10:00 PM
$$e = \lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}}.$$

@cassandra0 Euler's number, you mean

yeah

Eulers constant is $\gamma \approx 0.577$

@Argon not exactly :)

10:05 PM
@OldJohn No?

sorry - thought you had an = sign there :)

Oh

It really bugs me that we all know $\gamma$ is irrational - but nobody can prove it ...

Hahahaha!
Maybe it is not irrational!
That would be a cool surprise!

I need to show $$(1 + \frac{x}{n})^n < (1 + \frac{x}{n+1})^{n+1}$$

10:07 PM
@Argon that would be an amazing surprise
@cassandra0 could you prove that $f(z) = (1+x/z)^z$ is an increasing function of $z$?
(just a thought)

I'm not sure

might be possible to prove $f'(z)>0$ (it is clearly continuous)

I got it down to $\frac{(n+1)^2}{n} < \frac{(n+1+x)^2}{n+x}$

@OldJohn I had another programming contest Wednesday!

@Argon how did it go?

10:20 PM
@OldJohn Very good! My team was 50th out of a few hundred!

@Argon that is good

4th best in my school
And we have some 21 teams
But we didn't use C :)

What was the contest problem?

We got all but 1
#5, that is

@Argon okay.

10:26 PM
@JayeshBadwaik You are a master programmer, I'm sure!

@Argon Amateur might be a better description. :-)

@cassandra0 look at $$-n\log\left(1+\frac{x}{n}\right)=n\log\left(1-\frac{x}{n+x}\right)=-\frac{nx}{n+x}-\frac12\frac{nx^2}{(n+x)^2}-\frac13\frac{nx^3}{(n+x)^3}-\dots$$

@Argon C++. C I use only for hardware programming. And python for quick prototyping.

hmmm

10:29 PM
@JayeshBadwaik Do you know assembly language?

@Argon Cool. :-)

@Argon pic?

@robjohn PIC microcontroller

@Argon ah

@Argon Some of them, x86_64, ARMv7, now learning ARMv8. PIC I did once some two years ago, forget the commands now.

10:30 PM

@Argon :-) Which ones do you know?

I can't think what to do with that robjohn

Only asm programming I ever did was on 68xxx - it was a cool processor in its day

@OldJohn ahh, the reverse endian motorola. :-)

@JayeshBadwaik C/C++ (poorly), Java, PHP... Some FORTRAN IV, Python and Perl too

10:31 PM
@OldJohn I started on the 6502, then moved to the 680x0.. I wrote assemblers for each

@OldJohn I used to know some MASM

@Argon FORTRAN is a surprise there. :-)

@JayeshBadwaik big-endian. reverse is a matter of perspective...

@JayeshBadwaik From old numerical analysis books

@robjohn I had a student who wrote an assembler for Z80 - I would have hated to have done that :(

10:32 PM
I sadly can't find any compilers for it!!!

@robjohn yes.

@robjohn That is neat, sire.

does Linux not still have Fortrtan compilers?

@Argon gfortran77, gfortran90.

@OldJohn when Apple moved to the PowerPC, I stopped doing assembly and stuck with C, then moved to Java

10:33 PM
@OldJohn it does, HPC would die without it.

@robjohn I have a problem for you
I just did a lot of work
And I can use a little extra neurons.

I don't think they go that old

@robjohn yeah - after 68K, I moved to C - after a brief flirtation with Forth

@OldJohn I have been doing more and more C stuff for microcontrollers - it is very nice!

@OldJohn I played a bit with Forth and Lisp

10:35 PM
@Argon I guess they have a backward compatibility option with gfortran77.

@PeterTamaroff what? :-)

@JayeshBadwaik I will check... Hold on

@robjohn I thought Forth was fascinating - but ridiculously hard to read /debug at a later date :)

@OldJohn I was used to programming my HP calculators, so Forth was pretty easy after that

@Argon BTW, I strongly recommend to drop FORTRAN programming at this very instant and instead switch to C, C++.

10:37 PM
I did actually once write my own Forth compiler/interpreter in assembler - ran like a bat out of hell, but was never really finished

@JayeshBadwaik Hahaha! I know some C/C++, I only know FORTRAN for old books

@JayeshBadwaik I prefer Java, but that may be because I have not written anything big in C++

I don't think old FORTRAN works
@robjohn I like the "forced" OOP

@robjohn I have worked only on HPC as of now, so I prefer C++. I have read a lot of JAVA, never written it though.

nowadays I find that pari-gp does everything I need for number theory, so I have not done any real programming for some years

10:39 PM
@Argon hmm, drop it altogether, there are enough new books with much better algorithms

@JayeshBadwaik so we're closely opposite :-)

@JayeshBadwaik What's wrong with my old ones?

@robjohn :-)

@OldJohn I have used pari-gp, but I wrote an arbitrary precision math package in C that handles rationals and polynomials, so for anything more than that, I use Mathematica

Bleh, costly

10:41 PM
@Argon Priorities have changed since the FORTRAN times. Compilers have gotten much better than humans at optimizing register usage. Caches have begun playing an important part.

Well, I have proven the following things:
$(a)$ Let $s$ be a step function with parittion $P$ and constants $s_i$. Set $\int_a^b s=\sum_{i=1}^n s_i (t_i-t_{i-1})$. Then the integral doesn't depend on the partition $P$.
$(b)$ We say a function is **ruled** over $[a,b]$ if it is the uniform limit of a sequence of step functions over $[a,b]$. In this case there exists for each $\epsilon>0$ an $N$ such that for every $m,n>N$ we have $|s_m(x)-s_n(x)|<\epsilon$ for each $x$ in $[a,b]$
$(c)$ The sequence $$\int_a^b s_n$$ will be Cauchy.

years ago, I wrote a set of arbitrary precision maths routines in Pascal - I still shudder when I look back on the time it took to debug the long-division part - it was HIDEOUS

@Argon Some algorithms which were not much useful in times of costly memory and comparatively cheaper CPU are now quite useful and vice-versa.

@OldJohn One should be glad of the FPUs then!!

10:43 PM
@JayeshBadwaik I didn't have one at the time - back in the late 80s, I think

@OldJohn Yes, hence I meant one should be glad that we have FPUs now.
@Argon Interesting

@OldJohn God bless the Fast Fourier Transforms.

I seem to have done most things the hard way - I recall installing Slackware linux in the 90's from a stack of 70 floppy discs :)

@PeterTamaroff what does $\int t_a^b f$ mean?

10:45 PM
@robjohn It was supposed to be $$\int_a^b f$$

Ah, I was just thinking that there was one too many t's

@robjohn Yeah

@PeterTamaroff This looks a lot like the definition of the Lebesgue integral

@robjohn ME LOVES IT.
@robjohn Could you tell me about it?
I have seven legal size pages with all that work

@PeterTamaroff The Lebesgue integral is about the same, but the step functions are replaced by simple functions, which are linear combinations of indicator functions.

10:50 PM
@robjohn But a step function is a linear combiation of indicator functions

@JayeshBadwaik IF (A .LE. 0) GOTO 2

@PeterTamaroff are your step functions combinations of indicators of intervals? or any measurable set?

@robjohn Precisely!

@PeterTamaroff how did that reply happen before the comment it replies to?

@Argon I am off to call Dijkstra.

10:52 PM
@robjohn I have my secret ways

@PeterTamaroff If it were edited, I could understand, but it doesn't appear edited.

@robjohn You did edit it
But let's focus on the maths!

@Argon Wrath

@PeterTamaroff I edited mine, but that shouldn't change the order. Like this will stay before Argon's comment

10:53 PM
Hahahaa!
I was doing it to bug you
I never use GOTOs ever

@Argon :P :P

ever ever ever

In assembly, you are going to need it though, there is no option there.

@robjohn It didn't swap in my page

@JayeshBadwaik Very true. But higher level languages can just have functions

10:54 PM
@PeterTamaroff It shouldn't.
@Argon procedures and functions (and procedures are essentially functions that don't return anything)

@robjohn Subroutines

@robjohn Paraphrasing Milhouse "I'm more preoccupied about the Lebesgue integral".

Yes, even then, goto should be restricted to very specific parts of assembly.
It should almost never cross the boundary of the routine except for the very basic initialization etc.
For everything else, you should prefer to CALL.

@JayeshBadwaik Java gets by pretty well without goto

@robjohn I think goto is reserved but not functional

10:57 PM
@Argon yes

Who invented GoTo

@robjohn Same with const

@robjohn Yes.
@Argon No. Const on the other hand is very important in C++ at least.

@JayeshBadwaik I mean it is reserved but not functional in Java
In Java, const = final

@Argon okay.

10:58 PM
But "const" is still a keyword that is reserved

@Argon static final is used instead of const. "static final" is used for actual compile time constants.

yup.

In the Java programming language, a keyword is one of 50 reserved words that have a predefined meaning in the language; because of this, programmers cannot use keywords as names for variables, methods, classes, or as any other identifier. Due to their special functions in the language, most integrated development environments for Java use syntax highlighting to display keywords in a different color for easy identification. List The following is a list of the keywords in Java, along with brief descriptions of their functions: ;abstract :The abstract keyword is used to declare a class or...
"Although reserved as a keyword in Java, const is not used and has no function"

@robjohn Have you forsaken me?
What I want to prove now is every continuous function is ruled.

@robjohn $\text{const}\neq \text{final}$
Const is useless
Final is a constant

11:01 PM
@PeterTamaroff I didn't know there was more about your integral. Was there a question?

@Argon In C++11, there is even a constexpr which evaluates a possible expression at compile time itself. .

@Argon yes, I know. I mistyped and was correcting, but I needed to reply to Peter :-)

@JayeshBadwaik Hm.... So could you do the following:
constexpr int a;
a = 5;
@robjohn :)

3 mins ago, by robjohn
@Argon static final is used instead of const. "static final" is used for actual compile time constants.

@robjohn Yep.

11:03 PM
Bad practice in C++. :-) RAII. Also, I think with constexpr it might be illegal.
you can do like this constexpr a = 1234567*23234234/342343243

Good night all!
@JayeshBadwaik Can you assign values to constants after declaration, like Java?

@PeterTamaroff was there?

Like I wrote

@Argon good night!

@Argon That would be simple const, yup.
constexpr is for compile time.
@Argon its night at your place?

11:05 PM
@robjohn Yes
@robjohn I have to show every continuous function is ruled.

@PeterTamaroff can't you just use a partition?

@robjohn Hmm?

@PeterTamaroff can't you make a partition and set the value on each interval to be the maximum of the continuous function?
@PeterTamaroff then refine the partition.

@robjohn Guess so. Spivak proposes the following:
To find a step function $s$ over $[a,b]$ with $|s(x)-f(x)|<\epsilon$ consdier the set of all such $y$ for which there exists such step function over $[a,y]$
The idea is that if $A$ is such set, then $a\in A$ (we just take $s=f(a)$) and $A$ is bounded whence it has a supremum
What I guess I should show is that $\sup A=b$
and $b\in A$?

@PeterTamaroff what does the set of $\{x:f(x)>a\}$ look like?

11:12 PM
@robjohn What do you mean by "look like"?¡

@PeterTamaroff does it look like a union of open intervals?

@robjohn Well, yes

@PeterTamaroff so could you use that to base your step functions? uniformly?

@robjohn base? Like "is based on"?

@PeterTamaroff use those open intervals to construct your step functions.
You can just mimic the Lebesgue integral definition using continuous functions instead of measurable ones.

11:19 PM
@robjohn Could you hint me how? I mean, I guess it shouldn't be too hard, but I'm not sure.

@PeterTamaroff divide the domain of $f$ into $S_n(k)=\{x:\frac kn< f(x)\le\frac{k+1}n\}$

@robjohn Am I crazy in saying that $f$ is ruled iff it is integrable?
@robjohn Oh, I see.
I'm going to make some pretty drawings now =)

@PeterTamaroff I am still uncertain of what it means for a function to be ruled. I thought I knew, but looking back at your comment, it does not tell how $f$ relates to the $s_i$ and $x_i$.

@robjohn Ruled means it is the uniform limit of a sequence of step functions.
No more, no less.
(over a closed interval)

I have to go afk for a while,

11:26 PM
@robjohn OK. I'll try and do something for when you get back
@robjohn are you assuming $f>0$ there?
@robjohn OK, I think I'm starting to get it!

@PeterTamaroff Cool! I stopped by before heading up to clear the gutters.

@robjohn Oh. Autumn?

@PeterTamaroff pre-Winter, and it has been raining.

@robjohn I will assume $f>0$ for simplicity.

@PeterTamaroff yeah then split it into positive and negative

11:35 PM
Then split I split $[\inf f,\sup f]$ into $n$ equal parts for each $n$
and consdier the intervals I get
over the $x$ axis

@PeterTamaroff that's the idea

then define the step function as the max or min over those intervals
now this pic makes sense

user19161
@PeterTamaroff Although you don't need me, I have decided to forsake you.

@WillHunting On what premises?

@PeterTamaroff yes, but I was thinking of making the vertical deltas equal in size and making the step functions be linear combinations of the indicator functions

11:39 PM
@robjohn yes, OK.

@PeterTamaroff yours will approximate the Riemann integral, whereas what I am suggesting will approximate the Lebesgue integral.

user19161
Wow, Zhen Lin used the word epiphany in his question, so deep.

@PeterTamaroff Riemann partitions the domain, and Lebesgue partitions the range
4

@WillHunting lol

user19161
@GustavoBandeira He then uses aka, to spoil it all.

11:41 PM
@robjohn but... my indicator functions will be on the $x$ axis right?
You mean your indicator functions will be over the $y$ axis?
@WillHunting where?

@WillHunting He should have started it with: "Yo nigga!"
2

user19161

@PeterTamaroff no, the functions are on the x-axis like normal functions, but they are evenly spaced in $y$, but they look very messy in $x$.

@WillHunting bleh
@robjohn messy? because the $x$ axis gets partitioned strangely right?

@PeterTamaroff yes.
I am 76% to the Epic badge :-)

11:45 PM
Epi =(0.76)Epic

@robjohn so for each $n$ I partition $[m_f,M_f]$ into $n$ equal parts and consdier the indicator functions over the sets $\{x:k/n <f(x)\leq (k+1)/n\}$
@robjohn What is that for?

user19161
Wow @amwhy I see you got 275 points today!

I didn't know that question would raise all that upvotes.

@WillHunting Yeah, I've been doing better with "accepts"; frustrating, though, to miss the points from upvotes once you exceed 200 upvote-points.

user19161
@amWhy Hehe, I got 300 today if there is no cap!

11:48 PM

There's a cap for daily rep?

user19161
@amWhy Sort of ridiculous!

This is conspiracy!

@WillHunting Yeah, I'd have 375 points if no cap! It is ridiculous.

user19161
@amWhy Oh no, I think if there is no cap Arturo would have 500k by now.

11:50 PM
@GustavoBandeira Yes, you can earn 200 points per day, max, from upvotes. "Accept" points don't count in that cap!
@WillHunting Yes, he would. What happened to him?

user19161
@amWhy Well, just left the site I guess. Nothing to explain.

Burn out?

@WillHunting Hmmm. Interesting. It seems Brian M has taken his spot!

user19161
It's not like this is some job where you get paid for answering questions!

user19161
11:52 PM
@amWhy Also, the other 100k guys.

@WillHunting Yes, we need to get a "gal" in their ranks!

user19161
@amWhy It will be A M Why!

@WillHunting Hahahahaha. A long long way off...

user19161
@amWhy Hmm, no doubt you will attain 10k by this year.

@PeterTamaroff It gives you a uniform approximation of your function

11:54 PM

@WillHunting too bad there weren't a few days left this month =)

@PeterTamaroff facepalm It is for reaching 200 points on 50 days

user19161
@amWhy Too bad we can't reverse time and right the wrongs.

@WillHunting So many wrongs to right!

user19161

11:55 PM
@robjohn =)
@robjohn 200 points of what?

user19161
@amWhy Did you watch the movie Butterfly Effect?

user19161
@PeterTamaroff Reputation.

Oh, like keeping up 200 each day?

@amWhy: Wow
@PeterTamaroff It does not need to be consecutive

@robjohn I wanna see mine!

user19161
11:57 PM
@PeterTamaroff You can navigate yourself!

@robjohn Never saw the graph before =)

@PeterTamaroff go to your profile. Select "network profile" then "reputation"

user19161
@PeterTamaroff Let me show you mine.

@robjohn I took a long time off, as you can see.

user19161
I have three lines there. =)

The zooming is pretty dope

@amWhy that explains the long plateau

@amWhy How young are you?