@s.harp I have. The problem is if I approach the limit delta z approaches $0$ along the real/imaginary axis or along a line, the function should ideally not have $z$ right? I mean shouldn't it just be in terms of delta z?
? for differentiability you should show that the limit $\lim_{h\to0}\frac{f(z+h)-f(z)}{h}$ is well defined. If you evaluate it at zero you get $\lim_{h\to0}\frac{|h|}{h}$, which does not exist/depends on how $h\to0$ is taken.
@s.harp yes exactly. So if we evaluate it at zero, we get $\lim_{h\to0}\frac{|h|}{h}$? I mean $f(z)$ is simply eliminated? It's a silly question but I wasn't sure if I could do that
I don't know what you mean with $f(z)$ is eliminated. You are evaluating an expression at the specific values $z=0$ and $z=h$, you will not have any $f(z)$ terms for general $z$.
So i need to include the sign as well. But if I also add the condition that the real number is positive then I don't have to worry about the sign right?
Sure, but what you want to show is that the function is not differentiable (at $0$ in this consideration), so you wnat to get different limits for the difference quotient. One way is eg $h=1/n$ then the quotient goes to $1$, but if you take $h=-1/n$ it goes to $-1$.
Ok, if I have $f(z)=z'^2$ where $z'$ is the conjugate of z, then can I say that $\lim_{h\to0}\frac{f(z+h)-f(z)}{h}$ becomes $\lim_{h\to0}\frac{h'^2}{h}$?
Oh yes I forgot to square it. So then I simply take $\lim_{h\to0}\frac{\overline h (2\overline z+ \overline h)}{h}$ and then approach $0$ via different paths? So for example I first take it along the real axis?
If $h$ is real then $\lim_{h\to0} \bar h/h (2\bar z + \bar h) = 2 \bar z$ (since $\bar h = h$). But if $h$ is pure imaginary, you get a different result
@s.harp if $h$ is real then how can the answer be $2 \bar z$? I mean the expression was $\lim_{h\to0}\frac{\overline h (2\overline z+ \overline h)}{h}$ so even if the $h$ and $\bar h$ are eliminated as $h= \bar h$ when $h$ is real, we still have $2 \bar z + \bar h$ in the parentheses
But if the real and imaginary components' partial derivatives are continuous but they don't satisfy the cauchy riemann equations then it will not be differentiable anywhere?
Guys, is there a good book which explores questions like why the square root of x^2 is mod x, or why (x^a)^b is only x^ab when x is positive and so on? Looks like i have a lot of confusion in this and was wondering what book i should read to finally put that confusion to rest.
@KarlKronenfeld I was working with a definition, which I could not understand. But in the mean time I have found out where the problem in my understanding was (the definition was that $F$ is a prefilter (on $X$) if The collection of all subsets which contain some element of $F$ is a filter, and I had understood the subsets to be subsets of $F$, but they are subsets of the space on which $F$ is to be the prefilter which clarifies my confusion)
@Danu I don't know why I couldn't prove this when you first mentioned it. Probably I wasn't completely awake then. Of course, if $\tilde{X}$ is compact, $p : \tilde{X} \to X$ is a covering map, then $p(\tilde{X})$ is compact as continuous image of compact sets are compact. But $p(\tilde{X}) = X$, as $p$ is surjective. :P
I just rethought about this and realized this was stupidly easy.
@Karl in a filter you have $A, B \in F \implies A\cap B \in F$, here you just have $\exists C \in F$ st $C\subset A\cap B$. Also if $A\in F$, $A\subset B$ does not need to imply $B \in F$ as it does for filters.
_the collections of all subsets_ is $\mathfrak P (X)$, I had thought it to be $\mathfrak P(F)$ which was my misunderstanding (taking that second one you always get a filter with above construction).
But proper containment is given for the construction for free (lets call it $\tilde F$), since if $A \in \tilde F$, $B \in \mathfrak P(X)$ and $A\subset B$ then $\exists C \in F$ st $C\subset A\subset B$ and $B \in \tilde F$.
This is equivalent to the above definition and probably simpler: For all $A_1, A_2 \in F$, there exists $A_3 \in F$ such that $A_3 \subset A_1 \cap A_2$
Anyhow, @Balarka, I put the problem — lacking the Hausdorff hypothesis on $X$ — on a graduate topology qual many years ago and no one (including me, apparently) caught the error. I just realized it a few days ago whilst writing my midterm exam for my two topology students.
@TedShifrin We're just using that if $A, B$ are two open sets in $X$, $a\in A$ and $b \in B$, then there exists open nbhds $U$ and $V$ of $a$ and $b$ resp. such that $U \cap V = \emptyset$, right? Is that equivalent to normality?
Well, @PVAL, if the problem specifies that one space is Hausdorff and doesn't mention it for the other, I consider that edict of yours complied with. :)
Anyhow, I am so used to working in the context of manifolds for that problem, I messed up. I'm just amused that I only realized I'd messed up after retiring :D
Actually, let me start thinking about those extra problems.
@TedShifrin Oh, and by the way, here's a problem I tried to solve a few days ago. $f_t : B \to B$ be a sequence of contractions, $t \in I$ so that $F : B \times I \to B$, $F(s, t) = f_t(s)$ is continuous. Then $g : I \to B$, $g(t) = x_t$ ($x_t$ is the unique fixed point of $f_t$) is continuous. $B$ here is a closed ball in $\Bbb R^n$.
My idea was the following: Choose a point $a \in B$. Then define the sequence $\{a_k^t\}$ where $a_{k+1}^t = f_t(a_k^t)$. $\{a_k^t\}$ converges to $x_t$ for each $t$. Here $a_0^t = a$.
$g_k : I \to B$, $g_k(t) = a_k^t$ is then a continuous path for each $k$. I tried to prove that $g_k$'s uniformly converge to $g$.
There was an old qual question here asking people to show that eps morphisms in Top were not necessarily surjections. Nobody realized, apparently, that this is false.
Hi. It appears that the product of the differences between 3 consecutive integers in whatever order always equals 2. However, I can't find the pattern for 5 integers...
I would seek to be a good problem solver.
So, below are the two courses from which I would like to pick one.
Course1:
https://www.coursera.org/course/algs4partI
https://www.coursera.org/course/algs4partII
Course2:
https://www.coursera.org/course/principlescomputing1
https://www.coursera.o...
I would seek to be a good problem solver.
So, below are the two courses from which I would like to pick one.
Course1:
https://www.coursera.org/course/algs4partI
https://www.coursera.org/course/algs4partII
Course2:
https://www.coursera.org/course/principlescomputing1
https://www.coursera.o...
