TITLE: Bound on the derivative of a holomorphic function at $0$
QUESTION [1 upvotes]: Suppose $f(z)$ is holomorphic in $|z|< 1$ and $\operatorname{Re}f(z) > 0$ in this region. Furthermore, suppose $f(0) = a > 0$. Then why does $|f'(0)| \leq 2a$?
The hint I have is to consider $g(z) = \frac{f(z) - a}{f(z) + a}$ and then show $|g'(0)| \leq 1$. If I do have $|g'(0)| \leq 1$, then the result follows as $|f'(0)| = 2a|g'(0)| \leq 2a$, but I have no idea on how to show $|g'(0)| \leq 1$.

REPLY [5 votes]: It follows from the Schwarz Lemma after you observe that
$$T \colon w \mapsto \frac{w-a}{w+a}$$
is a Möbius transformation that maps the right half plane to the unit disk with $T(f(0)) = T(a) = 0$.
The Schwarz Lemma says that for a holomorphic $f \colon\mathbb{D}\to\mathbb{D}$ with $f(0)= 0$, you have $\lvert f'(0)\rvert \leqslant 1$.
You can easily derive that from the Cauchy integral for the derivative,
$$f'(0) = \frac{1}{2\pi i} \int_{\lvert \zeta\rvert = r} \frac{f(\zeta)}{\zeta^2}\,d\zeta,$$
for which the standard estimate yields $\lvert f'(0)\rvert \leqslant \frac{1}{r}$. Letting $r \to 1$ yields $\lvert f'(0)\rvert \leqslant 1$.


TITLE: Evaluate $\iint \text{curl} (y\,\mathbf{i}+2\,\mathbf{j})\cdot n\; d\sigma$ using Stokes Theorem
QUESTION [1 upvotes]: Evaluate $\displaystyle\iint \text{curl} (y\,\mathbf{i}+2\,\mathbf{j})\cdot n\; d\sigma$ where $\sigma$ is the surface in the first octant made up of the plane $2x+3y+4z=12$, and the triangles in the $(x,z)$ and the $(y,z)$ planes.
I cannot even out how to even attempt this problem. Any help in getting started is much appreciated.

REPLY [2 votes]: Stokes' Theorem, for your case, says that
$$\iint_S \nabla\times\mathbf{F}\;\cdot\;d\mathbf{S}=\oint_{\Gamma}\mathbf{F}\cdot d\mathbf{\Gamma}.$$
Now here $\Gamma$ is the boundary of the non-closed surface $S$. And you've got to use the right-hand rule to figure what's "outside" and what's "inside", and therefore which direction along the boundary of $S$ makes up $\Gamma$. Also, notational issue: $\text{curl}(\mathbf{F}):=\nabla\times\mathbf{F}.$ The $\cdot\; n \, d\sigma$ in your problem is equivalent to $\cdot \; d\mathbf{S}$ in Stokes' Theorem. So, if we figure out the path $\Gamma$ making up the boundary of your surface described by the plane and the two triangles, we should have
$$\iint \text{curl} (y\,\mathbf{i}+2\,\mathbf{j})\cdot n\; d\sigma=\oint_{\Gamma}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\Gamma}.$$
To figure out $\Gamma$, you need to draw a picture. I think the triangles being talked about are the "projection" of the plane onto the axes mentioned. That is, the plane is a triangle in the first octant. To get the "triangle" in the $(x,z)$ plane, imagine a flashlight shining from somewhere on the $y$ axis, towards the origin, and the shadow it casts on the $(x,z)$ plane will be a triangle. I think that's the triangle they're talking about. If you draw the plane, you'll find it intersects the $x$ axis at $x=6$, the $y$ axis at $y=4$, and the $z$ axis at $z=3$. So the triangle on the $(x,z)$ axis would be comprised of the vertices $(0,0,0), (6,0,0),$ and $(0,0,3)$. Similarly, the triangle on the $(y,z)$ plane would be comprised of the points $(0,0,0), (0,4,0),$ and $(0,0,3)$.
So, if the surface is made up of the plane and these two triangles, that leaves a triangle in the $(x,y)$ plane as the boundary, and that triangle is comprised of (in the correct order of traversal as given by the right-hand rule) $(0,0,0), (6,0,0),$ and $(0,4,0)$. From $(0,0,0)$ to $(6,0,0)$ we'll call $\gamma_1;$ it has length $6$. From $(6,0,0)$ to $(0,4,0)$ we'll call $\gamma_2$. It has length $2\sqrt{13}$. Finally, from $(0,4,0)$ to $(0,0,0)$ we'll call $\gamma_3$, and it obviously has length $4$. So, we have
$$\oint_{\Gamma}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\Gamma}
=\int_{\gamma_1}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\gamma}+
\int_{\gamma_2}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\gamma}+
\int_{\gamma_3}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\gamma}.$$
$\gamma_1$ and $\gamma_3$ are the easiest:
$$\int_{\gamma_1}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\gamma}
=\int_0^6(y\,\mathbf{i}+2\,\mathbf{j}) \cdot (\mathbf{i} \, dx)=\int_0^6y \, dx=0,$$
since $y=0$ on $\gamma_1$.
For $\gamma_3$, we have
$$\int_{\gamma_3}(y\,\mathbf{i}+2\,\mathbf{j})\cdot d\mathbf{\gamma}
=\int_4^0(y\,\mathbf{i}+2\,\mathbf{j})\cdot(\mathbf{j} \, dy)=-2\int_0^4dy=-8.$$
For $\gamma_2$, we need a vector from the start point to the finish point, which would be given by $(0,4,0)-(6,0,0)=(-6,4,0)$. And here, in the grand tradition of leaving the harder parts to the student, I will let you finish.


TITLE: If I flip a coin and get 1 billion heads in a row, is tails more likely in the next flip?
QUESTION [0 upvotes]: If I flip a coin and get 1 billion heads in a row, is tails more likely in the next flip?
Is it 50% or does the poisson distribution say it would be a good to bet on tails?

REPLY [1 votes]: If the coin is fair, then heads and tails have equal chance. Take your pick.
If you were able to flip the coin $1$ billion times in a row, then safe to bet on heads, probably biased.


TITLE: Properties of Arbitrary Cartesian Products
QUESTION [3 upvotes]: Let $X_\delta$ and $Y_\delta$ be index families of sets with the index set $\Delta$.
Show the following:
$$\begin{align}
\prod_{\delta \in \Delta} X_\delta \cup \prod_{\delta \in \Delta} Y_\delta \subseteq \prod_{\delta \in \Delta} X_\delta \cup Y_\delta & \tag{i}\\
\prod_{\delta \in \Delta} X_\delta \cap \prod_{\delta \in \Delta} Y_\delta = \prod_{\delta \in \Delta} X_\delta \cap Y_\delta &\tag{ii}
\end{align}$$
When working on this problem I arrived at a small issue.  For finite intersections and unions the proofs are straightforward.  However, when using arbitrary indices, how would my strategy change. I'm a bit gun shy about starting proofs for these because I'm starting to think that the arbitrary index changes everything.

REPLY [2 votes]: This is really just about chasing after the definitions.
Recall that $f\in\prod_{\delta\in\Delta} X_\delta$ if $f$ is a function whose domain is $\Delta$ and for every $\delta$, $f(\delta)\in X_\delta$.
So suppose that $f\in\prod X_\delta\cup\prod Y_\delta$ then $f$ is a function whose domain is $\Delta$ and $f(\delta)$ is in $X_\delta$ for all $\delta$; or $f(\delta)\in Y_\delta$ for all $\delta$. In particular this means that $f(\delta)\in X_\delta\cup Y_\delta$ for all $\delta$, and so $f\in\prod(X_\delta\cup Y_\delta)$.
To see that equality may fail to holds, note that the union of the product does not contain "mixed" choice functions, that is functions that part of their images are elements from $X_\delta$ and part are from $Y_\delta$ and so if $X_\delta\cap Y_\delta=\varnothing$ equality might not hold.
The part about the intersection is very similar. One has to trace back what does it mean to be in the intersection, and what does it meant to be in the product of the intersections, but it's more or less as above.


TITLE: Let $A\subset R^{n}$ . Then $A$ is disconnected iff there exists a continuous and surjective functon $f:A\to${0,1}
QUESTION [4 upvotes]: Let $A\subset R^{n}$ . Then $A$ is disconnected iff there exists a continuous and surjective function $f:A\to${0,1}
How can I prove this?
To prove $\rightarrow$, I know that if $A$ is disconnected, then there are two open, non empty and disjoint sets $U,V\subset R^n$ such that $A=U\cup V$ , $A\cap U \ne \emptyset $, $A\cap V \ne \emptyset$ and $A \cap U \cap V=\emptyset$ .
Then we define a function $f(x)=\begin{cases} 0, & x \in U \\ 1, & x \in V\\
\end{cases}$
We can see that this function is surjective because both sets are non empty and there is not an $x$ such that $f(x)=0$  and  $f(x)=1$ (the sets are disjoint). Is this correct?
Then I don't know how to prove the continuity of $f$, I've tried using the fact that $f$ is continuous iff the inverse image of an open(closed) set is open(closed) in $A$, but I still struggle using the concepts of relative open and closed sets(my professor didn't explain them very well) so I don't know how to finish the proof.
Any help will be apprecciated, thanks.

REPLY [1 votes]: $'\implies '$
You have chosen the right function $f$ .To justify its continuity note that $\{0,1\}$ is equipped with discrete topology. So what are the open sets  available ?
They are $\{0\},\{1\},\{0,1\},\emptyset $ Then $f^{-1}(\{0\})=U;f^{-1}(\{1\})=V;f^{-1}(\{0,1\})=A;f^{-1}(\emptyset)=\emptyset$ which is open in each case.Thus $f$ is continuous.

REPLY [0 votes]: Forget $\mathbb{R}^n$ and stay in the induced topology of $A$.
A space $A$ (by its own right) is connected iff (by definition) there does not exist two non-trivial, disjoint, proper, open subsets $U,V$ of $A$ such that $A=U \cup V$ .
Supposing $A$ is not connected, then there are such $U,V$. Your function $f$ defined in the way you did is continuous, because the pre-image of every open subset of $\{0,1\}$ (I assume you are considering the discrete topology) is either $\emptyset, A, U$ or $V$, hence open.


TITLE: Is a single element linearly independent in an abelian group?
QUESTION [3 upvotes]: This is my definition of linearly independent elements in an abelian group:

Let $A$ be an abelian group, let $X \subseteq A$ be a subset and $x_1,\dots,x_k \in X$ with $x_i \neq x_j$ for all $1 \leq i \neq j \leq n$. The elements $x_1,\dots,x_k$ are defined to be linearly independent if they satisfy the following condition:
\begin{equation} n_1x_1 + \dots + n_kx_k = 0 \, , \, \text{with} \hspace{2mm} n_1,\dots,n_k \in \mathbb{Z} \implies n_i = 0 \hspace{2mm} \text{for all} \hspace{2mm} 1 \leq i \leq n\end{equation}

I'm proving theorem 1.6 of Hungerford's Algebra and got stuck at "Either $G \cap H = \{0\}$, in which case $G = \langle d_1x_1 \rangle$ and the theorem is true [...]". Why is $d_1x_1$ linearly independent? I need this in order to say that $\{d_1x_1\}$ is a basis of $G$. I'm pretty sure this comes from the fact that $d_1x_1 \neq 0$, but linearly independence does not seem to be true because $nd_1x_1 = 0$ might be true even though $n \neq 0$. For example, if $d_1x_1$ is an element of order $n$. So, is the implication
\begin{equation}
 n(d_1x_1) = 0 \quad \wedge \quad d_1x_1 \neq 0 \quad \implies \quad n = 0
\end{equation}
true or false in the context of abelian groups? What if we consider free abelian groups instead? If it's always false, then how is the linearly independence proven in Hungerford's proof?

REPLY [3 votes]: You’re correct that, for general abelian groups, any set containing finite order elements is not linearly independent.
For example, in $\mathbb{Z}/4\mathbb{Z}$ I can take $n = 2$ and $x_1 = 2$. Then neither element is zero, but $2\cdot2 = 0$. Thus $\{2\}$ is not linearly independent.
However, in the free abelian case, this fails. If $nx_1 = 0$ for $n\ne 0$ then this implies $x_1$ has finite order. No nontrivial element of any free abelian group has finite order, so this won’t occur in your case.


TITLE: prove the trigonometry identity:
QUESTION [2 upvotes]: prove the trigonometry identity:

$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$

of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.

REPLY [1 votes]: I'll begin by saying that when you want to prove an identity like $ L=R $  you cannot multiply both sides by any value, since then it wouldn't be an identity.
Anyway, Lets say that $L$ is LHS, and $R$ is RHS. Let $t=\tan(A)$. We can then say that $\frac{1}{t}=\cot(A)$.
Therefore, RHS becomes
$$ R = \frac{t^3 + \frac{1-t}{\frac{1}{t}}}{\frac{1-\frac{1}{t}}{t} + \frac{1}{t^3}} $$
Simplifying the top and bottom (separately) shows us that
$$ R = \frac{t^3-t^2+t}{\frac{t-1}{t^2} + \frac{1}{t^3}}$$
Multiply the top and bottom by $ t^3 $, then
$$ R = \frac{t^3(t^3-t^2+t)}{t(t-1)+1} = \frac{t^4(t^2-t+1)}{t^2-t+1} = t^4 $$
Then, substituting back $ t=\tan(A) $, we get
$$ R = \tan^4(A) = L $$
Like we wanted.