the no-normie zone - 2017-10-15

Anonymous

07:46

Let $f(z)=z^2$ if $z\neq i$ and $f(z)=0$ if $z=i$. I need to show that the limit as $z\to i$ is $-1$. So, basically I need to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $0<|z-i|<\delta\implies |z^2+1|<\epsilon \implies |z-i||z+i|<\epsilon$. In the next step in my book they write $0<|z-i|<1$,$|z+i|<3$ and $\delta=\min\{1,\epsilon/3\}$. I can't what they are doing @BalarkaSen

Anonymous

*understand

Anonymous

How is $0<|z-i|<1$ ?

Anonymous

Or are they saying if $0<|z-i|<1$

Anonymous

Then $|z+i|<3$

Balarka Sen

So suppose $|z^2 + 1| < \epsilon$ for some $\epsilon > 0$

Anonymous

07:50

@BalarkaSen Okay

Balarka Sen

That means $|z - i||z + i| < \epsilon$

And $|z + i| < |z - i| + 2$, right?

Anonymous

@BalarkaSen Right

Anonymous

Triangle

Balarka Sen

Mhm

So given a $\delta$ such that $|z - i| < \delta$, we have

$|z^2 + 1| < \delta(\delta + 2)$

We need to choose $\delta$ such that $\delta(\delta + 2) < \epsilon$. That should suffice.

@Blue Here is what they are doing. We need to show for every $\epsilon > 0$ there is a $\delta > 0$ such that ...

Look at the two cases

(1) $\epsilon \leq 3$, (2) $\epsilon > 3$

If (1) holds, choose $\delta = \epsilon/3$. Then $|z^2 + 1| < \delta(\delta + 2) < \epsilon/3(1 + 2) = \epsilon$, as desired.

If (2) holds, choose $\delta = 1$. Then $|z^2 + 1| < \delta(\delta + 2) = 3 < \epsilon$, as desired.

Sum the two cases up. You're really choosing $\delta = \text{min}\{1, \epsilon/3\}$

Anonymous

@BalarkaSen I think you are framing it in the wrong way. We are given an $\epsilon$. We need to find a delta (say) $D$ such that if $|z-i|<D$, then $|z^2+1|<\epsilon$. But $|z^2+1|<(|z-i|+2)(|z-i|)<(D+2)D$ If we can find a $D$ such that $D(D+2)<\epsilon$ we are done. We can then solve that inequality to get the allowed range of $D$.

Anonymous

07:58

@BalarkaSen Ah, I get it now!

Balarka Sen

4 mins ago, by Balarka Sen

@Blue Here is what they are doing. We need to show for every $\epsilon > 0$ there is a $\delta > 0$ such that ...

6 mins ago, by Balarka Sen

We need to choose $\delta$ such that $\delta(\delta + 2) < \epsilon$. That should suffice.

Nothing you said is different from whatever I said.

Anonymous

Right right. I was pointing towards the phrase "given a $\delta$" :P

Anonymous

But I get what you wanted to say

Balarka Sen

Well I'm starting with a generic $\delta$ to see what conditions we need to impose on it for the implications to hold. Just doing the necessary calculations.

Anonymous

@BalarkaSen Yup, got it naouw :)

Anonymous

08:03

I'm revising this complex differentiability, limits and continuity stuff today

Balarka Sen

That's good stuff

Anonymous

Yeah :P I'll ping you if I get stuck again. :D

Balarka Sen

OK

By complex differentiability you mean Cauchy-Riemann too I presume?

Anonymous

@BalarkaSen Yup

Anonymous

CR equations

Anonymous

08:11

And the proof of that which you had taught me

Anonymous

(Will have to search the transcript)

Balarka Sen

Cool, cool

Complex analysis is a gem

Anonymous

@BalarkaSen Heard there's something called "germ" in maths =P

Anonymous

Germ (mathematics)

In mathematics, the notion of a germ of an object in/on a topological space is an equivalence class of that object and others of the same kind which captures their shared local properties. In particular, the objects in question are mostly functions (or maps) and subsets. In specific implementations of this idea, the sets or maps in question will have some property, such as being analytic or smooth, but in general this is not needed (the maps or functions in question need not even be continuous); it is however necessary that the space on/in which the object is defined is a topological space, in...

Anonymous

"The name is derived from cereal germ in a continuation of the sheaf metaphor, as a germ is (locally) the "heart" of a function, as it is for a grain."

Anonymous

08:18

Oh lol

Anonymous

I want to learn topology someday soon

Balarka Sen

germs are great

they have roots in complex analysis

Anonymous

@BalarkaSen wow

Balarka Sen

Revisit complex differentiability/C-R equations then if you want I can tell you about analyticity of complex functions

Anonymous

Okay! Analytic functions is the next chapter only :)

Anonymous

08:27

I'm trying to prove $\lim_{z\to z_0} z^2=z_0^2$. For a given $\epsilon$, we need to find a $\delta$ such that $|z-z_0|<\delta$ implies $|z^2-z_0^2|<\epsilon$. But $|z^2-z_0^2|=|z-z_0||z+z_0|<|z-z_0|(|z-z_0|+|2z_0|)<\delta^2+2\delta|z_0|$. We need a $\delta$ such that $\delta^2+2\delta|z_0|<\epsilon$. Should I leave it at that?

Anonymous

$(\delta+|z_0|)^2-|z_0|^2<\epsilon$

Anonymous

$(\delta + |z_0|)^2 <\epsilon +|z_0|^2$

Anonymous

$ -(\sqrt {\epsilon+|z_0|^2}) -|z_0|<\delta < \sqrt {\epsilon+|z_0|^2} - |z_0|$

Balarka Sen

I think that's making things too hard.

Anonymous

But $\delta>0$

Balarka Sen

08:32

You need to find a $\delta$ such that $\delta(\delta + 2|z_0|) < \epsilon$, right?

Why not use the same trick as before?

Partition into the case $\epsilon > 3|z_0|$ and $\epsilon < 3|z_0|$, I expect

Anonymous

@BalarkaSen Oh. That will work, yes

Balarka Sen

In the second case you can take $\delta = \epsilon/3$, right?

Anonymous

I was thinking I can't partition because I don't know the value of $|z_0|$

Anonymous

But here we need a generic $|z_0|$

Balarka Sen

All you need to know is that it's positive.

Hm, I am having trouble making my argument work

But I'm off to dinner now. I'm sure you can figure that out

Anonymous

08:38

Yes, I'm trying

Anonymous

Have good dinner

Anonymous

(Lunch :P)

Balarka Sen

all the same...

Anonymous

09:25

@BalarkaSen What's the correct way to show that $\Re(z)$ and $\Im(z)$ are not differentiable at any point?

Balarka Sen

Cauchy-Riemann :)

Anonymous

I mean without that

Anonymous

From first principle

Anonymous

(Definition of derivative)

Anonymous

Should I plug in $x+iy$ ?

Balarka Sen

09:26

I mean you'll end up with a proof equivalent to C-R. Approach the limit from the real and the imaginary axis, show that they don't agree.

Anonymous

Oh, I see

Anonymous

Here's what I'm doing:

Anonymous

Let $z=x+iy$. Then by definition, $f'(z)=\lim\limits_{h \to 0}\frac{f(z+h)-f(z)}{h}$.

Anonymous

Let $h\to0$ along the real axis. Then $z+h=(x+h)+iy$, so $f(z+h)=x+h$. Thus we have $$f'(z)=\lim\limits_{h \to 0}\frac{x+h-x}{h}=1$$

Balarka Sen

That is correct

Anonymous

09:30

Now let $h\to0$ along the imaginary axis. Then $z+h=x+i(y+h)$, so $f(z+h)=x$. Hence $$f'(z)=\lim\limits_{h \to 0}\frac{x-x}{h}=0$$

Anonymous

Now I'm stuck

Balarka Sen

That's pure garbage

Anonymous

Wait

Anonymous

That's 0

Anonymous

$h\to 0$

Balarka Sen

09:31

The final answer is correct, but your derivation is not.

You meant $\lim_{h \to 0} (x - x)/(ih)$

Anonymous

Oh. Yes. Either I should take $h$ as imaginary there or take $ih$

Balarka Sen

Right.

But yes, this is it.

You proved $f'(z)$ does not exist for any $z = x + iy$. What do you mean?

Anonymous

So, along real axis I got 1 and for imaginary axis I got 0. Is showing that the derivative doesn't match along two directions sufficient? I guess it is

Balarka Sen

Yes, this is just like multivariable calculus.

If limit along various directions don't agree, it doesn't exist.

Anonymous

Let me get some things cleared. In multivariable calculus if the limit along two or more directions doesn't match the double limit does not exist. But, can all the directional derivatives exist at that point, or no?

Balarka Sen

09:42

Yes, but complex derivative is not directional derivative.

$f'(z)$ is the limit of the sequence $(f(z + h_n) - f(z))/h_n$ where $h_n$, $n = 1, 2, \cdots $ is any sequence of complex numbers approaching $0$.

Anonymous

Ok. So what is the basic condition for complex differentiability? (I'm not asking about CR eqns...I want to derive that from sratch)

Balarka Sen

What do you mean? The condition is $\lim_{h \to 0} (f(z + h) - f(z))/h$ exists in the complex plane. What else do you want?

Anonymous

Alright. So $h$ is a complex number there. Meaning, it can approach from any direction and all of them have to match for it to be differentiable at that point

Anonymous

I showed that the derivative does not match along two axes. So, that showed that it cannot be differentiable

Balarka Sen

Not just any direction/line. It can approach from any sequence of complex numbers that approach $0$.

@Blue Yes.

Anonymous

09:47

@BalarkaSen Is it similar to double limit $(x,y)\to(0,0)$ in multivariable calculus?

Balarka Sen

It is, yes.

Look, all you are saying is there is some constant $C$ such that $|(f(z+h) - f(z))/h - C|$ tends to $0$ as $|h|$ tends to $0$.

And then you call $C = f'(z)$

Here everything is happening in the absolute value in the complex plane, $|x + iy| = \sqrt{x^2 + y^2}$.

Anonymous

Then it means that it will be much more difficult to prove that the derivative does exist than to prove the derivative doesn't exist. For example along all straight line paths it might exist but along all spiral paths it might not exist. We can perhaps find the bound of the function and use the shrinking disc concept we use to find double limit

Anonymous

But there are C-R eqns for that

Anonymous

I should now look at the derivation of that

Anonymous

Sep 15 at 18:42, by Balarka Sen

Then $f'(z) = \lim_{h \to 0} (f(z + h) - f(z))/h$, if you approach $h \to 0$ along the real axis, this becomes $\lim_{h \to 0} (f(x + h, y) - f(x, y))/h$

Balarka Sen

09:52

@Blue This is correct, on face value. But surprisingly, it turns out it is enough to check it along the real and imaginary axes (so you're wrong, the phenomenons you speak of cannot happen). That is what the Cauchy-Riemann equations say.

$f$ is holomorphic if and only if the Cauchy-Riemann equations (conditions obtained from taking the limits of the first principles derivative along real and imaginary axes) are satisfied.

That's an iff!!!

I don't think I told you the proof of the converse, though.

Anonymous

I checked the proof you did on Sept 15. In that you equated the derivatives along the two axes and said that is the necessary and sufficient condition for complex differentiability. But you didn't prove that that is the necessary and sufficient condition for same value of derivative along all directions/paths of approach

Balarka Sen

Yes, I indeed did not.

Anonymous

So,....what's it? :P

Balarka Sen

No wait. "In that you equated the derivatives along the two axes and said that is the necessary and sufficient condition for complex differentiability"

No, I proved it is a necessary condition, not sufficient.

Anonymous

@BalarkaSen Oh, what else is there

Balarka Sen

09:57

??

Anonymous

Continuity of partial derivatives is also needed for complex differentiability?

Balarka Sen

No, you are not understanding me.

I proved complex differentiability $\implies$ Derivatives along the two axes must be equal $\implies$ Cauchy-Riemann equations hold.

Anonymous

Okay, I'm listening. Go on

Balarka Sen

I did not prove the converse, which is also true.

Anonymous

Right. I'm interested in the converse proof

Anonymous

10:00

Will that proof be too difficult for me to understand?

Balarka Sen

No it's just a small technicality

So suppose $f : \Bbb C \to \Bbb C$ is a complex function

Decompose it into real and imaginary parts $f(z) = u(x, y) + i v(x, y)$

Anonymous

Alright so far

Balarka Sen

So let's assume $u, v$ are continuously differentiable and and they satisfy the C-R equations

Anonymous

Okay

Balarka Sen

Given that, write $u(x + h_1, y + h_2) = u(x, y) + u_x h_1 + u_y h_2 + |h|\varphi_1(h)$ and $v(x + h_1, y+h_2) = v(x, y) + v_x h_1 + v_y h_2 + |h|\varphi_2(h)$

Where $\varphi_i(h) \to 0$ as $h \to 0$.

And by $h$ I mean the vector $h = (h_1, h_2)$.

Anonymous

10:07

mhm

Balarka Sen

And $u_x, u_y$ as usual means the partial derivatives of $u$ wrt $x$ and $y$ at the point $(x, y)$.

Anonymous

yup

Anonymous

But one thing

Anonymous

How do you know : $u(x + h_1, y + h_2) = u(x, y) + u_x h_1 + u_y h_2 + |h|\varphi_1(h)$

Anonymous

I mean how do you know it can written like that

Balarka Sen

10:11

Ah, that's because $u$ has continuous partial derivatives, hence is differentiable as a multivariable function because of that famous lemma, and $[u_x, u_y]$ is it's Jacobian.

So just write down the definition of multivariable differentiability, and you'll get that formula

You know, the one which says $\|u(x + h_1, y + h_2) - u(x, y) - Ju(x, y)(h_1, h_2)\|/\|h\| \to 0$ as $h \to 0$.

Anonymous

That's like the Jacobian and Taylor's combined

Anonymous

The linear approximation one

Balarka Sen

No, it's the definition of the Jacobian.

Anonymous

@BalarkaSen J is the jacobian, that's what I mean

Balarka Sen

Sep 12 at 16:02, by Balarka Sen

$f$ is said to be differentiable at $\mathbf{a} \in \Bbb R^n$ if there is a linear map $A : \Bbb R^n \to \Bbb R^m$ between the vector spaces such that $$\lim_{\mathbf{h} \to \mathbf{0}} \frac{f(\mathbf{a}+\mathbf{h}) - f(\mathbf{a}) - A\mathbf{h}}{\|\mathbf{h}\|} = \mathbf{0}$$

The definition of Jacobian has first order Taylor's approximation in-built in it.

Anonymous

10:14

$u(x + h_1, y + h_2) = u(x, y) + u_x h_1 + u_y h_2 + |h|\varphi_1(h)$

Anonymous

That last term is the error term (if I am not wrong)

Balarka Sen

That is true.

Anonymous

So it's basically taylor (first order)

Anonymous

Got it

Balarka Sen

I mean it is but you're just invoking the definition of $Ju$. This is not an application of Taylor's theorem.

Anonymous

10:16

Anyhow, continuous partials gives a stronger than differentiability condition there

Anonymous

5 mins ago, by Balarka Sen

Ah, that's because $u$ has continuous partial derivatives, hence is differentiable as a multivariable function because of that famous lemma, and $[u_x, u_y]$ is it's Jacobian.

Balarka Sen

Just like by definition derivative in single variable calculus means $f(x + h) = f(x) + f'(x)h + o(h)$.

@Blue Sure, I don't care.

Anonymous

@BalarkaSen Yes yes, I got it now

Anonymous

Anyhow, I'm not sure why continuous partials is needed for complex derivability

Anonymous

Perhaps the proof will show it

Anonymous

10:18

Go on then

Balarka Sen

@Blue You don't...

In fact complex differentiability implies continuous partial derivatives, continuous double partial derivatives, continuous whatever bullshit, infinitely differentiable, and analytic

Anonymous

@BalarkaSen mathworld.wolfram.com/ComplexDifferentiable.html

Balarka Sen

Don't give me that Wolfram bullshit.

In many places it is assumed for convenience.

Anonymous

Are you sure? Even our prof said the partials need to be continuous. But he may be wrong...

Balarka Sen

Complex differentiability is much stronger than any regularity condition to bother about this.

Anonymous

10:20

I see

Balarka Sen

Complex differentiability $\implies$ smooth.

i.e., derivatives of all orders exist

So in particular it is continuously differentiable :P

Anonymous

@BalarkaSen And derivatives of all order are continuous too?

Balarka Sen

@Blue ... if something is twice differentiable it is continuously once differentiable ...

Anonymous

@BalarkaSen Oh, yes

Anonymous

You're right

Balarka Sen

10:22

So in particular thinking about regularity in complex analysis is empty pedanticism, "phaka atlamo"

It is true, however, that many proofs become simpler when assuming continuous differentiable beforehand. But it is not needed, and careful books mention these points.

Anonymous

Hmm. Can you show why just the CR eqns are sufficient to prove complex differentiability ?

Anonymous

@BalarkaSen Lol

Balarka Sen

Oh, I mean you need $u$ and $v$ to be differentiable + CR equations to have that $f = u + iv$ is complex differentiable. (Because that's why this proof works)

But you don't need continuous differentiability or any sort of regularity in the definition of complex differentability

Anonymous

@BalarkaSen Yes, I agree

Anonymous

@BalarkaSen Yeah, I wanted to know why

Anonymous

10:28

"Derivatives of all orders are continuous" if "complex differentiable" is too shocking to believe

Balarka Sen

@Blue Well, I don't know what you want as an answer. Definition of complex differentiability is $\lim_{h \to 0} (f(z + h) - f(z))/h$ exists, in which you don't need any assumption on $f$ nowhere.

Oh, you want to know why $f$ is complex differentiable $\implies$ $f$ is infinitely differentiable?

Anonymous

@BalarkaSen Yes

Anonymous

How can you show infinite differentiability if CR eqns are satisfied and u,v are differentiable ?

Balarka Sen

Right, that takes some work.

You don't even need "u, v are differentiable" if instead of saying CR eqns are satisfied you say "$f$ is holomorphic"

So if $f = u + iv$ in general no conditions on $u$ and $v$ are imposed. Just that $f$ needs to be once (complex) differentiable.

It is not a simple proof. It is basic and is crucial, but not simple. I can tell you the idea.

Anonymous

@BalarkaSen Is it on the net?

Anonymous

10:32

I'd like to read the full proof. Literally every article I read has the condition "first order partials need to be continuous".

Anonymous

You can tell me the basic idea though

Balarka Sen

Look in the textbook of Stein-Shakarchi

They do everything carefully so as to not assume continuity of first partials

Page 47

(A pdf's available on the internet)

Anonymous

kryakin.org/am2/Stein-Shakarchi-3-Real%20analysis.pdf

Anonymous

This ^ ?

Balarka Sen

Oh no sorry I mean the book "Complex analysis"

Anonymous

10:35

checking

Balarka Sen

That's the third book in their analysis trilogy

Anonymous

libraryqtlpitkix.onion.link/library/Math/…

Anonymous

Lol...it's an .onion link

Anonymous

XD

Anonymous

@BalarkaSen Page 47 deals with Cauchy's integral formula

Balarka Sen

10:37

Look at the corollary at the bottom of the page

Anonymous

Umm..."Corollary 4.2 If f is holomorphic in an open set Ω, then f has infinitely
many complex derivatives in Ω."

Balarka Sen

Correct.

Anonymous

But that isn't what we're looking for...are we?

Balarka Sen

Huh?

Anonymous

In mathematics, a holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of every point in its domain.

Balarka Sen

10:40

I mean that's the theorem which says once differentiable implies infinitely differentiable.

What else do you want

Anonymous

I wanted to know 1) Why continuity of partial derivatives is not necessary for complex differentiability? 2) Without continuity of partial derivatives how can we show that existence of partials and C-R are sufficient to prove infinite differentiability?

Balarka Sen

You're mixing things up. The answer to (1) is "obviously" and (2) is "not sufficient".

Do spend some time trying to ask the right questions, otherwise you'll just be exhausting the answerer by asking the same things again and again...

Anonymous

25 mins ago, by Balarka Sen

In fact complex differentiability implies continuous partial derivatives, continuous double partial derivatives, continuous whatever bullshit, infinitely differentiable, and analytic

Anonymous

Ok. Lemme frame my question now:

Anonymous

1) "Why continuity of partial derivatives is not necessary for complex differentiability?" You say "obviously". Means it is not necessary. But then you claim in 2) "not sufficient"...meaning CR eqns and existence of partials is not sufficient for infinite differentiability. Just sometime ago you said that CR equations and existence of partials is sufficient for complex differentiability (here)

Anonymous

10:49

Apparently you are contradicting yourself

Balarka Sen

If you can read carefully, please read my message again. It said it is sufficient for $u, v$ to be differentiable and to satisfy the CR equations to guarantee complex differentiability.

I did not say just existence of partials and CR equations is sufficient.

Anonymous

@BalarkaSen What type of differentiable? In the multivariable sense?

Balarka Sen

What else can differentiability mean when speaking of multivariable functions?

differentiable means differentiable means differentiable

@Blue I hope you'll take into account the possibility that you can misread alongside "This guy is speaking garbage" now.

And stop accusing people of contradicting themselves when they are indeed saying the correct thing, and you are wrong.

In any case, to answer (1). That continuity of partials is not necessary for complex differentiability is "obvious" because to define the complex derivative, the first principles one, you do not need any form of regularity assumption on $f$.

So you just need to spend 2 seconds recalling the definition to convince yourself.

In any case, the correct question here should be Why does $f$ being complex differentiable imply it is infinitely complex differentiable?, because that in particular means $f$ being once complex differentiable is enough to guarantee the real and imaginary parts of $f$ are smooth functions, a very very special case of which is $u_x, u_y, v_x, v_y$ exist and are continuous.

Anonymous

@BalarkaSen Oh don't get angry now =P I am renowned for misinterpreting stuff. So get used to being called "you are wrong" :) Anyhow, yes. I fully understand what you meant, now.

Balarka Sen

I am not angry, just annoyed. I don't usually falsely call out people for being wrong when they are explaining things and I don't understand what they are saying.

I don't think it's a good practice in general.

Anonymous

11:02

@BalarkaSen Okay baba _/_ I'm sorry :P

Anonymous

I just mentioned you are "apparently contradicting" yourself though

Anonymous

Not that you're wrong

Anonymous

"apparently"= "from my point of view"

Balarka Sen

Whatever, your tone came out to be quite confrontational, and I hope you realize that.

I am still prepared to talk about math in the next couple minutes I'm here, if you want to

Anonymous

@BalarkaSen I don't know what made you feel that especially when you know me for so long. BTW, I'll have to leave at the moment for my father is calling me. See you at night. :)

Balarka Sen

11:32

Here's the most general result along the lines, by the way. If $f$ is complex differentiable on an open disk $D \subset \Bbb C$ centered at some point $z_0$, of some radius $\epsilon_0$, then at every point $z \in D$ we can have

$$f(z) = \sum_{k = 0}^\infty a_k z^k$$

Where the right hand side is a power series which converges absolutely (google absolute convergence if you want) to $f(z)$.

I meant $(z - z_0)^k$ there, not $z^k$. And you can do this for any $z_0$ (i.e. you can find a disk around any $z_0$ and get a power series there...)

This is stronger than infinite differentiability, notice. Because by differentiating multiple times termwise, you can conclude $a_k = f^{(k)}(z_0)/k!$.

So that means $f$ is infinitely differentiable at any $z_0 \in \Bbb C$ (because there exists a power series around any $z_0$)

As to why "$f$ is complex differentiable $\implies$ $u$ and $v$ are infinitely differentiable", just take $f'(z)$, $f''(z)$, etc and take the limit $h \to 0$ along the real and imaginary axis to see this.

So the whole point boils down to showing in the complex analytic world, 1st order derivative implies infinite order derivative implies infinite order Taylor series. Of course this is counterintuitive, because it is FALSE in real analysis.

I hope that's a useful summary.

Anonymous

12:16

@BalarkaSen Indeed. I get the basic idea now.

Anonymous

Absolute convergence

In mathematics, an infinite series of numbers is said to converge absolutely (or to be absolutely convergent) if the sum of the absolute values of the summands is finite. More precisely, a real or complex series ∑ n = 0 ∞ a n {\displaystyle \textstyle \sum _{n=0}^{\infty }a_{n}} is said to converge absolutely if ...

Anonymous

I need to read through the absolute convergence though.

Anonymous

@BalarkaSen I read through the conversation again. I realize I was being both annoying and dumb from your point of view as I was making you repeat the same points over and over again (which I was misinterpreting). Even I feel annoyed when I have to explain the same thing again and again to someone else. I totally agree with you and re-apologize for saying that you're contradicting yourself without being fully certain myself. I will try to never repeat that again. I hope that all's good between us.

Balarka Sen

14:40

@Blue It is fine, I was not angry, just momentarily annoyed. We can move on :)

I wasn't annoyed at being pointed out wrong, to be honest, just the parsing of it.

It is a fairly possible scenario that I can be wrong.

(Eg, when I was doing spherical coordinates computations yesterday)

Balarka Sen

15:20

@Blue So, any more thoughts on complex analysis?

Anonymous

@BalarkaSen No, I more or less got it. I'm revising a bit of linear algebra and QM today. Tomorrow I need to convince/impress that professor to teach me, so. Lol. He was asking me questions from Linear Algebra and Functional Analysis over phone yesterday. I could answer a few (thanks to you). Also, have to revise programming for tomorrow's viva. I'll let you know if I have further doubts regarding Complex Analysis. I'll probably study Analytic Functions over the weekend.

Balarka Sen

Aw nice man

Anonymous

He asked me if I know what Hermite polynomials are

Anonymous

I said I heard of it XD

Anonymous

But forgot what it means

Balarka Sen

15:26

Oh right there's some shit

If you write $\cos(nx)$ in terms of $\cos(x)$ or something

Anonymous

Hermite polynomials

In mathematics, the Hermite polynomials are a classical orthogonal polynomial sequence. The polynomials arise in: probability, such as the Edgeworth series; in combinatorics, as an example of an Appell sequence, obeying the umbral calculus; in numerical analysis as Gaussian quadrature; in physics, where they give rise to the eigenstates of the quantum harmonic oscillator; in systems theory in connection with nonlinear operations on Gaussian noise. in random matrix theory in Wigner-Dyson ensembles. Hermite polynomials were defined by Laplace (1810) though in scarcely recognizable form, and studied...

Anonymous

Ya, sounds like cool stuff

Anonymous

I should read it someday

Balarka Sen

Ah so they are orthogonal in some sense

I dunno this

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♦ $Mathematics$ the no-normie zone

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♦ $Mathematics$ the no-normie zone