I am drunk right now...why am I here?

dunno

I am actually trying to prove Collatz conjecture.

Unwise to attempt when drunk.

Thanks for the feedback, @skullpatrol.

np pal @DonLarynx

Hello,everyone,the last day problem is nice:2014China's Olympic mathematics problem:math.stackexchange.com/questions/1076363/…

@Venus,Hello, yesterday integral have simple closed form?

in other words, the AMM integral have simple closed form?

Cool pic @skullpatrol, but this happened long after I realized most of society is pretty dumb

@skullpatrol generalization

What about "generalization"?

@skullpatrol i am disinterested in others based on the fact that the majority drink and 99.999% of the students smoke.

Then you're in the right subject my friend.

@skullpatrol Thanks. I tried smoking every day and doing math when I was a sophomore and could barely manage a 2.5 GPA those two semesters. Downright embarrassing, but it is what it is (or was)

Smoking is not cool.

Indeed. I don't really do anything when I smoke.

I have not done any research in smoking. I am now not sure if smoking is really harmful.

Wow, that awkward moment when your pen runs out of ink.

That awkward moment when your colleagues realize you do your math in pen.

@Jasper: It certainly is not not harmful. Does this answer your question?

@DonLarynx And whatca we conclude from that? :)

@DonLarynx Could you maybe explain me the proof?

@DonLarynx: You're still here? O_o

I've been doing meth all day @Huy

*math

@DonLarynx: Which time zone are you located in?

@DonLarynx What can be harmful?

@Huy -5

@evinda Smoking

@DonLarynx: It's high time you went to bed!

You can do some more maths after you've had a good night's sleep.

@DonLarynx And what's with the proof? :p

@Huy: I'm not tired yet

@evinda: Prove smoking is harmful.

@DonLarynx: That's because you messed up your sleeping pattern.

@Huy: correct, so I will wake up at 7:30am tomorrow, drink coffee, and hopefully sleep by 11pm tomorrow or sooner

@DonLarynx How shall I prove it? :p

@evinda I don't know

@DonLarynx Neither I do.

:D

@DonLarynx: I'll take you by you word. If you don't, you owe me two sheep.

@Huy I will give you two lamb, lamb is good according to Bart

@DonLarynx: Deal.

@Huy when you are drunk time passes 1000x faster you have no idea

@DonLarynx: I was drunk the previous two evenings, which is why I stayed up until after midnight. But at some point, I decided that it was time to sleep, still.

@Huy: I woke up yesterday at 1:38pm

The result of drinking 4 16oz natty ice

don't drink

it makes you sleep more

@DonLarynx: That does not sound like a healthy condition.

I'm unfamiliar with anything other than SI-units.

@Huy: I had not drinked in 19 days.

@Huy I drank 475mL of beer last night

LOL

@Huy Typo: 12 days

and you're drunk of it?

@DonLarynx Why does it hold that $x_l \equiv x_n \mod{p^{n+1}}$ and $y_l \equiv y_m \mod{p^{m+1}}$ ?

I am more tipsy than drunk, more like above-average tipsy

@evinda define l, n, and m for me please

Well, I guess it is time for sleep

See you all in the AM

@evinda I will attain to you in the AM

goodbye

@DonLarynx Goodbye :)

@r9m Sorry. Real busy day. Finally back. Let me look.

The question is weirdly worded, @KajHansen.

@evinda Basically, one writes $x = p^{n'}\cdot u$ and $y = p^{m'}\cdot v$ with $u,v$ units in $\mathbb{Z}_p$ and $n' \leqslant n,\; m'\leqslant m$, then $x\cdot y = p^{n' + m'}\cdot (uv)$, where $uv$ is a unit (as a product of two units) and $n' + m' < l$, so it follows that $xy \neq 0$.

Morning @BalarkaSen.

@DanielFischer How do we choose $n'$ and $m'$? How do we know that $n' \leqslant n,\; m'\leqslant m$ ?

@math110 I believe it has, but I haven't found it yet. I think it's too difficult for me. How about you? Have you found it?

@evinda $n$ and $m$ were picked such that $\overline{x_n}\neq 0$ and $\overline{y_m} \neq 0$. That means that $x_n = p^{n'}\cdot u_0$ for an $u_0$ which is not divisible by $p$ and $n' \leqslant n$ - for if we had $n' > n$, then $x_n \equiv 0 \pmod{p^{n+1}}$ and hence $\overline{x_n} = 0$, contradicting the choice of $n$. Similarly for $m',m$ and $y$.

@venus,I think maybe we must use sum define

@DanielFischer I haven't understood why if $n'>n$ then $x_n \equiv 0 \mod{p^{n+1}}$.. Could you explain it further to me? :/

@math110 I did do that but I'm getting nowhere.

@evinda If $n' > n$, then $x_n = p^{n'}\cdot u_0 = p^{n+1 + (n'-n-1)}\cdot u_0$, and $n'-n-1 \geqslant 0$, so then $x_n$ is divisible by $p^{n+1}$.

Hello @DanielF

@DanielFischer A ok.. Is it $(\overline{x_n})=(x_1, x_2, \dots, x_n)$ ?

@Huy Guete Morge

Guete Morge, @Venus. Where did you learn that from :D

@evinda No, $\overline{x_n}$ is the residue class of $x_n$ modulo $p^{n+1}$.

@Huy So you're using Swiss German language, I presume?

@DanielFischer So $\overline{x_n}=x_n \mod{p^{n+1}}$, right?

@Venus: When I'm talking with Swiss German friends, yes, I do.

Yes.

@DanielFischer Then what will be the form of $x=(\overline{x_k})$?

@Huy How many languages do you speak?

@Venus: Not many. Swiss German, normal German, English, French, Vietnamese, and I'm studying Hungarian a bit.

@evinda That's the representation of elements of $\mathbb{Z}_p$ you're using in your course.

I can understand basic Italian too, but I'm not good at speaking it.

@Huy Jeez! That's too many for average people

Who said I was average people. :P

brb, need to make some lunch.

@Huy Show-off!

@DanielFischer So is it $\overline{x_n}=(x_1, x_2, \dots, x_n)$ where $x_{k+1} \equic x_k \mod{p^{k+1}}$?

@evinda No. Look at the definition of $\mathbb{Z}_p$ you're using. You need to be familiar with that, knowing what everything designates, without that, you won't understand the proofs of properties $\mathbb{Z}_p$.

@DanielFischer Do you mean this one: $\mathbb{Z}_p=\{ x \in \mathbb{Q}_p | |x|_p \leq 1 \}$ ?

@math110 The solution of Problem 11796 hasn't been published yet

@evinda No, the one with the sequences, $$\mathbb{Z}_p = \left\{ (\overline{x_n}) \in \prod_{0}^\infty \mathbb{Z}/p^{k+1}\mathbb{Z} : x_{n+1} \equiv x_{n} \pmod{p^{n+1}} \right\}.$$

Hi guys, this is my first time in a chat room of SE so I don't know if what I'm doing is according to rules (I read the rules but they are a bit vague). But I was wondering if someone can help me with some little questions on group theory. I have my retake tomorrow and I want to make sure that I understand every part of it

@RubenMeijs Ask, and we will see if somebody is able and willing to help.

I have to give an example or disprove the following statements.

@DanielFischer So does $x=(\overline{x_k})$ mean that $x=(x_1, x_2, \dots, x_k)$ with $x_k \in \mathbb{Z}/p^{k+1} \mathbb{Z}$ such that $x_{k+1} \equiv x_k \mod{p^{k+1}}$ ?

@Venus: Back. How are you today? What are you up to?

THe first statement is: a cyclic group which is the direct product of two non-trivial cyclic groups. I think $\mathrm{Z}_2 \times\mathrm{Z}_3$ is a proper example

@evinda You're always writing finite tuples. That's wrong. If you mean $x = (x_0,x_1,\dotsc,x_k,\dotsc)$, then that's okay.

@Huy Good. I'm doing nothing since it's Sunday. I spend all the day on the bed :P

Lazy Sunday, you say? :P

@RubenMeijs It is. Do you know the general condition when the product of two cyclic groups is cyclic?

@DanielFischer I do know when a product of $\mathrm{Z}_m \times \mathrm{Z}_n$ is cyclic, namely if gdc(m,n)=1. But I wouldn't know such a statement for all groups

@Huy Yep. Holiday is lazy time, haha

Good for you, @Venus, I hope you enjoy it!

@DanielFischer A ok... At the beginning we take $x_n \not\equiv 0 \mod{p^{n+1}}$ because we know that $x=(\overline{x_k}) \in \mathbb{Z}_p \setminus \{ 0 \}$ and thus at least one component is non-zero, let $x_{n+1}$, so $x_{n+1} \not\equiv 0 \mod{p^{n+1}}$ and since $x_{n+1} \equiv x_n \mod {p^{n+1}}$ we conclude that $x_n \not\equiv 0 \mod{p^{n+1}}$. Or am I wrong?

The second statement is to give an homomorphism $\phi$ from G to H, between finite groups G and H, such that |ker$\phi$| doesn't divide $|G|$

@RubenMeijs Every cyclic group is isomorphic to exactly on $\mathbf{Z}_n$ (if finite) or to $\mathbf{Z}$ (if infinite). In the finite case, $n$ is of course the order of the group.

@Huy I always enjoy my life.

That's good!

@RubenMeijs You asked a question on main not long ago, in the course of which you used the answer to that.

@DanielFischer yes I thought the statement was false, because ker $\phi$ is a normal subgroup. But I

@evinda You're off with the indices. The component $\overline{x_n}$ is an element of $\mathbb{Z}/p^{n+1}\mathbb{Z}$.

But I just want to know for sure, it's the fourth time taking this exam. Want to get a good grade right now...

@RubenMeijs You don't even need normality. The kernel is a subgroup of $G$, hence its order divides the order of $G$ (Lagrange).

Where on earth can you take one exam four times?

So, $x_n \equiv 0 \mod{p^{n+1}}$, @DanielFischer ?

@Huy This is one the reasons why I always grateful to my life

@evinda Well, no, we picked $n$ so that $x_n \not\equiv 0 \pmod{p^{n+1}}$. Typo?

@DanielFischer how does this statement about every cyclic group help with the direct product of two cyclic groups?

@Venus: I only look out of my window and I'm grateful to my life. :)

@Huy well I failed the course last year, but I did make the retake. So I had to redo the course. I passed on the first exam, but I want to get a higher grade so I was allowed to do the exam agian

@RubenMeijs If $C, \tilde{C}$ are two cyclic groups, isomorphic to $\mathbf{Z}_n$ and $\mathbf{Z}_m$ respectively, then $C\times \tilde{C}$ is isomorphic to $\mathbf{Z}_n\times \mathbf{Z}_m$, so one product is cyclic if and only if the other is.

No, you're right.

@DanielFischer And how do we deduce that $x \cdot y \neq 0$?

@evinda We look at the component with index $l = m+n+1$. We know $x_l \not \equiv 0 \pmod{p^{n+1}}$ and $y_l \not\equiv 0 \pmod{p^{m+1}}$, which means that the highest power of $p$ dividing $x_l$ is $p^{n'} \leqslant p^n$, and ditto the highest power of $p$ dividing $y_l$ is $p^{m'} \leqslant p^m$. So the highest power of $p$ dividing $x_l\cdot y_l$ is $p^{n'+m'}$, and $n'+m' \leqslant n+m < l$, so $x_l\cdot y_l \not\equiv 0 \pmod{p^{l+1}}$.

Would work with $l = n+m$, It might be that that was used, I forgot.

@DanielFischer At the proof it was taken $l=n+m+1$. :)

@DanielFischer Do we conclude that the highest power of $p$ dividing $x_l$ is $p^{n'} \leqslant p^n$ since $x=p^{n'} \cdot u$ where $u$ is a unit?

@evinda We conclude it because we saw that $x_l \not\equiv 0 \pmod{p^{n+1}}$.

@DanielFischer And how do we conclude from this that the highest power of $p$ dividing $x_l$ is $p^{n'}$ ?

@evinda That's how we define $n'$. And from $x_l \not\equiv 0 \pmod{p^{n+1}}$ we conclude $n' \leqslant n$.

I only see one answer there, not 16 and there's no a bounty given to the answer

@Venus: Because the other answers have been deleted, probably.

@Venus Get 10k, then you can see the deleted answers there. It's truly impressive.

@Venus Yes, yes, stay away from integrals!

@DanielFischer Are you serious??

@DanielFischer Hello Daniel Senpai.

@PedroTamaroff That's my new hobby

@Venus Yes, one of the good things of reaching 10k is that you can see deleted posts.

Bonjour @PedroTamaroff.

@DanielFischer I'm reading in French now, might actually learn something, eventually.

Bonjour, messieurs.

@DanielFischer So, if we have 10K rep, we can see deleted posts. What else privilege do we have on this site? How if I get 20K, 50K, or 100K?

Mostly, I have learned "Supposant," "Considerons," "On consdiere un diagramme commutatif,"

So much for my talking skills.

@Venus No additional privileges above 20k, that's the last step.

math.stackexchange.com/help/privileges

@DanielFischer So, any thoughts about the upcoming diamond?

@PedroTamaroff The election is not over yet, let's wait what it brings.

@DanielFischer Maybe it takes a year for me to get 10K, provided I can maintain my perform here like the past 1-2 months

If every message gets a star, no message gets a star.

If you want to highlight a piece of conversation, there's a tool for that.

@Venus Well, you have a goal now, to see the deleted answers of the thread mentioned above. That may speed things up ;)

Go to the transcript and have fun.

I seriously think to leave this place, it's not anymore to me as it was before.

@DanielFischer Ya, I guess so. I wonder, what kind of 16 deleted answers are there to answer an impossible integral.

@Chris'ssis Why do you wanna leave this place?

@Venus Too many negative thoughts related to my work in chat.

@Chris'ssis Well, you can just ignore the people you don't like.

@Venus Mostly un-thought-out starts, some troll posts, and a bit of completely wrong mathematics. The un-thought-out starts were deleted by owner, the others by 20k users or moderators.

@DanielFischer Why does $\overline{x_n}$ mean that $x_n=p^{n'}u_0$ for a $u_0$ whihc is not divisble by $p$?

@Chris'ssis Come on. This is internet, don't let the one who is far away makes you upset. ^^

@Chris'ssis But I can understand why you want to leave this site. It's just like me. I want to leave my country because I can't stand the people here anymore. And I cannot just ignore them, because they are all around me all the time.

@DanielFischer OK, I'll reach 10K in less than a year to see them all. Ganbatte!! :D

@Venus Actually, I find it a bit funny that people distinguish internet from real life. For some people, the internet is real life.

@evinda $n'$ was defined so that $p^{n'}$ is the highest power of $p$ dividing $x_n$. Then of course the cofactor of $p^{n'}$ in $x_n$ cannot be divisible by $p$.

@JasperLoy I mean it's a matter of good sense for someone that doesn't do very well in my area not to express things about my work. I mean one should learn for some years (many) and then say things about my work.

@JasperLoy For me, they're completely different.

@DanielFischer I haven't understood why $\overline{x_n}$ \in \mathbb{Z}_p \setminus \{ 0 \} implies that $x_n=p^{n'}u_0$... :/

@Venus I have learnt more about life online than offline. For me, the internet is real life and off the internet is fake life.

@Nick I think you only waste your rep. I don't think this integral has a closed form, IMHO.

@Chris'ssis Hmm. Are you upset by the people who said you were bragging yesterday? Or is it the people who rejected your article for publication?

@JasperLoy I agree this statement: "I have learnt more about life online than offline. " since you're older than me :D

@Venus: O_O Oh? Then atleast provide an answer using numerical analysis. I don't know how to evaluate integrals without closed forms.

@JasperLoy I was annoyed by the part with bragging. I worked a lot to get the results I got so far. No such results fall from the sky for free.

@DanielFischer * $\overline{x_n} \neq 0$

@Venus Never agree with people because they are older or are more senior. Most people grow more stupid as they grow older.

@JasperLoy What it is all about? Tell me, tell me. I don't know

Any ideas with the question above to do with injective maps? ^_^

Hey, @Nick!

@KhallilBenyattou Greetings, bud :)

How've you been, @Nick?

@Chris'ssis I think they just did not understand why you kept announcing it in this chat. But I do, because I like to repeat things in this chat too.

@Nick The hardest part is evaluating$$\int_{0}^{\pi/2}\frac{\tanh x}{1+\tan x}\mathrm dx$$

Okay, does the following integral have a closed form? $$\int \frac{\cos x \ \mathrm dx}{1 + e^x}$$

My attempt of $\int \frac{\cos x}{1 + e^x}$

@Venus Well, not much. Just that she kept announcing her result yesterday without the details, so some people got tired and said she was bragging.

@JasperLoy Starred! I saw many cases recently about that specially people with higher education :D

You could consider that $\cos x = \Re ( e^{ix} )$, @Nick. Then you have that $$\int \dfrac{\cos x}{1+e^{x}} \text{ d}x = \Re \left( \int \dfrac{e^{ix}}{1+e^{x}} \text{ d}x \right)$$ I'm not sure if that makes it any simpler.

@KhallilBenyattou I've been dwelling in things beyond me.

@KhallilBenyattou Um, where do I learn about $\Re ( e^{ix} )$ . I've honestly never seen it before.

@Venus lol, is it possible to do this question? At all?

@KhallilBenyattou Ah, looking at the complex part would be complicated? lol

@Venus Yes, most of my politicians are highly educated, yet they talk rubbish.

@JasperLoy Hhmmm, maybe they were just kidding

@Venus Partly yes, partly no, I think.

@JasperLoy You own politicians. Good for you :D

@evinda I don't understand what your confusion is. We have an integer $x_n$. It is not divisible by $p^{n+1}$. Hence, if we write it in the form $x_n = p^{n'}\cdot u_0$ where $u_0$ is not divisible by $p$ - and every nonzero integer has a unique such decomposition - the exponent $n'$ must be smaller than $n+1$, i.e. $n' \leqslant n$.

@Nick Not funny, lol.

@Nick Let me study your answer first

I came prepared. =P

@Venus It's just repeated use of by-parts. I can't integrate yet. I'm just a dumdum. It isn't an answer. I'm stuck at $\int \ln(1 + e^x)\ \mathrm dx$

Oh, do you just need to integrate by parts again and again, @Nick?

Anybody? ^_^

@KhallilBenyattou No, the method is probably more complicated than that. But all I did was by-parts. Typical me.

@Nick I doubt it's correct

Wolfram Alpha gives this

@KhallilBenyattou That's not a map with domain $B$. Its domain is $\{e,f\} \subsetneq B$. It is injective, however.

So does every element of the domain need to be mapped to some element in the codomain, @DanielFischer? (Thanks for the reply!)

@KhallilBenyattou It's the definition of the domain of a function that every element of the domain (and only these) are mapped by the function.

@Venus Well, ofcourse it's not the $F(x)$ for whom $F'(x) = \frac{\cos x}{1 + e^x}$ but I was halfway there. The thing is too complicated for me. I'll try it again in a few years when I learn enough.

Oh, so if a domain $A$ and codomain $B$ have the same number of elements and the mapping from $A$ to $B$ is injective, does that automatically mean that the mapping is also surjective (and therefore bijective), @DanielFischer?

@Nick How old are you Nick?

@KhallilBenyattou If the number of elements is finite. For infinite sets, it does not follow.

Ah, and considering only finite sets, then the only way for a mapping from $A$ to $B$ to be injective and not surjective is for $|B|>|A|$, right @DanielFischer?

@KhallilBenyattou Right.

@Venus: I am 6255 days old. But I started learning math with interest only 2 years ago... not half the interest you have though.

@Nick Where are you from?

@Venus ... My social security number is 31415926535. Seriously, what's with all the questions?

Favourite song, @Nick? =P

@Nick Nothing, I'm just curious. Am I bothering you with my questions? ^^

Thank you for the help, @DanielFischer!

@KhallilBenyattou You are God - Tryad

@Venus Nah, this information is publicly available at my profile page + my exact location is tagged on the mse-chat dwellers map.

@Nick I haven't see your profile page

@Nick What is mse-chat dwellers map?

@KhallilBenyattou Busy now, I have guests. I'm also sneaking and doing physics work.

No prob! That's dedication, @Nick! ^_^

@KhallilBenyattou: I'll listen to it later. I'll ask you some other time.

Cool!

I'll add to the 'About Me' section of my profile so that you can listen to it whenever you wish, @Nick. ^_^

Someone pin back the map!

I've lost the link

Wait, I've got it.

@Venus

=_= no, don't star the transcript post.

Pin Your Location | Instructions

Star that, @Nick!

(We can save the mods some work. =P)

Great :)

(Sorry, I thought it'd star the original post, but it starred the transcript post!)

^_^"

@Nick I see it

@KhallilBenyattou wow, everyone's location is there

@Venus Do you know a good book or something online where I can learn Lebniz formula in detail with examples? (I'm on that chapter in the book Interesting Integrals, and I can't understand it much)

@TheArtist Yes, HERE! :D

@Venus didn't understand what you mean :)

It's a pretty cool feature that was here a while ago and always used to be pinned to the starred message board, @TheArtist. ^_^

@TheArtist If you need online sources, you can learn it from MSE

hi

Or try Google "Feynman Integration" @TheArtist

i was grounded!

@Venus I like to ask questions over here :) it's hard to post questions asking people to esplain theory

@TheArtist What kind of questions?

did you miss me Venus??

@Venus with a particular definite answer

@Venus okay okay :D I guess I can ask here :)

@KhallilBenyattou Yep it is :D

@TheArtist I'm not good at theory but just ask away, maybe someone else can answer your questions here

I'll try my best to help out as well, @TheArtist!

@KhallilBenyattou @Venus Thank you both so much :)

Okay now :

The formula is :

@beginner How are you?

I'm sorry if I forgot you

There are too many people here

$$\frac{dI}{da}=\int_b^c \frac{\partial f}{\partial a} dx$$

Namaste @Integrator

This is the formula , since the last two terms in the RHS, of what I'm about to show cancels

@Anastasiya-Romanova秀 You haven't yet answered this dear!

@Venus Hi!

Wait wait :) I'm writng the question :)

^^^^ I don't understand how this result came, the author shows no step :/ and this is the first time I'm exposed to Leibniz formula

@TheArtist Just differentiate both of sides w.r.t. $a$

I don't know how the second result came, but I know the second integrand is the derivative of the first one with respect to a

It holds provided $a$ is a real number, not integer

Oh, this is differentiation under the int. sign. I've only seen it a few times, but the main idea is to introduce a new parameter ($t$) into the integral to generalise it, then differentiate it according to the rule $(\psi)$ and finally substitute a value of of the parameter in such that the integral matches up to the original. $$(\psi) \qquad I(t) = \int f(x,t) \text{ d}x \implies I'(t) = \dfrac{\text{d}}{\text{d}t} \int f(x,t) \text{ d}x = \int \dfrac{\partial}{\partial t} f(x,t) \text{ d}x $$

@TheArtist $$\frac{\partial}{\partial t}\int_a^b f(x,t)\mathrm dx =\int_a^b \frac{\partial}{\partial t}f(x,t)\mathrm dx$$

I'm not sure of how to prove the relation you've written down, but the general idea of how to go about questions like that is above, @TheArtist. This might be worth reading as it gives a few examples of how to go about using this technique to integrate things in a different way. I'm not sure if it has a proof of the relation though. ^_^"

@TheArtist You should take a look at answers by @Xpaul and @RandomVariable

Also, as a quick aside, the expression on the left hand side of your comment shouldn't be a partial derivative w.r.t. $t$. Instead it should just be an ordinary derivative $\text{d}/\text{d}t$, @Integrator.

@KhallilBenyattou yes @KhallilBenyattou thank you, this is the formula which I mentioned :) I just don't know how it has been applied to what the author has written :/ (the example I have shown)

@Integrator thanks :)

like he has just differentiated the first equations, both right and left side with respect to a. And that's what the second equation is :/

It might be clearer if we knew which integral the author was trying to evaluate, @TheArtist.

Is the author trying to find the integral at the bottom of the page image?

@KhallilBenyattou yes :)

@KhallilBenyattou though he doesn't mention it :) but yes

Really? It seems like a smaller part of a bigger problem.

In the next page ^^^ he puts down this result, and differentiates it again to get another result :)

Oh, the author's just showing the results but not the intermediate steps.

But if we were finding the second result, we would go from the second to finding the answer. He's doing it the opposite way :) I would never know how to start with the third and go to the first to find it :p

I'm slightly confused now. All the author is doing is integrating the result in the grey box at the top of the image w.r.t $a$, using the rule above that @Integrator wrote down, and is rearranging to get the result in the next grey box, @TheArtist.

@KhallilBenyattou Feynman method is nothing but an application of leibniz rule!

@KhallilBenyattou me too

@Integrator so all that's happening here it's basically just differentiating a known result to get more results? :)

@KhallilBenyattou Isn't $$\frac{\partial }{\partial a}f(a,b)=\frac{\mathrm d}{\mathrm d a}f(a)$$ if we assume $b$ as constant?

Ah, gotcha. I haven't seen that version of Leibniz's rule. The only one I've seen is to do with differentiating a product of two functions $n$ times and getting a summation that looks eerily like the Binomial expansion, @Integrator.

@TheArtist Wait!

Yep, it is @Integrator.

@TheArtist Where exactly do you have a problem?

@Integrator okay :)

Ummmmm

@TheArtist in 3.1.2 ?

@Integrator yep :) if I know that , il know the next

Hi folks

does anybody have some courses about waves propagation (plane waves, transverse, etc) in a graduate level