Mathematics - 2014-12-21 (page 1 of 6)

Daniel Fischer

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.

Jasper Loy

Hello @anon how is your exam?

anon

pretty good

Mathy Person

Is $\frac{x^2}{x^4-3x^2+1}$ a rational function?

Jasper Loy

@anon Will you consider telling me your last name?

anon

it doesn't mattress

@MathyPerson do you know what a rational function is?

Don Larynx

00:07

@Mathy a rational function is in form $\frac{p}{q}$ where $p, q$ are poly's, $q \neq 0$

skullpatrol

q not zero

Don Larynx

what do we call $\frac{ln(x)}{x}$?

skullpatrol

same

Jasper Loy

We call it a function, lol.

Don Larynx

but $ln(x)$ isn't poly, I agree with Jasper haha

Mathy Person

00:09

@DonLarynx @anon does that mean it is not a rational function? since $x^2$ is not a poly?

Don Larynx

What makes you say that @Mathy

anon

@MathyPerson ....... do you know what a polynomial is?

Jasper Loy

@MathyPerson Do you know what a polynomial is?

Don Larynx

A polynomial is of form $c_1x^1 + c_2x^2 + \dots + c_nx^n$

c's coeffs and x's vars

skullpatrol

do you know what a ratio is

Jasper Loy

00:10

@DonLarynx So is x^2 a polynomial?

Mathy Person

@DonLarynx @JasperLoy @anon Wouldn't it have to have more than two terms?

Unless you count it as x^2+0x+0?

anon

@MathyPerson does the definition say it has must have more than two terms?

Don Larynx

@Jasper x^2 has form $c_0x^0 + c_1x^1 + c_2x^2, c_1, x^1 = 0, c_2 = 1, x^2 = x^2$

anon

also your definition is missing the constant coefficient

Jasper Loy

@MathyPerson No, it can have 1 term. 0,3,x,-2x^2 are all polynomials.

Mathy Person

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)."

Jasper Loy

@DonLarynx Sorry, I asked the wrong person. I do know what a polynomial is, lol.

Don Larynx

@Mathy, you are correct. x^2 + 0x + 0, but x^2 is much simpler to write

Mathy Person

@DonLarynx So it is a polynomial, then.

Jasper Loy

@MathyPerson Yes.

Don Larynx

@Mathy: Yes. Just ask yourself if it fits the form I wrote.

skullpatrol

00:13

a polynomial is a sum of monomials

Jasper Loy

@MathyPerson It is very important to read your textbook with understanding.

Mathy Person

@DonLarynx Yes, it fits

skullpatrol

do you know what a monomial is?

Mathy Person

@skullpatrol it only has one term

Jasper Loy

@skullpatrol The word is not often used.

Mathy Person

00:15

@skullpatrol like a number or 3x or something like that

skullpatrol

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

Mathy Person

mono means one

Mike Miller

@Behaviour How are hats organized on the leaderboard? Does it show the ten rarest hats you have, in order of scarcity?

skullpatrol

exactly, one term in a polynomial

Mathy Person

anybody have the chameleon hat?

skullpatrol

00:17

@MathyPerson me, if you look closely i'm wearing it :-)

Mike Miller

Yeah.

Behaviour

@MikeMiller Yes, the rarest first: by the number of users who earned the hat, in the ascending order.

Mike Miller

I'm surprised treasure hunter is so common. What are people getting it for?

Mathy Person

@MikeMiller I don't have that hat

@skullpatrol oh, haha, i see that now

Jasper Loy

@MikeMiller What is the reason some people prefer to use their middle name instead of their first name, like you?

Mathy Person

00:19

@JasperLoy maybe they just like their middle name more? or their first name is embarassing?

Mike Miller

@JasperLoy I can't speak for everyone. My parents always called me by my middle name. I can't explain why they did so.

skullpatrol

@MathyPerson just edit your profile discription to get one

Mathy Person

@skullpatrol the about me?

Jasper Loy

@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.

skullpatrol

@MathyPerson yes

@JasperLoy have or had?

Jasper Loy

00:21

@skullpatrol Well, it depends on what have means, lol. I still know him, so I use have, lol.

Don Larynx

I have chameleon, how do you think you get it?

skullpatrol

icic

Mathy Person

@skullpatrol i edited it. no hat

skullpatrol

@MathyPerson soon

Mathy Person

@DonLarynx it just said that the chameleon was a "secret hat"

Don Larynx

00:22

chameleon means you have more than one stack account. interesting/

Mathy Person

@skullpatrol oh ok

Mike Miller

No it doesn't, @DonLarynx

Don Larynx

maybe it is change in display picture. the world will never know

Behaviour

292

Q: Winter Bash 2014 Secret Hats

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

skullpatrol

@MathyPerson BINGO!

you got it :D

Behaviour

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.

Mathy Person

@skullpatrol got what?

the secret hats?

skullpatrol

yep

Mathy Person

@DonLarynx " Chameleon - for joining a site for editing your profile, including linking a new site to your account "

Mike Miller

@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).

Don Larynx

why is number theory so beautiful you guys

even more than real analysis, by miles :o

Behaviour

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. :)

Mike Miller

Fair 'nuff.

MarcGato

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 ?

Mike Miller

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.

Behaviour

Yes, the Abby hat is ridiculous. They could have had something difficult to find, and still meaningful; like posting a community ad last year.

Don Larynx

what is $argth, th$ @MarcGato

anon

00:30

arctanh

MarcGato

@DonLarynx ok in english is arctanh as anon said. sorry ^^

Khallil Benyattou

Wait, is $y'$ supposed to be $y'(t)$, @MarcGato?

Huy

No, $y'(x)$.

$\dot{y} = \dot{y}(t)$, obviously.

Khallil Benyattou

Why does $t$ appear later on in Marc's comment, @Huy? :/

Huy

He is using wrong conventions.

He must leave this place immediately.

anon

00:33

@KhallilBenyattou of course

Ted Shifrin

Hi @anon

Khallil Benyattou

You're killing me, @Huy. >_<!

Mike Miller

I think your past activity shouldn't stop you from getting hats (other than reputation requirements, like ability to close and edit)

anon

@TedShifrin hi

Jasper Loy

@BalarkaSen Do return to this chat. It will be alright!

Khallil Benyattou

00:34

Has @Balarka left, @Jasper?

Huy

@KhallilBenyattou: i.imgur.com/yIUdrWV.jpg

2

Jasper Loy

@KhallilBenyattou Well, he felt he was being ignored and disliked in chat, so he sort of left.

MarcGato

@Huy I prefer french conventions ;-)

Jasper Loy

@Huy Not funny, lol.

Huy

@JasperLoy: You're not funny, lol.

Khallil Benyattou

00:37

Following 'not funny' with 'lol' is kind of contradictory, @Jasper.

2

Ted Shifrin

Not funny, LOL, @Khallil

Jasper Loy

@KhallilBenyattou Well, it's called wave particle duality.

Huy

@JasperLoy: Do you behave like a particle or like a wave?

Don Larynx

How are you supposed to solve a function of the form $y'(x) + xy^2(x) = 0$?

Khallil Benyattou

That's not a function, I don't think.

Jasper Loy

00:39

@Huy Both, hence duality. It's called Pontryagin duality.

Khallil Benyattou

I think it's a non-linear homogeneous ordinary differential equation with non constant coefficients.

anon

@JasperLoy ?

Don Larynx

@MarcGato: HINT: Re-write it as a separable function.

Jasper Loy

@anon Nothing, just trying to show off my vast knowledge, lol.

Mike Miller

no it's not, @Jasper...

Huy

00:41

@DonLarynx: It's actually not too hard to guess a solution to that one.

Don Larynx

@Huy: I haven't done partial fractions in forever.

Huy

@DonLarynx: Same here.

Don Larynx

@Huy: wat. go on....

wait

I got it

that is also a separable equation

its just non-linear

Khallil Benyattou

It seems that way, @DonLarynx. ^_^

Don Larynx

Thanks @Khallil

Jasper Loy

00:44

@DonLarynx Your username sounds familiar. Is it a movie or something?

Khallil Benyattou

When you integrate the equality w.r.t $t$, that's what you get I think: $$\displaystyle \dfrac{1}{1-y^2} \dfrac{\text{d}y}{\text{d}t} = 1 \implies \int \dfrac{1}{1-y^2} \text{ d}y = \int \text{d}t$$

Mathy Person

Any advice on writing $1-x-x^3+x^4+x^5+x^6-x^7+.....$ as a power series representation?

Don Larynx

@Jasper: I took a hiatus from math.SE for a year and a half or so

@Khallil correct

Jasper Loy

@DonLarynx I see. I deleted several past accounts, lol.

skullpatrol

I remember you now :D

Don Larynx

00:46

Yay! @skullpatrol

Huy

@DonLarynx: I had to solve $$-f''(x) - \frac{2a^2}{\cosh^2(ax)}f(x) = k^2f(x)$$ in an exam this summer.

skullpatrol

@DonLarynx welcome back!

Khallil Benyattou

Are $k$ and $a$ constants, @Huy?

Huy

@KhallilBenyattou: $k > 0, a \in \mathbb{R}$.

skullpatrol

@DonLarynx welcome back

Khallil Benyattou

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

it really doesn't

anon

sure it does

Huy

took me almost half of the exam's time to solve it

Mike Miller

@skullpatrol started humming when I saw the bluelink

skullpatrol

@MikeMiller classic

Jasper Loy

00:52

@MikeMiller What do you look like? I tried to find your pic on the internet, lol.

Mike Miller

I don't, @Jasper.

skullpatrol

-_-

anon

@Huy did you use matrix exponentials?

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.

Jasper Loy

@Huy I guess you use ansatz there very often.

Don Larynx

00:53

@Huy: Is the solution of the form $cosh^2(ax)$?

Huy

@JasperLoy: No. We have seen many many different ansatzes in the lecture and that particular ansatz was used exactly once.

Jasper Loy

@Huy I mean the word ansatz.

Huy

@DonLarynx: It does involve hyperbolic functions.

@JasperLoy: I don't think I use it more often than I have to.

Khallil Benyattou

The correct ansatz would be great, @Huy!

Also, welcome back @DonLarynx!

Jasper Loy

Going to sleep in 1 hour.

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.

skullpatrol

@Huy no way to sneak in?

:D

Khallil Benyattou

Actually, not yet!

Not yet, @Huy! I'll give it a go on my own first! ^_^

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

Don Larynx

How do I make a text see-able only if it's moused over?

skullpatrol

@Huy I'm talking about the other room

Don Larynx

00:57

on chat

Huy

@skullpatrol: You can always sneak in, but the conversation will be in German

skullpatrol

that's what I mean pal :-)

user134177

@skullpatrol hi

Jasper Loy

@skullpatrol Is everyone a pal?

Huy

@skullpatrol:
Step 1: Learn German.
Step 2: Join the German room.
Step 3: ???
Step 4: Profit.

skullpatrol

00:58

:O

@blondblau welcome back

3 mins ago, by Khallil Benyattou

Also, welcome back @DonLarynx!

:D

Don Larynx

@skull the song was good!

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?

skullpatrol

@DonLarynx its a classic

user134177

@skull but i will go now. i am very tired

skullpatrol

@blondblau see you later

Jasper Loy

01:01

After Jonas left, not many people use the word bro in this chat.

skullpatrol

now it is "smack"

Jasper Loy

Maybe I should start using it again, just like @skullpatrol uses pal.

skullpatrol

np

Jasper Loy

Old John has also vanished.

skullpatrol

people come, people go

user134177

01:03

@skullpatrol, @all bye bye

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?

skullpatrol

@blondblau bye

Jasper Loy

@Huy That should be closed as duplicate.

Khallil Benyattou

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

Okay then. Good luck and don't forget to sleep!

Khallil Benyattou

01:04

I won't! Thanks. Good night, @Huy!

skullpatrol

later

p

a

L

Don Larynx

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 $y = f$

Jasper Loy

Hello @Did welcome to this chat.

Don Larynx

Strangely enough $k$ is equal to the reciprocal of $a$. What gives?

@Huy

skullpatrol

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

Jasper Loy

01:15

What have you never seen before? What's so weird about a breakup?

skullpatrol

Who comes into a math chat room 5 days before Christmas with that kind of news?

Khallil Benyattou

The boyfriend breaking up with the girlfriend, @skull?

skullpatrol

yes

Jasper Loy

Well, I don't see what is so weird about it.

@skullpatrol Life is full of tragedies.

Khallil Benyattou

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.

Jasper Loy

01:19

@KhallilBenyattou Now now, you have learnt from @skullpatrol to quote messages.

Don Larynx

themarysue.com/wp-content/uploads/2010/11/…

3

@skull

Khallil Benyattou

Mondays, @DonLarynx! ^_^

The shortage of break-ups on Christmas day isn't surprising. =P

skullpatrol

@DonLarynx :D

Don Larynx

Does the statement $log_2^2$ make sense??

This is from sosmath.com/CBB/viewtopic.php?t=24832

Khallil Benyattou

I'd say the argument of the logarithm is missing, @DonLarynx.

Don Larynx

01:26

Hi @Venus

Khallil Benyattou

It's log of something to base 2 all squared.

Don Larynx

@Khallil: yes, but it wasn't immediate to the posters on that topic. Maybe we are missing something?

evinda

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

Don Larynx

Remember: $\mathbb{P}$ is the set of primes.

@evinda

Jasper Loy

math.stackexchange.com/questions/1073609/… I am unhappy my answer got less votes than two others, lol.

Kaj Hansen

01:31

That's a really complicated proof that $\mathbb{Z}_p$ is an integral domain.

Jasper Loy

And I was the first one to post an answer, lol.

Kaj Hansen

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.

Venus

Hi @DonLarynx

Jasper Loy

@Venus I don't think Jack will be elected.

Venus

@JasperLoy Why so?

Kaj Hansen

01:33

And all fields are principal ideal domains, so just show $\mathbb{Z}_P$ is a field :P

Jasper Loy

@Venus Just judging from the primary phase votes.

Don Larynx

meh, I should have been moderator. everyone gets free hats.

Jasper Loy

@KajHansen Are you done with exams?

Kaj Hansen

Yeah @JasperLoy

Venus

We never know that for sure. How about Anna? In what rank do you think she would end this election?

Jasper Loy

01:36

@Venus I think very low rank, again judging from the primary phase.

Kaj Hansen

I don't think I've voted yet. How does one vote?

Jasper Loy

@KajHansen Go to /election and use common sense.

Mathy Person

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)?

Venus

@KajHansen vote for Jack. It makes sense

Kaj Hansen

Well, once I figure out how

Venus

01:38

@JasperLoy Maybe you're right about her

Ilmari Karonen

@KajHansen Here's the link, and there are instructions in the side bar.

Don Larynx

$\frac{1}{(1+x)(1−x)(1+x2)(1−x2)(1+x3)(1−x3)....}$

That's a term.

What term number?

Mathy Person

@DonLarynx power series representation

Don Larynx

What term is that, is that term 3, 4, 352, etc?!

never mind.

Mathy Person

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 ^

Venus

01:42

@JasperLoy it's still 9.40 am & it's Sunday. Why do you wake up so early?

Jasper Loy

@Venus I woke up at 12 am. I am going to sleep soon.

Mathy Person

@DonLarynx You can also see this: $title$ for reference

Venus

I wanna sleep again. My head is dizzy

Don Larynx

I thought your name said Mathy Petersoh

*Peterson

gcy-rolle

Hello,everyone

Jasper Loy

01:44

I am going to sleep. I hope I get a miracle in my sleep.

skullpatrol

later

Jasper Loy

@Venus Yes, you should go and sleep.

Mathy Person

@DonLarynx It's Mathy person, haha

Kaj Hansen

@JasperLoy: The Bedtime Police

Jasper Loy

@KajHansen I hope you find a new girlfriend this Christmas.

Kaj Hansen

01:47

I do too @JasperLoy. I do too.

Mathy Person

@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. $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)....}$

Don Larynx

@Mathy The pattern isn't immediately obvious...there's 1 +, 2 -, then 3 +'s, then the -'s and +'s oscillate.

In regards to "title"

Mathy Person

@DonLarynx yes, that is what i noticed as well

@DonLarynx was the representation generated by vadim correct?

Don Larynx

02:03

I'm not very sure @Mathy, as for $k = 2$ I got as LHS = $\frac{-(x-1)^2}{x}$ and RHS $\frac{1}{1+x}$.

Mathy Person

@DonLarynx hmm.... i asked it on math.se as well, but no responses yet. text

Don Larynx

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

Khallil Benyattou

It's getting pretty late where I am. Good night all!

skullpatrol

later

Khallil Benyattou

^_^

Don Larynx

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

Don Larynx

02:45

yay number theory

skullpatrol

welcme back @robjohn

Don Larynx

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

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 !

Don Larynx

04:11

Thank you again math.SE for a beautiful day of learning!

Oh and Tycho as well

Mathy Person

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

JohnMerlino

Why does cross-multiplication work?

anon

04:41

you mean a/b=c/d iff ad=bc?

just multiply a/b=c/d by b and d to clear denominators

Don Larynx

@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

anon

multiplying the reciprocal by its resciprocal?

Don Larynx

@anon, yes multiply $\frac{1}{56}$ by $\frac{1}{\frac{1}{56}}$

JohnMerlino

@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

anon

@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

Don Larynx

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.

anon

it's horrible word choice

Don Larynx

@anon: Multiply the reciprocal by the reciprocal's reciprocal?

anon

yes

Don Larynx

ok

JohnMerlino

@anon I understood your explanation, thanks

Algebra really helps in expanding the brain

Chris's sis

07:47

Greetings (kind of busy today, I need to check some documents)

Alpha

HEllo

Anyone here knows about the Luna 1 sodium gas experiment?

Transcript for

$Mathematics$ Mathematics