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13:34
Anyone with some knowledge in coding\information theory?
13:49
@Tunk-Fey for use in the chatroom, "start ChatJax" is better. It automatically updates your chat window rather than requiring you to click "render MathJax" repeatedly.
@robjohn You are up early.
@JasperLoy It's 6:50... is that that early?
@robjohn Yup. You should sleep till 10 am.
@JasperLoy Ick! think how much math I'd miss out on.
@robjohn Maybe you should quit your job and go back to math again, lol.
13:52
@JasperLoy I think you'll find a consensus that it is hard to get back into academia after an extended absence. It has been since 1988 since I taught at UCLA.
@robjohn Fortunately, I haven't entered academia yet, so I can take my time now before entering, lol.
@Studentmath why would the even $k$ be easier than the odd $k$?
14:27
@robjohn Thanks for suggestion. I use it now.
14:47
@robjohn What is your job, if you don't mind me asking?
14:58
He's the imperial overlord of MSE, @rehband. ^_^
Hey, @Huy. Do you want to play some FIFA?
How are you doing, @skullpatrol?
@Robjohn easier for the odd $k$, harder for the even $k$. But anyhow managed to prove what I wanted without proving this.
Turned to linear algebra instead
Were you trying to find $\displaystyle \sum_{k=1}^{n} \binom{n}{k}$, @Studentmath?
Nah, was trying to show that the weight of every code-word in the simplex code is $q^{r-1}$
Managed without the counting argument I tried to prove above
Oh. Tricky stuff!
The proof became much simpler when I turned from combinatorics to Linear Algebra, actually.. so I was as usual stupid trying to prove things the 'wrong' way
15:13
@Khallil True :)
Huy
Huy
@Khallil: Maybe later, in a few hours. Going through some proofs for my exam tomorrow, atm.
@Khallil: I have to warn you though. I am terribly bad. Out of practice, of course. :D
I always sucked with co-ordination when it came to FIFA. I can manage black ops and all these games, but FIFA..
Ah, ok. I'm out of practice too, @Huy. I got relegated to the third division in Ultimate Team: Seasons!
FIFA's always been my favourite, @Studentmath. Black Ops was pretty boring. The last good Call of Duty game (in my opinion) was Modern Warfare 2. World at War was ridiculous. You could sweep a whole team with the MP40 but the maps were pretty good.
Battlefield 4 is probably better than every single Call of Duty game.
@rehband wrote a program to teach students sentential logic. It is used in about 2 dozen universities worldwide (US, Japan, Canada mainly)
now I maintain and add to the program as needed
@robjohn Awesome!
15:27
@Studentmath so you know that for $n\gt0$, $$\sum_{k=0}^n\binom{n}{k}=2^n$$ and $$\sum_{k=0}^n(-1)^k\binom{n}{k}=0$$
I didn't know the latter. That would've made it much easier.
@Studentmath Expand $(1-1)^n$
Awesome
off to the store... BBL
Yeah that makes it much easier. Maybe I will try to go back to the original proof I had in mind just for practice.. thanks @robjohn!
15:37
The point with that "arrogant" thing is big misunderstanding. First of all, I'm not obliged to answer the questions of someone that didn't even greet me at that point, but just "solve this, solve that" and then he did things I asked him no to do them.
Things are very simple as regards myself: you respect me, I respect you, always.
Or I respect you, you respect me. (as you wish, the order you want)
I've never ever had a small, but a very small issue with @robjohn , @r9m and other persons that were nice.
The second point, I think everyone that comes here should focus on math, not on people's life, to characterize persons, the way they are. I don't think this is the right place for doing that.
Since between me and china math is a pretty long story, please do not enter in this game of labelling people, you do not know me as I do not know you.
I repet, I come here ONLY for math, never to judge someone, to get in others' business since I'm sure I'm not good enough for doing that and I might miss a lot of details.
Be sure in the real life I helped with math tutoring lots of persons, and many times for free.
I hope now this subject is completely closed.
@rehband However, I'll give you a point for that integral in Ovidiu's book.
15:54
hello, have you an idea about this: math.stackexchange.com/questions/908826/…
@rehband use this fact and then integrate by parts $$\frac{\log(1-x)}{1+x}=\frac{\partial}{\partial x}\int_0^ x\frac{\log(1-t)}{1+t} \ dt$$ After that you bring all in $2$ dimensions and there you need to recognize some type of integrals that can be tackled by some known formulas.
You've just got to be more open to how other people try to help you, @Chris'ssis.
@Khallil I don't "have", "must"," ought", but I am simply the way I am. I come here for math not to hear, read the others opinions about me. Anyone has the possibility to ignore me anytime.
16:10
This is a nice one: Define $A_i$ as the number of words of weight $i$ in a given code $C$ of length $n$. For a binary Hamming code, show the following holds for $1\le i<n$:
$$(i+1)A_{i+1}+A_i+(n-i+1)A_{i-1}={n\choose i}$$
Simply based on that reply, I'm going to ignore you for a while, @Chris'ssis.
@Khallil Thank you! Great decision! Go ahead. (+1) I only hope you are not the only one but all those that starred the message.
I just entered the drama club again. I could swear this was a math chat last time I checked.
@robjohn there is something really nice I wanted to show you.
$$\sum_{n=1}^{ \infty}\frac{1}{ n}\left( 1+\frac{1}{2}+ \cdots +\frac{1}{n} \right) \left( 1+\frac{1}{2}+ \cdots+ \frac{1}{n}\right)\left(\zeta(2)-1-\frac{1}{2^2}-\cdots-\frac{1}{n^2}\right)$$
$$\sum_{n=1}^{ \infty}\frac{1}{ n}\left( 1+\frac{1}{2}+ \cdots +\frac{1}{n} \right)\left( 1+\frac{1}{2}+ \cdots +\frac{1}{n} \right) \left( 1+\frac{1}{2}+ \cdots+ \frac{1}{n}\right)\left(\zeta(2)-1-\frac{1}{2^2}-\cdots-\frac{1}{n^2}\right)$$
@Chris'ssis Sry for the silly inquiry, but I'm not sure what you did there. What do we now integrate by parts? :) Just ignore this if u don't wanna give anymore details on this one
16:22
@rehband Did you get an integral in 2 variables? The point is to get a double integral over $[0,1]^2$ first.
@Chris'ssis No :)
@rehband hehe, get that first. I like your attitude, you always talk about math and seem very interested in learning. :-)
@Chris'ssis I'm still a newbie, but try to learn as much as I can every day! In a few years, I'll catch up :P
@rehband I'm sure you'll be great one day! Very serious! I rarely see persons such focused on math things.
@Chris'ssis Thanks! Your work ethic is an inspiration :P
16:30
@rehband And you know what? I'd be proud to learn things from you one day!
Huy
Huy
@Khallil: I see you had a short dialogue with Chris'sis. xD
@Chris'ssis: How are you today?
@Huy In a very great shape, I did some important discoveries in the series area. :-) How about you? :-)
@rehband Thanks! :-)
Huy
Huy
@Chris'ssis: So, so. I have an exam tomorrow and I'm starting to understand what's going on in the subject I studied but I need more time to really grasp it. I hope the professor won't be too strict. :)
@Chris'ssis: I am not sure but I think I read that you are self-taught and did not study maths, or am I confusing you with someone else?
@Huy Did you have enough time to learn?
Huy
Huy
@Chris'ssis: I was busy with my private life, unfortunately, and thus didn't manage to spend as much time with learning as I wanted to.
16:34
@Huy It's true, I'm self-taught, I have no background in mathematics.
Huy
Huy
@Chris'ssis: May I ask where your great interest for evaluating these kinds of series/integrals comes from?
@Huy Sure. Some time ago I learned mathematics very much for entering an university, and at that moment I felt in love with this stuff.
Huy
Huy
@Chris'ssis: How come you didn't enter that university, then?
@Huy I entered with a perfect score, but I dropped out after a while.
Huy
Huy
@Chris'ssis: Do you mind sharing the reason?
@Khallil: What's your PSN? I have never played online before, so it might take a while until I configured everything properly.
@Khallil: I assume I'll have to add you outside of FIFA first?
16:38
Who know something about homotopy invariance on homology ?
@Huy There were more reasons, but I prefer not to share them.
Yea, @Huy.
Huy
Huy
@Chris'ssis: Fair enough. Are these series and integrals the only things in maths that fascinated you that much or are there also other areas you love, but just don't have enough time at the moment to work in?
@Huy I love analysis, and I also like geometry pretty much, but at the moment I prefer to work more on some limits, series and integrals. Actually, I have some interesting ideas of solving pretty hard series by using geometry, but I need to work some more to develop and finalize my tools.
It's going to be something pretty new I think.
Huy
Huy
@Chris'ssis: Looking forward to it.
16:44
OK
Huy
Huy
@Khallil: Where art thou?
@Chris'ssis I think you're awesome, just figure I'd throw that out there
2
@MickLH Thanks! :-)
I am of high social value in real life as I can "talk smooth" with new people. I also never insist on something I have not *very* recently proven, so I am well trusted in real life as well (since I extremely rarely provide false information). On top of all of that I've never had a problem making any amount of money I need, because I can create you a computer from sand. I know the details from computer science and software all the way down to the chemistry and physics.

You are free to form your own opinions, and I understand if you don't believe me. I do hope that you value my positive opin
Huy
Huy
@MickLH: I am very proud of you.
16:53
fun problem, don't google : choose two random points on the unit circle. what's the probability that the staightline joining those two is of length 1?
@MickLH!
How's life?
@Huy I hope these are genuine sentiments :) I do not have any preconceptions of you, even though my words may have suggested it.
@BalarkaSen It's good! Things are going pretty well, I just did a consultation yesterday for someone and I might be starting another gig today
great. what of the CAS stuff?
I've put the math program we talked about on the "back burner" because it needs a support system now, but if you have a use for it in the near future I can separate the code into library
Hello
So I came across this problem
But it's still cooking along at a reasonable pace! It's just that I'm not implementing solvers right now until the User Interface is completed, @BalarkaSen
hey @VibhavPant
16:57
@MickLH it's a pretty cool stuff. if you have the time, do that. i'll happily have it stored in my machine.
Hi @MickLH
Huy
Huy
@MickLH: Let's just say I don't particularly like it when people feel the need to emphasise their skills - whatever they may be. It feels to me as if they need to compensate something. In my opinion, if someone is very skilled in some particular area, they wouldn't need to talk about their skills explicitly but people around them would notice themselves. But that's just my opinion. :)
@BalarkaSen Alright! I'll package it up and put it on GitHub, expect it within a week or two
gah coding.
cool though it is, frustrating when i can't get it right
I'll give a command line interface most likely
I'll most likely make it a front-end to MACSYMA / Maxima, so that the lack of good integration support doesn't get in the way of using the numerical computation
16:59
@rehband $$\int_0^1 \underbrace{\log(1+x) }_{f(x)}\frac{\partial}{\partial x}\underbrace{\int_0^ x\frac{\log(1-t)}{1+t} \ dt}_{g(x)} \ dx$$. So, you have there $f(x)\cdot g'(x)$
for those who are wondering, i am writing up a code for calculating a first hundred zeros of zeta on the 1/2-line
"If $f$ is a polynomial, prove $f(\bar x) = \overline{f(x)}$
@BalarkaSen ooh, interesting
@MickLH i look forward to it.
@Huy I find this view to be highly effective, but I believe it breaks down on the internet. Since you all do not see me in day to day life, you could never see the extent of what I do
@VibhavPant $\bar{z}$ is the complex conjugation, right?
Huy
Huy
17:01
@MickLH: That is of course true. Then again, I believe, people on the internet should behave as if it was real life, and not differently, just because it's the internet. I do so, but I know many other people don't.
@BalarkaSen yes
@BalarkaSen also, have you put up the code somewhere?
@VibhavPant what have you tried?
@Huy Oh well, I still believe I made a good choice as it was an emotional setting where the information would be irrelevant by the time it can be proven "the slow way"
@VibhavPant i am still preparing it.
are you interested?
So I consider information delivered sub-optimally, to be more valuable than information not delivered.
17:02
yes
Huy
Huy
@MickLH: OK.
will ping it to you when i finish it. it's a PARI/GP code, which you'll find available online @Vibhav
@BalarkaSen I tried writing $f(z) = \sum_{i=0}^n a_iz^i$, but couldnt get any futher
@BalarkaSen thanks for the motivation, I've been moving slowly recently but I feel reinvigorated!
Unless, it is a result that $\overline{z^i} = \overline{(z)}^i$
17:04
@VibhavPant wait a sec. are you sure there weren't any condition of $f(z)$ imposed?
@BalarkaSen oops, $f$ is a polynomial
polynomial over what?
$\mathbb{R}$
Huy
Huy
@Khallil: Where on earth are you =_=
right, then it's true, @VibhavPant
17:05
However, it can have roots in $\mathbb{C}$
@Chris'ssis Ahh, right. Thanks! I'll have a look at that in a bit. I'm trying to compute $$\lim_{x\to1} \frac{ f_n(x) - f_{n-1}(x) }{(1-x)^n}$$ where $$f_1(x) = x,\,\, f_2(x)=x^x,\,\, f_3(x)=x^{x^x},...$$ at the moment.
@VibhavPant sure. and that means...?
@BalarkaSen Yeah, I needed to prove that
@rehband AWESOME!!! It's from Ovidiu's book, right? :-)
@BalarkaSen $f$ is a polynomial with real coefficients. Show $f(\overline z) = \overline{f(z)}$
17:07
@Chris'ssis Yep :)
@VibhavPant yes, ok, i got that much. i want you to prove that if $f$ is a real polynomial with root $z_0$ then $\overline{z_0}$ must necessarily be another root.
@rehband The beautiful challenge is to compute that without pen and paper. It's a very nice question. :D
@r9m can you help me on homology please ?
@Chris'ssis Already using pen & paper :P
:173334 hehe, it's OK :-)
17:09
@VibhavPant actually you need something called the fundamental theorem of algebra to start with this.
That all polynomials have roots in $\mathbb{C}$?
(IIRC)
yes.
all polynomials over C, btw
polynomials over floopskywhoomskitunks may or may not have roots in C
i have to run. gotta finish coding.
$\mathbb{F}$ ?
r9m
r9m
17:11
@Vrouvrou I don't know Homology ^^ .. its too advanced for me ..
So, anyone to help me with my question?
@rehband I don't know how many copies he sold so far, but I can imagine it's about tons of copies (taken into account the quality of the content).
r9m
r9m
@Chris'ssis sweeet !! :D you have Elementary solutions ? ;)
thank you, then who can help me ?
@Chris'ssis It would be well deserved
r9m
r9m
17:13
@Vrouvrou idk :( .. you can check who all answered the homology tag in the main .. and see if one of them is available in a chat room :) ;)
@rehband You know, if you love these things and see that book, you definitely want to publish such a book, especially if you like to create math stuff.
Does proof by Mathematica count as viable these days?
Sometimes I ask it to refine, and it just goes "true"
@Chris'ssis I can imagine! :)
@rehband It also works by using the elementary limit (I think) $$\lim_{x\to 0} \frac{e^{x}-1}{x}$$
Suppose $n=3$, then $$\lim_{x\to 1} \frac{x^{x^x}-x^x}{(1-x)^3}$$
@Chris'ssis I just got the other solution. I love these unexpected applications of the Mean Value Theorem
17:23
@rehband Using the exponential function combined with that limit one sequencially get the final limit after some applications of it in cascade.
@Chris'ssis Okay
@rehband Something like that was given on one of the MIT-Harvard tournaments. I don't remember the limit exactly.
@Chris'ssis Haha nice. Very nice problem indeed
@rehband It was a small case anyway (a small tower I mean).
@rehband Indeed, it's really nice.
It has been a nice morning, but I've gotta close Stack Exchange to boost productivity. Have a good night everyone!
17:37
@MickLH Have a nice day! You're full of positivity! ;-)
@rehband I think I'll do some research on 3.27.
18:01
@rehband $$\sum_{n=2}^{\infty} \frac{n+2}{n(n+1)2^{n+1}}(n-\zeta(2)-\zeta(3)-\cdots -\zeta(n))=\log(2) +\frac{1}{2}\gamma-\frac{3}{8}-\frac{1}{2}\log(\pi)$$
2
Wait, I come with another ...
@Chris'ssis weirdo
@rehband $$\sum_{n=2}^{\infty } \frac{n+4}{(n^2+3n+2) 2^{n+2}}( n-\zeta(2)-\zeta(3)-\cdots -\zeta(n))=\frac{3}{2}\log(A)+\frac{1}{8}\gamma+ \frac{19}{48}- \frac{19}{24}\log(2)-\frac{1}{4}\log(\pi)$$
still more weirdo
@blue
here comes absolute blue
I can't believe 9 people starred arrogant, they must be mad.
2
@JasperLoy truth is harsh
@Chris'ssis don't kill me!!!! =)
18:15
I think there is a problems with the second series.
I do not think she is arrogant. It seems that people on the internet are stupid and evil just like the people around me.
@Chris'ssis Wow crazy! What's A? Is it that weird constant?
Glaisher Kinkelin
@rehband Glaisher-Kinkelin constant
Right
18:17
@JasperLoy everyone is evil except you
@rehband I need to recheck all I did there. Something might be possibly wrong.
@BalarkaSen That sounds sarcastic.
@JasperLoy it is a sarcasm
@BalarkaSen Since I know you, even if you starred that you probably did it for fun, but I'm sure you don't believe I'm arrogant. :-) We never shared such thoughts.
@Chris'ssis i don't as i have already said i didn't star it
i was joking
18:19
I no longer care about what people think about me. Because they are not fit to judge me.
3
@BalarkaSen hehe, OK =)
let it go. don't care about it anymore, @Chris'ssis
@JasperLoy Well-said! :-)
@JasperLoy it was probably all a misunderstanding.
but if @Chris'ssis wasn't joking, what she said to chinamath was pretty mean
In real life, I got into trouble because I said a certain 5 letter word. This incident showed me that many people around me are just extremely stupid and evil.
I no longer feel that I belong to the society I live in.
18:22
@JasperLoy Exactly the same feeling here.
@JasperLoy better not say it if you don't want to get in trouble.
why fall in a trap deliberately?
@Sawarnik!
Huy
Huy
@JasperLoy: Good luck getting through life.
@BalarkaSen Hey.
I think I have not done any maths in a week now, and the month ahead offers no possibilities :(
CBSE nonsense stuff. And then SA.
@Sawarnik i have got another interview sigh
just can't cope up with these stuffs
lol :P
You have no school pressure, and all the free time in the world. sigh
18:25
@Sawarnik Do you still talk to Rachel?
@JasperLoy Rachel?
@Sawarnik math101
@Huy Is your preparation done? :)
Huy
Huy
@JasperLoy: You know, playing League of Legends, I have noticed that a lot of people think "all people except for me are stupid". I think those people should start thinking about themselves, at some point, but that's just like my opinion.
@JasperLoy Nopes.
Huy
Huy
18:26
@Sawarnik: Probably for today, I'll get up at around 4am to study some more before the exam.
@JasperLoy I'm unable to understand the evilness some feed themselves with daily. I think it's a genetic issue.
Oo .. getting up at 4am to study ew .. I thought people sleep at 4am after studying
@Sawarnik RACHE!
HAHAHA
Huy
Huy
@Sawarnik: Most people, including me, can't really soak up new information in the evening but in the morning. That's why I stop studying at around 8pm and rather wake up really early.
18:28
I don't talk with Rachel over email either these days.
@BalarkaSen That's what came to my mind first! LOL!
:D :D
So what is so funny?
@Huy Most people? I don't think so. @Balarka Is it?
@JasperLoy HAHAHAHA
18:29
@JasperLoy You won't understand :P :P :P
@BalarkaSen Why do you laugh? Is it a funny name?
I am really rolling on the floor.
Huy
Huy
@Sawarnik: I'm pretty sure there are some clear statistics.
@BalarkaSen Great work!
@JasperLoy RACHE, my dear Watson, is the German for "Revenge".
18:29
@JasperLoy I am laughing still! :P
@BalarkaSen Oh OK, but not really funny.
@JasperLoy Read A Study in Scarlet. Then tell :P
Funny if you have read Sherlock Holmes
OK. I thought she changed her hairdo or something.
@rehband It looks like both results are OK.
18:34
@BalarkaSen I'm really trying to keep on pushing forward, but I keep getting distracted with finite element analysis
@MickLH Finite element analysis?
Yes I love simulating things of the real world by numerical approximations
I dream of making a 3D soft-body physics engine which operates directly on triangle meshes
Hmm. Examples?
@MickLH I didn't even get what you just said.
Say you're designing something and need to model the heat flow
18:40
The way I like to do this is by tessellating the geometry and approximating the flux over time
I'm back, @Huy.
What's up?
@MickLH I am not familiar with that stuff.
No worries, lol I'm probably one of the worst around here at math, but I sure do spend a lot of time applying it
Huy
Huy
@Khallil: I was waiting for you for ages. Maybe tomorrow then.
You're interested in CS, right @MickLH?
18:42
Oh, for FIFA.
Yea, tomorrow!
@BalarkaSen Yeah, CS and EE have been my hobbies literally since as long as I can remember (and before then too according to friends / family)
Good luck for the exam, @Huy!
@MickLH cool. you do seem to have an expertise in that area
:D I try to not suck lol
@MickLH Have you tried coding up an efficient algo for calculating zeros of zeta?
18:44
@MickLH I miss Alyssa.
@BalarkaSen I don't think so, all my endeavors have been application oriented. I don't know how I would apply the zeta zeros
@JasperLoy :(
@MickLH no, I mean for your CAS
@JasperLoy same here man, it seems like she just disappeared
@MickLH I just emailed her my new email addy.
18:46
who are you guys talking about? meer2kat?
@MickLH Due to me :(
@BalarkaSen Yeup.
guessed right
@Khallil was asking me the same thing too.
@BalarkaSen ah, no there's only the basic solvers and basic symbolic manipulation at the moment
18:48
You kids like to (removed)
@JasperLoy i do that as less as possible
remember that kids are the most dogpiled
@JasperLoy Yes.
The most complex thing it can do without user assistance is rearranging expressions into a simplified polynomial, and then compute the roots of it using closed forms :P
@MickLH what about $z^5 - z - 1 = 0$?
@MickLH i thought you were implementing a lot of numerical computation stuffs like hypergeometric functions and whatnots. do you mean you haven't developed the symbolic manipulation part yet?
Yes, the symbolic manipulation is extremely weak, only supports basic algebra
I just tested that expression, it triggers a numerical root finder
18:51
ah that makes sense.
@MickLH i knew it'd =P
@Jasper Have you messaged her on Fb? :)
gives ~1.167
@Sawarnik I do not have FB, lol.
@MickLH $z^5 - z - 1$ has no closed-form root. you need to implement soem hypergeometric stuff to have a closed form (see Glasser's "Quadratic Equation Made Hard : A Less Radical Approach")
Ok.
Bye.
18:53
@BalarkaSen I figured but I couldn't remember if it stopped at 4 or 5 :P
@MickLH how about $z^5 + z + 1$
check that one. it's solvable.
falls under the same numerical method
Basically it just does bisection for those
ah, then it stops at 5 altogether.
@MickLH it's factorizable though.
$x^5 + x + 1 = (x^3 - x^2 + 1)(x^2 + x + 1)$
you might want to implement factorization in your CAS
I actually removed all the symbolic stuff for now, because I had a better insight into how to organize the data
good luck!
18:56
thanks, also earlier I was referring to the basic numerical solvers that I'll make available in the near future, I don't think I'll have re-implemented the symbolic stuff by then
@Mick talking of factorization, have you implemented something for integer factorization?
or primality tests? AKS should be easy enough to code.
yes there's a prime factor routine, I forget how it even works now it's been so long
interesting.
I remember it starts with trial divisions to try and deal with most common problems, or at least make them smaller
I think it's some shady thing I made up at the time, I'm pretty sure it does some voodoo with computing greatest common factors
either way, I'll need to re-write it, I wanted to do a sieve but I never implemented the prime counting function
Jesus, what an identity I just derived!!!
19:09
@Chris'ssis :O I wanna see :D
@Hippalectryon btw, I posted above 2 very nice series ... :-)
@Chris'ssis I got the starred one, what is the other ?
@Hippalectryon $$\sum_{n=2}^{\infty} \frac{n+4}{(n^2+3n+2)2^{n+2}}(n-\zeta(2)-\zeta(3)-\cdots -\zeta(n))=\frac{3}{2}\log(A)+\frac{1}{8}\gamma+\frac{19}{48}-\frac{19}{24}\log(‌​2)-\frac{1}{4}\log(\pi)$$
Now, I'm going to combine my identity with a result by Ramanujan. Let me see what happens ... (brb)
chat is going off-line
19:22
How do you know?
ELL just went
Ah, maybe only the beta sites, lol.
I seriously don't know, of course.

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