Also has quantum statistics

Have you guys read Galois theory?

not myself, no. didn't go down that route.

You're into physics?

yep

i'm a physics grad student

no I am doing galois theory next semester

@robjohn Perhaps hippa was sad some days ago because of my problem. This is the way I perceived him 2 days ago or so.

theoretical physics? @Semiclassical

yeah

i tend to be on the math-physics side of things

i think the following is a pretty good hint for how to actually compute the integral: split up the kinetic energy into the 1-particle part and the $(N-1)$-particle part, and then treat $E-K_{1ptcl}$ as though it's the total energy. at that point, you should just be able to use the original result.

$$\Huge{\text{OOOOOOOO MYYYYYYYY GOOOODDDD!!!!!}}$$$$\Huge{\text{AN AMAZING RESULT HERE!!!!!}}$$

What a result guyssssssss!!!!!!!!!!1

@robjohn re: your comment, it's a sphere in momentum space. so i don't see anything objectionable about the units

I'M DONE WITH THE RESULTS FOR THE WHOLE MONTH!!!

If you want to delete the messages above I'm fine with that, but I'm simply done with the marvellous results for the whole month, or even for the next few months.

I can't ask for more!!!

a simple "EUREKA" would probably have sufficed :)

@Chris'ssistheartist a toast for your result

btw do you need a result every month @Chris'ssistheartist

congrats @Chris'ssistheartist

@zed111 Thank you! I have results more times a day, but not one like this one.

@KarimMansour Thousands thanks! :-)

I'm out for a couple of minutes, I simply wanna recover myself.

BBL

hey @Semiclassical you know I was solving this problem in point set topology it was asking me prove that points are accumulation points and determine if they are open or closed sets rigorously $X = \{ (-1)^n + \frac{1}{m}: n,m \in \mathbb{N} \}$ we can easily see that -1 and 1 are accumulation point but to prove that other points aside from those aren't accumulation points requires some thinking

very nice problem

the problem will come from constructing the right r for interval (-1,1)

eh. that's the kind of problem which i'm okay with not having to do as a physicist. finding a rigorous argument by applying the dfns is just not something i find that interesting anymore.

no I still like physics but I want the good rigour in it not just hand wavy

though if i had to try, i guess i'd start from observing that $X$ is contained in $[-1,-1/2)\cup [0,2]$

with the first subset consisting of points with $n$ odd and $m>1$, and then everything else

I did it as follows I considered the set that X is contained in which is (-1,1) and considered points outside that set and points inside

that is

@Freeze_S I don't think people are too interested in mathematical rigour in a physics class

$(-\infty,-1] or [1,\infty)$ and (-1,1)

the hard thing is (-1,1)

i wanted to pull out that interval $[-1/2,0)$ so that the two sets are disjoint.

Ahh I am having difficulty understanding generators and relations

oh I see

And using them in subgroups

but yeah. by and large, physicists are quite happy to be lazy about rigor

@evinda Hi

so you have generators and you have relations that those relations are the minimum amount of relation to generate others

for example

there are exceptions, of course, but it's a fair rule of thumb

@Ramanewbie Hi!!! How are you?

one way to represent dihideral group is $X_{2n} = <x,y | x^n = y^2 = 1, xy = yx^{-1}>$

Hi@Ramanewbie

Hi @Rememberme

because those relations will generate a group of order 2n and we have already found that by construction

if your reading DF

@evinda To answer your question I have been to Crete indeed 3 years ago

it's equivalent to $y^{-1} x yx=1$

So let me give you some arbitrary dihedral group....

Just for a few weeks

@Ramanewbie And did you like it in Crete?

you can see $xy = yx^{-1}$ by geometry btw

what is

xy

Applying x then y

is that it orients the rotation anti clock wise

@evinda Its rather warm in comparison of France but I liked it a lot

that is why $yx^{-1}$ does

i like it as $y^{-1} x y x=1$

@Ramanewbie Did you go with your family?

@evinda yes for 3 weeks

rotate, flip, rotate, flip, and nothing changes

yeah that is another way to represent it @Semiclassical

yeah exactly

@Ramanewbie Nice :)

though since $y^2=1$, you could also view it as $(yx)^2=1$

@evinda And yes, I can manage to talk with my brother throw the net

then n becomes 176

and you still have

@Ramanewbie Good!

y^2 = 1

the same thing....

yeah

the only thing that changes is the order of rotation

which makes sense if you think about it geometrically

@DanielFischer I also could have evaluated this one using the mysterious Maclaurin series for $\exp \left(\frac{z-1}{z+1} \right)$. But instead I used contour integration, basically modifying an approach we talked about a long time ago. achille hui's answer is similar to his other answer.

am i correct in saying log(4)2 + log(4)6 would be log(16)8?

@RandomVariable At least the way you did it, we have no unknowns in it. Nice job.

@DanielFischer Thanks.

@Dave nope. you need to factor out the common $\log(4)$ term before you add

that should be $16\log(2)$ ... @Dave Do check it I am in half sleep condition

that's correct

how do you get (2) ?

i understand the 16 being (4)*(4)

it's not. $4\times 4=16$ is really not relevant to the problem

at least not in the way you're saying.

hmm

@SohamChowdhury kids are easily amused. On a serious note : more generally, $F[x]/\text{min} \alpha \cong F(\alpha)$ for a field $F$, an algebraic $\alpha$, and it's minimal polynomial $\text{min} \alpha$

@Semiclassical so wouldn't it be log(4)2+log(4)6 = log(4)6*2 = log(4)12 ? because this website im looking at seems to suggest thats how you add them

no.

though, actually, i may be misinterpreting your notation

So what does it mean why it says: logb(mn) = logb(m) + logb(n)

ahhhh

granted they are using brackets different to me

@Semiclassical okay... but the form of the expression is just not right. I've commented on the question and zed111's self-answer.

when you say log(4)2, which of those is the base of your log?

x^4 = 2

so, base 4

yes :)

so people don't normally use brackets on the base?

eh, i'd use log_4(m)

but i'd also use chatjax, per being able to use latex in chat with that up

hence $\log_4(m)$.

@Semiclassical I have answered my question based on your suggestion earlier

but in that case, your logic is sound.

But there is something @robjohn says which I seem is correct

i was panicking i hadn't understood it

i think there's really two issues

one is that you've put the product symbol to the left of the integration parens, rather than (say) $\int \prod dp_i$ where i'm being lazy about subscripts and limits

what does it mean to "take logs" to solve an equation ?

the other is that your summations inside the integral should be taken with respect to a variable other than your integration index

say, $i$ instead of $j$

though i'll confess that that kind of abuse of notation is something which doesn't bother me a heck of a lot.

and i wouldn't be surprised if that's how it showed up in lecture notes.

Generally when we find the volume of phase space, we keep the integrals nested, right?

right. you've got a multiple integral of the form $\int_{-\infty}^\infty dp_{k+1}\int_{-\infty}^\infty dp_{k+2}\cdots \int_{-\infty}^\infty dp_{kN}F(p_{k+1},p_{k+2},\cdots p_{kN})$

with the differentials being to the left of the integrand being typical of physics convention

myself i'm fond of functional notation, say $\int_{-\infty}^\infty D[p]_{k+1}^{kN}$ in writing the measure

Ok thanks

@pjs36 Thanks!

$\hskip -1.2in\longrightarrow$ This is really cool!

Anyone here familiar with prolog?

@anon d'you know of classification of $k$-algebras $A$ such that taking tensor $A \otimes_k L$ with a sep. extension $L/k$ splits it into a finite product of smallish sep. extensions, i.e., $A \otimes_k L \cong \prod K_i$ where $K_i/K$ are all separable.

@Studentmath What is your question?

I have to do it recursively, but in order to avoid infinite loops I have to somehow mark the the slots I have already visited.

I have no idea how to mark them, as prolog doesn't carry on the marking, and anyway that will carry on the marking won't carry them specifically for the slots I want.

I rather need the fact that there exists such an algebra $A$ for each sep. ext $L/k$, than a classification, though, which seems easy enough : Let $A = k[x]/(f)$ where $f$ is a polynomial which splits completely in $L$. then the tensor becomes isom to finite product of $L$s.

but I want to know if these guys can be classified.

in particular, let $A$ be a $k$-algebra such that $A \otimes_k k^{alg}$ is a direct product of $k^{alg}$s. can we classify such $A$s?

apparently fields are a subset of all such k-algebras.

Hey

did anybody use C* algebra murphy before ?

I almost forgot about DRE today

I don't understand the following statement

If A is a normed algebra, then it is evident from the inequalities that ||ab - ab|| $\leq$ ||a|| * || b - b` || + ||a - a|| ||b|| that multiplication (a,b) $\mapto$ ab is jointly continous

any remark ?

Other than the definition of jointly continuous, that is (which I assume means that the map $A \times A \to A$ is continuous?)

yeah

so that is by definition of what it means for the map $A \times A \mapsto A$ is continous ?

I think you meant to write $\|ab - a'b'\| \leq \|a\| \cdot \|b-b'\| + \|a-a'\| \cdot \|b'\|$. And yes, it will follow from some epsilons.

yeah

I am supposed to read about this GNS construction stuff but I don't know anything about C* algebra

so reading this murphy's book on C* algebra

it is hard read

I could just read take some results for granted but I don't feel good when I do that