Mathematics - 2013-05-18 (page 4 of 8)

DanZimm

08:00

@Faust7 night

heh nice

Zolomon

@DanZimm: Why isn't it $x^2=4\Longleftrightarrow x=\pm \sqrt{4} = \pm2$ ?

$$x^2=4\Longleftrightarrow x=\pm \sqrt{4} = \pm2$$

Hmm.. I am missing something!

DanZimm

it is equivalent but was just making the point the implication you gave wasn't correct

Zolomon

Aha!

Thank you. :)

DanZimm

np

Zolomon

Okay, so back to my original questions; which operators do I need to look out for that changes the range. Ooh.. I think I just answered my own question!

DanZimm

08:05

i dont understand "range of equality" still, but none should, as long as you properly use the theorems and such

Lord_Farin

@Zolomon In principle, almost any function will change the range; you should regard the occasions where it does not change as a mere coincidence.

DanZimm

properly

what is range of equality?

Zolomon

As soon as I apply some theorem or operator that changes the range/output of my statement(?), that's when I should use implication/equivalence to show that the application of the operator/theorem yields this new expression that is true?

DanZimm

yes

I still don't understand what @Lord_Farin said

wants to understand!

Lord_Farin

@DanZimm I confused this range with the range of a function. Let me try to reformulate the question as I think @Zolomon intended it.

DanZimm

08:10

ok

Lord_Farin

"Given $\{x:a(x) = b(x)\}$, which functions $f$ have $\{x: f(a(x)) = f(b(x))\} \ne \{x: a(x) = b(x)\}$?"

DanZimm

that should be always true, no?

meaning all $f$ those two sets will be the same

or should be

Lord_Farin

@DanZimm What if $f$ is constant?

Zolomon

sneaky!

DanZimm

aha interesting ok

otherwise, is it true?

actually no it won't be

Lord_Farin

08:15

@DanZimm It will in general only be true if $f$ is injective on a suitable subset of its domain (which relates to $a$ and $b$, of course). So it will hold at least if $f$ is injective, but otherwise it needn't.

DanZimm

injentive = one to one right?

Lord_Farin

Yes.

DanZimm

yep

@Lord_Farin are you sure? can you give an example of a function that's not injective (other than a constant function) that this holds?

Zolomon

I feel enlightened and illuminated! :D

Chris's wise sister

@Ethan not bad

DanZimm

08:19

oh wait, this is the set of all x

Lord_Farin

@DanZimm Suppose that $a$ and $b$ take positive values, but are never equal. Take $f$ injective on positive inputs, constant on negative ones.

Since $a$ and $b$ do not "reach into" the region where $f$ is not injective, the sets will still be equal (and empty, but we didn't really use that fact).

DanZimm

@Lord_Farin right I meant is there a non injective function where as $\{x : a(x) = b(x)\} \neq \{ x : f(a(x)) = f(b(x))\}$

Lord_Farin

@DanZimm Then just take $a$ and $b$ never equal, but $f$ constant. Then one is empty, the other is all of $\Bbb R$.

DanZimm

i said besides $f$ constant

then the step function

well I think why I'm getting so cognitively dissonant is because $a(x) = b(x) \implies f(a(x)) = f(b(x))$ is true, right?

@Lord_Farin ^

@Zolomon good question ;D

Zolomon

@DanZimm: Which one? :)

DanZimm

08:25

the "range of equality" one

never thought about that

Zolomon

I don't really understand what @Lord_Farin said, injective means there is 1:1 mapping between the inputs and outputs of a function?

Lord_Farin

@DanZimm Yes that is true.

DanZimm

phew ok

yea exactly

aight now i can go to bed, i was srsly reconsidering studying math if I had that wrong

Lord_Farin

@Zolomon Indeed, but there may be values not attained by the function.

Zolomon

I haven't read any set theory, so I barely understand your problem.

DanZimm

08:28

wishes he learned set theory and logic properly

I basically skipped over the foundation of mathematics, which seems like a VERY bad idea lol

Zolomon

So from what I understand it, any other function than f(x) = x will be true..?

@DanZimm: I think I skipped them by mistake while recovering from leukemia, or at least I forgot it all. :(

DanZimm

@Zolomon no, it's a case by case type thing

in general it'll hold as long as both the left and right hand side of the equation are in the domain of $f$ and $f$ is injective

otherwise it's unknown, and must be done on a case by case basis

Zolomon

Which equation are you talking about; a(x)=b(x) ?

DanZimm

yea

Zolomon

Someone should modify the bookmarks to rerender as soon as a latex expression is found in each new message. mutters

DanZimm

08:35

bookmarks?

Zolomon

Yeah, for the LaTeX?

Aha, now I read more carefully!

start ChatJax doesn't seem to render my latex.

DanZimm

theres already a bookmark that will just continually render latex

Lord_Farin

@Zolomon Then you've just picked the wrong bookmark, I'm afraid.

DanZimm

yea works for me

Lord_Farin

@Zolomon You'll have to enclose your TeX statements in dollar signs.

DanZimm

08:36

or double dollas

Zolomon

Is it double or single $?

$x=4$

$$x=4$$

DanZimm

you see the difference?

Zolomon

Yes!

DanZimm

$this is inline$ $$this is larger and brings focus to it$$

$\begin{proof} \end{proof}$

Zolomon

$$I like focus$$

DanZimm

08:38

xD

Zolomon

However, the chatjax loop doesn't run for some reason even after I start it.

DanZimm

do you get some sort of js error?

Zolomon

$$This is a proof of this sentence existence.$$

No, no errors.

I am using chrome, might be what's wrong?

DanZimm

just realized my "define -1*-1 = -1" comment is in the starred thing

look at the web inspector console

you sure nothing is popping up?

copy and paste the bookmark into pastie.org

Zolomon

pastie.org/private/sfugvmeywko8yqr8nmj8w

DanZimm

08:43

hrm it looks like it should work

Zolomon

It's probably just something temporary

DanZimm

yea

id try to restart

restart chrome that is

rest art

Zolomon

Anyway, now I'm going to spawn a few lines of Scala and then I'm heading out to uni to induce some marvels into the range of existing proofs.

Oh, I like that expression; rest art.

DanZimm

xD

gl have a nice day

Zolomon

You too (evening/night!), and thanks for the help again!

DanZimm

08:46

np

BenjaLim

09:02

@robjohn I can use cauchy - schwarz but I feel that is a little too heavy handed

Babak S.

09:29

@Lord_Farin: Hi

Lord_Farin

@BabakS. Hello.

Babak S.

How are you? What are you doing with your thesis?:)

Everything's OK?

Lord_Farin

@BabakS. Fine, thanks.

I'm currently preparing to go into the final stage, finishing off the preliminary work and getting to the actual paper that's the subject of my thesis.

But presently I have to work on a hand-in assignment for a different course, Set Theory.

Babak S.

Do you think we can use Laplace Transformation about math.stackexchange.com/q/395327/8581?? @Lord_Farin

@Lord_Farin: I wish you a very good times full of prosperity. :)

Lord_Farin

@BabakS. Thanks, same. I don't see immediately how Laplace transform could be used there.

Babak S.

09:33

Ok. Thanks for the time. @Lord_Farin

Lord_Farin

@BabakS. Welcome. Good luck with your thesis :).

robjohn

@BenjaLim Does C-S work here?

TheNotMe

Hi everyone.

robjohn

@TheNotMe kinda quiet here right now.

$$
\begin{align}
\int_0^1\int_0^{x^2}\frac{|f(t)|}{1-t}\,\mathrm{d}t\,\mathrm{d}x
&=\int_0^1\int_{\sqrt{t}}^1\frac{|f(t)|}{1-t}\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\frac{1-\sqrt{t}}{1-t}|f(t)|\,\mathrm{d}t\\
&=\int_0^1\frac1{1+\sqrt{t}}|f(t)|\,\mathrm{d}t\\
&\le\int_0^1|f(t)|\,\mathrm{d}t
\end{align}
$$

Rajesh D

10:16

Hi @robjohn

robjohn

@RajeshD hey there

somaye

10:38

@BabakS.

are you here?

Babak S.

Hi somaye.

Long time

somaye

@BabakS. how are you?long time for what?

Babak S.

Did you send an email to Will Jagy? @somaye

somaye

@BabakS. not yet why?

Babak S.

He asked me that, you did. Sending a paper to him?

somaye

10:40

no

i did not

@BabakS. he wanted from you to do it?

Babak S.

See this @somaye.

somaye

what?

yes he is right

ok

i sent him my article

Babak S.

@somaye: Age frestadidi, bahash tamas begirid. Un mikhast motmaeen beshe

somaye

mano negaran kerdid moshkel chiye?

kardid

Babak S.

Hichi. diruz didam ino. harchi montazer shodam nabudid begam

somaye

10:43

@BabakS.

Babak S.

behesh email bezanin

akse bala ro bardarid. mikham pakesh konam

@somaye

Jayesh Badwaik

@somaye Hi. Wassup?

Did you get my mail?

@BabakS. I have added you on FB. (If you don't mind.)

somaye

@BabakS. ok thank you

Babak S.

Really @JayeshBadwaik. :-)

Jayesh Badwaik

:-)

somaye

10:45

@BabakS. then every thing is ok ?

Babak S.

Negaran nabash. Dr tawakoli tu jaryane?

somaye

@BabakS. chera?

Babak S.

Mage kare moshtarak nist. Khode will goft. Akso bebin.

@somaye

@somaye: Omidwaram movafagh bashid. :)

Bajaj

hello all

somaye

@BabakS. chera vali be har hal on ye maghale chap shodast'

Babak S.

10:48

@somaye: Ahan. man fek kardam hanus submit nashode. Ok I got it

@somaye: Ok thanks for your time. :)

somaye

@BabakS. :) thanks i am at work right now i'll see you tonight (@babak @Jayesh and all of freinds)

your welcome:)

bye

Gustavo Bandeira

11:03

@somaye Hello. =)

Gustavo Bandeira

11:17

@Lord_Farin Do you know The Joy of Cats? Is it too bad to be atracted by this book just because of dem cats?

Lord_Farin

11:33

@GustavoBandeira Heard of it; didn't read it. I'd say the title has done its job (I've always read it as a pun on The Joy of Sets by Devlin).

Gustavo Bandeira

@Lord_Farin Oh really? I thought the author had somehow managed to see a cat in the covers notation.

Vrouvrou

12:19

@Lord_Farin hi

please if you can help me

Question on a third order boundary value problems

please @Lord_Farin

just an idea

BenjaLim

@robjohn Hey

I deleted my question

robjohn

@BenjaLim thanks

BenjaLim

@robjohn I don't understand how you used Fubini

are the hypothesis satisfied?

@robjohn Oh wait

robjohn

@BenjaLim you know that your answer requires that $f\in L^2$. If you only know that $f\in L^1$ your answer does not work

BenjaLim

yea that's the thing I realised damn

robjohn

12:23

@BenjaLim $|f|$ and $\frac1{1-x}$ are positive. we can switch order of integration

BenjaLim

@robjohn I can write $g(x)$ as $\int_0^1 f(t)/(1-t) \times \chi_{[0,\sqrt{x}]}(t) dt$

damn my lecturer just sent an email saying that the integral is from $0$ to $\sqrt{x}$ and not $x^2$

robjohn

that is okay:
$$
\begin{align}
\int_0^1\int_0^{\sqrt{x}}\frac{|f(t)|}{1-t}\,\mathrm{d}t\,\mathrm{d}x
&=\int_0^1\int_{t^2}^1\frac{|f(t)|}{1-t}\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\frac{1-t^2}{1-t}|f(t)|\,\mathrm{d}t\\
&=\int_0^1(1+t)|f(t)|\,\mathrm{d}t\\
&\le2\int_0^1|f(t)|\,\mathrm{d}t
\end{align}
$$

BenjaLim

@robjohn Doesn't Fubini require knowing that the integral of $1/(1-x)$ is finite?

robjohn

@BenjaLim nope, just that everything is positive.

BenjaLim

@robjohn You see the thing for me is that what I'm asked to show is $\int_0^1 g(x) dx = \int_0^1 (1+t) |f(t)| dt$

And to do that we first need to show that $g(x)$ is integrable

so in some sense your manipulations above don't apply to my question

You see what you did above I already did

I was just having trouble proving that $g(x)$ is integrable

robjohn

12:28

@BenjaLim If $f\in L^1$ my proof shows that the integral is absolutely convergent, so you can include the signs without problem

BenjaLim

@robjohn I just read wiki saying we can exchange the integration if everything is finite

robjohn

Fubini's theorem

In mathematical analysis Fubini's theorem, named after Guido Fubini, is a result which gives conditions under which it is possible to compute a double integral using iterated integrals. As a consequence it allows the order of integration to be changed in iterated integrals. Theorem statement Suppose A and B are complete measure spaces. Suppose f(x,y) is A × B measurable. If :\int_{A\times B} |f(x,y)|\,\text{d}(x,y) where the integral is taken with respect to a product measure on the space over A × B, then :\int_A\left(\int_B f(x,y)\,\text{d}y\right)\,\text...

BenjaLim

@robjohn Hmm now I'm confused

It seems to me that I need to use some other method other than Fubini to show that $g(x)$ is integrable

because otherwise it would be circular reasoning

robjohn

$F(x,t)=\frac{f(t)}{1-t}\chi_{[0,\sqrt{x}]}(t)$

BenjaLim

yes that is what I wrote above yea

robjohn

12:33

@BenjaLim why?

BenjaLim

and I want to show that $\int_{[0,1]^2} F(x,t) $ is finite

robjohn

@BenjaLim It is not circular unless you are proving Fubini

Babak S.

@robjohn: Hi Rob. ;-)

BenjaLim

@robjohn But you're saying we can exchange the order of integration because of positivity. I don't think that's enough

robjohn

@BabakS. hello

Babak S.

12:36

@robjohn: May I ask you to remove the link above I attached. Will Jagy wanted me to do something and I did. That isn't need to be on anymore. Thanks and sorry for asking that.

robjohn

@BenjaLim See Theorem 1. The continuity is to ensure that the functions are measurable.

skullpatrol

Just kidding :D

robjohn

@BenjaLim If the integral of the absolute value is finite, the integral is finite

@BabakS. where?

BenjaLim

@robjohn You're saying we can apply Fubini because $\int_{[0,1]} f(t) dt < \infty$ is enough?

Babak S.

@robjohn: Above the page. Almost up.

robjohn

12:40

@BenjaLim neither needs to be finite. only the product needs to be finite.

@BenjaLim they only need to be measurable otherwise

$f$ and $\frac1{1-t}$ are measurable

BenjaLim

That is correct

integrable implies measurable

Babak S.

@skullpatrol: Hello skull

BenjaLim

You see my assignment says "prove $g$ is integrable and $\int_0^1 g(x) dx = \int_0^1 (1+t)f(t) dt$

it seems we have to prove that first $g(x)$ is integrable before doing those manipulations above

I also don't see why your first line is justified before knowing that $g(x)$ is integrable.

skullpatrol

@BabakS. Hello Babak, how are you?

BenjaLim

@robjohn Sorry for being like a stupid idiot

Babak S.

12:44

Fine! and you? Buy the way, why do you like skull? :-)

skullpatrol

Fine thank you.

robjohn

@BenjaLim It is a bit confusing. The concepts of integrable and measurable are often confusing.

@BenjaLim the key thing is that $\left|\int f(x)\,\mathrm{d}x\right|\le\int|f(x)|\,\mathrm{d}x$

BenjaLim

@robjohn The text we are using is Stein and Shakarchi. You see, I can see that $g(x)$ is integrable once I know that $F(x,t)$ is on the whole of $[0,1]²$

Babak S.

@skullpatrol:

robjohn

@BenjaLim so once we showed that $\int|f(x)|\,\mathrm{d}x\lt\infty$ we are good

skullpatrol

12:48

@BabakS. skullpatrol

BenjaLim

@robjohn In your answer above the first line you exchanged the order of integration

Babak S.

@skullpatrol: Thanks I got that

BenjaLim

How is that justified without knowing that $g(x)$ is integrable

robjohn

@BenjaLim everything is positive

BenjaLim

that's my problem now @robjohn

@robjohn Ahh I understand what you're doing now

robjohn

12:51

It's just like with series. If the series converges absolutely, you can rearrange terms

BenjaLim

we just need one of $\int dx dt$ or $\int dt dx $ to converge @robjohn

If at least one of them does

then we know the two are equal

robjohn

@BenjaLim and one does

BenjaLim

@robjohn So really in your answer above we cannot put an equals sign in the first line

robjohn

@BenjaLim the one on the right side of my first line is finite

BenjaLim

@robjohn Because a priori we don't know that they are equal

robjohn

12:54

@BenjaLim yes we can because the one on the right is finite, but if it weren't then neither would the one on the left, so we do know a priori that they are equal

BenjaLim

i don't understand

robjohn

If either is finite, then they are equal, right?

BenjaLim

yea ok I guess but aren't we trying to show that the right hand side is finite?

robjohn

suppose that one were not finite, then the other could not be finite... right?

BenjaLim

@robjohn Give me some time to relax my mind. I think it is clouded now :(

@robjohn Theorem 4.1 here is what I need

robjohn

12:57

so whether one is finite so that both are equal, or both are infinite, they are both equal

BenjaLim

@robjohn yea. But actually what you said about positivity is true. It is Tonelli's theorem

@robjohn I think I got confused by reading Stein and Shakarchi

@robjohn So I think all is ok now :D

robjohn

@BenjaLim Huzzah!

BenjaLim

@robjohn Fuck I can't believe I got confused over this shit

robjohn

@BenjaLim new stuff can get confusing.

BenjaLim

@robjohn If you haven't noticed my algebra is a lot more advanced than my analysis :D

robjohn

13:02

@BenjaLim I am the opposite

BenjaLim

@robjohn :D

@robjohn I should go now

@robjohn Thanks for your help.

robjohn

@BenjaLim np have a good day!

BenjaLim

@robjohn bye!

Peter Tamaroff

13:37

@JonasTeuwen Reminded me of you

9gag.com/gag/anqpvBz

gekkostate

14:55

Does anyone know where I can find a list of relatively prime numbers? E.g. The number of relative primes to 2,3,4 etc?

Peter Tamaroff

@gekkostate You mean $k$ such that $(k,n)=1$ yes?

Vrouvrou

15:51

hello

Gustavo Bandeira

16:34

My girlfriend made this drawing. Pretty much our actual situation.

Jayesh Badwaik

You are being tortured at the altar of mathematics?

Gustavo Bandeira

@JayeshBadwaik No. We're locked to it. Sad by not being able to solve some of the hardest problems.

Jayesh Badwaik

16:52

I see.

Dominic Michaelis

why are we locked naked ?

Gustavo Bandeira

@DominicMichaelis Because we're sexy as shit. We like girls looking us like that.

Jayesh Badwaik

@GustavoBandeira And why does the man have boobs on the back?

Gustavo Bandeira

@JayeshBadwaik Boobs? XD

Jayesh Badwaik

@GustavoBandeira Yes.

Gustavo Bandeira

16:58

Those are not boobs.

Jayesh Badwaik

Then?

Gustavo Bandeira

Dunno.

Must be some artistic resource to demonstrate those back bones.

Jayesh Badwaik

@DominicMichaelis How is quantum physics?

Dominic Michaelis

not that nice

I am confused

We surely know that the cantorset is closed hence $[0,1]\setminus C$ is open. So you can write it as a countable union of distinct open intervalls. How does that work when there is a point of the complement between every two points of the cantorset ?

Faust7

don't ask me m8

I just learned yesterday that -1*-1 = -1 lost in a whole new world

Dominic Michaelis

17:11

ah it is pretty much the same as with $\mathbb{R}\setminus \mathbb{Q}$ und $\mathbb{Q}$

Martin Sleziak

@DominicMichaelis They are precisely the intervals which you are omitting when you are creating the Cantor set. You omit 1 interval in the first step, 2 in the second, 4 in the third.

Countably many steps of the construction, finitely many intervals in each of them.

Faust7

i can't find the render latex in chat favorite?

Martin Sleziak

what yo mean by favorite?

Faust7

i was too drunk to figure it out last night n now i can't find it

it used to be starred on the right hand side fo this screen

Martin Sleziak

@Faust7 You can find those links on meta.

Jayesh Badwaik

17:17

Etiquette Guidelines | $\LaTeX$ support for chat

12

Star this, it went away earlier than I thought.

Martin Sleziak

BTW I've noticed that t.b. came back for election: see here.

Jayesh Badwaik

Yes, that seems true.

Martin Sleziak

He has not been here for a long time.

Faust7

i did star it

Jayesh Badwaik

@robjohn Please pin this. Thanks. :-)

@Faust7 Star this new post now. Stars are only valid of 15 days.

Faust7

17:29

done n done

skullpatrol

It would be so much better if the one of the owners put the link into the room description IMO

$\large\text{Mathematics}$

Associated with Math.SE; for both general discussion & math questions alike. Rarely if ever expressible as a ratio of integers. See below for guidelines.

Jayesh Badwaik

@skullpatrol We tried that before. It was a complete mess.

skullpatrol

@JayeshBadwaik It is better to have a mess than no link.

For new comers.

IMO

Faust7

start ChatJax

what does this mean?

skullpatrol

Click on it.

Faust7

17:41

that does not work

skullpatrol

Is it bookmarked on your bar?

Faust7

$x^{2}$

Dominic Michaelis

@faust7 you have to start it when you are on this site

not on another page

Faust7

i dont get it

skullpatrol

1 min ago, by skullpatrol

Is it bookmarked on your bar?

You must click on it to start it.

Faust7

17:45

my god that confusing

works now though

skullpatrol

It takes awhile to activate...

Faust7

not everyone is a programer needs more idiot proof instructions

skullpatrol

...in fact it freezes my browser every time :-(

Charlie

Hi @skull

skullpatrol

Hi @Charlie how are you?

Charlie

17:48

@skullpatrol fine fine

skullpatrol

@Charlie good good

Dominic Michaelis

When one writes id for identity, do you set it recto or italic ?

skullpatrol

In what context?

Dominic Michaelis

when you write a formula where id is a function

Charlie

@skullpatrol you?

Jayesh Badwaik

17:52

@DominicMichaelis I have seen it italics in some places.

skullpatrol

@Charlie Fine thanks.

Charlie

@DominicMichaelis hi herr Michaelis

Dominic Michaelis

@Charlie Hello Miss charlie :)

Charlie

@skullpatrol nice

skullpatrol

@DominicMichaelis Since it has a "special" meaning use italics.

Charlie

17:54

@DominicMichaelis wie gehts?

Dominic Michaelis

ganz gut und dir ?

Faust7

@ holy crap

skullpatrol

Who is "holy crap"?

Charlie

@DominicMichaelis ich weiss nicht

Transcript for

$Mathematics$ Mathematics