Mathematics - 2016-12-30 (page 5 of 6)

GFauxPas

18:03

sup guys

@DHMO you here?

DHMO

yes

GFauxPas

check out the new versions of the graph on PW

I'm very proud of my self

Mahmoud

Conductance vs conductivity ?

@DHMO

DHMO

@Mahmoud conductance = distance x conductivity

@GFauxPas nice

GFauxPas

ty

DHMO

18:06

You should totally deal with the theorems here

GFauxPas

right now I'm working on graphs of complex functions and on Laplace transforms

Mahmoud

@DHMO What does each of em' mean and why on earth do they have very similar names.

DHMO

@Mahmoud according to wikipedia‌, conductance is the inverse of resistance

and the definition of resistance is V/I

voltage / current

@GFauxPas maybe you would like to work on Lambert W function first?

GFauxPas

I dunno, depends on my mood I guess

Semiclassical

here's something I always find funny.

First, the unit of resistance (voltage per unit current) the ohm $\Omega$.

Now, the official SI unit for conductance (current per unit voltage) is Siemens (S).

DHMO

18:17

I hope "Siemens" is pronounced "Simons" but unfortunately it isn't.

Semiclassical

But there's an older name for that unit, though it's now discouraged.

Namely, since conductance is the reciprocal of resistance, it's said to have units of mhos ($\mho$)

3

Latex even has as a symbol for that :)

Secret

For any number people currently in the chat, I strongly encourage you to check out this proof of mine because it is wrong:

If this proof holds, it means $2$ will be transcendental, which is not true. However I have yet to see anythign that suggest cracks thus I am kinda stumped

in Algebraic/Transcendence Theory, 10 mins ago, by Secret

Therefore for $a > 1$ (to be analysed):
$$(p-1)!\sum_{k=0}^m|b_0(\ln a)^{-k}|(n!)^p<|J|<\max (|b|)B\sum_{j=0}^n\left|\frac{a^{j}}{\ln a}\right|$$
and for $0 < a < 1$(?)
$$(p-1)!\sum_{k=0}^m|b_0(\ln a)^{-k}|(n!)^p>|J|>\max (|b|)B\sum_{j=0}^n\left|\frac{a^{j}}{\ln a}\right|$$
and for $a$ (???)
$$(p-1)!\sum_{k=0}^m|b_0(\ln a)^{-k}|(n!)^p? |J| ? \max (|b|)B\sum_{j=0}^n\left|\frac{a^{j}}{\ln a}\right|$$
To be analysed...

Secret

Joke: $contradiction\circ contradiction = ?$

Mahmoud

$T$

GFauxPas

18:37

where are the crests?

happening at which horizontal lines?

DHMO

@GFauxPas explicit version: shit's happening.

clean version: bug's happening.

Semiclassical

Nah, nothing buggy there.

Mahmoud

@GFauxPas What's the simplest way to explain how do you do that ?

Semiclassical

Take $z=i y$. Then arg gives iy mod 2\pi.

GFauxPas

@Mahmoud I'm using a language called R. The environment is RStudio

DHMO

18:39

@Semiclassical right

GFauxPas

If you want I can find an old guide my professor wrote if you want to learn the basics

Semiclassical

Since arg restricts to (-\pi,\pi], it increases up to pi and then jumps down to -pi.

GFauxPas

Oh, so, at which imaginary values does it peak at? let me think

can I have a clue

Mahmoud

@GFauxPas I'm trying to learn how to use an animation tool called ''manim'', written in python.

Semiclassical

Pretty sure I've said all that's necessary.

If nothing else, plot Arg(e^z) along the positive imaginary axis.

GFauxPas

18:46

wrong one

DHMO

@GFauxPas bye

GFauxPas

bye dhmo

image of imaginary positive axis

Semiclassical

right. should be easy enough to see from that where the jumps are.

GFauxPas

odd multiples of pi?

Semiclassical

right.

GFauxPas

18:48

nice

thanks Semiclassical :)

Semiclassical

np

GFauxPas

@Mahmoud R isn't the typical program used to graph things, I like it because it's versatile and free and I've used it for lots of things already. there's probably a better language to learn

but let me know if you want me to dig up my professor's old R guide

Mahmoud

@GFauxPas github.com/3b1b/manim This is it.

GFauxPas

I don't know anything about manim

Semiclassical would it be more meaningful to plot Arg and Mod of complex valued function using spherical coordinates?

change of coordinates to spherical

Mahmoud

@GFauxPas Very few people do, 3 at most.

GFauxPas

18:53

then you should learn a language that's common

like R or Python

Mahmoud

@GFauxPas An example of what it can do youtube.com/watch?v=sD0NjbwqlYw

GFauxPas

cool

I never tried animated plots in R but I don't think it's meant for it

Semiclassical

@GFauxPas Do you mean polar coordinates?

In any case, I dunno if that's terribly useful.

GFauxPas

I menat polar

Semiclassical

I like domain coloring myself

some examples thereof: wismuth.com/complex/gallery.html

GFauxPas

18:59

hm, this looks interesting

DHMO

P(20) = 4849781/4849845

Semiclassical

so that's a total of 64 exceptions.

@GFauxPas I suspect what you'd get for the Lambert-W function using domain coloring is rather similar to what they get for the Gamma function on the linked page (near the bottom)

GFauxPas

I'm not sur eI understand this semiclassical

Semiclassical

Well, take a look at Figure 1 in the gallery of pictures

GFauxPas

you're coloring the argument or the image?

AccidentalFourierTransform

19:02

Hi peeps :-)

any physicist here?

Semiclassical

physics grad student here

AccidentalFourierTransform

I asked this question yesterday, and I could really use some feedback

Semiclassical

Ah, QFT.

GFauxPas

isn't the argument of the image just.. whereever the spot on the plane it is?

Semiclassical

I'm not a master of functional integrals, so I may not be able to offer much feedback.

AccidentalFourierTransform

19:03

welp, I had to try

thanks anyway =)

Semiclassical

good luck

AccidentalFourierTransform

thanks, and happy holidays everybody

Sophie

suppose $\alpha_1$ is an irrational algebraic number and $\alpha_1,\alpha_2\dots \alpha_n$ are the roots of the minimal polynomial of $\alpha_1$. Is the splitting field of that polynomial equal to $\{q_0+q_1\alpha_1+q_2\alpha_2\dots+q_n\alpha_n|q_0,q_1\dots q_n \in \mathbb{Q}\}$?

Semiclassical

@GFauxPas Yeah, that's all. You also see variants where regions with larger modulus are made brighter than others, but that's not done in that gallery.

GFauxPas

I'm missing what information is added by coloring

Semiclassical

19:07

It's just a different way to plot Arg. It has the advantage that it doesn't suddenly 'jump' from pi to -pi

GFauxPas

ah I see

I think r can do that, let me see

Kasmir Khaan

hi chat

i got question about the jacobian , when i change u = x+y , v=x-y i get the jacobian = 2

but what do i use d(x,y) /d(u,v) or the inverse of that 1/2 ?

Semiclassical

some more examples of domain coloring @GFauxPas

@KasmirKhaan Use the version which seems consistent with how u-substitution works.

Kasmir Khaan

am comfused because i worked 2 exercices one i used x and y as functions of u and v

and other u and v as functions of x and y

Semiclassical

when you do that, you have for instance $\int f(u)\,du=\int f(u(x))\frac{du}{dx}\,dx$

Kasmir Khaan

19:13

but never took the invers of jacobian and got both right answers

Mithlesh Upadhyay

-1

Q: What is special about the number 2017?

A new year is coming, I know that $2017$ is a prime number but - Are there other nice properties to make it special? EDIT: I read the comments, and I agree that the question is too vague. Whit other proprieties I mean something like : $2017$ is the smallest natural numbers such that... or other ...

Semiclassical

following the 1-variable example, if you have $\iint f(u,v)\,du\,dv$ then how would you write this in terms of $x,y$?

(feel free to write u(x,y) as u to avoid tedious expressions)

Kasmir Khaan

d(x,y) / d(u,v)

correct?

Semiclassical

Write out the entire expression and compare it to the 1-variable case I just wrote.

GFauxPas

did I do it

Kasmir Khaan

19:17

$\iint f(u,v)\,du\,dv$ d(x,y) / d(u,v) so that d(u,v) "cancel out" ?

Semiclassical

No.

Or, well.

GFauxPas

no to me or no to him?

Kasmir Khaan

let me type the 2 examples i worked out and show you where am lost

Semiclassical

Well, both. But I meant @KasmirKhaan :)

With a proper domain coloring, it shifts around the color wheel.

GFauxPas

oh I can do that

Semiclassical

19:18

here it just goes from blue to green.

GFauxPas

can't I choose my own colors though

or do I want a color wheel

Kasmir Khaan

double integral of (x^2-y^2) ^10 dx dy here i used u = x+y , v = x-y

double integral of ln(1+x^2+y^2 ) dx dy here i used x= rcost , y= rsint

GFauxPas

I guess the color wheel gives you an order

ROYGBIV and all that

Kasmir Khaan

first jacobian is = 2 and second = r

Semiclassical

It's advantageous to use all of them, since you can shift smoothly as red -> purple -> blue -> green -> yellow -> orange ->red

Kasmir Khaan

19:19

but as you see i had in first u and v as function of x and y ,

Semiclassical

if you only use a subset, you'd have to jump back to red at some point.

Kasmir Khaan

but on second x and y as function of u and v

Semiclassical

@KasmirKhaan Let's go back. You want to write a double integral in two different ways.

GFauxPas

Semiclassical

Yep @GFauxPas

GFauxPas

19:20

hmm interesting

Semiclassical

You might check to see whether you're getting the full range of the color wheel---ideally it's red at -pi and pi---but that's basically right.

GFauxPas

well here's what my code is doing

I'm choosing how many colors to use, using an argument n

Then I used rainbow(n), which creates $n$ colors from the beginning of the color wheel to end

Semiclassical

@KasmirKhaan Suppose I start with $\iint f(u,v)\,du\,dv$. How would I write it in terms of $(x,y)$ if I have $u=u(x,y)$ and $v=v(x,y)$?

@GFauxPas Ah. You might try scaling it so that pi and -pi both get the same color, but I'm not sure how practical that is.

GFauxPas

wismuth.com/complex/gallery.html these guys use only 4 colors

Semiclassical

Yeah. I don't like their's as much as others for that reason :)

GFauxPas

19:24

stat.columbia.edu/~tzheng/files/Rcolor.pdf I can create my own order as well

Semiclassical

I grabbed that link from Wikipedia rather hastily. A different source would've been better.

Kasmir Khaan

f(u(x,y) , v(x,y) d(x,y) / d(u,v) du dv

GFauxPas

pi and -pi get the same color in this case, because we're dealing with argument?

Semiclassical

No. That's still an integral in (u,v).

Kasmir Khaan

sorry i meant
f(u(x,y) , v(x,y) d(u,v) / d(x,y) dxdy

double integral f(u(x,y) , v(x,y) d(u,v) / d(x,y) dxdy

so the jacobian is d(u,v) / d(x,y)

Semiclassical

19:27

Yes, that's correct: $$\iint f(u,v)\,du\,dv=\iint f(u(x,y),v(x,y))\left(\frac{\partial(u,v)}{\partial(x,y)}\right)\,dx\,dy$$

But u,v,x,y are just labels. I can rename them if I want. So I also have

$$\iint f(x,y)\,dx\,dy=\iint f(x(u,v),y(u,v))\left(\frac{\partial(x,y)}{\partial(u,v)}\right)\,du\,dv$$

Mahmoud

@Semiclassical What field of Math is that ?

Semiclassical

What I just said? Multivariable calculus.

@GFauxPas ya.

Mahmoud

@Semiclassical Is it as hard as it seems to be ?

Kasmir Khaan

thanks semi i think my book is wrong

please confirm this with me

double integral (x^2 -y^2 )^10 dx dy

Semiclassical

@Mahmoud Eh, if you've got some one-variable calculus it isn't so bad.

Kasmir Khaan

19:31

over abs(x)+abs(y) <= 1

Semiclassical

Okay. Which scenario are you in---first or second integral I wrote?

I should also note

Kasmir Khaan

second

u = x+y , v= x-y

Semiclassical

If I define $g(u,v)=f(u,v)\left(\frac{\partial(x,y)}{\partial(u,v)}\right)^{-1}$

Kasmir Khaan

and we want d(x,y) / d(u,v)

Semiclassical

then the first integral I wrote becomes

$$\iint g(u,v)\left(\frac{\partial(x,y)}{\partial(u,v)}\right)^{-1}\,du\,dv=\iint g(u(x,y),v(x,y))\,dx\,dy$$

Kasmir Khaan

19:34

yes exactly

Semiclassical

Which now looks rather like the second integral. There remains a difference: In this example, I have $u,v$ as functions of $x,y$. In the second integral I wrote above, you've got $x,y$ as functions of $u,v$.

Kasmir Khaan

they claim the answer is 2/121

Semiclassical

In the present case, you're doing $u,v$ as functions of $x,y$. So therefore you're not in the second, you're in the one I just stated.

Typo in what I just wrote: the left-hand side should not have had that -1.

Can't fix it now.

In any case, though, we're getting mixed up.

The rule I usually follow when doing Jacobians is write $x,y$ as functions of the new variables $u,v$.

Kasmir Khaan

yes :(

I got comfused with that inverse too

Semiclassical

Here, that's easy: $u=x+y,v=x-y\implies u+v=2x,u-v=2y$

So therefore $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$.

The relevant Jacobian is then d(x,y)/d(u,v). What's that give you?

Kasmir Khaan

19:41

1/3

1/2*

Semiclassical

Yes, 1/2.

Kasmir Khaan

i dont understand how they got 2/121

i mean we now changed to problem to

1/2 u^10 v^10 *

u^11 / 11 * v^11 /11

Semiclassical

So therefore the integral I wrote above becomes $$\iint f(x,y)\,dx\,dy=\iint f\left(\frac{x+y}{2},\frac{x-y}{2}\right)\left(\frac{\partial(x,y)}{\partial(u,v‌)} \right)\,du\,dv$$

Except, uh, formatted correctly.

Kasmir Khaan

yes yes i got it now :D

thanks alot @Semiclassical

Semiclassical

np.

Kasmir Khaan

19:45

i came up with easy way to rememver it

Semiclassical

In principle, you can compute (d(u,v)/d(x,y))^{-1}.

Kasmir Khaan

u can pretend as if they cansel out like fractions

yes if its easiar

:)

Semiclassical

But in practice it's easier for me to remember d(x,y)/d(u,v)

It is rather similar to the old "treat differentials like you can cancel them" way of remembering the chain rule (dy/dx) = (dy/du)(du/cx)

While not without its hazards, it does get the order of things right.

Kasmir Khaan

yes its alot clear now for me thanks again !

il lkeep working on them

Semiclassical

glad to help.

One practical point, extending what I indicated above.

There's always two ways to get the relevant Jacobian. You either express x,y in terms of u,v and compute d(x,y)/d(u,v), or you compute (d(u,v)/d(x,y))^-1 in x,y and then express the result in terms of u,v.

Kasmir Khaan

19:51

yes in this example it was not needed to express x and y in u and v

Semiclassical

I typically find the former easier to think about, but if you're given u,v in terms of x,y then the latter may be simpler.

Kasmir Khaan

u=x+y and v = x-y

we could compute the natural thing d(u,v) / d(x,y) and take inverse

Semiclassical

aye. (d(u,v)/d(x,y))^-1 = (2)^(-1)

If it's a linear transformation, it doesn't matter much either way.

Kasmir Khaan

:)

Semiclassical

If it's not linear, though, it may be that getting x,y in terms of u,v is a real pain.

In which case, you definitely want to do the inverse.

Liad

20:01

someone here familiar with simple groups ?

Mariano Suárez-Álvarez

20:21

does the operation $S_p\times S_q\to S_{p+q}$ that "concatenates" permutation have a standard name?
Here the image of $(\sigma,\tau)$ is the map $\mu$ such that $\mu(i)=\sigma(i)$ if $i\peq p$ and $\mu(i)=p+\tau(i-p)$ if $i>p$.

Semiclassical

Got a question for someone who knows Lie algebras.

Mariano Suárez-Álvarez

Just ask the question

Semiclassical

I'm trying to remember a diagram that shows up in there. I thought it was a 'weight diagram' but that doesn't seem to be correct judging from Google.

(was typing)

Mariano Suárez-Álvarez

It is very rare that anyone who knows about a subject will say "I know about the subject"

Semiclassical

Yes, and I've harped on people myself re: "don't ask to ask"

Mike Miller

20:24

@MarianoSuárez-Álvarez if anyone has a name for it, the symmetric spectra people probably would

Semiclassical

To be clear, I wasn't going to wait until someone responded to type up the question.

The diagram I'm trying to remember, in any event, has the various weight vectors connected together in a layered tree-like structure.

Mike Miller

Schwede doesn't seem to have a name for it

Semiclassical

Hrm.

It's like this, but without as many decorations: goo.gl/images/a7c4zL

Mariano Suárez-Álvarez

I looked in MacDonald's book on symmetric functions and he also does not hat a name :-/

Liad

I have proved that $H \le G $ subgroup is normal iff $H $ is the union of Conjugacy classes of G.
Now, using this i need to show that all of the normal subgroups of $S_5 $ are $A_5 $ , $\{e\} $ , $S_5 $ .

anyone see a way how should i do that ?

Mike Miller

20:27

I would probably be tempted to say that it's the homomorphism induced by the inclusion $\{1, \dots, n\} \sqcup \{n+1, \dots, n+m\}$ into $\{1, \dots, n+m\}$

Semiclassical

Judging from this graphic on Wikipedia's page on root systems, the term "hasse diagram" is the appropriate one. (and it's also the one I started with :/ )

Thought there was a particular name for it in the Lie algebra context, but maybe not.

arctic tern

long time no see mariano

there's the usual "canonical embedding" and "the obvious map" phrases

Mahmoud

@Semiclassical

Mariano Suárez-Álvarez

@arctictern I'm writing something that has way too many of those :-)

This is in the middle of a computation of a $d_3$ in a specral sequence. Everything is canonical and obvious heh

And a mess, of course.

Mahmoud

If $f$ and $g$ are functions, defined on $\Bbb R$, and are increasing and decreasing respectively on $\Bbb R$, how can we prove that $f(x)=g(x)$ has at least one solution ?

Mike Miller

20:33

what are you calculating?

GFauxPas

guys someone told me I should make the colors more distinct so I need your opinion, which one isbetter:

which one better gives you the idea that it starts low and comes up at the sides

Semiclassical

@Mahmoud Going to have a hard time proving that since it's false.

GFauxPas

towards the viewer

Mariano Suárez-Álvarez

@MikeMiller cohomology of Joyal's combinatorial species

Semiclassical

Unless by "defined on R" you further mean that it's a surjection to R, which is a much stronger statement.

arctic tern

20:35

@GFauxPas you can link to images instead of pushing everyone else's discussion off the screen (protip)

GFauxPas

I don't know how to do that

arctic tern

[text](url)

GFauxPas

they're not hosted anywhere

Semiclassical

Yes, they are. That's how they get uploaded here.

Mahmoud

I'll rewrite it,

Mariano Suárez-Álvarez

20:36

@Mahmoud that's not true. Take $f(x)=\arctan x$ and $g(x)=1000-\arctan(x)$ (or the other way round)

arctic tern

@GFauxPas when you upload them they go to imgur, so you can just up them to imgur yourself

GFauxPas

oh, okay. sorry everyone

Semiclassical

I feel like we talked about this before, though @Mahmoud

With $\pm e^x$ as a simple counterexample.

Null

an injection from R to R that is no surjection? can't think of one out of my hat

or..

mmh, i mean from R as a domain

Mike Miller

@MarianoSuárez-Álvarez some cohomology of species or like HH of the category thereof?

Mariano Suárez-Álvarez

20:37

of individual ones

trying to deform them

Mahmoud

Supposing both f and g are surjections to $\Bbb R$

Mike Miller

cool

Mahmoud

@Semiclassical Is it still false ?

Semiclassical

@Null e^x is an injection from R to R, but not a surjection.

...uh.

Null

oh!

sorry

@Semiclassical ah, surely ;)

Semiclassical

20:39

@Mahmoud In that case, note that $f(x)$ and $-g(x)$ are both surjective and increasing functions from R to R.

What about the sum of those two functions?

I presume they're also continuous.

Alas, I see a problem with where I was going: If two functions $f,g$ are continuous and surjective, must $f+g$ also be a surjection? Not certain that's true.

(If it were true, you'd be done: 0 is in the codomain, so there must exist x such that (f-g)(x)=0)

I think it's true but I've forgotten enough of analysis that I don't remember how to prove it.

Mahmoud

@Semiclassical $(\forall y\in \Bbb R)(\exists x\in \Bbb R)| f(x)=y$

$(\forall y\in \Bbb R)(\exists x\in \Bbb R)| g(x)=y$

Null

is there a term for the phenomen that a smaller intervall gets mapped to a bigger one, while all elements of the bigger one are hit? Or is small and big nonsense for real intervals?

Mahmoud

$(\forall y\in \Bbb R)(\exists x\in \Bbb R)| h(x)=2y$ and $h(x)=f(x)-g(x)$ @Semiclassical ?

Semiclassical

no. @Mahmoud

Mahmoud

Is it the variables ?

Semiclassical

20:48

well, how did you take the difference and get 2y?

Mahmoud

Wait I didn't, it should be $0$

Semiclassical

but in any case there's a more important problem. it's true that, given a y, there must be an x such that $f(x)=y$.

and similarly for g(x)=y. problem is, there's no guarantee that it's the same x for each of these

for instance, $f(x)=x$ has $x=0$ as solution to $f(x)=0$.

but $g(x)=x+1$ has x=-1 instead.

Mahmoud

@Semiclassical It's hard

Semiclassical

so we can't conclude that there's some $x$ such that $f(x)=g(x)$. indeed, there is no such $x$ in this case.

What's missing is use of the fact that $f(x)$ / $g(x)$ are supposed to be increasing / decreasing respectively.

That needs to get used in here or we're stuck.

I think what's needed is to observe that, if $f$ is supposed to be increasing and surjective, then $f(x)\to\infty$ as $x\to \infty$ and $f(x)\to-\infty$ as $x\to-\infty$.

Whereas the opposite is true for $g(x)$.

Something like that.

Mahmoud

So they have to cross at least once.

Semiclassical

20:56

right.

for a proof, one probably notes that the above properties imply that f(x)-g(x) behaves like f(x) itself, i.e. crossing from negative to positive infinity

so as a continuous function, it had better cross the axis.

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