« first day (1451 days earlier)   

12:57 AM
I got a triangulation in 2d, I know what the convex hull is, but what is the affine hull? Is that the whole space?
 
 
2 hours later…
2:55 AM
 
 
3 hours later…
5:51 AM
Hello
 
6:21 AM
 
 
3 hours later…
9:37 AM
If $X$ is a metric space and $A$ is a subset of $X$, is the function $\rho(p)$, where $p\in X$ so well-known that $\rho $ need not be defined? My only guess is that $\rho(p)=1$ when $p\in A$ and $\rho(p)=0$ otherwise.
Sorry, replace $\rho$ with $\rho_A$
 
@TheSubstitute I would use $\chi_A$ for the characteristic aka indicator function. I would also explicitly call it that, and not just use the symbol without defining it. I don't think I've ever seen the letter $\rho$ used for the characteristic function.
 
10:09 AM
hi guys
i stuck in syupid problem
if $b^2x^2 + a^2y^2 =c^2$
and a,b,x,y,c are positive
find max of $x*y$
 
what have you tried?
 
well
b^2x^2+a^2y^2 is constant
so for maximum multipication value
b^2x^2=a^2y^2
so c^2=2*b^2x^2=2*a^2y^2
 
where are you getting that from?
 
i got there
x=$\frac {c\sqrt 2}{2b}$
y=$\frac {c\sqrt 2}{2a}$
 
where are you getting b^2x^2=a^2y^2 from?
 
10:14 AM
when xy is maximum
b^2x^2 * a^2y^2 is maximum
their sum is constant
 
and maximizing uv subject to u+v=const means setting u=v. (can you justify that, btw?)
 
yes
 
good, so it seems like you don't have any issues...
 
well i found maximum of xy is $\frac{c^2}{2ab}$
but answer key says its $\frac{c^2}{2a^2b^2}$
@blue where i made a mistake?
 
you didn't, the answer key is wrong
 
10:50 AM
If $E/F$ is normal, for each $e \in E$ must there exist an $f \in F[X]$ such that $e$ is a root of $f$?
 
normal implies algebraic
 
How does the proof of that go?
Actually no, I'll try to prove it first.
 
@Chris'ssis: I hope your work is going well. You haven't been on chat since Monday
I don't know if that will get through since she has been gone so long.
 
11:06 AM
@Alyosha the splitting field of a family of polynomials is obtained by adjoining all of their roots. adjoining roots yields an algebraic extension (no matter how many you adjoined).
 
@robjohn I seem to recall her saying she was going to move...
 
@skullpatrol Ah... I didn't see that in her last chat comments. Perhaps I need to look further.
 
@robjohn I think she mentioned it quite awhile ago in passing...
 
@skullpatrol It's definitely back further than that log.
 
@robjohn it was just a hint too
not a declaration.
 
11:21 AM
@skullpatrol She is still working on main
 
@robjohn Orly?
something may have happened
 
Her profile says about an hour ago.
 
@robjohn maybe leave a message on one of her questions/answers
 
@skullpatrol The only controversial thing I noticed was that she was discussing what seemed to be religious beliefs with some others. That can be messy on the internet with a lot of varied beliefs.
@skullpatrol I was thinking of doing that.
 
r9m
Jun 26 at 11:57, by Chris's sis
@r9m Well, the bad thing is that no one here likes to attend things I like. In the future I plan to move to Timisoara city.
 
11:25 AM
@r9m nice work
 
r9m
I remembered the name of the city coz I googled it right after she said she wished to move there .. that made the search easier ! :)
 
@r9m I've always attempted her problems. I found them interesting. Perhaps she wants more people to find them interesting.
@r9m that would indeed :-)
@r9m However, she is still on the site, just avoiding chat.
 
r9m
@robjohn ya .. I saw that :!!
@robjohn definitely ! .. she's writing a book after all !!
 
27 mins ago, by robjohn
@Chris'ssis: I hope your work is going well. You haven't been on chat since Monday
@r9m That's why I wrote that
 
r9m
@robjohn :-)
 
11:32 AM
We miss you on chat :-( I hope your work is going well. — robjohn ♦ 19 secs ago
 
Hello
 
@VibhavPant Hey
@Alyosha There is no normal transcendental extension.
 
hi @Balarka
 
Please come back @Chris'ssis :-)
 
@skullpatrol What happened to her?
 
11:38 AM
dunno
 
Come to think of it, I haven't seen her for quite some days.
 
r9m
@robjohn :D !! nice !!
 
@VibhavPant What problems have you been practicing lately?
For the entrance exam?
 
@Balarka continuity
@Balarka I started with polynomial approx. to functions today
 
Ah.
 
11:51 AM
The lack of a pattern in CMI's entrance confuses me, so I dont know what problems to practise
and I try to do everything then
Someone recommended to Tom Apostol's Calculus, which till now seems to be a pretty good book
 
Apostol is good for hard analysis.
For something softer, I'd go for Rudin.
 
@BalarkaSen Im able to do questions in it though
 
r9m
@VibhavPant oh! there is a pattern ! .. my advice is to go through the olympiad books they recommended ! .. 80% of the questions are framed keeping olympiad problems in mind !
 
@r9m Im recently did number theory from "Challenge and Thrill of Pre-College mathematics", and I plan to finish geometry and combinatorics soon
 
@r9m Such pity.
 
r9m
11:54 AM
@BalarkaSen what's pityful about that ?
@VibhavPant ! good :D (I guess by 'doing' you mean 'exhausting all contents' :-) )
 
yeah
sure
 
@r9m Exams are for testing your skills and understanding. Olympiad problems are for testing your tricks.
 
r9m
@Vibhav In case you don't have it, get yourself the TOMATO (Test of mathematics in 10+2 level, published by ISI, Kolkata) .. its pretty useful :)
 
Now wonder guys from CMI learning galois reps can't classify groups of order 6
 
r9m
@BalarkaSen hehe :P .. where did you hear about that ?
 
11:59 AM
@r9m My professor. He is in contact with many people in Chennai.
 
r9m
@BalarkaSen :P .. what can we do .. we are asked to classify groups of order 72 in exam :P
 
@r9m 2^3 * 3^2.
 
r9m
good for you
how many are there ?
 
@r9m I haven't counted yet. Let me do that.
 
r9m
that was not my point :P .. its a good exercise for home/assignments .. but what the hell is the meaning of putting that as an exam Q ? :O
 
12:03 PM
@r9m Any idea how good are the NCERT books?
 
@r9m well, they test your skills in problem solving, no?
 
r9m
@Vibhav I dunno' .. :|
 
oh, np
 
one of the basic stuff in group theory is to count how many subgroups there are of order something.
i believe you know Sylow?
3^2. # 3-sylow is one of 1, 2, 2^2 or 2^3. 2 and 8 gets ruled out since they aren't 1 mod 3. So you have either 1 or 4.
 
r9m
@BalarkaSen nope .. its testing how fast you can write stuff down in exam hall .. there should be 14 subgrps to the naked eyes, proving that its actually 13 is slightly trickey business (especially when the prof hasn't proved the required lemmas in class .. so that one has to write it down in exam hall .. which IMHO is 'sad' )
 
12:07 PM
@r9m what book in group theory are you reading?
the tricks in there should be pretty standard.
 
r9m
@BalarkaSen D&F ..
 
that book.
it's only any good if you are solving problems. try Artin for the theory.
it's a lot better to explicitly write down the representation than using semidirect products.
@r9m can you tell me what happens if you have 4 3-sylow subgroups?
 
Oops, make "Polynomial approx. to functions" "differential equations" :)
 
r9m
wait .. it think it was 56 ! :P lol (not 72)
 
@r9m 56 is easier, heh.
but 72 shouldn't be a great deal too.
 
r9m
12:13 PM
it was 2nd sem ... I forgot what the q was :P
 
56 = 7 * 2^3
there must be either 1 or 8 7-sylow subgroups.
 
Greetings all
 
heya @Chris'ssis!
 
@BalarkaSen Hi :-)
 
r9m
@Chris'ssis :D (couldn't make the smile bolder)
 
12:15 PM
@r9m :-))))))
 
@Chris'ssis where have you been?
 
@BalarkaSen Well, I was around, but I needed to be alone for some time ... Now, I'm back. :-)
 
yeah, a self-ban sometimes is good.
 
@Chris'ssis GREEEEETINGS :DDD
 
@skullpatrol Hey, how are you doing? :-))))
 
12:18 PM
@Chris'ssis Fine thanks, how are you?
 
Guys, thanks for asking about me. It really made me feel better. :-)
 
We missed you.
 
OK, I'm fine, thanks.
 
@r9m The trick (pretty standard) is to note that 7-sylow subgroups intersect trivially.
 
@skullpatrol Glad to hear that, I missed you too, but I needed to be alone for a while.
 
12:20 PM
understandable
 
r9m
@BalarkaSen indeed :) .. not just that though :|
 
@r9m well, once you have one of the sylows to be normal your are pretty much done.
cause that'd lead you to a split short exact sequence, which makes things easier to classify.
 
@robjohn @r9m I created some nice stuff in the meantime.
 
r9m
ya son .. I know my group theory :P
 
$$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}\sum_{j=1}^{i} \frac{(-1)^{j+1}}{j}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{j+1}}{j}\right)=\frac{1}{2}\zeta(2)\log(2)+\frac{1}{4}\zeta(3)+\frac‌​{1}{6}\log^3(2)$$
The next step is to consider the following form
$$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}\sum_{j=1}^{i} \frac{(-1)^{j+1}}{j}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{j+1}}{j}\right)^2$$ and also this form $$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}\sum_{j=1}^{i} \frac{(-1)^{j+1}}{j}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{j+1}}{j}\right)^3$$
 
12:22 PM
@r9m if you are reading dummit and foote all the time, then you hardly know the fun behind group theory, even if you know group theory.
study short exact sequences.
idiotic definitions of semidirect products are the last things i have ever read from dummit-foote.
 
r9m
This is how @robjohn was since monday .. 'I worry about 'her' constantly' !!
 
@robjohn My work is OK. Still creative. :D
 
@r9m ?
 
@r9m lollll, come on! :-)
 
@Chris'ssis cool. I'll take a look.
 
12:27 PM
aha.
 
@robjohn hehe, glad you like it. How are you doing?
@r9m How are you doing? Did you create some other problems? :D
 
@Chris'ssis No, he is busy with his group theory =P
 
@BalarkaSen aaaaaaa, you steal the people working on integrals, series and limits ...
 
Heh.
 
:-)
 
r9m
12:29 PM
@Chris'ssis :P nothing exciting :| .. being lazy is what I do :P
 
@Chris'ssis pretty well. Answering some questions, dealing with some flags, getting some work done in between :-)
 
But in the whole, @r9m, I won't really bug you with group theory. I am not a group theorist. I have my own favorite galois theory.
 
@r9m :D
 
@Chris'ssis I probably spent too long on a problem that dealt with attractive and repulsive fixed points.
 
r9m
@BalarkaSen if you are trying to relieve me .. doing you are doing great ! .. good job :D
 
12:33 PM
@robjohn Which one?
 
@robjohn Interesting.
 
Talking of nested radicals, someone asked me to evaluate $$\sqrt{1 + 2\sqrt{3 + 4\sqrt{5 + 6\sqrt{7 + \cdots}}}}$$
He says that it was on an article about Ramanujan. So I guess it was done by Ramanujan after all.
 
r9m
@Chris'ssis @robjohn This looks interesting :)
 
@r9m Probably there ain't any closed form.
 
r9m
12:39 PM
@BalarkaSen I just saw it :) .. I still haven't formed an opinion yet :-)
 
@BalarkaSen @r9m then you should also see this one math.stackexchange.com/questions/455939/…
 
Galois theory is definitely not enough to prove that it is transcendental.
OEIS returns nothing about that constant. Darn.
 
@BalarkaSen I think I may have worked on that here at one time.
 
@robjohn Great! Can you give me the link?
Are you sure you are not confusing with $$\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + \cdots}}}}$$ ? @robjohn
 
@BalarkaSen I don't get a closed form, but I prove convergence in the $n=1$ case of this answer
 
12:53 PM
@robjohn I don't see how it relates to my question.
 
@BalarkaSen Oh.. I missed that there were multiplications in your question.
@BalarkaSen looking back, I saw the $n=1$ case of that question, which is not what you wrote.
 
@robjohn n = 1 case?
 
$\sqrt{1+\sqrt{2+\sqrt{3+\dots}}}$
 
Oh, that is well-known.
But, AFAIK, it has no closed form.
 
@BalarkaSen Yours converges, but I don't know if it has a closed form.
I did work on the Ramanujan nested radical in this answer
 
1:04 PM
If it could be proven that for certain sequences of complex numbers $A_n$ and $a_n$:
$$\prod _{n=2}^{\infty} \frac{1}{1-\frac{1}{A_n}} = \lim_{n\to \infty } \, \frac{A_n}{a_n}
$$

would it be progress?
 
r9m
@robjohn $\sqrt{1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}} = l_n$, then $\sqrt{n}(l - l_n)^{1/n}$ has a nice closed form :D .. where $l = \lim\limits_{n \to \infty} l_n = l$ (problem by @Chris'ssis :-) )
 
@r9m Yeah, this is nice. :D
 
@r9m I know I worked on some nested radical questions by Chris's sis. Did I miss that one?
 
r9m
Jul 8 at 15:03, by Chris's sis
@r9m Have you ever seen this one? $$\lim_{ n\to \infty} \sqrt{n}\left(l-\sqrt{ 1+ \sqrt{2 + \cdots +\sqrt{n}}}\right)^{1/n}$$ where $l$ is the limit of the nested radical.
 
http://math.stackexchange.com/questions/877841/complex-numbers-product-and-ratio-prove-this-relation
Is the question understandable?
 
r9m
1:10 PM
my touchpad is rebelling against my authority :P .. I need to teach it a lesson or two :P
@Chris'ssis How did you do it ? :D
 
@r9m Didn't you do it?
 
@r9m Oh! I was searching the main questions. I know that I worked on something with Chris's sis on main a while ago.
@r9m I missed that one.
 
r9m
@Chris'ssis ya :-) .. but you know how I'd like a Chris'ssis style solution :D
 
@r9m lol :-)
 
r9m
@robjohn It is nice :) .. I found $\sqrt{e}/2$
2 days ago, by r9m
AMM problem 4305 (April 1950) by H. F. Sandham $$\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} = \dfrac{17\pi^2}{360}$$
 
1:23 PM
@r9m That looks familiar...
 
$$\pi = \int_{-\infty }^{\infty } \frac{1}{x^2+1} \, dx$$
 
r9m
@robjohn How ? :D
 
@r9m hmmm, that's quite interesting. Did he have an elementary solution to it?
 
r9m
@Chris'ssis Idk .. I came upon it by chance while I was downloading random old AMM articles from scribd :P
 
$$\text{Ei}(2)-\gamma -\log (2) = \int_0^{\sqrt{2}} \left(\int_0^{\sqrt{2}} e^{a b} \, da\right) \, db$$
$$\text{Si}(2) = \int_0^{\sqrt{2}} \left(\int_0^{\sqrt{2}} \cos (a b) \, da\right) \, db$$
 
1:33 PM
@r9m Looking at the derivative of $\sqrt{1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}}$ I get $$\lim_{n\to\infty}\sqrt{n}\left(\frac1{2^n\sqrt{n!}}\right)^{1/n}=\frac{\sqrt{e‌​}}{2}$$
 
$$-\text{Ci}(2)+\gamma +\log (2) = \int_0^{\sqrt{2}} \left(\int_0^{\sqrt{2}} \sin (a b) \, da\right) \, db$$
 
Off to the park... BBL
 
r9m
@robjohn :D Awesome !! .. but derivative :o ?
 
1:53 PM
Hi everyone
 
Has anyone studied Brezis book on functional analysis?
 
2:22 PM
@r9m Yep... $$\frac1{2^n\sqrt{1+\sqrt2+\dots+\sqrt{n}}\sqrt{2+\sqrt{3+\dots+\sqrt{n}}} \cdots\sqrt{n}}$$
 
yeah...
 
@r9m Then I use the result of this answer to get the terms are between $\sqrt{k}$ and $\sqrt{k}+1$, which leads to the $\sqrt{n!}$
 
@r9m my way is ugly, much to write ...
 
@r9m The next term to be added to the $n$ will be between $\sqrt{n+1}$ and $\sqrt{n+1}+1$ which when multiplied by the derivative and the $n^\text{th}$ root is taken is insignificant.
@Chris'ssis for the question I am talking about?
 
@robjohn Yeap.
 
2:36 PM
@Chris'ssis Do you have your approach written up? I'd like to see another way of doing this.
 
@robjohn No, I don't have it yet.
 
Have you studied some Sobolev spaces @cirpis
 
My downvoter has reawakened... I've started getting downvotes again. I wonder if a suspension has just ended :-)
 
@Hakim I did it :D i found a way to get admin, it's ok now
 
r9m
2:55 PM
@robjohn :D WoW !! This is super slick way !! :D
 
@r9m I am curious about how you found the limit.
 
r9m
@Chris'ssis I used a bit of calculus where ever I could :)
 
@r9m I also used a bit of calculus. :-)
 
So not so different from mine, since I used the derivative and the chain rule :-)
 
r9m
@robjohn long drawn .. but the idea is similar :-)
 
3:02 PM
How about finding the asymptotic expansion of $$\left(l-\sqrt{ 1+ \sqrt{2 + \cdots +\sqrt{n}}}\right)^{1/n}$$?
$$\left(l-\sqrt{ 1+ \sqrt{2 + \cdots +\sqrt{n}}}\right)^{1/n}\sim \frac{1}{2}\sqrt{e/n}$$
 
r9m
curse my touchpad :( .. I need a mouse or something now :(
 
Guys, how do I evaluate $\lim_{n \to \infty} (3^n + 5^n + 7^n)^{1/n}$
 
r9m
@Chris'ssis that didn't cut deep enough :P
@ShuklaSannidhya step 1: pull out the greatest term from inside ..
 
@r9m Like $7^n(3^n/7^n + 5^n/7^n + 1)$ ?
 
r9m
yes
 
3:11 PM
?
 
r9m
now can you find nice positive upper and lower bound for $(3^n/7^n + 5^n/7^n + 1)$ ?
then apply squeeze thm for limits
 
$(3^n/7^n + 5^n/7^n + 1)$ is greater than 1 for all n, right?
 
r9m
right
 
@r9m I used the derivative of the nested radical for another question (well, that proof is an overkill)
 
that thing approaches 1 as n approaches infinity but I also have a $7^n$
 
3:15 PM
It's an old proof.
 
r9m
@Chris'ssis ya .. definitely god of overkill ! :P
@ShuklaSannidhya you have a ${()^{1/n}}$ too right ? so ..
 
oh is it 7?
 
r9m
@ShuklaSannidhya yes :)
 
$$\text{li}(k+1)+k-\log (k)-\gamma = \int_0^k \left(\int_1^2 \frac{(s+1)^{n-1}+s-1}{s} \, dn\right) \, ds$$

$$n = \lim_{s\to 0} \, \frac{(s+1)^{n-1}+s-1}{s}$$
 
3:18 PM
@r9m You the man!
 
r9m
^^' ..
@Chris'ssis I didn't know sos440 used to come here ! :)
 
@r9m I think I saw him here only once (a long time ago).
 
r9m
@Chris'ssis I'm reading the old transcript ! ^_^
 
@robjohn did you manage to try any of the previous series I posted?
(by the way, I'm creating some new series now ...)
 
r9m
3:38 PM
@Chris'ssis I have started a blog :) .. without the faintest idea what good the blog is going to do to me :P
 
@r9m Nice ... :-)
 
r9m
any ideas .. ?! what do people usually fill up the blog pages with ? :)
@Chris'ssis do you have a blog ? :)
 
@r9m No, I never had a blog.
 
r9m
@Chris'ssis why ?
 
@r9m That might be an idea to use somewhere in the future, but not now.
 
r9m
3:47 PM
@Chris'ssis okay :-)
 
If I had obtained the job I applied for, my time for doing this kind of activities would have been dramatically reduced. At any rate, sooner or later, my time won't permit doing this, and a blog really requires a lot of time.
(that's why I said it's an idea to use somewhere in the future)
 
r9m
okay :)
 
@r9m Then, there is the book where I'm working ...
:D
 
r9m
@Chris'ssis :D ya ,. I'm waiting for that :D
 
how can i ask a question about something which i have no idea about it on mathse without the question becoming on hold or down voted?
 
r9m
3:52 PM
bbl ((:D)
 
Hmmm I like this answer
 
@N3buchadnezzar lolll (+1). That was funny. :-))))))
 
4:20 PM
@Chris'ssis =)
 
Hi everyone, I'm trying to understand the lonely runner conjuncture but I do not undertand the meaning of "pairwise distinct". So.. what is it?
 
@Kartik all of them are diferent from one another
 
@cirpis So what is "pairwise"
 
@Kartik there is no pair such that both members of the pair are equal
 
@cirpis Thanks
 
4:26 PM
8
Q: Difference between pairwise distinct and unique?

n1kh1lpI've come across the term "pairwise distinct" in many research papers. But, I don't understand how it differs from just saying that the elements of a set are unique instead of saying that they are pairwise distinct. Can someone please explain the difference, if any, to me?

 
4:36 PM
Meh
I failed an integral, bleh
 
@Chris'ssis Anything interesting recently?
 
@Alizter I posted some nice series above.
 
@Chris'ssis Ok I will have a look
 
4:51 PM
Darude sandstorm
 
5:11 PM
@Chris'ssis Neat observation
 
@Alizter hehe, a different way of putting things. :-)
 
@Chris'ssis I will be back. Damn mathematica making things hard
 
@Alizter OK :-)
 
$$\sum _{i=1}^\infty\frac{(-1)^{i+1}}{i} \sum _{j=1}^i \frac{(-1)^{j+1}}{j} \left(\sum _{k=1}^j \frac{(-1)^{k+1}}{k}\right)^m$$
I conjecture that it converges with $m\to\infty$
@Chris'ssis Of course this is very difficult. Or is it?
 
@Alizter I also thought of some generalizations, but not too much. I'll do this after I have finished some particular cases I'm working on.
 
5:19 PM
@Chris'ssis I am trying to get some evidince with mathematica
Here are the partial sums for $1\le m\le 20$
 
@Alizter Cute.
 
I bet they converge
 
@Alizter Indeed.
 
Now to see what kind of values
 
@Alizter By the way, I also created $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{n+1}}{n}\right)^2= \frac{3}{2}\zeta(2)\log(2) + \frac{1}{3}\log^3(2) - \frac{1}{2}\zeta(3)$$
 
5:26 PM
@Chris'ssis as $m\to\infty$ we have $\approx 0.693123$
 
@Alizter This is close to $\log(2)$, and this might be the limit.
 
Okay doing values for around $m,n=1000$
 
5:43 PM
The growth rate on these computation times are crazy
 
@Chris'ssis What is the technique you used for the m=1 case?
 
@sarah the technique that uses no pen and paper.
 
You did it in your head?
 
@sarah Yeah. You can do it as well.
 
or on the computer?
 
5:50 PM
@sarah I did it in my head. But also you can do it ...
(you only need to apply something ...)
Abel's summation + symmetry
 
and we all know you are very "able" :)
 
@skullpatrol Thanks, but that way really makes things beautiful. All you need is to apply it. :D
 
speaking of "able" hi @robjohn
 
@sarah Wait, you didn't refer to my last series ...
 
@skullpatrol UEUUUEEAEAAAAGGHHHH
 
5:55 PM
@sarah You meant this $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{n+1}}{n}\right)$$?
 
@Chris'ssis No I was talking about the one @Alizter was drawing pictures for
 
@skullpatrol back :-)
 
@sarah Ah, that is a different story.
 
@robjohn wb :)
 
@sarah $$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}\sum_{j=1}^{i} \frac{(-1)^{j+1}}{j}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{j+1}}{j}\right)=\frac{1}{2}\zeta(2)\log(2)+\frac{1}{4}\zeta(3)+\frac‌​{1}{6}\log^3(2)$$
 
5:59 PM
Yes that is the one
 
@sarah Did you work on it?
 
@Chris'ssis I am trying
 
@sarah OK. Then I let you trying it. :-)
 
@Alizter s conjecture caught my eye
 
@Chris'ssis I think It is $\log(2)$ with some other stuff but I cannot quite find it
 
6:05 PM
@robjohn I'm inclined to believe we can make use of $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(1-\frac{1}{2}+\cdots + \frac{(-1)^{n+1}}{n}\right)$$ to elementarily compute $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} H_n}{n}$$
 
@Chris'ssis I will have to look at my approach to $A(1,1)$ My approach looks pretty elementary
 
@robjohn Yeah, true.
 

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