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00:04
$$\nabla \cdot \nabla f(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4|( \nabla(f(z))|u(x,y) + iv(x,y)||( \nabla(f)|u(x,y) + iv(x,y)|$$
       hmmmmmm This is sloopy ^
$$\nabla \cdot \nabla f(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4|( \nabla(f(z))||( \nabla(f)||$$
                                                                        ^ ok so this looks a little better
@Zophikel No I do not sadly D:
@Zophikel And yes, that does look much better
@Simply I can give you the intuition :)
hold on let me fetch the propostion
Sure, go ahead lol
$$\nabla(|f|^{2})=4 |\partial f / \partial z|^2$$
Ok so bassically what this is saying is that laplican of two complex functions is equal to the partial deratives of those two complex functions
taking away rigour the Laplican gives us the average values of around a point on a surface
@Simply does that make sense
00:09
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.
        ^ Better intution for the laplican
Oh my, I wonder what brings someone like @JohnHughes around to these parts
@Simply think you can prove it
my method of going about it is way to mechanical
@Simply now the student in $\mathbb{C}$ has become the master lol
$$\nabla \cdot \nabla f(|u(x,y) + iv(x,y)||u(x,y) + iv(x,y)|) = 4|( \nabla(f(z))||( \nabla(f)||$$
                                                               ^ Now much of the key resluts you see in the approach use the Laplacian
One challenginge is to proof this without using the Laplacian
You mean use definition of Laplacian operator?
@simply yeah
Welp, that looks tedious af
00:16
@Simply one thing I challenging you to do is try to find a proof with out using the Laplacian Operator
@Simply think you can do it :D
I'll try
@Zophikel You know the Riemann zeta function?
@Simply I've heard of it
I know it's defined by an infinite sum
$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$$
00:19
yes that's the one
doesn't it converge
Prove that:
$$\sum_{n=2}^\infty(-1)^n\zeta(n)x^n=x^2\int_0^1\frac{1-t^x}{1-t}~\mathrm dt$$
@Simply ummmmmmmmmmmm
I have a couple idea's of going about :)
it
@Simply but i'll have to read up on the function first and get some background and motivation first
:)
I'll tell you the tools I used to prove that equation
a) geometric series: $$\frac1{1-t}=\sum_{n=0}^\infty t^n,|t|<1$$
b) Maclaurin expansion: $$f(x)=\sum_{k=0}^\infty\frac{f^{(n)}(0)}{n!}x^n$$
00:23
Awesome i'll to finish up the problem i'm working on
More or less, you want to try and take the Taylor expansion of that integral
00:35
@Simply what's your notebook look like can you post an image :)
LMAO... uh... >.> yeah no
My "notebook" is composed of the 100 or so scraps of paper lying on my desk plus the 5+ notebooks lying around as well. I'd rather not show you what it looks like lol
@Simply so your disorganized as well interesting
it seems like our styles are similar
:)
@Simply how does it look close to the final reslut :)
Man I love the Laplacian it's so elegant
00:50
@Simply time to post the question I made some slick attacks can't wait for that rep :P
0
Q: Proving $\nabla(|f|^{2})=4 |\partial f / \partial z|^2$ via a "Laplacian Appoarch"

ZophikelIn 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 b...

01:06
@Simply what made you start what do doing math, what's your story
$e^{\pi i}+1=0$
Had to prove that for the sake of my 9th grade curious self
Which is how I got into complex anlaysis
And why hello @Dibbs
Right now I'm trying to see if it is possible to prove that $x$ is transcendental where $x^x=\pi$
@Simply nice, wanna know how I started
@Simply so I start of failing Algebra 1, honors kids were making fun of me for not being on their level of 'math'
Lol, I think I may have found the longest mathematical proof on Wikipedia
In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states that if α1, ..., αn are algebraic numbers which are linearly independent over the rational numbers ℚ, then eα1, ..., eαn are algebraically independent over ℚ; in other words the extension field ℚ(eα1, ..., eαn) has transcendence degree n over ℚ. An equivalent formulation (Baker 1975, Chapter 1, Theorem 1.4), is the following: If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over t...
@user400188 Good morning I assume?
01:16
stuggled for nearly 1.5 years, finally ended up in my mentor classroom he begins helping me out
Morning o/
@Zophikel Dang, that sucks
@Zophikel Did you know number theory uses integrals?
@Simply yeah there's lots of integrals :)
@Simply so he begins training me teaching me how to solve problems introduced me to calculus
By the end of the course I was doing what pre-calc kids were doing, after I had him I began gulping down parts of the k-12 math curriculum
@Simply during this time when I was in Geometry 2 I was doing BC and AP Calc
I shocked some of the Calc kids
and AMT
then by the summer of that sopheremre year I began exploring proof-based math
I tried to learn measure theory XD thinking it was the first cousre you take as a math major
That was one rough week XD
@Simply now teacher's are hearing about my situation
@SimplyBeautifulArt hmm I did not know about
ζ(-0) ≠ 1 + 1 + 1 + ...
ζ(-1) ≠ 1 + 2 + 3 + ...
ζ(-2) ≠ 1² + 2² + 3² + ...
ζ(-n) ≠ 1ⁿ + 2ⁿ + 3ⁿ + ...
Can you explain? I'm not too knowledgeable about the Riemann-zeta function as I'm sure you're aware by now
01:39
I think it has to do with the fact that the real part of n must be greater than 1, but I'm not quite sure why this constraint must hold true
@Dibbs Do you know calculus?
Nvm, you know integration, what am I saying?
So we have the Riemann zeta function:
@Dibbs If you don't have chatjax installed, see here:
https://chat.stackexchange.com/transcript/0?m=38299166#38299166
$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s},~s=\sigma+\rho i,\sigma>1$$
And anywhere else, we have:
$$\zeta(s)=\sum_{n=0}^\infty\frac{\zeta^{(n)}(a)}{n!}(s-a)^n,~|s-a|<|1-a|,a\in\mathbb C\setminus\{1\}$$
@Dibbs A.k.a. Taylor expansion of the Riemann zeta function expands the Riemann zeta function outside its original domain
Other nicer forms of the Riemann zeta function include:
$$\zeta(s)=\frac1{1-2^{1-s}}\lim_{x\to-1^+}\sum_{n=1}^\infty\frac{x^{n+1}}{n^s}$$
And now I shall be heading to bed. Good night everyone!
01:56
I'll look into it. Thanks and good night!
Good Night @SimplyBeautifulArt
wow it must be late if it were morning when I talked to you last night
 
13 hours later…
15:16
@user400188 lol, yeah
Good morning @SBM
SBM
SBM
@SimplyBeautifulArt good evening
 
6 hours later…
20:55
Oi @Dibbs

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