Mathematics - 2012-11-27 (page 1 of 4)

Peter Tamaroff

12:08 AM

@Argon That is awful notation man.

Argon

@PeterTamaroff Meh

What would you do?

Peter Tamaroff

You mean $$\left(\frac{i+1}{\sqrt 2}\right)^2=i$$

Argon

My lack of $\pm$ implies (to me, at least), that I am choosing the positive branch of the square root.

robjohn

12:20 AM

$$
\begin{align}
\int_{-\infty}^\infty e^{2\pi i\theta^2}\mathrm{d}\theta
&=2\int_0^\infty e^{2\pi i\theta^2}\mathrm{d}\theta\\
&=2\frac{1+i}{\sqrt{2}}\int_0^\infty e^{-2\pi\theta^2}\mathrm{d}\theta\\
&=\frac{1+i}2
\end{align}
$$

Argon

@robjohn Why don't the bounds become complex?

robjohn

@Argon The integral out the real axis is the same as the integral along the line at $45^\circ$ to the real axis....

Argon

@robjohn Curious. Why is that?

robjohn

Because there are no singularities in the wedge and the integral over the eighth circle at infinity is 0

Argon

Does that mean the integral can be rotated by any number of degrees?

Because no singularities exist anywhere

user19161

12:29 AM

@PeterTamaroff Yours is terrible too!

robjohn

@Argon if the integral along the piece of the circle at infinity vanishes.

@PeterTamaroff: Hey there! haven't seen you in a while

Peter Tamaroff

@WillHunting ORLY?

@robjohn I had some exams going on, but I'm free now.

user19161

@PeterTamaroff Yes, like the last time you used Greek letters when there are so many English ones available!

Peter Tamaroff

Free to study whatever I like.

robjohn

@PeterTamaroff freedom is good :-)

Peter Tamaroff

12:30 AM

@WillHunting You mean latin?

user19161

@PeterTamaroff Good good, let's go on a date. =)

Peter Tamaroff

@WillHunting NO, thanks?

Argon

@WillHunting $\omega i \lambda \lambda$

@PeterTamaroff Or nothing at all :)

Peter Tamaroff

@Argon Hmmm....

I have been playing Portal 2 and Assasins' Creed II,

...and Napoleon Total War and Civilization V

PROCRASTRINATE, PROCRASTRINATE, PROCRASTRINATE!

user19161

@PeterTamaroff Games are boring! I rather do math 24/7...

Peter Tamaroff

12:34 AM

@WillHunting Portal 2 is quite nice

Argon

I like Rise of Nations, though I play seldom...

@WillHunting You must take some breaks!

Peter Tamaroff

@Argon That one is nice.

user19161

@PeterTamaroff I prefer to play with my balls.

Argon

@PeterTamaroff You know it?!?! Nice!

@WillHunting I hope no flaggers are present...

...

user19161

@Argon Basketball, football, etc.

Peter Tamaroff

12:36 AM

@WillHunting Chuck Norris disapproves.

peoplepower

@Argon One moment, I have to log out.

Peter Tamaroff

You better run, pal.

Argon

@WillHunting Hahaha :)

Hi @people

peoplepower

@Argon Hello world.

Argon

printf ("Hello World!");

user19161

12:38 AM

@PeterTamaroff I don't know who Chuck Norris is, sorry.

Argon

@WillHunting From the movies?

user19161

@Argon That is one name I keep forgetting.

Argon

@WillHunting Intentionally?

user19161

@Argon Nope, just doesn't register.

user19161

You know, I can't figure out why I hate computer games so much...

peoplepower

12:42 AM

@WillHunting I am too lazy for that stuff. :)

Peter Tamaroff

@robjohn HOw are you?

robjohn

@PeterTamaroff doing pretty well. Busy at work, and not a lot of time to answer questions here

Peter Tamaroff

@robjohn Oh. I have something to ask =)

robjohn

@PeterTamaroff aure

Peter Tamaroff

I take it one uses induction to prove there is one and only one alternating multilinear function $f:K^{n\times n}\to K$ such that $f(I_n)=1$?

robjohn

12:48 AM

like determinant?

Peter Tamaroff

There is a Lemma in my book that one can obtain a alt.mult. function over $K^{n\times n}$ given one over $K^{(n+1)\times (n+1)}$

@robjohn Hehehe yes.

user19161

Sounds like the determinant function.

robjohn

@PeterTamaroff that should be true.

Peter Tamaroff

@WillHunting It is the determinant function

robjohn

and pretty easy

user19161

12:49 AM

Yeah, Pedro should do this himself.

Peter Tamaroff

@robjohn Yes. One makes a cross of zeroes with a one in the center

over the $i$th row and $i$th column one likes.

robjohn

@PeterTamaroff I was thinking of an $n\times n$ minor in the upper left and a $1$ in the lower right

Peter Tamaroff

@robjohn Well, yeah, that is nthe idea

robjohn

@PeterTamaroff You define a general one using Gaussian elimination to bring the "matrix" into diagonal form.

Peter Tamaroff

robjohn

1:01 AM

@PeterTamaroff You can put the 1 anywhere

sonicboom

If I roll 4 dice what is the probability that I get exactly one pair?

Peter Tamaroff

@robjohn What do you mean by anywhere?

Say $ij$?

user19161

Hey @amwhy have you deleted your extra account?

amWhy

@WillHunting - re a comment you left a day or two ago: How did you learn I have two accounts? I didn't even know that.

sonicboom

I got $$\frac{{4 \choose 2} {2 \choose 1} {1 \choose 1} {6 \choose 1} {5 \choose 1} {4 \choose 1}}{6^4}$$ which is obviously wrong as the numerator is bigger than the denominator

user19161

1:10 AM

@amWhy Ah, I found it by searching the user list. math.stackexchange.com/users/38542/amwhy

amWhy

@WillHunting Hey, I was just asking about that. I had no clue. I'll have to talk to a moderator about deleting it!

user19161

@amWhy Well, @rob is here to delete it for you!

user19161

@amWhy You know what's amazing? That account was last logged in to on my birthday! Haha!

amWhy

@robjohn - how convenient: it seems I have two accounts? Maybe when I first used my netbook?

Argon

@WillHunting Aug 23? Were you born at exactly 14:27, because that would be a coincidence!

(but, as you say, nothing is by chance!)

amWhy

1:14 AM

@robjohn math.stackexchange.com/users/38542/amwhy Same username, same gravatar, just be careful that if/when it's deleted, I don't lose my "main" account!

@WillHunting How amazing!

user19161

@Argon Not the exact time, but the date yes. 23 is also my favourite number and it appears on watch advertisements as well as T-shirts.

Argon

@WillHunting "There were 23 problems on David Hilbert's famous list of unsolved mathematical problems"

amWhy

@WillHunting I like "9" 8)

@WillHunting I like "5" too!

@WillHunting Now you just have to figure out which is month and which is day ;-)

user19161

@amWhy Erm OK I will figure out some time in the future...

user19161

@amWhy Do you use Windows or Mac or Linux?

user19161

1:23 AM

@Argon Also, 23 is a prime number.

Argon

@WillHunting And the fifth factorial prime, the third Woodall prime

Eisenstein prime...

Sum of three other, consecutive, prime numbers

"The sum of the first 23 primes is 874, which is divisible by 23, a property shared by few other numbers."

Yay!

user19161

@Argon This is a sign that miracles are about to happen...

robjohn

@PeterTamaroff anywhere in the matrix

@WillHunting what, are we here to cater to your every whim?

amWhy

@WillHunting Windows: I do have a "virtual" machine installed which runs Linux.

Argon

robjohn

1:29 AM

@Argon Taken by itself, it was a great movie. The other two should not have been made.

user19161

@Argon You must be the hot guy standing in front.

Argon

@WillHunting ...

robjohn

@amWhy I can merge your accounts.

John Smith

anyone good with modular arithmetic? if i want to find (a+b+c+d...)/6 mod 10^5 for instance, how can i modify the equation to be modular by somehow applying it to a,b,c,d...etc

amWhy

@robjohn Thanks! Take your time on it. The accidental account was never used.

robjohn

1:31 AM

@amWhy It's already done.

amWhy

@robjohn Thank you!

user19161

@amWhy You can check that searching for amwhy now yields only one amwhy, not two! Existence and uniqueness!

amWhy

@WillHunting Hahaha...

@WillHunting I just never thought to search for myself. But that raises the question: who am I? Maybe I should change my user to amWho!

user19161

I was surprised that qiaochu commented on my answer here. He's just clarifying but I found it a little weird. math.stackexchange.com/questions/245363/…

Argon

@amWhy Hahahaa

user19161

1:36 AM

@amWhy Hmm, I think you are set for 10k this year...

amWhy

@WillHunting Maybe. 8-)

@WillHunting Yeah, that is kind of wierd. At least the OP backed you up!

user19161

@amWhy I think maybe he just interpreted the question too literally!

amWhy

@WillHunting That could be...But it's not like you solved a homework problem for someone who asked for only a hint!

user19161

@amWhy I must say that sometimes there is no way to help the asker other than giving the answer itself for certain kinds of problems...

amWhy

@WillHunting You're right, and I think most OP's don't mind that at all!

user19161

1:43 AM

@amWhy Anyway, I am sick of this internet language "OP". I will just use asker and answerer now!

user19161

@amWhy For a moment I thought qiao was pissed with me, but he's not I think.

amWhy

@WillHunting Oh, btw, be careful when editing: I edit a lot of my posts, and some are near the "wikification" point if edited again. When I've overdone editing (of my posts), I usually know, and the mods are good about de-wikifying, but if I don't know it's been turned CW...oh well!

@WillHunting I never knew "OP" until frequenting math.se: you're right, I prefer "users" in general.

@WillHunting I don't think he was pissed at all; it's his manner, in general.

user19161

@amWhy Ah OK. I can edit much more actually, but I also want to preserve the style of the answerer. So I have been reserved in my edits already. Also, I find that you not only use many ellipses but also many commas and quote marks.

user19161

@amWhy Yes, which is why I did not over-react! I think in weird ways myself too.

amWhy

@WillHunting Indeed, I do!

user19161

1:47 AM

@amWhy Honestly, I find it hard to understand some of your writing. Some sentences seem too convoluted to me with all that extra punctuation!

amWhy

@WillHunting It's so hard to "read" feelings; particularly - like me - for those of us who are sensitive.

user19161

@amWhy Ultimately we can only approximate what another is thinking. The more info we have, the better the approximation. QED.

Argon

$\blacksquare$

amWhy

@WillHunting Well said. And such approximations can also be impacted by projection.

user19161

@robjohn In the ELU room, they call the mods janitors...

amWhy

1:50 AM

@WillHunting hahahaa!

user19161

@amWhy Yes, our own experiences determine much how we perceive things.

amWhy

@WillHunting Absolutely. Again, well-said!

user19161

I usually try to look at things "from first principles" so that I don't fall prey to subconscious biases.

amWhy

@WillHunting I am pretty good about not judging others, but not so good with respect to judging myself. It's weird, because I rarely get angry, but I easily feel hurt.

Argon

anon, it's @anon!

anon

1:54 AM

Egads!

Argon

:)

anon

Zoinks, even!

Argon

Halloa!

user19161

@Argon I found that long ago dude!

user19161

The great @anon is here!

Argon

1:58 AM

@WillHunting "Found?" I guess you don't read old enough books then! Hahahaa!

user19161

@Argon The only books I read are math books.

Argon

@WillHunting Really? No other literature?

user19161

You know what I hate? The higher rep user usually gets way more votes in his answer than a lower rep user, even though the answers are equally good...

anon

Googing "what is dp" won't get you a useful answer oh man qchu is stealthy

Argon

@WillHunting I upvoted, btw!

user19161

2:01 AM

And sometimes a middle rep user posts a comment to suck up to a higher rep user saying "+1 This is brilliant!" when the lower rep user posts a more brilliant answer.

amWhy

@WillHunting Exactly. That's what I point out in my meta post. Just because high rep = upvote

user19161

@amWhy Well, we just need to solve the millennium problems and put our real name there, then all mouths will be shut! =)))

Argon

Good e'en chaps, I am off anon to watch my favourite show on the tele!

amWhy

@WillHunting I noticed that too. I've been feeling snubbed lately!

@WillHunting Let's go for it!

@WillHunting I shouldn't say "snubbed" but "passed-over"? Oh well. All in good time...

user19161

@amWhy Are you going to change your username soon?

user19161

2:07 AM

@Argon Hehe, you are off anon to watch TV, but I am off TV to watch anon. =)

amWhy

@WillHunting Haven't been planning on it. Maybe. I'll think about Why I might want to.

user19161

@amWhy OK, I will change mine on Dec 04...

amWhy

@Argon Congrats, Argon! over 6K!

@WillHunting I'll keep a look out.

user19161

@amWhy Hehe!

user19161

@amWhy I will keep the steelblue for identification, but others could try to impersonate me...

amWhy

2:19 AM

@WillHunting Thanks for the tip! I'll watch for blue.

robjohn

@WillHunting well, I can tell you that I am not cleaning the toilets here.

user19161

@robjohn Hehe, but I am cleaning toilets on the MIT campus!

robjohn

@WillHunting my sympathies

John Smith

2:42 AM

is there a mathematial way to express ceil()

for instance can i rewrite this without ceil(): x=int(ceil(((v-4*N)**.5-3)/4))-1

user19161

Hmm, nothing that I know of.

Argon

@amWhy Thanks for the final upvote!

amWhy

@Argon Enjoy the +6K club!

Argon

@amWhy Maybe soon enough I will make it to the 8k+ club like you!

amWhy

@Argon In no time at all: just keep at it!

Argon

3:01 AM

Good night all!

user19161

3:17 AM

@Argon See you in your dreams.

user19161

@amWhy You will be in the 10k club by Dec 31 I suspect! Are you gonna sleep soon?

amWhy

@WillHunting probably going to sleep soon, yes. How 'bout you?

user19161

@amWhy Hmm, not so soon. OK good night!

amWhy

@WillHunting Not heading to bed quite yet, but I'm yawning :-O

user19161

@amWhy The O represents a big mouth!

amWhy

3:26 AM

@WillHunting I didn't know that! oops...

user19161

@amWhy Well, it just means you are yawning, so it fits! =)

amWhy

@WillHunting Hmmm. So if I change my username to amWhyNot, would that be a denial of my existence?

user19161

@amWhy Well, I prefer to see an ordinary name there, but it's up to you of course!

amWhy

@WillHunting That's true. But I've learned that being obviously female (using a girl's name) can be problem. But then again, I get tired of "Thank you, sir!"

user19161

@amWhy Oh, why is it a problem?

amWhy

3:33 AM

@WillHunting problem in the sense that assumptions are made, and some users (askers) are not as apt to take a woman seriously, in math.

user19161

@amWhy Hmm OK. Well, I am reminded of Emmy Noether now.

Mariano Suárez-Alvarez

well, it is not a bad memory to have

user19161

As usual, you pop in suddenly!

amWhy

@MarianoSuárez-Alvarez Yes indeed!

Mariano Suárez-Alvarez

A non-negliglible part of the problem with people not taking women seriously in math is that women that ought to be taken seriously more often than they might do actually not make their womanhood explicit :-)

user19161

3:38 AM

I am trying to parse your sentence!

amWhy

@MarianoSuárez-Alvarez Yes, very true!

Mariano Suárez-Alvarez

hahaha

user19161

Erm, still doesn't parse.

amWhy

@WillHunting I got it quickly, Will. You need to read more than math!

user19161

@amWhy Is it grammatical?

user19161

3:39 AM

I think some words are missing there...

Mariano Suárez-Alvarez

probably s/might/are/

or s/might/& be/

amWhy

@WillHunting the "do" between might and actually could/should be omitted.

user19161

@amWhy I have difficulty parsing your sentences too!

amWhy

@WillHunting I know, so you've told me! ;-)

user19161

@amWhy Yes, and perhaps I might share my secrets with you later this year...

amWhy

3:43 AM

@WillHunting I'll be waiting! (Ellipses again, I'll influence you more than you care!)

user19161

@amWhy OK, I have made some plans with regard to the mode of revelation, you will see...

amWhy

@WillHunting Now I'm holding my breath in suspense!

user19161

I have a confession to make: I like my rep to be a multiple of 5, so if it is not I will look for some poor answers to downvote!

amWhy

@WillHunting I thought it was just me! (...liking my rep to be 5n for some n!)

user19161

@amWhy But I see that yours is often not that!

user19161

3:46 AM

@amWhy Well if you are going to say for some n you might as well say for some integer n.

amWhy

@WillHunting right now there's an ugly trail of ...39. Hmmm, two downvotes in my pocket.

user19161

@amWhy Wrong! When you downvote a question victim -2 you 0. For an answer victim -2 you -1.

amWhy

@WillHunting Ohhh! I knew that there's no point "penalty" for down-voting questions, I forgot that down-voting an answer is only -1.

@WillHunting It's a toss-up: I like my daily rep to be a multiple of 5, as well.

@JayeshBadwaik Hello!

Jayesh Badwaik

@amWhy Hi!! Wassup?

@WillHunting Hello!

user19161

@amWhy I think I will take a shower now and go to bed, see you in your dreams! =)

user19161

3:53 AM

@JayeshBadwaik Hi and bye!

amWhy

@WillHunting night-night. Me too!

Jayesh Badwaik

@WillHunting Good afternoon. (Time zones messed up.) :P

@MarianoSuárez-Alvarez Its a cycle, people tell young girls that they are not to take math seriously, and hence, as they grow up, they do exactly that, and then people think girls don't take math seriously, and hence, people don't take the girls in math seriously, and then those who do take math seriously feel social pressure and hence, do not reveal their identities resulting in the perception that girls don't take math seriously.

Mariano Suárez-Alvarez

Well, I know a lot of people who do take women in math seriously

Jayesh Badwaik

Yes, I know too, but I know equally well of people who don't (which are sadly in majority), leave aside math, any other technical field. Its kind of stupid.

Mariano Suárez-Alvarez

it is the same thing with gay people in high visibility positions coming out

amWhy

4:01 AM

@MarianoSuárez-Alvarez Thank you, Mariano. It is clear you do, too.

Mariano Suárez-Alvarez

well, no t the same

one thing that bothers me is that I can never tell if an author of a paper, who is chinese, or generally asian, is a man or a woman from the name

but I cannot tell the first name from the last name in chinese authors, either!

amWhy

@MarianoSuárez-Alvarez Maybe I'll use my real name. My first name is sort of a give-away from my username, phonetically, it spells out my first name.

Mariano Suárez-Alvarez

(that starred comment by Khromonkey is very funny, and sort of relevant in an evil way)

Jayesh Badwaik

-@MarianoSuárez-Alvarez Hmm, with India, generally, if the name is pronounced in a way such that the ending is in "i" as in confetti or 'a' as in algebra, then it is most probably a girl.

Mariano Suárez-Alvarez

that's a useful tip!

amWhy

4:06 AM

@MarianoSuárez-Alvarez I actually thought @Qiaochu was a woman, when I first came to math.se

Jayesh Badwaik

If the endings are in consonants, it is most probably a guy.

@amWhy Hmm, I first emailed Qiaochu, and then came to math.se. :P

Mariano Suárez-Alvarez

that someone is a man or a woman is the sort of irrelevant detail that, despite that irrelevance, helps me turn online people into more real people

amWhy

Yes, I agree. In my case, just replace "Why" with "y": amy

Mariano Suárez-Alvarez

clever :-)

user19161

4:48 AM

@JayeshBadwaik You mean in Indian names?

Jayesh Badwaik

@WillHunting Yes.

user19161

@JayeshBadwaik The thing about Chinese names is that in Chinese names are written in last name first name format, and after transliteration into English they are turned into first name last name format.

Jayesh Badwaik

There are few exceptions, like words ending in "um" like Neelam, Sonam etc.

user19161

Also, there are two kinds of transliteration in use, hanyupinyin and Wade.

user19161

I find the Wade system rather silly. It's hanyupinyin for me all the way.

user19161

4:51 AM

And yes you can't really tell the sex of the person given only the Chinese name. In fact even if you know Chinese it's very hard to tell!

Jayesh Badwaik

@WillHunting Hmm, in southern parts of India, there is no last name. you have names like VVS Laxman, where Laxman is the first name and then VVS is an acronymt that has different meanings in different regions. Sometimes, it is names in lineage (father, grandpa, great grandpa etc). Sometime, it is the name of the village the family originally came from. Argh.

@WillHunting There are a few names in Indian languages which are unisexual, but I forget the examples.

user19161

Also, CHinese first names can be one character or two characters while last names are usually one character.

user19161

Also a CHinese name can be mixed with an English name and this is the practice in some countries.

Jayesh Badwaik

Hmm..

user19161

Where some countries means you know where. =)

Jayesh Badwaik

4:53 AM

@WillHunting I somewhere read that there is a traditional chinese and then there is a simplified chinese with about 5000 and 3000 characters respectively.

user19161

@JayeshBadwaik Yes I think Taiwan still uses the traditional writing system and speak the Hokkien dialect, while in Chine they use the simplified writing system and speak Mandarin. There are too many Chinese dialects and these are not mutually intelligible.

user19161

Whereas English is pretty much the same everywhere in the world =)

user19161

Just a few difference in spelling like color colour and other things.

Mariano Suárez-Alvarez

except in texas

Jayesh Badwaik

@WillHunting Hmm. I heard on a Russell Peters show that there are two main dialects, Cantonese and Mandarin, with Mandarin spoken near Beijing and Cantonese spoken near Hong Kong, or may be vice versa.

Mariano Suárez-Alvarez

4:56 AM

and most of england

and auustraliaa

the only people that speak understandable english are the swedes and finns, really!

user19161

Wow, Mariano seems to be an expert in Eng!

Mariano Suárez-Alvarez

«speaking the same language» can be a red herring :-)

user19161

Honestly, I can't understand New Zealand accent.

Mariano Suárez-Alvarez

I've seen films made in Spain which I wish had been subtitled!

I don't think NZ speaks English, really :-)

Jayesh Badwaik

@MarianoSuárez-Alvarez Hahaha.

Mariano Suárez-Alvarez

4:59 AM

just like only USians think Canadians speak french

Jayesh Badwaik

Hindi is a strange language, "Abhiyanta" means an Engineer and "Abhiyantriki" means engineering. However, by those rules "Abhiyan" should mean an engine or something, but it means a campaign, drive, expedition etc etc. Drive is the closest word, but still. :P

And I think, people now use "Ingjan" as a slightly modified word of "Engine" and no one use "Abhiyan" to denote an engine. :P

Mariano Suárez-Alvarez

almost no one can complain aabout hindi, because it begat pretty much all of the others :-)

Jayesh Badwaik

@MarianoSuárez-Alvarez yeah, that's true. :D

user19161

5:49 AM

@JayeshBadwaik Yes, Beijing Mandarin, Hong Kong Cantonese, and Taiwan Hokkien. QED.

Jayesh Badwaik

8:44 AM

Interesting

a Dangerous Idea

9:16 AM

@JayeshBadwaik "What matters is that the person is fully engaged in an activity that is optimally challenging and feels in control of the situation."

robjohn

@aDangerousIdea who told you to say that?

:-D

a Dangerous Idea

@robjohn I think that is a contradiction in terms to be engaged in an activity that is optimally challenging and feels in control of the situation."

robjohn

For some, it is a challenge to be in control.

a Dangerous Idea

@robjohn Does that make it "optimally challenging"?

Jayesh Badwaik

@aDangerousIdea Control of the situation refers to the control over the activity he has taken up. In the sense that the activity is not only challenging, but that he actually wants to do that activity.

robjohn

9:24 AM

@aDangerousIdea I don't exactly know what optimally challenging means. Not too hot and not too cold?

For Goldilocks, choosing the right bed was optimally challenging...

3

a Dangerous Idea

lol

Jayesh Badwaik

@aDangerousIdea for example, a reason I have seen for a person prefer not to take up a 600,000 INR per year job instead of 1,200,000 INR per year job is that people will think he was not good enough to get the 1,200,000 INR per year job.

this is when the student does not even have any intention of doing a job right now, and instead wants to go for post grad.

a Dangerous Idea

@robjohn Do you think Goldilocks felt in control of the situation after choosing the right bed and the bears came home? ;-)

Chris's sister

Hi

a Dangerous Idea

hi

Chris's sister

9:35 AM

Any idea for computing $\int_0^{\pi/2} \sin^2 x \ln (\sin x)\ dx$?

robjohn

@aDangerousIdea In her situation, she was alone, so she did have to run faster than the bears.

@Chris'ssister I would try the substitution $u=\sin(x)$

Chris's sister

@robjohn: I started from the form you suggest to return to. :) I see no way there.

Jayesh Badwaik

@Chris'ssister What was the original form?

Chris's sister

@JayeshBadwaik: $\int_0^1 \frac{x^2 \ln x}{\sqrt{1-x^2}} \ dx$. The initial form gets me nowhere. Then I've decided to use the variable change.

a Dangerous Idea

9:55 AM

hi @OldJohn

Old John

Hi there

a Dangerous Idea

@JayeshBadwaik What I'm saying is that you can not be "optimally challenged" by an activity and still feel that you are completely in control of the situation. It's all about pushing the limits of your ability, no?

robjohn

I am working on the full integral, but do you know that
$$
\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=-\frac\pi2\log(2)
$$

I am using this in the full integral

Chris's sister

@robjohn: yeah, I know this result. I tried to use it but I failed.

robjohn

10:55 AM

$$
\begin{align}
\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x
&=\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\\
&=\frac12\int_0^{\pi/2}\log\left(\frac12\sin(2x)\right)\,\mathrm{d}x\\
&=\frac12\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x-\frac\pi4\log(2)\\
&=-\frac\pi2\log(2)
\end{align}
$$
$$
\begin{align}
\int_0^{\pi/2}\cos(2x)\log(\sin(x))\,\mathrm{d}x
&=\frac12\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}\sin(2x)\\
&=-\frac12\int_0^{\pi/2}\sin(2x)\frac{\cos(x)}{\sin(x)}\,\mathrm{d}x\\
&=-\int_0^{\pi/2}\cos^2(x)\,\mathrm{d}x\\

Chris's sister

@robjohn:: ooo, nice! Thank you!

Hello, everyone!

hi

@sos440 Howdy

@robjohn Not so good as usual, I'm still struggling with my grad. application. :(

Chris's sister

11:01 AM

@sos440: hi

robjohn

@sos440 I remember that time. It sort of puts a damper on the holidays

sos440

@Chris'ssister Hi!

@robjohn Yeah, I was half-mad then.

@Chris'ssister Every time I see you here, you have a nice problem :)

Chris's sister

:-)

@sos440: did you see my last problem? @robjohn just provided with a beautiful answer you may find above.

sos440

@Chris'ssister Yes, I saw that right now and it was very enjoyable!

a Dangerous Idea

She is unique.

Chris's sister

11:05 AM

Hmmm, I also think at a solution that doesn't use trigonometric functions (I'm working right now on it). If I get such a solution then I'll post it here.

brb in 15 minutes

sos440

@Chris'ssister How can you evaluate something involving trigonometric functions not using trigonometric functions?

3

Oops

robjohn

@Chris'ssister with $\pi$ in the answer, I bet a trig sub will enter into the mix somewhere.

sos440

I know some solutions using mean value property or Poisson kernel-like thing, but I can't say they do not use trigonometry...

a Dangerous Idea

11:21 AM

robjohn

Trig or Treat!

I don't know if that translates well...

a Dangerous Idea

Old John

Just been reading Disquisitiones Arithmeticae - I reckon Gauss pretty much invented group characters back in 1801 ... before groups had been properly defined :)

robjohn

@aDangerousIdea but those are supplementary angles...

@anon welcome back

user19161

I tried to find a way to distinguish between complementary and supplementary angles.

user19161

11:30 AM

The best I have come up with is that c goes before s like 90 goes before 180.

anon

@OldJohn an English translation? Is it funky because the notation and machinery is pre-modern?

Old John

@anon Yeah - I bought the Springer translation into English years ago - and finally got round to reading the chapter about roots of unity and the 17-gon

anon

A complementary angle measures the angle of the reflected ray.

Old John

at the time, it was the most I had ever paid for a book :)

anon

Oh boy, sounds calculation-painstaky.

user19161

11:33 AM

@anon What has complementary got to do with reflected?

user19161

Oh I see, that was not exactly a response to my line.

anon

The ray coming into a plane and the reflected ray combine to form a unified trajectory in a "billiard's" sense, so here one part complements the other in the sense that they combine to make one whole.

user19161

Ah, billiard! We should have a billiard date one day. =)

anon

I don't play pool though.

where is row 52 in the end table? wtf

it skips many $n$ in fact, hrm

Chris's sister

Back.

user19161

11:42 AM

Front.

anon

OEIS only goes to 47 >:

user19161

I only learnt about OEIS on MSE.

sos440

Middle

Chris's sister

@sos440: this is a nice example where you may compute the integral without using trigonometric functions, but only some elementary stuff like $\int_0^1 sqrt{x-x^2} \ dx$. I'm trying to find such a case in my last problem.

a Dangerous Idea

43 mins ago, by a Dangerous Idea

She is unique.

Chris's sister

11:48 AM

@aDangerousIdea: who is unique? :-)

a Dangerous Idea

@Chris'ssister You are.

Chris's sister

@aDangerousIdea: :D

a Dangerous Idea

@Chris'ssister :D

Chris's sister

@sos440: you let x = t+1/2 and you're done by geometric interpretation.

sos440

@Chris'ssister If you decided to do so, I think the first thing you have to do is to transform the integral to another one which does not involve trigonometric functions...

user19161

11:50 AM

Some people say that unique is more than special, but I say that special is more than unique, because we are all unique, but perhaps not all of us are special to someone.

Chris's sister

@sos440: the initial integral was $\int_0^1 \frac{x^2 \ln x}{\sqrt{1-x^2}} \ dx$

sos440

@Chris'ssister Ah, I understood. I didn't know that it was your initial problem.

Chris's sister

@sos440: I plan to resort to the integration by parts as follows $\int_0^1 (-\sqrt{1-x^2})' x^2 \ln x}{\sqrt{1-x^2}} \ dx$

sos440

I love the word idiosyncracy. It reminds of me a nerd-like thing.

@Chris'ssister Do you mean $$\int_{0}^{1}\frac{x^2 \log x}{\sqrt{1-x^2}} \, dx = \underbrace{\left[ -x \sqrt{1-x^2} \log x \right]_{0}^{1}}_{0} + \int_{0}^{1} \sqrt{1-x^2}\left( 1 + \log x \right) \, dx$$?

Chris's sister

If I go this way and combine it with the trigonometric way (which I don't want) then I finish very fast the whole story.

@sos440: exactly!

sos440

11:57 AM

Oh, now I see some solutions using beta function identity, but unfortunately it does not fit in elementary range.

Chris's sister

@sos440: you get the same integral also in the right side but with the minus in front. Then you may move it to the left and you're done.

@sos440: because $\int_0^1 \frac{\ln x}{\sqrt{1-x^2}} \ dx$ this is already computed.

(if we let $x=sin u$)

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$Mathematics$ Mathematics