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2:23 AM
^ Strong/bad language. So, don't click if easily offended.
 
2:34 AM
@anorton looks all clear?
 
I might get banned here, but this guy is being a real douche.
@IceBoy
@IceBoy I deleted the comment. "Retired professor." LE SIGH. I was thinking this guy was some 13 y.o. with too much free time.
 
I agree, pal.
 
3:17 AM
Retired professor? Could be me :D
Hi @anorton @skull mr @pedro
 
Hi! :)
 
@TedShifrin Hello Professor
Hi @robjohn
 
@IceBoy hello
 
@TedShifrin Hello!
I answered a counting question today.
I couldn't count, so I had to use variables, and then be thankful I only had to count up to 26.
 
LOL ... You can take over my class. Doing expectation of a random variable tomorrow :)
 
3:22 AM
Sorry Pedro, but I just had to star that one. :)
 
@anorton: I presume that profane person is banned, but will likely return with a new moniker.
 
@TedShifrin Unfortunately, I think that will probably be the case. :( But, we can keep banning his new monikers every time they turn up.
 
@Pedro: I didn't see doucheness.
 
@TedShifrin Oh, you missed it.
I counted 25 letters in the phrase, but there were 26.
Some guy commented saying "n=25, really?"
And then when I forgot to correct something else say "Getting there... 26-4+1=22" (I corrected the 25 but not the 22.)
 
I didn't read anything carefully. How do you know he's a retired prof?
 
3:25 AM
SIGH,
@TedShifrin Profile said so.
 
Well, you've made the occasional obnoxious comment. It's karma.
Speaking of obnoxious, I haven't been around much, so I've avoided running into René. Does @Mike's absence indicate that he's off drinking martinis to celebrate?
 
He could be drowning his sorrows also.
Either way, he needed a break.
 
Algebra does that to some people :)
 
@TedShifrin I'm having tons of fun with Complex Analysis.
 
3:42 AM
It's beautiful stuff, @Pedro ... And it has beautiful interplay with topology and diff geo (e.g., Schwarz Lemma)
 
Finally back after a year :D
 
We've forgotten you ever existed, @missiledragon
 
:-OMG where have you been?
 
lol
/sarcasm
 
@skull: Who is this ?
 
3:46 AM
@TedShifrin he used to come in here all the time
 
yeah dood
Im a retired professor so yeah
JKS
 
Dood? No wonder I don't know him.
 
Im 11 years old calm down
jks
 
jks = jokes
 
if you have tan(pi/2-x)=tan2x
when are you allowed to say pi/2-x=2x
 
3:49 AM
Are you supposed to be right?
 
yes
0
A: Finding the exact value of $\tan(\pi/5)$

houdiniFor first part use $\tan(a+b)= \dfrac{\tan a+\tan b}{1-\tan a \tan b}$ Put $a=x$ , $b=2x$ Further evaluate $\tan 2x$ using addition property described above by substituting $a=x$ and $b=x$ For second part use $\tan 3x$ and convert $\cot 2x$ into $\dfrac{1}{\tan 2x}$ Using the second result ...

I was looking at this question
and noticed that egreg did that
 
Then that's never right. It's only one possibility.
 
which is why I'm confused
 
Do we know a range on $x$ (like first quadrant)?
 
The question doesn't state anything, so no.
Well, we are finding the exact value of pi/5, so the first quadrant.
exact value of *tan(pi/5)
 
3:53 AM
I think you're missing something. He knew $\pi/2 -3x =2x$ to start, then took tan.
 
How does he know that?
 
Because he set $x=\pi/10$. Reading helps.
Night, all.
 
Night
 
@TedShifrin Good night Professor.
 
leo
4:13 AM
Is it standard to call the set in this Ass?
 
 
4 hours later…
8:32 AM
Is it necessary to do all exercises in my book ? (Baby Rudin) ?
 
8:42 AM
start with the odd numbered exercises
 
@IceBoy
 
if you have time later, come back and do the even numbered exercises
 
that's a good approach
 
:-)
 
I totally agree it's a good one ! Because this way is more probable to touch upon all topics in a chapter rather than just doing the first half of exercises ! Brilliant
 
8:52 AM
@ZeroCool lower level math books teach this by providing answers to only the odd numbered exercises in the back of the book.
remember math builds upon itself
 
@IceBoy Interesting
@IceBoy what do you mean math builds upon itself ?
 
one method can be used in many different kinds of applications further into the book
so the applications are built upon the method
 
9:09 AM
ah I see
 
9:21 AM
thanks @IceBoy have a good day
 
thanks for asking :-)
 
10:11 AM
@IceBoy i don't think that's a good idea
 
Greetings
Today I was comparing the Riemann zeta function with the human being, and I was thinking that $\zeta(0)$ is to the Riemann zeta function for $s>1$ as the soul is to body.
I wonder how crazy this comparison is. So, also mathematics "talks" about a world beyond what we see. :-)
The only difference is that by mathematical means we can see all the parts.
@robjohn I finished that question posted here integralsandseries.prophpbb.com/topic453.html
 
@Chris'ssis You haven't posted an answer?
 
@robjohn I won't post it.
 
@Chris'ssis why not?
 
@robjohn take them
 
10:25 AM
@Chris'ssis got it. Why don't you want to post the answer?
 
@robjohn I don't feel the need to do that.
 
@Chris'ssis I computed $\zeta(0)$ for a question... did I find a soul?
 
@robjohn lol :-) Where exactly? Did you use the product between Riemann zeta function and Gamma function?
 
@Chris'ssis here
 
@robjohn That is very nice. By the way, is there a nicer way? I mean I doubt I can see something nicer.
 
10:30 AM
@Chris'ssis Thanks... Using the same method, I remember computing $\zeta(-1)$ for another answer.
@Chris'ssis I don't know... This is the only way I know to compute it, but there are probably others.
 
 
1 hour later…
11:38 AM
@robjohn Hello, you are up early. =)
 
@WillHunting who says I'm up?
 
12:02 PM
@Chris'ssis Hello, how is progress with your book?
 
@WillHunting Hi. I'm working on the stuff that is going to be added in it.
@WillHunting I might publish a book even these days but I don't want since I have greater expectations from myself ...
(there is much work to do though)
 
12:37 PM
For f : A -> R with A subset of R. If f is bounded bellow and monotonicly non-increasing, f has a left limit as x tends to any left limit point of A. If f is bounded above and monotonicly non-decreasing, f has a right limit as x tends to any right limit point of A. Are these correct ?
Or is any assumption unnecessary ?
 
12:50 PM
Hi all
Need some help here
 
1:22 PM
@Chris'ssis That's just BS.
@Chris'ssis Yes, there are lots of ways to compute zeta special values. Whip up the analytic continuation at the places you want to.
Take out the functional equation if you want to calculate $\zeta(-1)$.
Or just use Euler-Maclaurin. Abel summation. Lots of them.
 
@BalarkaSen I see. So, what's your way to compute $\zeta(0)$?
(not now, but when you have time)
 
Oops, didn't see that you pinged me.
@Chris'ssis Using the functional equation is standard.
 
@BalarkaSen Yeah, that's clear. I was thinking you have some more as you said above.
 
@Chris'ssis Yes, I have two or three. The best I know of is the functional equation one.
 
@BalarkaSen OK
 
1:34 PM
You want to see it?
 
@BalarkaSen No, thanks.
 
@Chris'ssis Well, why not?
You asked me what my method was!
 
@BalarkaSen Because I know how to do it by using that way.
 
@Chris'ssis Ah. Cool then.
 
1:36 PM
There is also the one with Bernoulli numbers. Another coming from Weierstrass product.
 
Hey, where is @robjohn? It seems he suddenly disappeared ...
@BalarkaSen I see.
 
Pretty much that is all I can think of off the top of my head, @Chris'ssis. I am not solely devoted about computing zeta particular values, so I guess you can think of much innovative methods than the ones I know.
 
@BalarkaSen hmmm, such methods usually arise when you work on something else and then ... the magic thing happen .... :-)
Well, it often happens to me to come across some results from other problems.
 
¯\O_o/¯ I don't know.
 
@BalarkaSen For instance, while I was working on something else I found a cute proof to $$\sum_{n=1}^{\infty} \frac{H_n^2}{n^3}$$
This is not an easy question, especially if you wanna finish it by real methods.
 
1:46 PM
Well, generating functions always does the trick by real methods.
Perhaps you mean "not by special functions"?
 
@BalarkaSen No, I meant what I wrote.
 
It's very easy by generating functions.
At least I believe so.
 
@BalarkaSen Do you really believe that? Show me someone that can do that by generating functions.
 
OK, let me try it first.
 
@BalarkaSen OK. Good luck! :D
 
1:49 PM
mumbles shouldn't have made the comment before trying it mumbles
 
:-))))))
 
2:12 PM
err @Chris'ssis, $7/2\zeta(5) - \zeta(2)\zeta(3)$?
 
@BalarkaSen Yeap
 
Phew.
@Chris'ssis I did it through the identity $$H_n^2 - H_n^{(2)} = \frac{2}{n!} \left [ n+1 \atop 2 \right]$$
 
@BalarkaSen Did you?
 
Multiplying both sides by $x^n$, summing over appropriate limits, diffing like mad.
Yes, @Chris'ssis
 
@BalarkaSen Great
 
2:19 PM
I didn't do the integrations by hand, of course. ;)
That's why it took less time.
 
@BalarkaSen I think you get some ugly integrals in polylogarithms.
 
Yes, @Chris'ssis, it's ugly :(
and tedious
Do you have a nice method, @Chris'ssis?
 
@BalarkaSen Yeap, one that comes from my research.
 
Can you post it?
 
@BalarkaSen Not yet.
 
2:25 PM
Ah, well. OK.
The generating function approaches are useful, nonetheless very tedious.
The A-Y series can also be done by generating functions.
It just gets too ugly.
 
Yeah
 
3:27 PM
@rehband Hey!
 
@Sawarnik Hi there!
How are u doing?
 
:)
I was studying sequences.
@rehband Are you German?
 
Nice, just sequences in general or any ones in particular?
@Sawarnik Yep. U?
 
3:51 PM
Do you guys know where I can get gifs of a semihemisphere being rotated and turned around.
semihemisphere means quarter of a sphere.
Meaning, if the center of the sphere was on the origin of the cartesian coordinate system, then the I'm referring to the part bounded by a quadrant.
0
A: Surface area of quarter of a Sphere

NickLet's first try to imagine the quarter of the solid sphere: It has a curved surface and $2$ mutually perpendicular plane surfaces $$\text{T.S.A} = \underbrace{\frac{1}{4} 4\pi r^2}_{\text{curved surface area}} + \overbrace{\frac{1}{2}\pi r^2 + \frac{1}{2}\pi r^2}^{\text{Plane surface surface}...

The image I have in my answer is obsolete.
@rehband: ich wünsche einen guten Tag! Let's do the Polka!
 
4:08 PM
@Nick Hi.
@rehband In general :)
 
@Sawarnik: Write the expression for the volume of a thick crust pizza with height "a" and radius "z".
 
@rehband I m from India, I asked because we were studying Nazi in school.
@Nick Cylinder?
@Nick Hmm...?
 
@Sawarnik: Yes.
 
@Nick Then?
 
Just write it here. Do you get the joke?
 
4:12 PM
No and no :|
@rehband Here?
 
$\text{Vol(Pizza)} = \pi \cdot z^2\cdot a \equiv \left(\text{pi}\cdot z\cdot z \cdot a\right)$
 
@Sawarnik Hahha, really?
 
Ah oh :D
@rehband Yes, why not? :)
 
@Sawarnik We studied that in History class from grade 8 until grade 11 or something lol
What grade are you in?
 
9
 
4:15 PM
Cool
 
I remember the story of Helmuth from Democratic Politics.
 
@Nick Helmuth?
 
The poor kid couldn't eat from his home because he was afraid his mother might poison the food and kill the entire family.
World War 2 was such harsh times.
 
Really was
 
@rehband: and we will never know how hard it really was. We can only imagine.
Well, but today is today :D
 
4:18 PM
@Nick You sure are right about that
 
16 mins ago, by Nick
@rehband: ich wünsche einen guten Tag! Let's do the Polka!
 
My grandfather was about to be killed by Nazi ... This world should never forget those days and never forgive ...
 
@Chris'ssis Sorry, agree with you
 
Ah my connection :(
@Chris'ssis Really?
 
A Nazi GI Joe was crushed by my grandfather clock. @Chris'ssis: But your story's better.
 
4:22 PM
@Sawarnik Yeah.
 
@Chris'ssis How did he escape?
 
He ran.
 
Good technique.
 
@Sawarnik: What @Chris'ssis meant to say was that his/her grandfather integrated the Nazi.
 
@Nick :/
 
4:25 PM
One of my grandfathers was a priest in Germany and the other fought for the USA
 
@rehband Against Germany?
 
Crazy to think if I had been born 70 years earlier, I would probably be learning to shoot a rifle
@Sawarnik Yea, I have some American ancestry
 
@rehband Yes, the children were put in youth orgs where they learn to glorify war, aggression ... and worship Hitler. Right?
@rehband The German words in that chapter was too tough btw, uuf.
 
@Sawarnik Yes. utter madness
 
Babi Yar (Russian: Бабий Яр; Ukrainian: Бабин Яр, Babyn Yar) is a ravine in the Ukrainian capital Kiev and a site of a series of massacres carried out by German forces and local collaborators during their campaign against the Soviet Union. The most notorious and the best documented of these massacres took place on September 29–30, 1941, wherein 33,771 Jews were killed in a single operation. The decision to kill all the Jews in Kiev was made by the military governor, Major-General Kurt Eberhard, the Police Commander for Army Group South, SS-Obergruppenführer Friedrich Jeckeln, and the Einsatzgruppe...
bbl
(I'm not a Jew, by the way, although many people ask me if I'm a Jew)
 
4:45 PM
I've got a very quick question!
I have a function (below) and I've been asked to show that $f_0 (x)$ is a polynomial in $x$. $$f_n (x) = x^{2(n+1)} e^{1/x} \dfrac{\text{d}^{n+1}}{\text{d}x^{n+1}} e^{-1/x}$$ I ended up with $f_0 (x) = 1$. Is that a polynomial in $x$?
 
@Khallil Very quick? :.
 
Yep. The last part is my question, @Sawarnik!
 
Ok.
@kha Are you any good at epsilon delta proofs?
 
Not at all. I haven't seen them before.
I'm sure I'll see 'em soon though, @Sawarnik.
Have you had a go at my question?
It's a simple sub in of $n=0$ and a bit of algebra.
 
@Khallil Alright. I m seeing them and its ... tough.
@Khallil The equation's too big, I m frightened out already. So no.
 
4:57 PM
Fair enough!
Don't worry about it, @Sawarnik.
^_^
 
hi all
:3
 
Hey Sab!
 
Hey Sawarnik :D
 
Which new movies you watched recently @sab
 
;D
The Guardians of the Galaxy
 
5:00 PM
@Khallil Yes it is! And $1$ looks correct to me
 
Cool. Thanks, @rehband! ^_^
Awesome!
 
@Sabಠ_ಠ Bollywood?
 
nop
bollywood none'
I'm thinking of getting some books from india, are the quality of the books any good?
I mean printing and paper quality
 
alright.
@Khallil I miss your jokes.
I mean the one-liners.
 
bah
 
5:08 PM
@BalarkaSen Hello.
@Sabಠ_ಠ It depends.
 
wiley india, pearson, ubs printing specfically
 
@DanielFischer Is it possible to have a Riemannsurface which contains a sheet on which points looping the ramification point are left on that sheet? I guess no, but I can't come up with a proof.
 
I got one from india and the paper is very thin and it can easily tears if not handled VERY carefully
 
Which one?
 
Maybe it's super-obvious.
 
5:10 PM
introduction to algorithms CLRS
2. The C Programming Language ANSI
3. Apostol both Volumes
The one I bought was Apostol
just volume 1
the rest I'm looking to buy but I'm not sure about the qualiy
 
Huy
Good evening, everyone.
@Sawarnik: Hey, long time no see!
@Khallil: How's your FIFA15 career going?
 
Hi pal :-)
 
Huy
Hey, @IceBoy.
 
I haven't played it in so long, @Huy. It's been 3 days but it feels like an eternity.
 
Huy
@Khallil: Why the hell would you stop?
 
5:14 PM
I won't be playing it for a while either.
 
Huy
@Khallil: But... WHY.
 
I couldn't bring the PS4 to university.
T_T
 
Huy
@Khallil: WHY NOT???
 
The rest of my family are playing it at home. ;-(
 
Huy
:(
 
5:17 PM
Publisher?
 
Is Mike mourning over his sorrows on doing the quals badly?
He didn't even come to MSE.... for 2 days!
 
Perhaps he just needs to take a break.
 
Hope so.
 
Wiley
 
in Mathematics Educators, 49 mins ago, by Shaun
Hello :)

I'm a mathematics student and I'm pretty low on money at the moment. Signing up as an online tutor is becoming quite tempting. However, I can't bring myself to do it. I don't want to be a part of private education; it annoys me: I think education should be free for everyone. It doesn't make sense to me that children of affluent parents should be given extra help, simply because their parents are can afford it.

Are there any private tutors here that feel the same way? Should I change my mind?
 
5:21 PM
Charge those who can pay give free for those who can't
simple
My high school teacher did that.
he still does I'm sure
 
what about those in the middle?
 
Huy
@IceBoy: Those only pay every other time.
 
good idea :D
 
Sorry @Sab I got disconnected.
 
The Apostol book is made by Wiley India
 
5:26 PM
@Sabಠ_ಠ I too have a Wiley, its thin, but still the quality is good.
I don't have much problems.
@Huy Hey.
@Sabಠ_ಠ I have Bartle and Sherbert of Wiley.
 
the apostol wiley has a bad printing quality paper very thin (already tore 2 pages)
 
@Sabಠ_ಠ Ah.
@Huy Hey! You here?
 
Huy
@Sawarnik: Yeah, what's up?
 
@BalarkaSen
Can you find the last two digits of ${}^{2012}3$?
 
@Huy I took up the real analysis book again, and crawled up the first section of the sequences chapter.
Still not sure I understood the whole.
 
5:32 PM
Nevermind @DanielFischer it was super-obvious. If there was such a sheet then one of the multivalues of the function would be continuous over it and thus all of the conjugates would be automatically continuous over $\Bbb C$ (the isomorphic copy of the sheet)
So the RS would just be $\Bbb C$
@Alizter $7$.
 
@BalarkaSen last two
 
Oh last two digits.
 
jinkz
 
Not sure. the last digit is an easy consequence of $a^{b^c} = a^{b^c \pmod {\phi(n)}} \pmod{n}$
perhaps you can use this one to compute modulo 100 too
 
@BalarkaSen :)
 
5:38 PM
Bye @BalarkaSir
 
byes
 
7 hours ago, by Balarka Sen
@IceBoy i don't think that's a good idea
 
@Alizter $87$, it seems.
 
@BalarkaSen why?
 
@IceBoy hmm?
oh
@IceBoy well it's just that many book has interrelated exercises
you can't do some if you haven't done something posed many chapters earlier
that was my point.
@Alizter 87 is correct, non?
 
5:41 PM
@BalarkaSen ok
 
Guys, I'm trying to write an algortihm that gives me {1,0,0; -1,0,0; 0,1,0; 0,-1,0} for 0,0,1 but can't think of a nice way
It is for getting upvectors for a given lookdirection.
Want them ortho
why the star?
 
Huy
Let $a_0 = 2$, $a_{n+1} = 2 a_n + 1$ and $A = \bigcup_{n=0}^\infty a_n$. Is it trivial that $a+b \notin A$ for all $a,b \in A$?
 
not that ugly with ifs but annoying.
 
@Alizter to reduce 3^3^...^3 modulo 100, you need to reduce 3^3^....^3 modulo 40(=phi(100)) and to do that you need to reduce 3^3^...^3 modulo 16 (=phi(40)) for which you need to reduce 3^3^...^3 modulo 8 (=phi(16)). But then any power of 3 is 1 or 3 modulo 8. use that.
 
Looking at this question I can easily think Cleo knows to manipulate CAS pretty well ...
42
Q: An integral involving Airy functions $\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx$

CleoI need your help with this integral: $$\mathcal{K}(p)=\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx,$$ where $\operatorname{Ai}$, $\operatorname{Bi}$ are Airy functions: $$\operatorname{Ai}\,x=\frac{1}{\pi}\int_0^\infty\cos\left(x\,z+\frac{z^3}{3}\right)\,\mathr...

 
5:56 PM
i like OL's answers
that guy knows his math
 
but since I don't know math :)
 
No one answered my question yet ...
5
Q: Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$

Chris's sisI remember that some time ago I was asking this question Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$ , and now, while I was making a review, I asked myself if we can get the closed form of $$\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$$ by using the similar tools as in that proof. The probl...

 

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