Mathematics - 2017-06-04 (page 4 of 4)

Fargle

18:00

Welp, just found out my prospective research topic was already studied about 60 years ago.

Faraad Armwood

oh wow. hopefully I'll be able to travel this year. I miss going to conferences and seeing new places.

Ted Shifrin

Dinner now. Bye :)

Faraad Armwood

Later :)

Mats Granvik

Is there a way to separate the real part from the imaginary part of a complex number?

I know one way for special cases but it involves several 1000 iterations and it is not systematic so it can't be used for arbitrary complex numbers.

Fargle

@MatsGranvik $\frac{z + \overline{z}}{2}$?

Mats Granvik

18:14

@Fargle Is over line the absolute value.

Fargle

@MatsGranvik It is the complex conjugate.

Mats Granvik

@Fargle Yes but that is cheating. It won't do. I need something algebraic, a formula.

The complex conjugate requires knowing where to put the minus sign.

SteamyRoot

What do you know about the complex number?

Mats Granvik

@SteamyRoot I know that it is the value of the Riemann zeta function.

I need a formula for the imaginary part of the Riemann zeta function: $\Im(\zeta (s))$

Akiva Weinberger

@MatsGranvik Neither the complex conjugate function $\bar z$ nor the real part function ${\rm Re}(z)$ are holomorphic (aka complex differentiable). Nearly every interesting function is, and combining holomorphic functions gives you more holomorphic functions.

Sha Vuklia

18:24

@Hippa you're a physics guy, right?

Hippalectryon

@ShaVuklia Kind of :-)

Sha Vuklia

haha, well can I ask a question? I didn't find an answer on the physics chat, so I might try here

SteamyRoot

@MatsGranvik That pretty much tells you nothing about the complex number itself.

Hippalectryon

@ShaVuklia sure. The physics chat is usually too advanced for "simple" questions >.> I never find answers there

Sha Vuklia

@Hippa haha true

So we have a magnetic field (the green area) with points into the page

and we move this loop with velocity $v$ to the right

now they say the following:

SteamyRoot

18:27

And if a formula for the imaginary part of the Riemann-Zeta existed that wasn't just "$\operatorname{Im} \zeta(s)$", it'd be widespread by now.

Sha Vuklia

Akiva Weinberger

Do you already know some complex analysis?

@MatsGranvik

Sha Vuklia

So $f_\text{pull}$ is the pulling force

Mats Granvik

@AkivaWeinberger Not really.

Sha Vuklia

and what I don't understand is, why we don't integrate over the entire path, i.e. $\oint f_\text{pull}\cdot dl$

they seem to only integrate over the path the charge follows as it goes from $a$ (left corner) to $b$ (right corner)

While I would think that we would have to consider a closed integral

Their whole point is that it equals emf:

And in the case of emf, obviously we only have to consider $a-b$, because on the horizontal segments the magnetic field doesn't have the "potential" to do work, and on the right vertical segment there is no magnetic field at all

So it makes sense to take $vBh$ as emf

However, in the case of the pulling force, I would think that we need to consider the other parts too (horizontal segments, and right vertical segment)

Hippalectryon

18:31

@ShaVuklia But the operator only applies the force on electrons on the AB segment ?

Sha Vuklia

that is true

but surely the electrons keep on moving after that?

because of the electric field that ensues

Hippalectryon

@ShaVuklia They keep moving, but they're not affected by the force anymore after they leave AB

Sha Vuklia

Yea, but why don't we take their movement after AB into account in the energy?

i.e., why don't we consider a closed integral?

Hippalectryon

@ShaVuklia Why does their movement after AB influence the work done by the force ?

Sha Vuklia

not by the force, but by the pulling

we're calculating the work done by the pulling force after all

Hippalectryon

18:35

Exactly, so why does their movement after the segment on which the pulling force is exerted affect the work done by the pulling force ?

@ShaVuklia Said another way, imagine that the circuit is now longer in the area that has no magnetic field: does anything change ? (assuming constant overall resistance)

Sha Vuklia

well yes, I would say the work is bigger then, because we calculate how much the electrons displace

and their displacement is now larger

As to your question why; well we simply integrate over the path the electrons take I would say, as they go from $a$ all the way back to $a$

Hippalectryon

@ShaVuklia ? the pulling force doesn't affect at all the electrons on, for instance, the horizontal segments, or the rightmost vertical segment

Sha Vuklia

of course it does?

Hippalectryon

It's a force, not a velocity

Sha Vuklia

Look, it the electrons go from the left upper corner to the right upper corner

Hippalectryon

18:39

Let me break it down in several steps, maybe it will help

Horizontal segments and rightmost vertical segment: the electrons are unaffected by the magnetic field.
Leftmost vertical segment: the electrons undergo a magnetic force to the left. That force is countered by a force of the same magnitude to the right by the operator

The operator only acts on the leftmost electrons

Sha Vuklia

The operator acts on all the electrons simultaneously, no? Because it asks on the entire loop

So it also acts on the electrons that are not affected by the magnetic field

And therefore it also does work on the electrons that are in the horizontal segments, or the right vertical segment?

because those electrons are still moving

apparently I'm really interpreting something wrong here

I might as a friend of mine maybe

Hippalectryon

@ShaVuklia I kind of get why that sounds weird to you - let me try to reformulate it again

Suppose the operator does nothing

Sha Vuklia

okay

Hippalectryon

What happens ?

Sha Vuklia

the loop will rotate?

Hippalectryon

18:45

Why rotate ? What's the force of the field on the loop ?

Sha Vuklia

or at least, the horizontal component of the magnetic field won't be counteracted

so the electrons will be slowed down

Hippalectryon

What about the loop's movement ?

Sha Vuklia

it will slow down?

Hippalectryon

Exactly. That's because a force will be applied on the loop, to the left.

Sha Vuklia

right

Hippalectryon

18:47

That force only comes from the leftmost electrons, but is applied on the whole loop. However, does the size of the loop in the fieldless area change anything to that force ?

Sha Vuklia

no

Hippalectryon

Exactly. That's because the magnetic force only works in the AB segment.

The operator's force is exactly opposite to that force. It only acts on the leftmost electrons, but is applied on the whole loop

Sha Vuklia

but if the horizontal component of the magnetic field works on the loop

then why doesn't it also work vertically on the loop? Is it because the electrons can move relatively freely, so the vertical force only induces a current, and it doesn't affect the loop as a whole?

Hippalectryon

@ShaVuklia By horizontal force you mean the force applied on the horizontal parts in the field ? Those cancel out since the current's direction is opposite in both branches

Sha Vuklia

I mean uB

actually

I remember a very similar example in Griffiths

where I didn't realise this thing about work

let me quickly quote it

We had a similar situation

where we said we applied a current

such that the (upward) Loretnz force exceeded the gravitational force

And in considering the work done by the battery, they also only consider the part where the charges moved along the upper horizontal segment of loop

instead of considering the other segments too

So I guess my question is why we only consider the charges that are affected by the magnetic field, and not the charges in the other segments, or the loop as a whole

oh wait

in the case of a battery (or generator) I can imagine that they only work on the upper segment

by construction, so to speak

maybe I shouldn't interpret the pulling force as a mechanical pulling force

Hippalectryon

18:56

It's like a mechanical pulling

There's a very simple mechanic equivalent of our problem.

Sha Vuklia

I know

Hippalectryon

Imagine you're pulling a block of mass m1 with a force F

Now imagine you add a block of mass m2 to the first one. The work is the same

Sha Vuklia

right, the work is the same because the force stayed the same

and we're still considering the same displacement

Hippalectryon

Exactly

I have to go eat dinner now :-) good luck

Sha Vuklia

bon appétit!

Leullame

19:08

The functions $f_n$ converge uniformly on a set $E$ to $f$. Then $f_n^2$ converge uniformly on $E$ to $f^2$?

Hippalectryon

19:21

@Leullame Have you tried on simple examples ?

Mats Granvik

@Fargle On second thought that will work as one can write $$\overline{z}=\frac{|z|^2}{z}$$

Leullame

@Hippalectryon In our case both sequences converge, so not many easy examples of sequences that converge not uniformly. The famous $f(x)=x^n$ does not work because its root does not converge uniformly either

Hippalectryon

19:36

@Leullame Let's try a more simple example, $f(x)=x$ and $f_n(x)=x+1/n$. What's $f_n^2-f^2$ ?

Mats Granvik

But I don't know how to compute the absolute value of z.

Leullame

@Hippalectryon Ah, okay, so no uniform convergence on the whole of $\Bbb R$. What about compact subsets?

Mats Granvik

With out resorting to $|z|=\sqrt{a^2+b^2}$

I was hoping that abs(z) could be computed through Taylor series.

Hippalectryon

@Leullame It's the same. Do you agree that working on $\mathbb{R}$ and working on an open interval is the same ? (for instance on $]-\pi/2,\pi/2[$ one could use $f(x)=\tan(x),f_n(x)=\tan(x)+1/n$

Leullame

I agree that for every uncountable set we can just change the names of the points to get the same function from earlier

Julius

19:48

Hi, a simple question about math.stackexchange.com/questions/796750/…. The basis in the first answer has 2 elements, and 4 in the second. I see that $dim(W)=2$, but then why the example from the second answer isn't a basis for $W$? It does span $W$ and it's element seem to be independent.. but it cannot be a basis as it has 4>2 elements

Waiting

@ShaVuklia A party day? That's cool :P Me? Not that bad actually. Working on some interesting & fascinating (at least to me) stuff.

Leullame

@Hippalectryon And if $f_n$ are continuous and the set is compact? It ought to be true

Waiting

Hippalectryon

@Leullame Indeed. Can you find out why ?

Waiting

@Hippalectryon how is it going? :-)

Hippalectryon

19:52

@Waiting great :-) and you ?

Waiting

@Hippalectryon Cool. Well, as you know me, I'm developing some interesting stuff. :D

Hippalectryon

@Waiting :D

Waiting

Can't stay at all, restless soul. :D

@Hippalectryon Do you have some new cool soundtracks to listen to? Audiomachine like.

Hippalectryon

@Waiting Of course :D for instance TSFH released a new album, Vanquish, did I already talk about it ? for instance youtube.com/watch?v=6vCxBQy2SOk

Leullame

I cannot, since $f_n$ have a different global bound on the set E for each n. (So we cannot say $|f^2-f_n^2|<\epsilon|f+f_n| \leq \epsilon(M+|f_n)|$ is arbitrarily small )

Waiting

19:56

@Hippalectryon No, you didn't (or I simply don't remember?). I'm looking for it right now. :D

Hippalectryon

@Leullame Let's write $f_n(x)=f(x)+\epsilon_n(x)$. Then $|f^2(x)-f_n^2(x)|\le2\| f\|_\infty \cdot\|\epsilon_n\|_\infty+{\|\epsilon_n\|_\infty}^2 \to0$

@Leullame ($\| f\|_\infty=|\max_Ef(x)|$)

@Leullame Does that make sense ?

Leullame

Yes!

Thanks (and good night)

Hippalectryon

@Leullame Great :-) btw notice how you don't need $f$ to be continuous, or $E$ to be compact. You just need $f$ to be bounded

Leullame

Yep

Mats Granvik

$a^2+b^2$ can't be factored.

Meow Mix

20:10

it can in $\Bbb C$

Mats Granvik

@MeowMix How?

Meow Mix

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

Alessandro Codenotti

Hi @Meow

Mats Granvik

I see

zed111

20:27

What is $(v \cdot \nabla v)v$ where v is a vector? Is it just $v \cdot (\nabla v)$?

Hippalectryon

20:45

@zed111 It's $\alpha v$ where $\alpha=v\cdot(\nabla v)$. I'm not sure I see where your difficulty lies

Meow Mix

hi ale

zed111

@Hippalectryon Oh I wrote the wrong expression

The correct one is $(v \cdot \nabla )v$

Dodsy

Hey Zach.

Glad to see you doing math again

Meow Mix

have you studied any more math since i last talked to you?

Dodsy

Eh, a little bit of number theory.

Currently learning about greatest common divisors and euclid's algorithm.

Meow Mix

20:48

oh, cool

Dodsy

Yeah, but I basically have to buckle down this month and write a bunch of exams to go to school next year.

So most of july and all of august I should be available to study math more rigorously.

Hippalectryon

@zed111 You're right, it's $(v\cdot\nabla)w=v\cdot(\nabla w)$

Dodsy

And you must be on summer holdiay soon Zach?

zed111

@Hippalectryon Since $\nabla w$ is a 3 by 3 matrix, how to dot it with v?

Dodsy

Going into grade 9 next year, right?

Hippalectryon

20:52

@zed111 A 3x3 matrix ?? What size is $w$ ?

zed111

w is a 3-vector

And so is v

$[\nabla w]_{ij} = \partial_i w_j $

Hippalectryon

@zed111 Nvm I was confused because for me $\nabla\neq\vec{\nabla}$

Dodsy

@MeowMix Hallo

Meow Mix

huh

oh sorry

i was doing something

didnt mean to not reply

yeah i have vacation in 2 weeks

Dodsy

Is okay, you are busy.

Meow Mix

20:55

and grade 9 next year

Dodsy

Wow that's crazy.

I bet you'll take all IB level courses right?

I bet you graduate high school before I'm done 2 years of university

Meow Mix

well the thing is

i cant skip any more years of math

TheGreatDuck

why?

Dodsy

Really?

Meow Mix

mhm

district policy

Dodsy

20:57

How many years have you skipped?

Meow Mix

only 1

TheGreatDuck

what...

that makes no sense

Dodsy

When I was taking biology in grade 11

there were two grade 10's in the class

who took grade 10 biology during the summer.

I'm sure you'll figure something out like that.

TheGreatDuck

so you have to waste time taking classes you don't need because you can only skip once?

Hippalectryon

@zed111 to me $v\cdot\nabla=v_1 d/dx+v_2d/dy+v_3d/dz$

TheGreatDuck

20:57

that's insane.

Dodsy

I'm not sure how beneficial it would be though.

Meow Mix

well

the kids who go to private schools get to take a placement test

Adam Warlock

Could someone take a look at this, please? I'd like a detailed solution, since I'm kind of stuck at this homework problem. Help a brother out, please.
https://math.stackexchange.com/questions/2301609/find-the-periodic-solutions-of-the-following-differential-equations

Meow Mix

to see where in math they belong

Dodsy

Oh that's cool!

Meow Mix

20:59

im going to my local public high school

Dodsy

No problem with that, Zach.

Meow Mix

pfft

of course

i dont mind

im just saying

Dodsy

I have a lot of regrets about high school. I had fun and people liked me, but I didn't do much.

I failed a lot of courses and skipped a lot of classes.

zed111

@Hippalectryon Thats what it means to me unless the operator is acting on a vector itself.

TheGreatDuck

i didn't really get to know anyone in high school

Dodsy

21:03

@AdamWarlock That's a really well asked question, and you've shown your work well, I'm sure somebody will help you shortly.

But I always stood up for people who i thought were good and I always made sure to be nice to everyone I met. So from a character perspective, I'm happy with the experience. From an academic aspect, I am not.

Meow Mix

my experience with high school was

NULL POINTER EXCEPTION

TheGreatDuck

lol

my experience with high school is that everyone is best to stay alone forever

XP

Hippalectryon

@zed111 Well the way I see it (as a physicist, not a mathematician), the RHS is just an abuse of notation. Taking the gradient of a vector doesn't make sense, and (as far as I know) nabla is a 1xn dimensional vector (n=3 here)

TheGreatDuck

nah

that's not abuse so long as you define $v = <v_1,v_2,v_3>$

zed111

Gradient

In mathematics, the gradient is a multi-variable generalization of the derivative. While a derivative can be defined on functions of a single variable, for functions of several variables, the gradient takes its place. The gradient is a vector-valued function, as opposed to a derivative, which is scalar-valued. If f(x1, ..., xn) is a differentiable, real-valued function of several variables, its gradient is the vector whose components are the n partial derivatives of f. Like the derivative, the gradient represents the slope of the tangent of the graph of the function. More precisely, the gradient...

Meow Mix

21:11

pretty sure thats accurate

and not abuse of notation

TheGreatDuck

@Hippalectryon a function that takes in a point and returns a point can have a gradient. Think second derivative in comparison to second gradient...

Hippalectryon

@TheGreatDuck Well yeah you're right >.>

@TheGreatDuck But then there's the dimension issues with the dot product ? Because I suppose that in your case $\nabla$ becomes a 3x3 matrix and not a 1x3 vector

TheGreatDuck

idk anymore

never thought about this

perhaps it is matrix multiplication?

oooh i know

it's a vector valued function

so it's derivative is a vector

google the frenet seret formulaw

consider it in the context of a vector input giving a vector output.

Paweł Kusz

Hi everyone :) I have a problem with integral. How calculate by parts $\int_{0}^{ \infty } \frac{\sin x}{(1+x)^2}dx$ ?

SteamyRoot

Are you sure that's the function you want to integrate? The integrand doesn't have an easy primitive...

Hippalectryon

21:21

@PawełKusz Isn't your function $\sin(x)/(1+x^2)$ ?

TheGreatDuck

@Hippalectryon no. That's not what they posted.

Hippalectryon

@TheGreatDuck I know :P but what they posted doesn't have any easy solution as far as I know, whereas the latter might

Semiclassical

Neither gives a terribly nice answer.

TheGreatDuck

@Hippalectryon fair enough. I don't like to make such assumptions or ask such things.

SteamyRoot

You can probably express both in terms of sine and cosine integrals, but those aren't very appealing.

TheGreatDuck

21:23

:p

Hippalectryon

:-)

TheGreatDuck

...

Semiclassical

Mathematica returns an answer in terms of the Ei(x) integral for the sin(x)/(1+x^2) case.

Paweł Kusz

I know but i want to show that $\int_{0}^{ \infty } \frac{\sin x}{(1+x)^2}dx=\int_{0}^{ \infty } \frac{\cos x}{1+x}$

Hippalectryon

Ah damn :( @Semiclassical

Semiclassical

21:24

well, note that $\frac{d}{dx}\frac{1}{1+x}=-\frac{1}{(1+x)^2}.$

So the first integral is the same as $\int_0^\infty \sin x\frac{d}{dx}\left(-\frac{1}{1+x}\right)\,dx$.

Doing integration by parts on that is then pretty much immediate.

zed111

@Hippalectryon Here is a post that answers my question physicsforums.com/threads/…

Hippalectryon

@zed111 great :-)

Meow Mix

semi

how do you even notice things like that

Semiclassical

eh, experience?

Paweł Kusz

@semi

@Semiclassical Thank you, but i dont understand :/

Astyx

21:34

He uses black magic and sacrifices kittens to Bourbaki

Semiclassical

Not sure what else to tell you, then.

For integration by parts, you always start by writing the integral as $\int u\,dv=\int u \frac{dv}{dx}\,dx$.

Paweł Kusz

I know..

Semiclassical

Then there's really nothing else for me to say.

user193319

21:51

This may seem like a strange question, but is it possible to do partial fraction decomposition on $\frac{\sin(x)}{(1+x)^2}$?

Astyx

What do you mean by that ? @user193319

user193319

Not entirely sure...I want to write it as a sum of two fractions, one having denominator $(1+x)$, the other having $(1+x)^2$.

Astyx

What about the $\sin$ ?

user193319

...Um not sure. That's why I am asking if it is possible.

Astyx

I guess you could use the taylor series for $\sin$ and get ${x\over 1+x^2}+\sum_{k=0}^{+\infty}a_k x^{2k+1}$

Astyx

22:11

Hi Dami

Long time no see

Daminark

22:31

Hey, how's it going?

@Astyx

Akiva Weinberger

22:52

@BalarkaSen Hey here's a video youtu.be/LO1mTELoj6o

Dodsy

Hey

Daminark

How's it going guys?

Fargle

23:07

Alright @Daminark, just devoured some Zaxby's. Yourself?

Daminark

Ontology is slowly becoming the strong p-p estimate and Radon-Nikodym

Fargle

B-huh-wha?

Daminark

Analysis final tomorrow

Fargle

Ah. You feel ready?

Daminark

Absolutely not

Ready to die, maybe

Fargle

23:21

Hoo boy

Well I'm rooting for you at any rate

Daminark

Thanks

Fargle

I know that totally helps

Daminark

triple plot twist: in the final, I think about your rooting for me, and something about that sentence leads me to figuring out the answer

Fargle

@Daminark You realize the answer requires taking roots.

Daminark

Lel

Julius

23:35

Hi. Any references where some generalised metric (where distance is not a real number) is used?

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$Mathematics$ Mathematics