Mathematics - 2015-07-27 (page 3 of 3)

robjohn

9:06 PM

Mathematica seems to make good use of the multiple cores

Cristopher

9:32 PM

Hi. How can we evaluate $\displaystyle\lim_{s\to\infty}\zeta(s)^s$ ?

robjohn

@Cristopher $\zeta(s)\sim1+2^{-s}$ as $s\to\infty$

Thus, $s\log(\zeta(s))\sim s2^{-s}\to0$ as $s\to\infty$

Khallil

How do people know about derivatives but can't square terms? :-/

robjohn

@Khallil The Binomial Theorem can be hard...

Cristopher

@robjohn How could we prove that $\zeta(s)\sim 1+2^{-s}$ as $s\to\infty$ ? I think that is not mentioned in Wikipedia

robjohn

@Cristopher It is $1+2^{-s}+3^{-s}+4^{-s}+\dots$

Chris's sis the artist

9:41 PM

@robjohn Great! :-) I also need to replace my computer.

robjohn

@Chris'ssistheartist I imagine that if I upgrade my Mathematica, it might be faster, too.

I am running 8

Chris's sis the artist

@robjohn Maybe not. Newer versions require more resources, but it might be.

Cristopher

@robjohn Oh, that's right. Thank you

robjohn

@Cristopher so $\lim\limits_{s\to\infty}\zeta(s)^{2^{\large s}}=e$

Cristopher

@robjohn That's beautiful. First time I've seen it

robjohn

9:49 PM

@TedShifrin where are you?

Vrouvrou

Hello, have you an idea about this please: math.stackexchange.com/questions/1372939/… Thank you

Ted Shifrin

San Diego. Waiting for movers Fri ....

robjohn

@TedShifrin So you're now in California! Welcome!

Ted Shifrin

Tanx ... Just had a credit card denied at Best Buy, but finally fixed :)

robjohn

@TedShifrin yeah, you need to tell them you will be using it in a different location. I had that problem when I went to visit my son at school in Chicago.

Ted Shifrin

9:52 PM

Well, I put the new address in 5 days ago ...

Balarka Sen

hi @TedShifrin.

Ted Shifrin

Off for now ... Back later... Hi/bye Balarka.

bye Ted.

Bye Ted

Hi @TedShifrin

10:01 PM

I find the physics way of writing integrals to be so confusing.

Sawarnik

@robjohn That is the first time I have seen it that way.

robjohn

@Sawarnik I have been running into it more and more frequently. I have to really work to try to figure out where the integrand ends. It is supposed to start at $\int$ and end at $\mathrm{d}x$ :-)

Sawarnik

:D

pjs36

I think we can all agree that it's pretty gross.

robjohn

In this form, it is written as an operator, but with no parentheses around the argument, it is hard to parse.

Cristopher

10:08 PM

I find it confusing too. I'm so used to seeing $\int f(x)\,dx$, this way looks nicer to me

Vrouvrou

10:23 PM

@robjohn you have an idea please: math.stackexchange.com/questions/1372939/…

Khallil

May I ask what you mean by an operator, @robjohn?

robjohn

10:50 PM

@Khallil $\color{#C00000}{\int\mathrm{d}x}\,[\cdots]$ is meant as an operator on $[\cdots]$, but without the brackets or parentheses, it is hard to tell where the intended integrand ends.

Khallil

May I ask how the $\text{d}x$ works as operator, @robjohn?

Is it just that the $\text{d}x$ tells us to sum up infinitesimal increments of the integrand?

Mike Miller

'operator' just means 'takes thing, spits out number'

robjohn

@Khallil No, not the $\mathrm{d}x$, but the $\int\mathrm{d}x$ is considered an operator.

@MikeMiller or takes thing, spits out another thing.

Mike Miller

fair enough

Khallil

Oh! I guess it does make sense to leave the $\int \text{d}x$ on the LHS.
Just like how we write functions as $f(x)$ instead of $xf$.

Do algebraists do something like that? I recall reading that they (without possibly being any more vague) write things on the other side.

Mike Miller

11:00 PM

it's a thing in Herstein's algebra book but not really anywhere else.

He thinks $fg$ should mean "do $f$ then $g$", so you have $x(fg) = (xf)g$

Khallil

Mike Miller

it'd be helpful if you could say where that's from

"Many" of the abstract algebra books... is just wrong, period

Khallil

Oh, it's a set of notes that one of the lecturers at my university wrote.

Mike Miller

It happens in a couple. Herstein is the most prominent one

Khallil

(In particular, they are abstract algebra notes)

Mike Miller

11:02 PM

in particular, look up a paper on the arxiv. I guarantee you $f(x)$ will be written with $f$ on the left (vs $xf$)

Khallil

Written as $xf$ or $f(x)$?

Oh, gotcha.

Mike Miller

there is precisely one place where I know of writing things on the right is actually more common in modern literature than on the left, which is a certain place in differential geometry. I don't even know why they do it

Khallil

They got bored I guess.

Mike Miller

Some historical reason no doubt. I do it their way because it's standard but it trips me up sometimes

(And this is a group action, rather than a function - in the same context you'll see $f(x)$. It's just that they'll have a point $p \in P$, and multiplying it by an element of a group $g \in G$ will be written $p \cdot g$ instead of $g \cdot p$)

Khallil

What's a group action?

(I mean, I haven't defined them yet, but what was it that you were referring to as a group action?)

Chris's sis the artist

11:07 PM

@robjohn Related to what I got now, I don't know if the present computational systems can do anything about it.

Mike Miller

you have a set $X$. a (left) action of $G$ on $X$ is a map $G \times X \to X$ (written $g \cdot x$) such that $h \cdot (g \cdot x) = (hg) \cdot x$ and $1 \cdot x = x$

ahhh, this happens because in the "canonical example" of the structure I'm thinking about, the "obvious choice" is a right action, not a left action

even in this one example, it all ends up coming from the "things should be written on the left" philosophy... which ends up forcing this one thing to be written on the right

Chris's sis the artist

11:26 PM

What kind of world this one belongs to? $$\int_0^{\infty} \frac{\text{PolyLog}^{(1,0)}(1,-x)}{1+x^2} \, dx$$ $$=\frac{3}{8} \pi \log ^2(2)-\frac{1}{4} \gamma \pi \log (2)-\frac{1}{16} \text{StieltjesGamma}^{(0,1)}\left(1,\frac{1}{4}\right)+\frac{1}{16} \text{StieltjesGamma}^{(0,1)}\left(1,\frac{3}{4}\right)+2\log (2) G$$

I just derived it.

robjohn

@Chris'ssistheartist Finding situations where those special functions arise is difficult to think of.

Khallil

Does anything in particular happen if a user who asks a question never accepts an answer?

Chris's sis the artist

@robjohn I think so.

robjohn

@Khallil They get looked at sideways.

Mike Miller

if they reaches ten questions without accepted answers, a group of people come to their house and say mean things to them

3

robjohn

11:33 PM

@Khallil actually, people may not answer their questions if they don't accept any answers.

Khallil

@MikeMiller's alternative is somehow more satisfying ...

Mike Miller

no idea how stackexchange funds this policy but it's not my business

robjohn

@MikeMiller I got a neat mug for something like that.

Mike Miller

;)

Chris's sis the artist

11:51 PM

@robjohn $$\int_0^{\infty} \frac{\text{PolyLog}^{(2,0)}(1,-x)}{1+x^2} \, dx$$
$$=\frac{1}{48} \left(3 \zeta ^{(2,0)}\left(2,\frac{3}{4}\right)-3 \zeta ^{(2,0)}\left(2,\frac{1}{4}\right)+6 (\gamma -1+\log (4)) \zeta ^{(1,0)}\left(2,\frac{1}{4}\right)-6 (\gamma -1+\log (4)) \zeta ^{(1,0)}\left(2,\frac{3}{4}\right)-6 \gamma \text{StieltjesGamma}^{(0,1)}\left(1,\frac{1}{4}\right)+6 \gamma \text{StieltjesGamma}^{(0,1)}\left(1,\frac{3}{4}\right)+96 G \left(\gamma -2 \log ^2(2)+\log (4)\right)+24 \pi \gamma _1 \log (2)-28 \pi \log ^3(2)+36 \gamma \pi \log ^2(2)\right)$$

robjohn

@Chris'ssistheartist Well, that's just swell ;-)

Chris's sis the artist

:D

@robjohn There is an amazing thing. I can calculate the whole family of these integrals no matter the order of the partial derivative of the polylogarithm.

robjohn

@Chris'ssistheartist I don't think I could even tell you what $\mathrm{PolyLog}^{(2,0)}(1,-x)$ even means, much less integrate it over $1+x^2$

Chris's sis the artist

@robjohn Well, consider polylog(s,p), differentiate twice with respect to s and then let $s\to 1$.

@robjohn That $0$ from the other side means that with respect to the other variable there is no differentiation.

robjohn

@Chris'ssistheartist Yes, the various things in the parens are partial derivatives... however I only know PolyLog for integer values of the first argument, I would have to think hard on how to generalize it to be able to take derivatives.

Chris's sis the artist

11:58 PM

@robjohn Why hard? It's simple, consider $$\sum_{k\ge1} \frac{x^k}{k^s}$$

robjohn

@Chris'ssistheartist Oh, yeah... $$\sum_{k=1}^\infty\frac{x^k}{k^s}$$

Chris's sis the artist

@robjohn Precisely.

Transcript for

$Mathematics$ Mathematics