Mathematics - 2014-08-21 (page 3 of 4)

Nick

9:00 PM

I'm having a bit of trouble solving for $x$

:) Math blokes care to help?

Khallil

Did you manage to find that integral, @Nick?

1 hour ago, by Khallil

To make it slightly simpler, @Nick. $$ \begin{aligned} \int_{1}^{\infty} \dfrac{1}{\lfloor x \rfloor} \text{ d}x \ \ & = \lim_{n \to \infty} \left( \int_{1}^{n} \dfrac{1}{\lfloor x \rfloor} \text{ d}x \right) \\ & = \lim_{n \to \infty} \left( \int_{1}^{2} \dfrac{1}{\lfloor x \rfloor} \text{ d}x + \int_{2}^{3} \dfrac{1}{\lfloor x \rfloor} \text{ d}x + \dots + \int_{n-1}^{n} \dfrac{1}{\lfloor x \rfloor} \text{ d}x \right) \end{aligned} $$ Now follow @robjohn's hint, I think.

2 hours ago, by robjohn

@Nick For $n\in\mathbb{N}$, do you see what $$\int_n^{n+1}\frac1{\lfloor x\rfloor}\,\mathrm{d}x$$ is? What is $\lfloor x\rfloor$ on $[n,n+1]$?

Balarka Sen

@Nick wat

Khallil

Also, yea. What's the question, @Nick?

Nick

@Khallil: ... Yes. But I didn't. Actually, I gave up. I need to study more.

Question..

Balarka Sen

@Khallil $\lfloor x \rfloor = n$ for $x \in [n, n+1)$

Nick

9:02 PM

$$\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$$

Balarka Sen

@Nick groan

i have to recall what those cots are again

1/cos, ain't they?

Khallil

I'd square the whole expression and multiply through by $\cot x - a^2$.

$\cot(x) = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)}$

Balarka Sen

ah

there is really just a bit of manipulation there.

Nick

@BalarkaSen: I already said what you said earlier

2 hours ago, by Nick

@robjohn: $$\lfloor x \rfloor = n \quad,\forall\space x \in [n, n+1)\\
\lfloor x \rfloor = n + 1\quad if x = n + 1$$

@Khalil, @BalarkaSen: I forgot to mention. $a^2 \neq 0$

Khallil

Try and plot $f(x) = \lfloor x \rfloor$. Then try and plot $g(x) = \dfrac{1}{\lfloor x \rfloor}$. It becomes really simple after you know what the graph looks like.

Balarka Sen

9:04 PM

@Khallil Oh, and the article about RH is almost complete. At least I have pushed it to the statement of the conjecture

@Khallil just triangles. darn triangles.

Khallil

Triangles?

Who said anything about triangles?

Balarka Sen

ah, no i thought it was $\{x\}$

it's a bunch of rectangles then.

Khallil

Yea, exactly.

(I've never seen the graph of $\{ x\}$.)

Balarka Sen

it should be easy enough to convert it into a sum of their areas.

Khallil

Mhm.

Balarka Sen

9:07 PM

$y = \{x\}$

Khallil

Ah, you spoiled it for me. I was just working it out for myself.

-_-

Each of those vertical drops are at the integers which have no fractional part.

Balarka Sen

right

it's a periodic function

with period being $1$.

Khallil

It's basically a bunch of repeated $y=x+c$ curves that are cut off between the integers.

How does the graph look for $x<0$?

I'd think that it'd have some sort of rotational symmetry about the origin.

Balarka Sen

same

it's the same

$\{-x\} = \{x\}$

so there is a reflectional symmetry across $x = 0$

Khallil

Oh.

Ah.

So they've just defined it to be positive for all $x$?

Balarka Sen

9:11 PM

what is $1.12$ and $-1.12$ with stuffs before the decimal chopped off?

oops.

i was thinking of a different function.

Khallil

$\{ x \} = x - \lfloor x \rfloor$

Balarka Sen

$\{-x\} = -x - \lfloor -x \rfloor $

Jinx

Nick

@Khallil: Yup, I finally got the integral. The answer is $\infty$

Khallil

Nick

Wow, people still remember Jinx

Balarka Sen

9:13 PM

Jynx, that is.

Nick

.... I seriously got to get back on the old stuff. (refers to pokemon episodes like amphetamines)

You guys by chance didn't get the $x$ I was struggling with did you?

Balarka Sen

nearly finished. mathhelpboards.com/math-notes-49/….

Khallil

$\{-1.12\} = -1.12 - \lfloor -1.12 \rfloor = -1.12 - (-2) = 0.88$?

Balarka Sen

@Nick i didn't try it.

@Khallil right, yeah

Khallil

Oh, so it's exactly the same graph for all $x$.

That's pretty neat.

Yea, the integral diverges, @Nick!

Balarka Sen

9:17 PM

yeah. not the reflection symmetry though. i was wrong on that.

Khallil

Yea, you were probably thinking of some cosine stuff.

Maybe $\{ |x| \}$?

Nick

@BalarkaSen: Even if you do, you can't explain to me why $ x = -\frac{\pi}{2}$ is a valid answer to it.

Balarka Sen

@Khallil maybe

Khallil

I'll give that question a shot, @Nick.
(The one with the cosines and cotangents.)

Balarka Sen

it's always less confusing to think of frac parts as $x \bmod 1$

Khallil

9:18 PM

Or more confusing if you've never done modular arithmetic!

Nick

@Khallil: In my defense, what I finally ended up with is $\gamma \cdot \lim_{x \to \infty} \ln(x)$

Khallil

What's the $\gamma$?

Balarka Sen

E-M constant

Nick

Euler–Mascheroni constant

The Euler–Mascheroni constant (also called Euler's constant) is a mathematical constant recurring in analysis and number theory, usually denoted by the lowercase Greek letter gamma (). It is defined as the limiting difference between the harmonic series and the natural logarithm: Here, represents the floor function. The numerical value of the Euler–Mascheroni constant, to 50 decimal places, is 0.57721566490153286060651209008240243104215933593992…. == History == The constant first appeared in a 1734 paper by the Swiss mathematician Leonhard Euler, titled De Progressionibus harmonicis observationes...

Khallil

Oh, is that for the integral or the trig stuff?

Nick

9:20 PM

Facepalm

.. that was for the integral

...Now, let me resurface my finding $x$ question:

19 mins ago, by Nick

$$\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$$

NOTE: $$a \neq 0 \text{ and } a in \mathbb R$$

Khallil

Ok.

I'll give it a go.

Nick

:D Have fun

Khallil

I've managed to reduce it down a fair bit, @Nick.

Nick

@Khalil: ... I'm guessing you got to $a^2 = - \tan x$?

Khallil

Yea.

Now to solve that.

Nick

9:27 PM

Lol

Khallil

Urm ...

Nick

stuck?

Khallil

Yep. I don't know what $a$ is, but I know that $\tan x$ lies in the second and fourth quadrants of a circle with radius $\sqrt{1+a^4}$.

Nick

Yup, I got there too.

Khallil

I guess you could just say that the general solution is $x = -\arctan(a^2) + n\pi$, where $n \in \mathbb{N}$.

Such a wishy washy question.

Nick

9:31 PM

Sure, but which quadrant would you say $x$ lies in?

Khallil

I actually meant to write that '$x$ lies in the second and fourth quadrants of a ...'.

Nick

Sure, but can you decide which one?

Khallil

Well, that's the general solution. With restrictions on $x$, I'd be able to tell you which, if any are solutions to that original equation.

Nick

Nope, the only info I have is what I've already given you

Khallil

Then I can't.

At least I don't think I can.

Nick

9:34 PM

Here's what the Wolf has to say

Khallil

Oh wait.

I forgot to tell you my other solutions.

I actually whittled it down to $\cos x \cdot (\sin x + a^2 \cos x) = 0$.

So my general solution for $x$ was $\frac{\pi}{2} + n\pi$ and $-\arctan(a^2) + n\pi$ where $n\in\mathbb{N}$.

Sorry for omitting that earlier, I thought you'd already taken the solutions to the cosine of $x$ being equal to $0$ before proceeding to deal with the tangent of $x$ being equal to $-a^2$.

Nick

Still, it doesn't explain how $x = - \frac{\pi}{2}$ is an answer

Khallil

Yea.

That's strange.

Have you got a better look at the graph?

It seems like there are loads of other solutions, but it appears as though they're taking them in the range $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$.

Nick

It is sure ugly ugly

Khallil

It's one strange thing.

Not purty at all.

Nick

9:44 PM

It isn't that strange when we take into consideration that $\cos x $ is an even function...

MrWho

@Chris'ssis I discovered the spider man function :D

Nick

@MrWho: Would you mind explaining to us how we can rationally explain why $x = - \frac{\pi}{2}$ is an answer to our question.

MrWho

@Nick Where is the question?

@Chris'ssis See my art : i.imgur.com/fcWE00b.png

Nick

26 mins ago, by Nick

19 mins ago, by Nick

$$\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$$

Khallil

I'm thinking that $\cos x = \sqrt{ ... }$ might have some significance.

Nick

9:50 PM

I've learned something today. Any monkey can solve equations. Explaining why the answers are correct. That takes a mathematician.

skullpatrol

So true^

Khallil

I'm thinking that a different type of manipulation might help us out here.

$\cos x = \sqrt{ \dfrac{\cot x}{\cot x - a^2} } \implies \cos x = \sqrt{ \dfrac{\cos x}{\sqrt{1+a^4}\cos (x + \arctan(-a^2))} }$

MrWho

@nick $$\frac {sin(2x)}{sin^2(x)+1}=2a^2$$

Simplified - Anyone has clue?

Nick

@MrWho: It's so scary that that was my first simplification as well.

MrWho

@Nick Wait

Nick

9:55 PM

Hey, The Morning is still night :D

MrWho

@Nick Indeed, I was in the middle of an integral that you came !

and asked ! :|

Nick

The Wolf tells me this is rocky terrain

@MrWho: The answer is $\infty$ to that.

MrWho

to what?

you said that it's $x=-\frac {\pi}{2}$

@Nick No, I know that, I'm talking about the integral I'm solving at the moment!

Nick

Ohk

Khallil

What's the integral you're solving right now, @MrWho?

(I'm still looking at the trig problem, @Nick.)

Nick

10:00 PM

@Khallil: Me too. Ping me if you get a lead.

MrWho

@Khallil The one Chris gave me.

Khallil

Is it this one, @MrWho?

MrWho

@Khallil Yeah

@Nick I solved it.

@Nick @Khallil $a=0$

Khallil

The initial condition is that $a \neq 0$, @MrWho. Also, $a \in \mathbb{R}$.

Chris's sis

@MrWho nice

MrWho

10:05 PM

@Chris'ssis You enjoyed spider man function , I discovered it while solving your integral :D

Chris's sis

:D

MrWho

@Khallil I wish you could tell me that sooner

Khallil

Also, we're solving for $x$ for a variate $a$.

We aren't solving for $a$.

MrWho

@Khallil If $a \neq 0$ then $x=\frac {-\pi}{2}$ doesn't hold as an answer!

@Khallil I know, I just wanted to eliminate $a$ somehow..

Indeed, it is $$\frac {sin(2x)}{sin^2(x)-1}=2a^2$$

Plug in the $x=\frac {-\pi}{2}$, it doesn't hold for $a \neq 0$

Khallil

How did you get that equality from the original one?

Yea, I'm more interested in how you got that equality.

MrWho

10:10 PM

$$cos(x)=\sqrt {1+\frac {a^2}{cot(x)-a^2}}$$

@Khallil You see?

Khallil

Kinda.

No, actually I don't.

MrWho

$$cos^2(x) = 1+\frac {a^2}{cot(x)-a^2}$$

Khallil

Actually, I think I do.

MrWho

then play.

play with equation.

Khallil

How can you guarantee that you haven't divided by $0$ throughout?

MrWho

10:13 PM

Did I?

Khallil

You've divided your equality by $\sin^2 x - 1$ (which is essentially the same as $-\cos^2 x$), which may be equal to $0$ depending on the value of $x$. I still stand by the way I did it. Wanna see it?

Nick

For $\cot x$ to be defined $\sin x \neq 0 \implies x \neq 0$

MrWho

@Khallil Then we should consider the expression under square root to be positive and see what happens.

Khallil

That's all good and well, but @MrWho has divided by $-\cos^2 x$, not $\sin x$.

MrWho

@Khallil We've already answered the question by reverse engineering

:D

Nick

10:15 PM

Gentlemen, I bring salvation

Khallil

So that's what I did.

Nick

4

Q: if $a$ is are real number that $a \neq 0$, and $\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$, $x$ is on which trigonometric quadrants?

if $a$ is are real number that $a \neq 0$, and $\cos x = \sqrt{\frac{\cot x}{\cot x -a^2}}$, $x$ is on which trigonometric quadrants? Things I have done so far: this problem is mostly different from that I previously solved.My Idea was to powering up both sides,take all to one side and then...

trigonometry

Khallil

39 mins ago, by Khallil

So my general solution for $x$ was $\frac{\pi}{2} + n\pi$ and $-\arctan(a^2) + n\pi$ where $n\in\mathbb{N}$.

MrWho

if $cos(x)$ is zero the equation holds perfectly.

Khallil

MrWho

10:18 PM

@Khallil The problem could be solved at the first sight.

Nick

Sure, now the only explanation I have to the OP of that question to why $x = - \frac{\pi}{2}$ is... because the equation holds

Khallil

How so, @MrWho?

MrWho

@Khallil Wait.

Chris's sis

@robjohn I derive here one amazing identity after another ... :-)

They simply flow all over. :D

MrWho

@Khallil Just find the inequality for the expression under the square root, it's hard to type it in Latex :-(

robjohn

10:20 PM

@Chris'ssis cool! Is this using the same idea as the others?

Chris's sis

@robjohn The same idea.

Nick

@Khallil: Why so glum chum, you're answer was right from the beginning. It's just difficult to explain why -pi/2 works

Chris's sis

@robjohn Just wait a bit ...

robjohn

@Khallil see no evil, see no evil, see no evil, see no evil, see no evil, see no evil, ...

3

Khallil

Are you referring to this, @MrWho? $$ \dfrac{\cot x}{\cot x - a^2} \geqslant 0$$

MrWho

10:21 PM

Yeah!

Khallil

Not glum, @Nick, just tired. ^_^

That inequality would be reduced to $\cot x \left( \cot x - a^2 \right) \geqslant 0$, @MrWho.

robjohn

@Nick $|\cos(x)|\le1$ and that means that $\cot(x)\le0$. Since the square root is positive you know that $\cos(x)\ge0$. Thus, $x$ is in the fourth quadrant.

Khallil

Could you explain the first part, @robjohn?

The $|\cos x| \leqslant 1 \implies \cot x \leqslant 0$ part.

(Also, I see evil, but I reject it. Much like Santen Kesshun.)

robjohn

If $\cot(x)\le0$ then $0\le\frac{\cot(x)}{\cot(x)-a^2}\lt1$

MrWho

@Khallil I said the same thing.

robjohn

10:29 PM

If $\cot(x)\gt a^2$ (so that the quotient is positive), then $\frac{\cot(x)}{\cot(x)-a^2}\gt1$

If $0\lt\cot(x)\le a^2$ then $\frac{\cot(x)}{\cot(x)-a^2}\lt0$ and the square root doesn't exist

Balarka Sen

@MrWho spider man function?

robjohn

@Khallil does that make sense?

Balarka Sen

say waaat

robjohn

@BalarkaSen is that related to the batman function?

Balarka Sen

@robjohn i dunno.

ah so this is MrWho's spider-man function

looks like a tan plot.

Chris's sis

10:36 PM

@robjohn Just look at that ... it's breathtaking, newly created

$$\int_0^1\sqrt{x}\tan(2\arctan(x))^{1/2^2}\tan( 2^2\arctan(x))^{ 1/2^3}\tan( 2^3 \arctan(x))^{ 1/2^4} \cdots \ dx = \log(2)$$

Khallil

It makes a bit of sense, @robjohn.

I'm still making sense of it.

11 mins ago, by robjohn

@Nick $|\cos(x)|\le1$ and that means that $\cot(x)\le0$. Since the square root is positive you know that $\cos(x)\ge0$. Thus, $x$ is in the fourth quadrant.

Oh, I got it.

I was wondering why $\cot x \ngeqslant a^2$, but I now see that it can't because of the very fact that $|\cos x| \leqslant 1$.

Chris's sis

Ooo, I have so fun today. Thanks God I'm so creative!

Khallil

Lucky you, @Chris'ssis!

Alec Teal

Hey @Ted you here?

MrWho

@BalarkaSen Yeah, I discovered it :D

@BalarkaSen I haven't read alot about NT

Chris's sis

10:40 PM

I think @BalarkaSen hates me much looking at that ... :-)))

(just kidding here :-))

skullpatrol

@AlecTeal he was here earlier, but classes have begun so...

Khallil

He's a busy man now.

Alec Teal

I've found more of his work

It's really odd how close I am to that man stalker face.

Khallil

Hahahaha!

Alec Teal

I found his differential geometry notes in 2009, in 2011 and 2012 I found more of his stuff

Khallil

10:46 PM

That's how I imagine you look right now!

Alec Teal

Then in 2013 I meet him, and have books he authored to read (recommended texts)

Khallil

Woah, that's pretty cool.

Alec Teal

In 2014 I find a paper in a book edited by my tutor that has his paper as the first paper in the boo.

Khallil

Not stalkery in any way in my opinion.

Alec Teal

He's the one author I keep accidentally finding

Take for example Serge Lang - great author, but I've never accidentally found stuff he's written

skullpatrol

10:48 PM

In my opinion, he would make a great mentor...

Alec Teal

@skullpatrol WTF?

He's dead.

skullpatrol

...Professor Shifrin

Alec Teal

Oh

Khallil

Always jumping to conclusions ...

sighs

Alec Teal

Serge Lang is a brilliant author at all levels though

His books on subjects tend to be surprisingly thin, however they are not lacking, it's an impressive thing to achieve

skullpatrol

10:50 PM

He's from Yale, right?

Alec Teal

I don't know @skullpatrol

Anyway the only downside to Lang's books is they lack pictures

Like in the front it'll say "With 3 figures"

I always read that in a sarcastic voice

skullpatrol

Some authors want the reader to make mental pictures.

Alec Teal

And the pictures tend to be crap.

"The arrow indicates direction" and it's just a line

Daniel Fischer

@skullpatrol He was.

Alec Teal

@skullpatrol @DanielFischer are any of you two at school now?

Khallil

10:56 PM

Are you in the UK, @DanielFischer?

Alec Teal

I am.

Daniel Fischer

No. Neither at school, nor in the UK.

Alec Teal

When you do return @DanielFischer can you email me any/all assignment sheets.

That applies to any of you!

Khallil

Do you teach, @DanielFischer?

Alec Teal

(I try but you'd be surprised how many are login only)

Daniel Fischer

10:58 PM

@Khallil I hope I teach a bit with my answers here.

4

Khallil

You do! ^_^

Alec Teal

You do @DanielFischer like recently you gave a hint to a book problem I was stuck on, I now know another way to look at things (I prove what you used) and I wouldn't have done that (confidently if at all) before.

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$Mathematics$ Mathematics