text:
what is the least dense layer of the earth
The Planet have the right to be split into 4 main layers: the solid crust on the external, the mantle, the outer core and the inner core. Out of them, the crust is the thincolony layer of the Planet, amounting for less than 1% of our planet's volume.
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byMihai Andrei
The Earth have the right to be separated right into four primary layers: the solid crust on the exterior, the mantle, the external core and the inner core. Out of them, the crust is the thinswarm layer of the Planet, amounting for less than 1% of our planet's volume.
The Earth's structure
Artistic depiction of the Earth's structure. Image via Victoria Museum.
The Earth's structure deserve to be classified in two methods – based on mechanical properties, and based upon the chemisattempt. Here, we're simply going to discuss a straightforward classification, without going right into a lot of details. The primary focus right here is to understand just how the Earth's crust really is, and why it is the thinswarm layer.
The crust arrays from5–70 kilometres (~3–44 miles) in depth and also is the outermost layer. The thincolony parts are oceanic crust, while the thicker parts are continental crust. Many rocks at the Earth's surconfront are relatively young (much less than 100 million years old, compared to the Earth's age, which is approximately 4,4 billion years), yet considering that we uncovered some rocks which are a lot older, we understand that Earth has had a solid crust for at leastern 4.4 billion years.The mantleextends from wright here the crust ends to about2,890 km, making it the thickest layer of Planet. The mantle is additionally created of silicate rocks, but the mantle in its entirety is exceptionally viscous – the high temperatures therereason the silicate product to be sufficiently ductile that it deserve to circulation (in a really long time). The mantle is mainly separated intothe upper and the reduced mantle.The core, generally divided intotheexternal core and also theinner core. The external core is related to as viscous, though a lot less so than the mantle, while the inner core is solid.Schematic view of the inner of Planet. 1. continental crust – 2. oceanic crust – 3. upper mantle – 4. reduced mantle – 5. external core – 6. inner core – A: Mohorovičić discontinuity – B: Gutenberg Discontinuity – C: Lehmann–Bullen discontinuity.
The Earth's crust
Our planet's crust is on average about 40 km deep – which is much thinner than the mantle, the external core and the inner core – you can think of it favor the peel of an apple. The crust below has been created via igneous processes, which describes why the crust has actually much more incompatible facets than the mantle.
Oceanic vs continental crust. Image by means of USGS.
At the bottom of the seas and also some seas, tright here is oceanic crust. Oceanic crust is exceptionally thin (generally under 10 km), and also is written of dense, commonly dark (mafic) rocks: basalt, gabbro, diabase. The continental crust is thicker than that – usually it's roughly 40 kilometres deep, yet deserve to go approximately 70. The two types of crust are likewise sometimes referred to as granitic (continental) and also basaltic (oceanic).
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The crust is not one rigid layer, yet is damaged into fifteen tectonic plates, all in loved one motion one to the various other. This is dubbed worldwide tectonics. The plates themselves are thicker than the crust alone, and also also consist of the shenable mantle beneath the crust – this together is called the litospbelow. The crust is where rocks communicate via the hydrospright here and also more importantly, the atmosphere. New rocks, minerals and products are created here. Here's the necessary part: every one of the range and also phenomena that we deserve to watch with our own eyes take area in the crust. Everything, from mining ores to oil to forming mountains to thick deposits, faults and whatever you ever before heard about geologists observing directly takes location inside the crust (or at the very surface). The deepest drill ever is just over 12 kilometres, and also we won't be seeing the bottom of the crust via our own eyes for a very long time.
How we know
OK, so there's a crust, it's thin, there's additionally a mantle and a core… however if we can't go tbelow, how perform we know?
That's a very good question – but this is where scientific research comes in. We recognize every one of this (and also we recognize it through a really high degree of confidence) via indirect monitoring.
Waves propagating from Earthquakes via the Earth. Image using Brisith Geological Survey.
A century ago, civilization didn't understand the Planet had actually a crust. Some theoretized it did, however tright here was very little proof. The initially ideas came from astronomic indications, but a lot of of what we know this particular day about the Earth's framework comes from seismological observations.Seismic waves from big earthquakes pass throughout the Earth, and also they carry with them information from the environments they passed via. Just like rays of light, seismic waves can reflect, refract and diffract.Since the rate of the seismic waves counts on density, we have the right to use the travel-time of seismic waves to map readjust in thickness via depth. Also, bereason some waves just propagate with solid environments, we know that some atmospheres (like the the external core) areviscous– because the waves don't propagate via them.
In 1909, the brilliant seismologist Andrija Mohorovicic foundthat about 50 kilometers deep in the Planet tbelow is a sudden adjust in seismic velocity – and knew that it have to be a really substantial discontinuity. He likewise observed that seismic waves reflect and also refract at that depth, which confirmed his principles.That discontinuity, called todaythe Mohorovicic discontinuity (or simply "Moho") is concerned this particular day as the limit in between the crust and the mantle.
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Mihai Andrei
Andrei"s background is in geophysics, and also he"s been fascinated by it ever before given that he was a kid. Feeling that there is a gap between scientists and also the general audience, he started ZME Science -- and also the outcomes are what you watch this day.
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Cambridge, Massachusetts, 1876-1880; Brooklyn, New York, c. 1886; Meriden, Connecticut, c. 1891-c. 1899.
Patent #222,866; December 23, 1879; reed organ action.
Patent #345,673; July 20, 1886; organ bellows valve.
Patent #570,200; October 27, 1896; suction valve for reed organ.
Patent #568,310; September 22, 1896; exhaust bellows or feeder.
Patent #627,570; June 27, 1899; valve for player organ.
Patent #445,044; January 20, 1891; improvement to key action for reed organ.
Patent 445,044: Joseph Herbert Chase and James B. Tracy, of Meriden, Connecticut, Assignors to the Aeolian Organ and Music Company, of New York, N. Y.
Patent 570,200: Joseph H. Chase, of Meriden, Connecticut, Assignor to the Æolian Company, of New York, N. Y.
Patent 627,570: Joseph H. Chase, of Meriden, Connecticut, Assignor to the Aeolian Company, of New York, N. Y.
Cambridge, Massachusetts, city directories, Cambridge Public Library, downloaded on various dates from Internet Archive, links at https://archive.org/details/cambridgepubliclibrary.
U.S. patent documents, United States Patent and Trade Office, accessed July 4, 2018, through the search page, http://patft.uspto.gov/netahtml/PTO/patimg.htm.
For further information, see: The Aeolian Co.
There are no entries in the database that describe organs by Joseph H. Chase.


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text:
\section{Introduction}

All graphs in this paper are finite and \DEF{simple}, i.e.\ no loops or multiple
edges are allowed. We use standard terminology and notation.
We denote by $\Delta(G)$ and $\delta(G)$ the maximum and the minimum degree of $G$,
respectively. If $H$ is a subgraph of $G$, we write $H\subseteq G$.
For a graph $G$, let $\rho(G)$ denote its \DEF{spectral radius}, the largest eigenvalue of
the adjacency matrix of $G$. More generally, if $M$ is a square matrix,
the spectral radius of $M$, denoted by $\rho(M)$, is the maximum modulus
$|\lambda|$ taken over all eigenvalues $\lambda$ of $M$.

If $T$ is a tree, then it is a subgraph of the infinite $\ensuremath{\Delta}(T)$-regular tree.
This observation implies that the spectral radius of $T$
is at most $2\sqrt{\ensuremath{\Delta}(T)-1}$. Similar bounds have been obtained for arbitrary
planar graphs and for graphs of bounded genus \cite{DM}. In particular,
the following result holds.

\begin{theorem}[Dvo\v r\'ak and Mohar \cite{DM}]
\label{thm:planar}
If\/ $G$ is a planar graph, then $$\rho(G)\le \sqrt{8\ensuremath{\Delta(G)}} + 10.$$
\end{theorem}

The proof in \cite{DM} uses the fact that the edges of every planar graph
$G$ can be partitioned into two trees of maximum degree at most
$\ensuremath{\Delta(G)} /2$ and a graph whose degree is bounded by a small constant.
A similar bound was obtained earlier by Cao and Vince~\cite{CV}.

Whenever a result can be proved for tree-like graphs and for graphs of
bounded genus, it is natural to ask if it can be extended to a more general
setting of minor-closed families. Indeed, this is possible also in our case,
and a result of Hayes~\cite{Hayes} (see Theorem \ref{thm:Hayes} below)
goes even further.

A graph is said to be \DEF{$d$-degenerate} if every subgraph of $G$ has
a vertex whose degree is at most $d$. This condition is equivalent to
the requirement that $G$ can be reduced to the empty graph by successively
removing vertices whose degree is at most~$d$.

A requirement that is similar to degeneracy is existence of an orientation of
the edges of $G$ such that each vertex has indegree at most $d$. Every such graph
is easily seen to be $2d$-degenerate, and conversely, every $d$-degenerate graph
has an orientation with maximum indegree $d$.

\begin{theorem}[Hayes~\cite{Hayes}]
\label{thm:Hayes}
Any graph $G$ that has an orientation with maximum indegree $d$
(hence also any $d$-degenerate graph) and with $\ensuremath{\Delta}=\ensuremath{\Delta(G)}\ge 2d$
satisfies $$\rho(G)\le 2\sqrt{d(\ensuremath{\Delta}-d)}.$$
\end{theorem}

It is well-known that each planar graph $G$ has an orientation with maximum
indegree $3$. Theorem \ref{thm:Hayes} thus implies that
$\rho(G)\le \sqrt{12(\ensuremath{\Delta}-3)}$, which is slightly weaker than the bound of
Theorem \ref{thm:planar} (for large $\ensuremath{\Delta}$).

The above results suggest the following definitions. We say that a graph $G$
is \DEF{spectrally $d$-degenerate} if every subgraph $H$ of $G$ has
spectral radius at most $\sqrt{d\ensuremath{\Delta}(H)}$. Hayes' Theorem \ref{thm:Hayes}
shows that $d$-degenerate graphs are spectrally $4d$-degenerate.
The implication is clear for graphs $G$ of maximum degree at least $2d$.
On the other hand, if $\ensuremath{\Delta(G)}\le2d$, then $\rho(G)\le \ensuremath{\Delta(G)}\le \sqrt{2d\ensuremath{\Delta(G)}}$.
The main result of this paper is a rough converse of this statement.

\begin{theorem}
\label{thm:almost degenerate}
If\/ $G$ is a spectrally $d$-degenerate graph, then it
contains a vertex whose degree is at most $\max\{4d,4d\log_2(\ensuremath{\Delta(G)}/d)\}$.
\end{theorem}

The proof is given in Section \ref{sect:main}.
If it were not for the annoying factor of
$\log(\ensuremath{\Delta})$, this would imply $f(d)$-degeneracy, which was our initial
hope. However, in Section~\ref{sect:LB} we construct examples
showing that the ratio between degeneracy and spectral degeneracy
may be as large as (almost) $\log\log \ensuremath{\Delta(G)}$.

In the last section, we consider computational complexity questions
related to spectral degeneracy. First we prove that for every integer
$d\ge 3$, it is NP-hard to decide if the spectral degeneracy of a given
graph $G$ of maximum degree $d+1$ is at least $d$. Next we show that
the problem of deciding if a graph is spectrally $d$-degenerate is
co-NP-complete.

\section{Spectral radius}

We refer to \cite{Bi,CDS,GoRo} for basic results about the spectra of finite graphs
and to \cite{HJ} for results concerning the spectral radius of (nonnegative) matrices.
Let us review only the most basic facts that will be used in this paper.
The spectral radius is monotone and subadditive. Formally this is stated in the following lemma.

\begin{lemma}
\label{lem:1}
{\rm (a)} If $H\subseteq G$, then $\rho(H)\le \rho(G)$.

{\rm (b)} If $G = K\cup L$, then $\rho(G)\le \rho(K) + \rho(L)$.
\end{lemma}

The application of Lemma \ref{lem:1}(a) to the subgraph of $G$ consisting of
a vertex of degree $\Delta(G)$ together with all its incident edges gives
a lower bound on the spectral radius in terms of the maximum degree.

\begin{lemma}
\label{lem:2}
$\sqrt{\ensuremath{\Delta(G)}} \le \rho(G)\le \ensuremath{\Delta(G)}$.
\end{lemma}

A partition $V(G) = V_1\cup \cdots \cup V_k$ of the vertex set of $G$ is called
an \DEF{equitable partition} if, for every $i,j\in \{1,\dots,k\}$, there exists an integer $b_{ij}$ such that every vertex $v\in V_i$ has precisely $b_{ij}$
neighbors in $V_j$. The $k\times k$ matrix $B=[b_{ij}]$ is called
the \DEF{quotient adjacency matrix} of $G$ corresponding to this equitable
partition.

\begin{lemma}
\label{lem:3}
Let\/ $B$ be the quotient adjacency matrix corresponding to an equitable
partition of\/ $G$. Then every eigenvalue of $B$ is also an eigenvalue of $G$,
and $\rho(G)=\rho(B)$.
\end{lemma}

\begin{proof}
The first claim is well known (see \cite{GoRo} for details).
To prove it, one just lifts an eigenvector $y$ of $B$ to an eigenvector
$x$ of $G$ by setting $x_v = y_i$ if $v\in V_i$. By the Perron-Frobenius
Theorem, the eigenvector corresponding to the largest eigenvalue of $B$
is positive (if $G$ is connected, which we may assume), so its lift
is also a positive eigenvector of $G$. This easily implies (by using the
Perron-Frobenius Theorem and orthogonality of eigenvectors of $G$)
that this is the eigenvector corresponding
to the largest eigenvalue of $G$. Thus, $\rho(G)=\rho(B)$.
\end{proof}

We will need an extension of Lemma \ref{lem:3}. As above, let
$V(G) = V_1\cup \cdots \cup V_k$ be a partition of $V(G)$, and let
$n_i=|V_i|$, $1\le i\le k$. For every $i,j\in\{1,\dots,k\}$, let
$e_{ij}$ denote the number of ordered pairs $(u,v)$ such that $u\in V_i$,
$v\in V_j$ and $uv\in E(G)$, i.e.~$e_{ij}$ is the number of edges between
$V_i$ and $V_j$ if $i\ne j$, and is twice the number of edges between the
vertices in $V_i$ if $i=j$. Let $b_{ij} = e_{ij}/n_i$ and let $B=[b_{ij}]$
be the corresponding $k\times k$ matrix. This is a generalization from
equitable to general partitions, so we say that $B$ is
the \DEF{quotient adjacency matrix} of $G$ also in this case.
If a matrix $B'=[b'_{ij}]_{i,j=1}^k$ satisfies $0\le b'_{ij}\le b_{ij}$
for every pair $i,j$, then we say that $B'$ is a \DEF{quotient sub-adjacency matrix}
for the partition $V_1\cup \cdots \cup V_k$.

\begin{lemma}
\label{lem:4}
If\/ $B'$ is a quotient sub-adjacency matrix corresponding to a partition
of\/ $V(G)$, then $\rho(G)\ge\rho(B')$.
\end{lemma}

\begin{proof}
By the monotonicity of the spectral radius, $\rho(B')\le\rho(B)$, where $B$ is
the quotient adjacency matrix. So we may assume that $B'=B$.
The matrix $B$ is element-wise non-negative. By the Perron-Frobenius Theorem,
its spectral radius $\rho(B)$ is equal to the largest eigenvalue
of $B$ (which is real and positive) and the corresponding eigenvector
$y$ is non-negative.
Let us define the vector $f \in \ensuremath{\mathbb R}^{V(G)}$ by setting $f_v=y_i$ if $v\in V_i$.
Then
$$
  \Vert f\Vert^2=\sum_{v\in V(G)}f_v^2
         =\sum_{i=1}^k n_i y_i^2.
$$
Furthermore, if $A$ is the adjacency matrix of $G$, then
\begin{eqnarray*}
  \langle f|Af\rangle &=& 2\sum_{uv\in E(H)} f_uf_v\\
      &=& \sum_{i=1}^k \sum_{j=1}^k e_{ij}y_iy_j \\
      &=& \sum_{i=1}^k n_iy_i \sum_{j=1}^k b_{ij}y_j \\
      &=& \rho(B)\sum_{i=1}^k n_iy_i^2 \\[1mm]
      &=& \rho(B)\Vert f\Vert^2.
\end{eqnarray*}
Since the matrix $A$ is symmetric, $\rho(A)$ is equal to the numerical radius
of $A$. Thus, it follows from the above calculations that
$\rho(G)\ge \frac{\langle f|Af\rangle}{\Vert f\Vert^2} = \rho(B)$,
which we were to prove.
\end{proof}

\section{Spectrally degenerate graphs are nearly degenerate}

\label{sect:main}

In this section we prove our main result, Theorem \ref{thm:almost degenerate}.
For convenience we state it again (in a slightly different form).

\begin{theorem}
\label{thm:almost degenerate1}
Let\/ $G_0$ be a spectrally $d$-degenerate graph with $r=\delta(G_0)>4d$.
Then $r \le 4d\log_2(\ensuremath{\Delta}(G_0)/d)$.
\end{theorem}

\begin{proof}
Suppose for a contradiction that $r > 4d\log_2(\ensuremath{\Delta}(G_0)/d) \ge 4d$.
Let $G$ be a subgraph of $G_0$ obtained by successively deleting edges $xy$
for which $\deg(x)\ge\deg(y)>r$, as long as possible. Then $G$ has the following
properties:
\begin{itemize}
\item[(a)] $\delta(G) = r > 4d\log_2(\ensuremath{\Delta}(G_0)/d) \ge 4d\log_2(\ensuremath{\Delta(G)}/d)$.
\item[(b)] $G$ is spectrally $d$-degenerate.
\item[(c)] The set of vertices of $G$ whose degree is bigger than $r$ is
an independent vertex set in $G$.
\end{itemize}
Our goal is to prove that $r \le 4d\log_2(\ensuremath{\Delta(G)}/d)$. This will contradict
(a) and henceforth prove the theorem.

Let us consider the vertex partition into the following vertex sets:
$$V_0 = \{v\in V(G)\mid \deg_G(v)=r\},$$
and for $i=1,\dots,l$,
$$V_i = \{v\in V(G)\mid 2^{i-1}r < \deg_G(v) \le 2^i r\},$$
where $l = \lceil \log_2(\ensuremath{\Delta(G)}/r) \rceil \le \log_2(\ensuremath{\Delta(G)}/d)$.
Let $B = [b_{ij}]_{i,j=0}^l$ be the quotient adjacency matrix for the
partition $V_0,V_1,\dots,V_l$ of $V(G)$. Since all vertices in $V_0$ have
the same degree $r$, it follows from the definitions of the entries of
$B$ that $r = \sum_{i=0}^l b_{0i}$. Thus it suffices to estimate the
entries $b_{0i}$ in order to bound $r$.

For $i=0$, let $H\subseteq G$ be the induced subgraph of $G$ on $V_0$.
Since $G$ is spectrally $d$-degenerate, we have that
$\rho(H)\le\sqrt{d\ensuremath{\Delta}(H)}\le \sqrt{dr} \le \sqrt{r^2/4} = \tfrac{r}{2}$.
On the other hand, since $H$ has average degree $b_{00}$, we have
$\rho(H)\ge b_{00}$. Thus, $b_{00} \le \tfrac{r}{2}$.
This shows that $\sum_{i=1}^l b_{0i}=r - b_{00}\ge r/2$, and thus
it suffices to prove that
\begin{equation}
\sum_{i=1}^l b_{0i}\le 2d\log_2(\ensuremath{\Delta(G)}/d).
\label{eq:zero}
\end{equation}

From now on we let $B'$ be the matrix obtained from $B$ by setting the entry
$b_{00}'$ to be $0$. This is the quotient adjacency matrix of
the subgraph $G'$ of $G$ obtained
by removing edges between pairs of vertices in $V_0$.

We shall now prove that
\begin{equation}
  \sum_{i=1}^t 2^{i-1}b_{0i}\le 2^t d
\label{eq:first}
\end{equation}
for every $t=1,\dots,l$. Let us consider the subgraph $G_t$ of $G'$
induced on $V_0\cup V_1\cup \cdots\cup V_t$ and the corresponding matrix
$$
   B_t = \left[\begin{matrix}
            0 & b_{01} & \dots & b_{0t} \\
            r & 0 & \dots & 0 \\
            2r & 0 & \dots & 0 \\
            4r & 0 & \dots & 0 \\
            \vdots & \vdots & \ddots & \vdots \\
            2^{t-1}r & 0 & \dots & 0
         \end{matrix}\right]
$$
Let us observe that the entries $2^{i-1}r$ ($i=1,\dots,t$) in the first column
of $B_t$ are smaller than the corresponding entries in $B'$ because every
vertex in $V_i$ has degree more than $2^{i-1}r$. Therefore, $B_t$ is a
quotient sub-adjacency matrix for the subgraph $G_t$.
By expanding the determinant of the matrix $\lambda I - B_t$, it is easy to
see that
\begin{equation}
  \rho(B_t)^2 = \sum_{i=1}^t 2^{i-1}r b_{0i}.
  \label{eq:2a}
\end{equation}
Using Lemma \ref{lem:4} and
the fact that $G_t$ is spectrally $d$-degenerate, we see that
$\rho(B_t)^2\le \rho(G_t)^2\le d\cdot2^t r$. This inequality combined with
(\ref{eq:2a}) implies~(\ref{eq:first}).

We shall now prove by induction on $s$ that
\begin{equation}
  \sum_{i=1}^s b_{0i}\le (s+1)d
\label{eq:firstx}
\end{equation}
for every $s=1,\dots,l$. For $s=1$, this is the same as the inequality (\ref{eq:first})
taken for $t=1$. For $s\ge2$, we apply inequalities (\ref{eq:first}) to get
the following estimates:
\begin{equation}
  2^{s-t}\sum_{i=1}^t 2^{i-1}b_{0i}\le 2^s d
\label{eq:second}
\end{equation}
and henceforth
\begin{equation}
  \sum_{t=1}^s 2^{s-t}\sum_{i=1}^t 2^{i-1}b_{0i}\le s \cdot 2^s d.
\label{eq:third}
\end{equation}
Finally, inequality (\ref{eq:second}) (taken with $t=s$) and (\ref{eq:third}) imply
\begin{eqnarray*}
 2^s\sum_{i=1}^s b_{0i}
   &=& \sum_{i=1}^s \Bigl( 2^{i-1} + \sum_{j=i}^s 2^{j-1} \Bigr) b_{0i} \\
   &=& \sum_{i=1}^s 2^{i-1} b_{0i} +
       \sum_{t=1}^s 2^{s-t}\sum_{i=1}^t 2^{i-1}b_{0i} \\[2.5mm]
   &\le& 2^s d + s \cdot 2^s d = 2^s(s+1)d.
\end{eqnarray*}
This proves (\ref{eq:firstx}).  For $s=l$, this implies (\ref{eq:zero}) and completes the proof of the theorem.
\end{proof}

\section{A lower bound}

\label{sect:LB}

In this section we show that the $\log(\ensuremath{\Delta})$ factor in the bound given by Theorem~\ref{thm:almost degenerate} cannot be eliminated entirely.

Let $\alpha\in \ensuremath{\mathbb R}_+$. We say that a graph $G$ is \DEF{$\alpha$-log-sparse},
shortly \DEF{$\alpha$-LS}, if every subgraph $H$ of $G$ has average degree at
most $\alpha \log(\ensuremath{\Delta}(H))$. Observe that being $\alpha$-LS is a hereditary
property and that every $\alpha$-LS graph $G$ is $\alpha\log(\ensuremath{\Delta(G)})$-degenerate.

Pyber, R\"odl, and Szemer\'edi \cite[Theorem 2]{PRS} proved that there exists a
constant $\alpha_0$ such that every graph $G$ with average degree at least
$\alpha_0\log(\ensuremath{\Delta}(G))$ contains a 3-regular subgraph. On the other hand, they
proved in the same paper \cite{PRS} that there exists a constant $\beta>0$ such
that, for each $n\ge1$, there is a bipartite graph of order $n$ with average
degree at least $\beta \log\log n$ which does not contain any 3-regular
subgraph (and is hence $\alpha_0$-LS). These results establish the following.

\begin{theorem}[\cite{PRS}]
\label{thm:PRS}
There exist constants $\alpha_0,\beta_0>0$ such that for every integer
$\tau>1$ there exists a bipartite graph $G$ with bipartition $V(G)=A\cup B$ with the following properties:
\begin{itemize}
\item[\rm (a)] $G$ is $\alpha_0$-LS.
\item[\rm (b)] $|A|\ge |B|$ and every vertex in $A$ has degree $\tau$.
\item[\rm (c)] $\beta_0\log\log |A|\le \tau$.
\end{itemize}
\end{theorem}

We will prove that graphs of Theorem \ref{thm:PRS} have small spectral degeneracy.
The proof will use the Chernoff inequality in the following form (cf. \cite{mat}, Theorem 7.2.1)

\begin{lemma}\label{lem:chernoff}
Let $X_1, \ldots, X_n$ be independent random variables, each of them
attaining value $1$ with probability $p$, and having value\/ $0$ otherwise.
Let $X=X_1+\cdots+X_n$.  Then, for any $r>0$,
$$\mbox{{\rm Prob}}\bigl[\,|X-np|\ge r\,\bigr] < \exp\Bigl(-\frac{r^2}{2(np+r/3)}\Bigr).$$
\end{lemma}

We can now prove the following lemma, showing that a bipartite graph
whose bipartite parts are ``almost'' regular cannot be log-sparse.

\begin{lemma}
\label{lem:log-dense subgraph}
Let\/ $T\ge 10$ and\/ $t>0$ be integers such that $$6\alpha_0 \log(20T)\le t\le T.$$
Let\/ $H$ be a bipartite graph of maximum degree $\ensuremath{\Delta}\ge 2T$ with bipartition $V(H)=A\cup B$ satisfying the following properties:
\begin{itemize}
\item[\rm (a)] $t\le \deg v\le T$ for each vertex $v\in A$.
\item[\rm (b)] Each vertex $v\in B$ has degree at least $\ensuremath{\Delta}/2$.
\end{itemize}
Then $H$ is not $\alpha_0$-LS.
\end{lemma}

\begin{proof}
Choose a subset $A'$ of $A$ by selecting each element uniformly independently with probability $p=2T/\ensuremath{\Delta}$, and let
$H'$ be the subgraph of $H$ induced by $A'\cup B$.  The expected size of $A'$ is $a'=2T|A|/\ensuremath{\Delta}$.
Note that $T|A|\ge |E(H)|\ge |B|\ensuremath{\Delta}/2$, thus $a'\ge |B|$.  Furthermore, $|A|\ge \ensuremath{\Delta}/2$, and thus
$a'\ge T\ge 10$.  By Lemma~\ref{lem:chernoff},
$$\mbox{Prob}\bigl[\,|A'|\le \tfrac{1}{2}a'\,\bigr] < e^{-3a'/28}< \tfrac{1}{2}.$$
Therefore, we have $2|A'|\ge a'\ge |B|$ with probability greater than $\tfrac{1}{2}$.

Consider a vertex $v\in B$.  The expected degree of $v$ in $H'$ is between $T$ and $2T$, and
the probability that $v$ has degree greater than $2cT$ is less than $e^{-cT}$ for any $c\ge 10$,
by Lemma~\ref{lem:chernoff}.  Let $z=0$ if $\deg_{H'} v\le 20T$ and $z=\deg_{H'} v$ otherwise.
The expected value of $z$ is
\begin{eqnarray*}
\sum_{j=20T+1}^\infty Pr(\deg_{H'} v=j)j&=&\sum_{j=20T+1}^\infty\sum_{i=1}^j Pr(\deg_{H'} v=j)\\
&=&\sum_{i=1}^{20T}\sum_{j=20T+1}^\infty Pr(\deg_{H'} v=j)+\sum_{i=20T+1}^\infty\sum_{j=i}^\infty Pr(\deg_{H'} v=j)\\
&=&20T Pr(\deg_{H'} v>20T) + \sum_{i=20T+1}^\infty Pr(\deg_{H'} v\ge i)\\
&=&20T Pr(\deg_{H'} v>20T) + \sum_{i=20T}^\infty Pr(\deg_{H'} v>i)\\
&\le&20Te^{-10T}+\sum_{i=20T}^\infty e^{-i/2}.
\end{eqnarray*}
We conclude that the expected number of edges of $H'$ incident with vertices of degree greater
than $20T$ is less than
$$|B|\Bigl(20Te^{-10T}+\sum_{i=20T}^\infty e^{-i/2}\Bigr)<|B|(20T+3)e^{-10T}.$$
By Markov's inequality, it happens with positive probability that $H'$ has less than $2|B|(20T+3)e^{-10T}$ edges incident with
vertices of degree greater than $20T$ and that $2|A'|\ge |B|$.

Let us now fix a subgraph $H'$ with these properties.
Let $H''$ be the graph obtained from $H'$ by removing the vertices of degree greater than $20T$.
Clearly, $\ensuremath{\Delta}(H'')\le 20T$.  Also, $H''$ has at most $3|A'|$ vertices and more than
$$|A'|t - 2|B|(20T+3)e^{-10T}\ge |A'|(t-4(20T+3)e^{-10T})\ge \tfrac{1}{2}|A'|t$$
edges, thus the average degree of $H''$ is greater than $t/6$.  Since $t/6\ge \alpha_0 \log(20T)$,
this shows that $H$ is not $\alpha_0$-LS.
\end{proof}

\begin{theorem}
\label{thm:main example}
Suppose that a bipartite graph $G$ with bipartition $V(G)=A\cup B$ satisfies properties (a)--(c)
of Theorem \ref{thm:PRS}, where $\tau\ge 10$ and $6\alpha_0\log(20\tau)\le \tau$. Then $G$ is spectrally $d$-degenerate,
where $$d = 48(3+2\sqrt{2})\alpha_0 \, \log(20\tau).$$
\end{theorem}

\begin{proof}
Suppose for a contradiction that $H$ is a subgraph of $G$ with maximum degree $D=\ensuremath{\Delta}(H)$ whose spectral radius violates spectral $d$-degeneracy requirement,
\begin{equation}
    \rho(H) > \sqrt{dD}.
\label{eq:P1}
\end{equation}
We may assume that $H$ is chosen so that $D$ is minimum possible.
Since $G$ is $\alpha_0$-LS, the same holds for its subgraph $H$. In particular, $H$ is $\alpha_0\log(D)$-degenerate and hence $\rho(H) \le 2\sqrt{\alpha_0\log(D)\cdot D}$. By (\ref{eq:P1}) we conclude that
\begin{equation}
    4\alpha_0\log(D) > d.
\label{eq:P2}
\end{equation}
This implies, in particular, that
\begin{equation}
    D \ge 2\tau.
\label{eq:P3}
\end{equation}
Let $\gamma = (3-2\sqrt{2})/8$. Let us partition the edges of $H$ into three subgraphs, $H=H_0\cup H_1\cup H_2$, such that the following holds:
\begin{itemize}
\item[\rm (a)] Each vertex in $V(H_0)\cap B$ has degree in $H_0$ at least $D/2$.
\item[\rm (b)] Each vertex in $V(H_0)\cap A$ has degree in $H_0$ at least $\gamma d$.
\item[\rm (c)] $H_1$ is $\gamma d$-degenerate.
\item[\rm (d)] $\ensuremath{\Delta}(H_2) \le D/2$.
\end{itemize}
Such a partition can be obtained as follows.  Let $H_0$ be a minimal induced subgraph of $H$
such that $E(H)\setminus E(H_0)$ can be partitioned into graphs $H_1$ and $H_2$ satisfying the
conditions (c) and (d) and $V(H_0)\cap V(H_1)\cap A=\emptyset$.  We claim that $H_0$ satisfies (a) and (b).  Indeed, suppose that
$H_0$ violates (a).  Then, there exists a vertex $v\in V(H_0)\cap B$ of degree at most $D/2$.
Consider the graph $H'_2$ obtained from $H_2$ by adding all edges of $H_0$ incident with $v$.
Clearly, $\ensuremath{\Delta}(H'_2)\le D/2$, since $v$ has degree at most $D/2$ and all vertices in $A\cap V(H'_2)$
have degree at most $\tau\le D/2$ by (\ref{eq:P3}).  Thus, there exists a partition of $E(H)\setminus E(H_0-v)$
satisfying (c) and (d), which contradicts the minimality of $H_0$.  Similarly, suppose that
$H_0$ violates (b), so there exists $v\in V(H_0)\cap A$ of degree at most $\gamma d$.
Since $V(H_0)\cap V(H_1)\cap A=\emptyset$, $v\not\in V(H_1)$, and thus the graph $H'_1$ obtained
from $H_1$ by adding all edges of $H_0$ incident with $v$ is $\gamma d$-degenerate.
Furthermore, $V(H_0-v)\cap V(H'_1)\cap A=\emptyset$, so we again obtain a contradiction with
the minimality of $H_0$.

Suppose that $H_0\ne\emptyset$. Then we use properties (a)--(b) of $H_0$ and
apply Lemma \ref{lem:log-dense subgraph} to conclude that $H_0$ is not
$\alpha_0$-LS. This contradicts our assumption that $G$ is $\alpha_0$-LS and
shows that $H_0$ must be empty.

Thus, $H = H_1\cup H_2$. Since $H$ was selected as a subgraph violating spectral degeneracy with its maximum degree smallest possible,
we conclude that $H_2$ is spectrally $d$-degenerate. By applying Lemma \ref{lem:1}(b) and using Theorem \ref{thm:Hayes} on $H_1$, we obtain
\begin{eqnarray*}
  \rho(H) &\le& \rho(H_1)+\rho(H_2)\\
  &\le& \sqrt{4\gamma d \ensuremath{\Delta}(H_1)} + \sqrt{d \ensuremath{\Delta}(H_2)} \\
  &\le& \sqrt{4\gamma d D} + \sqrt{d D/2} \\
  &\le& \bigl(\sqrt{4\gamma} + \sqrt{1/2}\,\bigr)\sqrt{d D} = \sqrt{d D}.
\end{eqnarray*}
This contradicts (\ref{eq:P1}) and completes our proof.
\end{proof}

By Theorem~\ref{thm:PRS}, there exist constants $\beta$ and $n_0$ such that we can
apply Theorem~\ref{thm:main example} to graphs on $n$ vertices with $\tau\ge \beta\log\log n$,
for any $n\ge n_0$.  Then, $d=O(\log\log\log n)$, and thus the ratio between the degeneracy
and the spectral degeneracy is at least $\Omega(\log\log n/\log\log\log n)\ge \Omega(\log\log \ensuremath{\Delta}/\log\log\log \ensuremath{\Delta})$.

Let us however remark that this does not exclude the possibility that the degeneracy is bounded
by a function of the spectral degeneracy.  Answering a question we posed in the preprint version
of this paper, Alon~\cite{alon} proved that that is not the case.

\begin{theorem}
For every $M$, there exist spectrally $50$-degenerate graphs with minimum degree at least $M$.
\end{theorem}

\section{Computational complexity remarks}

Our results raise the problem of how hard it is to verify spectral degeneracy
of a graph.

\begin{quote}
	{\sc Spectral Degeneracy Problem} \\
	{\sc Input}: A graph $G$ and a positive rational number $d$. \\
	{\sc Task}: Decide if $G$ is spectrally $d$-degenerate.
\end{quote}

\noindent
Below we prove that this problem is co-NP-complete.
To demonstrate this, we need some preliminary
results.  First, we show that distinct roots of a polynomial
cannot be too close to each other. For a polynomial $p(x)=\sum_{i=0}^k a_ix^i$ with integer coefficients,
let $a(p)=\log \max_{0\le i\le k} |a_i|$.

\begin{lemma}
  \label{lemma-distroots}
  Let $p(x)$ be an integer polynomial of degree $k$.
  If $p(u)=p(v)=0$ and $u\neq v$, then $-\log |u-v|=O(k^3(a(p)+\log k))$.
\end{lemma}

\begin{proof}
Mahler~\cite{mahler} proved that if $y$ and $z$ are two roots
of a polynomial $s(x)$ of degree $d$, then $-\log |y-z|=O(-\log |D| + d\log d + da(s))$,
where $D$ is the discriminant of $s$.
To apply this result, we need to eliminate the roots of $p$ with multiplicity greater than one.
By Brown~\cite{brown}, there exists an integer polynomial
$q(x)$ that is a greatest common divisor of $p(x)$ and $p'(x)$ such that
$a(q)=O(k(a(p)+\log k))$.
Let $c$ be the leading coefficient of $q$ and let $r(x)=c^kp(x)/q(x)$.
Note that $r(x)$ is an integer polynomial, all of whose roots are simple,
$r(u)=r(v)=0$, and $a(r)=O(k^2(a(p)+\log k))$.
Since $r$ is an integer polynomial with simple roots, the absolute value of its
discriminant is at least $1$. Using the afore-mentioned result of
Mahler~\cite{mahler}, we conclude that $-\log |u-v|=O(k^3(a(p)+\log k))$.
\end{proof}

Cheah and Corneil~\cite{regsg} showed the following.
\begin{theorem}\label{thm-regsg}
For any fixed integer $d\ge 3$, determining whether a graph of maximum degree $d+1$
has a $d$-regular subgraph is NP-complete.
\end{theorem}

We need an estimate on the spectral radius of graphs where the vertices of maximum degree
are far apart.

\begin{lemma}\label{lemma-near}
Let $G$ be a graph of maximum degree $d+1$ such that the distance between
every pair of vertices of degree $d+1$ is at least three.  Then
$$\rho(G)\le \sqrt[3]{(d+1)(d^2+1)}.$$
\end{lemma}

\begin{proof}
We may assume that $G$ is connected, since the spectral radius of a graph
is the maximum of the spectral radii of its components.
We use the fact that $\rho(G)=\limsup_{n\to \infty} \sqrt[n]c_n$,
where $c_n$ is the number of closed walks of length $n$ starting at an arbitrary vertex $v$ of $G$.
For any vertex $z$ of degree $d+1$, $G$ contains at most $(d+1)[(d-1)d+(d+1)]=(d+1)(d^2+1)$
walks of length $3$ starting at $z$, including those whose second vertex is $z$ as well.
Similarly, the number of walks of length $3$ from a vertex of degree at most $d$
is at most $(d+1)d^2$.  We conclude that $c_n\le \left[(d+1)(d^2+1)\right]^{\lceil n/3\rceil}$,
and the claim follows.
\end{proof}

We will also use the following result which shows that the spectral radius
of a connected non-regular graph of maximum degree $d$ cannot be arbitrarily close to $d$.

\begin{lemma}[Cioab\u{a} \cite{Ci}]
\label{lemma-nearlw}
Let $G$ be a connected graph of maximum degree $d$ and with diameter $D$.
If $G$ has a vertex of degree less than $d$, then
$$\rho(G) < d - \frac{1}{D|V(G)|}.$$
\end{lemma}

We can now proceed with examining the complexity of spectral degeneracy computation.

\begin{lemma}
\label{lem-coNP}
The {\sc Spectral Degeneracy Problem} is in co-NP.
\end{lemma}

\begin{proof}
To verify that the spectral degeneracy of $G$ is
greater than $d$, guess a connected subgraph $H$ of $G$ (on $k\le|V(G)|$ vertices)
such that $\rho(H)>\sqrt{d\Delta(H)}=b$.  To prove that $H$ has this property, first
compute the characteristic polynomial $p(x)=\det(xI-M)$, where $M$ is the adjacency matrix of $H$.
Note that the absolute value of each coefficient of $p$ is at most $k!$
and that $p$ can be computed in polynomial time using, for example,
Le Verrier--Faddeev's algorithm~\cite{leverrier}.
Then, we need to show that $p$ has a real positive root greater than $b$.  This is the case if
$p(b)<0$ and this condition can be verified in a polynomial time, since $b$ is a square root of a rational number.
Hence, we may assume that $p(b)\ge 0$. Let us recall that $\rho(H)$ is a simple root of $p$.
Hence, if $\rho(H) > b$, then there exists a root $y$ of $p$ such that
$b \le y < \rho(H)$ and $p(x)<0$ when $y<x<\rho(H)$.  To prove that $b < \rho(H)$, it suffices to guess
a value $x>b$ such that $p(x)<0$, say any value between $y$ and $\rho(H)$.  By Lemma~\ref{lemma-distroots},
$-\log (\rho(H) - y) = O(k^4\log k)$, and thus such a number $x$ can be expressed in polynomial space.
\end{proof}

For the hardness part, let us first consider a related problem of deciding whether the spectral degeneracy
is greater or equal to some given constant.

\begin{theorem}\label{thm0-hard}
For any fixed integer $d\ge 3$, verifying whether the spectral degeneracy of a graph is at least $d$
is NP-hard, even when restricted to graphs of maximum degree $d+1$.
\end{theorem}

\begin{proof}
We give a reduction from the problem of finding a $d$-regular subgraph in a graph $G$ of
maximum degree $d+1$, which is NP-hard by Theorem~\ref{thm-regsg}.  Let $G'$ be the graph
obtained from $G$ by replacing each edge $uv$ by a graph $G_{uv}$ created from a clique on $d+1$ new vertices
by removing an edge $xy$ and adding the edges $ux$ and $vy$.  Consider a connected subgraph $H\subseteq G'$.  If $H$ is $d$-regular and
$z\in V(H)$ belongs to $V(G_{uv})\setminus \{u,v\}$, then $G_{uv}\subseteq H$.  It follows that
$G'$ contains a $d$-regular subgraph if and only if $G$ contains a $d$-regular subgraph.

Furthermore, if $\Delta(H)=d+1$, then by Lemma~\ref{lemma-near} we have
$$\rho(H)\le \sqrt[3]{(d+1)(d^2+1)}<\sqrt{d\Delta(H)},$$ and if $\Delta(H)\le d$,
then $\rho(H)\le\sqrt{d\Delta(H)}$, where the equality holds if and only if $H$ is $d$-regular.
Therefore, $G$ has a $d$-regular subgraph if and only if the spectral degeneracy of $G'$ is
at least $d$.  Since the size of $G'$ is polynomial in the size of $G$, this shows that
deciding whether the spectral degeneracy of a graph is at least $d$ is NP-hard.
\end{proof}

A small variation of this analysis gives us the desired result.

\begin{theorem}\label{thm-hard}
The {\sc Spectral Degeneracy Problem} is co-NP-complete.
\end{theorem}

\begin{proof}
By Lemma~\ref{lem-coNP}, the problem is in co-NP, so it remains to exhibit a reduction
from a co-NP-hard problem.

Consider the graph $G'$ from the proof of Theorem~\ref{thm0-hard} and its connected subgraph $H$.
If $H$ has maximum degree $d+1$, then the spectral radius of $H$ is at most
$$\sqrt{\sqrt[3]{\frac{(d^2+1)^2}{d+1}}\Delta(H)}$$
by Lemma~\ref{lemma-near}.
If $H$ has maximum degree at most $d-1$, then $$\rho(H)\le \sqrt{(d-1)\Delta(H)}.$$  Finally,
if $\Delta(H)=d$ and $H$ is not $d$-regular, then
$$\rho(H)\le \sqrt{\left(d-|V(H)|^{-2}\right)^2}\le \sqrt{\left(d-|V(H)|^{-2}\right)\Delta(H)}$$
by Lemma~\ref{lemma-nearlw}.

Let $n=|V(G')|$.  Let $b$ be a rational number such that
$$\max\left\{\sqrt[3]{\frac{(d^2+1)^2}{d+1}},\ d-1,\ d-n^{-2}\right\}\le b<d.$$
We conclude that either $G'$ has spectral radius at least $d$ or at most $b$.  Thus, deciding whether
the spectral degeneracy of a graph is at most $b$ (where $b$ is part of the input) is co-NP-hard.
\end{proof}

However, this does not exclude the possibility that the spectral degeneracy could be approximated
efficiently. Let $\varepsilon>0$ be a constant.

\begin{quote}
	{\sc Approximate spectral degeneracy} \\
	{\sc Input}: A graph $G$ and a rational number $d$. \\
	{\sc Task}: Either prove that $G$ is spectrally $(1+\varepsilon)d$-degenerate,
or show that it is not spectrally $d$-degenerate.
\end{quote}

Does there exist $\varepsilon$ such that this problem can be solved in a polynomial time?
Or possibly, is it true that this question can be solved in a polynomial time for every $\varepsilon>0$?
Both of these questions are open.

\bibliographystyle{siam}

redpajama_set_name: RedPajamaArXiv


The 111 Club is run to raise funds towards capital projects in the school as well as giving members a chance to win some cash. For just £12 for the year, you have the chance to win £25 each month except in June and December when the prize money is £200. There are currently 10 'spare numbers.' If you would like to sign up for The 111 Club just pop into the office and ask for a form.


"I believe the Holy Spirit fanned the spark of a small beginning into a flame that spread across the country," said O'Brien, director of the The Global Center at Samford University. "The same Holy Spirit protected the movement from being organized and institutionalized. It was a 'kairos' moment. No one could ultimately take credit for what God was doing."
Now, a half-century later, with the original evangelists in their 70s, O'Brien is coordinating Revival Revisited, a free public symposium on the youth revival movement June 18-20 at Samford. At least a dozen of the original revival team members will be present to share their experiences.
O'Brien remembers that the youth evangelists were "heroes to my age group . . . their commitment and charisma captivated us.
"I was in Junior High and High School in west Texas during the first six years of the youth revival movement," he recalled. "My brother, Chester O'Brien, was a preacher in the original wave, but was not in the Baylor group. Chester, Warren Hultgren, and singer Asa Couch were at Hardin-Simmons University and Jimmy Allen from Howard Payne College. These guys were nine to ten years my elders. We wanted to be like them and someday do what they did."
Dr. W.F. Howard, director of Baptist Student work for the Baptist General Convention of Texas, captured the spirit of the moment, screening and training youth revival teams from colleges and universities across the state, according to O'Brien. Training occurred each spring at Mt. Lebanon Baptist encampment near Dallas. Churches that calendared youth revivals sent their requests to Howard's office, where team assignments were made for the entire summer.
"Even so," said O'Brien, "with this more organized approach to enlarging the movement, it never moved into the next stage of 'institutionalizing' the effort."
He quoted Dr. Eric Lincoln, Duke University historian, who said the sad fate of an idea is that it becomes an ideology. Then we institutionalize it, he said, and spend the rest of our lives defending the institution.
"I am not for sure that even an organized approach to youth revivals could have kept them from falling prey to the mood of the late 1960s and 1970s," O'Brien said. "But now my wonderment is not so much about the forms of the past. Rather, in our day of great confusion, would God do it again?"
The evangelists will share their experiences during two sessions on Friday and Saturday, from 9 a.m. until 4 p.m. A fellowship banquet (requiring a fee and reservations) will be held Friday night. The symposium will conclude with a Service of Celebration and Renewal on Sunday evening at 7 p.m. in Beeson Divinity Chapel on the Samford campus.


text:
Interview with Kate Larkindale, Author of Stumped
I was privileged to invite Ms. Larkindale for an interview with BooKecCenTric to discuss her latest book and her inspiration behind it. Read the interview below. Enjoy!
Thank you @KateLarkindale for giving me a bit of your time today. Let's start off with the obvious question – What's your latest book about, and what inspired you to write it?
KateLarkindale: Stumped is about a 17-year-old guy struggling to figure out how to be a man after losing his legs in an accident. I was inspired to write it after meeting the subject of the film Scarlet Road, a documentary about a sex worker who works almost exclusively with the severely disabled.
Who's your target audience, and why do you think this book will appeal to them?
KateLarkindale: It's definitely for older teens. And obviously I'd love for disabled teens to read it and recognize some of their own struggles and fears in Ozzy's journey. But Ozzy is a very funny guy, and some of the situations he finds himself in are hilarious, even while others are devastating. I think this book will appeal to anyone who enjoys books that will make you laugh and cry, sometimes both on the same page.
What message do you wish to pass across to your readers with this book?
KateLarkindale: It's really about the importance of intimacy, and the difference between sex and intimacy. And that having sex doesn't automatically make you a man.
What was the hardest thing about writing this book?
KateLarkindale: The hardest thing about writing Stumped was accepting that the thing that sparked it ended up being the one thing readers couldn't believe would happen and I ended up having to re-write the last third of the book. It still makes me sad to have lost that aspect of the story...
As a writer, is there anything you've learned about yourself while writing this book?
KateLarkindale: I really love writing boys' POV! Ozzy was so much fun to write. I can't wait to write another boy book.
For your own reading, do you prefer eBooks or traditional paper/hard back books, and why?
KateLarkindale: I like both. I always have one paper book on the go because I like to read in the bath and e-readers and bathwater don't go well together. But I always have an e-book going too, so I have something to read on public transport.
What is your niche genre, and if you were to write in a different genre, what would it be?
KateLarkindale: YA contemporary is where I seem to always end up. I have tried a number of different genres though, especially with my short fiction. I've published stories from horror to erotica to sci-fi and back again!
What books and authors have most influenced you?
KateLarkindale: I love books that manage to have both a great story and beautiful writing. Jandy Nelson's The Sky Is Everywhere and Janet Fitch's White Oleander are two of my favorites. I also like really powerfully masculine writers like Russell Banks, Sam Shepherd and Cormac McCarthy. But in terms of influence, I have to mention S.E. Hinton. I read The Outsiders for the first time when I was 12 and it completely changed my life. That was the book that made me want to be a writer, and the many years I spent writing copy-cat Hinton stories made me into a writer.
Do you have any more books in the works?
KateLarkindale: I do! I have three or four that are close to being ready for publication, so I just need to figure out which one I should submit to my publisher next! I'll let you in on a secret: I think it's going to be my sex, drugs and rock n roll book...
Do you read your book reviews? How do you deal with bad or good ones?
KateLarkindale: I do read my reviews. I know you're not supposed to, but I can't help myself. As long as they are thoughtfully written, I learn something from every review, good or bad. Sometimes it's just that the book wasn't the right book for that reader... I review books too, and sometimes I don't like books other readers are raving about, so I know all books aren't for all readers. It's easier to deal with bad reviews when you remember that.
What is your preferred method to have readers get in touch with or follow you (i.e., website, personal blog, Facebook page, Goodreads, etc.)?
KateLarkindale: I absolutely love hearing from my readers. I have a blog (www.katelarkindale.blogspot.com) where I review books and talk about what I'm working on, and I also hang out on Twitter more than I should... You can find me there @vampyr14.
Having spent a lifetime travelling the globe, Kate Larkindale has settled in Wellington, New Zealand. A marketing executive, film reviewer and mother, she's surprised she finds any time to write, but doesn't sleep much. As a result, she can usually be found hanging out near the espresso machine.
She is the author of contemporary YA novels An Unstill Life and Stumped, along with several others that no one is allowed to see. Yet. She has also written one very bad historical romance, which will likely never see the light of day. She is working on several more YA novels that may or may not ever be finished…
Her short stories have appeared in Halfway Down The Stairs, A Fly in Amber, Daily Flash Anthology, The Barrier Islands Review, Everyday Fiction, Death Rattle, Drastic Measures, Cutlass & Musket and Residential Aliens, among others.

redpajama_set_name: RedPajamaCommonCrawl


Based in Leicester, George Walker (Tree Care) Ltd is an established tree surgery business that undertakes all manner of tree work across Leicestershire and the surrounding counties. With an experienced team of highly qualified and motivated personnel, we aim to deliver a professional service to all of our clients, whether you are a commercial customer, a domestic customer or a local authority. Our standards and procedures are independantly verified on a regular basis, which reinforces our commitment to best practice. Among others, we are long standing ARB Approved Contractors accredited by the Arboricultural Association.
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text:
Products For Gardeners
Whizbang
Down-To-Earth Books, Tools & Inspiration
CLOTHESPINS
The Story Behind The Name & The Logo
By: Herrick Kimball
Back in 2000 I developed plans for a homemade chicken plucking machine. It worked so well that I named it the Whizbang Chicken Plucker. Whizbang is a dictionary word that means conspicuous for speed, excellence, or startling effect. It was the perfect name.
In 2002 I self-published the book, Anyone Can Build A Tub-Style Mechanical Chicken Plucker, a.k.a., the Whizbang Plucker plan book. I named my little publishing company, Whizbang Books.
More plan books followed. Then, in 2009 I developed plans for a garden wheel hoe with an oscillating stirrup blade (Click Here to learn about the wheel hoe). naturally, I called it the Whizbang wheel hoe.
My wheel hoe was a modern incarnation of the old Planet Jr. wheel hoe produced by Samuel Leeds Allen back in the late 1800s and early 1900s. I happened to own a Planet Jr. wheel hoe and a few old Planet Jr. catalogs.
Samuel Leeds Allen was an amateur astronomer as well as an inventor. One of his first inventions was a planter made with a couple of wash basins bolted together. The planter reminded Mr. Allen of the planet Saturn. So he adopted Saturn as the Planet Jr. company logo. Exactly how the name name, Planet Jr. came to be is a complete mystery to me. Planet Jr. was once a very big company with a worldwide market. That unusual (but memorable) name and logo was as well known and recognized as John Deere is today.
Inspired by the Planet Jr. name, I decided to rename my wheel hoe the Planet Whizbang wheel hoe. Then it occurred to me that I should rename my whole business from Whizbang Books, to Planet Whizbang, and that's what I did.
When I considered the possibility of a unique logo for my new business name, I wanted it to have some out-of-this world, or planetary theme in deference to the old Planet Jr. logo. But I also wanted my Planet Whizbang trademark to have a decidedly down-to-earth aspect. In other words, I wanted a totally oxymoronic logo.
This was quite a conundrum. But in a flash of inspiration, I came up with the idea of a leafy beet surrounded by golden Saturnal rings.
What, I ask you, is more down to earth than a beet? Perhaps a potato but, as much as I respect and appreciate potatoes (after all, my grandfather was a potato farmer), they do not compare to the beet for visual appeal.
I strongly considered a carrot because, truth be known, I'm especially fond of carrots. But your average carrot is simply not globular enough for the job.
Turnips came to mind. I love the two-tone color of the root. But turnips are a little too obscure.
Rutabagas (known by the old timers as "swedes") are more obscure than turnips but they are probably the one root crop that looks more like a planet than any other. Nevertheless, rutabagas are not nearly as recognizable (or respected) as the beet.
Radishes? Well, radishes are certainly earthy. But it's hard for me to take radishes seriously.
And so it was that I chose a lovely beet with its delightful red-veined green foliage and, of course, the Saturnal rings.
When I first showed my wife the new Planet Whizbang logo, she was speechless. I watched her face intently for some clue to her thoughts. I sensed a hint of a smile as she finally mumbled: "It's a nice looking beet." I took that as affirmation of the "rightness" of the design.
Later, my wife made some comment about "that beet with a halo." I immediately corrected her: "That's not a beet with a halo. Those are Golden Saturnal Rings!"
I wonder what S.L. Allen's wife thought of the name "Planet Jr." and his planet Saturn logo?
Whatever the case, let me make it clear now and for future generations that there is no symbolism or hidden meanings in the Planet Whizbang logo. It is simply a whimsical variation of a unique old trademark. In the end, I can tell you that I settled on this new logo for the same reason I like old-time banjo music.....it makes me smile.
So that's the story. Thanks for reading about it. But wait, there's more...
I'd like to tell you about S.L. Allen. He was a remarkable fellow. This is him and his wife, Sarah, in 1901...
Now some history... Samuel Leeds Allen was born May 5, 1841 into a prominent Quaker family in Philadelphia, Pennsylvania. After graduating from a Quaker boarding school in 1859, S.L. went to live and work at his father's farm in New Jersey.
Young Sam applied his natural problem-solving intellect and mechanical skills to the work at hand. When he was 27 years old he patented his first agricultural tool— a seed planter fashioned from two wash basins.
Shortly thereafter he established the S.L. Allen manufacturing company. Allen was awarded almost 300 patents in his lifetime, most of which were for different farm and garden tools sold with the Planet Jr. name attached to them.
But Samuel Allen's most famous invention was not an agricultural implement. In 1889 he invented a snow sled with steel runners that could be easily steered. He called his new sled the Flexible Flyer.
Allen died in 1918 at 77 years old. The S.L. Allen Company and Planet Jr. were sold off in 1968. Planet Jr. is now a part of history, but the Flexible Flyer sleds are still being made (in China).
S.L. Allen became wealthy from is many inventions and his manufacturing company. But he was also known as a caring and innovative employer. Allen was one of the first industrialists to provide his workers with disability insurance and retirement plans. It is reported that he developed and produced the Flexible Flyer sled as a way of keeping his employees working in the winter months when demand for agricultural tools was slow and employees were often laid off.
Back in 2015 I blogged about Sam Allen ,and shortly thereafter I received an e-mail from his great, great granddaughter. She informed me that there was a biography about the once-famous S.L Allen. I found a pdf copy online (check it out HERE). Right in the beginning of the book are the "Precepts of Samuel L. Allen." They provide a very good insight into the man's character, and his "secrets" to success. Here they are...
Acquire the Habits of :
Punctuality in everything. Attention. Observation. Patience. Doing things systematically. Finishing everything undertaken. Untiring industry.
Cultivate the Habits of :
Thoroughness in every study. Doing everything well. Learning something from everyone. Thinking deeply, powerfully, and comprehensively. Reviewing — remembering that next to perseverance it is the great secret of success as a student.
Cultivate the Habits of a Gentleman:
Politeness. Cheerfulness. Good humor. The memory, by observation, reading, conversation and reflection. Command over my temper. The conscience.
Cultivate the Habits of:
Daily prayer. Self-control of the tongue. Self-control of the feelings. Self-control of the thoughts. Self-control of the heart. Soundness of judgment. Humility and liberality of heart.
Beware of: Temptations:
Light reading (which enfeebles the mind and corrupts the heart). Silly speeches. Silly acting. Fault finding. Bad company. The first step in sin. Secret sins. Bad books. Indulging in reveries of imagination. Contracting the habit of procrastination. Levity upon sacred subjects.
Do not refuse to walk in a difficult path of duty. Never neglect any opportunity of self-improvement. Strive to improve thoughts when alone. Have a plan laid beforehand for every day. Have regard to the position of the body. Be simple and neat in personal habits. Treat properly my parents, friends and companions. Seek to " know thyself." Form fixed principles on which to think and act. Faithfully review my conduct at stated intervals. Live to do good and make this my aim in company and conversation. Do not waste the company's time or my own by talking trifles. Do not endeavor to be a wit or punster. Do not view words in an unnatural light for the sake of smart sayings. Beware of severe speaking. Be careful in introducing topics of conversation. Say as little as possible about myself, friends, deeds, etc.
If you are still reading this, I'm going to assume you'll like this rare old video showing the S.L Allen factory in Philadelphia, PA. I find it fascinating...
And finally, here's a great photo of S.L. Allen IV with a magnificent antique 8-foot-long "Airline Commander" sled that was made in his great, great grandfather's factory.
© 2019 by Herrick KImball and Planet Whizbang
Herrick@PlanetWhizbang.com


PETER BOODY PHOTO | A new house takes shape on site of the century-old Artemas Ward home.
Richard Hogan's new house is taking shape on the shoreline in Shorewood, incorporating the watchtower of the original shingle-style house on the site built by advertising and marketing entrepreneur Artemas Ward a century ago.
Ward (1848-1925) was one of the founders of modern advertising as well as an author and a publisher.
Compromised by extensive rot, it was demolished in 2009. Mr. Hogan's American Direct LLC won a special exception permit from the Town Board to build a house larger than 6,000 square feet last June. It will have a total of 7,524 square feet of habitable living area , slightly less than the Ward house because it will not have bedrooms on the third floor.


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(-) Abuse
(-) Race / Intersectionality
What's Right With Feminism
By Gretchen Gaebelein Hull | October 30, 1995
As Christian women confront the complex (and often negative) baggage carried by the word "feminist" today, these women can often feel ill-equipped to sort out the many social and theological issues regarding women's roles.
Women Caught in Conflict
By Frances Hiebert | October 30, 1996
Groothuis clearly defines and describes Evangelical Feminism in contrast to other forms of feminism and in distinction from "traditionalism." Two other areas treated in the book make significant contributions from my point of view. One is the historical evidence that the church has accepted in its view of the role of women from the culture, rather than constructing a truly biblical view. The second is the role that Satan plays in restricting women's use of their Spiritgiven gifts in ministry to the church and to the world.
What Does She Want From Me Anyway? Honest Answers to the Questions Men Ask About Women
By James R. Beck | March 04, 1997
Holly Phillips has written her book from the heart of the Promise Keepers movement (literally and figuratively). Holly is the wife of founding president Randy Phillips, has been a PK staff member from its early days, and was the first woman to address a PK rally. Her book gives us a fascinating glimpse into the homes of PK staffers, especially the Phillips' themselves.
Women in the Church: A Fresh Analysis of 1 Timothy 2:9-15
By John Stanley | April 29, 1998
Women in the Church is a dangerous book which should not have been published because, while it appears to be scholarly, it actually teems with historical and theological errors and also emotional subjectivity. Alan G. Padgett has provided a critical rebuttal to Women in the Church in the Winter 1997 issue of Priscilla Papers.
Domestic Violence: What Every Pastor Needs to Know
By Joanne Nystrom Janssen | September 04, 2000
The only thing wrong with Domestic Violence: What Every Pastor Needs to Know is the title. This book contains information essential to every person, not just pastors.
No Place for Abuse: Biblical & Practical Resources to Counteract Domestic Violence
By Kriss Erickson | September 05, 2001
"When abuse strikes, there is no home." So say Catherine Clark Kroeger and Nancy Nason-Clark in their book, No Place for Abuse. This quote struck me, as I grew up in a fundamentalist church where mentioning some personal abuse brought blame to me and sympathy to my father. This book is refreshing in its directness as it addresses the ticklish issue of how churches have traditionally dealt with abuse.
Feminist Theology Through the Ages: Why We're Equal
By Joseph Modica | January 30, 2002
Val Webb has written an engaging, readable, and mostly historical approach to feminist theology. Her thesis is straightforward and often restated: "The goal of this book is to look at the diversity of the feminist movement and show how limited and inaccurate negative stereotyping is."
What Science Says About Superiority: Shattering the Myth of Race
By Kriss Erickson | March 04, 2003
Shattering the Myth of Race by Dave Unander is a thoughtful discussion of the conflict of race and ethnicity against the backdrop of the history of Western Europe and the United States.
Rise Up: A Call to Leadership for African American Women
By Sylvia Rose | March 04, 2005
Rise Up is primarily about African American women discovering their capacity to lead, whether in large or small ways. In the introduction, Rose gives a wake-up call to African American women, claiming that there is a crisis in church leadership and African American women need to step up into those roles.
Africa and the Bible
This responsible analysis of Africa's presence in the Bible should be must reading for all thinking Christians who want to deepen their knowledge of the ethnic equality of Christians throughout the ages and in particular the true presence of Africans in our sacred Scripture.


DiveIn: Celebrating Diversity | Atlanta CPCU Society Chapter, Inc.
Please join the Atlanta CPCU Society, Hall Booth Smith, AIG and CNA in our innagural Dive In Event!
A: Dive In is the international festival for diversity and inclusion in insurance.
Dive In is helping insurance get fit for the future, highlighting the business case for diverse and inclusive workplaces and providing practical ideas and inspiration for how to bring about positive change.
This year will see events in a record 27 countries, returning to global cities such as London, New York City, Zurich, and Bermuda, as well as new cities including Amman, Tokyo, Riyadh and São Paulo, between 25-27 September 2018.
This year there are themes such as mental health, modern families and carers and multi-generational workplaces to add to gender, LGBT, multicultural and workability. The themes are spread out across the 3 days so that each day there is something for everyone.
Our sponsors have organized an event that celebrates diversity.
Starting at 2pm, we will offer a 1 hour presentation on unconcious bias followed by a 1 hour diversity panel discussion.
At 4pm we welcome everyone to stay for golfing, drinks and appetizers at our Diversity Happy Hour Celebration.


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1 dead, 1 injured in Kansas City shooting
Teen dies after being shot near 10th Street, Monroe Avenue
Updated: 7:49 PM CST Dec 28, 2015
David Pierce
Digital Managing Editor
A teen was killed and another teen was injured in a shooting Monday evening in Kansas City, police said.A male teen was found in a car around 5:45 p.m. near 10th Street and Monroe Avenue. He was taken to a hospital, where he was pronounced dead.A female teen arrived at a hospital a short time later with a gunshot wound to her foot. She was listed in fair condition.Four black males were seen leaving the scene of the shooting, police said.Anyone with information about the incident is asked to call the TIPS Hotline at 816-474-8477.
A teen was killed and another teen was injured in a shooting Monday evening in Kansas City, police said.
A male teen was found in a car around 5:45 p.m. near 10th Street and Monroe Avenue. He was taken to a hospital, where he was pronounced dead.
A female teen arrived at a hospital a short time later with a gunshot wound to her foot. She was listed in fair condition.
Four black males were seen leaving the scene of the shooting, police said.
Anyone with information about the incident is asked to call the TIPS Hotline at 816-474-8477.