Mathematics - 2014-12-20 (page 2 of 8)

@math110 OK, I'll have a look if I have done with answering one of problem here

@math110 (+1)

A: How find this sum closed form $I=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos{(2kx)}x^{m-1}e^{-ax}dx$

11:58 AM

@math110 That one is a simple application of the polygamma function. The higher order of $m$ are obtained by considering higher orders of polygamma function. It should flow naturally by considering the derivatives with respect to $a$ ....

@math110 is my answer good enough?

0

For $m=1$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^1}+\frac{1}{(a+2 i k)^1}\right)=\frac{1}{2} i \left(\psi ^{(0)}\left(1+\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1-\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1+\frac{i a}{2}\right)+\psi ^{(0)}\left(1-\frac{i a}{2}\right)\right)$$ For $m=2$ $$\sum _{k=...

math110

I guess have this but I think the AMM problem have closed form

mat.uniroma2.it/~tauraso/AMM/amm.html

hahahahahahahahaha

Look at that

mat.uniroma2.it/~tauraso/AMM/AMM11771.pdf

How about find an elementary solution to that limit? :-)))

No need for Stirling!

$$\lim_{n\to\infty} \sqrt[n]{(2n-1)!!}\left(\tan\left(\frac{\pi \sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)-1\right)$$

COOL!!! It's going to be the problem (better say, the limit) of the day to me!!!

12:35 PM

Done!!! Without pen and paper!!! :-)))

I'll explain in my proof why a solution without pen and paper is possible. Well, I made use of a celebre limit for computing it.

1:10 PM

@BalarkaSen my final attempt pal.

You got an answer, no, @skullpatrol ?

@Huy yes :(

What are you studying these days, @skullpatrol?

@huy a little bit of this , a little bit of that :-)

Care to specify, @skullpatrol?

1:22 PM

@Huy not really, if I discover anything interesting I'll let you know my friend.

@skullpatrol: You should study GR with me.

@Huy I'm still struggling with SR :(

@skullpatrol: Why?

@Huy I think I'm using the wrong book

@skullpatrol: You're using a book. :P

1:25 PM

@Huy relativity: the special and general theories

@skullpatrol: Never heard of it. I didn't read many books during university time.

@Huy google the title it is freely available

@skullpatrol: Are you studying in Germany or am I mistaking you for someone else?

@Huy this is how he wanted his theory taught to high schoolers

@skullpatrol: i.imgur.com/1ZcPEQX.jpg This is the only book about relativity I own. :P

1:30 PM

:D

@skullpatrol: Are you studying in Germany or am I mistaking you for someone else?

not me

Ok. Are you a high schooler, @skullpatrol?

no

Then why are you reading something you state is intended for high schoolers, @skullpatrol?

1:32 PM

@Huy the author suggests it

@skullpatrol: That doesn't really answer my question

@Huy I know, but I can't say much more...sorry

If you're actually trying to study SR with that text, imo it's way to much text without much content.

in general, I like to use the book closest to the source

@skullpatrol: Why? What advantages do you see in using a book "close to the source"?

1:40 PM

@Huy more content and less text

@skullpatrol: But your book for SR is the opposite.

@Huy have you read it?

@skullpatrol: I have had a look at it.

@Huy skimming over it and actually trying to read it are two different things my friend :-)

@skullpatrol: Okay. Just trying to give you some advice, but it's your time of course.

1:43 PM

@Huy np pal

thanks for the advice @Huy

:-)

2:04 PM

Well thanks for trying @skullpatrol.

@BalarkaSen np pal

:-)

@r9m I got naruto before you.

:P

I'm shell-shocked :D

@BalarkaSen :P good job ! ;)

how did you get fascinating ma'am, @r9m?

@BalarkaSen upvoting an answer with atleast 10k views and 25 upvotes (or sth like that :P)

2:12 PM

ah

2:33 PM

@Chris'ssis Have you seen this integral before $$\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$

Do you think it has a closed-form?

Morning @Mike.

Hi @robjohn, any progress for the integral problem that I asked you yesterday?

Do you think it has a closed-form?

I don't think I wanna get those hats which have anything to do with me downvoting some post :-(

Your loss, @r9m

@Venus You can use the paper I gave you yesterday.

2:36 PM

@Chris'ssis What paper?

@BalarkaSen I know ,... but I don't downvote :|

@Venus arxiv.org/pdf/1201.3393.pdf. Anyway, that one can be approached in more ways.

I downvote the posts that deserve downvotes.

Plus, every post of Dr. Sonnhard.

:P

I kid.

2

@Venus It should work wswapping a bit the integration order and using the differentiation under the integral sign.

I just finished a gem (on paper)!

Hallo @DanielFischer!!! :D I want to describe an algorithm that given an unsorted array $B$ that stores $m$ integers, and any integer number $y$, determines if there are two elements of the array of which the quotient is equal to $y$. The time complexity of the algorithm should be $O(m \log m)$.

Couldn't we sort tha array with the algorithm Heapify?

How could we check if there are two elements of the array of which the quotient is equal to y?

2:42 PM

@Chris'ssis I don't get it

^_^

what OP @r9m?

^_^

@Venus First integrate with respect to $y$ and $z$ ...

@evinda Sorting is good. Whether you use heap sort or any other algorithm with a guaranteed $O(m\log m)$ bound is your choice. Now think about how you would check whether a sorted array contains two elements with quotient $y$ in $O(m)$ time.

2:46 PM

morning

hi

Morning, @Mike.

@Venus Wait, I have far nicer ideas ... (I need to finish a proof I'm working on and then I'll try yours)

How's your Saturday, @DanielFischer?

@Chris'ssis OK

2:47 PM

@MikeMiller Rainy.

@Venus It should be $$\frac{1}{4} (-2 \gamma -5+2 \log (2 \pi ))$$ Well, I'd start with the integration by parts.

We had some rainstorms here, but they stopped a few days ago.

@BalarkaSen math.stackexchange.com/questions/1075355/… :)

Wish they hadn't - maybe the drought finally would have ended.

@Chris'ssis How do get it?

2:49 PM

@Venus I'll show you my proof after I finish the proof I'm working on.

@DanielFischer Do we have to check the numbers as pairs from the beginning of the array till the end ?

@Chris'ssis Take your time

@Mathy I see Mr. Gruber helped you, that method certainly is better than my tedious solution considering $4,5,6$ edges distinctly (really tedious. More tedious than that.)

Mr. Gruber, @Studentmath?

:P

I felt official, stop judgine me!

2:53 PM

@evinda That depends on how you interpret "check as pairs from the beginning till the end". You need not look at all pairs (of course, that would be $\Theta(m^2)$), and if you find a pair, you can stop.

@Mike I think someone passed you with the number of hats

I saw, @Studentmath

who?

But his new hats are easy ones, like 'post today', or 'answer five questions today'

I'll be eternally grateful if someone does not drop the Cheby-Shell on that OP ! .. I like the fact it has a binomial coeffn tag as primary ! ^_^

2:54 PM

@DanielFischer How can we know at which pairs we have to look? Should they satisfy some conditions?

@r9m sounds like an interesting question, which is it?

@Studentmath this :)

Also @Balarka @Mike I've some question I am struggling with since yesterday

OK

@evinda Yes. You want the quotient to be $y$. (By quotient, is the integer quotient $\left\lfloor \frac{a}{b}\right\rfloor$ meant or the rational quotient $\frac{a}{b}$? That decides whether you look for $yb \leqslant a < (y+1)b$ or $a = yb$.)

2:58 PM

Sheesh. So many of Dr. Sonnhard's posts flagged.

@BalarkaSen how do you see flags ? :o I thought moderators were the only one who could see them ..

@r9m here

So I would appreciate some hint. Anyhow, I have a continuous unto function $u:X\to Y$. Given that for every topological space $Z$ and every function $w:Y\to Z$, if $wu$ is continuous then $w$ is continuous, I want to show that $X/u$ is homeomorphic to $Y$. I've been thinking of setting $Z$ to be $X$, then I have that $f^{-1}$ is continuous, or $Z$ to be discrete space, but I have no clue how that helps me (either)

What is $X/u$?

@BalarkaSen ah low quality posts ! ;)

3:02 PM

@DanielFischer I assume that the rational quotient is meant.. :/

Quotient topological space of $X$ according to $u$

I.e. every class is all the $x$'s in $X$ so that $u(x_1)=u(x_2)$

oh ok

@BalarkaSen @r9m I'm just wondering till this day, how did he got those answers? He can answer every problems appears on timeline within minutes, probably less than 3 minutes.

I'm talking about Dr. S' answers

@Venus he's good at what he does

@Studentmath fantastic way of putting it !! :D

3:06 PM

:P

@Studentmath Some of his answers are good enough but mostly ... (no comment)

Oh no, @DanielFischer. The questions this morning are crummy again.

Sanity check : Isn't it true that every such group must also satisfy that every subgroup is norma;.

Well no I think the converse holds.

@Balarka I really have no clue how to build it up. I've been trying to think what do I need to know about $u$ that will allow me said isomorphism - but it's so general I don't think it's gonna help. I mean, I need the distinct $u(x)$'s to fit open groups in $Y$ according to sources in $X$ and vice-versa, but it's not helpful as I said..

I haven't looked at the question really @Studentmath.

Busy trying to get a hat.

3:11 PM

Not you too, brutos

Hi folks

Heya @ccorn

@DanielFischer Can you still cool off when dealing with so many raised flags from Dr. SG's answers if you're elected as a mod?

Hi @Balarka. BTW, I am glad Tito Piezas is posting again.

yeah, i've noticed that.

3:13 PM

why did he stop?

@Venus That sort of flag puts things in the review queues, moderators don't need to deal with them if enough people review ;)

no idea @Mike. i mailed him a few months ago, and he replied.

@MikeMiller I don't know. It just seemed so to me

so i guess he had been taking a break

@DanielFischer Ohh, lucky you then

3:15 PM

@DanielFischer Is it possible to know which are candidate elements, before looking at each pair? :/

Titus comes up with interesting questions often.

@evinda That doesn't have an important impact, it just means we write the test as a < y*b in the loop and check for a == y*b, otherwise we'd test a/b < y and check a/b == y.

by Titus you mean Tito?

Yes, his full name is Titus Piezas III

ah

3:17 PM

It seems some mods undeleted an answer of mine I'd rather see deleted. It's about this question. Actually, it has two answers of mine, and I am quite sure to have deleted both. The other deleted one has undelete votes too, but as the question has changed, undeleting would be bad.

Originally, the question was as interesting as indicated by its title. I gave an answer, then found that it did not exactly answer the question (which seemed to not allow a nontrivial multiplier system), so I deleted it and gave an updated reference in the comments. The OP than included the criteria given in my early answer and changed the question into asking why one would need integrality of the exponents in there.

@ccorn Mods don't undelete anwers; this would have been done by the community. Anyway, can you not just delete again?

@evinda Well, due to the array being sorted, we can rule out a huge number of pairs without looking at them. Just in case: If the array can contain negative as well as positive numbers, we need to take care of that and treat the part with negative numbers and the one with positive numbers separately.

I gave an answer to the updated question too. That's the currently undeleted one. The answer is really un-enlightened and involves guessing; and the question it answers stems from a change that actually moved it out of my field of interest. I felt that it did not look good, particularly not in my answers list, and so I deleted it. Now I find it undeleted. It might be stupid to enter another delete-undelete cycle. Suggestions?

Q: does reopening a questio bump it?

@MikeMiller I suppose only a privileged community can see deleted posts?

3:22 PM

I got my 20th hat ! Where's my Aztec Crown ?! :(

@ccorn Since it was undeleted by ordinary 10k users, you can delete it yourself. You should then edit in a note "I'd rather this be deleted because ...". Another option is to flag for moderator attention and ask for the answer to be disassociated from your account [hmm, or does one need community team involvement for that and hence use the "contact us" link?].

@DanielFischer OK, doing the edit+delete

@ccorn 10k+.

robjohn

@Venus I have it down to a sum of residues, but I don't know about computing that sum yet.

@ccorn If the answer is not really bad, just not up to your personal standards, disassociating may be preferable.

3:23 PM

@DanielFischer trying disassociaton, thanks

@robjohn To disassociate a post from one's account, does one flag for moderator attention or contact SE?

(Concerning ccorn's enquiry above)

Thinking... no. I'd rather explain in an edit and leave it undeleted. Perhaps I can avoid bumps by framing it between a delete and an undelete.

@DanielFischer I think that the array could also contain negative numbers.. How could we rule out a huge number of pairs? With what criterion?

@robjohn It seems it's a joke because I found this interesting fact $$ \int_{-1}^{1}\frac{\mathrm dx}{\left ( x^2+1\right )\left ( e^x+1\right )}=\frac{\pi}{4} $$

robjohn

@Venus That is a lot easier since the domain is symmetric and $\frac1{e^x+1}+\frac1{e^{-x}+1}=1$

3:27 PM

@robjohn Indeed, but I doubt the previous one has a closed form

@Venus Seems you are chris's sis? (Have not been following the chat for some time.)

@ccorn No, I'm her sister. So, I'm Chris'ssis's sister :D

@ccorn Nope, two different people

And good morning everybody

@teadawg1337 whatever :-)

@ccorn :(

3:29 PM

@robjohn Yes, we can use substitution $x=-x$ and add the two integrals

robjohn

@Venus indeed.

Oh goodness, someone's asking for help with their dissertation....

@evinda To get the idea, suppose all entries are positive. You have a $y > 1$ (the case $y = 1$ is clear), and look at two array elements A[k] and A[r] with r > k. Where do you look next if A[r] < y*A[k]?

@MikeMiller If $G$ is a nonabelian group with every subgroup normal, isn't it true that there exists a nontrivial subgroup $H$ of $G$ such that every nontrivial subgroup of $G$ contains $H$?

@BalarkaSen Consider your favorite non-cyclic abelian group.

3:32 PM

@robjohn could you please look at this page and tell me if 0 has a representation in scientific notation?

cheater

:P

ain't that the truth

oh dear

@DanielFischer I love The Doors!!!! My dad got me into 60's and 70's music in middle school and I still haven't grown out of it! Although, my musical taste is currently more diverse than it was back then...

I digress.

3:38 PM

music is waste

@DanielFischer So do we have to find the mid of the array and if A[mid] < yA[mid+1], we will check if $A[p] == yA[p+1]$ for $p \in [mid+1,m-1]$ and elsewhise we check if $A[p] == y*A[p+1]$ for $p \in [0, mid-1 ]$ ?

@evinda No. If we have negative and positive entries, we need to find $0$ or the position of the first positive entry. If we have only positive entries, we begin at the start, and if only negative entries, we begin at the end.

janos

hey @MikeMiller, do you have an idea why your Eureka hat was awarded to you on Mathematics and not on MSE like all the other people I see?

@janos I talked about how to get Chameleon in this chatroom, IIRC. I assume the folks in charge of Eureka found me.

If only Abby Hairboat would post on MSE, I could have them all here... :)

janos

I probably shouldn't tell you this, but.... if you send Abby a plausible question, he'll probably kindly post it here ;-)

3:45 PM

@DanielFischer We can this position like that:

i=0;
while (i<m and B[i]<0){
i++;
}

right? How can we use this position? :/

hehehe... devilish

wonder how to get time lord

become a lord of time

you can get warm welcome by upvoting a post by new user, i think

i'll try it out

G.T.R

@MikeMiller I turned my cameleon upside down and didn't get a waffle hat. How come ?

3:48 PM

Oh no! I've become the local hat expert.

:P

@MikeMiller tell me how to get time lord

No.

Google it!

janos

:)

LeGrandDODOM

@MikeMiller so ?

Because that's not how you get the waffle hat!

You get it for eating five waffles on a Saturday.

Or wait...

3:50 PM

@evinda It gives you a split of the array into two parts. [Aside, I haven't mentioned it yet, if the desired quotient $y$ is negative, you do things a little differently; if $y = 0$, it basically comes out to checking whether $0$ is in the array but not all elements are $0$.] For $y > 0$, you look at both parts separately.

Q: Closed form of $\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$

@robjohn @Chris'ssis In case you're interested in answering it, I post the question on the main page

0

While trying to find several references to answer Pranav's problem, I encounter the following multiple integrals $$I=\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$ Question : Does the above integral have a closed form...

calculus integration definite-integrals improper-integrals closed-form

LeGrandDODOM

I got mixed up

@voldemort what's on your profile picture ?

yes, my guess works

i got warm welcome

@Venus I didn't give me the possibility of working on it.

@DanielRust Hey.

Daniel Rust

3:53 PM

@BalarkaSen hey

You should post your answer to the question and get my bounty, really.

@Chris'ssis I think it's better to post it on the main so that you can answer it. Besides, your work will be more valuable there than in chat room.

@Venus I don't wanna share all my work (I just finished the one below)

Daniel Rust

@BalarkaSen Ha, maybe I will. I'm pretty busy atm with christmas shopping though

@DanielFischer Ok.. But when we consider the two parts of the array, don't we have to look at the first element with the second, the second with the third and so on? So, won't the time complexity be again O(n^2)? Or am I wrong?

3:58 PM

@Chris'ssis Well, it's okay then. You may not answer it. :-)

@DanielF Can you guess what the OP is having trouble with here? I tried to be a little bit more explicit than the author of their book, but I doubt my answer will help much without knowing where they're actually stuck.

I shouldn't be the one to talk here either though ! :o ^^'