TITLE: Question in proof that $\dim(U_1+U_2)=\dim U_1+\dim U_2-\dim(U_1 \cap U_2)$
QUESTION [3 upvotes]: In the proof that Dim$(U_1+U_2)=DimU_1+DimU_2-Dim(U_1 \cap U_2)$
In "Linear Algebra Done Right" by Sheldon Axler:

Theorem 2.18:
If $U_1$ and $U_2$ are subspaces of a finite dimensional vector space,
  then:
$\dim(U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1 \cap U_2)$

PROOF (in short):
Let $(u_1,u_2,...)$ be a basis of $U_1\cap U_2$. This can be extended to a basis $(u_1,u_2,...,u_m,v_1,v_2,...,v_j)$ of $U_1$. Also, it can be extended to a basis $(u_1,u_2,...,u_m,w_1,w_2,...,w_k)$ of $U_2$.
Clearly $\text{span}(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$ is $U_1 + U_2$. To show that this list is a basis of $U_1+U_2$ we just need to show that it is linearly independent.
I cannot understand why $\text{span}(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$ is $U_1 + U_2$. Can someone help explain this to me?

REPLY [1 votes]: In very simple terms:

by construction every vector in $U_1$ is a linear combination of $(u_1,...,u_m,v_1,...,v_j)$ and every vector in $U_2$ is a linear combination of $(u_1,...,u_m,w_1,...,w_k)$.
By definition, every vector in $U_1+U_2$ is of the form $u+u^\prime$ with $u\in U_1$ and $u^\prime\in U_2$.

Putting things together every vector in $U_1+U_2$ can be written as a linear combination of $(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$.


text:
\section{Definition of Minimal Obstructions} \label{sec:definition_of_obstructions}

In this section, we formally define all twelve classes of minimal obstructions which make a partial
representation non-extendible.

\heading{Definition.}
Every \emph{obstruction} consists of a graph $H$ and a non-extendible partial representation $\calR'_H$.
This obstruction is \emph{contained} in $G$ and $\calR'$ if (i) $H$
is an induced subgraph of $G$, (ii)~the pre-drawn vertices of $H$ are mapped to pre-drawn vertices of
$G$, and (iii) the endpoints in $\calR'_H$ have the same left-to-right order as the endpoints of the
corresponding pre-drawn vertices in $\calR'$. For instance, the partial representation in
Fig.~\ref{fig:int_example}b contains a $1$-FAT obstruction, given in Fig.~\ref{fig:minimal_obstructions}a.

We give $H$ by \emph{descriptions} using finitely many vertices, edges, and induced paths. For inner
vertices of the induced paths, we specify their adjacencies with the remainder of $H$. Since these
induced paths do not have fixed lengths, each description having at least one induced path defines
an infinite class of forbidden subgraphs $H$. Unlike LB obstructions, most classes of minimal
obstructions need infinitely many different descriptions. For instance, each FAT obstruction has $k$
induced paths, and different values of $k$ need different descriptions.

If $H$ contains an induced path $P_{x,y}$, and $x$ and $y$ are allowed to be adjacent, then
$P_{x,y}$ can be a single edge.  When $N[x] = N[y]$, we allow the length of $P_{x,y}$ to be zero,
i.e., $x=y$.

\heading{Minimality.} An obstruction is \emph{minimal} if $\calR'_H$ becomes extendible when any
vertex is removed or any pre-drawn interval is made \emph{free} by removing it from the partial
representation $\calR'_H$.

\subsection{List of Minimal Obstructions}

In Fig.~\ref{fig:LB_obstructions}, we have already described minimal LB obstructions
of~\cite{lb_graphs} with $\calR'_H = \emptyset$.  There are eleven other other classes of minimal
obstructions we describe now.

\begin{figure}[t!]
\centering
\includegraphics{definition_of_obstructions_SE.eps}
\caption{The SE obstructions, on the left with $u=v$, on the right with $u \ne v$.}
\label{fig:SE}
\end{figure}

\begin{figure}[b!]
\centering
\includegraphics{definition_of_obstructions_k_FAT.eps}
\caption[ ]{(a) A $k$-FAT obstruction is created from a $(k-1)$-FAT obstruction. It consists of the
vertices $x_1,\dots,x_k$, $t_2,\dots,t_k$, $y_k$, $z_k$, and the induced paths $P_1,\dots,P_k$. The
adjacencies are defined inductively.\\
(b) In every representation $\calR_H$, the pre-drawn interval $\inter{x_k}'$ together with $P_k$ and
$t_k$ forces $\inter{x_{k-1}}$ to be placed on the right of $\inter{z_k}'$. Therefore, the induced
$(k-1)$-FAT obstruction is forced.\\
(c) The global zig-zag pattern forced by a $k$-FAT obstruction, with $k$ nested levels going across
$\inter{y_k}'$ and $\inter{z_k}'$. It is an obstruction since $P_1$ going from $x_1$ to $z_1$ with
all inner vertices non-adjacent to $y_1$ cannot be placed.}
\label{fig:k_FAT}
\end{figure}

\heading{SE obstructions.} We start with two simple \emph{shared endpoint obstructions} which deal
with shared endpoints in $\calR'$; see Fig.~\ref{fig:SE}. We have two pre-drawn vertices $u$ and
$v$ such that $r(u) = \ell(v)$ (possibly $u=v$, so only one interval may be pre-drawn). Further,
there are two non-adjacent vertices $x$ and $y$, both adjacent to $u$ and $v$.  If $u \ne v$, the
minimality requires that $\ell(u) < \ell(v) = r(u) < r(v)$.

\heading{$\boldsymbol{k}$-FAT obstructions.} The class of \emph{forced asteroidal triple obstructions}
is defined inductively; the first two obstructions $1$-FAT and $2$-FAT are depicted in
Fig.~\ref{fig:minimal_obstructions}a and b, respectively.

A $1$-FAT obstruction consists of three pre-drawn non-adjacent vertices $x_1$, $y_1$ and $z_1$ such
that $\inter{y_1}'$ is between $\inter{x_1}'$ and $\inter{z_1}'$. Further, $x_1$ and $z_1$ are
connected by an induced path $P_1$ and $y_1$ is non-adjacent to the inner vertices of $P_1$. See
Fig.~\ref{fig:minimal_obstructions}a.

A $k$-FAT obstruction is defined as follows; see Fig.~\ref{fig:k_FAT}a.  Let $H_{k-1}$ be the graph
of a $(k-1)$-FAT obstruction.  To get $H_k$, we add to $H_{k-1}$ two vertices $x_k$ and $t_k$
connected by an induced path $P_k$.   Concerning edges, $t_k$ is adjacent to all vertices of
$H_{k-1}$, except for $x_{k-1}$. All vertices of $H_{k-1}$ are non-adjacent to $x_k$ and to the
inner vertices of $P_k$.  Further, for $k > 1$, we allow $P_1$ to be a single edge, so $x_1$
can be adjacent to $z_1$.

We put $y_k = z_{k-1}$ and $z_k = y_{k-1}$.  A $k$-FAT obstruction has three pre-drawn vertices
$x_k$, $y_k$ and $z_k$ such that $\inter{y_k}'$ is between $\inter{x_k}'$ and $\inter{z_k}'$.  The
role of $\inter{x_k}'$, $P_k$ and $t_k$ is to force $\inter{x_{k-1}}$ to be placed on the other side of
$\inter{z_k}' = \inter{y_{k-1}}'$ than $\inter{y_k}' = \inter{z_{k-1}}'$, thus forcing the
$(k-1)$-FAT obstruction of $H_{k-1}$; see Fig.~\ref{fig:k_FAT}b. The global structure forced by a
$k$-FAT obstruction is depicted in Fig.~\ref{fig:k_FAT}c.

\begin{figure}[b!]
\centering
\includegraphics{definition_of_obstructions_overview_of_obstructions.eps}
\caption[ ]{Nine classes of obstructions derived from $k$-FAT obstructions.}
\label{fig:overview_of_obstructions}
\end{figure}

The remaining nine classes, depicted in Fig.~\ref{fig:overview_of_obstructions}, are derived from
$k$-FAT obstructions. Let $H_k$ denote the graph of a $k$-FAT obstruction. With exception of the
last class of $(k,\ell)$-CE obstructions, we create the graphs $\widetilde H_k$ of these
obstructions by adding a few vertices to $H_k$. For $k=1$, when one of $x_1$ and $z_1$ is not
pre-drawn, we also allow $x_1$ to be adjacent to $z_1$. We also consider all of these
obstructions horizontally flipped.

It is possible that the added vertices already belong to $H_k$; for instance, a $k$-BI obstruction
may have $u = t_k$ or $v = t_k$.  Also, we do not specify in details the edges between the added
vertices and $H_k \setminus \{x_k, y_k, z_k\}$.  An accurate description would be too lengthy and
the reader may derive it from Fig.~\ref{fig:k_FAT}c.  We describe all minimal $(k,\ell)$-CE
obstructions in details.

\heading{$\boldsymbol{k}$-BI obstructions.} The class of \emph{blocked intersection
obstructions} is shown in Fig.~\ref{fig:overview_of_obstructions}a. To create $\widetilde H_k$ from $H_k$, we add
two vertices $u$ and $v$ adjacent to $x_k$, $y_k$ and $z_k$. Then the partial
representation contains four pre-drawn vertices $x_k$, $z_k$, $u$ and $v$.
We have $\ell(u) \le \ell(v) < r(u) \le r(v)$, $\inter{x_k}'$ covering $\ell(v)$, and
$\inter{z_k}'$ covering $r(u)$. We allow $u = v$.

The minimality further implies that $k \le 2$. Indeed, a $k$-BI obstruction with $k>3$ contains a smaller
$(k,1)$-CE obstruction by removing $v$ and freeing $\inter{x_k}'$ (this follows from
Lemma~\ref{lem:k_fat_contains_1_fat}). Concerning $1$-BI, we allow $x_1 = z_1$. We illustrate all possible cases only for $1$-BI
obstructions, so the reader can understand the complexity of these classes; see
Fig.~\ref{fig:BI_obstructions}.  For $k=2$, we know by Lemma~\ref{lem:k_fat_contains_1_fat} that
$x_2$ is adjacent to $t_2$. The pre-drawn intervals are as follows:
\begin{align*}
\ell(x_2) \le \ell(u) \le r(x_2)&<\ell(z_2) \le r(u) \le r(z_2), & \text{for $u=v$,}\\
\ell(u) < \ell(x_2) \le \ell(v) \le r(x_2)&<\ell(z_2) \le r(u) \le r(z_2) < r(v),&
		\text{for $u \ne v$.}
\end{align*}
The position of $\inter u'$ and $\inter v'$ forces $\inter{y_2}'$ to be placed between
$\inter{x_2}'$ and $\inter{y_2}'$ in every extending representation, which forces a $2$-FAT
obstruction.

\begin{figure}[t!]
\centering
\includegraphics{definition_of_obstructions_BI_obstructions.eps}
\caption{All minimal 1-BI obstructions are depicted.  Each gives several $1$-BI obstructions with
different $\calR'_{\widetilde H_1}$, since there are several possible orderings of the endpoints
satisfying the given inequalities.  We have $x_1=z_1$ for (a) and (b), and $x_1 \ne z_1$ otherwise.
We have $u=v$ for (a), (c) and (e), and $u \ne v$ otherwise.}
\label{fig:BI_obstructions}
\end{figure}

\heading{$\boldsymbol{k}$-FS obstructions.} The class of \emph{forced side obstructions} is shown in
Fig.~\ref{fig:overview_of_obstructions}b.  To create $\widetilde H_k$ from $H_k$, we add a vertex
$u$ adjacent to $y_k$ and $z_k$. The partial representation contains three pre-drawn vertices $x_k$,
$y_k$ and $u$. We have $\ell(y_k) \le \ell(u) \le r(y_k) < r(u)$ and $\inter{x_k}'$ is on the
left of $\inter{y_k}'$.

\heading{$\boldsymbol{k}$-EFS obstructions.}
The class of \emph{extended forced side obstructions} is similar to $k$-FS obstructions; see
Fig.~\ref{fig:overview_of_obstructions}c. To create $\widetilde H_k$ from $H_k$, we add $u$ adjacent
to $x_k$, $y_k$, and $z_k$, and $v$ adjacent to $y_k$ and $z_k$. The partial representation contains
four pre-drawn vertices $y_k$, $z_k$, $u$ and $v$ pre-drawn as follows:
$$\ell(u) < \ell(v) < \ell(y_k) \le r(y_k) < \ell(z_k) \le r(z_k) < r(u) \le r(v).$$

\heading{$\boldsymbol{k}$-FB obstructions.}
The class of \emph{forced betweenness obstructions} is similar to $k$-BI with $u=v$; see
Fig.~\ref{fig:overview_of_obstructions}d. To create $\widetilde H_k$ from $H_k$, we add $u$
adjacent to $y_k$ and $z_k$. The partial representation $\calR'_{\widetilde H_k}$ contains three
pre-drawn vertices $x_k$, $z_k$, and $u$.  We have $\ell(u) < \ell(z_k) \le r(u) \le r(z_k)$ and
$\inter{x_k}'$ is pre-drawn on the left of $\inter u'$.

\heading{$\boldsymbol{k}$-EFB obstructions.}
The class of \emph{extended forced betweenness obstructions} is similar to both $k$-BI and
$k$-FB; see Fig.~\ref{fig:overview_of_obstructions}e. To create $\widetilde H_k$ from $H_k$, we add $u$
adjacent to $x_k$, $y_k$ and $z_k$, and $v$ adjacent to $y_k$ and $z_k$. The partial representation
contains four pre-drawn vertices $x_k$, $z_k$, $u$, and $v$.  We have
$$\ell(u) < \ell(x_k) \le r(x_k) < \ell(v) < \ell(z_k) \le r(u) \le r(z_k) < r(v).$$

\heading{$\boldsymbol{k}$-FDS obstructions.}
The class of \emph{forced different sides obstructions} is shown in
Fig.~\ref{fig:overview_of_obstructions}f. To create $\widetilde H_k$ from $H_k$, we add $u$ adjacent
to $x_k$, $y_k$, and $z_k$, and $v$ adjacent to $y_k$ and $z_k$. The partial representation
contains three pre-drawn vertices $y_k$, $u$ and $v$ pre-drawn as follows:
$$\ell(u) < \ell(y_k) \le \ell(v) \le r(y_k) < r(u) \le r(v).$$

\heading{$\boldsymbol{k}$-EFDS obstructions.}
The class of \emph{extended forced different sides obstructions} is similar to $k$-FDS
obstructions; see Fig.~\ref{fig:overview_of_obstructions}g. To the construction of
$k$-FDS, we further add $w$ adjacent to $y_k$ and $z_k$.  The partial representation
contains four pre-drawn vertices $y_k$, $u$, $v$ and $w$ as follows:
$$\ell(u) < \ell(v) < \ell(y_k) \le \ell(w) \le r(y_k) < r(w) < r(u) \le r(v).$$

\heading{$\boldsymbol{k}$-FNS obstructions.} The class of \emph{forced nested side obstructions} is
constructed similarly as $k$-EFDS obstructions, but with $z_k$ pre-drawn instead of $y_k$; see
Fig.~\ref{fig:overview_of_obstructions}h. In $\calR'_{\widetilde H_k}$, we have
$$\ell(u) < \bigl\{\ell(v), \ell(w)\bigr\} \le \ell(z_k) \le r(w) < r(u) \le r(v),$$
where $\ell(v)$ and $\ell(w)$ are ordered arbitrarily.

\begin{figure}[t!]
\centering
\includegraphics{definition_of_obstructions_kl_CE_obstructions.eps}
\caption[]{(a) There are three minimal $(1,1)$-CE obstructions, and they are all finite graphs. The
reason is that if, say, a path $P_1$ was very long, we could replace $x_1$ by an inner vertex of
$P_1$; so such an obstruction would not be minimal.\\
(b) There are three minimal $(2,1)$-CE obstructions. Due to the minimality, $P_2$ is a path of length at most
two.  When $x_2$ and $t_2$ are non-adjacent, then $P'_1 = y_2t_2z_2$ is a path
avoiding $N[x_2]$. When $x_2$ and $t_2$ are adjacent, there are two cases, namely, $u \ne t_2$, and
$u = t_2$.  The path $P'_1$ is of length at least two.\\
(c) There are four minimal $(2,2)$-CE obstructions, and they are all finite graphs. Necessarily
$x_2t_2$ and $y_2t'_2$ are edges, and we can choose $x_2$ in such a way that it is adjacent to
$y'_1$ (similarly, $y_2$ is adjacent to $x_1$). There are four different graphs, because the
vertices $u$, $t_2$ and $t'_2$ might be distinct or not.}
\label{fig:kl_CE_obstructions}
\end{figure}

\heading{$\boldsymbol{(k,\ell)}$-CE obstructions.}
The class of \emph{covered endpoint obstructions} is created from a $k$-FAT obstruction glued
to an $\ell$-FAT obstruction; see Fig.~\ref{fig:overview_of_obstructions}i.  To create
$\widetilde H_{k,\ell}$, we glue $H_k$ with $H'_\ell$.  We put $z_k = z'_\ell$, $x_k = y'_\ell$ and
$y_k = x'_\ell$, and some other vertices of these obstructions may be also shared.  We add $u$
adjacent to $x_k$, $y_k$, and $z_k$. The partial representation contains two pre-drawn intervals
$\inter{z_k}'$ and $\inter{u}'$ such that $\ell(u) < \ell(z_k) \le r(u) \le r(z_k)$. We always
assume that $k \ge \ell$.

In $(k,\ell)$-CE Lemma~\ref{lem:kl_ce}, we show that the only $(k,\ell)$-CE obstructions which are
minimal are $(k,1)$-CE obstructions and $(2,2)$-CE obstructions; so either $\ell = 1$, or $k = \ell
= 2$. For $k \ge 3$, a minimal $(k,1)$-CE obstruction consists of the subgraph $H_k$ of a $k$-FAT
obstruction together with a vertex $u$, where $u$ is either adjacent to all vertices of $H_k$, or $u
= t_k$.  The remaining $(k,\ell)$-CE obstructions, where $2 \ge k \ge \ell$, are depicted in
Fig.~\ref{fig:kl_CE_obstructions}.

\subsection{Proofs of Non-extendibility and Minimality}

We sketch proofs that the defined obstructions are non-extendible and minimal. This implies the
first part of Theorem~\ref{thm:repext_obstructions}.  We establish the harder implication in
Sections~\ref{sec:strategy}, \ref{sec:leaves}, \ref{sec:p_nodes}, \ref{sec:q_nodes}, and
\ref{sec:main_result}.

\begin{lemma} \label{lem:correctness_k_FAT}
Every $k$-FAT obstruction is non-extendible and minimal.
\end{lemma}

\begin{proof}
We prove the claim by induction. For $k=1$, non-extendibility and minimality are clear. For
$k>1$, assume that $\inter{x_k}'$ is on the left of $\inter{y_k}'$, and $\inter{y_k}'$ is on the left of $\inter{z_k}'$. In every representation of
$k$-FAT, $\inter{t_k}$ covers $[\ell(y_k),\ell(z_k)]$.  We know that
$\inter{x_{k-1}}$ cannot be on the left of $\inter{x_k}'$, since $H_{k-1}$ is connected and $x_k$ is
non-adjacent to all vertices of $H_{k-1}$.  Therefore $\inter{x_{k-1}}$ has to be placed on the
right of $\inter{z_k}'$. We get a $(k-1)$-FAT obstruction, which is non-extendible by
the induction hypothesis.

It remains to argue the minimality. If one of $\inter{x_k}'$, $\inter{y_k}'$ and $\inter{z_k}'$ is
made free, we can place them in such a way that $\inter{z_k}$ is between $\inter{x_k}$ and
$\inter{y_k}$. This makes the partial representation extendible: It works for $k=1$, and for $k>1$,
we can place $x_{k-1}$ on the right of $\inter{y_k}$, which makes the induced $H_{k-1}$ extendible.
If we remove one of the vertices or induced paths, the argument is similar.\qed
\end{proof}

\begin{lemma} \label{lem:correctness_all}
The following obstructions are non-extendible and minimal:
\begin{packed_itemize}
\item SE, $k$-FS, $k$-EFS, $k$-FB, $k$-EFB, $k$-FDS, $k$-EFDS, and $k$-FNS obstructions,
\item $k$-BI obstructions for $k \le 2$,
\item $(k,\ell)$-CE obstructions where either $\ell = 1$, or $k=\ell=2$.
\end{packed_itemize}
\end{lemma}

\begin{proof}
For SE obstructions, the proof is trivial. For the remaining classes (aside $k$-BI and
$(k,\ell)$-CE), we proceed as follows. Non-extendibility follows from the fact that, in all cases,
$\inter{y_k}$ is forced to be placed between $\inter{x_k}$ and $\inter{z_k}$. To show minimality, we
use the minimality of $k$-FAT obstructions. Then it is easy to show that freeing any added pre-drawn
interval or removing any added vertex results in the possibility of placing $\inter{z_k}$ between
$\inter{x_k}$ and $\inter{y_k}$.

Consider a $k$-BI where $k \le 2$.  Non-extendibility follows from the fact that $\inter{y_k}$ has
to be placed between $\inter{x_k}'$ and $\inter{z_k}'$, thus forcing the k-FAT obstruction,
which is non-extendible by Lemma~\ref{lem:correctness_k_FAT}.  By removing a vertex or an induced
path of $H_k$, it becomes extendible as argued in Lemma~\ref{lem:correctness_k_FAT}. By freeing
$\inter{u}'$ or $\inter{z_k}'$, we can place $\inter{y_k}$ on the right of $\inter{z_k}'$ which
makes the partial representation extendible. By freeing $\inter{v}'$ or $\inter{x_k}'$, we can place $\inter{y_k}$ on the
left of $\inter{x_k}'$, which also makes it extendible because $k \le 2$ and $x_2$ is adjacent to
$t_2$.

For $(k,\ell)$-CE, either $\inter{y_k}$ is between $\inter{x_k}$ and $\inter{z_k}'$ (non-extendible
due to the $k$-FAT obstruction), or $\inter{y_\ell}$ is between $\inter{x_\ell}$ and
$\inter{z_\ell}'$ (non-extendible due to the $\ell$-FAT obstruction). Minimality is also easy:
Removing or freeing $u$ allows to place $\inter{z_k}'$ between $\inter{x_k}$ and $\inter{y_k}$,
which is extendible. And removing anything from one of the FAT obstructions allows one of the
orderings of $\inter{x_k}$ and $\inter{y_k}$ to be extendible.\qed
\end{proof}

We note that the list of minimal obstructions is unique. Indeed, every minimal obstruction itself
corresponds to a valid input, which cannot be obstructed by a distinct obstruction due to the
minimality. Therefore, it is not possible to construct a smaller list of minimal obstructions, or to
argue that if the partial representation contains a particular obstruction, then it also contains an
additional one.

config: arxiv


TITLE: Question about a Maclaurin series of an elementary function
QUESTION [0 upvotes]: The Maclaurin series expansion for $(1+z)^\alpha$ is as follows:
$$(1+z)^\alpha = 1 + \sum_{n=1}^\infty \binom{\alpha}{n}z^n$$ with $$|z|<1$$
What I don't understand is why is $|z|<1$?

REPLY [1 votes]: For $\alpha \in \mathbb{N}$, the series contains only finitely many nonzero terms, and is hence everywhere convergent.
For $\alpha\notin\mathbb{N}$, the power function $w \mapsto w^\alpha$ has a pole in $0$ if $\alpha\in\mathbb{Z}$, and for $\alpha\notin\mathbb{Z}$, it has multiple branches (finitely many possible values if $\alpha\in\mathbb{Q}$, infinitely many if $\alpha\in \mathbb{C}\setminus\mathbb{Q}$). In a small enough neighbourhood of every $w \in \mathbb{C}\setminus \{0\}$, there are holomorphic branches of the power function, but in no neighbourhood of $0$ can a continuous branch of the power function be defined.
So a power series expansion of $(1+z)^\alpha$ cannot have a radius of convergence $R > 1$ unless $\alpha\in\mathbb{N}$, since a power series defines a holomorphic function in its disk of convergence, and by the above, the disk of convergence cannot contain $z = -1$. On the other hand, a holomorphic function can be expanded into a power series in every disk contained in its domain, so the radius of convergence of the binomial series cannot be smaller than $1$.
So for $\alpha\notin \mathbb{N}$, the power series expansion
$$(1+z)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n}z^n = 1 + \sum_{n=1}^\infty \binom{\alpha}{n} z^n$$
holds in the largest disk with centre $0$ that does not contain $-1$, that is the unit disk $\mathbb{D} = \{z : \lvert z\rvert < 1\}$.


TITLE: Why is $\sum_{n=1}^{\infty} \frac{1}{n^4} = -\frac{1}{2} \operatorname{Res}\Bigl( \frac{\pi \cot \pi z}{z^4} \; ; \; 0 \Bigr) = \frac{\pi^4}{90}$?
QUESTION [0 upvotes]: The text I'm reading that states $\sum_{n=1}^{\infty} \frac{1}{n^4}= -\frac{1}{2} \operatorname{Res}\Bigl( \frac{\pi \cot \pi z}{z^4} \; ; \; 0 \Bigr) = \frac{\pi^4}{90}$, without explanation. Where does the $-\frac{1}{2}$ come from? I thought the first equality should be $\sum_{n=1}^{\infty} \frac{1}{n^4}= - \operatorname{Res}\Bigl( \frac{\pi \cot \pi z}{z^4} \; ; \; 0 \Bigr)$.

REPLY [0 votes]: Hint: Consider the integral
\begin{eqnarray*}
\int_{C_N} \frac{\pi \cot \pi z}{z^4} dz
\end{eqnarray*}
where the the contour is a circle with center zero and radius $N$, now consider the limit as $N$ tends to infinity.


TITLE: Falling - is the gain in KE linear?
QUESTION [2 upvotes]: When a ball falls from a high place, it experiences a gravitational force. Forces make objects accelerate (in this case, it is constantly increasing the velocity). Because $KE = \frac{1}{2}mv^2$ this should mean that Kinetic Energy should grow quadratically (please correct me if wrong) because of the increasing velocity right?
But also, $GPE=mgh$ where the potential energy is a linear equation. How can this happen? If energy has to be conserved wouldn't both equations have to change linearly?
Can you please explain how the $KE$, $GPE$ and conservation of energy are reconciled in this system? Could you also confirm the shape of the graph of $KE$ and $GPE$ against time?
(I had initially come up with this problem for electric fields but I think that it might've been easier to answer the question in terms of gravitational fields)

REPLY [3 votes]: Assume that something is dropped from a certain height and no air resistance etc.
Consider the following two graphs:

The left hand graph has height $h$ as the abscissa.
${\rm pe} = mg \cdot h +0$ and ${\rm ke} = - mg \cdot h + {\rm constant}$, both linear relationships.
The sum of the potential energy and the kinetic energy is constant for all heights.
The right hand graph has time $t$ as the abscissa.
Now the velocity $v$ does depend linearly on time $t$ with $v = gt$ but the relationship between kinetic energy and time is a quadratic ${\rm ke} = \frac 12 mg^2t^2$ as is the potential energy ${\rm pe} = {\rm constant} - \frac 12 mg^2t^2$.
Again the sum of the potential energy and the kinetic energy is constant for all times.


TITLE: The physics behind The Great Flood
QUESTION [2 upvotes]: The book of Genesis floats (pardon the pun) some interesting numbers when discussing the Great Flood.  For example, it rained for 40 days and 40 nights, and at the end of that time, the entire planet was covered in water.
I think we can deduce how much water that would have had to be, estimating that the highest peaks in the Himalayas were covered with water. (8,848 meters above sea level)
My questions are, how fast would the rain have had to come to raise the ocean level that high in 40 days and nights, how much would the mass of the earth have changed for this event, and would that significantly alter the strength of gravity on earth?

REPLY [4 votes]: A typical tropical storm drops about 40inches/rain in 24hours (sorry for the medieval units!)
So 40days/nights = 1600inches, or 40metres of water. If you want to cover even reasonable mountains you have to rain a lot harder than that = 200x harder.

REPLY [2 votes]: The closest to a physical discussion of the Great Flood may be
http://www.talkorigins.org/faqs/faq-noahs-ark.html#flood