@MartinSleziak When I learned how to represent recurrent sequences using this technique many years ago, the technical term was Lamere Ladder. Interestingly, it's not widely used (the term) these days, it took me some time to find a link highlighting the term books.google.co.uk/…

@rtybase As I learned later, the term cobweb plot is also used.

I have also mentioned this here: chat.stackexchange.com/transcript/14150/2016/6/21

Hi @arjafi and @Mithrandir24601! I am glad to see that some users are interested in this room enough at least to peek in :-)

This room definitely has bigger chance of getting useful if more users start coming here. We will see what happens after the project planned by heather and Simply starts - that might bering at least a few more users here.

*bring (sorry fo the typo)

@MartinSleziak I'm actually here because of that project :P

I see. Although look at the room schedule, it's still a bit of time before it really starts.

I am slightly surprised by the sheer number of chat rooms on MSE though - even other similar sites such as physics only have a few and other sites in involved with only have somewhere between 1 and 3

@MartinSleziak Oh yeah, I know that (I won't actually be here for the first few times), but I've been told to 'peek my head in' so I'm pingable and have a habit of leaving leaving tabs open for such things

@Mithrandir24601 I am partly the culprit responsible for this - at least for some of them. I have tried to describe my view on specialized chat rooms in this answer: Increasing chat use, pros, cons and the tour.

@MartinSleziak If people use them, I do agree it's a good idea. I just don't get why people don't use them :P

Well, people do use some of them. And several rooms had some periods of higher activity and then periods of almost complete inactivity. For example, in functional analysis room most of the messages have been posted in the past two weeks. But maybe later it will back to the previous state - when it was virtually inactive and eventually frozen.

Or Set Theory was rather active until December 2012, but then there was rather low activity for a very long time.

Then also a bit more activity in 2016 when johnny09 asked a few questions there. We will see whether the activity picks up a bit thanks to the study group proposed not too long ago by Alessandro Codenotti: chat.stackexchange.com/transcript/2318/2017/8/14

Hi @supinf! Having seen your post on the main I'd say that you would be able to contribute a lot both here and in the functional analysis chat room. So I think that people who come here to ask questions from these areas will be grateful if you occasionally have a look whether something interesting is going on here.

In functional analysis chat room, two users asked quite recently a series of questions - one of them from Rudin's Functional Analysis, the other one from Limaye's Functional Analysis. In fact, you can see some of those discussions here.

I wonder whether the increased presence of mods in this room (both mixedmath and arjafi are here) is just an accident or whether it has to do with the calculus "class" which was heavily discussed both in chat on meta in the last few days.

@MartinSleziak I would imagine so

@MartinSleziak thanks but i am not sure if i have enough time to regularly check these rooms.

No problem.

Since I saw you here and I have noticed your contributions in questions on functional-analysis, it was hard for me to resist and not to invite you.

Hello all , i have a question regarding the continuity of Arg z on the non-positive real axis.

I am not really the right person for questions from complex analysis, but here it seems to me that the limit does not exist. You can get close to $\pi$ and to $-\pi$.

Related question on main: Complex Analysis: Show that $Arg(z)$ is discontinuous at each point on the nonpositive real axis.

Yep i can prove either way

BTW are both definitions of arg commonly used - with values between $0$ and $2\pi$ and with values between $-\pi$ and $\pi$?

sorry i mistyped , should be Arg z

But my question is, to prove that $f(z)$ is not continuous on $z_0$, can i simply show $\lim_{z\to z_0}f(z)\neq f(z_0)$ although the limit does not exist?

or should i prove that $\lim_{z\to z_0}f(z)$ does not exist, and hence $\lim_{z \to z_0} \neq f(z_0}$

It depends on what you mean by $\ne$. I guess some people might argue that this notation does not make much sense if the LHS is not a nubmer (does not exist).

But yes, showing that limit for $z\to z_0$ is not equal to $f(z_0)$ is sufficient.

This implies that the limit either does not exists or if it exists then it has a different value.

I used the negation of the definition of limit to show $\lim_{z\to z_0}f(z) \neq f(z_0)$. Is that okay?

But it is not necessary to prove that the limit does not exist. (Although I'd guess it is quite easy, so I do not see any important reason to avoid this.)

@LittleRookie Yes that should be ok. Although I would imagine that sequential definition of limit might be quite handy here.

By which I mean that it is enough to find one sequence such that ...

i used the epsilon-delta definition.

But essentially both approaches boil down to almost the same thing.

I have no idea how to use sequential definition on complex function though =/

If a function $f$ is continuous and $x_n\to x$, then also $f(x_n)\to f(x)$. This is true for complex functions, too.

So if you exhibit a sequence when the above property fails, that is enough to show that $f$ is not continuous.

Oh i see.

I guess i have to look at my lecture notes on real analysis again. It's not mentioned on complex analysis textbooks

In fact, the above is true for any metric space. (I am not sure whether you have learned about metric spaces already.)

Not yet =/

I dont think i will learn about it during undergrad, because im on Statistics track.

And it is true also in topological spaces, but there this property is no longer equivalent to continuity. (Sequences do not suffice to determine topology.)

But that is only a digression.

Depending on what you are studying, it might be enough for you to work with continuous functions on $\mathbb R^n$; at least for some time.

Is it okay?

Well, it seems that you used $Arg(z_0-i\cdot\frac\delta2)=-\pi$, which is not true.

Still, it is close to $-\pi$.

my bad =/ let me edit it

But still you should be able to show that the difference is bigger than $\epsilon=1$. (Even if it is not equal exactly to $2\pi$.)

$|Arg(z_0 - i\cdot \frac{\delta}{2}) - Arg(z_0)| \geq |Arg(z_0-i\cdot \frac{\delta}{2})|-|(-Arg(z_0))|> \pi+\pi = 2\pi > \epsilon$

I do not think this one is correct either.

Ahhh sorry

Based on what do you claim that $|Arg(z_0-i\cdot\frac\delta2|>\pi$?

Its wrong =/

I think $Arg(z_0-i\cdot\frac\delta2)<0$ should be completely sufficient for what you need here.

Together with the fact that you know the value of $Arg(z_0)$.

ahh okay, so i dont have to explicitly write out the form of $Arg(z_0-i\cdot \frac_{\delta}{2})$

I'll have to go. See you later.

so i can just conclude $|Arg(z_0-i\cdot \frac{\delta}{2})-Arg(z_0)| > \pi > \epsilon$

See you

@LittleRookie The is exactly in the direction I meant.

Im troubled on showing continuity of $Arg z$.

It is a piece-wise defined function, so im not sure how to prove $Arg z$ is continuous on points that are purely imaginary.

Should i use the epsilon delta definition of continuity? Or there is another better method instead?

In the case of real piece-wise function, Eg: $f(x) = x^2$ if $x\leq 1$, $f(x) = 1$ if $x>1$.

Instead of using the epsilon-delta definition, i can observe that both $x^2$ and $1$ are continuous everywhere. Hence, $lim_{x\to 1-} f(x) = lim_{x\to 1-} x^2 = 1 = lim_{x\to 1+} f(x) \implies lim_{x\to 1} f(x)= 1 = f(1)$.