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12:52 AM
@N3buchadnezzar I get $\frac12\log(2\pi)$. Was this from a question?
I will add my method, if it isn't the same.
 
@robjohn Yes, It was fro a question on a site. It used Eulers reflection formula
 
@N3buchadnezzar Ah, so mine is different.
@N3buchadnezzar was it on mse?
 
Yeah, I could find it but it is almost 3am here..
 
@N3buchadnezzar Okay, I will look
 
1:34 AM
Gey guys, why is 0.999.. = 1 ?
*hey
 
Because their differrence is infinitely small?
 
Hello all
 
@Nick I don't want to come across wrong... but you aren't a troll, right?
 
@Nick Do you know about geometric series? Specifically that when $-1<r<1$ $$\sum_{n=0}^\infty r^n =\frac 1{1-r}$$
Then let $r=9/10$.
 
1:50 AM
@PeterTamaroff I just saw. Suffice it to say that a lot has been done on roots related to partial sums and tails of the exponential series, but I don't see a quick way to use it to get that result.
 
@AntonioVargas Ah. Well, note that we are looking at roots of $r_n+r_{n-1}=0$; I think.
 
Right.
 
@PeterTamaroff For some reason, I'm having a hard time parsing your question. Either I'm very tired, or I don't know enough. In either case, I think I'll leave it to the experts tonight
other than that, how goes it all?
 
@mixedmath Good. I'm tired too but I've been too lazy today so I'm forcing myself to work.
And of course getting distracted by M.SE.
 
@Antonio: Tomorrow, I'll be meeting with lots of people with a lot more credentials and initials after their names than I have. So today, I've been working working working so as to not embarrass myself.
So I reward myself by going to MSE and answering a few old but easy unanswereds
2
the easy part - very important though
 
2:02 AM
Congrats on your productivity :)
@PeterTamaroff You might find some useful ideas in Section 1.2 of my msc thesis and Appendices A and B of Mallison's phd thesis. Like I said the results can't be applied directly but it's where I'd start.
 
@mixedmath Good, thank you. I wish you luck and success tomorrow!
 
@PeterTamaroff haha don't worry about that downvote. Personally I like it when good though incomplete ideas are posted as answers.
 
2:25 AM
hi guys
What is the connection between mathematical induction and implication? I always see that mathematical induction is about

P(k)->P(k+1).
From what I know, mathematical induction works by finding a way to transform P(k) into P(k+1) and if you can do it, then you prove it.

Now, speaking in terms of logic formality, we can treat P(k) as statement A, and treat P(k+1) as statement B. We know that an implication is true if statement A is true and Statement B is true; A is false and B is true/false (implication is false if A is true and B is false).
 
@AntonioVargas Ah! You're a pro!
 
I'm heading home. G'night all.
 
@vvavepacket If A is true, and A implies B, and B implies C, and C implies D, and D implies E, and so on, then obviously all of the statements are true.
Given A's truth, the first implication yields B's truth, given B's truth, the second implication yields C's truth, and so on.
 
@anon ok. btw, where did you get your profile picture? I like it
 
don't remember
 
2:55 AM
@anon, what if we try to derive P(k+1) from P(k), but we failed to derived exactly, yet when we plug in different values of k, it still holds. what do you think, will that be still considered proved?
 
huh?
if you fail to to create a valid induction argument but P(k) seems to hold for all k anyway, the most you can say is "I failed to create a valid induction argument but P(k) seems to hold for all k anyway."
 
@AlexanderGruber Dude!
 
ahh. so does that mean P(k) -> P(k+1) is still true?
 
it's true whenever P(k) and P(k+1) are true.
 
how do we know if P(k+1) is true?
how do we know if P(k+1) is true?
 
3:17 AM
your questions are too broad. that's why I've been forced to only give broad answers. you can't just ask "how do we know if a statement is true" (absolutely no description of what statement is being referenced) and expect a precise or concrete answer
you might as well ask me "how do we know if something is true?"
entire bookshelves are necessary to answer something like that in detail
simply because there are so many "somethings," so many possible statements, so many possible P(k+1)'s, that one could be talking about
 
 
1 hour later…
4:33 AM
ERMAGHERD! @anon
 
go on
 
what
 
@anon I'm second in the weekly rep chart.
@anon What time is it there?
 
midnight minus 20 min
 
4:39 AM
@anon Ah, two hours less.
 
 
1 hour later…
5:58 AM
@PeterTamaroff are you around?
 
@AntonioVargas Yep.
Where are you from?
 
I'm in Halifax, NS right now.
3am
 
I mean, originally.
 
heh
Oh I'm from California
 
@AntonioVargas We have the same hour zone. Take a step further, drop, and you'll fall on my head.
@AntonioVargas So that is where the hispanic name comes from.
 
6:01 AM
Indeed, my dad's family is from Mexico.
 
@AntonioVargas So, what's the news?
 
I wanted to ask about the edit on your answer
I really do think the estimate requires the alternating sign property of the remainder
 
@AntonioVargas How so?
 
you wrote that $r_n(x)=(-1)^ne^{x'}\frac{x^{n+1}}{(n+1)!}$ implies $e^{-x}|r_n(x)|\leq e^{-x}\frac{x^{n+1}}{(n+1)!}$, but I think it only implies that $e^{-x}|r_n(x)|\leq e^{x'-x}\frac{x^{n+1}}{(n+1)!}$, where $0 \leq x' \leq x$.
 
@AntonioVargas Sorry, I mean $e^{-x'}$. This is clearly less than one.
 
6:05 AM
oh, derp, yeah you're right. I wonder what julien was going on about in his comment on my answer then
 
@AntonioVargas Heh, no clue!
I took a look at your thesis. The images are mesmerizing!
Of course, I don't think I can read it now.
Read "now" as a mathematical moment.
 
Glad you like the pictures. I had a lot of fun making them.
 
@AntonioVargas Did you see Grüber's question?
 
where?
 
6:08 AM
Oh yeah. Pretty neat.
 
@AntonioVargas "I like studying the zeros of functions." LOL
(zeroes?)
 
Zeros is correct :)
 
@AntonioVargas Is it an American vs. British thing?
 
Not sure. I've only ever seen "zeros".
 
I am more on the Bs. side, using colour, flavour, programme, &c.
 
6:10 AM
28
Q: What is the plural form of "zero"?

Doctor JonesI tried looking on Google, but there are some fairly contradictory results. I thought I'd ask you guys so we could get an authoritative answer on the subject!

Everything is on stackexchange.
 
@AntonioVargas Heh. Let's not presume.
@AntonioVargas How often have you used Laplace's Method?
 
"It's interesting to note that the Oxford Dictionaries site's sole definition of zeroes is related to zero as a verb, e.g. 'watch as he zeroes his sights on the target'; not as the plural of zero." --- this aligns with my experience
@PeterTamaroff A few times successfully.
Its big brother is the so-called "saddle point method" which I recently used to find an asymptotic for the Euler numbers. You can get an integral representation for them by using Cauchy's integral formula on their generating function.
I think I've answered a couple of questions on here on Laplace's method, I'll find you the links...
Hrm, this does not deserve nearly so many upvotes.
 
@AntonioVargas Ah, I can guess where that is going, but having seen the proof for the unidimensional case, I fear drowning in variables.
@AntonioVargas Tell me about it - but don't upvote me! I've capped!
@AntonioVargas (I did upvote, BTW).
@AntonioVargas Sheesh! Euler numbers do grow rapidly.
I like Eulerian numbers.
@TobiasKildetoft Do people call you Toby?
 
6:28 AM
@PeterTamaroff I found six...
 
@AntonioVargas Six what?
 
@PeterTamaroff a few, but only people I almost never see
 
@PeterTamaroff Six of my answers on integral asymptotics I thought you might like.
@PeterTamaroff first, second, third, fourth, fifth, and sixth
You might also take a look at the wiki page on Watson's Lemma. I wrote the example there.
It needs a proof though. Maybe I'll write one for it this week.
 
@AntonioVargas Nice! =)
@TobiasKildetoft Ah.
 
@PeterTamaroff Alright, I really should get to sleep. Or else I'll be an irritable tutor tomorrow. G'night.
 
6:38 AM
@AntonioVargas Leave no man standing.
Or let the men stand. But not their brains.
Or neither.
 
6:53 AM
what is a linear set ?
 
@DominicMichaelis in what context?
 
Maybe some context could help, I am reading measures and categories and he introduces the Baire classes in the very beginning
And the sentence is "In fact the closure of a linear set $A$ is of first category if and only if $A$ is nowhere dense.
 
@DominicMichaelis none of that looks familiar to me
 
A set is of the first Bairy category when you can write is as the countable union of nowhere dense sets
He is currently just talking about $\mathbb{R}$, could that mean the linear ? But I think that would be kind of strange.
 
He might mean linear as in a set where the topology comes from a linear (ie, total) ordering
 
7:12 AM
oh ok
 
@DominicMichaelis But you'll never know. MUAHAHAHAHAHAHA.
 
7:52 AM
Greetings
Here is a delightful solution by robjohn -> math.stackexchange.com/questions/167206/…
 
8:19 AM
Greetings, thanks for sharing :D
 
@skullpatrol :D
 
9:17 AM
Do groups of order 768 have some nice shared properties? For example, they are all solvable, but what about stronger properties? The reason I ask is that 768 is the smallest composite order where GAP does not have a full library of the group.
 
 
1 hour later…
10:20 AM
I'm a Physics major. Is it a waste of time to become a linux system administrator?
is it like manual labour? or is it important to mathematicians/physicists working at colleges
 
10:40 AM
@Raindrop Don't waste your time, do what is important to you, in my opinion.
 
10:57 AM
@skullpatrol it was a time when I thought life has a meaning, but it has no meaning, excepting the meaning you give it.
A run to nowhere.
 
11:46 AM
Once again, you have lived up to your user name @Chris'swisesister
 
@Chris'swisesister Hi
 
12:06 PM
I don't know if this is the right place to ask this but can anyone tell me why, in logical reasoning, when the original statement is true, nothing except the contrapositive of that statement is true
 
1:00 PM
@Nick I am not sure I understand what you mean. What statement? (Contrapositive does not make sense for an arbitrary statement)
 
IMX
Hey Tobias
I have a question you might know how to answer
I have this here: e^(xy)*y (x^2 + y^4 + 2xy^3) + e^(xy) (2xy^4+2y^3)
When I simplify this, I get e^(xy)(x^2y^5+4xy^4+2y^3)
But shouldn't it be ..4xy^7..?
since adding 2xy^3 and 2xy^4 somehow shouldn't make 2xy^4
 
note that you have an extra y in the first part
so your 2xy^3 term there becomes 2xy^4 which gives the 4xy^4 when added to the 2xy^4 from the other part
not sure where you get an x^2y^5 term from
 
IMX
is y^4 + y^4 not y^8?
 
no
it is 2y^4
 
IMX
Why not?
 
1:06 PM
to get y^8 you need to multiply them
 
IMX
Ah ok
Thanks mate
Much appreciated
 
@N3buchadnezzar hi
I'm working on $$ \sum_{k=1}^{\infty} \text{Ci}(2 k\pi)$$
 
Any progress?
 
@N3buchadnezzar I wonder if this is to be found in Table of Integrals, Series, and Products (I need to check that right now). At the moment I admire it more than doing something concrete. :-)
 
1:24 PM
:-)
 
1:38 PM
@N3buchadnezzar some people are born mathematicians, physicists, but this wasn't my case ever. I needed to work so much (hard to measure in words) to play with these beautiful things (and I'm aware I need to work much more than that). My dream is to be able one day to find amazing things as Ramanujan did.
 
Dito, I love mathematics but yeah, it is very hard..
 
 
2 hours later…
3:18 PM
@skull hello
@argon $\huge \text {AARON?}$
 
@Charlie Marilia?
 
@Argon YES
 
@Charlie HI!
 
@Argon HI
 
I will be back...!
 
3:23 PM
@Argon YAY!
YOU
hi @jayesh
Today is Jonas' Birthday!
 
3:51 PM
@Charlie Hi.
 
4:07 PM
hey
 
@N3buchadnezzar hey
@N3buchadnezzar how are you $\varnothing \text {instein}$?
 
Fine, doing integrals, swinging at parties.
And you ?
 
@N3buchadnezzar I'm fine, too
 
=)
Schools out
 
@JayeshBadwaik how is it going? :-)
 
4:21 PM
@N3buchadnezzar here too
@Chris'swisesister hies
 
@Charlie hi :-)
 
@Chris'swisesister how are you?
 
@Charlie working and creating some beautiful series. You may see one above. How about you? I didn't see you around for some time.
 
@Chris'swisesister I'm fine, thanks. I was taking a time .....
 
@Charlie Who did you take it from ?
 
4:31 PM
@N3buchadnezzar everyone
some people just piss me off
 
@Charlie =(
 
yeah
 
Life is just a competition in eating enough **** without giving up.
 
what **** stands for?
 
It deppends on your view of life.
 
4:36 PM
hmm
i can't think of anything that fits
@carol what part of Brazil are you from?
 
hi :) brasilia!
 
hey guys, can I ask a dumb question about eigenvectors?
 
@Carol nesse Brasil lugar melhor não há
@Clash do it
 
4:52 PM
I'm quite stuck on how to calculate eigenvectors. I've been trying to do $(A-\lambda I)x=0$ and failing miserably. I have for the example the Identity Matrix 2x2, $I_2$. The eigenvalues are $1$ and $-1$. For the eigenvalue $1$ I will get the null matrix. How am I supposed to get a eigenvector from that?
 
well yes, it works pretty well with matrices that end up with linear dependent rows
however I'm stuck when the lines aren't multiple of eachother
 
@Clash give us your matrix :)
 
I want to calculate the eigenvectors from the identity matrix
 
ok
 
4:57 PM
For the eigenvalue 1, it gets me the null matrix. For -1 I get independent lines, which results on the eigenvector null. I must be doing something wrong
 
no
it is really simple what I want
i want to calculate the eigenvectors of the identity matrix, it shouldn't get any simpler than that
 
@Chris'swisesister In this math.stackexchange.com/questions/285130/… nice answer, how do you prove the second equality on the second line?
 
I've posted it here
http://math.stackexchange.com/questions/444319/calculating-eigenvectors-from-the-identity-matrix
 
@Clash any linearly independent vectors are eigenvectors
 
5:17 PM
Charlie, thanks for the help... i've found my mistak
wasted 1 hour on this crap and it was right on my face
 
@Clash oh!
@Clash thanks for sharing your problem
 
5:43 PM
hola @peter
 
@Alyosha did you hear of the incomplete beta function?
 
@Charlie $\sup$?
 
@PeterTamaroff not much. and you?
 
Hi @Charlie how are you?
 
@skullpatrol hi, Skull, I'm fine... and you?
 
5:50 PM
@Charlie Fine thanks.
 
@skullpatrol :)
 
@Charlie :D
 
@Charlie Trying to see when I will sign up for the finals.
23rd? 30th? 6th? Who knows!
 
@PeterTamaroff weird...
I can't believe that Flinn from Glee died...
 
5:53 PM
@PeterTamaroff I can't choose
 
@Charlie Really?
 
@PeterTamaroff yes....
 
@Charlie He abused drugs, it seems. Pretty usual amongst "famous" people.
 
yep .....
 
@Charlie So the above was only a saying?
 
5:55 PM
@PeterTamaroff yes....
 
Ah, the whole crew is here!
 
@Argon glee crew?
what are they doing there?
 
@Charlie Nope, math.SE crew
glee?
 
@Argon aah!
@Argon that tv show
 
@Alyosha First of all let the variable change t^2=u, and then combine the identities (1.1) & (1.3) here 129.81.170.14/~vhm/papers_html/final11.pdf. Done!
 
6:00 PM
Hi @Chris'swisesister
 
@Argon hi :-)
How are you?
 
@Chris'swisesister Very well, and you?
 
@Argon I'm thinking at one of the most beautiful things I met in the last period of time
 
@Chris'swisesister New sum/integral/limit?
 
@Argon $$\sum_{k=1}^{\infty} \text{Ci}(2 k\pi)$$
You guessed! :-)
 
6:02 PM
I've seen it in relation to the digamma (?)
 
I think
But I may be mistaken
Where do you get these from??
 
This came from my mind. Hard to explain how they come to me.
 
Ah, very nice
I asked a question a while ago about a sum with Si(x) before, did you see it?
No closed form exist (that I can see), but cool alternate forms did
 
@Argon no. Where is it?
 
6:05 PM
1 sec...
 
ok
(no hurry)
:-)
 
17
Q: Evaluation of $\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)$?

ArgonI would like to evaluate the sum $$ \sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right) $$ Where $\operatorname{Si}$ is the sine integral, defined as: $$\operatorname{Si}(x) := \int_0^x \frac{\sin t}{t}\, dt$$ I found that the sum could be also written as $$ -\sum\limits...

 
2 seconds later...
 
Ooooooo :-)
 
@Chris'swisesister Hey
 
6:08 PM
@N3buchadnezzar back
 
$$ \int_0^a \frac{f(x)}{f(x) - f(a-x)} \,\mathrm{d}x = \frac{1}{2}a $$
 
@Argon very nice question with very nice answers!
 
Nicer answers! :)
 
I upvoted!
 
Yay!
@Chris'swisesister arxiv.org/pdf/1008.0040v2.pdf , Corollary 7
 
6:11 PM
@N3buchadnezzar I think you have "+" sign in denominator.
 
@Chris'swisesister Yeah. Anyway does this holds for absolutely every single function ?
 
@Argon oh, amazing. How do you find these papers? :-)
 
@Chris'swisesister Completely incedental. I saw this paper recently and remembered the cool sums with Ci. I just remembered it once yo brought it up again just now :)
Check out Proposition 4
 
hehe, nice
 
The proof has hypergeometrics too. All your favourite things packed into one paper! :D
hehe
 
6:16 PM
@Argon yeah, it's an awesome pack! I need to study the paper in details.
 
@Chris'swisesister So do I. I didn't look at the proofs really
 
@anon hi blond boy avatar guy
 
@Charlie blond?
 
@Argon I called the paper as: mindblowing_stuff :)))))))))
 
@Charlie hello purple eyehand avatar girl
 
6:18 PM
@Chris'swisesister An apt name! :D
 
@anon how are you doing?
@JayeshBadwaik yes
 
or blonde
 
@Charlie I am tired. Worked with uncles to take down three trees. And you?
 
6:19 PM
@anon i'm fine.... no trees
@JayeshBadwaik he got it, okay? @anon got it
 
yes
 
@anon Its better than working with a tree to take down three uncles.
 
both are accepted
 
@JayeshBadwaik $\huge \text {YEEEEEEEEEEES}$
 
6:20 PM
no
 
@anon You should be taking down Cayley graphs. Eventually you will get to taking down trees.
 
blond or blonde
 
herpaderp
 
@JayeshBadwaik my yes is stronger than your no. SHUT UP!
 
6:21 PM
yes
1 min ago, by skullpatrol
both are accepted
blond or blonde
Go to the OED
 
@JayeshBadwaik AAAAAAh
 
@skullpatrol I was not correcting her spelling. :P
@Charlie :-)
 
oh. pardon me :(
 
@skullpatrol he likes to irritate me
 
My uni network admin so silly, he blocked wget. :P
On the other hand, it seems arxiv is silly and not my uni network admin, they blocked wget! :P
 
6:25 PM
 
70's copy cats^
 
@skullpatrol True dat.
 
@JayeshBadwaik what this all mean?
 
The Texas Chainsaw massacre!
 
I didn't understand what is a class
in set theory
 
6:33 PM
public class SetTheory{ public static void main (String[] argv){} ....}
 
@Argon ohes nohes
 
@Charlie hehe don't worry... you aren't learning Java in programming :)
 
@Charlie Something that's too big to be a set.
 
@PeterTamaroff it makes no sense to me
 
6:35 PM
@Charlie What are you reading?
 
@PeterTamaroff set theory, by T. Jech
 
@Charlie That's a mighty book, M.
 
@PeterTamaroff and, P?
 
my understanding is probably off, but here it is: "properties" can be thought of as either descriptions that are in some formalish language, or alternatively as the set of all things having said property (so, a "property" is ultimately just a set, and having property = being in the set). classes correspond to predicates without reference to the idea of "all things this predicate describes," making no assumptions about things like that.
 
@Charlie Let me rephrase: that is a good book.
 
6:37 PM
I have heard it is notorious :)
 
@PeterTamaroff so what? "it's a good book" itdoesn't mean i have the power to understand everything that is in it!!!!
I'm good
it doesn't mean you understand me
 
notorious B.I.G.
 
@Charlie You have a particular deductive system.
 
@anon I think now i get it
 
@PeterTamaroff Better watch it Pedro: chat.stackexchange.com/transcript/36?m=9169690#9169690 :)
 
6:40 PM
hahahaha YES
 
@PeterTamaroff How are things?
 
@Argon Haven't heard from them in a while. Personally, I am trying to choose a date to give my finals. I can't start studying if I don't commit.
 
Heh i read that totatly wrong. Like you need a date for your finals.
 
hahahahahahhahaah
 
@N3buchadnezzar Wouldn't be bad, right? ;)
 
6:43 PM
@PeterTamaroff Very true. Godspeed.
 
@PeterTamaroff Could only make it better ;)
 
@N3buchadnezzar But not for the final mark!
 
Hi @amWhy how are you?
 
@Argon Deppends, could be a motivational factor to finish fast.
 
@N3buchadnezzar But do poorly nonetheless :)
 
6:44 PM
@skullpatrol Hello! I'm okay. How are you doing?
 
@Argon Yeah, finish fast is not always good
 
@amWhy chillin'
 
@skullpatrol Like what, for example?
 
@Argon listenin' to the above, for example...
 
6:47 PM
@skullpatrol don't get too "cool" ;-)
 
@skullpatrol The right answer: A villian
 
@Argon notorious B.I.G. is the original gansta villain* :D
 
Although
He is not as cool as Coolio
 
True dat^ Coolio is no foolio
 
As I walk through the valley of death
 
6:59 PM
@skullpatrol so....
 

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