@JoshuaCiappara doesn't seem too difficult. We have $B = \bigcup_{n=1}^\infty B_n$ where each $B_i$ is closed and nowhere dense. $X$ is a Baire space, so it's not the countable union of nowhere dense closed sets. If $A$ were such a union, $A = \bigcup_{n=1}^\infty A_n$, then we'd have that $X = \bigcup_{n=1}^\infty C_n$ where $C_n = \begin{cases} A_k & \text{when } n = 2k \\ B_k & \text{when } n = 2k+1\end{cases}$ contradicting that $X$ is a Baire space.
(I didn't do the calculation, but it seems that the other cases should be easy)
@cyclochaotic have you tried looking at only the ones for n divisible by 4 (so put in $4n$ instead of $n$) and seeing if the smallest prime divisor seems to stay bounded?
I am given $f=u+iv$ with $u,v\in C^1$ over $A=\{|z|<1\}$, $f$ continuous on $\bar A=\{|z|\leq 1\}$ and $u(x,y)=x,v(x,y)=y$ over $\{|z|=1\}$, and $J_f(z)>0$ if $z\in A$. Let $B=f(A)$. I have to show that $f$ is an open mapping, $B$ is an open disk of radius $1$ and for each $u_0+iv_0\in B$ there are but finitely many$z\in A$ such that $f(z)=u_0+iv_0$!
@robjohn
Have you faced this one before? I wanna think about it, though.
there is theorem which says that if a continuous function $f$ defined on a region of the complex plane has a primitive $F$ then $$\int_\gamma f(z) dz$$ is 0, where $\gamma$ is a continuous closed curve. Now consider $1/z$ and $\gamma$ a circle around the origin so it's a closed curve. If you take a domain $D$ so that $\ln z$ is well defined and analytic, then $\gamma$ would be piecewise continuous and closed curve. But $$\int_\gamma \frac{dz}{z}$$ is $2\pi i$ which is very different from 0.
@leo I am given $f=u+iv$ with $u,v\in C^1$ over $A=\{|z|<1\}$, $f$ continuous on $\bar A=\{|z|\leq 1\}$ and $u(x,y)=x,v(x,y)=y$ over $\{|z|=1\}$, and $J_f(z)>0$ if $z\in A$. Let $B=f(A)$. I have to show that $f$ is an open mapping, $B$ is an open disk of radius $1$ and for each $u_0+iv_0\in B$ there are but finitely many$z\in A$ such that $f(z)=u_0+iv_0$!
@leo Hmm... I have theorem 13.3 which says that if $f$ is continuous and has finite partial derivatives over an open set $A$, $f$ is one one over $A$ and $J_f(x)$ doesn't vanish on $A$, then $f(A)$ is open....
So, take an open set $\Omega$ in the disk.
Since $J_f(z)>0$, the inverse function theorem says $f$ is locally one-one. But I cannot ensure that $f$ is one-one over $\Omega$ but rather, on an open set inside it.
@PeterTamaroff Well I have a theorem (6.4) which says that if $F:W\subseteq\Bbb R^n\to \Bbb R^n$ with $W$ open and $F$ of class $C^r$ and $a\in W$ with $J_F(a)\neq 0$, then there is an open nhbd of $a$ where $F$ is a diffeomorphism of class $C^r$ (hence an homeomorphism)
@PeterTamaroff indeed. But you have that $f$ is of class $C^1$ in the whole open. And that the Jacobian is not zero in each of its points. So aply the theorem to each point. The image set would be an union of open sets. One for each point. Union of opens is open. Or something along those lines...
P Since $f$ is of class $\mathcal C^1$ over $D$ and the Jacobian doesn't vanish, for each $z\in D$ there exists an open nbhd $N_z$ such that $f\mid_{N_z}$ is a diffeomorphism of class $C_1$. Let $\Omega\subseteq D$ be open, and choose $z\in\Omega$. Then there exists $N_z$ as above. Let $\widetilde N_z=N_z\cap D$. This is open (...)
(...) and lies inside $N_z$ so its image under $f$ is open by the above. Then $f(\Omega)=\bigcup_{z\in\Omega}\widetilde N_z$ is the union of open sets, thus is open. $\blacktriangle$.
@user1 I have studied a bit of group theory before, also a bit ring theory for a number theory project, but this will be my first time seriously going through a book.
@Bageer The reason why I ask is that there are lots of critics of this approach. I am not one, but I would still like to see how smoothly it goes for you.
@PeterTamaroff is $f$ differentiable in the complex sense. Sorry, you said CR equations and those are only satisfied if the function is differentiable (holomorphic)
"Let $u,v$ be real valued functions defined in s subset $S$ of the complex plane, and assume $u,v$ are differentiable in a point $s\in S^{\circ}$, and that their partial derivatives satisfy the CR eqns. Then $f=u+iv$ admits a derivative at $c$ and moreover $f'(c)=D_1u(c)+iD_1v(c)$."
@TedShifrin Oh, well. He has the following theorem.
Let $B$ be an $n$-ball, and suppose $f=(f_1,\dots,f_n)$ is continuous on $\bar B$ and all partial derivative exists if $x\in B$; while $f(x)\neq f(a)$ if $x\in\partial B$, and that $J_f(x)\neq 0$for each $x\in B$, Then $f(B)$ contains an $n$-ball with center $f(a)$.
I don't see where we ought to use the Jacobian is always positive (i.e. not only non-zero).
Right. I know lots of ways to prove this, but I will have to think how to do it without any topological stuff. I'll get back to you tomorrow (i.e., today). My fault for thinking it was easy ...
@robjohn I did this (that is straightforward) $$\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{\log k}{2^n}=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\frac{\log k}{2^n}=\sum_{k=1}^{\infty} \frac{\log k}{2^{k-1}}$$
@robjohn I think that I saw such a series in the past that related to a nice integral (as appearance) but I'm not sure yet if it's so. I need to check that.
I think it was a multiple integral. (again, not sure)
@robjohn in a way I agree with you. If I find something interesting around it then I let it you know.
No new question posted by Laila Podlesny (I'm waiting for some new entries)
Another question I thought of yesterday is $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_n}{n^2}$$ that works nicely by integral method. However I was wondering if I can come up with some proof that only containS sums manipulation (I didn't manage to do it yet). In a more general sense, I was wondering how far one can get by only using multiple sums when manipulating Euler sums.
This one evaluates $\displaystyle \frac{5 \zeta{(3)}}{8}$.
@Lord_Farin I was asking myself if I may evaluate 2 series by a single shot: $$S_1=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}$$ and $$S_2=\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}$$
@Chris'swisesister Of course. But I'm a bit bifocal these days; an exam next week demands some time as well.
And while it is tiresome, I need to work on it regardless, otherwise it'll remain being tiresome forever, and never become a source of some pride and satisfaction.
