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09:01
Greetings
Very nice this answer here by xpaul
7
Q: Showing that $\int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ (-\pi < a < \pi)$

Random VariableI want to show that $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ ( -\pi < a < \pi) . $ So I let $f \displaystyle (z) = e^{ibz} \frac{\sinh az}{\sinh \pi z} $ and integrated around a rectangular contour with vertices at $R, R...

@robjohn I wonder if it'a just a coincidence the fact that I'm downvoted at 2 min after entering the chat. :-)
I couldn't live without a strong personality, character, just staying in the shadow and biting people.
@Chris'ssis could it be that you've just become active?
@robjohn I'm active for more than half an hour I think.
lol, he couldn't accept that requirement here ...
0
Q: Another limit related to pi number

Chris's sisFind the value of the limit: $$\lim_{n\to\infty} \sum_{k=0}^n \frac{{k!}^{2} {2}^{k}}{(2k+1)!}$$ I'm trying to find out if this limit can be computed only by using high school knowledge for solving limits. Thanks.

r9m
r9m
@Chris'ssis something should be done about this anonymous downvote campaigners ! like making it compulsory to give a reason (by a comment) for downvoting/upvoting ..
@r9m I think the accounts that are mainly used for downvoting specific people should be warned and then deleted.
@r9m I mean I create an account and begin to downvote $X$ every day. What's the meaning of that? Should such accounts exist?
Hi guys i am new here I need a bit of help
r9m
r9m
09:12
@Chris'ssis account just for downvoting is outright childish :| .. I pity them .. (besides I have only dowvoted once .. on china math's question .. that too was an accident when I was accessing M.se from tab :P)
Does any of you know what fixing of triangle mean
@r9m But think about it, you as a person don't have the courage and the dignity to do that from the original account, but from a fake one. Cowards ... :-)
r9m
r9m
@Chris'ssis thats why I pity their puny mortal souls :P
@Chris'ssis was there another downvote on the question just now ? :O
Can any of you answer the quedion that i asked
@Akshay I don't. What is the context?
09:22
@Committingtoachallenge I have a seminar on tomorrow and the subject that i got was fixing of triangle. I have no idea what that means i googled it but did not get the answer.
@Chris'ssis Although I was quoted in Pedro's answer, I don't think I ever saw that question.
r9m
r9m
WATTTT !! it just rained upvotes !!! :P LOL what the heck was that ?! :O
It would be really helpful if any of you can answer it
@r9m Mind answering my question now?
i think this is a beautiful solution math.stackexchange.com/questions/990174/… by raymond
r9m
r9m
@Akshay :O did you just do it ? I mean the upvotes ? :O I'm stupified ..
09:25
@r9m I really need some help now and i saw that you were sad about someone down votting you
r9m
r9m
@Akshay man .. I'd really answer you if I knew what fixing a triangle was !! there was no reason to upvote me ...
@Akshay and I was not unhappy about downvotes .. it was @Chris'ssis
@r9m I know what it feels like when someone down votes you for no reason
r9m
r9m
I feel bad too if I feel I'm being upvoted for no reason .. did you read all the answers that you just upvoted ?!!
@r9m A few of them. Can you look into my question? Please
This guy reminds me of a colleague of mine, when he came into campany was totally intrigued by my analysis and calculations and began to do some investigation, he thought I'm helped by someone else, that my work is not really mine, it was hard to him to accept that he with PhD couldn't understand my work or had some terribly difficulties. That was until one day when I gave him a lesson and explained some of my work in details. Since that day he never tried to do anything else but to respect me.
09:33
Who does this respectful guy remind you of Chris?
Manufacturing analysis can be terribly hard, it depends on what you analyze, it's hard to imagine the madness there. You don't analyze functions, but very complex processes.
Back a bit later. Some work to do now.
r9m
r9m
@Akshay I've got a class now !! gtg .. sorry .. I'll come back and try to check Google if I can find anything on fixing a triangle .. :-)
bbl
@r9m ok
Hi @Robjohn
0
Q: A question about pointwise convergence of Fourier transform in $N$-dimensions

Rajesh DThis is a question about pointwise convergence of a Fourier transform of functions of the form $f: \mathbb{R}^N \to \mathbb{R}$, which is potentially a $N$-dimensional generalization to pointwise convergence of $1$ dimensional Fourier transform. This question arose when I am trying to generalize ...

@RajeshD Hey there! Haven't seen you here for a while.
09:43
yep been a bit busy with home and office work
@robjohn
@RajeshD By "a curve of bounded variation" do you mean a rectifiable curve?
10:07
@robjohn Yes
'Rectifiable' would be more appropriate?
@robjohn Yes! Thank you for the explanation, can you help with some topology ? I would like to prove that if $f:\Bbb{R}\rightarrow \Bbb{R}$ is continuous then the graph is closed.
I am trying to prove that its complement is open, my problem is I would like to justify that if $y$ and $f(x)$ are distinct point in $\Bbb{R}$ then I can found to disjoint open set.
I made it, I just choose an open ball centered at $y$ and $f(x)$ with radius $d(y,f(x))/2$.
@robjohn edited.
For instance, if someone has a proof using the reciprocal of a function it would be cool :-)
(using sequence for this exercise it's easier but I just want to be able to find another proof)
@DanielFischer If you are here and if you have time, can you help?
10:28
@MarcGato If $f\colon X \to Y$ is continuous and $Y$ is a Hausdorff space, then the graph is closed. The proof is pretty much the one you used for $X=Y=\mathbb{R}$, just you can't get the separating neighbourhoods of $y$ and $f(x)$ explicitly in the general setting.
For $Y = \mathbb{R}$ (or similar spaces), you can also use the map $g\colon X\times \mathbb{R} \to \mathbb{R}$ defined by $g(x,y) = f(x)-y$, and see that the graph of $f$ is just $g^{-1}(\{0\})$, which is closed, since $g$ is continuous [prove that].
@DanielFischer great idea to denote $g=f-y$, I was thinking about the diagonal, your method is easier. For the fact that $g$ is continuous it's because g is a linear combination of continuous function, no
@MarcGato Yes, basically. You need to know or show that $(-) \colon \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ is continuous, and a few fundamental properties of the product topology.
(That the coordinate projections are continuous.)
10:45
@DanielFischer Ah yes the projections, I have seen this in few days ago. Thanks for your time. Enjoy your meal :).
11:00
Can anyone answer this?
in Coding, Graph and Design Theory, 23 mins ago, by Fred Kline
I have found a D3H infinite prism that has 0 volume. Is this a trivial discovery?
11:39
Hi @Com
12:26
@saw Oh I think I missed you - how are you doing friend?
13:02
I wonder how weird it would be if someone saw in my browser history "colon latex"
@UserX I have thought similar things for different latex searches while at uni
I just posted a very nice question
0
Q: Computing $\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$

Chris's sisI'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible? $$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$

@Chris'ssis Very nice (+1)
@Chris'ssis what would be your particular solution if you had $x\sin^2 x$ on your RHS?
Girlfriend needs to get up earlier than normal, so I guess it is an early night. Cya later friends
13:24
@Committingtoachallenge I also added a supplementary question, that is $$\sum_{n=1}^{ \infty} (-1)^{n+1} \left( \psi^{(0)}\left( \frac{1+n}{2} \right)- \psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n} \right)$$
hehe, that question is for having fun! Math should make people laugh and have fun!
How do I typeset \coloneqq (:=) correctly on mathjax
@Chris'ssis Im reading Interesting Integrals :D
@TheArtist Good! :-)
Here is a more advanced version $$\sum_{n=1}^{\infty} \left(\psi^{ (0)}\left(\frac{1+ n}{2} \right)- \psi^{(0)} \left( \frac{n}{2} \right)- \frac{1}{n}-\frac{1}{2n^2}\right)$$
@Chris'ssis If I get stuck in the middle :) can you help me clear it ? :)
@TheArtist In the middle of what?
13:32
@Chris'ssis I mean if I find something I understand less about :)
@TheArtist OK
@Chris'ssis ok cool thanks
@robjohn we did many such series in the last months, lots of them :-). It's said though there is not a culture of these series ... (the solutions hardly come to the surface as if they were inequalities)
Not said but sad.
@Chris'ssis inequalities? Which series are you talking about; the series just above or others?
@robjohn I know you briefly answer them. I'd also like to see other solutions ... :D
13:38
@Chris'ssis yes, the ones just above... but I am curious about the allusion to inequalities.
I wanna see the creativity at work, to see something I didn't see before.
@robjohn I mean the inequalities, in general, are hard, and they are not answered that easily. The realm of inequalities is maybe one the toughest areas known.
One needs many many years of learning for being pretty good, but not very good (in my opinion).
(well, out there might be some exceptions)
Oh ...
This one $$\sum_{n=1}^{\infty} \left(\psi^{ (0)}\left(\frac{1+ n}{2} \right)- \psi^{(0)} \left( \frac{n}{2} \right)- \frac{1}{n}-\frac{1}{2n^2}\right)$$ IS NOT more advanced.
(that's because we can sum separately over $1/(2 n^2)$)
@robjohn I also added the series $$\sum_{n=1}^{\infty} \left( \psi^{(0)}\left(\frac{1+n }{ 2}\right) -\psi^{(0)} \left( \frac{n}{2} \right)-\frac{1}{n} \right)^{2014}$$
as a supplementary question.
13:56
$$
\int \bigl[ \log(x^2+1)-2(1+x) \arctan x \bigr] e^{x} \,\mathrm{d}x
$$
Meh
Double integration by parts, any easier ways?
Somebody listening?
to what?
14:33
to?
@robjohn I can imagine that for the last series is even hard (maybe impossibe) to think of a result for that. :D
@robjohn just wait a bit ... :-)
@robjohn did you receive the message?
14:57
I just gave the answer to the supplementary question
0
A: Computing $\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$

Chris's sis$$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)^{2014}=$$ (-155015596184364440945084299975759662568630079642309524338450663434751218053297478865897807539291708732320332850003195271367671359884906941357534021938515281380329185...

@Chris'ssis Seems pointless and inelegant, really.
@Sawarnik For you seems pointless, but for me there is a flood of beauty that even takes your breath away ... :-)
That fraction?
Sure, that one didn't fall from the skies ... it's the expression of the sublime beauty and high creativity at the same time.
Well, ok.
15:07
@Chris'ssis If it had a reasoning or an explanation of how you got that result, it might look that way to others, too.
@ypercube My answer deserves downvotes? Well, I think you deserve to be ignored by me totally.
Seriously, what does the answer have, besides a large fraction? Does it have any formula, explanation, a clever idea, a proof?
Why should it be upvoted? Please ignore me if you can't accept criticism.
@ypercube For the same reason Cleo's answers are upvoted.
That's another riddle. Why are they?
r9m
r9m
15:23
@Akshay some context will be nice .. what seminar is it ? Googling spits out a variety of results that starts from elementary concept of geometric congruence, geometric algorithms to stuff that are totally out of my league =P .. It might be helpful if you said a little bit more about the target audience of this seminar :)
@Chris'ssis Insanity !!! :D
Might I get somw hints?
$$y'''+y=x\sin^2 x$$
Would the particular solution be of the form Ax^4+Bx\sin x+C\cos x$?
With every day I realize the knowledge is a sacred thing that cannot be share with anyone.
r9m
r9m
@DanielFischer 'cos the answers are insane one liners ! and people either completely ignore or take for granted (without questioning their validity) what they cannot relate to .. :P
@Chris'ssis wive/husband is another example that fits the profile :P
I talk too much .. maybe I need some of @TedShifrin smacks !
bbl
@r9m lol :-) Now, I agree that I don't explain things in my answer but still ... to some user things happen in a certain way and with other ones things happen completely differently ..
@Chris'ssis For the record, I find Cleo's answers downvotable, too.
Mere results, without any insight on a pattern thought or reasoning.
15:30
@ypercube Well, I didn't ask you to upvote me, but keep in mind that in this life you would have never known the answer to that question. Maybe it's not that important. Just for the record.
@Chris'ssis I upvoted the question and downvoted the answer. Because that's my opinion of whether they are helpful or not.
@Chris'ssis I can send you a copy of my class notes & sheets if you want >:c
@Hippalectryon did you see my last question?
8
Q: Computing $\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$

Chris's sisI'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible? $$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$ where $\psi^{(0)}(x)$ is digamma function Here is a su...

@Chris'ssis Do you imply that only you on this earth could have solved that problem?
@rehband you enjoy these questions ... ^^
15:33
@Chris'ssis I just did
@Hippalectryon OK :-)
@Chris'ssis Why do you always write $\psi^{(0)}$ for digamma ? :c
Why not use the shorter $\psi_0$
@Hippalectryon Yeah, but I learned to write things that way.
I usually laugh loud when I attend problems, I have a lot of fun, I'm not used to quarrel, this is weird to me. Mathematics is for fun, it must be like that.
@Chris'ssis I usually laugh loud when you believe I'm a pro <_<
@Hippalectryon LOLLL
15:45
0
Q: Social Choice Theory: Source of problems sets and practice questions.

Amy WilliamsFrom where can I practice questions related to the following topics: Arrow's Impossibility Theorem Nash's Axiomatic Bargaining. Harsanyi's Axiomatic characterization of Utilitarianism. ?

r9m
r9m
@Chris'ssis LOLL :P
@r9m LOL :-))))
@Chris'ssis Haha, yeah it looks like a crazy one :)
Can someone help me simplify $\frac{1}{2}\left(i-1 \pm 2i\sqrt{2i - 6}\right)$
It would be nice if I could get it into $a + ib$ form
The closest I've gotten is $\left(\frac{1}{2}\pm\sqrt{2(i - 3)} \right)i - \frac{1}{2}$
Maybe that's good enough?
@Nick Good enough for what ? What is it used for ?
16:00
@Hippalectryon: It's just the root of a complex quadratic. I'm trying to make it look "nice"
@Nick Sure
Use polar coordinates.
@BalarkaSen Splendid idea. But maybe a decimal approx would be enough. How do you do $\sqrt{i -3}$?
$x^2 = i - 3$. Set $x = re^{i\theta}$. $r\cos(2\theta) + ir\sin(2\theta) = i - 3$. Compute real and im parts
@BalarkaSen You have taught me something immensely useful. (You're expecting me to approximate using Taylor series right?)
Serious stuff going on :|
16:11
@Nick you may or may not use taylor.
oops, i gotta go byes
@BalarkaSen You have another way? (Ping it to me when you get back :D)
byes.
@Sawarnik Aww, c'mon. Don't be sarcastic.
@Nick Yes, I am not being sarcastic. I was serious :P
16:20
lol
Somebody listening?
@DigitalBrain: We copy you loud and clear, roger.
@DigitalBrain Batman is
I'll dance with batman but.........
16:24
Maybe a watchful protector, a silent guardian ...
@Nick I want to discuss something serious
@Hippalectryon Can I discuss something?
@DigitalBrain Of course !
you must!
@Hippalectryon Is it okay to abandon your MathSE profile and start over again?
@DigitalBrain It is a rule that you must!
@DigitalBrain: @JasperLoy does it all the time
16:26
@Nick I think you've heard that
@JasperLoy PING PING ^
@Hippalectryon There's no JasperLoy in users
He's away for now
He's usually there later on :)
Anyhow, several people have done so @DigitalBrain
Why would it be bad anyway ?
@Hippalectryon So it's okay right?
@DigitalBrain Of course ! Why were you thinking it wouldn't be ?
16:28
@Hippalectryon Because I did that and somebody noticed
@DigitalBrain If you're not trying to circumvent a suspension or question/answer ban, it's okay to do that.
@DanielFischer No not at all, It's just I want to get rid of number of downvotes from my profile
@Hippalectryon @BarryCipra Noticed
@DigitalBrain No need to do it though :)
@DigitalBrain Why?
goes away again... @Nick I leave on 6th morning :( :( :( :( :( :( :(
16:30
@DigitalBrain There is no user called @BarryCipra
I saw @Anastasiya-Romanova 's profile
@Sawarnik: For RBIQ?
yes
@DigitalBrain :D :D
@DigitalBrain Oh with a space
16:31
@Digital She is amazing, isn't it?
@Hippalectryon There is.
Potatoes
@Sawarnik I searched without the space, that's why I didn't find him/her
@Sawarnik Mayve she's younger than you
@DigitalBrain Not by much though :P
@Nick I know but it's okay
16:32
Few months, at most.
@Sawarnik Really?
Yes :D
@Sawarnik Okay
@Hippalectryon So, I need to start over but this name looks cool!
I mean sounds cool
@r9m do you see any awesome way to prove that $$\int_0^{\infty}\int_0^{\infty} \frac{\sin(y)}{y(x+y)(1+x^2)} \ dx \ dy=\frac{\pi}{2}\left(\gamma-e \operatorname{Ei}(-1)\right)$$?
Bye.
16:40
@DigitalBrain There are a lot of cool names out there :)
@DigitalBrain I own the accounts Ramanewbie and IntegralPicnic
@Chris'ssis Why is that result so important for you ?
@Hippalectryon I need to find some!
@Hippalectryon I proved it already. Well, I did some important discoveries these days (that integral was important for some reason)...
@Chris'ssis are you busy?
@Unnamed Why do you ask?
@Hippalectryon and What should I do with my rep??
@Chris'ssis I want to start fresh with new account, what should I do to my rep??
16:45
@Unnamed the best person to talk about it is @robjohn
@Unnamed : Give it away as bounties! If you can't decide where, I'll help you find questions.
@Nick I should ask robjohn
@robjohn
@robjohn Someone's calling you!
@DigitalBrain: He's busy... with real life and stuff.
@Chris'ssis his profile pic makes me think like he's mean squared
@Nick You know him???
@Unnamed Your old rep ?
16:48
@Unnamed I only know this for sure about him:
There is not a single person left in this room who has not tried to conduct a psychoanalytical study of @robjohn on the basis of his mean orange square avatar.
This is my old account
@Unnamed Give me some points. I'll squander it away like the child I am.
@Nick How?
Oh lol wrong se website profile
@Unnamed Well, if you can place a bounty on one of my questions... :D
@Hippalectryon My name is Aditya not Johnskeet
16:51
@DigitalBrain: Wait, as in prodigy @Aditya ? I though he was a different person
@Nick My profile's are @DigitalBrain and @Unnamed Aditya is just my name
#@Nick Not profile name
If someone can give a free 500 rep bounty :D math.stackexchange.com/questions/998535/…
@Amy ??
@Unnamed ohk. I was just guessing.
@Hippalectryon Maybe @DigitalBrain will be so kind. Ask him.
@Hippalectryon Done!
16:54
@Unnamed ⊂((・▽・))⊃
@Hippalectryon What's that?
〜(^∇^〜)
Me happy
Maybe that one is more explicit ヽ(⌐■_■)ノ♪♬
@Unnamed Thaaanks :D
1
Q: By what criteria do you judge whether mathy questions should or should not be considered off-topic?

DanielSankLet us take as an example this question. Consider whether you think it should be considered off-topic (for being purely mathematical in nature). Should one's opinion on whether a mathematical question be closed be based on what one finds interesting, the specific written rules of this site, a...

People contribute to this ^ The future of Physics Question inbetween MSE and PSE depends on this post. @robjohn
@Nick make your wish
r9m
r9m
@Chris'ssis LHS reduces to $\displaystyle \frac{\pi}{2}\int_{0}^{1} \frac{1-e^{-x}}{x}\,dx$ .. after that its all Merry go round the mulbury bush Puiseux Series of $Ei$ :D
16:57
@Unnamed Are you now a genie? ... or did someone collect all the dragon Balls.
@r9m How did you get that integral? ... just to be sure you didn't start from the right side :D
@Unnamed I was asking if any one has some practice questions related to the topic "Nash Axiomatic Bargaining". I need to practice for my upcoming college exams.
@r9m: You remember "Merry go round the mulbury bush"?
r9m
r9m
@Chris'ssis -_- of course I started from RHS !! who the hell wants to use pen and paper (in my case chalk and cupboard) after a 8 hour long class ?? :P
@r9m lol, OK :-)
17:00
@AmyWilliams: That sounds like a branch created in the 21st century.
r9m
r9m
@Chris'ssis can I add that to my blog ? :-)
@r9m my integral? Well, add it if you like it. :-)
r9m
r9m
@Chris'ssis yes !! (and I'll mention source + link to chatroom conversation) pretty please with sugar on top ? =)
:D
@r9m OK ;)
@Nick This is the topic I am referring to: ma.huji.ac.il/~mseyal/Nash.pdf
17:07
@AmyWilliams Oh, and I was under the impression it was something that could help me purchase my vegetables and fish at a cheaper price.
Good luck on it, anyhow :D
r9m
r9m
@Chris'ssis Ramanujan has got some really amazing results with the integral $\displaystyle I(y) = \int_{0}^{y} \frac{1-e^{-x}}{x}\,dx$ :D ..
@Nick I also wished it to be simple. That is why I want to be thorough before the exams by practicing a few problems.
@r9m hmmm, I missed that ... or I simply don't have in mind the proper results.
@r9m Do you have in mind a nice example?
r9m
r9m
@Chris'ssis oh ! don't miss entry 44: page 167 of Ramanujan's Notebook Vol II and subsequent results :D They look AWESOME !!!
17:32
@r9m I particularly like the limit from the page 178, that one is crazy awesome. I wonder if I can come up with a nicer proof (although that one is very nice).
$$\LARGE \text{It's simply marvellous!}$$
@Chris'ssis You probably can :P
@Hippalectryon lol, it's not easy to beat Ramanujan :D
@nick hi again.
@r9m I wonder if that one can be generalized in some way ... :D
(I feel a whole family of eta functions is waiting for being discovered ...)
r9m
r9m
@Chris'ssis hai !! subarashii !! :D
17:38
@Chris'ssis: He's being portrayed by the guy from Slumdog Millionaire... You know, Prince Zuko who works at the best exotic marigold hotel. He's pretty scrawny, you can definitely beat him up.
@Sawarnik: yellow :D
Hi @user130018.
@Nick I didn't get your point, but I'm not worried at all about that. :-)
r9m
r9m
@Nick are you talking about the upcoming movie 'Ramanujan' ?
@Chris'ssis: My point was if you want to beat Ramanujan, beat up Dev Patel instead.
r9m
r9m
googles who the heck is Dev Patel ..
17:41
.. finds image of guy from slumdog millionaire
@r9m says to himself, "Oh, that guy"
@Nick My point is that I do exactly what I want to do. Capisci? ;)
r9m
r9m
@Nick wasn't this guy there in the movie 'The last Airbender' ?
3 mins ago, by Nick
@Chris'ssis: He's being portrayed by the guy from Slumdog Millionaire... You know, Prince Zuko who works at the best exotic marigold hotel. He's pretty scrawny, you can definitely beat him up.
r9m
r9m
ahha !! thought so !!! "Oh, that guy" LOL
@r9m: Yes, he was the guy with the burnt-ass face. (No, not Darth Vader)
r9m
r9m
17:43
@Nick :P lol who cares ?
The king of the fire nation cares.
lol, I received 3 downvotes for this one
9
Q: Computing $\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$

Chris's sisI'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible? $$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$ where $\psi^{(0)}(x)$ is digamma function Here is a su...

r9m
r9m
yes yes that guy ..
@Nick prince of fire nation
smacks @r9m for good measure
But you know what? No need to share interesting answers and very nice questions ... :-)
Maybe it's time to delete some ... :-)
r9m
r9m
17:45
@TedShifrin (+1) support .. smack accepted !!
I have no idea why, though.
@r9m: You never saw the last episode? Lightning is now a fundamental element and Zuko is king!
OK, I have work to do ... See you guys later.
r9m
r9m
@Nick ah !! I stopped watching it so so long ago .. I almost remember nothing :P
@r9m: Me too. but I have memory like Aappa.
r9m
r9m
17:48
now what is Appa .. that flying bison ?
@Nick You think you're smart giving those replies, but you also miss another point: one day I hope to be like Ramanujan or far better in terms of integral, series and limits (although it's hard to make such appreciations, it's important to get the idea behind the statement).
@Nick I don't believe in humans that cannot be overpassed in terms of performance. You can believe that all as long as you want to.
Any limit can be broken, anything, impossible is just a word ... If you don't bilieve in yourself highly and if you cannot imagine yourself breaking all those limits you wanna break then you'll never break them.
@Chris'ssis: No, I think I'm delusional, high on prescription medication and pretty much bored and in search of cheap chuckles when I was giving those replies. Don't take it seriously, silly goose.
@Chris'ssis I believe you... and I do believe in you too
@r9m: Yes, the flying bison.
No one should fear of the limits posed by the others by their performance, no one needs to live within these limit-walls as if they were condemned. The attitude is important, not the fact that one might fail doing that.
The limits are there only for breaking them.
2
That's all.
Hello @MikeMiller !!! Could you maybe help me at this exercise?

http://chat.stackexchange.com/rooms/18400/p-adic-field

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