Mathematics - 2012-12-03 (page 2 of 5)

3:17 AM

"Congratulations! You just installed the world’s #1 BitTorrent client like a boss."

Then you can factor $a$, and try the possibilities of splitting 16a^4 into product of two numbers - but I'm not sure if that's what you want.

@Sanchez I mean "try a=1, b=2, c=3, doesn't work? try 1,2,4, then 1,2,5, etc"

i.e. the long naive way

let's just assume a is unbounded

@WillHunting, it seems very elegant. I don't fully comprehend it.

but a<b<c<2a

if that helps

@KaliMa, OK, so what I outlined above is a slightly less naive way.

3:19 AM

Lewasgrtrsgl.

user19161

@Limitless Well, I am not sure if it is totally correct, I checked through at least once though.

i also found empirically that b^2+c^2 = 5m^2

I'm headed to bed. I have school tomorrow, and it will be so mundane I will want to headdesk. Nonetheless, sleep helps that.

@PeterTamaroff, I've updated my about me and such if you're curious. You are the person who did inspire me to get a username.

The slightly more general problem of solving $b^2+c^2 = 5m^2$ is actually easier

@WillHunting, have fun mathing. I will be jealous in my dreams; although, I might just mental math.

3:20 AM

well what i mean is that we're trying to solve 4a^2(b^2 + c^2) = 5b^2c^2 and i know that the (b^2 + c^2) portion is a multiple of 5

does this help in any way?

don't think it helps much.

user19161

@Limitless No need to think about me in your dreams, just think about your girlfriend. =)

But two things: 1. What's the motivation for the problem?
2. Would solving $b^2+c^2 = 5m^2$ be happy enough for you?

user19161

Hmm I will be back later.

possibly

would prolly help

3:24 AM

do you know how to solve $b^2+c^2 = m^2$ in integers?

pyth. triples, there's a n & m paramterization thing for it

yes, but how do you get that?

?

How do you derive the parametrization

i mean i have code that generates triples

3:25 AM

Oh.

i dont know how

ok

parameterizing is a weak spot for me

So one way to think about it, is to move m to the other side, so that we are solving for $(b/m)^2 + (c/m)^2 = 1$, i.e. solving $x^2+y^2 = 1$ in rational numbers

does it make sense?

are we talking a^2 + b^2 = m^2 here? without the 5?

3:27 AM

yes

ok

So here's one way to find all rational solutions of $x^2+y^2 = 1$.

First, (1,0) is one such solution.

Second, If (x,y) is another rational solution, we draw a line from $(x,y)$ to $(1,0)$. You can check that this line has rational slope, and rational y-intercept.

i.e. The line would intercept $y$-axis at some point $(0,y)$, with $y$ being rational number.

Conversely, if you start with a point $(0,y)$ where $y$ is rational, we can form the line $(0,y)$ to $(1,0)$, and it will cut the circle at a rational point.

Does it make senes?

yeah

So let's say we start with the point $(0,t)$, $t$ being rational

Compute the line from $(0,t)$ to $(1,0)$

The slope is -t, so the line is y = -tx + t

We want to find where this line interescts the circle

$(1,0)$ is one such point

What is the other one?

dunno; i think i will need to try a different approach

3:38 AM

Alright then

If you understand this method, you can do exactly the same thing to $b^2+c^2 = 5m^2$.

Then to your original equation 4a^2b^2 + 4a^2c^2 = 5b^2c^2

let x = 2ab, y = 2ac, z = bc

You will ways to generate x,y,z, and (I guess you are trying to write an algorithm that generates solution), just try to check that the corresponding a,b,c would be integers

then that become x^2 + y^2 = 5z^2

Yes, and there is a way to generate solutions to this equation, similar to pythagorean triples

so once you generate solutions for this equation, try to solve for a,b,c, and it suffices to check if they are integers

this is again a pretty naive way for writing down solutions for ur original equation, but maybe it helps for your purpose

i am just trying to understand how to even generate the triples for that

wiki shows it for the general case

What's your background?

some college math

nothing insane

3:47 AM

I see

Try to read math.uconn.edu/~kconrad/blurbs/ugradnumthy/pythagtriple.pdf

Section 3 - it's a detailed version of what I tried to tell you earlier

If you understand how you can geometrically get the formula for pythagorean triple, you can duplicate that to handle the x^2+y^2 = 5z^2 case.

i think this is maybe too advanced for me

i'll have to try something else i think

Oh hmm, sorry.

i can't derive my own equations that easily

I see.

In that case may be you can try to post on main, and see if others can help you. There are many great expositors around that I'm sure can help you.

thanks

user19161

4:13 AM

Hey @amwhy yesterday you saw me in your dreams, today I will see you in mine then. =)

user19161

Hmm, just answered another low hanging fruit...

@Wil

user19161

@Sanchez Yes?

@WillHunting, sometimes I'm curious, what is your motivation to answer so many questions?

made a question on Main

4:16 AM

@KaliMa, cool, I'm sure that you will get a great answer soon.

user19161

@Sanchez Well, first of all, I don't answer that many questions. Second, I have the time now. Third, it's good to contribute to the site and help others. Fourth, it's good to do something productive online and keep a nice record of your online activities.

@WillHunting, lol I see.

user19161

@Sanchez You love to LOL don't you? LOL.

@WillHunting, sort of LOL.

4:33 AM

someone replied with the circle thing too

i dont understand

how on earth am i supposed to know what slope to choose? there are infinitely many t

and how do i know where else on the circle it hits

and how to translate that into a parameterization for general a^2+b^2=5c^2

user19161

4:52 AM

I am going to bed, over and out!

anon

5:28 AM

Main Chatroom Etiquette Rules | $\LaTeX$ support for chat

5

user19161

7:19 AM

Hey @rob you're always in the chat room.

@WillHunting the magic of the 24/7 online computer.

user19161

@JayeshBadwaik Hmm, I woke up from my sleep already, slept only two hours...

@WillHunting In NUS's admission thingie, they said they accept GATE score card, is this GATE the one conducted by IITs?

user19161

@JayeshBadwaik I never heard of GATE.

@WillHunting Here

user19161

7:23 AM

@JayeshBadwaik I rather not click on that for some weird reasons.

@WillHunting okay, no problem.

user19161

@JayeshBadwaik I think you should check with the department if in doubt.

@WillHunting yes, was going to do that.

user19161

@JayeshBadwaik Also, it is up to you. But I really see no reason for you to do MSc here.

@WillHunting Thanks. I am right now, just exploring every place I can get into.

user19161

7:25 AM

@JayeshBadwaik The grad course offerings are pretty limited actually.

@WillHunting As in? Courses you mean?

user19161

@JayeshBadwaik I don't know about your local situation, but maybe you can just try to do something there and then apply to other places for grad school.

user19161

@JayeshBadwaik Well, courses are courses, modules whatever you call them.

@WillHunting then, limited as in?

@WillHunting Yup, I am trying local too. Lets see.

user19161

@JayeshBadwaik In terms of the breadth and depth.

7:27 AM

@WillHunting okay.

user19161

@JayeshBadwaik Now I have taken a look at some Indian syllabi some time ago and they look pretty good.

user19161

@JayeshBadwaik I am not trying to be biased against things here, but I am really sick and tired of the things here. People and things here are extremely superficial...

@WillHunting Its okay, I understand and appreciate your opinion. I will keep it in mind.

user19161

I no longer listen to news because every time I hear it (local and global) I hear so many stupid things that upset me.

@WillHunting Yes, some are pretty good, but most institutes are very isolated you know.

Some good institutes in mathematics, do not have a physics dept.

user19161

7:31 AM

Like I said, often I feel I don't belong to this world anymore...

Some are algebra specific, and not much strong in analysis.

others are strong in analysis, not algebra

and by not strong, I mean pretty weak

user19161

Well, the big three are always algebra, analysis and geometry/topology.

Engineering and mathematics are almost totally separated in institutes. (which is a big negative for me)

user19161

And the smaller three are logic/set theory, combinatorics and number theory.

user19161

Essentially most of mathematics can be classified into the big three, or the big three and small three, depending on how you wanna do it.

7:35 AM

@WillHunting yes. I agree.

user19161

@JayeshBadwaik So tomorrow I change my username back.

@WillHunting Good.

8:07 AM

@WillHunting I was gone for almost all day yesterday.

csss

Guys can someone tell me if I have done this right...I have an exam in an hour! math.stackexchange.com/questions/249825/…

a Dangerous Idea

8:41 AM

@robjohn Did you see the documentary I posted?

►

8:59 AM

@aDangerousIdea I will have to look at it in a bit. It is over 40 minutes long.

user19161

@robjohn Did I tell you I capped a few hours ago? =)

@WillHunting only 9 hours into the "capping day"! pretty good.

@WillHunting That's a lot of answers

user19161

@robjohn Well, I am just lucky to have a few fans these days. =)

@WillHunting OMG, you are capping almost everyday now

user19161

9:14 AM

@JayeshBadwaik Not really, that would be Amy, not me. =)

@WillHunting woot? she is capping too? hmm.

user19161

@JayeshBadwaik Well, she answers a lot of questions. I only answer a few low hanging fruit.

@WillHunting You two are in top 5of this months rep!!! o.O

user19161

@JayeshBadwaik Well, for me just luck, for her it is called ability.

@WillHunting Don't sell yourself short mister.

user19161

9:16 AM

@JayeshBadwaik Thanks madam!

@WillHunting You're welcome.

user19161

Anyway, I think ELU is getting quite weird nowadays. Too many questions are getting downvoted and closevoted too quickly.

a Dangerous Idea

Are bananas considered a low hanging fruit?

user19161

@aDangerousIdea Hmm, not sure. My knowledge of such things is almost zero.

user19161

@anon Thanks for pinning this, very industrious.

user19161

9:21 AM

@JayeshBadwaik Well, hmm, I think I get the point now.

@WillHunting nice.

user19161

@JayeshBadwaik Why delete? Crazy.

@WillHunting Hmm, may be.

user19161

One motivation to keep answering after capping is to try to get accepts.

Heh, I figured out something cool

user19161

9:24 AM

@N3buchadnezzar That I am hot?

@WillHunting That is an oxymoron

$$\large f(x) = \sqrt[\large \alpha]{x-\beta}$$

user19161

@N3buchadnezzar I only know that I am a moron and I use the pimple cream Oxy.

@N3buchadnezzar that is tangent to the parabola. Why is this interesting? (genuinely asking)

The image shows $f$ in blue, and the tangent to $f$ at $x=a$. (Here $\alpha\geq 1$) and $a>\beta$

okay

ahh, did not notice $\alpha$

my bad

9:27 AM

The ratio between the red and blue area is independendant of both $\beta$ and $a$.

hmm

@JayeshBadwaik It is actually $$ \frac{\text{red}}{\text{blue}} = \frac{1}{2}(\alpha - 1)$$ or the reciprocal. Somewhat cool ?

Yup

Nicely cool.

Seems I made some kind of calculating mistake, anyway the ratio is only defendant on $\alpha$. And the areas are always alike given $\alpha=3$. Although it does not make that much sense intuitively yet.

hmm

it is kind of intuitive

9:33 AM

Linearity?

the farther the tangent, less the slope

longer the intercept

so directly proportional you can infer from the diagram

and there must be a sweet spot by IVP

or something

^^

user19161

It is raining heavily here.

its shining brightly here

It is -20 here

user19161

9:36 AM

Perhaps the rain is a sign that miracles are about to happen.

@N3buchadnezzar cool.

-20 C?

user19161

-20 Kelvins.

@WillHunting Rather bemused hahaha.

Nah, I was wrong. It was only -10 here now. Was -20C yesterday though, bloody cold.

user19161

@JayeshBadwaik -20 Kelvins is way too cool.

9:37 AM

@N3buchadnezzar -20C is freezing. Snow?

user19161

Of course if I were there, the temperature would rise because I am too hot.

@WillHunting -20 K is impossibly cool.

no one can reach that level of cool factor in his community.

@JayeshBadwaik No snow, thats the weird part.

@WillHunting If you are hotter than me, does that mean I am cooler than you?

@N3buchadnezzar Hmm. Weird.

@N3buchadnezzar Yes it does!!

Heh, tonight there was one place in Norway with -32.2

9:41 AM

But Will Hunting said he is too hot, and @N3buchadnezzar you are too cold.

@N3buchadnezzar Hmm. Kind of cool. At what temp do you expect to get frost bites? Or is it about the snow?

yr.no/place/Norway/S%C3%B8r-Tr%C3%B8ndelag/Trondheim/Trondheim

@JayeshBadwaik Well we just stopped using shorts and short-sleaved t-shirts.

I guess it is soon time to stop grilling too.

@N3buchadnezzar grilling?

bbq?

jacobsen.no/anders/blog/archives/2002/12/03/…

@JayeshBadwaik Indeed

@N3buchadnezzar hmm. Funny jokes! final barbecue!!

but in the jokes, the final barbecue comes before long sleeves

The first 4 are actually correct though ^^

9:46 AM

hahaha -273 is cool

and the cow one is also great

funnynorwayfacts.blogspot.no

@N3buchadnezzar Is the thing about liquor true?

@JayeshBadwaik Everything is true

Normal stores are only allowed to keep alcohol up to 4.7 percent.

@N3buchadnezzar Hmm.

@JayeshBadwaik We dont have a lot of alcohol problems in norway so to say:p

It is a bit irritating for us students though, having to travel quite a while to obtain liqour.

9:56 AM

@N3buchadnezzar Hmm. And whale hunting?

Kill those bloody whales..

I love whale meat

Of course it is not black and white, we do have hipsters that supports the whales too. Although most old people just want to watch the wales burn.

Hmm.

Somewhat disturbing I find it.

why?

Well, for one, whales are an endangered species.

satwcomic.com/…

10:00 AM

In India, there was a sport of tiger hunting for example.

We do have restrictions oon the number we kill though

@JayeshBadwaik Did learn something new about Norway now? ^^

Brown cheese!

@N3buchadnezzar Yes. Brown cheese. Hmm.

It is very popular, but I do not think it is the most popular one.

@N3buchadnezzar Bear hunting is illegal, but whale is legal? Woot?

Yeah, we have like under a 100 bears left in norway :p

10:05 AM

Norwegians are going to hate me now! :P

Situation is not better with whales either.

They are anyway dying because of the big ships and their low-frequency engine noise.

Whale meat tastes better than bear meat though

I bet.

And the size/population ratio is better for whales than bears.

@N3buchadnezzar It's not so surprising... write the equation as $f(x)=(x-\beta)^\gamma$ where $0<\gamma<1$. Then $f'(x)=\gamma\frac{f(x)}{x-\beta}$ so the point where the tangent intersects the $x$-axis is $x-\frac{f(x)}{f'(x)}=x-\frac{x-\beta}{\gamma}$ so the area of the triangle with vertices $\left(x-\frac{x-\beta}{\gamma},0\right)$ and $(x,0)$ and $\left(x,(x-\beta)^\gamma\right)$ is $\frac12\frac{x-\beta}{\gamma}(x-\beta)^\gamma=\frac1{2\gamma}(x-\beta)^{\gamma+1}$.

@N3buchadnezzar That's no excuse. In the world, bears and wovles are better off than whales.

10:07 AM

@N3buchadnezzar The area under the curve is $\frac1{\gamma+1}(x-\beta)^{\gamma+1}$ and the area in red is the difference: $\left(\frac1{2\gamma}-\frac1{\gamma+1}\right)(x-\beta)^{\gamma+1}$. The ratio of red/blue is $\frac12\left(\frac1\gamma-1\right)=\frac12(\alpha-1)$.

@JayeshBadwaik Nuuhuuuu

@N3buchadnezzar Yes sir.

@robjohn Indeed! But what happens to the ratio when $\alpha\in(0,1)$? =)

Hunting wolves is illegal in norway too, I think.

Hmm.

@N3buchadnezzar it's still a constant ratio, but the red area overlaps the blue area and is negative

10:18 AM

@robjohn Yes, somewhat cool =)

10:33 AM

@robjohn Most honorable.

@PeterTamaroff ?

@robjohn Have I baffled you?

@PeterTamaroff since I have no idea what you're talking about, yes.

@robjohn Man. Is it too early in the morning?

user19161

@PeterTamaroff Man.

10:38 AM

@PeterTamaroff too early in the morning for what?

user19161

@old I capped a few hours ago!

beer

@robjohn For jokes

@robjohn I was talking the other day with Mariano about Lebesgue

He told me the space of Lebesgue Integrable functions is quite nice

Whereas the space of Riemann integrable functions is quite crappy

user19161

@PeterTamaroff After you study Lebesgue integration you can read the paper I sent you for an alternative approach!

@PeterTamaroff Riemann integrable functions are those whose set of discontinuities has measure 0

10:43 AM

@robjohn Yes, OK.

@robjohn And when is a function Lebesgue integrable?

@PeterTamaroff I believe it only needs to be a measurable function

@robjohn Meaning?

@PeterTamaroff In both cases we are considering bounded functions on a bounded interval

@robjohn I still have to do the other day's work, and prove that continuous functions are ruled.

But I'd like to see if I can see what other functions are ruled.

of course knowing how to show that continuous functions are, may help finding what other functions are.

10:48 AM

@robjohn Right.

Wikipedia says we can define the Lesbesgue integral as an improper Riemann integral.

^^

Old John

@WillHunting I noticed - well done!

@PeterTamaroff That isn't right, because the indicator function of the rationals set is Lebesgue integrable, but not Riemann integrable

Old John

Hmm - I think that to make more progress with algebraic number theory, I ought to go over all the algebra that I used to know (and have at least partially forgotten) ... but where to start, withut going right back to the basics? :(

@robjohn See "toward a formal definition"

11:08 AM

@robjohn Is $$ f_n(x) = \left\{ \begin{array }{lll} 0 & \text{if} 0\leq x \leq \frac{n}{2(n+1)} \\ (n+1)x & \text{if} & \frac{1}{2(n+1)}< x \leq 1 - \frac{1}{2(n+1)} \\ 1 & \text{if} 1 - \frac{n}{n(n+1)} < x \leq 1 \end{array} \right. $$

$${f_n}(x) = \left\{ \matrix{
0{\text{ if }}0 \le x \le {n \over {2(n + 1)}} \cr
(n + 1)x{\text{ if }}{1 \over {2(n + 1)}} < x \le 1 - {1 \over {2(n + 1)}} \cr
1{\text{ if }}1 - {n \over {n(n + 1)}} < x \le 1. \cr} \right.$$

@N3buchadnezzar

Hmm, I can not delete my message

$$ f_n(x) = \left\{ \begin{array}{clrllrr} 0 & \text{if} & 0 \quad & \leq & x & \leq & \cfrac{1}{2}\cfrac{n}{n+1} \\ (n+1)x & \text{if} & \cfrac{1}{2}\cfrac{1}{n+1} & < & x & \leq & 1 - \cfrac{1}{2}\cfrac{1}{n+1} \\ 1 & \text{if} & 1 - \cfrac{1}{2}\cfrac{n}{n+1} & < & x & \leq & 1 \quad \end{array} \right. $$

@robjohn Is this function Lebesgue integrable on [0,1] ?

@N3buchadnezzar For each $n$?

@PeterTamaroff Yeah, it is a fairly cool function. It proves that the space of continous functions on [0,1] is not complete.

@N3buchadnezzar It is like a line joining two points which get's steeper and steeper?

11:18 AM

Indeed, draw it and see =)

@N3buchadnezzar But one can use $x^n$ for that!

Which is Lebesgue integrable .. ;)

@N3buchadnezzar Aha... but not Riemann integrable?

No, I think x^n is riemann integrable on (0,1)

@N3buchadnezzar I meant your function

11:20 AM

@PeterTamaroff Oh, I was hoping this function was Riemann integrable, but not Lebesgue.. But more likely it is the other way around.

@N3buchadnezzar yes

The function is a bit wrong, the midle part should be $(n+1)x - n/2$

Plotting in maple is weird

11:55 AM

@robjohn OK, I wrote it out for continuous functions

Gustavo Bandeira

12:08 PM

$Pi^{OPPA-GANGNAM-STYLE}$

@PeterTamaroff Nabucodonozor showed impressive formating skills on LaTeX...

@PeterTamaroff BTW