Mathematics - 2012-11-30 (page 6 of 6)

cassandra0

22:00

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

Argon

@cassandra0 Euler's number, you mean

cassandra0

yeah

Argon

Eulers constant is $\gamma \approx 0.577$

cassandra0

ah

Old John

@Argon not exactly :)

Argon

22:05

@OldJohn No?

Old John

sorry - thought you had an = sign there :)

Argon

Oh

Old John

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

Argon

Hahahaha!

Maybe it is not irrational!

That would be a cool surprise!

cassandra0

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

Old John

22:07

@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)

cassandra0

I'm not sure

Old John

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

cassandra0

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

Argon

@OldJohn I had another programming contest Wednesday!

Old John

@Argon how did it go?

Argon

22:20

@OldJohn Very good! My team was 50th out of a few hundred!

Old John

@Argon that is good

Argon

4th best in my school

And we have some 21 teams

But we didn't use C :)

Jayesh Badwaik

What was the contest problem?

Argon

@JayeshBadwaik There were 5

We got all but 1

#5, that is

Jayesh Badwaik

@Argon okay.

Argon

22:26

@JayeshBadwaik You are a master programmer, I'm sure!

Jayesh Badwaik

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

Argon

@JayeshBadwaik What is your weapon of choice? C?

robjohn

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

Jayesh Badwaik

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

cassandra0

hmmm

Argon

22:29

@JayeshBadwaik My Dad recently built a pic programmer :)

@JayeshBadwaik Do you know assembly language?

Jayesh Badwaik

@Argon Cool. :-)

robjohn

@Argon pic?

Argon

@robjohn PIC microcontroller

robjohn

@Argon ah

Jayesh Badwaik

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

Argon

22:30

@JayeshBadwaik Cool!!!!

Jayesh Badwaik

@Argon :-) Which ones do you know?

cassandra0

I can't think what to do with that robjohn

Old John

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

Jayesh Badwaik

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

Argon

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

robjohn

22:31

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

Argon

@OldJohn I used to know some MASM

Jayesh Badwaik

@Argon FORTRAN is a surprise there. :-)

robjohn

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

Argon

@JayeshBadwaik From old numerical analysis books

Old John

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

Argon

22:32

I sadly can't find any compilers for it!!!

Jayesh Badwaik

@robjohn yes.

Peter Tamaroff

@robjohn That is neat, sire.

Old John

does Linux not still have Fortrtan compilers?

Jayesh Badwaik

@Argon gfortran77, gfortran90.

robjohn

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

Jayesh Badwaik

22:33

@OldJohn it does, HPC would die without it.

Peter Tamaroff

@robjohn I have a problem for you

I just did a lot of work

And I can use a little extra neurons.

Argon

@JayeshBadwaik For FORTRAN IV?

I don't think they go that old

Old John

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

Argon

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

robjohn

@OldJohn I played a bit with Forth and Lisp

Jayesh Badwaik

22:35

@Argon I guess they have a backward compatibility option with gfortran77.

robjohn

@PeterTamaroff what? :-)

Argon

@JayeshBadwaik I will check... Hold on

Old John

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

robjohn

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

Jayesh Badwaik

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

Old John

22:37

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

Argon

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

@JayeshBadwaik gcc.gnu.org/fortran

robjohn

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

Argon

I don't think old FORTRAN works

@robjohn I like the "forced" OOP

Jayesh Badwaik

@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.

Old John

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

Jayesh Badwaik

22:39

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

robjohn

@JayeshBadwaik so we're closely opposite :-)

Argon

@JayeshBadwaik What's wrong with my old ones?

Jayesh Badwaik

@robjohn :-)

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

Argon

Bleh, costly

Jayesh Badwaik

22:41

@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.

Peter Tamaroff

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.

Old John

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

Jayesh Badwaik

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

Argon

@JayeshBadwaik Hmmm

Jayesh Badwaik

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

Old John

22:43

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

Jayesh Badwaik

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

@Argon Interesting

Old John

@JayeshBadwaik absolutely

Jayesh Badwaik

@OldJohn God bless the Fast Fourier Transforms.

Old John

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 :)

robjohn

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

Peter Tamaroff

22:45

@robjohn It was supposed to be $$\int_a^b f$$

robjohn

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

Peter Tamaroff

@robjohn Yeah

robjohn

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

Peter Tamaroff

@robjohn ME LOVES IT.

@robjohn Could you tell me about it?

I have seven legal size pages with all that work

robjohn

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

Peter Tamaroff

22:50

@robjohn But a step function is a linear combiation of indicator functions

Argon

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

robjohn

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

Peter Tamaroff

@robjohn Precisely!

robjohn

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

Jayesh Badwaik

@Argon I am off to call Dijkstra.

Peter Tamaroff

22:52

@robjohn I have my secret ways

Argon

@JayeshBadwaik Algorithm?

robjohn

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

Peter Tamaroff

@robjohn You did edit it

But let's focus on the maths!

Jayesh Badwaik

@Argon Wrath

robjohn

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

Argon

22:53

@JayeshBadwaik Ya, I know

Hahahaa!

I was doing it to bug you

I never use GOTOs ever

Jayesh Badwaik

@Argon :P :P

Argon

ever ever ever

Jayesh Badwaik

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

Peter Tamaroff

@robjohn It didn't swap in my page

Argon

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

robjohn

22:54

@PeterTamaroff It shouldn't.

@Argon procedures and functions (and procedures are essentially functions that don't return anything)

Argon

@robjohn Subroutines

Peter Tamaroff

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

Jayesh Badwaik

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.

robjohn

@JayeshBadwaik Java gets by pretty well without goto

Argon

@robjohn I think goto is reserved but not functional

robjohn

22:57

@Argon yes

a Dangerous Idea

Who invented GoTo

Argon

@robjohn Same with const

Jayesh Badwaik

@robjohn Yes.

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

Argon

@JayeshBadwaik I mean it is reserved but not functional in Java

In Java, const = final

Jayesh Badwaik

@Argon okay.

Argon

22:58

But "const" is still a keyword that is reserved

robjohn

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

Jayesh Badwaik

yup.

Argon

List of Java keywords

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"

Peter Tamaroff

@robjohn Have you forsaken me?

What I want to prove now is every continuous function is ruled.

Argon

@robjohn $\text{const}\neq \text{final}$

Const is useless

Final is a constant

robjohn

23:01

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

Jayesh Badwaik

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

robjohn

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

Argon

@JayeshBadwaik Hm.... So could you do the following:

constexpr int a;

a = 5;

@robjohn :)

robjohn

3 mins ago, by robjohn

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

Argon

@robjohn Yep.

Jayesh Badwaik

23:03

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

Argon

Good night all!

@JayeshBadwaik Can you assign values to constants after declaration, like Java?

robjohn

@PeterTamaroff was there?

Argon

Like I wrote

robjohn

@Argon good night!

Jayesh Badwaik

@Argon That would be simple const, yup.

constexpr is for compile time.

@Argon its night at your place?

Peter Tamaroff

23:05

@robjohn Yes

@robjohn I have to show every continuous function is ruled.

robjohn

@PeterTamaroff can't you just use a partition?

Peter Tamaroff

@robjohn Hmm?

robjohn

@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.

Peter Tamaroff

@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 $b\in A$?

robjohn

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

Peter Tamaroff

23:12

@robjohn What do you mean by "look like"?¡

robjohn

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

Peter Tamaroff

@robjohn Well, yes

robjohn

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

Peter Tamaroff

@robjohn base? Like "is based on"?

robjohn

@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.

Peter Tamaroff

23:19

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

robjohn

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

Peter Tamaroff

@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 =)

robjohn

@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 $x_i$.

Peter Tamaroff

@robjohn Ruled means it is the uniform limit of a sequence of step functions.

No more, no less.

(over a closed interval)

robjohn

I have to go afk for a while,

Peter Tamaroff

23:26

@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!

I made a pretty pic

robjohn

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

Peter Tamaroff

@robjohn Oh. Autumn?

robjohn

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

Peter Tamaroff

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

robjohn

@PeterTamaroff yeah then split it into positive and negative

Peter Tamaroff

23:35

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

robjohn

@PeterTamaroff that's the idea

Peter Tamaroff

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.

Peter Tamaroff

@WillHunting On what premises?

robjohn

@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

Peter Tamaroff

23:39

@robjohn yes, OK.

robjohn

@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.

robjohn

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

4

Gustavo Bandeira

@WillHunting lol

user19161

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

Peter Tamaroff

23:41

@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?

Gustavo Bandeira

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

2

user19161

@PeterTamaroff Here.math.stackexchange.com/questions/248316/…

robjohn

@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 $x$.

Peter Tamaroff

@WillHunting bleh

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

robjohn

@PeterTamaroff yes.

I am 76% to the Epic badge :-)

a Dangerous Idea

23:45

Epi =(0.76)Epic

Peter Tamaroff

@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!

Gustavo Bandeira

I've made a question that rendered me 8 upvotes.

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

amWhy

@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!

amWhy

23:48

I see your functional iteration question got 21 upvotes (last count)...

Gustavo Bandeira

There's a cap for daily rep?

user19161

@amWhy Sort of ridiculous!

Gustavo Bandeira

This is conspiracy!

amWhy

@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.

amWhy

23:50

@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.

a Dangerous Idea

Burn out?

amWhy

@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!

Gustavo Bandeira

@amWhy He died in the street

user19161

23:52

@amWhy Also, the other 100k guys.

amWhy

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

user19161

@amWhy It will be A M Why!

amWhy

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

user19161

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

robjohn

@PeterTamaroff It gives you a uniform approximation of your function

Peter Tamaroff

23:54

@robjohn I meant the badge

amWhy

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

robjohn

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

user19161

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

amWhy

@WillHunting So many wrongs to right!

user19161

@PeterTamaroff headdesk

Peter Tamaroff

23:55

@robjohn =)

@robjohn 200 points of what?

user19161

@amWhy Did you watch the movie Butterfly Effect?

user19161

@PeterTamaroff Reputation.

Peter Tamaroff

Oh, like keeping up 200 each day?

robjohn

@amWhy: Wow

@PeterTamaroff It does not need to be consecutive

Peter Tamaroff

@robjohn I wanna see mine!

user19161

23:57

@PeterTamaroff You can navigate yourself!

amWhy

@robjohn Never saw the graph before =)

robjohn

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

user19161

@PeterTamaroff Let me show you mine.

user19161

stackexchange.com/users/510689/will-hunting?tab=reputation

amWhy

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

Peter Tamaroff

23:58

I'm a steady guy stackexchange.com/users/1188301/peter-tamaroff?tab=reputation

¿?

user19161

I have three lines there. =)

Peter Tamaroff

The zooming is pretty dope

robjohn

@amWhy that explains the long plateau

Peter Tamaroff

@amWhy How young are you?

Transcript for

$Mathematics$ Mathematics