Mathematics - 2016-11-11 (page 6 of 7)

Ramanujan

10:00

@Brody find range

saturatedexpo

@Brody from which show is that?

@Brody pic-upload.de/view-32095947/Picture001.jpg.html

bigger image ;)

Brody

@saturatedexpo it's called "Community" I think. never saw it myself

SteamyRoot

Yeah, it's Community

Ramanujan

@DHMO got?

This is hint given

saturatedexpo

@Brody ah llol the last line isnt even right, but the 4 at the end wouldnt change the point.

also this method sucks if they are constants involved

(unknown constants)

Brody

10:06

@saturatedexpo the third equation down looks wrong

Oh nevermind, you're working backwards?

saturatedexpo

mmh

Arrow

@BalarkaSen still trying to cheat a little. The transition matrix between the standard basis to the spherical frame should (I think) have orthogonal columns since the latter is an orthogonal basis. This transition matrix is the Jacobian at a given point, so doesn't this prove the orthogonality of the columns?

Brody

@saturatedexpo You wrote the $n+1$ form in the second line and went from there?

saturatedexpo

@Brody i found the error

multiplied out wrong

$(n+2)(2n+1)=(2n^2+n+4n+2)$ and not what stands there

Brody

@saturatedexpo You want to show assuming the $n$-th case implies the $(n+1)$-th case is true

What did you conclude about the $(n+1)$-th case in your work?

saturatedexpo

10:11

i made of $4^{n+1}=4*4^n$ and inserted

thats the whole point or not?

then the RHS times 4 of the assumtion

and then try to show that 4*assumtion is less then the final term

but forget to check in this picture, its a mess anyways^^

Perturbative

Hey everyone, quick question on convention, when denoting subsets do you guys mainly use $\subset$ notation or $subseteq$ notation?

SteamyRoot

I use $\subseteq$

saturatedexpo

@Perturbative both is valid and \subset is just more seen

SteamyRoot

The problem is that $\subset$ usually means $\subseteq$ and very sometimes $\subsetneq$

saturatedexpo

if A is strictly lesser then B i for mine will use $\subset$

SteamyRoot

10:15

So in that sense I find it safer to use $\subseteq$

Brody

for (not necessarily proper) subset of, I use $\subseteq$

saturatedexpo

for most things

it really doesnt matter

like the whole 0 being positive thing

Brody

if you're nervous readers might misunderstand, you can briefly remark your notation to clarify $[A\subseteq B]\Longleftrightarrow [(A\subsetneq B)\lor (A=B)]$

@saturatedexpo Assume $\exists n>1,\; 4^n\cdot(n!)^2 < (2n)!\cdot (n+1)$

saturatedexpo

@Brody now i finally understand where the (n+1) in the nominator comes from -.-

Brody

@saturatedexpo What you do to one side, you do to the other!

saturatedexpo

10:22

he made $(n!)^2$ to $((n+1)!)^2$. therefore he had to multiply both nominator and denomintor by $(n+1)^2$

but since he already has one (n+1) for "free" of the LHS

...

no

Astyx

What is the issue here ?

saturatedexpo

i dont understand

Brody

@saturatedexpo Thats where $(n+1)^3$ came from, but the $n+1$ itself came from earlier

We have $\dfrac{4^n}{n+1}<\dfrac{(2n)!}{(n!)^2}$

saturatedexpo

@Astyx i dont understand this solution from super dooper math.stackexchange.com/questions/2008867/…

Brody

We want to make it look like the $(n+1)$-th case

So we go ahead and transform the LHS

saturatedexpo

10:25

how so?

times 4

Brody

$\dfrac{4^n}{n+1}\cdot\dfrac{4(n+1)}{n+2}=\dfrac{4^{n+1}}{n+2}$

that's the LHS, now it looks like the $(n+1)$ form we need

Astyx

Wow this looks overly complicated

Brody

but we multiplied it by $\dfrac{4(n+1)}{n+2}$, we have to replicate that on the RHS

Astyx

Do you want to understand the proof or specifically what he wrote ?

saturatedexpo

@Brody for 4^{n+1} also the 4

@Astyx i want to understand the steps he did

Brody

10:27

fixed, thanks

Astyx

And which ones don't you understand specifically ?

saturatedexpo

the first haha

but brody got it i think :D

Brody

@saturatedexpo We multiply both sides by $\dfrac{4(n+1)}{n+2}$. That's where those extra terms, including the $n+1$ factor, come from

saturatedexpo

and now we got to show

that the RHS is smaller then the RHS?

Brody

Yes, now we work with the RHS side to drive the proof

saturatedexpo

10:29

wow i just didn get it haha

finally

Balarka Sen

@Arrow I don't understand what you're trying to do. I mean, just computing the Jacobian shows they have orthogonal columns and you can do it on your head.

Astyx

@saturatedexpo See that $${{4^{n+1}}\over n+2} = {4^n\over n+1}{4(n+1)\over n+2}$$

saturatedexpo

yes we already worked that out :D

on the long way i guess xd

Astyx

Then use induction hypothesis

Balarka Sen

There's no need to compute the basechange matrix from the standard base to the spherical one.

That's the harder way around.

saturatedexpo

10:34

ok now, why does he use: "so $\frac{4(n+1)^3}{(n+2)(2n+1)(2n+2)} <1$" why now the 1?

Brody

$$\dfrac{4^{n+1}}{n+2}<\dfrac{(2n)!}{(n!)^2}\cdot\dfrac{4(n+1)}{n+2}=\dfrac{(2n+‌2)!}{((n+1)!)^2}\cdot\underbrace{\dfrac{4(n+1)^3}{(n+2)(2n+1)(2n+2)}}_{\text{less than } 1}<\dfrac{(2n+2)!}{((n+1)!)^2}$$

The quotient is less than 1. You can check that the difference in the polynomials is positive for $n>1$ @saturatedexpo

Astyx

You could also see that $(2n+2) = 2(n+1)$ thus $${4(n+1)^3 \over (n+2)(2n+1)(2n+2)} = {2(n+1)^2 \over (n+2)(2n+1)}$$

Brody

^ better

saturatedexpo

wow i mean its cool, but not that intuitive for me

Brody

@saturatedexpo There's a method to it

Astyx

10:38

And expanding the numerator and denominator, you get $${2(n+1)^2\over(n+2)(2n+1)} = {n^2+2n+1\over n^2 + {3\over2}n + 1} \ge 1$$

saturatedexpo

but 3/2<2 and therefore the fraction is >1?

wait

how does there a fraction get anyways

Brody

@Astyx huh?

>.>

Astyx

Wait ..

saturatedexpo

$2n^2+n+4n+3$ is the denominator

Brody

the middle term has coefficient $\frac{5}{2}$ in the denominator

Astyx

10:42

Oh yeah that's it

But I can't edit any more ...

saturatedexpo

yeah np ;)

ok

Brody

${2(n+1)^2\over(n+2)(2n+1)} = {n^2+2n+1\over n^2 + {5\over2}n + 1} \le 1$

saturatedexpo

but why we now take the fact that this is <1?

Brody

I think it's strict for $n>1$

saturatedexpo

ah ok

so for n=1 we have to show it by hand (which is trivial)

Brody

10:43

Move the denominator over, and take the difference of like terms

Astyx

I personally feel dealing with this problems with fractions is a lot of hard work for little benefit. It's much more natural to show that $4^n (n!)^2 \lt (2n)!(n+1)$

saturatedexpo

yes

i tried, but no fruits

Brody

Factorials and fractions go well together though

Though I'm sure the other way might be the same, if not easier

Astyx

It would probably be the same

saturatedexpo

@Astyx thanks for the edit

Brody

10:46

@Astyx Good editing over there

lol^

Balarka Sen

@Brody Isn't it midnight in your side of the world?

Brody

@BalarkaSen oh it's way past midnight...

Astyx

Haha I personnally couldn't read a thing, that's why I edited :)

Balarka Sen

I guessed.

Brody

I'm on Eastern time. Same as New York City and Washington, DC

Balarka Sen

10:47

Ah.

Brody

Coming up on 6 am

Balarka Sen

yikes lol get some sleep

Brody

What's the time over there if I might ask?

saturatedexpo

but we have to strictly prove for n=0 and n=1 for a complete prove or?

Balarka Sen

about 4 PM

Brody

10:48

@saturatedexpo No. We must prove the inequality for $n>1$

So we consider $n=2,3,4,\ldots $

saturatedexpo

@Brody original assignment at the bottom

reh.math.uni-duesseldorf.de/~koehler/Lehre/2016-17/Uebung04.pdf

Brody

(hopefully not extending to non-integral $n$ lol)

saturatedexpo

i mean we must show the whole thing for those 2 cases

because the prove only shows for n>1

Brody

@saturatedexpo $\mathbb{N}_0=\{0,1,2,\ldots \}$?

saturatedexpo

yes

my bad!

Brody

10:50

then yes, those must be considered with the non-strict inequality $\le$ as posted

saturatedexpo

did not include that in the question ;)

Brody

but you can just show the $n=0$ case true and use induction to do the rest

saturatedexpo

so at <1 we might as well post $\leq 1$

Brody

yep, think so

saturatedexpo

mmh better not

because the whole thing builds on that being strictly less then one

Balarka Sen

10:53

@Brody I'm going to give you some homework for the next time we talk : learn and understand cross product and it's geometric meaning from chapter 1, section 5. Then we can talk about volume of surfaces and it's relation with Riemannian metrics.

Brody

@saturatedexpo doesn't matter, the proof itself needs a non-strict inequality now, so we can use that

$$\dfrac{4^{n+1}}{n+2}\le \dfrac{(2n)!}{(n!)^2}\cdot\dfrac{4(n+1)}{n+2}=\dfrac{(2n+‌2)!}{((n+1)!)^2}\cdot‌\underbrace{\dfrac{4(n+1)^3}{(n+2)(2n+1)(2n+2)}}_{\le 1}\le \dfrac{(2n+2)!}{((n+1)!)^2}$$

Nothing really changes, we proved what we needed to

@BalarkaSen Will do. Might take a couple days though

Balarka Sen

Sure, time isn't a factor. Also, chapter 3 section 5 contains a proof of equivalence of the two definitions of arclength. I think you'd be able to understand it just fine.

Feel free to not bother with any of these if you don't care though! I am just giving you "homeworks" to get you started and put something in the back of the mind as a motivation.

Brody

I feel like my brain continually atrophies without proper stimulation, so that's all welcomed. I'm also very endeared by your care and effort to spur another's growth in math

Astyx

@saturatedexpo Some very nice proofs here

Balarka Sen

@Brody Yes, mine does that too. I think multivariable calculus, with appropriate care, can be very interesting 'cause it's the gateway drug to lots and lots of interesting mathematics. Ted does that a lot throughout the book, but since you already know some of the multivariable calc (although in a not very motivated way, as I gather), I think I can tell you a bit more.

And sure, I like to talk math with people who like math.

saturatedexpo

11:11

how do you use \underbrace{ properly? i want to have something under the brace (a small explanation)

Balarka Sen

$\underbrace{some text}_{text}$

oops

Brody

@BalarkaSen Multiple variables definitely make for more complicated math, but one can't ignore them.

Astyx

meta.math.stackexchange.com/questions/5020/… answer titled additionnal decorations

Balarka Sen

@Brody Not complicated, beautiful! Differential topology/geometry, Morse theory, differential forms/de Rham stuff, more advanced analysis, etc.

saturatedexpo

i dont feel envy for my corrector haha

the fuck is so small cant even read it on pc

Balarka Sen

11:18

Ted also has a lecture on Fourier theory that I like a lot (it's not in his book).

Brody

Ted seems to know a lot

saturatedexpo

@Astyx that goes to my favs ;)

Balarka Sen

@Brody Yep, he sure well does!

Brody

@BalarkaSen I also wish the higher hyper-operations were better understood, even though they're not quite useful

saturatedexpo

hard to imagine that Ted thinks of someone that way :s

and even harder to imagine that the "bigger Ted" thinks of someone the way Ted thinks of him :O

Brody

11:23

@saturatedexpo Are you referring to @BalarkaSen as Ted?

Balarka Sen

@Brody I am not quite the tetration enthusiast here :)

@saturated Huh?

saturatedexpo

mmh, i dont think Balarka speeks of himself in the third person^^

Brody

@BalarkaSen suppose I wish mathematics were more complete in general. We have solid theorems and understanding of various algebras, various variables and dimensions, various hyper-operations, altogether working

saturatedexpo

@Brody Gödel

Brody

@saturatedexpo Gödel what?

Balarka Sen

11:26

@Brody That it's not is why it's fun!

saturatedexpo

it is already pretty big, to the point where we got statements that are true but cant be proven. So what is exactly your wish? That all subfields of Math work together?

i mean that taking a course in set theory makes you pretty much ready for everything?

Brody

@BalarkaSen Actually, there's some leisure in looking at the established repertoire and marveling at the objects and relationships we already understand. But that's just one aspect of the job

Balarka Sen

That's true. Well put.

Brody

There's also lots of adventure in exploring new territory, but that takes time and effort (which sometimes does not feel so leisure-full)

saturatedexpo

@Brody Could one explore if he doesnt feel happy about it? :s

meh im done

i take a pause :) till later and thanks @Astyx and @Brody ;)

Astyx

11:32

My pleasure !

Brody

@saturatedexpo Think of sailing a ship to the New World. You have the shoreline in mind, but miles and miles of harsh sea life before you ever might reach it (like working an unsolved problem)

No problem :)

saturatedexpo

@Brody that analogy makes sense to me since Columbus had serious problems i guess with "point of no return" stuff

basicly suicide mission for glory? in math more like boredommission for knowledge^^

Brody

@saturatedexpo I'm sure some mathematicians work in part for the glory too

saturatedexpo

@Brody at least russians don't (well for glory maybe, but not for money)

is there a good lecture about ordered fields?

Balarka Sen

I think most people don't do math for fame or money.

Qwerp-Derp

11:37

Hello

saturatedexpo

@Qwerp-Derp welcome aboard :)

Qwerp-Derp

Uhhh can I request a formula for a thing?

I don't have any idea where to start

Astyx

Don't ask to ask, just ask :)

saturatedexpo

well, post your problem :)

Brody

@BalarkaSen Albeit, corporate and government mathematicians generally get paid pretty well here

But that's a different 'mathematician' arguably

Qwerp-Derp

11:39

Uhhh

So n people are in a ring

(This is not the Josephus problem)

So 4 people for example, 1 2 3 4

Balarka Sen

I used the sneak-word "most", not "all" :)

Qwerp-Derp

The 2nd person is removed

so it becomes 1 3 4

the person after 2 is 3

Astyx

^

Qwerp-Derp

And then because 2 is removed, we move forward 2 for the next person to remove

So the next person removed is 1 (3 -> 4 -> 1, moved forward 2)

And so on and so forth

saturatedexpo

isnt this equivalent to joesphus (i mean np, just saying)

Qwerp-Derp

11:41

Until one person is left, in this case 3

Brody

@BalarkaSen I often think about those career prospects for the income. It's a trade-off between academic pursuit and fiscal comfort

Qwerp-Derp

@saturatedexpo Is it?

saturatedexpo

just a different ruleset

Qwerp-Derp

True

It's really similar to Josephus

Astyx

So if I understand correctly, if you have 10 people, you would remove 2,5,1,4,... ?

Qwerp-Derp

11:42

@Astyx Correct

I don't seem to find any patterns with the thing either

saturatedexpo

is the next person to be removed always the "p" person where p=last person removed?

Balarka Sen

@Brody Sure, I am not saying that choice is "wrong" or anything.

Qwerp-Derp

@saturatedexpo Yup, instead of Josephus's p=1

Balarka Sen

It's perfectly fine if one wants to secure their financial position.

saturatedexpo

try find a pattern, and look for shinies i guess?

try some circs maybe 1 to 5?

Astyx

11:44

@Qwerp-Derp And if you have three persons, you would remove 2 then 3 right ?

saturatedexpo

also what is $p_0$?

Qwerp-Derp

@saturatedexpo I used matplotlib the formula for 1 - 10,000 I think

@saturatedexpo Only positives

Brody

@BalarkaSen I know. But I personally do seriously consider certain paths just for the money, even if to the potential diminution of learning rich, beautiful math

saturatedexpo

oh, so p_0 is unknown and n too?

Qwerp-Derp

@Astyx Yup

@saturatedexpo wait what n?

saturatedexpo

11:45

n=number of people at start

Astyx

@Brody And also just for the fame

Qwerp-Derp

Ah

Yup

saturatedexpo

mmh, such a formular would be a generalization of josephus or not?

Qwerp-Derp

IDK

Astyx

I don't think so

Qwerp-Derp

11:47

I think it's probably too random

Balarka Sen

@Brody Well, it's your decision after all. But be sure to always do what makes you happy!

Astyx

The rule depends on the eliminated person thus the induction formula would be more complicated

Qwerp-Derp

Would a question asking for a formula on Math.SE be downvoted?

Astyx

Not if you show your progress

(well one can never tell)

saturatedexpo

one upvote from me at least :P

Brody

11:48

@BalarkaSen Problem is feeling torn between the two. I'm probably not the only prospective who's dealt with this though

Qwerp-Derp

I honestly dont have much progress

I just have a plot and that's about it

(And a Python script)

saturatedexpo

is p_0 random or do you want a formular that calculates the person surviving f(p,n)=??

Qwerp-Derp

I kinda want a formula that calculates the person surviving (f(p))

saturatedexpo

yeah but that depends on the number of people too

Qwerp-Derp

I don't think there's any use for n, because we know that the person eliminated is always 2 at the start, and then it goes on from there

saturatedexpo

11:51

if p_0=1000, and the number of people $2^{100}$, what does it say?

Qwerp-Derp

Wait so we're starting with the 1001st person getting eliminated?

saturatedexpo

but we can't wait, because we have to (in josephus shoes) say at the beginning who has to start or not?

Balarka Sen

@Brody I don't have any useful suggestions for you because I haven't faced this situation (yet), unfortunately.

Qwerp-Derp

Oh

saturatedexpo

so the function has to (at first, we can probably express p_0 in n terms or other way around) be dependant of 2 variables, p_0 and n

Brody

11:56

@BalarkaSen That's alright. And perhaps better you never have money muddle your art

Qwerp-Derp

p_0 is where the "removing" process actually starts, right?

Brody

Isn't this a specific case of the Josephus problem?

Astyx

I don't think so

Brody

The initial point is equal to the number skipped

saturatedexpo

@Qwerp-Derp yes, if p_0=5, then the 5th person gets removed, or 4 further then josephus

at least how i understand it

well

makes not much sense because if p_0=1 joesephus gets killed always

so one might say p_0 means the p+1th person gets killed

Astyx

12:07

@Qwerp-Derp Could you give the winning positions for the first 20 tries please ?

saturatedexpo

12:21

@Brody money is an illusion. but math is forever there. matter of taste probably?

Danu

12:53

@TedShifrin Thanks, that makes sense.

Tal Shani

13:55

I need example of a probability space (Ω,β,p) such that Ω is an infinite sample space but |β| < ∞.
someone has any idea?

Astyx

@TalShani $\Omega = \Bbb N$ and $\beta = \{\emptyset, \Bbb N\}$ ?

Basically $\beta = \{\emptyset, \Omega\}$ will always work

Tal Shani

but then beta is infinite

Astyx

No, $|\beta| = 2 here$

One of the elements of $\beta$ is infinite (since it's $\Bbb N$)

But this is unavoidable since $\Omega \in \beta$ by definition of $\beta$

Tal Shani

and what do you think about this one?

example of a probability space (Ω, β, p) such that |β| = 3

is that possible?

Astyx

I'd be tempted to say no

Tal Shani

14:09

I'm trying to think what not

why*

Astyx

Because $\beta$ must be stable under complementation (not sure about the english term)

Tal Shani

one of the element is omega the second is the emptyset

Astyx

Let $\beta = \{\emptyset, \Omega, A\}$

Then $\overline A \in \beta$

Tal Shani

ok but what is

a

A

Astyx

Your third element in $\beta$

Tal Shani

14:11

but we said that is impossible

Astyx

Yes I'm trying to prove it to you :)

Tal Shani

haha thanks

Astyx

Imagine it is possible, then $\beta = \{\emptyset, \Omega, A\}$

Tal Shani

ok

Astyx

Thus $\overline A \in \beta$

However $\overline A \ne A$ (otherwise $\Omega = \emptyset$ and that mean $|\beta| = 1$)

Tal Shani

14:14

i think i know how to prove that

the same like you said

Astyx

If $\overline A = \emptyset$ then $A = \Omega$ thus $|\beta| = 2$

If $\overline A = \Omega$ then $A = \emptyset$ thus $|\beta| = 2$

Tal Shani

bc A is closed under complements

Astyx

Therefore $|\beta| \ne 3$

Yup

Tal Shani

and this one:

an example of a probability space (Ω,β,p) such that |β| is an odd integer or prove that no such a probability space can exist.

i think it must be even right?

Astyx

Yes

Do you know about equivalence relations ?

It would be suited in this case

Tal Shani

14:16

not really

Astyx

Mm then it's tricky

The idea is to say that you can group the elements of $\beta$ by pair consisting of an element and its complement

Can $\Omega$ be $\emptyset$ ? (I can't remember)

Tal Shani

i dont think so

so beta has to be even?

Astyx

Well then you can do what I'm telling you, group the elements of (the formal term is "partition") $\beta$ by pairs

And see that $|\beta| \in 2\Bbb N$

Tal Shani

exactly

but how can i prove that?

in formal way?

Astyx

Well what I would do is introduce the binary relation $\sim$ such that $$\forall A,B \in \beta, A\sim B \iff B \in \{A, \overline A\}$$

See that it is an equivalence relation

Tal Shani

14:23

ok

Astyx

Parition $\beta$ with it.

Tal Shani

thank you so much!!

Astyx

And see that any equivalence set has cardinal 2

It's my pleasure !

Astyx

14:47

Would one of you have any lectures concerning binomial coefficient (in)equalities and efficient approaches towards them ?

Semiclassical

hi chat

Astyx

Hi

Semiclassical

what's the inequality you're aiming for again?

Astyx

Not one particular

I'm just looking for tricks

Semiclassical

ah.

can't say I have a good source in general

Astyx

14:50

Thanks any way :)

teadawg1337

15:34

It's been a while, and I don't remember. Do badges on this site award additional Reputation Points?

I do recall that badges take forever and a day to be rewarded, due to the automated process that is used

shai horowitz

nope

teadawg1337

Good

Akiva Weinberger

16:16

So this question of mine from last year finally got answered:

4

Q: Writing $\sqrt[\large3]2+\sqrt[\large3]4$ with nested roots

Let $C\subset\Bbb R$ be the smallest set containing $0$ and closed under whole number addition/subtraction, whole number exponents, and whole number roots. That is, for all $c\in C$ and $n\in\Bbb N$, we have $c\pm n\in C$, $c^n\in C$, and $c^{1/n}\in C$. We know that $\sqrt2+\sqrt3\in C$, since ...

mercio

you should celebrate and throw a party

your question reminds me of another one

that only accepted, I think, square root and adding an integer

and it had a half-dozen non-trivial examples of things

Akiva Weinberger

@mercio Care to share a link?

mercio

sadly I don't think it's in my favorites

after giving them a glance

42

Q: Converting sums of square-roots to nested square-roots

When solving different equations, I have realised, that some roots containing only arithmetic operations and square roots (4th, 8th roots too, because they can be represented using only square roots) can be converted to nested square roots form. Examples (these are roots of equations of 2nd, 4th,...

radicals nested-radicals

(i didn't glance hard enough)

teadawg1337

16:35

Would it be acceptable for me to accept an answer on this question I posted a while ago?

3

Q: Summing of factorials to produce perfect cubes

I was playing around with factorials the other day, and I realized that $4!+5!$ is a perfect square. Perplexed by this result, I started looking for other pairs of factorials that produce a perfect square when added together (unbeknownst to me, I had stumbled across a well-known open problem in n...

number-theory factorial

Or has it been too long?

shai horowitz

feel free to accept an answer iff you feel it deserves it

teadawg1337

The answer I want to accept doesn't fully answer the question, but I think it's the closest I'm gonna get

shai horowitz

lol its true

this site is self moderated what is or is not acceptable is in a state of flux with active users

Andrew Thompson

@TobiasKildetoft mathoverflow.net/questions/254453/… This was a good question imo.

shai horowitz

i doubt anyone would want to do anything unless you choose a really wrong answer and even then probably people wont care but it brings the question back to the active list

Gridley Quayle

16:52

Can someone give me a hint for iii)?

I am not sure how those two functions will help me find all the congruency classes

teadawg1337

17:19

When would it be reasonable to start a bounty on an unanswered question? I asked this question a week and a half ago, and it still has no answer. I've been stuck on the problem for months, but I'm thinking I should wait another month or two to start a bounty

8

Q: Closed form for $\int_0^e\mathrm{Li}_2(\ln{x})\,dx$?

Inspired by this question and this answer, I decided to investigate the family of integrals $$I(k)=\int_0^e\mathrm{Li}_k(\ln{x})\,dx,\tag{1}$$ where $\mathrm{Li}_k(z)$ represents the polylogarithm of order $k$ and argument $z$. $I(1)$ evaluates to $e\gamma$, but $I(2)$ has resisted my efforts (wh...

calculus integration definite-integrals special-functions closed-form

Although, a bounty probably won't really do much anyway

teadawg1337

17:49

The more I look at it, the more I question whether a better form is even possible

saturatedexpo

what does double dollar mean in latex?

bigger equations?

i mean dd some text dd

teadawg1337

It displays the contained text on its own line

It also makes some symbols bigger and easier to read

saturatedexpo

thats good :)

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