Simply Beautiful Art's realm of calculus and analysis

Zophikel

01:16

@amWhy o/

Dibbs

04:07

Thanks for the answers to my -1/12 question Nilknarf and Simply!

So, Nilknarf, regarding that proof about the relationship between the Riemann zeta function and the integral from -1 to 0 of those functions, do you know for sure that such a proof is actually something that can be done or were you proposing the idea just as a possibility?

Simply Beautiful Art

11:39

@Nilknarf

@Dibbs Good morning

@Dibbs I believe I have found said relationship

user400188

12:14

$poof$ Im here

Simply Beautiful Art

Hello lol

Have you not done anything with set theory?

user400188

not much at all

Ive had the inklings on what it is from the rest of math and a few pointers on definitions in it but thats it

Simply Beautiful Art

Do you know any of the nature of ordinals and cardinals?

user400188

I tried to learn it once but eventualy i realised I was more a fan of logic itself and higher things like catagory thoery (sine it skips a lot of the work in set thoery)

I've covered ordinals before but it was a while ago and I havent studied them since

Simply Beautiful Art

Lol

user400188

12:17

so you can assume no knowlage

Simply Beautiful Art

You know of $\omega$ and all that?

Hm, well, the basic principle of how I'm using ordinals is as follows:

user400188

know of and know are two different things. I have the former.

Simply Beautiful Art

We have a class of functions $f_\alpha$. I define it as follows:
$f_0(n)=1$
$f_{\alpha+1}(n)=n\times f_\alpha(n)$
$f_\alpha(n)=f_{\alpha[n]}(n)$

And so,
$f_0(n)=1$
$f_1(n)=n$
$f_2(n)=n^2$
$f_3(n)=n^3$
$\vdots$

$f_\omega(n)=n^n$
$f_{\omega+1}(n)=n^{n+1}$
$f_{\omega+2}(n)=n^{n+2}$
$\vdots$

$f_{\omega a+b}(n)=n^{na+b}$

user400188

hmm ok

one question

if $f_α(n)=f_{α[n]}(n)$

then does $f_{α+1}(n)=n×f_α(n)=n×f_{α[n]}(n)$ ?

Simply Beautiful Art

The third rule is meant for limit ordinals.

One cannot take $\alpha[n]$ if $\alpha=\beta+1$, which means we would use the second rule

$\alpha[n]$ denotes the $n$th term of $\alpha$'s fundamental sequence.

The usual FS is given by:
$\omega=\sup\{ 1,2,3,4,\dots\}$

Thus, $\omega[n]=n$

$f_\omega(n)=f_n(n)=n^n$

$f_{\omega+1}(n)=nf_\omega(n)=n^{n+1}$

And so the goal from here on out is "what's the largest ordinal we can construct?"

user400188

12:27

so the $f_α(n)$ in the secont rule is different from the one in the third?
also, what does sup stand for?

Simply Beautiful Art

$f_{\omega^2}(n)=f_{\omega n}(n) = n^{n^2}$

Yes, the second and third rules cannot occur simultaneously

user400188

hmm ok

Simply Beautiful Art

Supremum

Kinda like a maximum + a limit

$\sup(A) = \min\{\gamma : \gamma\ge\delta\forall\delta\in A\}$

For example,
$1=\sup\{0.9 ,0.99, 0.999, 0.9999, \dots\}$

user400188

google defines it as the smallest quantity that is greater than or equal to each of a given set of quantities. So basicly the maximum element in the set; but defined using the concept of smallest greater element

Simply Beautiful Art

As $1$ is the smallest number larger than all elements of that set

@user400188 Ha, yes, that is my definition. But its not equal to the maximum element

user400188

12:30

ah yes your right

Simply Beautiful Art

Note that $\max\{0.9, 0.99, 0.999, 0.9999,\dots\}$ does not exist

As $\max$ finds the largest element in a given set

user400188

so when we say $\omega[n]=n$ the set $\{1,2,3,4,...\}$ does not contain $n$

Simply Beautiful Art

But clearly there is no largest, as the supremum of the set is not within the set itself

@user400188 No, we are saying $n$ is the $n$th term of $\omega$'s fundamental sequence

As you can see, $1$ is the first number in the set that defines $\omega$. $2$ is the second, $3$ is the third...

user400188

but if ω=sup{1,2,3,4,…}, and ω=n then shouldnt be not included in that set?

Simply Beautiful Art

No. By nature, $\alpha[n]$ is in the set, as $\alpha[n]$ is the $n$th term of the set

And $\omega\ne n$

@Mathmore Hello and welcome to my realm

user400188

12:35

so whats the difference here between ω=sup{1,2,3,4,…} and ω={1,2,3,4,…}
in both cases it appears that ω[1] would be 1 and ω[n] would be n
unless the []'s are not conventional brackets, and stand for something else

Mathmore

Thanks!

user400188

welcome @Mathmore

Mathmore

Hello!

Simply Beautiful Art

Its just different ways to write the same thing

Mathmore

Thanks!!

Simply Beautiful Art

12:36

Which is the problem when trying to use ordinals in this manner

Mathmore

supremum of the set {1,2,3,4,...} ?

Simply Beautiful Art

If we don't define how an ordinal is broken down i.e. we don't give it a fundamental sequence, then we can't use it quite well

@Mathmore Yes, we are doing ordinally stuff

user400188

1=sup{0.9,0.99,0.999,0.9999,…}. Doesnt that imply that a single set has just 1 value when we take the sup of it? As opposed to the multiple we obtain when we write ω[n]

Mathmore

hahah!

user400188

@Mathmore as you can see, I am rarther new to this stuff

Mathmore

12:38

Okay. So you say that the supremum of the set of naturals exist?

Simply Beautiful Art

@Mathmore Von Neumann ordinals. The supremum of every set of ordinals is another ordinal, except for the supremum of all ordinals

Mathmore

Okay. I will have to study this. I can't comment anything.

Simply Beautiful Art

No no, that's not what I meant. For example,
$$1=\sup\left\{ \frac12, \frac12+\frac14, \frac12+\frac14+\frac18, \dots \right\}$$

^ As you can see, there are infinitely many sets who's supremum is $1$.

Likewise, there are infinitely many set who's supremum is $\omega$

To make any sense of $\omega[n]$, we have to assign it a "fundamental sequence" to which we may pull values from

@Mathmore Its open discussion, we learn as we go/chat, so feel free to join in, even if you don't know much

user400188

so $\omega$ is a number and not a function, and $\omega[n]$ is something that starts at the first element of the set it is defined for and has a sequence associated with it

Simply Beautiful Art

Well, it is the case that:
$$\alpha=\sup\{\alpha[1], \alpha[2], \alpha[3], \dots\}$$

And yes, $\omega$ is a "number", as are all ordinals

user400188

12:43

hmm, so $\alpha[1]$ here would be $\alpha[1]$ and so forth for each element?

Simply Beautiful Art

Mhm

user400188

and $\alpha$ on its own is a number associated with the whole thing

Simply Beautiful Art

Yes

user400188

cool, Ive understood the notation

Mathmore

@SimplyBeautifulArt That's cool.

Simply Beautiful Art

12:44

Though for me to create any finite numbers out of this, I can't just say "$\alpha$" for some really large $\alpha$ without defining $\alpha[n]$

user400188

makes sense. $\alpha$ only appears to be a label for the thing on the right.

Simply Beautiful Art

Well, its mostly a label for me. For the purposes of large finite numbers, I mostly need syntactic ordinals

That is, ordinals defined by syntax. For example, I could've just said $\omega[n]=n$ without ever introducing the concept of sets and fundamental sequences

user400188

I see I'm going to get used to a lot of notation differences

hmm actualy, the notation doesnt idffer, it just leaves a lot out

Simply Beautiful Art

Anyways, these ordinals are useful for their well ordering as well:
$0<1<2<3<\dots <ω<ω+1<ω+2< \dots<ω2<ω2+1< ω2+2<\dots<ω3<\dots < ω^2<\dots$

$$\omega2=\sup\{\omega+n| n\in\mathbb N\}$$

$$\omega3=\sup\{\omega2+n|n\in\mathbb N\}$$

$$ω^2=\sup\{ωn|n\in\mathbb N\}$$

$$ω^2+ω=\sup\{ω^2+n|n\in\mathbb N\}$$

user400188

I feel like we got big really fast just then

Simply Beautiful Art

12:50

Nah. $f_{ω^2+ω}(n)=n^{n^2+n}$

So its not so large yet. It also makes it somewhat obvious how $n$ relates to the $ω$'s

user400188

isn't $\omega$ bigger than all the naturals by that first line?

then $\omega 2$ larger than $\omega$ and so on?

Simply Beautiful Art

Yup

Be back in a bit, breakfast

user400188

alright

enjoy it :)

Simply Beautiful Art

Welp, I'm back

user400188

to clarify; is $\omega 2=sup\{\omega+n|n\in N\}$ the act of adding any natural to $\omega$ then taking the supremum?
to me that means we could add the last natural then increment one further, and that this is $\omega 2$

that was a short breakfast

Simply Beautiful Art

12:56

I don't eat much

@user400188 There is no last natural number

user400188

simply: gulps air "I'm done"

Simply Beautiful Art

If there were a number directly before $ω$, it would be $x$, where $ω=x+1$

user400188

hmm ok

Simply Beautiful Art

However, there exists no such $x$. You can prove by contradiction by assuming there is such an $x$ and using $ω=\sup\{1,2,3,\dots\}$

Mathmore

Is omega a complex number?

user400188

12:58

oh; x would be in the set {1,2,3...}

Simply Beautiful Art

@Mathmore $ω$ is a transfinite number. Positive and larger than all finite numbers

user400188

and omega - 1 would not be in the set

Simply Beautiful Art

@user400188 And necessarily there will exist $y>x$ for all $x$ in the set.

Mathmore

Oh! interesting

user400188

so its implied that x is both in and not in that set

Mathmore

12:59

What is a transfinite number?

Simply Beautiful Art

Kinda like an infinite number

Mathmore

Okay.

Simply Beautiful Art

@Mathmore You know about supremums?

Mathmore

Yup!!

Supremum is the least upper bound

Simply Beautiful Art

Well, $ω=\sup\{1,2,3,\dots\}$

So $ω$ is the least number larger than any finite number

Mathmore

13:00

Yup that I figured out from this chat!

Simply Beautiful Art

:P

Mathmore

;)

Simply Beautiful Art

And... brb. Nature calls lol

Mathmore

Hahaha

I intend to read Halmos's Naive set theory soon. I have drastically less knowledge about set theory.

user400188

probaly more than me

that said; have you read any logic textbooks before?

Mathmore

13:03

@user400188 No. I haven't.

I just rely on the opening chapter of a standard Analysis/Topology text book for set theory.

user400188

I'm finding they make a lot of math easier. You only have to read one then you can understand so much more quickly. Particularly proofs and set theory.

Simply Beautiful Art

@Mathmore Fun

@user400188 Haha

Well, perhaps you've noticed the pattern by now

user400188

I used to find topology so much fun, I'm kind of sad that I'm finding it less interesting now than other topics.

Ill probaly return to it another day though

Mathmore

@user400188 Thanks! I will definitely take a look.

You have some suggestion?

user400188

a friend of mine would reccomend a book named Sweet Reason

Simply Beautiful Art

13:06

$$f_{ω^{ω^{ ω^3+3}5+ω}8}(n) = n^{n^{n^{ n^3+3}5+n}8}$$

Mathmore

You perhaps want an interesting problem to get your love for topology back.

user400188

its a paid for book; but you can access it from sites like library genesis

Mathmore

Okay will check out thanks!

user400188

Authors are Tom Tymoczko and Jim Henle

Simply Beautiful Art

@user400188 PBS's Infinite series has an episode on topology

user400188

13:07

I havent checked out much of PBS's stuff

Mathmore

I have heard about Infinite...

Simply Beautiful Art

Topology Riddles | Infinite Series

►

user400188

I watched a few physics vidoes from it and found them too opinionated

Mathmore

Detective using Mahts?

Maths*

Simply Beautiful Art

I like their math videos a bit

They also have a few on how ordinals relate to finite numbers in manners other than the ones I'm showing you

user400188

13:08

personly I like the vsauce video on them

you cant really learn much from vsauce, but it can certainaly give you an appreciation

and Ive yet to find a mistake in any of thier math videos. Its nice to know your not been fed incorrect stuff

Simply Beautiful Art

Yeah lol. All the awesome math xD

Mathmore

I suggest NPTEL video courses.

Real Analysis, Linear Lagebra, Measure and Integration, Functional Analysis, Complex Analysis covered very well there.

user400188

wow there page looks good

Mathmore

I am following Measure and Integration currently.

It's going good. :)

user400188

maybe Ill give some of there courses a listen tomorrow.

But for now I think Ill be heading off to bed. (Its 11:17 pm here)

Simply Beautiful Art

13:18

g'night

Mathmore

Oh! Sure. Where do you live anyway?

user400188

Australia

Mathmore

Good night. :)

user400188

Night @everyone
Thanks for the introduction to ordinals @SimplyBeautifulArt

Simply Beautiful Art

Np

Mathmore

13:19

@SimplyBeautifulArt Have you studies dense subsets?

studied*

Simply Beautiful Art

I know of them

Trying to solve this integral right now:

1

Q: Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$

Can someone compute $$ \int_0^1\frac{\ln^4(1+x)\ln x}x \,dx$$ in closed form? I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $\operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $\mathrm{Li}_n...

Mathmore

A question I encountered : Is the subset of set of all 2 \times 2 matrices with both eigenvalues real a dense subset?

@SimplyBeautifulArt Oh okay.

Simply Beautiful Art

@Mathmore Yeah, I'm not super familiar with that

Mathmore

Np

How are you tackling that integral?

Simply Beautiful Art

$$\int_0^1\frac{\ln^4(1+x)\ln(x)}x~\mathrm dx=\lim_{s\to0,t\to-1^+}\frac{d^4}{ds^4}\frac d{dt}\int_0^1(1+x)^sx^t~\mathrm dx$$

F.Tasos

13:32

hello people

Simply Beautiful Art

@F.Tasos @kingW3 Hello and welcome to my realm

F.Tasos

greetings from greece

Simply Beautiful Art

Hm, I can't seem to figure out how to take the derivative so nicely

Mathmore

So $...$ doesn't work here?

I mean MathJax.

Simply Beautiful Art

Oh, MathJax works, you just gotta install it

Mathmore

13:35

@SimplyBeautifulArt I can't see what you wrote there.

Can you tell me how to?

amWhy

Mathmore, it's chatjax that you need for using mathjax in any chatroom on site.; there's a bookmark you can add to your browser (find it on the general Mathematics chatroom).

Simply Beautiful Art

See this and below

More or less, follow the images I posted if you use Google Chrome

Mathmore

Thanks @amWhy @SimplyBeautifulArt

amWhy

$\LaTeX$ in chat: tinyurl.com/cfqcvpc

Simply Beautiful Art

@amWhy Put https:// in front

Mathmore

13:43

I added the bookmark to my browser.

I use mozilla. I am seeing your images in the link you've provided. @SimplyBeautifulArt

amWhy

Go to your bookmarks and click on "start Chatjax".

Mathmore

Awesome!

Worked.

amWhy

@Mathmore Yay!

Mathmore

:D

Simply Beautiful Art

$$\color{red}n\color{orange}i\color{yellow}c \color{green}e\color{blue}l \color{purple}y\text{ done}$$

Mathmore

13:51

This chat room seems so cool. Everybody is so friendly.

Simply Beautiful Art

:P

Mathmore

On the website though I have to be afraid while typing. :P

Simply Beautiful Art

lol

Mathmore

@SimplyBeautifulArt How did you do that? "Nicely done" ?

Simply Beautiful Art

\color{red}{n}

\color{orange}{i}

etc.

Also, you have to put spaces in the code or else it'll break.

$$\begin{align}\int_0^1\frac{\ln^4(1+x)\ln(x)}x~\mathrm dx&=\lim_{s\to0,t\to0^+}\frac{\mathrm d^4}{\mathrm ds^4}\frac{\mathrm d}{\mathrm dt}\int_0^1(1+x)^sx^{t-1}~\mathrm dx\\&=\lim_{s\to0,t\to0^+}\frac{\mathrm d^4}{\mathrm ds^4}\frac{\mathrm d}{\mathrm dt}\frac1t~_2F_1(-s,t;t+1;-1)\end{align}$$

Huh, surprised this is short enough not to break

Gosh dang it, why you no break? I try to make example out of you!

Mathmore

13:54

$\color{blue} a \color{red} w \color{yellow} e \color{purple} s \color{green} o \color{blue} m \color{red} e$ mate!

Simply Beautiful Art

xD

Mathmore

Xp

Simply Beautiful Art

Hm, where is @Nilknarf this morning. Missing out on the chatroom lol

Mathmore

Ah I see you using two variables. hmm

Simply Beautiful Art

Yeah, I doubt I'll be able to complete the integral

Mathmore

13:56

What is $_2F_1$ ?

Simply Beautiful Art

Hypergeometric function

Mathmore

Okay.

Simply Beautiful Art

Special function stuff

Mathmore

I see.

F.Tasos

pardon me for asking but is there a special room for functional analysis ?

Simply Beautiful Art

14:00

No idea

F.Tasos

thanks anyway

Mathmore

@F.Tasos I would be interested to join one!

Simply Beautiful Art

@F.Tasos I am unsure, but I believe functional analysis has been mentioned in this room:

$Mathematics$ Calculus and analysis

For questions about calculus, real analysis, functional analys...

F.Tasos

thanks @SimplyBeautifulArt i appreciate it

Mathmore

Subbranch of analysis. :D

Simply Beautiful Art

14:01

:P

F.Tasos

guys any ideas on this math.stackexchange.com/questions/2337725/… ?

Simply Beautiful Art

Not my area of knowledge sadly

kingW3

Hello,@SimplyBeautifulArt Did you try doing partials with $u=\ln^4(1+x)$ and $dv=\ln x/x$?

Simply Beautiful Art

No, I gave up on that integral lol @kingW3

kingW3

Oh nevermind :P

Mathmore

14:13

@F.Tasos I have done basics of Hilber space. But no idea on this. What did you try?

F.Tasos

someone answered

Mathmore

Hilbert*

Yup I can see that.

I think we can use following things : 1) In metric spaces, a space is separable iff it is second countable. 2)Gram-Schmidt process to generate basis (as suggested by the answerer on the OP). What say?

amWhy

@SimplyBeautifulArt Perhaps you can learn from others here?

Simply Beautiful Art

Yes, I do listen in on the conversations :-)

F.Tasos

im trying to figure out why the span of projections is dense in U^{\bot}

if i cant do it in 10 min ill grab pencil and paper

Mathmore

14:24

Since $L^2[a,b]$ is separable, it is second countable as well. So it has a countable basis. Now we have $e_n$ as an orthonormal set so it is linearly independent.

F.Tasos

that is true

Mathmore

From here on I'm thinking what can we get...

Either $e_n$ is the basis or we can extend it.

I give up on my way currently.

Zophikel

15:41

@Simply need help getting the intuition on something

mathb.in/147631

$$\nabla(|u(x,y)+ iv(x,y)|)^{p} = p(p-1)|)|u(x,y)+iv(x,y)|^{p-2}| \nabla f|^{2}$$

                                                        ^ If $f$ is harmonic and real-valued on $U \, \subset \mathbb{C}$ and if $f$ is nonvashing  then ^

Simply Beautiful Art

@Zophikel ?

You need to prove that?

Zophikel

@Simply yeah i'm trying to get the geometrical intuition, I already know a way to prove it @Simply

If you have a function f, think of the Laplacian of f at x as a measurement of how close you are the average value of points around x. For example, if the Laplacian of f is 0, then you have that u(x) is exactly equal to the average value of u on a ball around x.

^ I know that's the intution behind $\nabla$

@Simply you have any ideas, been wrapping my head around this

Simply Beautiful Art

Hm, I don't think I have a good explanation for you. :) You are extending to places I haven't gone.

Zophikel

@Simply :) I found a couple of ways to define but I feel like my feel on theorem is too formal :(

Nilknarf

16:52

Hello all!

@Dibbs In response to your message to me: yes, the proof can be done (in fact, it seems that @SimplyBeautifulArt has already done it for you, so there's not really much more I can do to help you).

@Typhon Hi!

Typhon

@Nilknarf what do you want? :-)

Nilknarf

Just saying hi since you entered the room

Typhon

no i didnt....

Nilknarf

Please don't tell me you didn't actually enter the room

Dammit

I need to start assuming that whatever the sidebar shows me is false

SBM

Hello and bye 22:25 today was really busy

Nilknarf

16:55

I'm so sorry. :(

@SBM Hi!

Typhon

@Nilknarf s'okay.

Zophikel

17:45

@Simply

0

Q: Geometrical Intution behind $\nabla(|f|^{p}) = p(p-1)|f|^{p-2}|\nabla f|^{2}$

In the text "Theory of Functions of One Complex Variable", I'm having trouble verifying the way I expressed the Laplacian Operator, and generalizing $(1.)$ to functions of more then one complex variable following Proposition in $(1.)$ Also I'm having trouble establishing the geometrical intuition...

multivariable-calculus complex-numbers

                ^ Can't wait for answers

Typhon

-1

Q: Is there a multivariate integer function f(x,y) that returns the number of factors of y in x with a closed form?

I'm looking for a simple function in terms of the elementary and exponential functions that when passed an integer x and integer y, it returns the number of times that y can be divided from x such that the result is an integer. I believe that it is something along the lines of $\lfloor\log_y(x)\r...

number-theory elementary-number-theory functions logarithms

need answer

plz

Zophikel

@Typhon I know nothing about the theory of numbers

Simply Beautiful Art

o/

Typhon

@Zophikel i didnt ask you to answer it

@Zophikel besides. it isnt number theory.

Zophikel

@Typhon you put it under number theory

Typhon

17:50

only so people would see it

Simply Beautiful Art

Gosh darn notifications lol. When you get a text in the middle of your game x.x

Typhon

lol

be back later

Simply Beautiful Art

God wtf

An enemy snuck behind my defenses T_T

Zophikel

lol @Simply

Simply Beautiful Art

I try to be good at video games lol

Dibbs

19:42

@SimplyBeautifulArt Could you share your proof with me? :D I'd really like to see how you did it.

Simply Beautiful Art

21:46

@Dibbs I've already posted it to your question:
https://math.stackexchange.com/questions/2252550/integrating-the-formula-for-the-sum-of-all-natural-numbers/2337019#2337019

Dibbs

22:06

Oh my bad. I misinterpreted your post to mean that you had found a different proof related to the topic. Sorry about that!

Simply Beautiful Art

No problem @Dibbs

And you welcome to stick around about whatever. I'll be unresponsive for a bit though, I'm a bit busy

Simply Beautiful Art

23:26

@Dibbs By the way, you do realize that:
ζ(-0) ≠ 1 + 1 + 1 + ...
ζ(-1) ≠ 1 + 2 + 3 + ...
ζ(-2) ≠ 1² + 2² + 3² + ...
ζ(-n) ≠ 1ⁿ + 2ⁿ + 3ⁿ + ...

Zophikel

@Simply how does this look

In the text "Function Theory of a Complex Variable", i'm having trouble proving the following relation in $(1.)$

$(1.)$

Prove that if $f$ is holomorphic on $\text{U} \subset \mathbb{C}$, then

$$\nabla(|f|^{2})=4 |\partial f / \partial z|^2$$

$$\text{Lemma}$$

The following observations can be made on $(1.)$

Recall:

$$\nabla(|f|^{2})=4 |\partial f / \partial z|^2$$

One can observe in $(2.)$

$(2.)$

$$\nabla^2f(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4 |\partial f / \partial z|\partial f / \partial z|$$

                       ^ Some very interesting problems in my CA Book that i'm reading out of

$$\nabla \cdot \nabla f(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4|( \nabla(f(z))|u(x,y) + iv(x,y)||( \nabla(f)|u(x,y) + iv(x,y)|$$

^ identified some errors i'll have to define some of the notiton I used to manipulate the differential operators used in the proof

Simply Beautiful Art

23:44

Hm, it looks a little messy, though I'd have to tidy myself up with more studying to understand everything you are doing @Zophikel

Zophikel

@Simply yeah I made some mistakes i'm tidying up some things, oh @Simply you've done lots of mutivarible calc/ mutivarible analysis

Simply Beautiful Art

Nah, I haven't done much serious diving into multivariable stuff. I just know lots of random things, per chance such random things happen to be related to multivariable calculus, but I haven't done such studying.

Zophikel

@Simply your eventually going to have to get serious in CA and it has lots of connections with DG and Multivariable Calc

:) it really makes the theory beautiful

Simply Beautiful Art

Yes, I can imagine

Zophikel

@Simply I've good a couple books you might love :)

@Simply it's been cleaned up a bit :) mathb.in/147666

                                       e                     ^ It looks look the conjecture may be true :)

@Simply do you have a good gauge on it

$Mathematics$ Calculus and analysis

Transcript for

$Mathematics$ Simply Beautiful Art's realm of calcu…