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00:01
first. Calling qsort is a quick implemetation of sorting, but it doesn't guarantee the required $O(m\log m)$ complexity. At least not officially.
Hello @anon how is your exam?
pretty good
Is $\frac{x^2}{x^4-3x^2+1}$ a rational function?
@anon Will you consider telling me your last name?
it doesn't mattress
@MathyPerson do you know what a rational function is?
00:07
@Mathy a rational function is in form $\frac{p}{q}$ where $p, q$ are poly's, $q \neq 0$
q not zero
what do we call $\frac{ln(x)}{x}$?
We call it a function, lol.
but $ln(x)$ isn't poly, I agree with Jasper haha
00:09
@DonLarynx @anon does that mean it is not a rational function? since $x^2$ is not a poly?
What makes you say that @Mathy
@MathyPerson ....... do you know what a polynomial is?
@MathyPerson Do you know what a polynomial is?
A polynomial is of form $c_1x^1 + c_2x^2 + \dots + c_nx^n$
c's coeffs and x's vars
do you know what a ratio is
00:10
@DonLarynx So is x^2 a polynomial?
@DonLarynx @JasperLoy @anon Wouldn't it have to have more than two terms?
Unless you count it as x^2+0x+0?
@MathyPerson does the definition say it has must have more than two terms?
@Jasper x^2 has form $c_0x^0 + c_1x^1 + c_2x^2, c_1, x^1 = 0, c_2 = 1, x^2 = x^2$
also your definition is missing the constant coefficient
@MathyPerson No, it can have 1 term. 0,3,x,-2x^2 are all polynomials.
00:11
@anon: The question just asks me this: "Write $F_{\text{even}}(x)$ as a rational function (that is, as a simplified quotient of polynomials)."
@DonLarynx Sorry, I asked the wrong person. I do know what a polynomial is, lol.
@Mathy, you are correct. x^2 + 0x + 0, but x^2 is much simpler to write
@DonLarynx So it is a polynomial, then.
@MathyPerson Yes.
@Mathy: Yes. Just ask yourself if it fits the form I wrote.
00:13
a polynomial is a sum of monomials
@MathyPerson It is very important to read your textbook with understanding.
@DonLarynx Yes, it fits
do you know what a monomial is?
@skullpatrol it only has one term
@skullpatrol The word is not often used.
00:15
@skullpatrol like a number or 3x or something like that
a monomial is a numeral, a variable, or the product of a numeral and one or more variables
2 mins ago, by skullpatrol
a polynomial is a sum of monomials
poly means many
mono means one
@Behaviour How are hats organized on the leaderboard? Does it show the ten rarest hats you have, in order of scarcity?
exactly, one term in a polynomial
anybody have the chameleon hat?
00:17
@MathyPerson me, if you look closely i'm wearing it :-)
@MikeMiller Yes, the rarest first: by the number of users who earned the hat, in the ascending order.
I'm surprised treasure hunter is so common. What are people getting it for?
@MikeMiller I don't have that hat
@skullpatrol oh, haha, i see that now
@MikeMiller What is the reason some people prefer to use their middle name instead of their first name, like you?
00:19
@JasperLoy maybe they just like their middle name more? or their first name is embarassing?
@JasperLoy I can't speak for everyone. My parents always called me by my middle name. I can't explain why they did so.
@MathyPerson just edit your profile discription to get one
@skullpatrol the about me?
@MikeMiller I see. I have a professor who also goes by his middle name. You may have heard of Jon Berrick, author of a few books.
@MathyPerson yes
@JasperLoy have or had?
00:21
@skullpatrol Well, it depends on what have means, lol. I still know him, so I use have, lol.
I have chameleon, how do you think you get it?
@skullpatrol i edited it. no hat
@MathyPerson soon
@DonLarynx it just said that the chameleon was a "secret hat"
00:22
chameleon means you have more than one stack account. interesting/
@skullpatrol oh ok
No it doesn't, @DonLarynx
maybe it is change in display picture. the world will never know
292
Q: Winter Bash 2014 Secret Hats

KevinSo far, I've unlocked one of the secret hats. I'm wondering what other secret hats are out there and how to earn them?

@MathyPerson BINGO!
you got it :D
00:25
I resolved that I'm not going to get either the pizza hat or the iOS hat; both are incompatible with my worldview.
@skullpatrol got what?
the secret hats?
@DonLarynx " Chameleon - for joining a site for editing your profile, including linking a new site to your account "
@Behaviour Well, one can get the pizza hat without outside input. Despite all the encouragement I gave to get mine, I think I got four answers from people outside this room in the first half hour (plus yours, plus Huy's).
why is number theory so beautiful you guys
even more than real analysis, by miles :o
00:27
I meant I'm not about to ask questions that can be answered within 30 minutes, let alone by 5 different people. My questions are supposed to be hard. :)
Fair 'nuff.
Hi, I have the following ode $y'/(1-y^2)=1$ wich implies $argth(y)=t+C$ then $y=th(t+C)$ so $y$ is bounded by $1$. Am I correct ?
I'm going to need to convince Abby to write a post on MSE... then, by the end, I'll have everything on this site.
Yes, the Abby hat is ridiculous. They could have had something difficult to find, and still meaningful; like posting a community ad last year.
what is $argth, th$ @MarcGato
00:30
arctanh
@DonLarynx ok in english is arctanh as anon said. sorry ^^
Wait, is $y'$ supposed to be $y'(t)$, @MarcGato?
Huy
Huy
No, $y'(x)$.
$\dot{y} = \dot{y}(t)$, obviously.
Why does $t$ appear later on in Marc's comment, @Huy? :/
Huy
Huy
He is using wrong conventions.
He must leave this place immediately.
00:33
@KhallilBenyattou of course
Hi @anon
You're killing me, @Huy. >_<!
I think your past activity shouldn't stop you from getting hats (other than reputation requirements, like ability to close and edit)
@TedShifrin hi
@BalarkaSen Do return to this chat. It will be alright!
00:34
Has @Balarka left, @Jasper?
Huy
Huy
@KhallilBenyattou: i.imgur.com/yIUdrWV.jpg
2
@KhallilBenyattou Well, he felt he was being ignored and disliked in chat, so he sort of left.
@Huy I prefer french conventions ;-)
@Huy Not funny, lol.
Huy
Huy
@JasperLoy: You're not funny, lol.
00:37
Following 'not funny' with 'lol' is kind of contradictory, @Jasper.
2
Not funny, LOL, @Khallil
@KhallilBenyattou Well, it's called wave particle duality.
Huy
Huy
@JasperLoy: Do you behave like a particle or like a wave?
How are you supposed to solve a function of the form $y'(x) + xy^2(x) = 0$?
That's not a function, I don't think.
00:39
@Huy Both, hence duality. It's called Pontryagin duality.
I think it's a non-linear homogeneous ordinary differential equation with non constant coefficients.
@JasperLoy ?
@MarcGato: HINT: Re-write it as a separable function.
@anon Nothing, just trying to show off my vast knowledge, lol.
no it's not, @Jasper...
Huy
Huy
00:41
@DonLarynx: It's actually not too hard to guess a solution to that one.
@Huy: I haven't done partial fractions in forever.
Huy
Huy
@DonLarynx: Same here.
@Huy: wat. go on....
wait
I got it
that is also a separable equation
its just non-linear
It seems that way, @DonLarynx. ^_^
Thanks @Khallil
00:44
@DonLarynx Your username sounds familiar. Is it a movie or something?
When you integrate the equality w.r.t $t$, that's what you get I think: $$\displaystyle \dfrac{1}{1-y^2} \dfrac{\text{d}y}{\text{d}t} = 1 \implies \int \dfrac{1}{1-y^2} \text{ d}y = \int \text{d}t$$
Any advice on writing $1-x-x^3+x^4+x^5+x^6-x^7+.....$ as a power series representation?
@Jasper: I took a hiatus from math.SE for a year and a half or so
@Khallil correct
@DonLarynx I see. I deleted several past accounts, lol.
I remember you now :D
00:46
Yay! @skullpatrol
Huy
Huy
@DonLarynx: I had to solve $$-f''(x) - \frac{2a^2}{\cosh^2(ax)}f(x) = k^2f(x)$$ in an exam this summer.
@DonLarynx welcome back!
Are $k$ and $a$ constants, @Huy?
Huy
Huy
@KhallilBenyattou: $k > 0, a \in \mathbb{R}$.
00:49
That looks pretty cool. It rearranges to $$f''(x) + \left( \frac{2a^2}{\cosh^2(ax)} + k^2 \right) f(x)= 0 $$ but that doesn't look any nicer! ^_^"
Huy
Huy
it really doesn't
sure it does
Huy
Huy
took me almost half of the exam's time to solve it
@skullpatrol started humming when I saw the bluelink
@MikeMiller classic
00:52
@MikeMiller What do you look like? I tried to find your pic on the internet, lol.
I don't, @Jasper.
@Huy did you use matrix exponentials?
Huy
Huy
@KhallilBenyattou: In case you are interested in solving it, it is a lot easier if you know a lot of ansatz-functions (still not trivial then, imo). I can give you the proper ansatz, if you want
@anon: No, just the correct ansatz.
@Huy I guess you use ansatz there very often.
00:53
@Huy: Is the solution of the form $cosh^2(ax)$?
Huy
Huy
@JasperLoy: No. We have seen many many different ansatzes in the lecture and that particular ansatz was used exactly once.
@Huy I mean the word ansatz.
Huy
Huy
@DonLarynx: It does involve hyperbolic functions.
@JasperLoy: I don't think I use it more often than I have to.
The correct ansatz would be great, @Huy!
Also, welcome back @DonLarynx!
Going to sleep in 1 hour.
Huy
Huy
00:56
@KhallilBenyattou: Are you sure you want it already? No worries, I doubt I could have found it without having encountered it before, either.
@Huy no way to sneak in?
:D
Actually, not yet!
Not yet, @Huy! I'll give it a go on my own first! ^_^
Huy
Huy
@skullpatrol: Maybe there is, but I don't know how I would sneak in because I'm by no means an expert in solving differential equations
How do I make a text see-able only if it's moused over?
@Huy I'm talking about the other room
00:57
on chat
Huy
Huy
@skullpatrol: You can always sneak in, but the conversation will be in German
that's what I mean pal :-)
user134177
@skullpatrol hi
@skullpatrol Is everyone a pal?
Huy
Huy
@skullpatrol:
Step 1: Learn German.
Step 2: Join the German room.
Step 3: ???
Step 4: Profit.
00:58
:O
@blondblau welcome back
3 mins ago, by Khallil Benyattou
Also, welcome back @DonLarynx!
:D
@skull the song was good!
Huy
Huy
@KhallilBenyattou: I'm making my way to bed. Do you want the ansatz, or do you want to try till tomorrow? Or should I mail it to you or something and you can look at it whenever you want?
@DonLarynx its a classic
user134177
@skull but i will go now. i am very tired
@blondblau see you later
01:01
After Jonas left, not many people use the word bro in this chat.
now it is "smack"
Maybe I should start using it again, just like @skullpatrol uses pal.
Old John has also vanished.
people come, people go
user134177
01:03
@skullpatrol, @all bye bye
Huy
Huy
@KhallilBenyattou: I'm making my way to bed. Do you want the ansatz, or do you want to try till tomorrow? Or should I mail it to you or something and you can look at it whenever you want?
@blondblau bye
@Huy That should be closed as duplicate.
Sorry for the late reply @Huy. I'll try it and if we both happen to be on at the same time tomorrow, I'll ask for it then!
Huy
Huy
Okay then. Good luck and don't forget to sleep!
01:04
I won't! Thanks. Good night, @Huy!
later
p
a
L
I've got it!!!
The solution has the form $y = k^2cosh(ax) - (ax)^2/2 + c_1x + c_2$
This was easily determined from $$\frac{cosh^2(ax)}{k^2cosh(ax) - a^2} = \frac{y}{y''}$$ where $y = f$
Hello @Did welcome to this chat.
Strangely enough $k$ is equal to the reciprocal of $a$. What gives?
@Huy
3 hours ago, by blondblau
my boyfriend broke up the relationship =(
3 hours ago, by blondblau
@skullpatrol yes, blond hairs, blue eyes
I have NEVER seen that^ here
01:15
What have you never seen before? What's so weird about a breakup?
Who comes into a math chat room 5 days before Christmas with that kind of news?
The boyfriend breaking up with the girlfriend, @skull?
Well, I don't see what is so weird about it.
@skullpatrol Life is full of tragedies.
With your 'vast knowledge', I suspect you see the bigger picture, as opposed to us mere mortals, @Jasper. =P
38 mins ago, by Jasper Loy
@anon Nothing, just trying to show off my vast knowledge, lol.
01:19
@KhallilBenyattou Now now, you have learnt from @skullpatrol to quote messages.
Mondays, @DonLarynx! ^_^
The shortage of break-ups on Christmas day isn't surprising. =P
@DonLarynx :D
Does the statement $log_2^2$ make sense??
I'd say the argument of the logarithm is missing, @DonLarynx.
01:26
Hi @Venus
It's log of something to base 2 all squared.
@Khallil: yes, but it wasn't immediate to the posters on that topic. Maybe we are missing something?
Could someone hep me to understand this proof?
http://math.stackexchange.com/questions/1076161/the-ring-is-a-principal-ideal-domain-especially-an-integral-domain
Remember: $\mathbb{P}$ is the set of primes.
@evinda
math.stackexchange.com/questions/1073609/… I am unhappy my answer got less votes than two others, lol.
01:31
That's a really complicated proof that $\mathbb{Z}_p$ is an integral domain.
And I was the first one to post an answer, lol.
If $x \cdot y \equiv 0 \pmod{p}$, this implies that $p|xy$.
In particular, $p|x$ or $p|y$.
But this is not possible since $x, y < p$.
So $\mathbb{Z}_p$ cannot contain any zero divisors.
Hi @DonLarynx
@Venus I don't think Jack will be elected.
@JasperLoy Why so?
01:33
And all fields are principal ideal domains, so just show $\mathbb{Z}_P$ is a field :P
@Venus Just judging from the primary phase votes.
meh, I should have been moderator. everyone gets free hats.
@KajHansen Are you done with exams?
Yeah @JasperLoy
We never know that for sure. How about Anna? In what rank do you think she would end this election?
01:36
@Venus I think very low rank, again judging from the primary phase.
I don't think I've voted yet. How does one vote?
@KajHansen Go to /election and use common sense.
how would I turn \frac{1}{(1+x)(1−x)(1+x2)(1−x2)(1+x3)(1−x3)....} into a generating function power series again (for powers past 15)?
@KajHansen vote for Jack. It makes sense
Well, once I figure out how
01:38
@JasperLoy Maybe you're right about her
@KajHansen Here's the link, and there are instructions in the side bar.
$\frac{1}{(1+x)(1−x)(1+x2)(1−x2)(1+x3)(1−x3)....}$
That's a term.
What term number?
@DonLarynx power series representation
What term is that, is that term 3, 4, 352, etc?!
never mind.
it is the partition generating function * $\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$
@DonLarynx ^
01:42
@JasperLoy it's still 9.40 am & it's Sunday. Why do you wake up so early?
@Venus I woke up at 12 am. I am going to sleep soon.
@DonLarynx You can also see this: $title$ for reference
I wanna sleep again. My head is dizzy
I thought your name said Mathy Petersoh
*Peterson
Hello,everyone
01:44
I am going to sleep. I hope I get a miracle in my sleep.
@Venus Yes, you should go and sleep.
@DonLarynx It's Mathy person, haha
@JasperLoy: The Bedtime Police
@KajHansen I hope you find a new girlfriend this Christmas.
01:47
I do too @JasperLoy. I do too.
@DonLarynx I am trying to find the truncation to degree 10
@DonLarynx Therefore, I need to expand out that power series representation into a power series
@DonLarynx still there?
i've noticed that this chat tends to slow down after 6
nobody types things anymore, hahah
@DonLarynx here it is in better form. i realized that some of the exponents didn't turn into LaTeX. $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)....}$
@Mathy The pattern isn't immediately obvious...there's 1 +, 2 -, then 3 +'s, then the -'s and +'s oscillate.
In regards to "title"
@DonLarynx yes, that is what i noticed as well
@DonLarynx was the representation generated by vadim correct?
02:03
I'm not very sure @Mathy, as for $k = 2$ I got as LHS = $\frac{-(x-1)^2}{x}$ and RHS $\frac{1}{1+x}$.
@DonLarynx hmm.... i asked it on math.se as well, but no responses yet. text
02:22
If $k$ is an odd divisor of $2b$, call $2k$ the *friend* of $k$. Split the divisors of $2b$ into pairs of friends. For example, if $b=45$, we have the following pairs of friends.

$$(1,2)\qquad (3,6) \qquad(5,10)\qquad(9,18)\qquad(15,30) \qquad (45,90)$$

We have split the divisors of $2b$ into pairs of friends. Each pair has one odd number and one even number, so $2b$ has exactly as many odd divisors as even divisors.
But my question is: How does this ensure we didn't leave out any even divisors?
$b$ is odd
It's getting pretty late where I am. Good night all!
Suppose $f$ is an even divisor we left out. Then we have $f = 2e$ where e is some odd number. But this is a friend, and so this is included in our original list.

Then suppose $e$ is even. Then $f = 2*2*g$ for some integer. but $b$ can't be divisible by 4. RAA
02:45
yay number theory
welcme back @robjohn
02:58
What would I do without you guys? I would go mad, that's what
At least until I got back into school
r9m
r9m
03:57
@robjohn sensei can you take a look at the second part of this question please if it is possible to derive the coefficients of the polynomial via combinatorial identities (avoid reference to Cheby polynomials) :-) I tried for a while but couldn't get anything useful !
04:11
Thank you again math.SE for a beautiful day of learning!
Oh and Tycho as well
04:24
@DonLarynx Here was the original question text
I'm stuck! How do I expand this out to get the truncation of degree 10? $(1−x+x^2−x^3...)(1+x+x^2+x^3+....)(1−x^2+x^4−x^6....)(1+x^2+x^4+x^6....)(‌​1−x^‌​3+x^6−x^9....)(1+x^3+x^6+x^9+....)....$
hulllooooo
anybody hereeeee
Why does cross-multiplication work?
04:41
you mean a/b=c/d iff ad=bc?
just multiply a/b=c/d by b and d to clear denominators
@JohnMerlino because cross-multiplication is simply multiplying the reciprocal by its reciprocal on each side. For example, $\frac{5}{7} = \frac{40}{56} \Longrightarrow 56*\frac{5}{7} = 40$ and so on.
@Mathy: one second
multiplying the reciprocal by its resciprocal?
@anon, yes multiply $\frac{1}{56}$ by $\frac{1}{\frac{1}{56}}$
@anon (3x + 1)/4*8x/(9x^2 – 1). You are allowed to cancel the 4 and 8 even though they are different fractions. I understand to cancel a fraction within itself, but being allowed to cancel two different fractions with each, it's hard to see why it's allowed
@JohnMerlino because (a/b)(c/d) equals (ac)/(bd); all the top things are up top and all the bottom things are down below, it doesn't matter in which fraction a term originated, only whether it's up top or down below
@DonLarynx your word choice is a bit ... strange
04:44
@Mathy: Just take the denominator, add $1$ to it, and then use the fact that $+ = (-)(-)$.
It's a very long, complicated power series..but it should work
@anon: It works.
it's horrible word choice
@anon: Multiply the reciprocal by the reciprocal's reciprocal?
yes
@anon I understood your explanation, thanks
Algebra really helps in expanding the brain
 
3 hours later…
07:47
Greetings (kind of busy today, I need to check some documents)
HEllo
Anyone here knows about the Luna 1 sodium gas experiment?

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