Mathematics - 2013-07-15 (page 2 of 3)

skullpatrol

7:02 PM

@Charlie yipyipyip?

Charlie

@skull Can you listen please?

skullpatrol

sure

Charlie

:D

IMX

Hi, how do you get x = -1 from x(x^3+x) ?

skullpatrol

@Charlie 7:43

Charlie

7:03 PM

@IMX what you mean? is $x(x^3+x)$ equal to something?

IMX

to = 0

Charlie

@IMX aaah then it is : $x(x^3+x)=0$

IMX

Yes. But how can it be x = -1?

Charlie

So we must now that $x=0$ or $(x^3+x)=0$

$x^3+x=0$

N3buchadnezzar

Heya

@Argon Mind taking a look at this?

22

Q: How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$

Evaluate the integral: $$ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$$ The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.

Charlie

7:05 PM

$$x^3+x=0$$

IMX

How do you come to tthat?

x^3 = -x then

Charlie

@IMX if $ab=0$ $a=0$ or $b=0$

IMX

Hmm, I don't see how that should explain x = -1 as a result

Charlie

@IMX isnt it -x ?

X^3 -x

IMX

it is

Charlie

7:08 PM

you wrote x^3+x

IMX

it is x^3+x = 0

But shouldn't I take +x over to the right side?

Charlie

well, if i'm not senile, you get $x(x^2+1)=0$

so, x =0 or x^2 + 1 = 0

that gives x^2 = -1

Peter Tamaroff

I'm your Algebro, man

Charlie

damn i can't feel my legs

skullpatrol

@Charlie Get up and walk around...

Charlie

7:10 PM

it's ok now, it's ok

Peter Tamaroff

@Charlie What did you do?

N3buchadnezzar

@PeterTamaroff Algebro, has all dem feels.

Charlie

@PeterTamaroff nothing... i am just on my sit

over my legs...

Peter Tamaroff

@Charlie India style?

@N3buchadnezzar LAWL

Charlie

yes..

N3buchadnezzar

7:12 PM

I would have used Taylor though. Taylor has so much more swift swag... Than Loli-tals..

Peter Tamaroff

GURRDAMN. I'm capped for the day.

IMX

@Charlie

x^2 = -1

Charlie

@IMX yes, child

@IMX that's why i think x^3-x=0

IMX

root(x^2) = root(-1)

Is the next step isnt it?

Charlie

@IMX yep

IMX

7:13 PM

root(x^2) = root(-1)

Is the next step isnt it?

yeah but root(-1) != -1

Charlie

yes.....

skullpatrol

@IMX What "number" squared equals -1?

1 min ago, by IMX

x^2 = -1

Charlie

if you are on IR none

if you are on $\mathbb {C}$

$i\text {.e two solutions}$

N3buchadnezzar

@Charlie WRONSKI

Charlie

@N3buchadnezzar oh

N3buchadnezzar

7:15 PM

Cough two solutions Cough

Charlie

two solutions

IMX

Dudes

I don't know if its me or the strange food I had today

But how can x^2 = -1 be x = -1????

"we get that x4 + x = 0 = x(x3 + 1), so x = 0 or 􀀀1"

-1"

That's the original solution (its ^3 and ^4 in quote)

Charlie

@IMX man... (-1)(-1 -1)= -1(-2)=2

it's not zero

@n3b

N3buchadnezzar

@Charlie Yeh

-i and i

Charlie

@N3buchadnezzar then -1 is not an answer....

N3buchadnezzar

7:20 PM

Yeh

Charlie

@skull I'm drawing skulls now

IMX

And I am leaving

skullpatrol

@Charlie Coooool :D

Charlie

@IMX look

@IMX don't walk away, stay here with aunt Charlie

look heeeeeeeeeeeeeeeeeeeeere

$x(x^3+x)=0$

now

skullpatrol

She told me to walk this way

Charlie

7:22 PM

make $x=-1$

@skullpatrol ;)

N3buchadnezzar

@skullpatrol Talk this way

skullpatrol

she knew what she was doin' :D

N3buchadnezzar

Life is pie, you take a slice; then you die.

Charlie

Hey @IMX

skullpatrol

:-)

N3buchadnezzar

7:35 PM

@skullpatrol sup?

skullpatrol

@N3buchadnezzar chillin' like a villain

you?

N3buchadnezzar

Stop chillin, its too warm for that!

Charlie

$\varnothing \text {instein}$ .....

N3buchadnezzar

Stahp, you are not a freezer what are you doing?

Charlie

that name sounds like old name

N3buchadnezzar

7:37 PM

Its set in stone

(Joke since stein = stone..)

Charlie

there are so many cool norwegian names

N3buchadnezzar

While many euopeans think that we have cool names, we think that Iceland has even cooler names. Usually they end their names with (fathers name + son or daughter)

Petra Halldórsdóttir

Charlie

@N3buchadnezzar yeah.. so awesome

N3buchadnezzar

For an example, here dóttir means daughter. So she is Petra son of Halldór...

Charlie

J.R.R had some inspirations in it

N3buchadnezzar

7:41 PM

We have that too, but not anywhere near as common as it is in Iceland.

Charlie

aah

@N3buchadnezzar what is your surname?

if you don't mind

N3buchadnezzar

Søvik

Charlie

cool!

like in Star Trek

that's volcano's name

Peter Tamaroff

A surname so cool it has the empty set in it.

Charlie

oh pedro...

N3buchadnezzar

7:50 PM

@PeterTamaroff Mind help me understand an answer?

Peter Tamaroff

@N3buchadnezzar OK.

N3buchadnezzar

I have been looking at this. I clearly understand the area argument.

Peter Tamaroff

@N3buchadnezzar OK.

N3buchadnezzar

But I have problems understanding the last part of SoS404. It does not look like normal substitution..

Argon

@N3buchadnezzar Where is your answer?

Peter Tamaroff

7:52 PM

@N3buchadnezzar It is integration by parts, kinda.

And substitution.

N3buchadnezzar

scratches head

Guess I will give it some more thought.

Peter Tamaroff

@N3buchadnezzar Let me see if I can clear it up.

We have $$\int_0^1 (1-x^n)^{1/m}dx$$

Now, let ${x^n} = 1 - {y^m}$

The intgral is now $$\int_1^0 y d\left( {{y^{ - 1}}} \right)$$

@N3buchadnezzar

Integrate by parts.

It gives $$yy^{-1}\mid_0^1-\int_1^0 y^{-1}dy$$

One can write it out explicitly, if you want. The integral is

$$\int_1^0 y \left( { - \frac{m}{n}{{\left( {1 - {y^m}} \right)}^{1/n - 1}}{y^{m - 1}}dy} \right)$$

N3buchadnezzar

So $(dy)' = d\left(y^{-1}\right)$, that is a very susect abuse of notation..

Peter Tamaroff

But what is the point of that?

@N3buchadnezzar Err, what does $(dy)'$ denote?

N3buchadnezzar

Oh, silly me forget it

You wrote $d(y^{-1})$ in the meaning of the derivative of $y^{-1}$ and not in the meaning of integrating with respect to $y^{-1}$. Just me confusing the two.

Peter Tamaroff

8:06 PM

@N3buchadnezzar Right.

N3buchadnezzar

I have seen Rob use the latter a bit in some of his answers =)

Charlie

8:24 PM

@anon can you just take a tree down, like that? I wanna take a tree down, I do?

anon

chainsaw

skullpatrol

2 hours ago, by Peter Tamaroff

►

Charlie

@anon oh! there's a whole burocracy here

you have to ask permission ....

it takes a life

Peter Tamaroff

@anon LOL

Charlie

@PeterTamaroff I don't get it

skullpatrol

8:29 PM

me neither

anon

haven't the foggiest what any of you are on about

skullpatrol

9000 years later...

;O

Charlie

;O

my head...

skullpatrol

:D

N3buchadnezzar

@anon I was imagine you just punching the tree down.

anon

8:32 PM

rawr

skullpatrol

and the trees all shaked and quivered as he robbed them of their gold...

anon

xmas trees eh?

Charlie

@N3buchadnezzar anon was like :"go , go , no no, left, to the left, more strenght"

skullpatrol

(-:

N3buchadnezzar

@skullpatrol While shaking and shouting Give me more points "I am not capped, I will cap you!"

@Charlie Just like a sports couch!

Charlie

8:35 PM

@N3buchadnezzar yes, exactly

@skullpatrol now, I'm adding the details

to the drawing

skullpatrol

hmmm...

N3buchadnezzar

Life is like a banana

Charlie

banana plant only produces bananas once

i love banana

i love its perfume

you can fry it

it tastes good

banana pie

N3buchadnezzar

=)

Charlie

so, life is like banana

you do whatever you wanna do with it

and it's just one

wild bananas are funny

they have seeds

N3buchadnezzar

8:48 PM

Life is yellow, and sort of bended. You do not have to make sense of it, or give it meaning. It just is life; a banana.

@Charlie And banans grow on trees that is wild.

Charlie

yes

timvermeulen

I've got a quick question, anyone willing to help?

Charlie

say it

N3buchadnezzar

@timvermeulen If it is about you beeing a tree and needing help hiding the bodies of Anons three uncles. I do not want to take any part in that.

Charlie

haha

timvermeulen

8:52 PM

In how many ways can 11 marbles be divided into a maximum of 4 groups?

So, $a+b+c+d=11$, $a,b,c,d \geq 0$ and $a,b,c,d \in \mathbb{Z}$

oh, that doesn't work?

oh well

Charlie

you want 11 combined 4 to 4

i think it is $\huge \frac {11!}{7!4!}$

skullpatrol

$\Huge\text{!}$

timvermeulen

I thought of stars and bars, but that only works when the four groups are distinct

Peter Tamaroff

@N3buchadnezzar Do you know about Lagrange multipliers?

timvermeulen

I forgot to add that $a \geq b \geq c \geq d$

N3buchadnezzar

8:54 PM

@PeterTamaroff Yes?

Charlie

@timvermeulen what does that mean?

timvermeulen

a >= b >= c >= d

because (3,4,0,4) and (4,0,3,3) are the same

Charlie

the groups have different number of marbles?

Peter Tamaroff

@N3buchadnezzar $\sin x\sin y\sin z\leq \frac 1 8$ when $0\leq x,y\leq \frac \pi 2$ and $x+y+z=\pi/2$.

Charlie

oh shit

timvermeulen

8:55 PM

not necessarily

N3buchadnezzar

@PeterTamaroff Why you bring up that inequality?

Charlie

hm.......

timvermeulen

listing them isn't very hard: (11,0,0,0), (10,1,0,0), (9,2,0,0), (9,1,1,0), etc

Peter Tamaroff

@N3buchadnezzar I was tempted to write $$4\sin x\sin y\sin z=\sin(x+y-z)-\sin(x+y+z)+\sin(x-y+z)-\sin(x-y-z)$$

Charlie

@timvermeulen that's boring, let's find a better way

timvermeulen

8:57 PM

@Charlie of course

N3buchadnezzar

@PeterTamaroff Just repeated use of the angle sum formula right

Peter Tamaroff

@N3buchadnezzar Aye-

Charlie

@timvermeulen are the groups different among them?

timvermeulen

@Charlie what do you mean?

Charlie

for example (9,2,0,0) means the same (0,0,9,2)

timvermeulen

8:59 PM

@Charlie yes

Peter Tamaroff

@N3buchadnezzar Then we just use $x+y=\dfrac\pi 2-z$,$x+z=\dfrac\pi 2-y$,$z+y=\dfrac\pi 2-x$

timvermeulen

@Charlie those are the same

Charlie

@timvermeulen okay, less bad

timvermeulen

@Charlie it's easy to see that a is at least 3

@Charlie otherwise it would have been an easy stars and bars problem, familiar with that?

N3buchadnezzar

Well you have a $\sin(x+y+z)$ in there as well that evaluates to $1$.

Peter Tamaroff

9:00 PM

@N3buchadnezzar Aha.

Charlie

@timvermeulen no

@timvermeulen ah yes

skullpatrol

See ya later guys
bye @Charlie

Peter Tamaroff

@Charlie So it all turns out to be $$\sin \left( {\frac{\pi }{2} - 2y} \right) + \sin \left( {\frac{\pi }{2} - 2x} \right) + \sin \left( {\frac{\pi }{2} - 2z} \right) - 1$$

Charlie

@skullpatrol Bye, Bye, Skull

Peter Tamaroff

Surprise, motherfucker!

Charlie

9:01 PM

@PeterTamaroff n3b

N3buchadnezzar

@PeterTamaroff Wolfram agrees with your expansion it gives $4 \sin x \sin y \sin z = -\sin(x-y-z) + \sin(x+y-z) + \sin(x-y+z) - \sin(x+y+z)$.

@PeterTamaroff Doakes!

Charlie

@N3buchadnezzar \sin

timvermeulen

@Charlie but that's not going to work, probably

Peter Tamaroff

@N3buchadnezzar I salute you, comrade.

N3buchadnezzar

I always knew something were of about you

@Charlie $\operatorname{sun}$!

Charlie

9:03 PM

@timvermeulen no, well we can make how many we can with 1 group, with 2 groups, with 3 groups and 4

@N3buchadnezzar $\text {moon}$

Peter Tamaroff

@N3buchadnezzar And it is $\cos 2z+\cos 2x+\cos 2y-1$.

N3buchadnezzar

@Charlie You are my sun, my moon, my stars Al-kazeeen.

@PeterTamaroff Yes

timvermeulen

@Charlie okay, there's only 1 way to do it with one group

Charlie

@N3buchadnezzar nice! If I'm a sun , I can become supernova and kill everything. Moons can be attracted to their planet, killing things, Star can become supernovas, dwarfs, if their mass is inferior to the sun's. otherwise they can become blackholes

timvermeulen

@Charlie 8 ways to do it with two groups, I believe

Charlie

9:05 PM

if too masive

@timvermeulen wait why?

N3buchadnezzar

snifs and weeps

Peter Tamaroff

@N3buchadnezzar Oh, noes.

timvermeulen

@Charlie (14,0),(13,1),(12,2),...,(8,6),(7,7)

N3buchadnezzar

When Peter makes popculture references everybody gets them, when I make them nobody gets them =(

@PeterTamaroff Yeah

Charlie

@timvermeulen but we have 11, no?

timvermeulen

9:08 PM

@Charlie oh, you're right, my bad

@Charlie so 5 ways

Charlie

@timvermeulen no problem

@N3buchadnezzar no one gets what i say, either. maybe one or another

N3buchadnezzar

Me and Peter should rent an old RV and start cooking meth.

Charlie

@N3buchadnezzar and get arrested and become a lady in jail

@timvermeulen are you getting it?

timvermeulen

@Charlie nope, are you?

N3buchadnezzar

@Charlie I will be crazy banana lady.

Charlie

9:14 PM

@timvermeulen I think so

N3buchadnezzar

@timvermeulen Should just ask on the site. There are some very clever guys and gals out there, who love to answer questions when they are not punching down trees.

Charlie

@N3buchadnezzar are you saying I'm not clever, mister?

timvermeulen

@N3buchadnezzar yeah, I thought it would be rather easy :p

N3buchadnezzar

@Charlie I am just saying that there are trees out there that needs to be punched.

Charlie

@N3buchadnezzar I'll punch your face

N3buchadnezzar

9:17 PM

You wanting to punch me in my face, would imply me being crazy banana treebeard lady... Dont sound good.

Another point is that when asking on the site, there are more eyes on the questionn. Hence more likely that someone finds the right answer.

Charlie

hunf..

►

timvermeulen

thanks for your help guys

I looked around math.se and it turns out there is no known closed form

so I guess I'll do it manually :p

Charlie

@timvermeulen Good luck. I'll be here, just in case

if you need

N3buchadnezzar

@Charlie We could have a slumber party while he is doing serious stuff, like math.

Charlie

@N3buchadnezzar it's up to you

N3buchadnezzar

9:28 PM

takes forth icecream and fried banans

Tries to give it to Charlie

Charlie

I can't have icecream

but i can have the bananas

N3buchadnezzar

=)

Allergic to milk ?

Charlie

@N3buchadnezzar I'm not good

N3buchadnezzar

Thats right

Charlie

sneezing, my throat is weird, icecream wouldn't be good

N3buchadnezzar

9:32 PM

You are Awesome, and dont dare to forget it.

Charlie

@N3buchadnezzar why you say? you don't know it

N3buchadnezzar

Yes I do=)

Charlie

no, I'm sorry, but you do not

N3buchadnezzar

gives more bananas and flowers

Charlie

I'm allergic

I don't want more bananas

i had enough

they could have a made a better google doodle

N3buchadnezzar

9:40 PM

:p

Charlie

My drawing looks horrible :(

fuck, i'll have to do it again[

mick

@Charlie watch your language lady :) hru btw ? :)

Charlie

@mick I'm fine, and you?

damn

Left is a male human skull, right is a female skull

:-/

it's shitty

@skullpatrol

mick

10:01 PM

@Charlie I wonder what you removed ? Inspired by ethan ?

@Charlie :)

Charlie

@mick it was horizontal

mick

lol

Peter Tamaroff

@Charlie Really? Interesting.

mick

you are almost a supersaiyan too Charlie !!

Peter Tamaroff

Is that how Bones does it? =)

mick

10:03 PM

( at 1000 rep you are a supersaiyan )

Charlie

@PeterTamaroff My brother says that I'm like her

mick

@Charlie your skull looks happy , like it prefers to be dead :)

Charlie

@mick hahah yes

I'm not a pro

mick

@Charlie you could draw for ren and stimpy xD

Charlie

?

mick

10:07 PM

its funny how ppl start to resemble their idols ...

like , i lost my notebook ;)

Charlie

?

mick

ramanujan's lost notebook @Charlie

Charlie

hmm

so?

mick

it was a joke

Charlie

ha ha

mick

10:08 PM

it that a real laugh or a polite one ?

Charlie

the latter

damn seems that there's something scratching my throat

mick

@Charlie are you seducing me ? :)

Peter Tamaroff

@Charlie In what sense?

mick

Im such a badass :)

Charlie

@PeterTamaroff her personality

@mick what??

@mick no, you're not

mick

10:13 PM

@Charlie I think you did not get my comment then :p

Charlie

@mick you don't get what you say either

mick

@Charlie duhhh whats my name again ?

Charlie

oh, please....

mick

:)

Charlie

Change my attempt good intentions
Should I, could I
Here we are with your obsession
Should I, could I

mick

10:18 PM

poetry ?

Charlie

a song

@peter did you decide?

Peter Tamaroff

@Charlie Decide what?

Oh.

No, I didn't

Charlie

:/

reptiles are fascinating

mick

@Charlie decide ?

Charlie

@mick yes

mick

10:29 PM

@Charlie decide what ?

Charlie

a thing

ask him

mick

@PeterTamaroff what do you need to decide ?

Chris Dueck

If I have a connected graph G, then I know a spanning tree for G exists. WLOG, can I assume the existence of two spanning trees that differ by one edge? I think I can since G is connected.

mick

bye

Charlie

@mick bye

Transcript for

$Mathematics$ Mathematics