Simply Beautiful Art's realm of calculus and analysis

Simply Beautiful Art

20:15

Good day @BrevanEllefsen

Typhon

@Nilknarf I wouldn't recommend reading it. It's old garbage from a few years ago.

Nilknarf

@Typhon Oh :(

Typhon

the math is a bit trivial in hindsight and the exposition is... not real good

idk

Nilknarf

@SimplyBeautifulArt I'm trying to do some analysis with the Lambert W function

Typhon

for me, hindsight is incredibly critical

Simply Beautiful Art

20:23

WHOA

Where'd everyone come from!?

x.x

@Nilknarf what of it?

Nilknarf

@SimplyBeautifulArt Just trying to prove some properties

Simply Beautiful Art

@Typhon Using 300 bytes, can you make a number larger than I can in 101 bytes?

Ah, okay @Nilknarf

Typhon

well of course

Nilknarf

@SimplyBeautifulArt Derivative, antiderivative

Typhon

i can add 1

XD

Simply Beautiful Art

20:26

@Typhon Oh, but I'm not going to show you my number until you show me yours

Typhon

@Nilknarf mine is more about ways to integrate piecewise constant functions.

Simply Beautiful Art

@Nilknarf Ah, okay

Nilknarf

But I'm stuck on
$$\int_0^\infty W\bigg(\frac{1}{x^2}\bigg)dx$$

Simply Beautiful Art

@55Cancri If you want, you can try the challenge as well. Make a larger number than I can, where you use 300 bytes and I'll use 101 bytes.

Typhon

@SimplyBeautifulArt nah. You are way better than me. I could probably improve functions, but you are the big numbers guy. I'm the "differential equations/geometry/logic/number theory" guy.

big numbers don't intrigue me in that sense

Simply Beautiful Art

20:27

@Nilknarf Oh deary me, one of those integrals I see

Typhon

there's no use for them aside from the thrill of the pursuit

Nilknarf

@SimplyBeautifulArt I know the answer thanks to wikipedia, but I can't figure out how to get it

Simply Beautiful Art

Well... technically, they can be used in proof theory, but I imagine you also don't care about that @Typhon

Typhon

@Nilknarf my paper would be helpful, but I honestly don't know the piecewise constant resulting. Read the section on implied integration. It would be quite helpful.

Nilknarf

@Typhon Ok

Typhon

20:29

@SimplyBeautifulArt actually, I like proofs a lot. Writing them and such. However, that's probably not what you mean.

Simply Beautiful Art

@BrevanEllefsen Do you know how to do the above integral?

Typhon

wait...

that's 1/x^2 isn't it?

well that's fine. Just focus on finding the antiderivative to the right of 0

Simply Beautiful Art

@Typhon Prove there exists a total computable function that cannot be proven total in Peano Arithmetic.

Nilknarf

@SimplyBeautifulArt Would it help if I told you what Wikipedia said the answer is?

Simply Beautiful Art

No thanks @Nilknarf

Typhon

20:33

@SimplyBeautifulArt I don't know "total computable" and I don't know "Peano Arithmetic"

regardless, I have to go soon

Simply Beautiful Art

Well, than I'll have to see you go soon @Typhon

Typhon

any thoughts on this though? math.stackexchange.com/questions/2325668/…

Simply Beautiful Art

@Typhon Nope, not my thing xD

Typhon

im thinking it might relate to some kind of curvature algebra rather than linear algebra

Nilknarf

@Typhon I think triangular matrices are a thing

Typhon

20:34

@Nilknarf umm...

those are just square matrices that have been reduced

they aren't literally a triangular grid

but im mostly thinking of ones shaped like surface geometries

Nilknarf

Well whatever

Oh.

Typhon

so like a 'matrix' that is literally shaped like a torus.

Simply Beautiful Art

9

Q: Evaluation of an integral involving the Lambert W function

Wikipedia claims that $$\int_0^\infty W\left(\frac{1}{x^2}\right) \,\text dx=\sqrt{2\pi}$$ and a numerical computation seems to confirm this. How can this result be proven?

Typhon

i surmise that they're algebrae will relate in some way to the geodesics of their respective surfaces

and also their concept of circles

in other words: i believe that the algebra of a matrix relates inherently to the surface geometry it is written upon assuming that you don't just project a euclidean one.

Simply Beautiful Art

Similarly,

17

Q: Does $\int_0^{\infty} \left( p + q W \left( r e^{- s x + t} \right) + u x \right) e^{- x} d x$ have a closed-form expression?

Does $\int_0^{\infty} \left( p + q W \left( r e^{- s x + t} \right) + u x \right) e^{- x} d x$ (with 6 variables) where W is the Lambert W function (also known as ProductLog in Mathematica) have a closed-form expression? If we drop the variable $s$ from the expression Maple is able calculate $$ ...

integration closed-form lambert-w

Typhon

20:37

i think that the rows and columns will distort in some way

shrugs

maybe its just a quack thought

the other guy used to think those a lot. I got rid of him. He was annoying.

Nilknarf

@Typhon A matrix shaped like a torus? How would that even work?

Typhon

@Nilknarf that's the point of the question essentially.

how would it work and has anyone studied it

there is an implied sense of being... creative.

i mean that in a healthy sense. I imagine that there is a logical means by which one could think out a way of interpreting matrix multiplication with toroidal matrices.

perhaps it might even lead to ways of expressing different exotic isometries

matrices do not represent the isometries of torus's very well

i mean... except a few trivial ones.

Nilknarf

@Typhon Toroidal matrices seems like a bit of a jump. Perhaps it would be best to try easier surfaces first.

Typhon

@Nilknarf the question is pretty general. So... to be fair one could try anything that they see as easy for them. I phrased it so that while it is definitively answerable, one can somewhat come up a general idea. :-)

Avantgarde

oh hey typhon

Typhon

20:44

matrices relate to vector which relate to points. So it might also be possible that a non-euclidean matrix gives a way of parameterizing points that map properly to other types of surfaces.

idk really

it might also end up like number theory:

Simply Beautiful Art

Hello and welcome to my realm? @Avantgarde

Typhon

totally and 100% useless

Avantgarde

@SimplyBeautifulArt Hello and thank you!

Typhon

@Avantgarde we are discussing math.stackexchange.com/questions/2325668/…

Simply Beautiful Art

Do I know you from somewhere? @Avantgarde

Just curious

Avantgarde

20:45

@SimplyBeautifulArt We talked the other day in the math chat

For a bit

Simply Beautiful Art

Ah, okay

You looked familiar :P

Avantgarde

yeah, I didn't know this room was yours

Simply Beautiful Art

:P

Typhon

@Avantgarde It isn't. It is the property of Stack Exchange.

Simply Beautiful Art

@Typhon Bleh! SHOO GOVERNMENT SPY!

Avantgarde

20:47

√_√

Typhon

@SimplyBeautifulArt hahaha. Typhon is the government.

Simply Beautiful Art

ಠ_ಠ

Avantgarde

time for new elections then

Simply Beautiful Art

I swear I already paid my taxes

Typhon

ROFLMAO

@SimplyBeautifulArt that face. It reminds of something from a while ago. Wow. That is an old reference

how did you get an image of True Valhalla's avatar from the old game maker forums?

Simply Beautiful Art

20:50

>_ಠ
He's still here...

Typhon

or was there a link to that image on the wiki I showed you?

Nilknarf

( ͡° ͜ʖ ͡° )

Simply Beautiful Art

in The Nineteenth Byte, 54 mins ago, by totallyhuman

ಠ_ಠ

Avantgarde

ß_ß

Typhon

...is that a common emote or something?

Simply Beautiful Art

20:51

Appears so

Typhon

oh...

i thought that was the guy's custom art. XD

nvmd then

@SimplyBeautifulArt even your taxes on the illegal ugg boot trafficking?

Simply Beautiful Art

Θ_Θ

Typhon

we know about that

@SimplyBeautifulArt ok that's just creepy

Simply Beautiful Art

* ::cocks the shotgun:: *

Typhon

im sorry nvmd

wait...

Simply Beautiful Art

20:52

* ::points to the door:: *

Typhon

grabs shotgun and breaks it in half

Nilknarf

Ok, I gotta go

Simply Beautiful Art

Oh, okay, cya @Nilknarf

Typhon

Shotguns. don't. work. on. me.

XD

alrighty. Enough goofing around.

Nilknarf

:: backs away slowly and slips out the back door::

Simply Beautiful Art

20:53

Cya ‮@Nilknarf

Typhon

@Nilknarf I said enough goofing around.

Simply Beautiful Art

:3

ε:

ε:3

Typhon

ew

quit making kissing faces

Simply Beautiful Art

xD

Typhon

you're a man

act manly.

Simply Beautiful Art

20:54

._.

Typhon

show off your big numbers

Simply Beautiful Art

I'm a horse

Typhon

wait...

Avantgarde

¢_¢

Typhon

is displaying big numbers like flexing big muscles?

anyway

bye

Avantgarde

20:55

50¢

everybody left

Simply Beautiful Art

21:11

Lol

Well, what do you like to do? @Avantgarde

Avantgarde

quite a few things @SimplyBeautifulArt Primarily diving into oceans

Simply Beautiful Art

xD What do you like to do that's math related?

Avantgarde

physics

Simply Beautiful Art

Ah, okay

Avantgarde

∆_∆

Simply Beautiful Art

21:16

∩_∩

Avantgarde

$\rightarrow_\rightarrow$

$\rightarrow$_$\rightarrow$

Simply Beautiful Art

._.

Avantgarde

$\mathcal N$_$\mathcal N$

Simply Beautiful Art

$$\huge\overbrace{\left(\ddot{\stackrel{\quad>}{\smile}}\right)}_{\begin{align}---‌‌\qquad\ \end{align}}$$

Avantgarde

God

Simply Beautiful Art

21:31

$$\huge \overbrace{\left(\frown-\overline{\overline{\boxed\circ}}\underline{\ >}\overline{\boxed\circ}\right)}$$

Brevan Ellefsen

21:57

@SimplyBeautifulArt You messaged me about an integral? Which integral?

Simply Beautiful Art

Nvm, we solved it

It was this one:

9

Q: Evaluation of an integral involving the Lambert W function

Wikipedia claims that $$\int_0^\infty W\left(\frac{1}{x^2}\right) \,\text dx=\sqrt{2\pi}$$ and a numerical computation seems to confirm this. How can this result be proven?

integration special-functions definite-integrals lambert-w

Brevan Ellefsen

@SimplyBeautifulArt I remember reading that post a while back. Integrals with W function are quite interesting.

Simply Beautiful Art

Yes they are...

Brevan Ellefsen

I think I am going to try to crack this integral next week

4

Q: An integral of a rational function of logarithm and nonlinear arguments

This problem was posted in I&S $$ \int_{0}^{1} \dfrac{\log x \log (1+x) \log (1+x+x^{2})}{(1-x)(1+x^{2})}\,dx \approx -0.223434$$ I am not sure if there exists a closed form but it seems worth trying. I am completely clueless on how to start with this beast. It is worth saying that $1-x^3= (1-...

calculus integration definite-integrals special-functions

Simply Beautiful Art

:-) Oh dear

Brevan Ellefsen

22:00

I can split it into 4 integrals trivially

8 of them with a little work

where I can reduce the denominator to a quadratic

in each integral

then I just have to find a way to express everything in terms of Beta Function and its derivatives

Simply Beautiful Art

Mhm...

Beta functions lead to Gamma functions which lead to digamma functions I presume?

Brevan Ellefsen

Yup

I've been splitting my time between researching integrals like it to study up on known techniques (such as learning a bunch of Euler sums)

as well as studying Complex Analysis rigorously

it is amazing what the Heine Borel theorem can do

Simply Beautiful Art

Btw, I had recently thought up a neat trick

Let $f(x)=\lim\limits_{n\to\infty}f(x,n)$ and $f(x,y)$ be a smooth function of two variables. Then,
$$\sum_{k=1}^nf(x,k)=nf(x)+\sum_{k=1}^\infty f(x,k)-f(x,k+n)$$

Now the RHS is continuous with respect to $n$, so you can differentiate or integrate it

Haven't found it particularly useful yet though.

Hm... one could try Taylor expanding the Harmonic numbers...

$$H_n=\sum_{k=1}^n\frac1k=\sum_{k=1}^\infty\frac1k-\frac1{k+n}$$

$$H_0=0\\H_0'=\zeta(2)\\H_0''=-2\zeta(3)\\\vdots$$

$$H_n=\sum_{k=2}^\infty(-1)^k\zeta(k)n^{k-1}$$

Huh, nifty...

Brevan Ellefsen

Ugh, not on a comp with ChatJax

gotta install it

Simply Beautiful Art

22:16

xD

More or less, I think I just solved the power series with $\zeta(k)$ as coefficients

Woo! That's... actually the Taylor series of the Harmonic numbers (or the related digamma function)

Brevan Ellefsen

Wow

Yeah

I recognized it immediately

Simply Beautiful Art

xD Well that was easy

Brevan Ellefsen

that is a clean proof right there

Simply Beautiful Art

Was honestly too easy :P

Brevan Ellefsen

Why the $nf(x)$ in the sum above

I don't see why a term would show up multiplied by $n$

did you perhaps mean $f(x,n)$? Or am I missing something obvious

Seems like the sum is just fine without that term

Simply Beautiful Art

22:31

@BrevanEllefsen Check how the telescoping works

If $n\in\mathbb N$, then you should get $$\sum_{k=1}f(x,k)-f(x,k+n)=\sum_{k=1}^nf(x,k) -\lim_{t\to\infty}\sum_{i=t}^{t+n}f(x,i)$$

If $f(x)=\lim_{i\to\infty}f(x,n)$, then the last sum comes down to $nf(x)$

Brevan Ellefsen

Odd. When I see $$\sum_{k=1}^\infty f(x,k)-f(x,k+n)$$ I think $$f(x,1)-f(x,n+1)+f(x,2)-f(x,n+2)+\cdots$$ which I would write as $$\left[f(x,1)+f(x,2)+\cdots f(x,n)\right]+\left[f(x,n+1)+f(x,n+2)+f(x,n+3)+\cdots\right] - \left[f(x,n+1)+f(x,n+2)+f(x,n+3)+\cdots\right] = \left[f(x,1)+f(x,2)+\cdots f(x,n)\right] = \sum_{k=1}^n f(x,k)$$

@SimplyBeautifulArt where did my logic go wrong there?

Simply Beautiful Art

@BrevanEllefsen You need to rewrite it as a partial sum

$$\sum_{k=1}^tf(x,k)-f(x,k+n)$$

This way, you won't be able to do the silly trick where you ignore the last few terms

Brevan Ellefsen

Ooooh, I see what you are saying

duh

I'm an idiot -_-

Simply Beautiful Art

* ::facepalm:: *

Brevan Ellefsen

Sounds like it is time for me to take a walk outside and get my head clear and working again

Simply Beautiful Art

22:42

Okay

But before you leave me...

I'm trying to extend this to cases when $f(x,n)$ doesn't have a limit.

Brevan Ellefsen

mmhmm

Simply Beautiful Art

Yeah... I don't have many good ideas...

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