Mathematics - 2014-12-04 (page 8 of 9)

@DanielFischer how would you recommend me to tackle this series above?

Sab ಠ_ಠ

so I'm not sure if it is cuz it's a furst

:P

Well it's past midnight and I'll go write up some algorithms in Java :)

night peeps

Night.

@robjohn Take a look at the series above and let me know how you would approach it (when you have time).

@Venus I also put my trust in you for the one above!

8:03 PM

@WillHunting My family tree is full of mental illness. I've been taking ADHD medication since kindergarten, antidepressants and anxiety meds since late elementary school, so I've pretty much been having issues for the past 7-10 years.

$$\LARGE \text{TYPO}$$

@Chris'ssis trust in what?

$$\sum_{k=1}^{\infty}\sum_{l=1}^{\infty} \sum_{n=1}^{\infty} \frac{(\sqrt{2}-1)^{2k+2l+2n-3}}{(2k-1)(2l-1)(2n-1) (2k+2l+2n-3)}$$

@Venus in your math abilities.

@DanielFischer @Venus @robjohn I just rewrote it, there was a type.

@Chris'ssis If you mean to evaluate this sum, I think I give up already

@Venus Why? :-)

8:05 PM

@Chris'ssis I only good at Taylor series

@teadawg1337 I see. I have felt pretty bad almost my whole life as well. Do you have any plans to "cure yourself completely", whatever that means? I have some plans for myself in this direction, and I hope I "get completely well" soon.

Not this monster

@WillHunting As far as I can tell, I won't be able to "cure myself completely"

@Hippalectryon I need a brilliant idea you might give to me ...

@Chris'ssis You're the one who usually finds brilliant ideas

8:10 PM

@Hippalectryon hmmm .... OK, let me do something brilliant then (first on paper)

@teadawg1337 OK. I hope you can some day, possibly by redefining the illness and redefining the cure.

@Chris'ssis :D

@WillHunting Maybe someday, but not anytime soon

@Chris'ssis You are so brilliant. =)

@WillHunting Let G be a collection of my illness. Also let $G\equiv\emptyset$.

8:12 PM

@UserX LOL, you must be on drugs, lol.

Nah. Just redefining my illness :P

Sven

Anyone heard of a "Triangle related to the asymptotic expansion of E(x,m=3,n)" before?

oeis.org/A163932

But i get what you mean

@JasperLoy hi

@user130018 Hello Bart. How are you these days?

8:22 PM

$$\sum_{k=1}^{\infty}\sum_{l=1}^{\infty} \sum_{n=1}^{\infty} \frac{(\sqrt{2}-1)^{2k+2l+2n-3}}{(2k-1)(2l-1)(2n-1) (2l+2l+2n-3)}=2\int_0^{\infty} \operatorname{arctanh}^3((\sqrt{2}-1)e^{-x}) \ dx$$

@Hippalectryon @Venus @DanielFischer @robjohn I just got its integral representation.

@Chris'ssis O_o how did you get that ?

@r9m and I think you help me finish the last integral :-)

@Hippalectryon "@Chris'ssis You're the one who usually finds brilliant ideas". I had to come up with something ... :D

@Chris'ssis :DD

@Chris'ssis wins

@Chris'ssis If one day you decide to tell me your name, please email me. =)

@WillHunting I'll tell it to you one day, it's not a big deal, I'm a simple person. :-)

Wait ... I'll tell you soon its closed form. ..

DONE

8:28 PM

Your name has a closed form?

2

1/256 (8 [Pi]^4-Log[2]^4+8 Log[2]^3 Log[2-Sqrt[2]]-24 Log[2]^2 (Log[2-Sqrt[2]]^2+Log[1+Sqrt[2]]^2-2 PolyLog[2,1-Sqrt[2]]+2 PolyLog[2,-1+Sqrt[2]])+32 Log[2] (Log[1+Sqrt[2]]^3+3 Log[2-Sqrt[2]] (Log[1+Sqrt[2]]^2-2 PolyLog[2,1-Sqrt[2]]+2 PolyLog[2,-1+Sqrt[2]])+6 PolyLog[3,1-Sqrt[2]]-6 PolyLog[3,-1+Sqrt[2]])+8 (Log[16] Log[2-Sqrt[2]]^3-2 (Log[2-Sqrt[2]]^4+Log[1+Sqrt[2]]^4+

6 Log[2-Sqrt[2]]^2 (Log[1+Sqrt[2]]^2-2 PolyLog[2,1-Sqrt[2]]+2 PolyLog[2,-1+Sqrt[2]])+4 Log[2-Sqrt[2]] (Log[1+Sqrt[2]]^3+6 PolyLog[3,1-Sqrt[2]]-6 PolyLog[3,-1+Sqrt[2]])-24 PolyLog[4,1-Sqrt[2]]+24 PolyLog[4,-1+Sqrt[2]])))

Q.E.D

@Chris'ssis OMG! You are a genius!

@JasperLoy Pretty good except too much workload

Well, this might be further simplified ...

@user130018 Have you done your Christmas shopping?

8:30 PM

@WillHunting The geniuses work now in places like NASA ...

@Chris'ssis I think you are smarter than them.

@WillHunting heh, that's an exaggerate statement :-)

@Chris'ssis No, I don't think so. =)

@WillHunting Just think about it: I failed in my country many jobs interviews for which the major part of you probably wouldn't accept as a job ... ;)

@Hippalectryon The cubic problem can be considered solved.

@Chris'ssis That was an interesting one :)

Send me the full proof if you can

8:36 PM

@Chris'ssis is that the closed form for the above integral ?!! O_o

@Hippalectryon There is much stuff to write. Wehn I do it, I'll send it to you.

@r9m I started from a series that gave me some headache until I was encouraged by@Hippalectryon :-)

Here is a simpliefied form

1/256 (8 [Pi]^4-Log[2]^2 (Log[2]^2-8 Log[2] Log[2-Sqrt[2]]+24 Log[2-Sqrt[2]]^2)+8 Log[2-Sqrt[2]]^3 Log[8 (3+2 Sqrt[2])]+8 (-2 ArcSinh[1]^4+6 ArcCosh[3]^2 PolyLog[2,1-Sqrt[2]]+3 ArcCos[3]^2 (ArcSinh[1]^2+2 PolyLog[2,-1+Sqrt[2]])+ArcCosh[3] (4 ArcSinh[1]^3+6 PolyLog[3,3-2 Sqrt[2]]-48 PolyLog[3,-1+Sqrt[2]]))+48 PolyLog[4,3-2 Sqrt[2]]-768 PolyLog[4,-1+Sqrt[2]])

:DD

@Chris'ssis ?! I can't even read the closed form ! its not human ..

@r9m this is a part of my proof to $$\int_0^{\pi/2}\arctan^3(\sin(x))dx$$

@Chris'ssis That closed form looks about right to me, I'm too tired to check the integral for myself

8:40 PM

@Chris'ssis ah !! I see .. this is just a part of a BIG picture ..

@r9m A tiny one.

@Chris'ssis got it !!! :)

@r9m I also added the improvements noted by @Hippalectryon and the result is marvellous. :-)

:-)

i'm celebrating right now with some chocolate

@Chris'ssis okay ! guess I have to read the transcripts for the past few days ! (I was travelling by train .. hence my absence)

8:43 PM

@Hippalectryon lol, don't make me think of chocolate now ... :-)

:DDD

@r9m By the way, where were you in the last days? :D

@JasperLoy I don't celebrate Christmas

@Chris'ssis I traveled form Chennai back to my home :) (28 hrs by train)

@user130018 You should

8:45 PM

@Hippalectryon How come

@r9m So many hours? It seems to me as long as a life especially when by train :-)))))

@user130018 Think to all the poor chocolate just waiting to be eaten !!

@Hippalectryon Who's gonna give me free chocolate

@user130018 I never said free :D

@DanielFischer $f$ and $g$ intersect transversally at $P=(0,0) \Leftrightarrow $ they don't have a common tangent. In this case, $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$. How can we check if they intersect transversally?

8:47 PM

@user130018 But it's really easy to get some really cheap

@Hippalectryon No chocolate for me then :(

@user130018 Step1: Buy one tablet Step2: Banach-tarski Step3: Two tablets

@Chris'ssis yes ! .. but I packed by laptop with new anime ! =P I was never bored .. Death Note

@r9m Is it nice (Death Note)? :-)

Can anyone explain to me why this is always true? $$\lim_{x\to0}\frac{\arctan{ax}}{\sin{bx}}=\frac{a}{b}$$

8:49 PM

@Hippalectryon I don't have that kind of money

:/

@Chris'ssis awesome ! the most cruel but at the same time intellectully stimulating anime I have ever seen :D

The most cruel? lol

@r9m I saw an episode, but I don't have any HD version :-(

There are far more cruel anime

Anyway, I'm out. Movie&physics studying till fallen asleep.

8:51 PM

@UserX are they intellectully stimulating as well ?! :-)

$$\lim_{x\to0}\frac{\arctan{ax}}{\sin{bx}}=\lim_{x\to0}\left(\frac{\arctan{ax}/(‌ax)}{\sin{bx}/(bx)} \cdot \frac{ax}{b x}\right)=\frac{a}{b}$$

@r9m Indeed, there are far more cruel ones

@Chris'ssis oo ! neither did I watch in HD ..

@r9m not really. You might like Code Geass too btw.

@UserX 'kay !! thanks ! I will check it out ! :)

8:53 PM

@Chris'ssis I must be super tired right now, that's pretty obvious in hindsight...

@Chris'ssis That's definitely better than partial fractions

@user130018 When do your exams end?

@robjohn Yeah, far better :-)

@JasperLoy I have my last final on December 23

@user130018 And then you can have a merry Christmas and happy new year, lol.

8:57 PM

@JasperLoy You too Jasper Loy

@user130018 I hope so. I am trying to start studying math on New Year's Day, as you know.

@teadawg1337 Just some practice needed.

@Hippalectryon I don't watch too many animes .. but Death Note was simply 'bang-on' with my taste ... that's literally everything that I could want from a good anime :)

Lucio D

@DanielFischer :) Yes that is the definition of bounded which is used.

@r9m I watched Death Note 1,2 and 3, lol.

9:02 PM

@WillHunting I haven't seen the movies ..

@r9m Oh, then what did you see?

@WillHunting 37 anime episodes .. imdb.com/title/tt0877057

@r9m OK. I watched the 3 Japanese movies.

@WillHunting are they nice too ? :D

@r9m Yes, all very good.

9:04 PM

@WillHunting okay ,, I'll surely watch them then !

I just missed 2 buses because they were full..sucks standing like an hour in this weather

@user130018 Just sit on the bus top, lol.

@JasperLoy I'm not that athletic :(

@user130018 Have you found a gf yet?

@JasperLoy I haven't spoken with a girl yet

9:10 PM

@WillHunting: Have you found a gf yet?

@Huy Nope, thanks, lol.

@WillHunting: You're welcome.

How are you doing these days, @WillHunting?

@JasperLoy Are SG girls nice in general or not

all girls who met me seems to have unanimously agreed to the fact that I am 'scary' :P rofl

@Huy I think I am slowly getting better. I hope to get completely well in a year from now.

@user130018 I don't like them and they don't like me.

9:12 PM

@WillHunting: I hope you achieve your goal.

@Huy How is that game theory class?

@r9m: What makes you scary?

@JasperLoy What don't you like about them

@WillHunting: It will start by the end of February, I didn't start preparing it yet.

@user130018 Well, I find most people here superficial, not just the girls.

9:14 PM

@WillHunting: You will find out people all over the world are superficial.

@JasperLoy Is SG based on Consumerism like the U.S.?

@Huy that's what makes me rofl :P they could not pinpoint what it was (objectively) ..

@r9m: And subjectively?

@evinda I can't make sense of that. The graphs are surfaces in $\mathbb{R}^3$, so their tangent spaces necessarily have nontrivial intersection. Find out what the definitions are.

@user130018 Certainly, money is the most important thing on many people's minds.

9:15 PM

@Huy well .. everyone had their own (s)opinions .. :P ..

@JasperLoy Sounds like the exact same as it is here

@r9m: Wanna tell me some? :D

@WillHunting Well, I ... got Friday on my mind.

@user130018: You don't like people around you either, then?

@DanielFischer Eberhard Freitag wrote two great books on complex analysis.

9:16 PM

@Huy No, I like those kinds of people

@DanielFischer $$f(x,y)=x^5+x^4+y^2$$
$$f_2(x,y)=y^2$$
$$f_4(x)=x^4$$
$$f_5(x,y)=x^5$$

So, the order of the point $P(0,0)$ is $2$.

To find the tangent of $f$ do we set $f_2(x,y)=0$ ? If so, the tangent would be $y=0$.

$$g(x,y)=x^6-x^5+y^2$$
$$g_2(x,y)=y^2$$
$$g_5(x,y)=-x^5$$
$$g_6=x^6$$

To find the tangent, do we set $g_2(x,y)=0 \Rightarrow y=0$

So, do $f$ and $g$ have the common tangent $y=0$ ?

@Huy hell .. that's embarrassing even considering the fact that its a chatroom where no one knows my real identity :P lol

@user130018: ???

Friday on My Mind

"Friday on My Mind" is a 1966 song by Australian rock group The Easybeats. Written by band members George Young and Harry Vanda, the track became a worldwide hit, reaching no. 16 on the Billboard Hot 100 chart in May 1967 in the US, no. 1 on the Dutch Top 40 chart, no. 1 in Australia and no. 6 in the UK, as well as charting in several other countries. In 2001, it was voted "Best Australian Song" of all time by the Australasian Performing Right Association (APRA) as determined by a panel of 100 music industry personalities. In 2007 'Friday on My Mind' was added to the National Film and Sound Archive...

@r9m: Nobody knows you. I tell secrets too. It's just a chatroom. :D

9:17 PM

@Huy They buy me free stuff since I can't afford it

@user130018: How does that make you feel?

@evinda I have no idea what you're talking about. What topic is it even? Differential topology?

@Huy I guess nice because I like getting free stuff

@Huy lul .. some other time perhaps :P I have assignments to finish ! ... its a mouthful of comedy :P

@user130018: What do you think they think about you?

@r9m: Good luck finishing your assignments then.

9:19 PM

@Huy (y) :-)

@Huy No idea, but I don't really care all that much as long as I'm getting stuff

@user130018: Okay, interesting.

@Huy What is interesting

@DanielFischer It is from an advanced algebra course...
I am trying to find the multiplicity of the intersection of the curves $f$ and $g$.

@evinda Not my area. For algebra, you might try Mike Miller or anon.

9:22 PM

@DanielFischer A ok... :)

@anon @MikeMiller @robjohn @KajHansen $$f(x,y)=x^5+x^4+y^2$$
$$f_2(x,y)=y^2$$
$$f_4(x)=x^4$$
$$f_5(x,y)=x^5$$

So, the order of the point $P(0,0)$ is $2$.

To find the tangent of $f$ do we set $f_2(x,y)=0$ ? If so, the tangent would be $y=0$.

$$g(x,y)=x^6-x^5+y^2$$
$$g_2(x,y)=y^2$$
$$g_5(x,y)=-x^5$$
$$g_6=x^6$$

To find the tangent, do we set $g_2(x,y)=0 \Rightarrow y=0$

So, do $f$ and $g$ have the common tangent $y=0$ ?

lol

I'm not sure I'm qualified to answer that :/

@DanielFischer For mental illness they might try me.

I'm having such a good laugh right now: adequacy.org/stories/2001.10.14.163749.94.html

I'm having a good laugh rewatching some Big Bang Theory episodes. =)

9:34 PM

@KajHansen lol

So good

I suspected it was satire, and it turns out I'm right, haha

I only like stupid jokes, not deep jokes, lol.

omg @KajHansen the comments are the best

My comment got 9 stars. I am waiting for the 10th one.

I can't even tell who's trolling who anymore @user130018

9:45 PM

@KajHansen Who's, not whose.

The $\nabla\left(x^5+x^4+y^2-z\right)=\left(5x^4+4x^3,2y,-1\right)$ Therefore, the tangent plane would be $\left(x-x_0,y-y_0,z-z_0\right)\cdot\left(5x_0^4+4x_0^3,2y_0,-1\right)=0$

Hush, I fixed it fast enough @WillHunting

@KajHansen Why did your gf break up with you, considering you are so hot?

@WillHunting, how did you know that happened? o_o
@robjohn, oohhhh, so that's what evinda's asking about.

@KajHansen You told me a while ago, lol.

9:47 PM

@KajHansen it seems... there was a surface and a tangent plane. I could be wrong

Ah. Well to answer your question, I'm not sure.

@KajHansen: Did she not tell you or did you not ask her?

@Huy, both

@KajHansen: Why? Did you not care or were you afraid?

@Huy, I'm not sure. You aren't usually in a talking mood when you've been hit with an emotional sledgehammer. At least I'm not. That was a while back though so I'm fine now.

9:52 PM

@KajHansen: Okay. I usually like to solve problems or at least understand them, so we're different in that regard.

There was once someone complained about me at work. I did not get fired but it was bad enough. And I did not even do anything I considered wrong. I was so mad I could not say anything.

@WillHunting: I didn't know you were such an emotional person.

@Huy Of course I am. I used to sing Italian opera, lol.

@WillHunting Liar. Bananas don't sing.

@WillHunting: I don't see how the two relate.

9:54 PM

@Huy Operatic people are emotional, QED.

@Hippalectryon I am a singing banana, lol.

@WillHunting: I don't think your proof is valid.

>.>

@Huy It is, by the conjecture theorem

@Hippalectryon: Bonsoir.

Bonsoir

Je m'appelle Huy.

9:56 PM

I have a terrible headache.

I think I am tired.

Have a nice day/evening, everyone.

I'm off to bed.

Twink

me too @WillHunting

:(

$$\Large \text{I created a crazy awesome question right now!}$$

@Hippalectryon ^^^

SHOW ME :DD

@Hippalectryon $$\int_0^{\pi} \frac{\arctan(\cos(x))}{\cos(x)} \ dx$$

10:00 PM

O_o

@Hippalectryon ^^^

Nice made-in-China Gaussian :D

no offense

@Hippalectryon hahahaha (+1)

:-)

:DDDD

@Hippalectryon You don't like Chinese products?

10:04 PM

It's a shame it goes a bit below zero near the edges

@WillHunting Depends

@WillHunting I don't like Chinese products when they break after 3 days

@Chris'ssis Can you help for lim of an intregral ? $\lim_{n\rightarrow \infty}\int_{0}^{1}\frac{e^t}{1+e^{nt}}$?

@Hippalectryon It has a very nice closed form!!!

@Chris'ssis tell me :-)

@Hippalectryon OK. I think that they are mostly alright, just like non-Chinese products.

dt*

10:05 PM

@WillHunting Indeed. Most. Not all.

@Hippalectryon Wouldn't you like to work a bit on it? :-)

@Chris'ssis I can't do everything you give me every day :/ Unlike you, I don't have as much free time for maths. I do save everything you send here, though

@MarcGato I think it goes to $0$.

@Chris'ssis I am not interested in limits, series and integrals, lol.

@Hippalectryon Do I have free time? I don't know what "free time" is (or it's rare to me) :D

Twink

10:07 PM

:(

@Chris'ssis Time to compute integrals :)

@Hippalectryon :-))))))))) I know, but the point is that I don't have free time, but I steal from the time assigned to other things to attend these things.

@Chris'ssis intuitively I a tempted to say that too, but how can I proceed?

@Chris'ssis Are you still doing accounting work?

@WillHunting yeah, not as much as before.

@MarcGato Think a bit of it, it's not hard at all.

10:12 PM

We have $e^x\ge x+1$ and so $\frac{e^t}{1+e^{nt}}\le \frac{e}{1+nt}$

it's not good enough. Do we need "taylor" series ?@Chris'ssis

@MarcGato Is this the simplest way? I don't think so.

How about this one $$\frac{e^t}{1+e^{nt}}\le \frac{e^t}{e^{nt}}$$?

@Chris'ssis Oh silly me!!

@MarcGato :D

now the integral is easy to compute ^^

Thanks @Chris'ssis :-)

@MarcGato Welcome ;)

@Hippalectryon OK ...

10:18 PM

:D

@Hippalectryon We have that $$\int_0^{\pi} \frac{\arctan(\cos(x))}{\cos(x)} \ dx=\pi\log(\sqrt{2}+1)$$

I like those $\sqrt{2}\pm 1$ that appear everywhere

@Hippalectryon I like them too. :-)

What's the mathematical meaning behind those ?

@Hippalectryon Depending on the context. Just think of the golden ratio ...

10:21 PM

I will note this answer in case people want to vote to reopen the question mentioned therein. That is, vote to reopen Finding $\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$

@Chris'ssis But why would it arise here ?

That's odd... I thought it had 4 upvotes. I wonder where I was looking insetad.

@Hippalectryon Well, I look at that and I think of the inverse hyperbolic sine (for some reason) ...

:D

>.>

No need for differentiation and other stuff like that ... (as I can see the differentiation makes it very easy unfortunately)

Zach Saucier

10:30 PM

For the problem "Integrate the function $G(x,y,z) = z$ over the cylindrical surface $y^2 + z^2 = 4, z \geq 0, 1 \leq x \leq 6$", I should do it in terms of $z$ and $\theta$, right?

Lucian

@Chris'ssis: How would we solve the above-mentioned integral ?

@Lucian I'd use the fact that $$\int_0^{\pi} \frac{\log(1+a \cos(x))}{\cos(x)} \ dx=\pi \arcsin(a)$$

and then set $a=i$

@Lucian We evaluate an integral, not solve it.

On the other hand, one might consider $$I(a)=\int_0^{\pi} \frac{\arctan(a\cos(x))}{\cos(x)} \ dx$$

and use D.U.I.S.

evaluate a proof

10:36 PM

@user130018 Are you living alone now?

@JasperLoy I have for a while

@Hippalectryon this is way nicer ...

$$\int_0^{\pi/2} \arcsin(\tan x) \ dx=2{\large\chi}_2\left(\sqrt{2}-1\right)-\frac{\ln\left(3-2 \sqrt{2}\right)}{2}\sinh^{-1}(1)-\frac{i\pi}{2 }\sinh^{-1}(1)$$

Looks nicer

Zach Saucier

I don't know how to solve my above problem since it's not a graph :P

heya @Zach, @Hippa, @mr eyeglasses, @Jasper ...

10:46 PM

Heya

Zach Saucier

hiya

Hello @Ted

heya @teadawg ... how you be?

@Hippalectryon Wait ...

?

10:48 PM

hi @Chris'ssis

@TedShifrin Hello! :-)

@Ted I'm exhausted today, I'm trying to make it through the day as best I can though

I'm sorry to hear that, @teadawg ...

@TedShifrin Ted, I wanted to ask you what the most horrible excuse for not doing the homework you've ever heard from your students. :-)

LOL ... I don't even want to think about that one :)

10:51 PM

@TedShifrin :-)))))))))))))

Death, disease, famine.

Famine ??

@TedShifrin In a class of mine, a guy said once he visited another planet for a couple of days and couldn't do the homework. :-)))))))

@Chris'ssis LOL

@TedShifrin Sorry prof Ted, my homework died of Ebola...

10:53 PM

@teadawg1337 lol, yeah, believe me, in a class you hear all kind of stuff ... :-)))))

oh, hell, I'm giving @Hippa ideas.

:DD

I plan on doing my homework one day and coming up with the worst excuse ever for not doing it, so that the professor will walk up and ask me why I didn't do my homework and I'd pull a complete 180

@Chris'ssis He visited another planet ? and so ?

@Chris'ssis Why couldn't he do his homework on the moon ?

Spend all that creative energy elsewhere, @teadawg :D

10:55 PM

@Hippalectryon lolllll, busy with the alien meetings perhaps :-)))))