Mathematics - 2017-06-05 (page 5 of 5)

Maks

6:07 PM

Hey !

Semiclassical

hiya

Maks

How can I write this two equations
$ ae^{2bx} = ye^{bx} $
$ a^2xe^{2bx} = axe^{bx}y $
In the form of a matrix ? Given that I have the find the value of $a$ and $b$

Semiclassical

I haven't a clue what you mean by that.

Maks

I want to write it in the form $AX = B$

Semiclassical

All that seems to have been done in the second equation is multiply both sides by $ax$.

More to the point, I don't see how on earth you'll have a unique value of $a,b$.

Maks

6:11 PM

Like, if I had $ ax = y $ and $ 2ax = y $ I can write a matrix A = [1 2], X = [a] Y = [1;1]

I'm doing approximation by least squares

Semiclassical

Huh.

I'm still perplexed, since the second equation you wrote differs from the first only in that both sides have been multiplied by $ax$.

Maks

@Semiclassical I know

Semiclassical

Um, $[1\; 2][a]=[a\; 2a]=[1\; 1]$ doesn't express $ax=y$,$2ax=y$

It expresses $a=1,2a=1$.

Maks

Ups, Y = [1]

Semiclassical

That's even farther off.

What you'd probably want for that one is $X=\begin{bmatrix} x \\ y \end{bmatrix}, A=\begin{bmatrix} a & -1 \\ 2a & -1 \end{bmatrix}, B=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.

Maks

6:17 PM

No wait, it was right A = [1;2], X = [a] Y = [1;1]

Semiclassical

Unless you mean something other than AX=Y, that's not a representation of your equations.

Maks

@Semiclassical Nope

Semiclassical

I really have no idea what you're doing, then.

Abcd

6:32 PM

Does anyone know Chemistry here? I have a small question.

Dodsy

very basic chemistry

Abcd

@Dodsy Stoichiometry?

Semiclassical

as in balancing equations?

Abcd

Nope

mole concept

SO here's my question:

Wait. How to get subscrip @Semiclassical?

subscript*

Semiclassical

Do you have chatjax enabled?

If not, use the link in the room desc.

Abcd

6:35 PM

]I have Chat Jax

Semiclassical

If you do, then it's just a_b in math mode

e.g. $H_2 O$.

Abcd

Test_message

Semiclassical

In math mode. You need the $'s.

Abcd

$Test_message$

Semiclassical

And you'll need to enclose the subscript with {}'s.

Dodsy

6:36 PM

This is like, equilibrium and products and reactants, right?

Like exothermic reactions and stuff like that

Abcd

$test_{message}$

@Semiclassical Thanks.

Now:

Semiclassical

Of course, if this were a real chemistry text one would have elements like H,O not italicized...but f*** it.

Abcd

Molarity = $w_{B}$ * 1000 / $m_B$ * $V_{ml}$

Dodsy

Ah, I might be wasting your time.

Abcd

I have understood this formula^^

But:

later my book says:

M = x*d*10/ $m_{B}$

Now the abbreviations:

mb = molar mass of solute

Vml = Volume in mL

wB = mass of solute

But d = density of solution in g/ml

Semiclassical

6:40 PM

Is that (1000/mB)*Vml, or 1000/(mB*Vml) ?

Abcd

I can't understand how did the author derive the second formula from the first @Semiclassical @Dodsy

Semiclassical

(warning: my laptop's battery is on its last legs until I can get to my charger. so i may disappear.)

Abcd

@Semiclassical Can you tell how to write fraction? I will re type

Semiclassical

it's \frac

e.g. \frac{a}{b} for a/b

Abcd

Molarity = $ \frac{$w_B$ * 1000}{$ m_B$ * $V_{mL} $}$

Sha Vuklia

6:43 PM

you forgot your dollar signs @Abcd

Semiclassical

^ not quite.

the dollar signs should be around -everything-, and not any inside \frac itself

Sha Vuklia

oh right :P I didn't notice them :P

okay just two dollar signs @Abcd

one at the beginning

one at the end

not separate dollar signs for each symbol

Dodsy

$Molarity = \frac{moles of solute}{litres of solution}$

Sha Vuklia

@Dodsy that is awful

Dodsy

Sorry.

Semiclassical

6:45 PM

use \text

Sha Vuklia

use \text

Abcd

I GIVE UP.

Semiclassical

Molarity $= \frac{w_B\times 1000}{m_B V_{ml}}$

Abcd

Can someone please type that?

Sha Vuklia

OMG SEMI

:P

semi what is this

:P

Dodsy

6:45 PM

anyways, that equation is for if you're given 4.0 m/5L

Take 4 and divide it by 5.

Semiclassical

Fail.

Abcd

@ShaVuklia MathJax is so complex!

Sha Vuklia

@Abcd lol you'll get used to it no worries

it's just a bit of a pain when you're trying to understand chemistry at the same time as figuring out latex

Abcd

@Semiclassical This is what I meant

Dodsy

When it comes to moles, there's different things for what you want to accomplish.

Different equations.

Semiclassical

6:47 PM

Better? I had to switch to my phone so mathjax won't render anyways

Abcd

Now how did the author derive the second formula from this @Semiclassical?

Dodsy

Obviously it has to do with density

the other equation doesn't involve density of the solute.

Abcd

@Dodsy How can you explain please

Semiclassical

What is x here?

Sha Vuklia

anyhow @Semi mind if I open a "private chat room" for us? You're one of my last resorts sometimes, and I'm struggling with sth for more than a week now:P But I don't want to disturb the math chat always. (I won't spam ya)

Dodsy

6:49 PM

That's what I'm trying to figure out..

Abcd

@Semiclassical percentage by mass of solute

Dodsy

But some variable x, multiplied by the density of the solution in g/ml , multiplied by 10, and divided by the molar mass of the solute

Semiclassical

You can set it up, but my laptop is dead now and I won't be able to charge it till evening

Dodsy

@Abcd if you're given the density of the solution, use equation #2

Sha Vuklia

yea that's exactly why a room would be good, because then I can just drop a question and you won't have to scroll through the chat and such. it will just kind of be tidy in one room

Semiclassical

6:50 PM

So I'll probably defer any detailed explanatiom till then

Sha Vuklia

yea sure!

Semiclassical

Makes sense

Sha Vuklia

I'll write out the question then, and feel free to answer whenever it suits you

Abcd

@Dodsy I wanna know how to derive that from the first equation.

Semiclassical

"Explanatom: atomic unit of explanation" :P

Dodsy

6:52 PM

Well, you know that $m/V$ = Density in g/mL

Abcd

yes

Semiclassical

Probably the conversion has to do with the factor of 100 in the percentage combining with the factor of 10

Abdullah UYU

Is there a faster way to find $\cos A \cdot \cos C$? I've found it via cosinus theorem but it takes a little bit long

Sha Vuklia

oh DAHM. @Semi After all those times, I suddenly see the answer. I was dealing with an "very long" solenoid, so I just treated it as infinite (of course) solenoid - but when you actually consider infinite space, then the finiteness of the solenoid plays a role again.

Semiclassical

How is x specified? I.e. Is 50 perecent given as x=50 or x=0.5

Dodsy

6:54 PM

And molarity divided by volume is equal to the mass of the solute multiplied by 100 divided by the molar mass of the solute.

Semiclassical

@sha yeah, the infinite solenoid is an odd beast

Abcd

@Semiclassical 50/100

Dodsy

Thus: $Density =M_s(1000)/M_b$

Semiclassical

The simplicity of the solenoid result belies the strictness of the assumptions

Dodsy

Meaning Density times volume is equal to molar mass.

Sha Vuklia

6:56 PM

yea I see. uh, this bothered me for such a long time:P I got too "physics-y" as soon as I read "very long" :P It's like a pavlov reaction; if something is very long or very big, you just immediately treat it as infinite without further questions :P

Semiclassical

It's a useful result, to be sure. But it is an artificial case, and should be understood as such.

Dodsy

molarity*

Sha Vuklia

@semi right, but Griffiths doesn't deal with the infinite case, and I don't mind to tell you the truth :P

Dodsy

anyways, by going around in these circles

Semiclassical

Are you sure? I'd have thought he'd have treated it when doing Ampere 's law

Dodsy

6:59 PM

its easy to see that $x(d)(10)/m_b$ = $(density)(volume)$

Semiclassical

Or as an exercise therein

Sha Vuklia

@Semi maybe as an exercise, but in the example he gives he treats it as a finite object, so that we can say that the magnetic field approaches zero at infinity, and hence is zero everywhere outside the solenoid

Semiclassical

Uh

Adeek

hi

@arctictern are you around ?

Semiclassical

That only works for the infinite solenoid

arctic tern

7:01 PM

yeah

Semiclassical

A finite solenoid will have field outside

Adeek

I was wondering are you familiar with the homotopy category ? the Algebraic one ?

not the topologic

Semiclassical

It'll be tiny, but it is there

Sha Vuklia

well they call it a "very long" solenoid

Semiclassical

eh, same thing

arctic tern

7:02 PM

not really

Sha Vuklia

but Griffiths says it approaches zero

it doesn't say it is zero

Adeek

@arctictern i.e the category formed by considering complexes of an additive category A together with morphism being equivalence class

Sha Vuklia

just make the solenoid larger if the small number bothers you and you want it smaller, I'd say

Adeek

oh ok

Dodsy

In fact making both equations equal to eachother

It becomes strikingly apparent.

Semiclassical

7:03 PM

Well. My point is that, for any finite solenoid, there will be a small but finite field outside (it'll decrease to zero as you move away, of course)

Sha Vuklia

yea right.

I think Griffiths just wanted to use the nice properties of both finite and infinite solenoids

without diving too much into it

Semiclassical

It's only in the infinitely-long case that you have B identically zero outside.

I wouldn't be surprised.

And, plus, it's useful as an approximation in any event

Sha Vuklia

yea so I think "very long" should be treated as infinite as we're considering some finite space, and when we consider infinite space, we can just treat it as finite as it suits us

@Semiclassical yea

Abcd

@Dodsy Alright. Got it. Thanks.

Adeek

@arctictern If A is a additive category, then we can consider the category of complexes that is morphisms the sequence $(X_n,d_n^{X})$ i.e objects are sequences $\ldots \rightarrow X_n \rightarrow X_{n - 1} \rightarrow \ldots$

and morphisms are maps $(f_n) : X \rightarrow Y$ making the whole diagrams commute

Semiclassical

7:06 PM

The main point in my brain is that, if the solenoid is infinitely long, then nothing can change as I move along the axis of the solenoid

Sha Vuklia

you mean the magnetic field can't change?

Semiclassical

right.

Sha Vuklia

but then you've only argued that the magnetic field is constant along the axis of the solenoid

you'd still have to argue that it is in fact zero

Adeek

Then we can do an equivalence class on morphisms in same way as we do in topology by considering a homotopy $s_n : X_n \rightarrow Y_{n + 1}$ such that $f_n - g_n = d_{n + 1}^Y \circ \s_{n - 1} + s_{n} \circ d_{n}^X$

then we can check this form equivalence class and we can form new category called homotopy category

Semiclassical

Actually, all I've argued is that B doesn't depend on z.

Sha Vuklia

7:08 PM

yea well, we already know it has no radial or circumferential component

but indeed

Semiclassical

It's pretty easy to argue that it only depends on r, of course.

But to then argue that it doesn't change outside requires an appeal to Ampere 's law

Sha Vuklia

huh, dependence on $r$?

Dodsy

semiclassical is a very helpful person.

Sha Vuklia

Griffiths didn't mention that

@Dodsy ts ts, too little praisal:P

Semiclassical

Well, sure. B=0 outside (r>R) and some fixed value inside (r<R).

Dodsy

7:11 PM

I chose my words carefully :)

Sha Vuklia

that is exactly my problem @Semi

Semiclassical

That is r-dependence, just of a very constrained form.

Sha Vuklia

the fact that $B=0$ is exactly what I'm trying to understand

because what Griffiths says

is that as we approach infinity, then $B$ goes to zero

and since $B$ is constant, we know that $B$ must be zero anywhere

so if we consider the infinite case, then we can't use this fact that $B$ goes to zero

so I still "don't know" that $B=0$ outside for the infinite solenoid

oh I see the $r$ dependence now btw

but we can chose our Amperian loop such that $r$ is constant

hm I donno, I think I'll just stick to what Griffiths doing. I've learned to appreciate handwavery:P

Semiclassical

I'm not seeing what you're visualizing.

What loop are you drawing?

Sha Vuklia

Semiclassical

7:14 PM

Right.

Do you see how that establishes B=const outside?

Sha Vuklia

yes

because the enclosed current is zero

Semiclassical

right.

Is your concern that, if I take the solenoid to be infinite, then it's not appropriate to say B must go to zero far away?

Sha Vuklia

exactly

like, if we consider an infinite plate or infinite wire, the electric field also doesn't go to zero

Semiclassical

Then I pretty much agree with you. It's an additional assumption that has to be made.

Sha Vuklia

maybe we could take a limit?

since we know that the longer the solenoid, the smaller the field

and then that would kind of be our argument

Semiclassical

7:19 PM

Probably. The one thing I'll say, though, is that while there is current at infinity here, it's all off at $z=\pm \infty$ here

Whereas the location we're interested in is r=infinity

Hippalectryon

Well there's a way to actually prove that the exterior field is zero by taking limits iirc

If we consider a very very very long toroid, its limit locally is an infinite solenoid

And the field outside is exactly zero

Semiclassical

ah, I like that

That lets you still use Ampere's law.

Sha Vuklia

@Semi why would that matter? In the case of an infinite wire, we also have charge at $z=\pm\infty$, while we consider $r=\infty$

Semiclassical

Well, in that case your electric field falls off as 1/r.

So the electric field still goes to zero at r=infinity in that case as well.

Sha Vuklia

right

Semiclassical

7:23 PM

The potential, by contrast, is genuinely problematic. But you actually run into a similar issue with the vector potential in the solenoid case.

Sha Vuklia

@Hippa I remember we were talking about something yesterday, but I don't remember 1) what it was, 2) if I solved it in the end :P

Hippalectryon

@ShaVuklia It was about work with laplace forces. I think we solved it, using a mechanical analog

Sha Vuklia

@Semi for some reason, my course skips the vector potential

but they do want us to understand magnetic fields in materials

Semiclassical

Meh.

Trent

Hi!

Sha Vuklia

7:24 PM

@Hippa oh yea I remember

Semiclassical

I mean, knowing vector potential isn't essential.

Trent

I'm trying write a sort algorithm based on multiple factors, I've a price factor, a distance factor, a boolean factor and a percentage factor, I've been said that I should try to normalise / scale my factors first, but I'm not sure what it means. Any hints?

Semiclassical

But I feel like one should see it, if only to see how much worse it is :P

Sha Vuklia

haha yea I think I'll just "brush it" slightly :P

Semiclassical

You also definitely need the vector potential if you want to understand charged particles in magnetic field in the context of quantum mechanics

Sha Vuklia

7:27 PM

right, but I'm kind of studying for a test on electromagnetism

Semiclassical

Sure.

Main reason I bring it up is because one of the big examples of that is the Aharanov-Bohm effect. Which basically amounts to: What happens if I stick a narrow infinite solenoid between the apertures in a double slut experiment?

The weirdness comes from the fact that, to the extent we think of the electron as a particle: its trajectory would entirely miss the solenoid regardless of which slit it passes through.

So it should experience 0 magnetic field and therefore be entirely unaffected

...and yet, the presence of the solenoid does affect the interference pattern. Weirdness!

Sha Vuklia

I won't try to understand it too much, but what do you mean by "its trajectory would miss the solenoid"? Does that mean it doesn't "move through" the solenoid?

Semiclassical

Right.

Sha Vuklia

hm right. I'm so perplexed by QM that I'm relatively not weirded out by this :P I still need to get used to the basic stuff that is already normal and accepted :P

Semiclassical

good call

Just pointing out that this solenoid example rears its head in a place you might not expect.

Sha Vuklia

7:36 PM

haha yea sure!:P

Semiclassical

And that the vector potential is really important once you start trying to talk about charged particles in magnetic fields

Which is pretty significant if you think about it.

Sha Vuklia

yea, I'll just try to read through it on a minimal level, just so that I know why a closed current feels a torque in a magnetic field.

because that's how the chapter on magnetised materials opens, and I think it's explained in the vector-potential part

Semiclassical

Well, one way to appreciate that is to think how two perpendicular current-carrying wires will act on each other

Sha Vuklia

I can't upload for some reason

but I wanted to post an image of what you just mentioned

which is indeed treated in the vector-potential bit of the book

Semiclassical

Or, for an even simpler case, how a small current loop will experience a field that's parallel to its surface

Ahah.

Of course, you'll also have to understand how (para/dia/ferro)magnetism works

So have fun with that :P

Sha Vuklia

7:45 PM

haha thanks :P

Semiclassical

The whole B vs H thing is more than a little liable to confuse

Especially when it comes to bound currents

John11

8:00 PM

wow I can only hope one day I can come here and help people out the way you, semiclassical and other frequent chat users do. Seems like my best right now is coming to ask lame questions

Dodsy

@John11 You'll get there, persistence is key. I'm in the same spot as you are. :)

Christian F. Madsen

How does one find the radius of convergence of $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}z^{4n+3}$? I feel like I only really understand how to do it when $z^n$.

Alessandro Codenotti

8:28 PM

You should know a formula that gives you the radius of convergence based on the coefficents of the series, does this sound familiar?

Christian F. Madsen

Yes however there is a condition that the following coefficient in the series must be non zero from some point onwards which doesn't happen in this case? (I guess you're referring to $R=\lim_{n \rightarrow \infty} \left | \frac{a_n}{a_{n+1}} \right |$)

I have to rewrite the series in some way so that I can use it, but I don't know how.

Only way I can think of is: $z^3\sum_{n=0}^{\infty}\frac{1}{(2n+1)!}w^n$ with $w=z^4$ but then I find the radius of convergence of another power series? Is it possible to relate the two?

Alessandro Codenotti

8:45 PM

I was thinking about the $\frac1{\text{limsup}|a_n|^{1/n}}$ way of calculating the radius

Christian F. Madsen

I have no idea what limsup of $\left ( \frac{1}{(2n+1)!} \right )^{\frac{1}{n}}$ is though I strongly suspect that it's 0 but I can't prove it.

robjohn

@ChristianF.Madsen try Stirling's Approximation?

Christian F. Madsen

@robjohn Don't know what that is unfortunately.

robjohn

Stirling's approximation

In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. It is a good-quality approximation, leading to accurate results even for small values of n. It is named after James Stirling, though it was first stated by Abraham de Moivre. The formula as typically used in applications is ln ⁡ n ! = n ln ⁡ n − n + O ( ln ⁡ n ) {\displaystyle \ln n!=n\ln n-n+O(\ln n)} or, for instance...

Christian F. Madsen

Surely there is another way? I can't refer to Stirling's approximation at the exam anyways.

nbro

9:04 PM

Hi!

Calculus question.

Does this always hold: $\frac{d f}{d g} \frac{d g}{d h} \frac{d h}{d x} = \frac{d f}{d h}\frac{d h}{d x}$? Does it even hold? If yes, why exactly?

where we have $f(g(h(x))$

user243301

@user1952009 congratulations 500 answers!

nbro

9:28 PM

Never mind. I found an answer to my question.

The answer is yes, it holds, as long as all functions involved are differentiable.

Waiting

10:06 PM

@robjohn hey! How are you doing? Long time no see. :-)

@Hippalectryon are you around? o/

Hippalectryon

yep @Waiting

Waiting

@Hippalectryon I find this song pretty nice youtube.com/watch?v=GL1hM1HeJPM

@Hippalectryon By Attila Áts (maybe not in the spirit of AudioMachine, but still nice to listen to)

Hippalectryon

@Waiting ty :-)

Waiting

@Hippalectryon np :-)

robjohn

10:32 PM

@Waiting been very busy. Usually, I come to chat only to have to leave right away.

Waiting

@robjohn I see. I don't come often/stay long either.

robjohn

@ChristianF.Madsen well, what tools do you have available? Do you know Jensen's Inequality? The Arithmetic Mean-Geometric Mean Inequality?

@Waiting that would reduce the probability of intersection greatly, then.

Waiting

@robjohn hehe, indeed. :-)

@robjohn I think I left a message some months ago to you but seeing no answer back I supposed you were pretty busy. Anyway, there was nothing of great importance there.

robjohn

@Waiting Sorry. Where did you leave it? I got pretty pissed at DropBox's policy change.

Waiting

@robjohn In DB, but you may forget that message, it was nothing important. Do you still have that account active? Asking just for the future in case I have something to send.

Semiclassical

10:44 PM

hi chat

Waiting

hi

robjohn

@ChristianF.Madsen You can take the logarithm and notice that $$\log(n!)\ge\frac{n}2\log\left(\frac{n}2\right)$$