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06:00
ya should
i read many books in my literature classes: i am the cheese is quite fucked up. number the stars is good, book set during nazi germany.
@Mike I am reading Starship troopers now.
i haven't taken lit classes past high school
i read for entertainment
well, those clases where during english highschool
they picked cool books
starship troopers film was far better than the book
book was unironic in its support of fascism
really?
there is this one book I never finished which is called "the giver", and also "brave new world"
two awesome books i'd like to finish up reading.
the first one was given to me by a teacher, kinda like a challenge
@Mike I don't remember the movie that much
06:03
giver was good if simplistic
@Mike you read it?
yes
was it worth it?
its a cute story
and an easy read
hehe, Wikipedia says it has been marked as "inappropriate for children"
06:07
that's odd. it's more a children's book than not
read brave new world?
yes, was pretty meh on the experience
agree with mike on the giver. never read brave new world. although it reminds me of this
it is certainly interesting.
@anon agree on what?
there was a joke in it about how they play riemann surface tennis
you'd like that
06:09
hahaha
@anon i've seen that, i'm not sure i buy it.
I love the short story called "there will come soft rains"
hey guys
based on the poem with the same name
would you mind helping me out real fast?
06:11
@user60887 yes
0
Q: Riemann Integration theory problem

user60887Let $P$ be a tagged partition of $[0,3]$. Show that the union $U_1$ of all the sub intervals in $P$ with tags in $[0,1]$ satisfies $[0,1-||P||]\subseteq U_1\subseteq [0,1+||P||]$. (||P|| is the norm of partition P). proof: We first show that $[0,1-||P||]\subseteq U_1$ with tags in $[0,1]$. Supp...

i mind most things
I can get rational fast
ive been stuck on this for a while now
but completion is a tough process
06:13
i don't buy it
you're pretty irrational @Pedro
@anon @mike I read count the stars by lowry
cannot remember it thoroughly, but was a damn good read
@user60887 I'm a little busy now, but will try to get back to you
ok thanks though.
WHAAAAAAAAAAA
taking a bite out of crime. delicious, delicious crime.
06:24
@Mike did you watch the departed?
ages ago, i don't remember it too well
fuck I had forgotten leo dies in this one
leonardo di caprio vs matt damon, everybody loses
everyone dies!
cept wahlberg
06:26
wahlberg is quite cool, but strikes me more as a funny guy
did you see "Pain & Gain"?
nah
wahlberg a funny guy?
his character in departed was quite funny
yes, totally
@Mike he has a movie with will ferrel
they also play cops
06:28
@Pedro have you seen american psycho
and pain & gain
then you have ted
@Mike yeah, I love it
@Pedro Do you like Huey Lewis and the News?
CREEPED OUT
"what are all those papers in the floor, Mike?"
"Do you have poodle or something"?
@Mike I wanted to read the book, actually.
but the movie seems waaaay better
No, @Pedro!
christian bale is an acting god, come on
This is prolly one of my favourite scenes of all times
06:31
I was responding to the poodle line. :P
@Mike Did I get the line wrong?
AH,.
That is Bateman's line.
Yah.
@PedroTamaroff Have you seen 'Adaptation'?
Poor Jared Leto
I friggin hate Witherspoon's acting there.
@Mike Sodium.
I guess that's a no.
oohhh bunny chews its ear
06:34
"I killed Paul Allen with an axe in the face!"
@Mike "Na", actually.
Hahaha,
You should.
eats Mike
@Mike KAY
takes note
not available
UGH.
Well just watch it.
Being John Malkovich too.
i will watch it tomorrow
I like this movie... ugh, what's the effing name
whatever works
that's the name
06:40
need munchies T_T eats @PedroTamaroff
lay off the drugs in here @usukidoll
rabbit hunting season
sorry @Mike ... HEY! I'm not on drugs...just waiting for my dinner to be cooked
someple please take this clip and make it into its own 3-second youtube video
@anon Actually wabbit.
06:41
wait shit
not that
tamest accidental hyperlink I've seen
@Pedro want to music?
Oh no you're movie atm.
I was expecting tentarape shit at least.
@Mike Movie's over.
06:43
or pr0n
mmm, grape soda
@anon then shall we?
@anon never tried that.
fuck
@Pedro then shall we?
ye
iam alone and scared now
guy garvey is there though
06:44
@anon you're free to join too
mmmm jalapeno poppers
@usukidoll what's that?
O_O it's well a jalapeno ... with cream cheese
@PedroTamaroff breaded jalapenos + cheese i think
stuffed jalapenos
06:47
aren't jalapeños ridiculously spicy?
depends
habanero...yeah you're screwed
I wonder what you guys sound when you say jalapeÑos
=D
"holla-pain-yos"
@Mike HA!
Well, it is not "yos" but "gnos", like "gnocci"
Do you know how the "eñe" (egne) sounds like?
i split the n and the y up, but we do put them together
painyos
i'm in california, i have a rudimentary knowledge of spanish @Pedro
06:49
@Mike I see.
also they're not spicy at all imo
We should have a conversation in spanish.
dear lord no
i.e. IRL
not here
do you intend to come to cali?
06:50
it should be amazingly rib cracking
@Mike you're on the other side, but I would really like to go to san diego
@Pedro Well, when are you planning? late enough and I'd probably be a reasonable distance
@Mike I'll be in New Jersey on July 29th, up to August 20th.
ah
i shant be anywhere near that
unless i'm in the city at some point in the summer, which seems unlikely but possible
whatcha doin there
hmm I could watch Adaptation right now
hold up
06:53
holds
@Pedro we could watch being john malkovich in the irony zone
think fast, you have 40 seconds to decide
"Heisenberg group".
@Mike @anon Nice name.
@PedroTamaroff That's not anon
Who is not anon?
Maybe I misunderstood
07:00
pedro is referring to heisenberg innie?
o.
Specially dis
you pick your lecture note sources from interesting places
Though I met H(F) in D&F before that.
I picked those from MSE, but KCd is known outside MSE surely.
07:02
yes
i just find it interesting that they're always number theorists
@Mike hey... I do like NT and I have sutided and itty bit of it.
Didn't you know?
i knew that you knew an itty bit :)
you don't do analytic nt though no?
nop
yeah, you don't like analysis.
neither does @anon
snif
07:06
i don't like real analysis
or more generally, the study of $\varepsilon$
07:21
@anon I'll try to prove (iv) now,
@PedroTamaroff math.stackexchange.com/questions/668291/… <---- am I on the right track on this?
@usukidoll $$h^{r+1}-k^{r+1}=hh^{r}-kk^{r}=(h+k)(h^r-k^r)-kh(h^{r-1}-k^{r-1})$$
If I am doing no silly mistake.
D: why r u guys doing it this way?
you're correct, senyor
@usukidoll Think Vieta.
07:32
so what should I fix?
yeah like did I do something wrong on my attempt or?
Well, let's start by changing $*$ to $\cdot$.
besides that...
That's would be an improvement. =D
07:34
I know that's X
another way of mulitplication
by what else
Well, you haven't arrived to an useful expression in your post.
ughhhhhhhh
sooo what should i do?
Read the answers?
huh?
but I want to learn the process not read answers and copypasta
well, it's all about pulling out an $h^r-k^r$ somehow, like pedro did
I think the clearest approach is writing $x^n-y^n=(x-y)(\rm blah)$.
Just note that if you evaluate at $y$, you get zero, hence looking at the poly $P(x)=x^n-y^n$ you get $x-y\mid P(x)$.
true
long division, and then stop when you get something familiar as a remainder
ok setting that aside.... math.stackexchange.com/questions/668272/… @PedroTamaroff does it look right
@usukidoll You need to use more words, gal.
It helps us to understand what you're thinking, what you're trying to get to.
Else I just see a bunch of numbers.
-________0 this is why I hate these assignments... If it's not all symbols it's too much words
MAKE UP YO MIND HOMIE!
07:43
who's said too many words?
those people are dumb and wrong
my prof
I wrote a lot of words on one o f my homeworks
da fuq
got slammed for it
if there's anything universally true about discrete math students it's that they don't write enough words
too used to being completely symbolic
yeah 13 pages worth of latex a two out of 10 COME ON!
well I'm going to put the words and whatever I wrote to make it complete @PedroTamaroff. I'm typing the latex as I speak
true story @Mike I was so confused...like wtf I put WORDS...my answers were CORRECT and I got a 2 out of 10 on it
ughhhhhhh
07:47
Greetings
This is very nice too $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} (-1)^{i+j} \frac{H_{i+j}}{i+j}$$
@Mike Amazing song in my list coming up.
@PedroTamaroff trance?
@Chris'ssis WAT
@Pedro Make sure to listen to the lyrics of the next one
I am watching some rice, so I'll grab and relisten.
08:02
Lol, not worth it
It's just a silly song
Hey guys, I needed some help with a math problem
it has been posted before but the author only wanteds answers for a aprticular part....and i am having difficulties with the remaining parts too
for part a)
i was thinking along the lines of proving that set of discontinuity has measure 0
now i have studied Thomae's functions continuity before....
Thomae's function, named after Carl Johannes Thomae, has many names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function, the Riemann function, or the Stars over Babylon (John Horton Conway's name). This real-valued function f(x) of the real variable x is defined as: :f(x) = \begin{cases} \frac{1}{q} &\text{if }x\text{ is rational, }x=\tfrac{p}{q}\text{ in lowest terms and } q > 0\\ 0 &\text{if }x\text{ is irrational.} \end{cases} It is a modification of the Dirichlet function, (which is 1 at ratio...
and i thought that it might be useful somehow
but was unable to connect dots
08:21
@Mike
agh
i overdid the spicy sauce
@PedroTamaroff Whimp.
there all in latex for the first two parts
@Mike I don't happen to b good at tolerating spicy or hot food.
I wish I had that problem.
Now I just can't taste spicy.
now how to do the c. If $h$ and $k$ are any two distinct integers, then $h^n-k^n$ is divisible bt $h-k$ the fast way
any shortcuts?
08:24
@usukidoll I just told you an hour ago!
yeah but what's the step by step process of it
I can't just put that... needs step by step induction thang
I know that if the basis is $n=0$ then yes $h-k$ holds
what's the difference between $hh^n-kk^n$ and $k(h^n-k^n)$? (literally, the difference)
for induction let's use $P(r)$ to prevent confusion then $h^r-k^r$
now $P(r+1) = h^{r+1} -k^{r+1}$
so there's $h^r \times h^1 - k^r \times k^1$
now what happens?!
like what gets yanked out? the $h^r-k^r$?
@usukidoll You start considering our advice?
hmm, I must be set on ignore
08:29
anon just tried to help you, @usukidoll
I didn't ignore you @anon I don't understand that part.. with the exception that $hh^n-kk^n$ is similar to $h^n \times h^1 - k^n \times k^1$
did you try taking the difference? i.e. subtract kh^n-kk^n from hh^n-kk^n
tell us what you get
but what did you do for $k(h^n-k^n?$ was it factored...... oh man nnn I see where this going
$kh^n-kk^n $from $hh^n-kk^n$
$kh^n-kk^n -(hh^n-kk^n)$
I smell typo
"subtract A from B" means do B-A
$hh^n-kk^n - (kh^n-kk^n)$
08:32
right
now simplify
$hh^n-kk^n-kh^n+kk^n$
$hh^n-kh^n$
$h^n(h-k)$
right, so $h^{n+1}-k^{n+1}=(h-k)h^n+k(h^n-k^n)$. now what?
? was the $k^{n+1} $subsituted
it's a sum of two things divisible by h-k
$h^{n+1}-k^{n+1} = h^nh^1-k^nk^1$
so... $(h-k)h^n+k(h^n-k^n)$ . . .
:/
08:36
2 mins ago, by anon
it's a sum of two things divisible by h-k
thinking about this?
know what "two things" I'm referring to?
$h^n-k^n$
@Mike Just started watching Adaptation.
sum of two things. that means there's a + sign between the two things. what are those two things?
@PedroTamaroff I'm heading to bed pretty soon.
My bed time has already passed by.
I hear birds chirping now.
Quite peaceful.
08:40
oh my bad x.x
$h^n+k^n$
no
that is not written anywhere
hmmm... $h^{n+1}+k^{n+1}$
no. do you see that written anywhere? go back and find the + sign!
$h^{n+1}-k^{n+1}=(h-k)h^n+k(h^n-k^n)$
where is the + sign? I spy with my little eye...
$(h-k)h^n+k(h^n-k^n)$
the first term is $(h-k)h^n$ and the second is $k(h^n-k^n)$. can you see that both of these are multiples of $h-k$?
08:42
well I do see $h-k$ on both sides of the problem
do you agree that $(h-k)h^n$ is a multiple of $h-k$?
@BalarkaSen Not seeing you around today?
do you understand that $h^n-k^n$ is a multiple of $h-k$, hence so is $k(h^n-k^n)$?
08:45
do you agree that a sum of two multiples of $h-k$ is another multiple of $h-k$?
how? oh right yes I do $(h-k)h^n+k(h^n-k^n)$

$(h-k)h^n+k(h^n-k^n)$
08:56
I don't even.
someone revived the question with a recent answer
once it gets downvoted we can delete it
09:17
Prove that the inequality $(1+ \frac{1}{n})^n < n$ holds for all $n \geq 3$

First we need to prove the basis. If we let $n=4$, then $(1+ \frac{1}{4})^4 < 4$

$(\frac{4}{4}+ \frac{1}{4})^4 < 4$

$(\frac{5}{4})^4 < 4$

$(\frac{625}{256}) < 4$

The inequality statement is true



We assume that $(1+ \frac{1}{n})^n < n$ is true for $P(n+1)$


$(1+ \frac{1}{n+1})^{n+1} < n+1$

And then I'm stuck afterwards ... arghhhh these induction problem have a ton of variety and different rules. x.x
your basis is $n=3$
really? But we need it to be greater than 3 at least
greater than or equal to
I'm sticking with the greater than ^^
then you won't have proved the theorem
09:20
:O
no way!
$(1+ \frac{1}{3})^3 < 3$
$(\frac{3}{3}+ \frac{1}{3})^3 < 3$
$(\frac{4}{3})^3 <3$
$\frac{64}{27} <3$
It is less than three since that fraction is 2.30 in decimal form
basis? base.
bases are for spaces.
I'm using induction
got to prove the basis first which I've done, but the $n+1$ is driving me nuts
I suppose "basis"="base case"
$(1+ \frac{1}{n+1})^{n+1} < n+1$ gotta prove through induction that it holds -_-
I can get the basis part... and the one for $P(n)$ but the $P(n+1)$ is a different story
o-o
09:49
You can prove it is always smaller than three.
Use the binomial expansion.
&@#((@*@&(*# you mean the binomial theorem X>X
the binomial theorem tells you how to expand a binomial..
$[(1+ \frac{1}{n+1})^n] [(1+ \frac{1}{n+1})^1 ]< n+1$
so that's ( n on the top 1 on the bottom ) if I'm going to do the binomial theorem
ughhhhhhhh punches this crazed problem
stop whining... you're making a big deal out of stuff.
forget about courses and names and scores and all that baloney and just solve problems
>:/ *chews on rabbit's ear *
10:22
@usukidoll Any update on that problem. Its very tough, especially that n+1 in the denominator.
@Sawarnik yeah I posted it on main
got some hints and one very umm not useful
@usukidoll the link
1
A: Prove that the inequality $(1+ \frac{1}{n})^n < n$ holds for all $n \geq 3$

David HThe following inequality will be needed: $$\frac{1}{n+1}<\frac{1}{n} \Leftrightarrow 1+\frac{1}{n+1}<1+\frac{1}{n}\\ \Leftrightarrow \left(1+\frac{1}{n+1}\right)^n<\left(1+\frac{1}{n}\right)^n.$$ From the induction hypothesis $\left(1+\frac{1}{n}\right)^n<n$ and the algebraic identity $\left(1+ ...

OOPS!
OOh! Just was going to try those inequalities.
just click on the title for the full discussion @Sawarnik
oh gawd help me D: Greg was so mean
10:24
Ya, I know.
"slow has nothing to do with it" psffffffffft
at least I USED LATEX AND ATTEMPTED THE PROBLEM UNLIKE THIS QUESTION THAT DESERVES TO BE DELETED math.stackexchange.com/questions/669350/…
@usukidoll Maybe you can edit that and gain 2 points.
Too fricking hard
that guy did massive damage
I did edit the tags
lolwut XD
10:58
@robjohn Hi

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