Mathematics - 2018-08-23 (page 1 of 2)

Semiclassical

00:00

"second verse, same as the first!"

manooooh

I don't want "because the sum of two squares is always positive"

Semiclassical

it's not.

it's never negative != is always positive

manooooh

@Semiclassical ahhh yes, exclude the case $x=y=0$. Sorry

Leaky Nun

finally

Semiclassical

So you want to prove that $x^2+y^2=0$ iff $x=y=0$.

Leaky Nun

00:02

@manooooh if you want to prove it using axioms, note that this is true more generally in ordered fields

Semiclassical

(the 'only if' direction is of course the only nontrivial one)

manooooh

No, I want to prove that $\forall x,y\in\mathbb R^2\setminus\{0\}$ then $x^2+y^2>0$

Semiclassical

ah, was reading your original version first

Leaky Nun

Ordered field

In mathematics, an ordered field is a field together with a total ordering of its elements that is compatible with the field operations. Historically, the axiomatization of an ordered field was abstracted gradually from the real numbers, by mathematicians including David Hilbert, Otto Hölder and Hans Hahn. This grew eventually into the Artin–Schreier theory of ordered fields and formally real fields. An ordered field necessarily has characteristic 0 since the elements 0 < 1 < 1 + 1 < 1 + 1 + 1 < ... necessarily are all distinct. Thus, an ordered field necessarily contains an infinite number of...

manooooh

@LeakyNun thanks! I didn't see fields, maybe using some properties of the real numbers?

Leaky Nun

00:04

I already told you

the relevant properties are called the ordered field axioms

manooooh

@LeakyNun ok, I will review it and if I have doubts I will post a question

geocalc33

00:30

remind me, why in complex analysis do we use the formula $e^{nlogx}=x^n$

mick

-2

Q: How is the g-factor $ 2.002319 $ computed?

I understand a lot of math but not so much physics. So I read about the feynman diagrams and QED and I started to wonder : How is the g-factor $ 2.002319 $ computed exactly ?? Do we use matrices , differential equations , ... ?

mercio

because that can be used for $n$ that is not an integer

mick

Help ?

Daminark

Sniped

Semiclassical

Feynman diagrams basically are a convenient tool for enumerating multiple integrals

mick

00:32

1

Q: Prime twins and $ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $

Let $p$ be a prime such that $p+2$ is Also a prime. Define $$ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $$ For $z \neq 1 $ and $re(z) > 1$ this $f(z)$ always converges. If there are infinitely many prime twins then clearly $f(1)$ diverges. If there are only fin...

number-theory analytic-number-theory conjectures zeta-functions prime-twins

geocalc33

so it's used to define $x^n$ for values of $n$ that are non integers. interesting

Semiclassical

so computing the g-factor amounts to writing out a lot of multiple integrals in the right way and computing them

being more detailed than that requires a lot more know-how than I actually possess, though

(also, this applies in the framework of perturbative QED. nonperturbative QFT is harder)

loch

it's the ring of regular functions on the variety yes! Let's say you're considering the variety in $\mathbb{C}^n$ defined by equations $f_1,\ldots,f_m = 0$, then if you think of the polynomial (regular) functions defined on $\mathbb{C}^n$ they're given by $\mathbb{C}[x_1,\ldots,x_n]$. Now if you restrict all such functions to the locus defined by $f_i=0$, then e.g. $g$ and $g+f_i$ will restrict to the same function on the zero locus of $\{f_i=0\}$. So the ring of regular functions is given by $\C[x_1,\ldots,x_n]/(f_1,\ldots,f_m)$.

geocalc33

so for example, the equation of a circle could be defined as $e^{nlogx}+e^{nlogy}=1$, which simplifies to $x^n+y^n=1$ right?

oh wait that's a pointless definition

it's like a pencil without a point

pointless

@geocalc

I want to ping myself I'm taking this up on META

Adam

01:21

They don't need the "hot network questions" panel on the RHS most of the questions I see there are definitely written by people under the influence of drugs and totally not related to mathematics!!!!!!!!!!!!!!!!!!!

geocalc33

@Adam is there any point in defining the equation of a circle as $e^{nlogx}+e^{nlogy}=1$

well its actually superellipses with $n=2$ being a circle

my name is Hertzcog Werner pronounced "Verner"

@XanderHenderson is there any point in defining the equation of a circle as $e^{nlogx}+e^{nlogy}=1$

Xander Henderson

Well, that won't generally be a circle, @geocalc33

geocalc33

drangus fandangus...

but for $n=2$

Xander Henderson

well, yes, that would be a circle for $n=2$

geocalc33

but why would anyone define it like that?

Xander Henderson

01:32

define what?

geocalc33

like you could just write $x^s+y^s=1$

instead of using the exponentials

Xander Henderson

A circle is typically defined to be the set of points that are a fixed distance from a distinguished point called the center.

geocalc33

$e^{nlogx}+e^{nlogy}=1$

Xander Henderson

It can be shown that that the set of solutions to the equation $(x-h)^2 + (y-k)^2 = r^2$ is a circle.

geocalc33

i know that

Xander Henderson

01:34

Okay, but that isn't the definition of a circle (usually) any more than the equation $y=mx+b$ is the definition of a line.

geocalc33

in complex analysis you define $x^n$ as $e^{nlnx}$

Xander Henderson

And just as we can write $Ax + By + C = 0$ or $y-y_0 = m(x-x_0)$, you could also write $\mathrm{e}^{2\log(x)} + \mathrm{e}^{2\log(y)} = 1$ if you wanted.

@geocalc33 Do I?

geocalc33

yes that's a common definition

i read about it

Xander Henderson

Do you understand why one might want to do that?

And do you also understand how that definition works if $n$ is an integer?

geocalc33

to extend $n$ to non integer values?

yeah you can differentiate $x^n$ and you get $nx^{n-1}$

Xander Henderson

01:38

I mean, another possible definition of complex exponentiation is that $x \mapsto x^s$ is the unique analytic continuation of the the usual real valued exponential function on $(0,\infty)$, continued to $\mathbb{C}$ (minus, perhaps, a branch cut).

Where we can define the real exponential as the unique continuous extension of the rational exponential, and the rational exponential can be defined in terms of integer roots and powers.

geocalc33

okay so what does all that mean in terms of the equation $e^{2logx}+e^{2logy}=1$

Xander Henderson

(where by "exponential," I really mean "power functions"; I should be more careful in my use of language)

@geocalc33 I don't understand what you are trying to get at with that equation.

The set of real solutions to that equation is a circle.

geocalc33

what about complex

Xander Henderson

And so?

It is more complex in complex land, since the exponential possesses some periodicity.

A circle in $\mathbb{C}$ is typically written like $|z| = 1$ (or, more generally, $|z-\alpha| = r^2$).

geocalc33

i don't know it's just a really interesting equation

Xander Henderson

01:43

@geocalc33 WHY is it interesting?

It is the usual equation for a circle, complexified, and written in a kind of funny manner.

geocalc33

because the domain and codomain vanish outside the unit square

for non negative $n$

so it basically gets rid of the other part of the circle

and just gives you the part that passes through the unit square

Adam

@geocalc33 are you referring the question I just posted?

geocalc33

which one is that?

Xander Henderson

You haven't defined any functions. You have given an equation. What domains and codomians are you talking about???

Adam

No it's just that you were talking about the unit circle and I did post a question that involves the typical identities that ends up having something to do with that

and you asked me specifically so I assumed it was for that post since it's probably my least effort one in a while so there is likely to be something stupid in ir

mainly it's because maple wont share their inbuilt code for how they compute floating point approximations, so every time they have inbuilt code that contradicts other inbuilt code, I'm just going to somehow find a way to complain about it as stack exchange question until they let me debug inbuilt code, so I no longer have a blind spot

its very frustrating to 85 percent of the time be able to trust that the approximations are accurate, and then all of a sudden have it tell you what you have done is false, when you are right, or vice versa

I got a free large iced coffee frappe from maccas drive thru this morning.

geocalc33

02:04

cool

Adam

it was a good day overall

geocalc33

nice mine was too

Xander Henderson

I did not think to coldbrew last night, and ended up getting terrible, terrible coffee on campus at Charbucks this morning. That sucked. :\

I need to get a second coffee maker for my second office so that I don't have to drink terrible coffee.

geocalc33

@XanderHenderson I was just trying to understand that equation

it seems it's not really that interesting

Xander Henderson

@geocalc33 I get that you are trying to understand the equation. I just don't understand what aspect of it you aren't understanding. It is just the equation of a circle, complexified, and written funny.

geocalc33

02:10

yeah I understand that part I was just trying to find somehting interesting about it that wasn't there

Xander Henderson

Just like $\alpha w + \beta z = \gamma$ is an equation for a line, complexified, and written funny.

geocalc33

that's what I was plotting

Silent

Is it true that set of all condensation points in a separable metric space $X$ is either empty or uncountable?

geocalc33

and I was thinking that the red line could be the identity element and on either side you have your elements and inverse elements

Xander Henderson

@geocalc33 of course that is what desmos plotted---desmos doesn't know about complex numbers

geocalc33

02:14

anyway..

okay but now I'm just talking about reals

Xander Henderson

wolframalpha.com/input/…

geocalc33

could it be a group though? with identity element as $(1-x)+y=1$

user10478

How can you find the summation of k = 1 to infinity of 3/(4^k) via telescoping?

Xander Henderson

@geocalc33 What is "it"?

Adam

anyway it shouldn't have a problem verifying $9\,\pi \,{{\rm e}^{3}}-180\,\pi =9\,\pi \, \left( {{\rm e}^{3}}-2
\right) -162\,\pi $

Xander Henderson

02:19

Could what be a group?

What is the underlying set, and what is the group operation.

Adam

Like the software does simplifications like that all the time why there would it suddenly need me change the settings

geocalc33

@XanderHenderson the set of elements consist of: $e^{slnx}+e^{slny}=1$. The set of inverse elements are those reflected about the identity element which is $e^{lnx}+e^{lny}=1$. I'm not sure what the group operation is though...hmm

addition? multiplication? composition?

Xander Henderson

@geocalc33 A group is defined by its elements and an operation; you don't need to specify inverse elements.

geocalc33

okay

Xander Henderson

Once you define the elements and the operation, you can determine the inverse elements from the definition.

geocalc33

02:25

I guess I'll go with addition

when the operation is applied to any two group elements it should give another element in the group

hooray

it works

do you concur @XanderHenderson

Drew Brady

03:24

Anyone feeling like helping me think about a measure theory problem? :)

Luc

04:52

I think I've reasonably described my question, but I find it hard to think of good tags. Looking through common tags, I'm having a "mhm, I know some of these words..." moment (the math education I had, which isn't much, wasn't in English). Could someone look over the tags of my question to see if there are any missing and whether the ones I chose are correct? math.stackexchange.com/questions/2891706

dude3221

05:06

hello

@Luc

sorry I am not qualified to help you properly

Alessandro Codenotti

07:19

@DrewBrady what's the problem?

Rudi_Birnbaum

08:00

Good morning all!

When I have two groups $G,H$ and I form new group by direct product formation, I do not see immediately that $|G\times H| = |G||H|$. Because its not immediately clear to me that $(g_1,h_1)=(g_2,h_2) \Rightarrow g_1=g_2 \land h_1=h_2$ ($\Leftarrow$ is clear). The new group potentially could have a different "identification operation", no?

Tobias Kildetoft

08:20

@Rudi_Birnbaum By definition, just as a set, we have that those orders agree

i.e. that property of when two pair are identical are part of the definition of the direct product as a set

Shobhit

What are the equivalent definitions for a set to be nowhere dense in R, all i know is the interior of the closure of the set is null

Abcd

Differential equation of all non horizontal lines in a plane?

Attempt:

$y = mx + c$, $m \ne 0$

$y' = m$

$y''= 0 $

But answer given is:

$\dfrac{d^2x}{dy^2} = 0 $

Why am I wrong?

Alessandro Codenotti

@Shobhit not dense in any open set

But I don't know how many characterisations are there or what are you looking for

Shobhit