Mathematics - 2020-08-13

Knight

3:16 AM

@Thorgott So, which one should I pick? Rudin is terse, and I thought classic books may be having clear explanation.

Yuvraj

4:10 AM

Hi everyone

Ted Shifrin

4:40 AM

@Knight: Rudin is way too hard for you. Are you taking an actual course that has a textbook? You were reading Spivak a week ago. That's an excellent source. Other options are Mattuck's Introduction to Analysis, which I like a lot, and Abbott's Understanding Analysis.

Hi @Yuvraj. What's up?

Yuvraj

@TedShifrin struggling with a small limit question!

Ted Shifrin

What is that?

The Substitute

In probability, what is the meaning of $(X,Y)\sim N_2(0,0,1,1,\rho)$? I believe this is a bivariate normal, but I don't see why we have 5 parameters.

Yuvraj

limit actually I need to find limit x tends to $0^+ $ [1+[x]]^{2/x}$

@TedShifrin

Ted Shifrin

@Yuvraj: If $x$ is a small positive number, what is $[x]$?

Oh, wait, you're using way too many brackets. What are you doing here?

I thought $[x]$ meant greatest integer $\le x$, but you have way too many brackets. Please make this clear.

Yuvraj

4:57 AM

It is gif

@TedShifrin

Each bracket is gif

Ted Shifrin

Oh, OK, it is what I thought.

But $[x] = 0$ for all small $x>0$.

So what is $[1+[x]]$?

Yuvraj

Is 1

Ted Shifrin

Yes. So the answer is ...

It's more interesting if you do $$\lim_{x\to 0^*} (1+x)^{2/x}.$$ Do you know that?

Yuvraj

Yes sir I wrote that $e^2$ but answer was 1

Ted Shifrin

LOL, the one I just gave you is $e^2$, but of course the answer is $1$ to your question.

Yuvraj

5:02 AM

How?

Ted Shifrin

You just told me that for $x>0$ small, we have $[1+[x]] = 1$. Always. So what is $1$ to any power?

Yuvraj

1

Ted Shifrin

What's the limit of the constant function $1$?

Yuvraj

Equal to constant

Ted Shifrin

So we're done.

Yuvraj

5:04 AM

Sir actually there is one more part

Where x was tended to $\infty$

For that also [x] =x

Ted Shifrin

So you're asking $\lim\limits_{x\to\infty} (1+x)^{1/x}$?

Yuvraj

Yes

Ted Shifrin

Is the $1$ relevant there?

You can do it without the $1$ first and then see if it changes anything. What's the answer?

Yuvraj

1 is not relevant right ?

Ted Shifrin

You'll check that in a minute. Without it, what's the limit?

Yuvraj

5:07 AM

1

Ted Shifrin

OK, how'd you do it?

Yuvraj

[x] will be equal to x as limit x tends to infinity

Adding one make no change

Ted Shifrin

No, that doesn't make sense. But you can use squeeze to deal with properly.

Yuvraj

Ah! I do not know how to apply it?

Ted Shifrin

$x-1<[x]\le x$.

Yuvraj

5:13 AM

Yes

abhas_RewCie

$\lim \limits_{x \rightarrow + \infty} (1+x)^{1/x} = \lim \limits_{x \rightarrow + \infty} e^{1/x \ln (1+x)}$

Ted Shifrin

So, if you can show that $x^{2/x}$ and $(1+x)^{2/x}$ have the same limit, you're done by squeeze.

abhas_RewCie

$1$ is not relevant here.

Yuvraj

@abhas_RewCie ok

@TedShifrin that was a easy to analyse how do we do it for a function like 1/n^2+2n function

abhas_RewCie

@Yuvraj If you understand $\epsilon - \delta$ definition, then you can make more sense of what's happening here...

Yuvraj

5:17 AM

@abhas_RewCie no I am not aware

abhas_RewCie

@Yuvraj brilliant.org/wiki/epsilon-delta-definition-of-a-limit

@Yuvraj I'd rather say prefer books like IA Maron's - Problems in Calculus of One Variable

They have some easy examples and good olympiad style examples.

Yuvraj

@abhas_RewCie got it

abhas_RewCie

:)

Yuvraj

@TedShifrin are you there?

Ted Shifrin

Yes, back.

I don't understand what your question is.

Yuvraj

5:27 AM

@TedShifrin limx tends to $1^-$

Ted Shifrin

Huh? You wrote stuff with $n$s.

Are you missing parentheses?

Yuvraj

I was asking about how do we apply squeeze theorem

For complicated function

Ted Shifrin

To the original question? I wrote that.

Yuvraj

Let $f$ be a function such that $-3 \le f(x) \le 5.$ Find $\lim_{x\to2}((x-2)^2f(x)+1)$

Like this one

Ted Shifrin

There's no universal answer. You have to think.

abhas_RewCie

5:32 AM

@Yuvraj if $f$ is continuous, then the answer is $1$

Ted Shifrin

Well, what can you say about $(x-2)^2 f(x)$?

Yuvraj

@abhas_RewCie f is continuous

Ted Shifrin

You don't need continuity at all.

Yuvraj

@TedShifrin ok?

Ted Shifrin

That's important.

Yuvraj

5:33 AM

It would be leess than 5(x-2)^2

Ted Shifrin

Use the inequalities they gave you.

Yuvraj

And greater than -3(x-2)^2

Ted Shifrin

Or $\le$, and ....

Yes, and the justification is that .... ?

Yuvraj

No wait it is 1

Ted Shifrin

Yeah, I'm ignoring the $1$.

What would you do if they had $x-2$ instead?

Yuvraj

5:38 AM

In the question?

Ted Shifrin

Yes. What made your inequalities work?

Yuvraj

It would not change anything I feel?

Ted Shifrin

Well, you didn't answer my question just above.

Yuvraj

What made your inequalities work?
This one

Ted Shifrin

Yes

Yuvraj

5:43 AM

Ok both the function just above and below have the se limit

As our function

Ted Shifrin

If $-3\le y\le 5$ is it always true that $-3u\le yu\le 5u$?

Yuvraj

Yes

I feel so

abhas_RewCie

nu, only $u \geq 0$

Ted Shifrin

Very important.

So if $u$ isn't known to be positive, what's the best estimate to give?

abhas_RewCie

$u=0$

Yuvraj

5:49 AM

@abhas_RewCie yes yes!

Ted Shifrin

No, @abhas.

Yuvraj

@abhas_RewCie it is absurd for u=0

Ted Shifrin

Hint: What is the biggest $|yu|$ can be?

Yuvraj

5u

@TedShifrin

abhas_RewCie

$+\infty$

That's biggest, I've heard...^

Yuvraj

5:52 AM

@abhas_RewCie there is no information about u

abhas_RewCie

okay

Yuvraj

If u can be anything if u=(-) then

abhas_RewCie

Honestly, I don't understand what's the whole conversation about...

will have to go... :)

bye :)

Yuvraj

@TedShifrin yes ?

Ted Shifrin

I want the answer in terms of $|u|$.

Yuvraj

5:55 AM

5u I said

Ted Shifrin

(Here $u$ mght be $x-2$ in our problem.)

What's the biggest $|y|$ can be?

Oh, $5|u|$ is right. good.

I missed it above, but you needed abs value. Now you can do squeeze with the abs value.

Yuvraj

What is the abs value sir ?

Sorry what does it mean?

Ted Shifrin

You know absolute value. I just got tired of typing.

Enough for me tonight. It's late here. Night!

Yuvraj

Good night sir thanks for helping me out

@TedShifrin

Balarka Sen

6:15 AM

@Lelouch You gave some example where the 4-point condition holds wrt a fixed basepoint but the guy isn't hyperbolic. Can you reiterate what it was? I am not sure if that works

Lelouch

8:58 AM

@BalarkaSen Sorry, in retrospect the "counterexample" is BS because if that holds for a particular point with $\delta$, then the space is $3 \delta$ hyperbolic. Maybe that's not $\delta$ hyperbolic, but that would be somewhat pathological counterexample ig.

I mean, if the basepoint condition holds with $\delta$, then the four point condition holds with $3 \delta$

Anyway, if you don't mind summarizing, can you please mention which theorems from BH he proved today ? I had algebra 2 endsem today and I am feeling bit sick so I didn't attend todays class.

There was a MSc First year guy I would have asked, but he's not attending the class today either.

Balarka Sen

9:28 AM

@Lelouch Right, I just noticed that. If that condition holds at a single basepoint then it's $2\delta$-hyperbolic, there is no counterexample.

He proved exponential divergence.

Lelouch

10:40 AM

and what did he do in the second half @BalarkaSen ?

Balarka Sen

hyperbolic => linear isoperimetric inequality

Suisse

@TedShifrin The word "lim" stands in english for limit, but in german we say Limes. And it comes from the romanian empire:

"von den Römern angelegter Grenzwall zur Befestigung der Reichsgrenzen." They for protecting the empire is called Limes.

Lelouch

ok cool, thanks

Edward Evans

11:15 AM

@Suisse du meinst "The Roman Empire"

Knight

11:44 AM

@TedShifrin Regarding this reply: Yes you said it quite correctly, it’s hard for me not terse. I shouldn’t blame a book which is so widely accepted just because I cannot get much from it. No, I’m not taking any Real Analysis course from any nationally/internationally recognised institutions,

I’m doing it on my own and take help from people over here.

Finally, thanks for that Mr. Mattuck’s book, I really like him I watched his lectures on Differential Equations (available on MIT.OCW) and in reality his way of teaching was very digestible.

(I’m still using Spivak, but it’s a first course Calculus book and I have quite completed it much)

Thorgott

Rudin definitely is terse

I think it's simply not suited for self-study anyhow

Knight

12:07 PM

Prof. Strang once said “We may not be Newton or Leibniz or Gauss or Einstein, but we can share some part of what they created”

Sometimes, it seems true!

feynhat

2:33 PM

So, the notion of zero-set (in $\mathbb{CP}^n$) of a polynomial in $(n+1)$ makes sense variables makes sense only when the polynomial is homogeneous. Apparently, the converse is also true. That is, if $F$ is a polynomial in $(n+1)$ variables and $F(x_0, \dots, x_n) = 0 \iff F(\lambda x_0, \dots, \lambda x_n) = 0$, then $F$ is homogeneous. Should this be easy? Any hints please.

Mike Miller

2:57 PM

I can't tell if I should write out my thoughts or not. What I did so far is write $F$ as a sum of homgeneous pieces and take derivatives in $\lambda$ and evaluate at $\lambda = 0$.

Oh whatever.

Write $F(x) = \sum F_i(x)$ where $F_i$ is homogeneous of degree $i$. Then $F(\lambda x) = \sum \lambda^i F_i(x)$. Supposing $F(\lambda x) = 0$ for all $\lambda$, plugging in 0 shows that $F_0(x) = 0$; taking derivatives in $\lambda$ and plugging in 0 you find that $F_1(x) = 0$; second derivatives show that $F_2(x) = 0$, etc. So $F(\lambda x) = 0$ for all $\lambda$ iff $F_i(x) = 0$ for all $i$. So your question is: "Suppose $Z(\sum F_i) = \bigcap Z(F_i)$. Is all but one $F_i = 0$?"

Anonymous

Is it possible to show by elementary means that the sum of two matrix commutators is a matrix commutator? That is, say $M = AB - BA$ and $M' = CD - DC$, can we somehow explicitly show that $M + M'$ is a commutator? $A, B, C, D$ are all $n \times n$ square matrices over $\mathbb R$.

Mike Miller

15

A: Does the set of matrix commutators form a subspace?

If the ground field $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, the following gives an elementary proof. Clearly every commutator has zero trace, so it suffices to show that every real or complex matrix with zero trace is a commutator. First, every traceless matrix is $\mathbb{F}$-similar to a ...

Anonymous

@MikeMiller That seems a bit complicated. Would it possible to express $M + M'$ as a commutator just in terms of $A, B, C$ and $D$?

2

Anonymous

Something like (A + B)(C + D) - (C + D)(A + B) (but that obviously doesn't work)

Mike Miller

No.

2

I'm not going to prove it though. You can try.

Anonymous

3:07 PM

@MikeMiller What's the reason?

2

Mike Miller

I believe I said that I'm not going to prove it for you, but that you could try.

Anonymous

@MikeMiller Oh, I thought that maybe you had some insight (short of a full proof) that allowed you to make the claim. :)

Anonymous

In any case, I don't know how to prove these sort of things so I'll probably have to wait till I learn some more algebra.

Knight

3:38 PM

I'm writing the actual keywords of the question:

Let $f$ be a polynomial with integer coefficients. Define $a_1 = f(0);~~a_2= f(a_1) = f(f(0))$ and $a_n = f(a_{n-1})$ for $n \geq 3$. If there exists a natural number $k \geq 3$ such that $a_k = 0$, then prove that either $a_1 = 0$ or $a_2= 0$

The $a_i$ s that are mentioned in the problem, are those the coefficients of the polynomial or just some random sequence?

Thorgott

I don't understand your question. They are defined in the very sentence you just wrote.

Knight

Okay! I thought they were the coefficients of the polynomial $f(x)$.

maths student

7:16 PM

Let A be a square diagonal matrix (n by n). Let the only elements on its diagonal to be +1 or -1. Alternating sign is allowed. For example, if n = 2. We consider A = [1,0; 0, -1] to be an ideal one. But I noticed A = [-1,0; 0,1] is as same as A = [1,0 ; 0, -1].

Question: If A (n by n) does not equal to the Identity matrix. How many distinct such matrix A exist? Please answer this in terms of n.'

This is question I saw in some website , I didn't understand what is meaning of this statement over here "I noticed A = [-1,0; 0,1] is as same as A = [1,0 ; 0, -1]".

Thorgott

that doesn't make sense

Nicholas Roberts

7:34 PM

does anyone know a good resource for learning about currents?

linear functionals on compactly supported differential m-forms

Thorgott

I'd assume Krantz & Parks is a good resource

Balarka Sen

7:49 PM

Federer :3

Thorgott

we all thought this, but you didn't have to say it

Alessandro Codenotti

Fremlin

Balarka Sen

@Alessandro uncannily rhymes with Gremlin

Lelouch

8:10 PM

what tf is wrong with the stars ?

Balarka Sen

someones starring crap

etc

Lelouch

the fault in our stars

lol

Balarka Sen

lmfao

Binod

8:41 PM

@robjohn Hi Sir .

robjohn

9:11 PM

@Binod hello

Binod

@robjohn Sir you told $x/2 +\sin(\theta)$ is not a one-one function.

I know it's correct , but how do I show it mathematically , without the help of a graphing calculator .

Leaky Nun

yeah cause you can't inject $\Bbb R^2$ into $\Bbb R$ continuously

@Binod any injective function must be strictly increasing or strictly decreasing

so its derivative, if it exists, must be always >= 0 or always <= 0

Binod

@LeakyNun I am not sure what you mean by that..

Leaky Nun

I was making a remark based on the fact that your function has two variables

Binod

@LeakyNun Okay ... but I have attempted questions where if f(x) is a polynomial function $f(x)+ sin(x)$ is one-one . Can you elaborate a bit more ?

Leaky Nun

9:23 PM

you probably mean $f(x) + \sin(x)$

Binod

@LeakyNun Yes ,edited .

Leaky Nun

so in that case see the actual message that I pinged you for the actual answer

5 mins ago, by Leaky Nun

@Binod any injective function must be strictly increasing or strictly decreasing

5 mins ago, by Leaky Nun

so its derivative, if it exists, must be always >= 0 or always <= 0

Binod

Okay , so it's not a one-one function as derivative changes it's sign multiple times.

Leaky Nun

exactly

Binod

But suppose if i have $Q(x) +sin(x)$ , after differentiating I will end up with a $Linear(x) + cos(x)$ , how can i solve it ?

Is it by default one-one ?

Leaky Nun

9:28 PM

well a linear polynomial must go to -infty in some direction and +infty in the other direction

robjohn

@Binod $f(0)=0$, $f\left(\frac{2\pi}3\right)=\frac\pi3+\frac{\sqrt3}2$, $f\left(\frac{4\pi}3\right)=\frac{2\pi}3-\frac{\sqrt3}2$. Note that $f\left(\frac{2\pi}3\right)$ is greater than either of $f(0)$ and $f\left(\frac{4\pi}3\right)$.

Hawk

I got a simple question. I am trying to show $cl(E) \subset E \cup E'$. To show $x \in cl(E)$ is in $E$ at least. I wrote that for any open set $N$, the intersection $N \cap E$ is non empty, but $N \cap E \subset E$ and I can take the open set to be as smalls possible. I just want to confirm if this step is right

robjohn

@Binod A continuous, injective function cannot have an interior maximum.

Binod

@robjohn Can I make an assumption if the value of f(x) does not decrease when x goes from 0 to $2\pi$ , it will be one-one in the domain R ?

(In general period of the trigonometric function )

robjohn

@Binod If it has period $2\pi$, then that sounds good.

You also have to make sure that it does not stay constant on any interval

so non-decreasing is not enough

Binod

9:37 PM

@robjohn Thanks .

Ted Shifrin

Hi @robjohn

Bhavay

Hi everyone. I was doing a question on limits .
$$\lim_{x\rightarrow0} [(x+1)/5]$$ []: greatest integer function, can i take 5 out of the limit ?

Ted Shifrin

Not in general, because the [] function is not a linear function.

Why would you want to, anyhow?

Bhavay

@TedShifrin no particular reason .

Ted Shifrin

Try doing this problem correctly and then doing what you want with the 5. Do you get the same answer?

Stan Shunpike

9:50 PM

@TedShifrin hey Ted!

how r u?

Ted Shifrin

hi @Stan

Hawk

I have another question, this one is a bit vague. I've been reading some theorems in measure theory and it looks like half the time the point of "almost xxxx" is there just because it can be inserted. Is it just to let everyone know "this can be made weaker"?

Ted Shifrin

Or stronger, you mean?

Depends if it's a hypothesis or a conclusion.

Hawk

hypthoesis

usually

Ted Shifrin

Then the result is stronger.

Bhavay

9:51 PM

@TedShifrin Sir is this the same as above :$[\lim_{x\rightarrow0} (x+1)/5]$?

Ted Shifrin

@Bhavay: Not necessarily.

Why don't you just work the problem as it's written?

Take $x$ with $|x|$ very close to $0$ and evaluate $[(x+1)/5]$.

Hawk

I got a simple question. I am trying to show $cl(E) \subset E \cup E'$. To show $x \in cl(E)$ is in $E$ at least. I wrote that for any open set $N$, the intersection $N \cap E$ is non empty, but $N \cap E \subset E$ and I can take the open set to be as smalls possible. I just want to confirm if this step is right. My def of closure is if x is in $\cl(E)$, then for any open set N(x), it also contains a point in E.

Ted Shifrin

What you said is just wrong.

Bhavay

@TedShifrin but then i don't understand what's the difference btw the two .

Ted Shifrin

"To show $x\in \text{cl}(E)$ is in $E$ at least." Is totally wrong.

@Bhavay: Because you're treating [] as a continuous function, which it is not.

Stan Shunpike

9:54 PM

@TedShifrin thank you for your comment about the bilinearity of the inertia tensor. really saved me a lot of mistakes

worked perfectly when i tried a few simple examples

Ted Shifrin

Try what you said if you remove the $1$ and the $5$. Is it true that $\lim_{x\to 0} [x] = [\lim_{x\to 0} x]$?

@Stan: It should work perfectly :P

Knowledge is powerful, eh?

Bhavay

@TedShifrin Sir is this the method for question 1 or question 2 ?

Hawk

is the latter stuff i wrote correct?

Stan Shunpike

@TedShifrin truly truly amazing. i had no idea how powerful vectors were til i did this project. i mean, i knew they were, but literally every single formula i learned in my calc class at caltech i've used on this project

Ted Shifrin

@Bhavay: I stated the question in that line, didn't I?

Bhavay

9:57 PM

@TedShifrin Ah , i see they are not . $\lim_{x\to 0} [x] = D.N.E $ , while for $[\lim_{x\to 0} x]=0$.

Ted Shifrin

@Hawk: I can't read the rest if you're trying to prove something wrong. Are you saying those words prove the wrong thing?

@Bhavay: Precisely. It's all about continuity.

Now, $[]$ is continuous at $1/5$, so it will work in the original problem, but you're making dangerous mistakes if you don't write a justification in every case.

Hawk

i wasn't going for a full proof.

Ted Shifrin

You don't understand what it means to prove $x\in A\cup B$. You can't prove it's in $A$.

Bhavay

@TedShifrin I see , thank you for the help. I have one more doubt though..

Ted Shifrin

@Stan: Of course vectors are powerful, both in physics and in math. Did you do some of those geometric proofs in my linear algebra book in the first chapter? I forget.

robjohn

10:01 PM

@Binod I hope that helps.

Hawk

okay so what i left out was the other part. I meant if $x \in cl(E) \implies N(x) \cap E \neq \empty$. If $x \not\in E$, then x \in $E'$. If $x \in E$, then we are done

robjohn

@TedShifrin hey there...

Ted Shifrin

@Hawk: If you want to prove that $A\cup B\subset C$, then you can do two cases. Show $A\subset C$ and show $B\subset C$, but this direction you just cannot separate.

Aha, @Hawk, now you're writing something that makes sense.

Stan Shunpike

@TedShifrin I did. Your book is terrific :)

Hawk

no i m thinking about the other inclusion $C \subset A \cup B$. i was trying to argue how it might also be in "A" if it is not in "B"

that's why i left out the accumulation set in the beginning

Ted Shifrin

10:02 PM

What you wrote did not make sense.

But now your thinking is clearer.

Hawk

is the new thing i wrote complete for $C \subset A \cup B$ ?

Ted Shifrin

You start with a point of the closure. If it belongs to the set already, we're fine. If not, you use your definition to conclude that it's a limit point. Therefore the point belongs to the union ...

Hawk

that's what i wrote

Ted Shifrin

No, go reread what you wrote.

Part of becoming a serious math student is being self-critical and rereading.

Hawk

"$x \in cl(E) \implies N(x) \cap E \neq \empty$. If $x \not\in E$, then x \in $E'$. If $x \in E$, then we are done"

bein in the limit point means still that N(x) \cap E\{x} is nonempty

Ted Shifrin

10:08 PM

That's what you wrote after I complained loudly.

Bhavay

$\lim_{x\rightarrow0}\sqrt{(x^2+5x+7)} -\sqrt{x^3+9x^2+8} $ , My book write since the effective degree of the polynomial is not same $\rightarrow \infty - \rightarrow \infty = D.N.E$ . But isn't $\rightarrow \infty - \rightarrow \infty $ an indeterminant form and we should tried manipulating it first before reaching to the conculusion that the limit d.n.e ?

Ted Shifrin

I wrote a summary in words, rather than in symbols.

Hawk

yeah so we are saying the same thing

Ted Shifrin

Yes, @Hawk. I just want you to be careful not to write the nonsense you wrote the first time. OK, we're done with this.

Hawk

yeah the first one i wrote i skipped some details

Ted Shifrin

10:09 PM

@Bhavay: That looks like garbage to me. That's in an actual textbook?

No, @Hawk, the first one was WRONG. Just admit it.

Hawk

cuz i wasn't sure if i need a longer argument for the set E. but thank you.

Ted Shifrin

I explained twice why it was wrong.

Hawk

no i know its wrong

Bhavay

@TedShifrin Tbh it was in a youtube video ..

Ted Shifrin

OK, so don't say "I skipped some details."

@Bhavay: You get what you deserve if you trust YouTube videos. (Perhaps that includes mine.)

Hawk

10:10 PM

youtube vids are gonna be what u have to depend on this semester....

Bhavay

Sir what's your take on the above question ?

Ted Shifrin

@Bhavay: Be patient. You wrote $x\to 0$. Do you mean $x\to\infty$?

Hawk

and probably the whole year....

Ted Shifrin

Even if you have remote learning, isn't your professor doing lectures or presentations of some sort, @Hawk?

Bhavay

opps , sorry i meant $x\to\infty\$

Hawk

10:12 PM

honestly the process is so confusing for me i don't even know what is going on

Ted Shifrin

OK, now is it legitimate to say the limit is $\infty$ or $-\infty$? Does your course allow that to be an answer?

For example, do you say that $\lim_{x\to\infty} x^2$ DNE or $=\infty$?

Hawk

anyways ciao @Bhavay gl with ur progress. might return later

Bhavay

@TedShifrin DNE .

Ted Shifrin

Because with the problem you posted, I would say the limit is $-\infty$.

DNE is very vague .... it allows for infinite limit, negative infinite limit, or no limit at all.

So what happens if you do your usual conjugate trick with the square roots?

Bhavay

@TedShifrin Are you saying limit is $- \infty$ , as rate of decreasing of the second polynomial is faster than the first one ?

Ted Shifrin

10:15 PM

Yes, if one refers to such things. You can see that with the conjugate stuff for sure.

Bhavay

@TedShifrin Thank you sir.

robjohn

10:30 PM

@TedShifrin: I have posted the egg with some more graphics.

Ted Shifrin

Cool, @robjohn. I'm in the middle of something right now but shall look.

Nicholas Roberts

@Thorgott springer.com/gp/book/9780817646769 is this the book you speak of?

Thorgott

yeah

Nicholas Roberts

Thanks, have you read it? is it any good?

@Thorgott

Thorgott

10:48 PM

I've read parts of it (though not the chapters dealing with currents) and another book from them and I usually find their exposition to be very clear and digestible. Your mileage may vary, of course.

robjohn

@Thorgott Krantz is a good teacher and writer. I had him for my undergrad analysis, complex analysis, and Fourier series courses.

Nicholas Roberts

very cool. thanks guys

Suisse

Hi there :)

How do you read this formula with the argmax?

nbro

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Q: What is the scope of real-world deep learning applications in 2020?

2015 was a milestone year for AI--"deep learning" was validated in a very public way with AlphaGo. However, at the time, the question was raised: "What else is deep learning good for?" 5 years later, I want to gauge: How is deep learning applied to real world problems in 2020? What real world ...

Suisse

11:03 PM

I know that this O with a belt is a "weak leaerner model"

the gay T is the parameter

soo how to read that? hmm and what does the star over the j on the O with belt means?

Thorgott

ah, that's cool

Suisse

@nbro wanted to upvote, but no reputation :( -its a "good" question. I think the next step is somehow to connect all those small AI models together.. they still don't get context.

nbro

@Suisse There is or was a lot of hype around deep learning, but is it really being used to solve important real-world problems? Well, I know of at least one important application, so I will provide an answer, but I would like to see more interesting answers and use cases.

Suisse

@nbro I only know Tesla self driving thing. And also recognizing cancer on CT images. I am currently doing "predictive maintenance" with random forest. Then you can count all speech recognition which is used.. but I don't see more realworld stuff there to be solved by AI.. there are a lot for sure

nbro

11:50 PM

I just provided an answer ;)

@Suisse Btw, feel free to create an account on AI SE! :)

Suisse

@nbro there is an AI SE?? :D cool

nbro

Yep! I am a moderator there :)

Suisse

cool :) .. why am I a noob everywhere I go? hehe

nice answer, I'm curious about more realworld things for deeplearning

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