\infty}^x \delta(t) dt$?

TheGreatDuck

I just told you

GFauxPas

it's constant except at zero

it's either $0$ or $1$

TheGreatDuck

@MikeMiller heavyside.

Mike Miller

woo

TheGreatDuck

4:17 AM

I don't need an explanation of it

I'm saying it cannot be defined as the derivative

GFauxPas

the derivative of heaviside is just the derivative of a constant function

except at its jump

TheGreatDuck

because it either is or it isn't.

Mike Miller

Whatever, I'll peace out here.

TheGreatDuck

@MikeMiller to be fair, I was merely discussing with @GFauxPas how to solve certain equations. I don't really wish to discuss the dirac delta function.

Balarka Sen

maybe I'll watch Mulholland Dr today

GFauxPas

4:18 AM

oh, my bad

that's just the way to go about such equations, in my experience

TheGreatDuck

@GFauxPas basically in the alternate system remember how I said c becomes C(x)?

GFauxPas

yes. and that C(x) means specifically piecewise constant?

TheGreatDuck

basically if you solve an equation in the alternate system (properly), choosing a C(x) such that the solution is continuous will yield a solution.

YES! :)

it will be piecewise constant

GFauxPas

if you say so

that it will yield a solution, I mean, that carries over

TheGreatDuck

the existence of such a C(x) can be argued. But while I cannot prove so I know that for the standard solutions the method works

and integrals in general as well

Mike Miller

4:22 AM

@BalarkaSen Good movie. But I've already suggested so much math I can't do more. :P

Balarka Sen

Ya, you did. This time it's just me being lazy.

Semiclassical

hi chat

Secret

Sometime later, when I try to wrote a program to do the embedding itself, I am going to explore it and try to find an answer to this question:

Ramanujan

Why writing $a^4 -b^4$ gives imaginary roots?(b=-ia)

TheGreatDuck

@GFauxPas it's plausible that you may be able to find an equation where C(x) does not exist due to some problem but I think one might argue that in math if a method works for useful situations, then it's good enough as a clever shortcut.

and to be honest, I think we would struggle for a while to find a situation where it fails to be true.

Secret

4:26 AM

TheGreatDuck

@Ramanujan Guess what. I got the airship working. I can hop in it and fly it.

Secret

The xw circle is confirmed by the animation in the yotube link

Ramanujan

@TheGreatDuck in game?

TheGreatDuck

yup

wanna see a pic?

give me a minute to load it back up

Ramanujan

Yes :D

Semiclassical

4:27 AM

@TheGreatDuck That's about how I think of stuff like $$y(x)-y'(x)=(1-D)y=f(x)\implies y(x)=(1-D)^{-1}f(x)=(1+D+D^2+\cdots)f(x)=f(x)+f'(x)+\cdots$$

Not helpful at all for getting the general solution (to the inhomogeneous ODE) but it can be cute for getting a particular solution

GFauxPas

I'm just not sure how your system is better than one already out there, but if it works for you...

Iwillnotexist Idonotexist

@Ramanujan Because when $b=-ia$, $(a)^4 - (-ia)^4 = a^4 - (i^2 a^2)^2 = a^4 - (-1)^2 a^4 = a^4 - a^4 = 0$.

Semiclassical

even though 'expanding in powers of the derivative' sounds pretty silly :)

TheGreatDuck

@GFauxPas huh? I'm not at all claiming the alternate system is correct. I'm saying converting to it and back is an easier method to solving an equation.

Ramanujan

But I was thought that for fourth degree equation,(for example $a^4 - b^4$) all four solutions will be "a"

Semiclassical

4:32 AM

(b-a)^4 =/= a^4-b^4

TheGreatDuck

@Ramanujan imgur.com/a/ZhE0J

docked airship

err...

actually randomly in space but whatever.

Ramanujan

@TheGreatDuck what's so interesting in that game?

TheGreatDuck

what do you mean by that? :)

Ramanujan

What makes you to play that game?

TheGreatDuck

It's fun.

:p

what makes you want to play whatever games you play?

Ramanujan

4:36 AM

I too play for fun (but sometimes by taking challenges :D)

TheGreatDuck

well I don't really play it anymore so much as enjoy adding stuff to it

Ramanujan

m.wolframalpha.com/input/?i=a%5E4+-+b%5E4&x=0&y=0

TheGreatDuck

and that's just cause it's fun to make sm64 like platformers and stuff

GFauxPas

the contour plot messes up there at the origin Ramanujan, way to break the internet

Ramanujan

Why can't all roots be "a" and why we take "i"?

GFauxPas

4:37 AM

Ramanujan what are "$a$" and '$b$'

real numbers?

Ramanujan

Yes

GFauxPas

Okay

you want to solve $a^4 - b^4 = 0$?

Ramanujan

Yes

GFauxPas

how many solutions would you expect, just by looking at it?

Ramanujan

4

TheGreatDuck

4:39 AM

isn't that just a^4 = b^4

GFauxPas

yes

so $a = b$ is only one solution

TheGreatDuck

um wait

there are infinite solutions

it's a function... sorta

GFauxPas

he means solving $b$ in terms of $a$ or the other way around

TheGreatDuck

oh

GFauxPas

anyway, $a = b$ and $a=-b$ are both solutions

TheGreatDuck

4:41 AM

how about you just say

GFauxPas

if you only want wholly real solutions, ignore the other ones.?

TheGreatDuck

(a^4)^(1/4) = |b|

GFauxPas

the great duck how would I solve $x^2 = 1$

but I like your idea.

TheGreatDuck

@GFauxPas |x| = 1

Ramanujan

TheGreatDuck

4:43 AM

you cannot get any more information

Ramanujan

How to write it in this form?

GFauxPas

right but to say just $x^{2/2} = 1$ is not so helpful

Ramanujan

For n=4

TheGreatDuck

@GFauxPas fractional powers do not commute like that

(x^2)^(1/2) =/= (x^(1/2))^2

one gives absolute value

one gives x itself

it's the one pet peeve I have with exponentiation and nobody catches often.

:p

GFauxPas

it works if you make the right branch cuts

but that doesnt help here

TheGreatDuck

4:45 AM

really?

x^2 gives a parabola

can you insert that function into a function and get back x?

GFauxPas

I meant fractional roots

TheGreatDuck

oh

GFauxPas

you can, but only half at a time

TheGreatDuck

well... in some cases it works and others not so much

GFauxPas

so you can define two "branches" of $x^{1/2}$

TheGreatDuck

4:46 AM

yeah

anyway it's late

GFauxPas

anyway ramunjan

if you know one root, you can factor it out

I dont know if it helps

so if you know $a = b$ is a solution then $a-b$ is a factor

so you can take $\dfrac{a^4-b^4}{a-b}$ and then work with what you get out of that

TheGreatDuck

@Ramanujan one other reason the game is awesome. It's basically a moddable version of sm64 and LOZ:OOT in one game. Why wouldn't one want to try that.

GFauxPas

oh wait

Ramunjan there's an easier way

I'm sorry I didn't see it before

it's a difference of squares

Ramanujan

$a^4 - b^4 =(a-b){a^3 + (a^2)b + a b^2 + b^3) $?

GFauxPas

let $u = a^2$ and $v = b^2$

TheGreatDuck

4:49 AM

Plus, it's a hilarious game. Also, I know the guy who made it so I'm a little biased as well @Ramanujan . It's actually fun to take some of the stuff in it and do crazy things. Much easier than working from scratch.

GFauxPas

then $a^4 - b^4 = u^2 - v^2 = (u+v)(u-v) = (a^2 + b^2)(a^2 - b^2)$

see what I did?

and now you have a difference of squares and a sum of squares

the difference can be factored into $(a+b)(a-b)$ and if you want you can factor the sum as ($a + ib)(a-ib)$

Ramanujan

How to expand sum of squares?

$(a+ib)(a-ib)$

Cool

yup

4:51 AM

2 mins ago, by Ramanujan

$a^4 - b^4 =(a-b){a^3 + (a^2)b + a b^2 + b^3) $?

@GFauxPas clever

thank you

Is it correct?

I think so

but easier to see it as a difference of squares

like I did the second time around after telling you a harder way :P

8 minutes to my birthday

TheGreatDuck

nope

1:07 to your birthday

Ramanujan

4:53 AM

And how old you will be?

TheGreatDuck

:-)

GFauxPas

TheGreatDuck I am in the time zone I was born in :P

30 :(

3

TheGreatDuck

O.O

GFauxPas

an old man

2

Ramanujan

Ooh

And what's your age @TheGreatDuck ?

GFauxPas

4:54 AM

anyway ramanujan you can always do a difference of squares if $a^n - b^n$ has $n$ even

TheGreatDuck

let a be the tens digit and b be the ones digit. a + b = 3 and a - b = 1

$n \ge 2$

@TheGreatDuck 21?

yup

ugh every birthday I'm sad I'm not married yet

Ramanujan

4:56 AM

Do you play counter strike game?

TheGreatDuck

nope

GFauxPas

like 20 years ago I did

Ramanujan

@GFauxPas are you married to your work?

GFauxPas

nope

TheGreatDuck

@GFauxPas 20 years ago video games like that didn't exist. I should know. I was an infant.

GFauxPas

4:57 AM

Counter-Strike (video game)

Counter-Strike (also known as Half-Life: Counter-Strike) is a first-person shooter video game developed by Valve Corporation. It was initially developed and released as a Half-Life modification by Minh "Gooseman" Le and Jess Cliffe in 1999, before Le and Cliffe were hired and the game's intellectual property acquired. Counter-Strike was first released by Valve on the Microsoft Windows platform in 2000. The game later spawned a franchise, and is the first installment in the Counter-Strike series. Several remakes and Ports of Counter-Strike have been released on the Xbox console, as well as OS X...

fine, 17 years ago

Ramanujan

We still exist who dont play counter strike and Pokemon go :D

TheGreatDuck

gotchya. ;-)

@Ramanujan i don't really play video games. I mostly do math, work on schoolwork, and work on that mod a bit for leisure. I don't have time for video games.

Ramanujan

@TheGreatDuck do computer science students want to become hackers?

Because they knew how to do stuff

TheGreatDuck

wat...

we don't even learn hacking.

Ramanujan

So it way different?

TheGreatDuck

5:01 AM

computer science is programming

making programs

GFauxPas

happy birthday sadly blows party noisemaker

TheGreatDuck

and modding is not considered hacking either if the program has in-game faculties for doing it.

Happy birthday to you. Happy birthday to you. Happy birthday @GFauxPas. Happy birthday to you.

GFauxPas

:(

TheGreatDuck

how does it feel to enter the 4th decade? Exhilarating?

GFauxPas

old and lonely

TheGreatDuck

5:03 AM

:-(

Kasmir Khaan

it feels the way you wanted it to feel

but your never old when you have a good company

GFauxPas

I don't have good company

Kasmir Khaan

thats what my great great great grandfather said

GFauxPas

how old was he

when he said that

Kasmir Khaan

how would I know

am just trying to be positive

GFauxPas

5:05 AM

good point

:)

Kasmir Khaan

:)

TheGreatDuck

great great great grandfather....

210?

Kasmir Khaan

i never stated he said it to me

GFauxPas

true

Ramanujan

Happy birthday to you. Happy birthday to you. Happy birthday @GFauxPas. Happy birthday to you!

Kasmir Khaan

5:06 AM

i miss Ted

ramanjuan i forgot to tell you that the aritmatic on infinite sets and finite sets are not the same thing

most laws dont work on infinite sets

if a series for example is conditionly convergent, you can manipulate it to get it to be any number u want

i seen you yesterday doing those kind of things if memory serves

Ramanujan

1-1/2+1/3-1/4-1/5+1/6-……=0

(aim just trolling) :D

GFauxPas

Same logic works with $1 + 1 + 1 + \cdots$ :P

Ramanujan

@GFauxPas 1=0?

GFauxPas

sure why not

Fargle

$1 + 1 + 1 + \cdots = 1 + (1 + 1) + (1+1+1) + \cdots = 1 + 2 + 3 + \cdots = -\frac{1}{12}$. Let's see who I can bother. :)

2

GFauxPas

5:21 AM

oh thank goodness, Fargle is here

now it's my birthday

Ramanujan

@Fargle pretty cool

GFauxPas

okay good night gents

Secret

$$1+1+1+\cdots=1+f(1)+g(1)+\cdots = \frac{1}{0}$$

(because why not, a divergent series is divergent anyway, thus divergent behavior is expected) >:D

Ramanujan

5:54 AM

@KasmirKhaan Ted is here now

Secret

6:11 AM

meh, Iam going to deal with this projective plane business later. Let me finsih up what I have not finished first

Kasmir Khaan

hi @TedShifrin :D

Ted Shifrin

heya @Kasmir

Kasmir Khaan

yeeey been waiting for you since 2 hours haha =p

we are supposed to proof the jacobian roll on double integral

Ted Shifrin

huh?

Kasmir Khaan

let me rephrase ><

we are supposed to proof on our exam , why we need to jacobian when we do change of variables in double integrals

i seen a proof that divides the region into a rectangle and it was very long

Ted Shifrin

6:15 AM

I don't know what you're supposed to do. Can you draw a picture to show where the $r$ comes from in $r\,dr\,d\theta$?

Kasmir Khaan

not just in case of polar coordinates but in general

Ted Shifrin

Cross product of $\partial f/\partial u\,\Delta u$ and $\partial f/\partial v\,\Delta v$ is a good thing.

Kasmir Khaan

Hmm you mean like the area of paralellogram ?

Ted Shifrin

Precisely.

Kasmir Khaan

in the proof in our book they used that idea

but with alot of subscripts

Ted Shifrin

6:20 AM

Subscripts?

Kasmir Khaan

it was very messy

yes like A_ij

Ted Shifrin

For only a double integral you don't need that.

Kasmir Khaan

A_i+deltay

Ted Shifrin

ohhh ... splitting into subrectangles

Kasmir Khaan

yes that =p

but i dont need perfect proof for that

i just want to say something about it if we do get that question

Ted Shifrin

6:21 AM

Just give the argument for one little rectangle with width $\Delta u$ and height $\Delta v$ in the domain.

And then mumble "add them all up"

Kasmir Khaan

hmm from what i understood the map has to be bijective

from D --> E

but a little area of D maps to somearea of E * Jacobian

we need to show why the determinant of jacobian is the correct factor to make it right

Ted Shifrin

If you compute the cross product I just wrote down, you will get the Jacobian times $\Delta u\Delta v$.

Kasmir Khaan

we dont pass if we get 0 on one of the two teori questions =p

Ted Shifrin

You won't get 0 for this.

Kasmir Khaan

yes yes i know :D

i just told you how strict they r

if we get more than half points and 0 on one of the theory questions

then we need to do the exam again for the theory

2 questions out of 11 =p

i only have this left and strokes and greens and cauchys =p

3 days left seems like impossible work but it can be done :D

Ted Shifrin

6:26 AM

Take $(x(u,v),y(u,v))$ and take partial derivatives with respect to $u$ and $v$. When you do the cross product you're doing the determinant. And so you have $$\left|\begin{matrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v\end{matrix}\right|.$$

Kasmir Khaan

thank you Ted again !:)

ill put that into latex so i can read it one second

do I have to say what the jacobian matrix is ?

Ted Shifrin

You should say that that is the determinant of the Jacobian.

Kasmir Khaan

yes yes i get that, but somehow the matrix must "code" something to make it work

if you get what i mean =p

Ted Shifrin

I don't know if your teacher says Jacobian for the matrix or for the determinant. But, either way, ...

Kasmir Khaan