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1:00 PM
but I wonder why not something more
 
user19161
@RajeshD Because they approximate many things up to that order. And many things are quadratic, like gravity etc.
 
yes
@robjohn : With me its generally the other way, "Theorems disappear and coffee appears"
 
@robjohn That is my oil 8-).
 
regarding the pinned comment, I wonder we are accountable for anything we do in the 61'st second?
 
user19161
@JonasTeuwen I would rather have a girl than oil. But I can survive without one.
 
user19161
1:09 PM
@RajeshD Depends on your definition of accountability. Not a real question.
 
user19161
And hey my Chrome still does not load chat properly.
 
stuck with an old version?
 
user19161
No, using latest chrome, but maybe it does not work so well on linux.
 
ubuntu ?
 
user19161
It's alright actually, only chat gives problems so I may see a message 9000 years after it is posted.
 
1:11 PM
how is the weather in 11012?
 
user19161
@RajeshD Nope. You should know what distro I use by now.
 
Maybe it was a message posted 9000 years ago.
 
Fedora!
 
To Tukulti-Ninurta, say thus, Abbashi-Bidathyur says, I asked you for twenty oxen and fourteen pots of beer to be sent to Ur, so why haven't you sent them?
 
user19161
@RajeshD And why are you still typing the space before the ? bro?
 
user19161
1:13 PM
@RajeshD Wrong.
 
Man are you Clark?
 
yes
 
I don't know any more distros.
 
user19161
@RajeshD You may get a hint if you read a little earlier in the transcript.
 
@JasperLoy Fedora?
 
1:15 PM
hahhaha
lol
 
user19161
@FrankScience Wrong.
 
@Jasper : I don't remember asking you 9000 years ago!
 
Fedora?
 
@JasperLoy Rajesh said that.
@JasperLoy I get used to linux.
 
guys you don't read transcript
 
user19161
1:17 PM
Is someone going to say Fedora a fourth time?
 
Okay, any other distro we herewith deem it as equivalent to Fedora
 
lol, you're asking about 90000000 years ago.
Ubuntu
 
I said that first
 
user19161
Haha, now you guys are just trolls!
 
ok suse? or redhat?
Jasper the trollifier!
like a mollifier
 
1:19 PM
Here is not a good place to discuss linux distro.
 
user19161
OK I use the distro named after a guy and his girlfriend.
 
Whoever always discuss off-topic might be ignored.
 
@FrankScience : When no other main topic is going on then no problem i guess.
 
user19161
Anyway I think you all know by now. If you still don't I'll tell you this Christmas. Remind me then.
 
@JasperLoy To tell you the truth, the first chapter of The Art of Computer Programming, in which some section is named after Mathematical Preliminaries, really confused me a lot.
 
1:24 PM
I'll freshen up bbl
 
So analysis of algorithms need a lot of mathematics.
 
Yesterday I was so bored that I decided to go to a random theatre and watch any movie that is playing there, too my luck it was Spiderman (the amazing) (just released that day). Its a good movie
 
really beyond my scope of math study in high school.
 
@FrankScience : Are you in a high school or an undergrad?
 
@RajeshD between.
 
1:28 PM
So you are going to Univ. this fall?
 
Yeah, this autumn.
 
By your attitude you kind a reminded me of the pre-university days, when I was just about to enter
Univ.
 
I can't get it.
 
Your fresh approach and enthusiasm reminded me of my pre-college days
 
Question: I want to learn knowledge about intergration and infinity series first, so can I only take an overview of applications of differentiation in geometry?
Any advice?
It seems that the applications of differentiation in geometry is not being interesting me a lot.
So I wonder whether it is important for me to prepare to learn the preceding topic: integration, infinity series.
Calculus
 
1:40 PM
integration, infinite series and sequences and calculus you will obviously get a good grip on them with ease. What do you mean by differentiation in Geometry?
@Frank
 
@RajeshD For example, curvature.
 
of curves in 2 and 3 dimensions?
 
you're gonna want to get the hang of basic analysis and calculus before you do anything multivariable or geometric
 
In my calculus book, multivariable is before integration and series.
 
No problem skip 'em
 
1:43 PM
I'm eager to learn integration and series, and the applications of multi-variable differentiation in the geometry doesn't attract me.
Thanks. I want to skim over.
 
Haven't you learned integration (to some extent) in high school?
 
Not systematically.
 
skimming screws one's mind -Anonymous
bbl
 
This means that I will leave some theorem unproved by myself and some examples skipped.
I have a problem now.
Is the function $f$ exists?
 
Does
 
1:57 PM
$\lim_{x\to x_0}f(x)$ exists whenever $x_0$ is arbitrary real number, but $f$ is discontinuous at every point.
Incidentally
 
hmm
 
How can I express $\forall$ in English?
 
"for all"
 
What about whenever?
and for each?
 
those both work too
 
1:58 PM
I found them in CMath
and what about the adj arbitrary?
 
it's sometimes ambiguous whether "arbitrary" / "fixed" is a universal or an existential quantifier
 
Well.
Now let's consider whether $f$ exists.
I found that such $x_0$ could be dense on $\Bbb R$, but I don't know whether it can be more.
The strongest is that, $x_0$ is everywhere on $\Bbb R$.
The weaker one is almost everywhere.
 
I always had the same doubt about 'arbitrary'
@Frank : What do you mean by $x_o$?
is it a set or a real variable?
 
As far as I see, arbitrary is a universal quantifier.
@RajeshD Well, I'll explain it rigorously.
 
density is usually defined for sets
ok go ahead
 
2:10 PM
Supposing that $D$ is the discontinuous points of $f$, and $\lim_{x\to x_0}f(x)$ exists whenever $x_0\in\Bbb R$.
 
just mentally interchange $x_0$ with "such $x_0$" (i.e. $\{x_0:\dots\}$), yeash
 
ok
 
I found that $D$ could be dense on $\Bbb R$. For example, $D=\Bbb Q$, and $f(x)=0$ when $x\not\in\Bbb Q$, and $f(x)=1/q$ when $x=p/q$, $q>0$ and $p\perp q$.
 
Oh I missed the 'such', it makes sense now
 
My question is whether $D$ could be more dense on $\Bbb R$, saying everywhere or almost everywhere?
Notation $a\perp b\overset{\textrm{def}}\iff\gcd(a,b)=1$.
 
2:14 PM
.
 
almost everywhere means that the probability is $1$.
 
confusing what you want to ask.
 
everywhere means $D=\Bbb R$.
 
Do you know what is meant by dense, i mean definition?
 
almost everywhere means its complement has measure zero
 
2:18 PM
$D$ is dense: For all $x_0\in\Bbb R$ and $\epsilon>0$, there exists $d\in D$ such that $|d-x_0|<\epsilon$. Am I right?
 
right, just to add it is defined for any metric space.
 
The word more dense is informal.
 
it is defined for any topological space, even in the absence of a metric
 
Now what do you mean by more dense, or almost/everywhere, If "for all x_o" means it is dense everywhere. for some $x_o$ means well its not dense oevr entire set but there are places where it is dense
@anon ok, i wasn't ture of that earlier
 
@RajeshD I've showed that it is possible that $D$ is dense. I wonder whether it is possible for $D=\Bbb R$ or $D$ is almost everywhere on $\Bbb R$.
Now is it clear enough?
 
2:24 PM
@FrankScience read separable space if you want to know about almost everwhere
i am having dinner
 
separability seems irrelevant to me. almost everywhere is measure-theoretic.
 
well yes
but lets not go there, for now
almost everywhere means here we define countable dense set
 
@RajeshD Well, but it's possible to prove some set $A$ is almost everywhere through elementary mathematics.
@RajeshD For example
 
General description of functions which have limit at every point is given in Exercise 7.K in van Rooij-Schikhof.
 
just read separable space @Frank it doen't take more t5han 10-15 min
 
2:27 PM
Let $L$ be the set of all functions $f\colon [0,1]\to\mathbb R$ that have the property that $\lim_{x\to a} f(x)$ exists for all $a \in [0, 1]$.
Show that:
(i) $L$ is a vector space. Each $f \in L$ is bounded.
(ii) For each $f \in L$, define $f^c(x): = \lim_{y\to x} f(y)$ ($x \in [0, 1]$). $f^c$ is continuous.
(iii) '$f^c =0$' is equivalent to 'there exist $x_1, x_2, \dots$ in $[0,1]$ and $a_1, a_2,\dots$ in $U$ with $\lim_{n\to\infty} a_n = 0$, such that $f(x_n) = a_n$ for every $n$, and $f=0$ elsewhere'.
(iv) Describe the general form of an element of $L$. Show that every $f\in L$ is Riemann integrable.
 
@RajeshD For all $\epsilon>0$, there exists an union of intervals, saying $A_\epsilon$, such that $A\subseteq A_\epsilon$.
 
@FrankScience Is the part (iv) the thing which you were asking about?
 
Hey guys =)D
Its a smiley with a manly beard.
 
With the exception that this is about real functions on $[0,1]$ and you asked about $\mathbb R$.
 
@MartinSleziak The difference is trivial.
 
2:29 PM
Does anyone have any intuitive explanation for fix-point iteration? I know what it is and I understand it. But I want to give some friends of mine a good exploitation =)
 
@MartinSleziak Seems good.
@N3buchadnezzar Y-combinator?
 
exploitation or explanation ?
 
@MartinSleziak I want to know whether it is possible that $f$ is discontinuous everywhere.
 
Every $f$, which has limit at each point, has the form: continuous function + function described in (iii).
 
$Y=\lambda x.f\ (x\ x)\ \lambda x.f\ (x\ x)$.
 
2:32 PM
So this is the same as asking whether a function which has the form described in (iii) can be discontinuous everywhere.
 
Coxeter came to Cambridge and gave a lecture [in which he stated a] problem for which he gave proofs for selected examples, and he asked for a unified proof. I left the lecture room thinking. As I was walking through Cambridge, suddenly the idea hit me, but it hit me while I was inthe middle of the road. When the idea hit me I stopped and a large truck ran into me....
So I pretended that Coxeter had calculated the difficulty of this problem so precisely that he knew that I would get the solution just in the middle of the road.... Ever since, I’ve called that theorem “the murder weapon.” One consequence of it is that in a group if $a^2=b^3=c^5=(abc)^{-1}$, then $c^{610}=1$.
---John Conway
 
@Frank I think the only discontinuities can be the points $x_n$. So it has at most countably many discontinuities.
 
@MartinSleziak Thanks
@MartinSleziak So $D=\Bbb Q$ is somewhat most dense.
@MartinSleziak Let me think how to prove (iii)
 
somewhat most dense sounds somewhat vague
 
@MartinSleziak I'll take a bath, and when finishing, I hope I'll get an idea.
 
2:39 PM
@MartinSleziak Explanation =)
 
guys what do you mean by "somewhat more dense"? I really do not understand. Please define it first before we have a huge confusion
 
wow : the consequence, I did not know group theory has such interesting problems
@anon
 
it's amazing how banal, mechanistic symbolic things like group presentations and diophantine equations can easily get very complicated, and even undecidability rears its head
 
2:57 PM
Hmm
Okay, is there any reason why Newton.Rhapsons approximation method fails for $$ f(x) = x - 2.5 \exp(-x/2) $$ ?
I know the $f(1.30392) \approx 0$ using fixed point iteration, and that $f'(2)=0$, but that should not cause problems ?
 
Hi people!
 
$f\,'(2)=0$?
$f\,'(x)=1+1.25 \exp(-x/2)>0$; the derivative never vanishes
 
@anon ops, I meant that given $g(x) = 2.5 \exp(-x/2)$ then $g'(2)=0$
I am supposed to approximate $g(x)=x$
 
@MartinSleziak Fortunately, it's not so complicated. Let me have a try to describe it.
@MartinSleziak I can only solve 1 through 3, because I've no idea about itegration.
 
3:03 PM
okay, what makes you think NR fails?
 
Sigh, once again I am horrible at typing in chat
 
First, we extend $[0..1]$ to $\Bbb R$ as this way: for $x_0<0$, we have $f(x_0)=\lim_{x\to0^+}f(x)$, and for $x_0>1$, we have $f(x_0)=\lim_{x\to1^-}f(x)$.
 
careful, you don't wanna use NR on $g$ when you need to use it on $f$
 
$g(x) = 2.5 x \exp(-x/2)$. Need to approximate $g(x)=x$. Solution: Set $f(x)=g(x)-x$, and use NR.
 
good. problem?
 
3:05 PM
@Frank Why don't I post it as the question on the main and you can write your solution there...?
 
@anon I wrote a script and tested it with a few initial values, it blows up every time.
 
It would be useful for other people.
And it would be much more readable than in chat.
 
@MartinSleziak Seems good.
 
What do you say?
 
@N3buchadnezzar you sure you have it as $\displaystyle x_{n+1}=x_n-\frac{x_n-2.5\exp(-x_n/2)}{1+1.25\exp(-x_n/2)}~?$
 
3:07 PM
@MartinSleziak How can I title it? I don't know what is appropriate.
 
@FrankScience I thought I am going to post the question and you the answer...?
 
@anon I multiplied the equation by $\exp(-x_n/2)$ to simplify it, obtaining.
 
@MartinSleziak and the concept vector space is out of my ability.
@MartinSleziak I don't know that.
@MartinSleziak Pretty well.
 
$\displaystyle x_{n+1} = x_n - \frac{1}{2} \frac{5x_n - 2 e^{x_n/2} }{ 5x_n - 10 + 4e^{x_n/2}}$
render
 
\displaystyle
or use double dollar signs
 
3:09 PM
@user34522 What do you want to ask?
 
why is there an $x_n$ on the bottom? why is there a constant 10?
 
@user34522 Your post
 
@N3buchadnezzar that shit ain't makin any sense a'all.
 
@anon You know I restated that $g(x) = 2.5 x \exp(-x/2)$ right?
 
yes
still looks like a trainwreck to me
 
3:11 PM
brb dinner
 
@anon How to use fleqn in $$..$$?
 
$g'(x) = 2.5 \exp(-x/2) - 1.25 \exp(-x/2)$
 
oh, there's an $x$ in front of exp
no wonder
I'm totally blind.
 
@FrankScience I've posted that problem here.
 
@MartinSleziak seen
 
3:14 PM
Please, could somene help me in doing this?
$$e^{x-|x^2-x|}\left[1-\frac{|x^2-x|}{x^2-x}\left(2x-1\right)\right] > 0$$
 
The exponential function is always positive.
Also $|x|/x=\operatorname{sgn}x$.
 
sgn?
 
@N3buchadnezzar what did you say you multiplied by?
 
So you are solving $\operatorname{sgn} (x^2-x) (2x-1) < 1$.
sgn stands for sign function.
It should be easy to plot the graph of $\operatorname{sgn} (x^2-x) (2x-1)$.
Also, don't forget that 0 and 1 are not in the domain of your function.
 
the way I see it, it should be $x_n-2.5x_n\exp(-x_n/2)$ in the numerator, so an $x_n$ factors clean out
@N3buchadnezzar also, $g\,'(x)=2.5\exp(-x/2)-1.25\color{Red}x \exp(-x/2)$
 
3:19 PM
@unNaturhal But basically you can do it separately for $x<0$, $x\in(0,1)$ and $x>1$. The whole expression is much simpler after that.
 
@MartinSleziak I'm sorry, but I haven't understand... I never used before the sgn function..
 
Then you can ignore the sign function.
Simply try to do the three cases.
And don't forget you can omit the exponential function.
 
four cases: (-inf,0), (0,1/2), (1/2,1), (1,inf)
 
@MartinSleziak I've already omitted it, 'coz it's always > 0
 
ok, or four cases
I've suggested the three cases which would help him to get rid of the absolute value.
And solve a linear inequality after that.
 
3:23 PM
Wait please
From where they came the four cases?
 
|x^2-x| is always positive (we assume x is not equal to 0 or 1) so it can be discarded
you're left with x(x-1)(2x-1), the zeros of which are x=0,1,1/2
wait, nevermind
forgot the 1- thing..
 
Almost always when you have some expression in absolute value, $|f(x)|$ then you can consider cases $f(x)>0$ and $f(x)<0$ to get a simpler expression.
Since for $f(x)>0$ you can write $f(x)$ instead of $|f(x)|$.
Similiarly, if $f(x)<0$ you can write $-f(x)$ instead of $|f(x)|$.
So, for example, if $x\in(0,1)$, you can write $-(x^2-x)$ instead of $|x^2-x|$.
The $x^2-x$ cancel out and the whole expression is much simpler.
 
Ok wait, I think there is an underlying problem :p
I try to explain: I'm studying this function: $$y = e^{x-|x^2-x|}$$
To study the monothony, I got the first derivatives
Which is what I wrote before
I put the first derivative > 0
And now I have the problem
Because before of this, I tried to split the function in three cases:
$$y = \left\{ \begin{array}{rcl}
e^{x^2} & \mbox{for}
& x < 0 \\ 1 & \mbox{for} & x = 0 \\
e^{2x - x^2} & \mbox{for} & x > 0
\end{array}\right.$$
So to leave the absolute value
 
But that's not correct.
 
Hi @BrianMScott
 
3:31 PM
But I really don't know how to continue, because I never used the method before
 
Hullo, Rajesh.
 
been a while since you were here
 
@unNaturhal You should consider cases $x<0$, $x\in(0,1)$ and $x>0$.
 
@RajeshD Yes, life’s been a bit complicated.
 
If you want to divide it into three cases.
Life of Brian - i think I've seen a movie about it.
 
3:33 PM
@MartinSleziak Why $\in (0, 1)$?
 
Complicated, indeed.
You have to look when $x^2-x$ is $>0$ and when $<0$.
 
@MartinSleziak Yeah, exactly
 
But maybe easier is this.
 
@MartinSleziak I’ve even seen it, though only because a young Dutch mathematician insisted that had to!
 
You have $f(x)=e^{g(x)}$. So $f'(x)=f(x) \cdot g'(x)$.
 
3:35 PM
@MartinSleziak Yes..
 
So in fact you only have to find the derivative of $g(x)=x-|x^2-x|$.
 
And it's what I do :p
 
You still probably would have to do three cases, but at least you don't have to write it in exponent.
@unNaturhal Very well, then.
 
@MartinSleziak ...?
 
If that's what you do, you're doing exactly what you should be doing.
That was a nice sentence.
I just wanted to say that you're on the right path.
 
3:37 PM
@MartinSleziak No I mean that I got the derivatives of the exponend multiplied for $f(x)$
 
ok
And do you know how to find $g'(x)$?
The derivative of the exponent.
 
Yeah. It's a derivative of a sum. so I got:
$$D[x]-D[|x^2-x|]$$
That is:
$$1- D[|x^2-x|] = 1 - \frac{|x^2-x|}{x^2-x}(2x-1)$$
Right?
 
It seems correct.
With the exception of the point 0 and 1, of course. (Where this expression is undefined.)
 
So I have to do the domain of the first derivative as first thing?
 
Well, I just wanted to remind you that you have to think bout the points $x=0,1$ separately.
 
3:45 PM
@MartinSleziak Separately?
 
But I agree that the result should be that derivative of the original points does not exists.
@unNaturhal You have used chain rule. One of the assumptions of this theorem that both derivatives appearing in the formula exist.
Since absolute value does not have derivative at 0, chain rule cannot be used in the points where $x^2-x=0$.
 
Mmmh
So, I'm wrong..
 
Sorry if I'm confusing you. I am not too good in explaining things.
 
@JonasTeuwen ah, then there is no need for adage adjustment.
 
@MartinSleziak :p
 
3:50 PM
I still think that considering three function (for $x<0$, $x\in(0,1)$ and $x>1$) might be simpler.
But the final result should be the same.
 
The problem is that I haven't understood how to get the final result
 
Maybe I should have written $x\le0$, $x\in[0,1]$ and $x\ge 1$.
@unNaturhal Your result is correct (if I did not miss something.)
I.e. you have $f'(x)=f(x) \left(1 - \frac{|x^2-x|}{x^2-x}(2x-1)\right)$.
Now the question is whether this can be simplified.
 
@MartinSleziak Mmmmh
I tried to simplify, but it seems to not be simplified
 
Perhaps it is matter of taste, but I would consider the function written as $$f'(x)=\begin{cases} \ldots & x<0 \\ \ldots & x\in(0,1) \end{cases}$$ nicer.
You have $f'(x)=-2xf(x)$ for $x<0$.
You can simplify it for the remaining cases, too.
What is the exercise asking to do BTW. Just to calculate the derivative?
 
@MartinSleziak Nope, ty study the entire fuction.. I got: Domain, Simmetry, Intersection with x=0 and y = 0, Asymptotes, and now I'm trying to do the first derivative
 
3:58 PM
ok
You can write the first derivative in the form you wrote (with $|x^2-x|/(x^2-x)$), but I think that the form with 3 cases is nicer.
You can even graph the derivative much easier.
 
@MartinSleziak done
@MartinSleziak here
 
I'll have a look Frank.
 

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