Mathematics - 2012-08-28 (page 1 of 2)

12:03 AM

Who let the dogs out?

34567

Our A/C compressor made some weird noise about an hour ago and when I went out to check, the fan was stopped and the motor was making a hissing sound. I pulled the fuses, but it is getting hot in the house. :-(

The people who checked the unit out about a month or two ago are coming out tomorrow afternoon. It is going to get mighty hot in here tomorrow.

Jayesh Badwaik

Ohh. :-( How hot does it get?

hhh

How is the reduced-cost calculated here?

I mean the line: -12, -4, -5, 1, 1, -1, 0, 0, 0

robjohn

12:19 AM

@JayeshBadwaik tomorrow it is only supposed to get to 99° F (37.2° C)

hhh

(doing two-phase Simplex here, this is the same phase however in all simplexes -- so reduced cost?)

$\min \bf c ^T \bf{x}$ so that $A \bf x \leq \bf b$. Now the reduced cost, according to Wikipedia, is $\bf c - A^t \bf y$ where $\bf y$ is the shadow price but now trying to calculate this, this should be very simple...

Jayesh Badwaik

Hmm. Its quiet hot. Well, I have to go now, be back later.

user19161

@robjohn My AC is leaking water here.

robjohn

@WillHunting That is not good either. Is it a window unit?

user19161

@robjohn What? That's above body temperature!

user19161

12:26 AM

@robjohn I am not sure what a window unit is. I think I need to spray it with some kind of liquid to fix it.

robjohn

@WillHunting It is indeed, but a couple of weeks ago we had temps in the 104-108 range

user19161

@robjohn I usually have subnormal body temperature. The usual is 37, mine is 36.5. :-)

robjohn

@WillHunting does your A/C unit sit in a window, or is it central air conditioning?

user19161

@robjohn It sits in a window I think, yeah.

robjohn

@WillHunting a certain amount of condensation is normal. Is the drip pan overflowing or is it leaking?

user19161

12:29 AM

@robjohn Well, it dripped quite a lot a few hours ago, but it stopped. It just gets clogged up now and then but after it finishes dripping it stops dripping for a while. So somewhere in between the two.

user19161

I guess it is because of the great humidity here.

robjohn

@WillHunting The condensation increases with humidity, too.

user19161

@PeterTamaroff You should talk to your neighbour if they get too fierce.

Pantelis Sopasakis

Hi folks. A quick question. X is a n-by-n matrix and function of t (scalar). Then, first of all d/dt||X(t)|| returns a scalar. At the same time d/dt ||X(t)|| = d/dX ||X||*dX/dt. But: d/dX ||X||=X+X' which is a matrix and dX/dt is a matrix again. The result is a matrix, not a scalar. What am I doing wrong?

robjohn

@PantelisSopasakis Does ||X|| mean the determinant of X?

Pantelis Sopasakis

12:42 AM

No, it is the norm-2 (euclidean)

user19161

@rob Not math but does this sentence sound grammatical to you? "I did it once when I got the answer after very long myself on another site."

robjohn

@WillHunting It should probably be two sentences, but they could be joined by a semicolon: I did it once; I got the answer by myself on another site after a very long time.

user19161

@robjohn But is it "wrong" to you?

user19161

"I did it once" refers to posting an answer to my own question.

Pantelis Sopasakis

Actually, what I'm trying to calculate is the derivative of $V(t)=\|sI-e^{At}\|$ where s is a scalar and $A$ a square matrix. $I$ is the identity matrix.

robjohn

12:47 AM

@WillHunting I can understand what you mean, but it doesn't sound right. The usual idioms are "after a very long time" and "by myself". I think the ordering is a bit non-standard. All together it sounds very strange.

hhh

Moved the simplex -problem here, should not be hard.

robjohn

@PantelisSopasakis The square root of the sum of the squares of the elements?

Pantelis Sopasakis

Do you think it's worth posting a question on the site about the derivative of $V(t)=\|sI-e^{At}\|$ (w.r.t. $t\in\Re$)?

@robjohn Yes, this one

@robjohn You have the same problem if you assume it is the Frobenius norm. I was just reading on Wikipedia that the chain rule does not apply to such cases (where differentiation wrt matrices is involved) :S

robjohn

@PantelisSopasakis Then d/dX ||X|| = X/||X||, if I am not mistaken.

@PantelisSopasakis that is true...

Pantelis Sopasakis

@robjohn Are you sure? Isn't it d/dX ||X|| = d/dX X'X = dX'/dX * X + X'* dX/dX = X+X'

X is a matrix

@robjohn ...using the fact that : ||X||=X'X

robjohn

12:55 AM

@PantelisSopasakis I am looking at d/dX ||X||^2 = 2||X|| d/dX ||X|| and noting that each element is simply the corresponding element of 2X.

Pantelis Sopasakis

@robjohn ...you can't apply the rule of chain. It doesn't hold.

robjohn

Is each element of d/dX the partial derivative of ||X|| with respect to the corresponding element of X?

That is how I understand derivatives with respect to vectors and matrices.

Pantelis Sopasakis

@robjohn true

robjohn

@PantelisSopasakis Then the change of the square of something is twice that something times the change of that something.

for infinitesimal changes

So what I wrote is correct.

@PantelisSopasakis Wait, are you defining ||X|| to be the square root of the sum of the squares, or the sum of the squares?

Pantelis Sopasakis

@robjohn To me it feels like a logical leap. You differentiate wrt a matrix. It is d/dX ||X||^2. This assumes the chair rule which doesn't hold.

@robjohn I define ||X||=sqrt(X'X)

@robjohn Yes the square root of the sum sum of squares

robjohn

1:03 AM

I was confused by this

Pantelis Sopasakis

@robjohn Oh, my bad. Sorry, I forgot the square root.

@robjohn Yes, in that case it is d/dX ||X||^2 = X+X'. Isn't it?

robjohn

@PantelisSopasakis You are using the chain rule here with respect to matrices, which is the wrong place. I use it only with resepct to the function x^2 as a real function

@PantelisSopasakis No; try a small 2x2 example

@PantelisSopasakis Think of what the upper left element of the derivative is: it is the partial derivative with respect to the upper left element of the sum of the squares of all the elements. That is twice the upper left element

Pantelis Sopasakis

@robjohn Just a min

robjohn

@PantelisSopasakis: do you have ChatJax installed?

Pantelis Sopasakis

@robjohn No, what is that?

robjohn

1:11 AM

@PantelisSopasakis Ah, it allows you to read things like $\frac{\mathrm{d}}{\mathrm{d}X}\|X\|^2$

Pantelis Sopasakis

@robjohn Oh great!!!

@robjohn :) Much better! Thanks a lot!

robjohn

Pantelis Sopasakis

@robjohn OK, let's try to get back to the problem. So you say that $\frac{d}{dX}\|X\|^2=2\frac{d}{dX}\|X\|$

?

@robjohn By the way, what I wrote before is wrong: $\|X\|^2=X'X$. This is true only for vectors but not for matrices. The only thing we know about norm-2 for matrices is that $\|X\|=\sqrt{\lambda_{max}(X'X)}$...

robjohn

@PantelisSopasakis yes

@PantelisSopasakis I've never seen that. I'd have to look a bit to convince myself.

Pantelis Sopasakis

@robjohn So, it's not very easy to work with the Euclidean norm for matrices. There is however the Frobenius norm defined by $\|X\|_F=\sqrt{\operatorname{trace}(X'X)}$

robjohn

1:21 AM

@PantelisSopasakis Well, it depends on what you want to do with the norm. The derivative of the $2$-norm, as I mentioned, is $\frac{X}{\|X\|}$

Pantelis Sopasakis

@robjohn The Frobenius norm possesses the nice property: $\frac{d\|X\|_F}{dX}=2X$... but again... no chain rule. So, how do I calculate $\frac{dV(t)}{dt}$ where $V(t)=\|sI-e^{At}\|_F$? Note: Can you once again explain to me why is the derivative of norm-2 $X/\|X\|$?

robjohn

Okay. $\frac{\mathrm{d}}{\mathrm{d}X}\|X\|^2=2\frac{\mathrm{d}}{\mathrm{d}X}\|X\|$ since this is just the chain rule applied to the function $x\mapsto x^2$

Pantelis Sopasakis

@robjohn and then...

robjohn

Then it is a matter of taking the partial derivatives of the sum of a large number of squares with respect to each of the terms... $\frac{\partial}{\partial x}(x^2+y^2+z^2)=2x$

If you think about that, you see that each element of the derivative of the square of the norm is simply twice the corresponding element of the original matrix.

Pantelis Sopasakis

@robjohn Ok, got it. You firstly write $\|X\|^2=\sum_{i,j}x_{i,j}^2$

robjohn

1:31 AM

@PantelisSopasakis precisely

Pantelis Sopasakis

@robjohn and after that you take derivatives wrt $x_{i,j}$ and you group them properly in a matrix.

@robjohn In that case, isn't it $\frac{d}{dX}\sum_{i,j}x_{i,j}^2=...=2X$?

robjohn

@PantelisSopasakis Indeed it is :-)

Pantelis Sopasakis

@robjohn or am I missing sth?

@robjohn But, still.. what about my initial question about $V(t)$?

robjohn

@PantelisSopasakis The norm we have been discussing is the Frobenius norm (at least according to what I have read)

Pantelis Sopasakis

@robjohn Oh, one more thing - quite important. (It seems I need to be more careful). Trully, $\frac{d}{dX}\|X\|=2X$... but this norm we worked on although it looks like the norm-2 in $\Re^n$ it is actually the Frobenius norm (problem with terminology)

@robjohn Haha! Yes, true!

@robjohn Now two questions come up: 1. What is the derivative of the norm-2 defined by $\|X\|_2^2=\lambda_{max}(X'X)$ and 2. The initial question.

robjohn

1:37 AM

@PantelisSopasakis Did you mean $\frac{\mathrm{d}}{\mathrm{d}X}\|X\|^2=2X$

Pantelis Sopasakis

@robjohn yes

robjohn

Okay, to calculate $\frac{\mathrm{d}}{\mathrm{d}t}V(t)$...

It is again simpler to compute $\frac{\mathrm{d}}{\mathrm{d}t}V^2(t)$

Pantelis Sopasakis

@robjohn I agree. That would do...

robjohn

$\left(e^{At}\right)^{\mathrm{H}}=e^{A^{\mathrm{H}}t}$ is it not? (for real $t$)

Pantelis Sopasakis

@robjohn Yes, it is indeed.

robjohn

1:48 AM

So $\left(e^{At}\right)^{\mathrm{H}}e^{At}=e^{\left(A^{\mathrm{H}}+A\right)t}$

Hmm... only if $A^{\mathrm{H}}$ and $A$ commute.

Pantelis Sopasakis

@robjohn True, but why should you need this property. It is $\frac{d}{dt}e^{At}=Ae^{At}$ - but again doesn't help per se...

robjohn

I was thinking of the trace of that formula for the Frobenius norm

@PantelisSopasakis I will have to think about this. I have to go afk for a while.

Pantelis Sopasakis

@robjohn Thank you very much. I'll go to sleep in a while...

robjohn

@PantelisSopasakis ping me when you're back.

Pantelis Sopasakis

@robjohn OK. See you tomorrow. Thanks a lot again!

Alex Becker

1:57 AM

@robjohn If you don't mind me butting in, do you know what happened to Arturo?

robjohn

@AlexBecker I know nothing more than what it says in his profile

user19161

@AlexBecker If you don't mind an answer from an ignorant one, actually it is common for people to come and go. After all, this is not real life.

robjohn

Arturo made a grand gross... what more can one person ask?

2

Alex Becker

@WillHunting Yes, but Arturo is the top math user, and seemed to be on 24/7 for years.

user19161

@robjohn Grand gross? LOL. How about gross grand?

Jayesh Badwaik

2:00 AM

@robjohn grand gross?

grand i knew, gross i did not. hmm.

robjohn

@WillHunting I thought about that, I don't know which sounded better.

@WillHunting a gross is $144=12^2$

user19161

@robjohn Oh noes, what is 100 then?

Jayesh Badwaik

@robjohn en.wikipedia.org/wiki/1728_(number)

?

user19161

@JayeshBadwaik We all know that one...

robjohn

@WillHunting a hundred?

Jayesh Badwaik

2:02 AM

@WillHunting hmm, 100 is cent?

robjohn

@JayeshBadwaik okay, so they call that a "grand gross"...

Jayesh Badwaik

@robjohn yeah
@WillHunting or centum if latin is what you want.

robjohn

kilogross

Jayesh Badwaik

For me it will always be a (taxi number -1), the wikipedia grand gross.

user19161

Well, let's just say Arturo got $12^2\times 10^3$. :-)

Pantelis Sopasakis

2:07 AM

Regarding the discussion we had with @robjohn on the differentiation of that function, I placed a question on the site: math.stackexchange.com/questions/187715/…

robjohn

@PantelisSopasakis I usually see $\mathbb{R}$ instead of $\Re$. Is that not the standard there?

Pantelis Sopasakis

@robjohn Don't know. I prefer $\mathbb{R}$ too. I just type \Re instead of \mathbb{R}.

hhh

What formula is used with this big-M-simplex? Is it just substitution? Nothing special? (sorry interrupting)

Pantelis Sopasakis

@robjohn I changed the notation in my question. Looks better now :)

hhh

$$\sum_{j=1}^n c_j x_j +M \sum_{i=1}^m y_i$$
<--- I know actually that this is the formula but cannot now just dig into it, for some reason.

(p. 117 Bertsimas)

(Trying to understand the line $5M-13$, $-2M+5$, ... and $2M-4$.)

3. DUNNO: I cannot see how the scalars get there like $-13$.

user19161

2:19 AM

@AlexBecker Should we expect the same of Alex? :-)

user19161

@robjohn One guy on TeX also has 100,000.

anon

@PeterTamaroff careful there...

Gustavo Bandeira

2:48 AM

I've answered my first question - decently - yay!

robjohn

3:03 AM

@WillHunting Onee of the owners of this room has over 332,000.

Jayesh Badwaik

@robjohn I believe he is one of the primary developers of stackexchange

robjohn

3:42 AM

@JayeshBadwaik yes, but that doesn't give him extra reputation, does it?

Jayesh Badwaik

naah, the reputation is purely his own effort.

Jayesh Badwaik

4:40 AM

Lang is so confusing here. He says:
Let $A$ be a finite abelian p-group. Let $b \neq 0$ be an element of $A$. Let $k$ be than integer so that $p^{k}b \neq 0$ and $p^{m}$ be the period of $p^{k}b$!!!
How can I multiply $p$ with $b$ when one is an element of a group and other is an integer?

@ZhenLin ?

book: undergraduate algebra third edition page 68

Zhen Lin

An abelian group is also a $\mathbb{Z}$-module.

Jayesh Badwaik

So, basically whatever abelian group we may have, he is taking an isomorphic group, which is here basically a $\mathbb{Z}/ p \mathbb{Z}$, proving stuff for it and then going back to the original group. Am I correct?

Zhen Lin

That's not what I said. The point is, you can just define $n x = x + x + \cdots + x$.

Jayesh Badwaik

Ohh. Okay. Got it.

Have not done modules yet. Looking it up just now.

hhh

5:24 AM

Moved the point about Big M -simplex here.

Gigili

6:29 AM

Umm, @Rob have a minute?

Jayesh Badwaik

7:28 AM

There is this problem from an entrance examination to TIFR. True or False:
There exists polynomials $f(x)$ and $g(x)$, with complex co-efficients , such that
\begin{equation}
\left( \frac{f(x)}{g(x)}\right)^2 = x
\end{equation}

Now, my first reaction is $g(x) = 1$ and hence, the statement is true and the problems seems too simple to be asked in an exam. However, if we say that trivial polynomials are not allowed, then every polynomial will have a root and what happens when $g(x) = 0$ and again it is too simple, but the other way round. Am I overthinking the stuff?

Or am I missing the elephant in the room and getting it wrong somehow?

When I say $g(x) = 1$, I mean $g(x) \equiv 1$

But the second time I am referring to $g(x_{0}) = 0$ for some $x = x_{0}$

Matt

@JayeshBadwaik If $g = 1$ then the equation would imply that $f= \sqrt{x}$. That's not a polynomial since polynomials have exponents in $\mathbb N$.

Jayesh Badwaik

@Matt Got it.

Matt

@JayeshBadwaik Perhaps one could argue somehow like this:

Maybe something like this could work: Assume there are such polynomials. Say $deg(f) = n$ and $deg(g) = m$. Then $f^2 = x g^2$.

Jayesh Badwaik

@Matt Hmm. Let it be. I will study polynomials from Lang first. Then see, if I can get the problem.

Matt

Then you have $deg(f^2) = 2n = 1 + 2m$.

So $2(n - m) = 1$.

Then $n - m = \frac{1}{2}$.

But that's not possible.

Jayesh Badwaik

7:43 AM

What about infinite series. Infinite series is not considered a polynomial right?

Matt

Yes, that's right.

They are finite sums. That's why I can assume something like $deg(f) = n$.

Jayesh Badwaik

Yup, that was my doubt to begin with. I tried that method, but then was not sure about infinity, but then I was forgetting the basics again. Arrgh.

Matt

Just keep practicing : )

BBL

Jayesh Badwaik

@Matt Thanks. Yup. Bye.

BenjaLim

7:57 AM

@ZhenLin Hey

Kannappan Sampath

Is this on topic here?

Ilya

@KannappanSampath Grammar/notation border

but Grammar intersects Notation, so the part of the border is a subset of notation

user19161

Anything asking where to put the semicolon is off-topic.

Kannappan Sampath

8:16 AM

Hmm, I was about to vote to close that as off topic.

@Ilya Thank you for telling me

BTW, Hello! It's nice to see you!

user19161

@KannappanSampath Not nice to see me?

Ilya

@Kan: nice to see you as well

Kannappan Sampath

@WillHunting Sir, did I tell you so? It indeed is pleasant to meet people in this chat room.

Perhaps, not everyone but most people.

user19161

@KannappanSampath Oh now you are like Iyengar, calling everyone Sir, hehe.

Kannappan Sampath

AND, you're one of them.

@WillHunting everyone? :/

user19161

8:18 AM

@KannappanSampath Have you been in chat when Iyengar is here?

Kannappan Sampath

@WillHunting Yeah, I've been here.

user19161

I was wondering if he was just fooling us, but it seems he was quite serious when he called everyone sir.

Kannappan Sampath

@WillHunting Can we talk about something that interests people?

user19161

@KannappanSampath OK, like masala tosai. :-)

Kannappan Sampath

@WillHunting Heh!

user19161

8:21 AM

I don't usually eat Indian, but my favourite is that.

user19161

I like it with the yoghurt like dressing.

Kannappan Sampath

@WillHunting Hmm. I see. It's nice. @BenjaLim 's knowledge of Indian culture just was amazing.

user19161

So now after editing my username, I get even more spam!

Kannappan Sampath

-1

A: Point on the line twice as far from P as from Q

Hint: Find the equation of the line. Write your condition by making use of distance formula. Use $1$. Stir until done.

What's wrong with that answer?

That's ridiculous!

user19161

@KannappanSampath Maybe they don't like the stir until done?

Kannappan Sampath

8:28 AM

Well, but, if my answer answers the question, I won't edit it out.

user19161

Yeah, don't bother. People upvote non-answers that are good jokes and downvote answers that are bad jokes.

Kannappan Sampath

:/

Kannappan Sampath

8:49 AM

Nasty downvoters.

I hate them.

I should never help homework cases.

They(the problems) suck.

Jayesh Badwaik

9:10 AM

@WillHunting Big Surprise. :P But what kind of spam do you get?

Zhen Lin

9:36 AM

Probability questions are fun. I almost miss doing them...

user19161

10:18 AM

@JayeshBadwaik All kinds.

Jayesh Badwaik

10:44 AM

@JonasTeuwen If you already have a Masters degree, how come your Masters thesis is still "under construction" or so to say? Or is it slowly transforming into a PhD thesis?

BenjaLim

11:02 AM

Sigh..........

@KannappanSampath You should not let what other people do control your happiness

Krat

I have a question.. In triangle ABC rt angled at C, what will be the slope of AB if all sides, angles and points A and C are known? What formula should I use? Definitely not BC/AC

N3buchadnezzar

11:49 AM

Heya =)

skullpatrol

yo sup?

N3buchadnezzar

Laplace transformations sucky suck

N3buchadnezzar

12:04 PM

Any hints on finding the inverse laplacetransformation of $$ \frac{1}{s^2+1} - \frac{2}{\left(s^2+1 \right)^2} $$ ? I know what is is from CAS, but not how to calculate it manually..

user19161

@JayeshBadwaik Maybe he wants to improve on it?

user19161

@Krat It is not clear what you mean by slope.

user19161

12:20 PM

@BenjaLim The fact is, what others do do affect our worldly happiness. It only has no effect whatsoever when we decide to abandon all worldly happiness.

Matt

whatever that means^

Lawn mowers do affect my happiness.

In a very negative way.

In fact if I had a sledgehammer I'd go out there and bash up the f*cking mower.

user19161

Is it too noisy for you?

Matt

I am wearing f*cking ear protection to practice shooting, the windows are closed and it's so loud that it completely prevents any sort of thinking.

It means I cannot work.

I should send the idiotic letting agency a bill for my time.

user19161

Oh dear, I did not know they are so fuqin loud.

Matt

They fuqin are.

And this is just a normal sized one.

user19161

12:26 PM

Hmm, try to work out some solution then.

user19161

Like talk to the people doing the noise outside.

Matt

The other house has a gardener with a f*ckin tractor to sit on and drive around. WTF.

user19161

Or go to another place at certain times.

user19161

Ben's statement makes sense, and so does mine.

user19161

There is no contradiction.

user19161

12:28 PM

I just want to complete the picture.

Matt

@WillHunting No, have to enforce it over the letting agency of the house. Will tell them that they must require their gardeners to use one of these.

user19161

@Matt Yes, if you close the windows and wear ear plugs and it is still so loud, it must be fuqin loud.

user19161

I only wear ear plugs when I was shooting.

Matt

Otherwise....

user19161

That looks like something Pedro would post.

Matt

12:30 PM

@WillHunting No fuqin ear plugs. I am wearing the real deal. Fuq.

user19161

@Matt OMG!

user19161

And yes, ear plugs are very uncomfy.

Matt

Yes fuq. Also tried that plus ear plugs. Can still hear it.

robjohn

@Gigili Sorry, I was away until a bit ago.

Matt

Drives me mad.

user19161

12:32 PM

@Matt Well, the only thing I can say is, try to focus despite the noise. I can sort of do it.

user19161

I am already mad, so it does not drive me mad. :-)

robjohn

@Matt You say that you have hearing protection to practice shooting. That means you have a gun. Problem solved: shoot the mower.

user19161

@robjohn Maybe it is from the toy store. Mine was real. I was in the army.

Matt

@robjohn No. I bought these because I cannot work if there is noise. I don't have a gun : (

user19161

@robjohn That is not an elegant solution though.

user19161

12:34 PM

@Matt I have one in my underwear. :-)

Matt

@WillHunting lol, but that's not exactly wide-range...

Also, won't do much damage. Apart from stains on the gardener's shirt maybe.

user19161

LOL.

Matt

But now I think I have tmi.

user19161

It's OK. Just the right amount to stop here.

Matt

@robjohn I need to buy an island.

Problem solved.

Like in north Norway or Canada.

Then I'd solve two problems: noise by people and bad weather.

robjohn

12:38 PM

@Matt Great, as long as getting the money to buy said island does not require the aforementioned gun :-)

user19161

@Matt If you can buy an island, you can buy something else too, no need to buy an entire island.

user19161

There are some things money cannot buy though, like sanity.

Matt

@robjohn : D

user19161

But with lots of money, 99 per cent of problems can be solved.

Matt

So much rage makes me hungry. Let's see what the fridge has to offer...

robjohn

12:39 PM

@WillHunting Ah, but money can make the madness more cushy :-)

user19161

@robjohn Yes, indeed.

user19161

@Matt They say that a hungry man is an angry man.

Krat

@WillHunting Slope means gradient... how to find the gradient of the hypotenuse?

user19161

@Krat Well, if you mean gradient of a line on the xy-plane, then just take difference in y coordinates divided by difference in x coordinates.

user19161

So in the xy-plane, a vertical line has undefined gradient since division by zero is not allowed, and a horizontal line has zero gradient.

Matt

12:42 PM

@WillHunting It seems to work in the other direction too : )

user19161

Proposition 1: A man is hungry iff he is angry. Proof: See MSE chat. QED.

Krat

@Will You are right.. but in this case, AB is the hypotenuse.. and we know only the coordinates of A and 'C'.. and not B. There is another formula in this case..

skullpatrol

If a man is hungry, then he is angry.
If a man is angry, then he is hungry?

user19161

Nowadays, people ping and say absolutely nothing.

skullpatrol

Or they say absolutely nothing and don't ping.

BenjaLim

12:49 PM

@Matt Hey

@Matt Can you help here please? math.stackexchange.com/questions/187864/…

Matt

No, I need to work on my own stuff. Also, atm I cannot do anything because it's too loud here.

Matt

1:01 PM

@WillHunting It's finally stopped.

Feeling less hungry now.

Should go and do stuff.

user19161

@Matt Of course! See Proposition 1 above. Good luck!

Matt

See you later!

Michael Greinecker

1:15 PM

Am I the only one for whom M.SE became partially invisible?

skullpatrol

@MichaelGreinecker What do you mean?

Michael Greinecker

I don't see certain elements anymore., such as the "Mathematics"-title, or the arrows for voting.

skullpatrol

@MichaelGreinecker I can see both of those.

N3buchadnezzar

hu=)

Michael Greinecker

It has returned now.

gone again..,

skullpatrol

1:31 PM

@MichaelGreinecker Have you tried rebooting?

Michael Greinecker

I'll try.

Works for now, let's see whether it lasts.

skullpatrol

:-)

@MichaelGreinecker BTW I like the "tree" (gr)avatar.

Michael Greinecker

thanks

Jayesh Badwaik

@WillHunting Of course one can improve on it. I have been improving my bachelor's thesis whenever I get time, but I was wondering whether it would still be called a thesis then? If I pushed my improved version as my thesis, it might be considered a fraud, so I do not call it my thesis now, just a report/paper.

N3buchadnezzar

2:43 PM

Hey

Could anyone give me some hints on how to find $f: \mathbb{Z}\to \mathbb{N}$ ?

Jayesh Badwaik

@N3buchadnezzar $f : \mathbb{Z} \rightarrow \mathbb{N}$ is a notation for any generic function which has an integer domain and a natural number co-domain.

You must be misreading your problem.

N3buchadnezzar

@JayeshBadwaik math.ntnu.no/emner/TMA4145/2012h/exercises/oving1.pdf

Find a bijection $f : \mathbb{Z} \to \mathbb{N}$

Jayesh Badwaik

Okay, so you have been asked to find an example of a bijection $f : \mathbb{Z} \to \mathbb{N}$.

N3buchadnezzar

Yes, and I simply cant!

Jayesh Badwaik

One of the examples is $f = 2n$ for $n > 0$ and $f= 2(-n)+1$ for $n \leq 0$

Matt

4:03 PM

"...It follows years in which Iranian women students have outperformed men,...". But banning women from university will not make Iranian men less mentally retarded and backwards...

Srsly?

jrand

Hello. I am curious about the notation of definition of a random variable: a random variable $X$ is a function $X:\Omega \mapsto \mathbb{R}$. Is it correct to say: $X$ is a function which takes in any subset of $\Omega$ as input and produces a real valued real number, which is not positive or negative infinity, as output?

Specifically, I was confused about the input part.

$F:\mathbb{R} \mapsto \mathbb{R}$ means takes in any single real valued number, and produce a single real valued number.

Jonas Teuwen

4:24 PM

@JayeshBadwaik It is finished, but I am adding small improvements.

jrand

It appears function $X$ takes in a single element of the set $\Omega$ and produces a single element of the set $\mathbb{R}$

robjohn

@N3buchadnezzar what distinction are they making between range and codomain? Are they saying that to have range $B$ the function must be surjective?

Jonas Teuwen

I feel better now that I don't have to go to work :-).

N3buchadnezzar

@robjohn I do not remember, I have only been in one lecture yet. And the books have not arrived. However I do recall they made a distinction between the range and the codomain of a function.

robjohn

@N3buchadnezzar my guess would be that they consider $f(A)$, the image of $f$, to be the range of $f$

@N3buchadnezzar In this article, they say that the range is sometimes considered to be the codomain (which is how I learned it) and sometimes the image.

@OldJohn: Good day...

Old John

4:39 PM

@robjohn Hi there - and everyone else

@JonasTeuwen Did you mention an excellent beer recently?

Jonas Teuwen

@OldJohn I very much did. "Hel en verdoemenis".

Jayesh Badwaik

@OldJohn hi

Old John

@JonasTeuwen That looks like a serious beer :)

@JayeshBadwaik hi there

Jonas Teuwen

@OldJohn It is, pretty lovely it is as wll.

Those people only make serious beers.

Old John

@JonasTeuwen have you tried Guiness?

Jayesh Badwaik

4:43 PM

@robjohn I learned that range is always the image.

Old John

(but you need to taste it in Ireland - it never tastes as good elsewhere)

N3buchadnezzar

@robjohn yes

Jayesh Badwaik

@N3buchadnezzar did you find my solution helpful?

N3buchadnezzar

@JayeshBadwaik Very!

robjohn

@JayeshBadwaik interesting. I wonder if that is the common way it is taught these days. Definitely different than when I was in school. I don't even remember hearing the word codomain until I was in grad school.

N3buchadnezzar

4:45 PM

@robjohn Can you try to explain why the function $f(a,b) = a^{a-1}(2b-1)$ is bijection from $f:\mathbb{N} \times \mathbb{N} \to \mathbb{N}$

Jayesh Badwaik

@N3buchadnezzar good. I am always afraid of my mistake-prone clumsiness.

robjohn

@N3buchadnezzar $\mathbb{N}$ being the positive integers, I assume.

N3buchadnezzar

@robjohn 1,2,3, . .. yes =)

Jayesh Badwaik

@N3buchadnezzar the way you must prove it is if $f:X \rightarrow Y$ is a bijection, then
$f(x_{1})=f(x_{2}) \implies x_{1} = x_{2}$ and of course that N is a surjection.

N3buchadnezzar

Yeah, I just do not see how

robjohn

4:49 PM

@N3buchadnezzar But doesn't $f(3,1)=f(1,5)$?

Old John

@N3buchadnezzar Not sure I believe it is a bijection ...

N3buchadnezzar

The only way to obtain the odd numbers is if $a=1$, and I can clearly see that each odd number is obtained exactly once.

But I have problems showing that each even number is obtained exactly once.

Old John

@robjohn That would do it :)

N3buchadnezzar

Probably I wrote it down wrong ;(

robjohn

Now, $2^{a-1}(2b-1)$ I would believe

Old John

4:51 PM

@robjohn yep

N3buchadnezzar

@robjohn Yeah!

Jonas Teuwen

@OldJohn Yes, I did.

I was wondering if I was drinking beer.

Maybe because I had a good one before it.

I mean, I was seriously wondering that or if there was something wrong with me...

Old John

@JonasTeuwen I only drink it when I am in Ireland - it tastes better there :)

Jonas Teuwen

Because all the rest is shittier?

robjohn

@N3buchadnezzar Do you see that, or do you still have questions?

Old John

4:53 PM

@JonasTeuwen pretty much

back shortly ...

N3buchadnezzar

@robjohn It is the correct equation, and I can see that all the odd numbers are mapped exactly once (a=1), but I have problems seeing that all even numbers are mapped exactly once.

robjohn

@N3buchadnezzar can you see that each integer is the product of a power of 2 and an odd number?

N3buchadnezzar

@robjohn Yeah

robjohn

@N3buchadnezzar and each power of 2 is uniquely written as $2^{a-1}$ and each odd number uniquely as $2b-1$

@N3buchadnezzar Try working backwards... for what $a$ and $b$ does $f(a,b)=12$?

N3buchadnezzar

@robjohn f(3,2)

robjohn

4:57 PM

@N3buchadnezzar indeed. Now do you understand?

N3buchadnezzar

4 \cdot 3 = 12 ... Hmm

:5948909 Does that really proove it is a bijection ?

Jayesh Badwaik

@N3buchadnezzar You can do it like this
Assume $2^{a_{1}-1}(2b_{1} -1) = 2^{a_{2} -1}(2b_{2} -1) = k $
And now, take all the powers of two on one side and all the odd numbers of the other side. An odd integer divided by another odd integer is always odd. Hence, $2^{a_{1} - a_{2}}$ is always odd which means $a_{1} = a_{2}$

So, now you have shown the uniqueness. Now you have to show the surjectivity. That for every integer $N$, you can find at least one representation of the form $2^{a-1}(2b+1)$

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$Mathematics$ Mathematics