This is the Realm of Simply Beautiful Art - 2017-03-28

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1:00 AM

@SimplyBeautifulArt As amWhy said get some rest and lots of fluid in you! A home made recipe you make quick for rehydration is 5-6 teaspoon of sugar, 1/2-3/4 teaspoon of table salt mixed in a liter of water.

Drink and Repeat.

This will give a quick boost in energy levels and help restore electrolyte balance. Hope you feel better soon!

And make sure to get some high potassium food as soon as you can eat!

Don't over do the electrolyte drink thing as too much sugar or too much salt and even too much water is never a good thing...

3 hours later…

3:34 AM

@SimplyBeautifulArt It's Python 3. Didn't you use Python before? I did use some features, like list slicing and unpacking and duplication, which may be obscure to non-Python programmers. And get well soon! =)

3:49 AM

And it turns out that my program for the large Veblen ordinal was shorter mainly because I had used a trick, which also works to shorten the small one:

c=9**9
o=[]
i=[o]
def r(x):
	v=o
	y,*d=x
	z,*p=y
	t=z!=o
	if p==o:
		if t: v=[[r(z)]]*c
	else:
		v=[i]
		if t: v+=[[r(z)]+p]
		m=0
		while p[0]==o:
			m+=1
			_,*p=p
		a,*b=p
		for k in x*c:
			v=i*m+[v,r(a)]+b
			if v[-1]==o: *v,_=v
			v=[v]
	return v+d
def f(n):
	if n:
		v=[i*c+[[i]]]
		while v!=o:
			c*=c
			f(n-1)
			v=r(v)
			print(c)
for k in i*c: f(c)

3 hours later…

6:29 AM

@SimplyBeautifulArt get well soon.

3 hours later…

Simply Beautiful Art

9:23 AM

:-) thanks guys

5 hours later…

2:13 PM

0

Q: Inequalites withen Contours

$Proposition:$ $$\int_{0}^{\infty}\frac{1-cos(x)}{x^{2}}dx=\frac{\pi}{2}$$ Initially the function:$f(z)=\frac{1-e^{iz}}{z^{2}}$ was considered due to the fact we have a trigonometric term within our numerator.Our function $f(z)$ was integrated on the upper semicircle within our Contour,our Sem...

integration complex-analysis

@SimplyBeautifulArt I'll think you will love my question

Simply Beautiful Art

@Zophikel :-)

Simply ahh read your comment

but how does it involve the Triangle inequality ?

Simply Beautiful Art

|(a+b)/c| = |a+b|/|c| <= (|a|+|b|)/|c| @Zophikel

So the trinagle inquaility would be applied to our integrads

Simply Beautiful Art

2:28 PM

Well, by definition, you could argue that an integral is a sum + limit, and apply triangle inequality to the sum, but I don't think that's where the author was going

@Simply all right, nice point

because after letting $R \rightarrow \infty$ he then drops the inequality

Simply Beautiful Art

Actually, I would look at it like this: (give me a moment)

all right

Simply Beautiful Art

$$\left|\int_{\gamma\stackrel+R}f(z)\ dz\right|\le\pi R\max_{z=Re^{i\theta}}|f(z)|$$

@Simply interesting

but why bring the inequality with the integral where are you going with this

Simply Beautiful Art

2:34 PM

At most, the absolute value of any contour integral is less than or equal to the arc-length of the path multiplied by the maximum value of the absolute value of the function along the path

ahhh ok

but how does it apply in this case

Simply Beautiful Art

Notice that $|f(z)|\le\frac2{|z|^2}=\frac2{R^2}$ so that:

$$\pi R\max_{z=Re^{i\theta}}|f(z)|\le\frac{2\pi}R\stackrel{R\to\infty}\longrightarrow0$$

Thus, the integral tends to zero as $R\to\infty$ be squeeze or whatever

@SimplyBeautifulArt "be squeeze" ?

Simply Beautiful Art

oops, I meant to say "by squeeze" as in the squeeze theorem.

This trick works nicely for most limits of contour integrals as $R\to\infty$ or $r\to0^+$.

So basically our inequality equal to \frac2{R^2} since we are letting $R \rightarrow \infty$

Then you took the max of f(z)

Simply Beautiful Art

2:44 PM

Well, since $z=Re^{i\theta}$, it follows that $|z|^2=R^2$.

@Fawad Hello

Hi

@Simply you similar tricks like this for Contour Integrals ?

Simply Beautiful Art

@Zophikel Well, when you do it enough, you kinda just look at the function and the path and say "this obviously goes to zero"

true

Simply Beautiful Art

when its not so obvious, then I usually have to come up with better ways of doing it.

Which tends to be a case by case thing

@Zophikel Similar to Jordan's lemma: en.wikipedia.org/wiki/Jordan's_lemma

2:50 PM

@Simply sweet i'll have a look

Simply Beautiful Art

Note that Jordan's lemma is stronger than my simple example above, since my above example has the max value multiplied by $R$, which is increasing, making for a much looser limit.

For example, my above method cannot prove that the semicircle contour in the upper half plane tends to zero as $R\to\infty$ on the function $\cos(x)/(1+x)$ while Jordan's lemma can.

@Simply I think i'll be keeping Jordan's Lemma in my backpocket for latter

@Simply are there other tools like Jordan's Lemma

Simply Beautiful Art

I would imagine so, though my complex analysis isn't the best, so I can't make a lot of pointers.

@Simply your Complex Analysis is way better then mine

Simply Beautiful Art

Estimation lemma

In mathematics the estimation lemma, also known as the ML inequality, gives an upper bound for a contour integral. If f is a complex-valued, continuous function on the contour Γ and if its absolute value | f (z) | is bounded by a constant M for all z on Γ, then | ∫ Γ f ( z ) d z | ≤ M l ( Γ ) , {\displaystyle \left|\int...

The estimation lemma is the one I showed you above

2:57 PM

Ahhh ok

user image

@Simply your Goku in the Complex Pane

Is there a typo or they are really same?

@Fawad the integrals are the same answers are different

Simply Beautiful Art

@Fawad Hint: cosine and sine differ by a constant $\pi/2$.

3:00 PM

@Simply there's a complex analysis chat that nonone seems to be using

Simply Beautiful Art

well yeah I suppose

@Simply it's sadding :(

Simply Beautiful Art

Methods of contour integration

In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane. Contour integration is closely related to the calculus of residues, a method of complex analysis. One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. Contour integration methods include direct integration of a complex-valued function along a curve in the complex plane (a contour) application of the Cauchy integral formula application of the residue theorem One...

Methods of contour integration

In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane. Contour integration is closely related to the calculus of residues, a method of complex analysis. One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. Contour integration methods include direct integration of a complex-valued function along a curve in the complex plane (a contour) application of the Cauchy integral formula application of the residue theorem One...

I learned how to do the "square of the logarithm" method recently

@Simply how many methods are there to take a Contour Integral

Simply Beautiful Art

so many lol

3:05 PM

@Simply I remember searching arxiv for stuff on Contour Integration

@Simply aren't there research techniques involving Contour Integration

Simply Beautiful Art

I don't know

And what's with the sudden rush of people lol

@MithleshUpadhyay Hello and welcome to my realm

@Simply lol you should rename your realm

Simply Beautiful Art

To what?

@SimplyBeautifulArt how about Simply's Musesum

Simply Beautiful Art

Nah

3:13 PM

@Simply but I do have questions about the calculus of residues

Simply Beautiful Art

okie dokie then

What is a Residue ?

Simply Beautiful Art

Hm...

It is a value relating to the pole of a function

Residues and Cauchy's Residue Theorem

lol, can't decide if you want to stay @MithleshUpadhyay ?

Mithlesh Upadhyay

@SimplyBeautifulArt thanks; however, I am searching room to sleep ;) LOL

Simply Beautiful Art

Ah, okay

We have plenty of "homeless" folks stopping by here all the time

Mithlesh Upadhyay

3:30 PM

I will visit your this room for powernap.

@Simply you know any good books covering Calc of Residues

Simply Beautiful Art

No sadly @Zophikel

@Simply :(

Simply Beautiful Art

:-(

3:59 PM

Why the sad faces?

@Simply I have noticed that Contour Integration or at least the way you apply it can very

The examples I've seen have mainly been with Residue based tools

Simply Beautiful Art

@shredalert Probably because it is hard to smile when your stomach wants to kill you

@Zophikel Yup, because it is easy :P

@SimplyBeautifulArt You'll get better.

Simply Beautiful Art

Yeah...

@Simply aren't there programs that can do Contour Integration for you

Simply Beautiful Art

4:05 PM

Not that I know of

Other than WolframAlpha and Mathematica probably

You might want to try Sage @Zophikel. I heard it is very versatile.

all right @shredalert

Simply Beautiful Art

4:18 PM

Veblen function

In mathematics, the Veblen functions are a hierarchy of normal functions (continuous strictly increasing functions from ordinals to ordinals), introduced by Oswald Veblen in Veblen (1908). If φ0 is any normal function, then for any non-zero ordinal α, φα is the function enumerating the common fixed points of φβ for β<α. These functions are all normal. == The Veblen hierarchy == In the special case when φ0(α)=ωα this family of functions is known as the Veblen hierarchy. The function φ1 is the same as the ε function: φ1(α)= εα. If α < β , ...

Hm, I need to write a fundamental sequence for this. Tad bit annoyingly confusing as always lol

@SimplyBeautifulArt You can see my descriptions and programs for this. In other words you just take the syntactic form of the expressions involving the Veblen φ function, as the elements of the well-ordering itself, and then you can easily extract fundamental sequences from any element.

this chat room is scary

Simply Beautiful Art

4:33 PM

@abinjacob I agree

So what brings you here @abinjacob ?

@SimplyBeautifulArt, description says..Room for totally random people to hang, so i thought to try

Simply Beautiful Art

Lol, its supposed to be for the mathematically inclined btw

one big question for mathemeticians, how do random number works

Simply Beautiful Art

Lemme find that conversation in the transcript for you ;)

@abinjacob chat.stackexchange.com/transcript/51337/2017/3/20/12-22

@abinjacob If I understood Kolmogorov complexity right I can tell you this much. No computer which is programmed can ever produce truly random things. It's all deterministic according to Kolmogorov complexity.

4:45 PM

question should be asked by a mathemetician, because that conversion seem to be over my head

i have boookmarked it, may be in future

It all comes down to the definition of random.

We can't talk about if something is random or not unless we have a definition for it in the first place.

i once tried to program random number generated, it was deterministic according to my logic

According to the definition of random as in Kolmogorov complexity, @abinjacob you programmed a computer to produce something, therefore by definition, the process is deterministic.

Are we all part of a computer program which we cannot comprehend? Is everything therefore deterministic? DO WE HAVE NO FREE WILL???!?!?!!?

?

Don't mind me, I'm rambling.

4:51 PM

i think random number can be generated by taking random noise from atmosphere,

@shredalert I believe we have a will that is completely free in certain aspects and completely bound in others (such as due to the physical laws). This is necessarily non-mathematical and hence neither provable nor disprovable in any mathematical system whatsoever.

@abinjacob That is true if and only if the atmosphere is random.

@SimplyBeautifulArt Are you feeling any better today?

@user21820 is atmosphere not random, silence, catastrophe etc etc?

@abinjacob It is not a mathematical claim. For all practical purposes, a lot of it seems random. But a lot of natural chaotic phenomena also have orderly parts.

5:06 PM

We've got theories that describe these things and can make predictions about 'chaotic' systems, so there is some structure in this 'chaos'. Interesting philosophical point, I think.

Simply Beautiful Art

5:20 PM

@amWhy A tad bit

@SimplyBeautifulArt Bananas are good for an upset stomach.

Simply Beautiful Art

@shredalert Thanks, I'll try

5:38 PM

@SimplyBeautifulArt Especially the greener ones.

3 hours later…

Simply Beautiful Art

9:05 PM

Alexander Gruber ♦, Gainesville, FL

22.2k 24 95 159

group-theory abstract-algebra finite-groups linear-algebra combinatorics permutations reference-request

Great profile description

Huh, and someone upvoted me today

meta.stackexchange.com/users/-1/…

Heh, keeping a low profile I see...

9:51 PM

@SimplyBeautifulArt "a tad," or "a bit"... but "a tad bit" is redundant. (Those were the words I hear out of memory, words which with which Iwas repetitively called on, due to my former prevalent use of "a tad bit" many years ago!)

Simply Beautiful Art

lol

@Phyllotactic math.stackexchange.com/questions/2207682/proof-of-0x-0

Simply Beautiful Art

10:23 PM

math.stackexchange.com/questions/tagged/context-free-grammar

Context free grammar?

10:43 PM

@SimplyBeautifulArt ?

Simply Beautiful Art

@Phyllotactic If you (or anyone else who is actively in this room) post a question and I happen to see it, I link it here

how dare you laugh at my modest questions? :-)

Feel better today?

Simply Beautiful Art

Much better now

until I try to get on my two feet of course, but that's another story

Stomach bug or do you suspect something else?

Simply Beautiful Art

Probably that combined with some other stuff

10:57 PM

:-| that sucks... hope it resolves quick!

Simply Beautiful Art

yup

Broke my brain on that 0x=0 and I'm not satisfied, its so trivial I don't feel like I've done anything at all in the end even though I proved it

apparently it feels like a low hanging fruit because it is...

Simply Beautiful Art

That's the nature of the beginning of the foundations in all things

And I found a very extensive article on large numbers: sites.google.com/site/largenumbers/home/1-1

That's a great way to put it!

Very nice link!

I love the quote on top

Simply Beautiful Art

Haha, yes. Another quote I like is "most numbers are very very large"

11:05 PM

Haha nice! Wait...are there...more large numbers than small numbers?

the inverse of any large number is very small right?

Simply Beautiful Art

Well, suppose we restrict ourselves to the naturals

Then I agree with that quote

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and we use this concept called "density" when thinking about the reals or the rationals or something like that.

That quote used to be on the googology site before

does density means where numbers tend to "aggregate"

Simply Beautiful Art

Hm, I'm not entirely sure

11:08 PM

because the inverse of all those large numbers is in (0,1)

my intuition would be to say that all large positive integers are "spread out" evenly and their inverse are all "squezzed" in (0,1)

Simply Beautiful Art

lol

Most large positive integers have an inverse in (0,0.5)

makes sense to me since the smallest positive integer larger than 1 is 2 and the inverse is 0.5

should have thought about that

Simply Beautiful Art

lol

So there even more squezzed!

but are they?

since they're small numbers...

darn it, broke my brain again...

Simply Beautiful Art

lol

11:42 PM

but I see that it makes no sense for real numbers anyway...are all real numbers spread out evenly?

there are no holes in the real number lines so they're evenly distributed everywhere right?

Simply Beautiful Art

*shrugs*

Should have re-read the real number density theorem before asking...

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Transcript for

Mar '1728

$Mathematics$ This is the Realm of Simply Beautiful…

Room for totally bored people to hang. Open discussions.

abstract-algebra big-numbers geometry logic recreational-mathematics

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