text:
\section{\normalsize Introduction}\label{sec:bd}

For real numbers $x, y>0$, the gamma, beta and psi functions
are defined as
\begin{eqnarray}\label{Gam-Psi}
\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt,
~B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},
~\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)},
\end{eqnarray}
respectively. (For their basic properties, cf. \cite{1,2,5,14}.) Let
$\gamma=0.5772156649\cdots$ be the Euler-Mascheroni constant, and for $a\in(0,1)$, let $R(a)$ be defined by
\begin{eqnarray}\label{R}
R(a)=-2\gamma-\psi(a)-\psi(1-a)
\end{eqnarray}
which is called the Ramanujan constant in literature although it is actually a function of $a$ and probably better to call $R(a)$ the Ramanujan $R$-function (cf. \cite{11}). By the symmetry, we may assume that $a\in(0,1/2]$ in (\ref{R}).
It is well known that $R(a)$ is essential in some fields of mathematics
such as the zero-balanced Gaussian hypergeometric functions
${}_2F_1(a,1-a;1;z)$, the theories of Ramanujan's modular equations and
quasiconformal mappings, and the properties of $R(a)$ are
indispensable for us to show the properties of
 ${}_2F_1(a,1-a;1;z)$ and the functions appearing in generalized Ramanujan's modular
equations. On the other hand, $R(a)$ and the function
\begin{eqnarray}\label{B}
B(a)\equiv B(a,1-a)=\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(\pi a)} ~(0<a<1)
\end{eqnarray}
are often simultaneously appear in the study of
the properties and applications of $R(a)$, and we often need to compare $R(a)$ with
$B(a)$. In \citep[Section 1]{11}, such kind of importance and applications of $R(a)$,
and the relation between $R(a)$ and $B(a)$ were described in details. (See also \cite{2,4,5,6,7,8,9,10,12,15,16,17}.)

Some authors have obtained some properties, including lower and upper bounds, for $R(a)$. In \cite{11}, for instance, power series expansion, integral representation and bounds were obtained for the difference $R(a)-B(a)$.
Some related studies showed that $B(a)$ is one of good approximation functions for $R(a)$, and we often require the properties of certain combinations defined in terms of $R(a)$, $B(a)$, $a(1-a)$ and other polynomials. (See \cite{3,8,10,11,15,16,17}.)

The main purpose of this paper is to show some monotonicity, convexity and concavity
 properties of certain combinations defined in terms of $R(a)$, $B(a)$ and polynomials, by which sharp
approximations given by $B(a)$ and
polynomials are obtained for $R(a)$. In addition, we shall also show
some properties of the Riemann zeta function
\begin{eqnarray}\label{zeta}
\zeta(s)=\sum_{k=1}^{\infty}\frac{1}{k^s}, Re~s>1
\end{eqnarray}
and its related special sums \citep[23.2]{1}
\begin{eqnarray}\label{lam-beta}
\lambda(n+1)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{n+1}},
~\eta(n)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{1}{k^n},
~\beta(n)=\sum_{k=0}^{\infty}(-1)^k\frac{1}{(2k+1)^n}, n\in\IN.
\end{eqnarray}

For the later use, we record the following identities and special values
\begin{eqnarray}\label{lambda(n+1)}
\lambda(n+1)&=&\left(1-2^{-n-1}\right)\zeta(n+1), ~
\eta(n)=\left(1-2^{1-n}\right)\zeta(n), \label{lambda(n+1)}\\
\zeta(2)-\frac{\pi^2}{6}&=&\zeta(4)-\frac{\pi^4}{90}=\lambda(2)-\frac{\pi^2}{8}
=\lambda(4)-\frac{\pi^4}{96}=\beta(1)-\frac{\pi}{4}=\beta(3)-\frac{\pi^3}{32}=0.\label{zeta(2)}
\end{eqnarray}
(See \citep[23.2.19-20, 23.2.24-25 \& 23.2.28-31]{1}).

Throughout this paper, the ``zero-order derivative"
$\varphi^{(0)}$ will be understood as $\varphi$ itself for real one-variable function $\varphi$, and we always let
\begin{eqnarray}
&{}&
\begin{cases}
a_0=1, a_1=-2\eta(2)-1=-1-\pi^2/6=-2.644934\cdots, a_2=2\zeta(3)=2.404113\cdots,\\
a_n=\left[1+(-1)^n \right][\zeta(n+1)-\zeta(n-1)]+\left[1-(-1)^n\right][\zeta(n)-\eta(n+1)]
\mbox{ ~for ~} n\geq 3,\label{an}
\end{cases}\\
&{}&
\begin{cases}
b_0=5\log 2-\pi=0.324143\cdots, b_1=-\log
2+[35\zeta(3)-\pi^3]/8=0.690067\cdots,\\
b_n=5\lambda(2n+1)-\lambda(2n-1)-4\beta(2n+1) \mbox{ ~for ~} n\geq 2,\label{bn}
\end{cases}\\
&{}&~~c_n=1-\sum_{k=0}^nb_k \mbox{ ~and ~}
A_n=2^{n+1}\left(b_0-\sum_{k=0}^n2^{-k}a_k\right) \mbox{ ~for ~} n\in\IN\cup\{0\}.\label{cn}
\end{eqnarray}
Clearly, for $n\in\IN\cup\{0\}$,
\begin{eqnarray}
&{}&
\begin{cases}
a_{2n+2}=2[\zeta(2n+3)-\zeta(2n+1)]=-2\sum_{k=2}^{\infty}k^{-2n-3}(k^2-1)<0,\\
a_{2n+1}=2\left[\zeta(2n+1)-\left(1-2^{-2n-1}\right)\zeta(2n+2)\right]>4^{-n}\zeta(2n+2)>0,\label{SignOfan}
\end{cases}\\
&{}&~~~~c_{n+1}=c_n-b_{n+1}, ~A_{n+1}=2(A_n-a_{n+1}), ~n\in\IN\cup\{0\}.\label{(cn+1)+(An+1)}
\end{eqnarray}
Some properties of the constants $a_n$, $b_n$, $c_n$ and $A_n$ will be given in Lemma \ref{Lem1} and Corollaries \ref{Col1} and \ref{Col2}.
 We now state the main results of this paper.

\begin{theorem}\label{th1}
(1) The function $f(x)\equiv[1+x(1-x)]R(x)-B(x)$ has the following power series expansions
\begin{eqnarray}\label{SerOff(x)}
f(x)=\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}b_n(1-2x)^{2n},
~0<x\leq\frac12.
\end{eqnarray}

(2) $f$ ($f'$) is strictly decreasing and convex (increasing and convex) from $(0,1/2]$ onto
$[b_0,1)$ ($(a_1,0]$, respectively), and the function $F(x)\equiv b_0+b_1(1-2x)^2+B(x)-[1+x(1-x)]R(x)$ is strictly completely monotonic on $(0,1/2]$ with $F(0^+)=b_0+b_1-1=0.0142104\cdots$ and $F(1/2)=0$. Moreover, for each $n\in\IN\setminus\{1\}$, the function $F_{1,\thinspace n}(x)\equiv (-1)^{n+1}f^{(n)}(x)$ is strictly completely monotonic on $(0,1/2]$ with $F_{1,\thinspace n}(0^+)=(-1)^{n+1}n!a_n$, $F_{1,\thinspace n}(1/2)=0$ if $n$ is odd, and $F_{1,\thinspace n}(1/2)=-2^nn!b_{n/2}$ if $n$ is even.
In particular, for $x\in(0,1/2]$,
\begin{eqnarray}\label{Ineq1OfR}
\frac{b_0+(1-2x)P(x)+B(x)}{1+x(1-x)}\leq R(x)\leq\frac{b_0+(1-2x)Q(x)+B(x)}{1+x(1-x)},
\end{eqnarray}
with equality in each instance if and only if $x=1/2$, where $P(x)=\max\{0,1-b_0-b_1+b_1(1-2x)\}$ and $Q(x)=\min\{1-b_0,b_1(1-2x)\}$.

\end{theorem}

\begin{theorem}\label{th2}
For each $n\in\IN\cup\{0\}$ and $x\in(0,1/2]$, let $R_n(x)=\sum_{k=0}^na_kx^k$ and $S_n(x)=\sum_{k=0}^nb_k(1-2x)^{2k}$, $f$ as in Theorem 1.1, and define the functions $f_n$ and $g_n$ on $(0,1/2)$ by
\begin{eqnarray*}
f_n(x)=\frac{f(x)-R_n(x)}{x^{n+1}} \mbox{ ~and ~} g_n(x)=\frac{f(x)-S_n(x)}{(1-2x)^{2(n+1)}},
\end{eqnarray*}
respectively. Then we have the following conclusions:

(1) $f_0$ is strictly increasing and convex from $(0,1/2]$ onto
$(a_1,2(b_0-1)]$. Furthermore, for each $n\in\IN$, $f_{2n-1}$ ($f_{2n}$) is strictly increasing and concave (decreasing and
convex) from $(0,1/2]$ onto $(a_{2n},A_{2n-1}]$ ($[A_{2n},a_{2n+1})$, respectively). In particular, for each
$n\in\IN$ and all $x\in(0,1/2]$,
\begin{eqnarray}
\frac{B(x)+R_{2n+2}(x)+A_{2n+2}x^{2n+3}}{1+x(1-x)}&\leq& R(x)\leq
\frac{B(x)+R_{2n+1}(x)+A_{2n+1}x^{2(n+1)}}{1+x(1-x)},\label{Ineq2OfR}\\
\frac{B(x)+R_4(x)+A_4x^3}{1+x(1-x)}&\leq& R(x)\leq
\frac{B(x)+R_3(x)+A_3x^4}{1+x(1-x)},\label{Ineq3OfR}
\end{eqnarray}
with equality in each instance if and only if $x=1/2$.

(2) For each $n\in\IN\cup\{0\}$, $g_n$ is is strictly increasing and concave from $(0,1/2)$ onto $(c_n, b_{n+1})$. Furthermore, for each $m, n\in\IN\cup\{0\}$, the function $G_{n,\,m}(x)\equiv (-1)^{m+1}g_n^{(m)}(x)$ is strictly completely monotonic on $(0,1/2)$.
In particular, for each $n\in\IN\cup\{0\}$ and all $x\in(0,1/2]$,
\begin{eqnarray}
\frac{B(x)+S_{n+1}(x)+c_{n+1}(1-2x)^{2n+3}}{1+x(1-x)}&\leq&
R(x)\leq \frac{B(x)+S_{n+1}(x)}{1+x(1-x)},\label{Ineq4OfR}\\
\frac{B(x)+S_2(x)+c_2(1-2x)^5}{1+x(1-x)}&\leq& R(x)\leq
\frac{B(x)+S_2(x)}{1+x(1-x)},\label{Ineq5OfR}
\end{eqnarray}
with equality in each instance if and only if $x=1/2$.
\end{theorem}

By Theorem \ref{th1}, it is natural to ask whether the functions
\begin{eqnarray}\label{F1-F3}
F_1(x)\equiv
[1+x(1-x)]\frac{R(x)}{B(x)}, F_2(x)\equiv
\frac{R(x)}{B(x)}-\frac{1}{1+x(1-x)} \mbox{ ~and ~} F_3(x)\equiv R(x)-\frac{B(x)}{1+x(1-x)}
\end{eqnarray}
are monotone on $(0,1/2]$. Our next
result gives the answer to this question.

\begin{theorem}\label{th3}
(1) There exists a unique number
$x_1\in(1/4,1/2)$ such that the function $F_1$ defined by (\ref{F1-F3}) is strictly increasing on $(0,x_1]$, and decreasing on $[x_1,1/2]$, with $F_1(0^+)=1$ and $F_1(1/2)=(5\log2)/\pi=1.103178\cdots$. However, $F_1$ is neither convex nor concave on $(0,1/2]$. In particular, for $x\in(0,1/2]$,
\begin{eqnarray}\label{Ineq6OfR}
\frac{\alpha B(x)}{1+x(1-x)}<R(x)\leq \frac{\delta
B(x)}{1+x(1-x)},
\end{eqnarray}
with the best possible coefficients $\alpha=1$ and $\delta=F_1(x_1)$. Moreover,
\begin{eqnarray}\label{delta}
1.111592\cdots=\frac{19\sqrt{2}\log8}{16\pi}<\delta<
1.112146.
\end{eqnarray}

(2) There exists a unique number $x_2\in(0,1/2)$ such that the
function $F_2$ defined by (\ref{F1-F3}) is strictly
increasing on $(0,x_2]$, and decreasing on $[x_2,1/2]$ with
$F_2(0^+)=0$ and $F_2(1/2)=4b_0/(5\pi)=0.082542\cdots$. However, $F_2$ is neither convex nor concave on $(0,1/2]$.

(3) The function $F_3$ defined by (\ref{F1-F3}) is
strictly decreasing and convex from $(0,1/2]$ onto $[\rho,1)$,
where $\rho=4b_0/5=\log16-4\pi/5=0.259314\cdots$. In particular, for $x\in(0,1/2]$,
\begin{eqnarray}\label{Ineq7OfR}
\rho+\frac{B(x)}{1+x(1-x)}\leq R(x)\leq
\rho+(1-\rho)(1-2x)+\frac{B(x)}{1+x(1-x)},
\end{eqnarray}
with equality in each instance if and only if $x=1/2$.
\end{theorem}

\section{\normalsize Preliminaries}

In this section, we prove two technical lemmas needed in the proofs of our main results
stated in Section 1. Our first lemma shows some properties of $b_n$, $\lambda(n)$ and $\beta(n)$.

\begin{lemma}\label{Lem1}
(1) The functions
$\lambda(x)\equiv\sum_{k=0}^{\infty}(2k+1)^{-x}$ and $\varphi_1(x)\equiv \lambda(x)-\lambda(x+1)$ are both strictly
decreasing and convex on $(1,\infty)$, with $\lambda((1,\infty))=(1,\infty)$ and $\varphi_1((1,\infty))=(0,\infty)$. Moreover, for each $c\in[C_1,\infty)$, the function $\varphi_2(x)\equiv \lambda(x+c)/\lambda(x)$
is strictly increasing from $[2,\infty)$ onto $[C_2,1)$, where $C_1=\log(\pi^2/8)]/\log3=0.191166\cdots$ and $C_2=8\lambda(2+c)/\pi^2>8/\pi^2=0.810569\cdots$.

(2) The function $\beta(x)\equiv
\sum_{k=0}^{\infty}(-1)^k(2k+1)^{-x}$ is strictly increasing from
$[1,\infty)$ onto $[\pi/4,1)$, and concave on $[2/\log3,\infty)$.

(3) Let $\varphi_3(x)=5\lambda(x+1)-\lambda(x-1)-4\beta(x+1)$, $\varphi_4(x)=(x-3+1/\log5)\varphi_3(x)$, $\varphi_5(x)=x\varphi_3(x)$ and $\varphi_6(x)=\varphi_5(x+2)-\varphi_5(x)$, for $x\in[3,\infty)$. Then $\varphi_3$ is strictly increasing and
concave from $[3,\infty)$ onto $[\mu_1,0)$, $\varphi_4$ and $\varphi_5$ are both
strictly increasing on $[3,\infty)$ with $\varphi_4([3,\infty))=[\mu_2,0)$ and $\varphi_5([3,\infty))=[\mu_3,0)$, and $\varphi_6$ is strictly decreasing from $[3,\infty)$ onto $(0, \mu_4]$, where $\mu_1=\varphi_3(3)=(5\pi^4/96)-4\beta(4)-\pi^2/8=-0.116088\cdots$, $\mu_2=\mu_1/\log5=-0.072129\cdots$, $\mu_3=3\mu_1=-0.348265\cdots$ and $\mu_4=\varphi_5(5)-\mu_3=0.337348\cdots$.

(4) For $n\in\IN$, the sequences $\{b_{n+1}\}$ and $\{(n+1)b_{n+1}\}$ are both strictly increasing, while the sequence $\{(n+2)b_{n+2}-(n+1)b_{n+1}\}$ is strictly decreasing, with $\lim_{n\to\infty}nb_n=\lim_{n\to\infty}[(n+1)b_{n+1}-nb_n ]=0$. In particular, for $n\in\IN\setminus\{1\}$,
\begin{eqnarray}\label{IneqOfbn}
-0.027624\cdots=2b_2\leq nb_n<(n+1)b_{n+1}\leq \min\{0,nb_n+3b_3-2b_2\},
\end{eqnarray}
with equality in each instance if and only if $n=2$.
\end{lemma}

{\it Proof.} (1) The monotonicity of $\lambda$ is clear. Since the derivative
\begin{eqnarray}\label{lam'}
\lambda'(x)=-\sum_{k=1}^{\infty}\frac{\log(2k+1)}{(2k+1)^x}
\end{eqnarray}
is clearly increasing on $(1,\infty)$, the function
$\lambda$ is convex on $(1,\infty)$.

It is easy to see that $\varphi_1(1)=\infty$ and $\varphi_1(\infty)=0$. By (\ref{lam'}), we have
\begin{eqnarray}\label{lam'(x+1)-lam'}
\varphi_1'(x)=-2\sum_{k=1}^{\infty}\frac{k}{(2k+1)^{x+1}}\log(2k+1),
\end{eqnarray}
which is negative and strictly increasing on $(1,\infty)$, and hence the result for $\varphi_1$ follows.

Clearly, $\varphi_2(2)=\lambda(2+c)/\lambda(2)=8\lambda(2+c)/\pi^2=C_2$ and
$\varphi_2(\infty)=1$. By differentiation and (\ref{lam'}),
\begin{eqnarray}\label{DerOfphi2}
\lambda(x)^2\varphi_2'(x)
=\sum_{k=1}^{\infty}\frac{\log(2k+1)}{(2k+1)^{x+c}}\left[(2k+1)^c\lambda(x+c)-\lambda(x)\right]
>\varphi_7(x)\sum_{k=1}^{\infty}\frac{\log(2k+1)}{(2k+1)^{x+c}},
\end{eqnarray}
where $\varphi_7(x)=3^c\lambda(x+c)-\lambda(x)$, and by (\ref{lam-beta}),
\begin{eqnarray*}
\varphi_7(x)=\sum_{k=0}^{\infty}\frac{3^c-(2k+1)^c}{(2k+1)^{x+c}}
=3^c-1-\sum_{k=1}^{\infty}\frac{(2k+1)^c-3^c}{(2k+1)^{x+c}}
\end{eqnarray*}
which is clearly strictly increasing on $(1,\infty)$. Since $c\geq [\log(\pi^2/8)]/\log3=[\log\lambda(2)]/\log3$,
$$\varphi_7(2)=3^c\lambda(2+c)-\lambda(2)\geq \lambda(2)[\lambda(2+c)-1]>0.$$
Hence the result for $\varphi_2$ follows from (\ref{DerOfphi2}).

(2) Clearly,
$\lim_{x\to \infty}\beta(x)=1$ and $\beta(1)=\pi/4$. Set $\beta_1(t)=(\log t)/t$ for $t\in[3,\infty)$. Then by differentiation,
\begin{eqnarray}\label{DerOfbeta}
\beta^{\thinspace\prime}(x)&=&\sum_{k=1}^{\infty}(-1)^{k+1}\frac{\log(2k+1)}{(2k+1)^x}
=\sum_{k=1}^{\infty}\frac{\log[2(2k-1)+1]}{[2(2k-1)+1]^x}-\sum_{k=1}^{\infty}\frac{\log[2(2k)+1]}{[2(2k)+1]^x}\nonumber\\
&=&\frac{1}{x}\sum_{k=1}^{\infty}\left[\beta_1((4k-1)^x)-\beta_1((4k+1)^x)\right].
\end{eqnarray}
It is easy to verify that the function $\beta_1$ is strictly decreasing on
$[\e,\infty)$. Since $(4k-1)^x\geq 3^x\geq 3>\e$ for $k\geq1$ and
$x\geq1$, $\beta^{\thinspace\prime}(x)>0$ for $x\in[1,\infty)$ by (\ref{DerOfbeta}), so
that the monotonicity of $\beta(x)$ follows.

By the first equality in (\ref{DerOfbeta}) and by differentiation,
\begin{eqnarray*}
\beta^{\thinspace\prime\prime}(x)=\frac{1}{x^2}\sum_{k=1}^{\infty}\left[\beta_2((4k+1)^x)-\beta_2((4k-1)^x)\right],
\end{eqnarray*}
where $\beta_2(t)=(\log t)^2/t$ for $t\geq 3$. It is easy to show
that $\beta_2$ is strictly decreasing on $[\e^2,\infty)$. Hence if
$(4k-1)^x\geq 3^x\geq \e^2$, that is, $x\geq2/\log3$, then
$\beta^{\thinspace\prime\prime}(x)<0$. This shows that $\beta$ is concave on
$[2/\log3,\infty)$.

(3) It follows from (\ref{zeta(2)}) and [1,Table
23.3] that
$$\varphi_3(3)=\mu_1=5\lambda(4)-\lambda(2)-4\beta(4)=\frac{5\pi^4}{96}-\frac{\pi^2}{8}-4\beta(4)=-0.116088\cdots.$$
Clearly, $\varphi_3(\infty)=0$, and $k(k+1)-1+(-1)^k\geq 6$ for all $k\geq2$. By (\ref{lam-beta}), we have
\begin{eqnarray}
\varphi_3(x)&=&-4\sum_{k=2}^{\infty}\frac{k(k+1)-1+(-1)^k}{(2k+1)^{x+1}},\label{phi3}\\
\varphi_3'(x)&=&4\sum_{k=2}^{\infty}\frac{k(k+1)-1+(-1)^k}{(2k+1)^{x+1}}\log(2k+1).\label{phi3'}
\end{eqnarray}
Hence the result for $\varphi_3$ follow from (\ref{phi3}) and (\ref{phi3'}).

Next, for each $k\in\IN\setminus\{1\}$ and for $x\in[3,\infty)$, let
$\varphi_{8,\,k}(x)=(x-3+1/\log5)(2k+1)^{-x}$. Then $\varphi_{8,\,k}(3)=1/[(2k+1)^3\log5]$, $\varphi_{8,\,k}(\infty)=0$,
and by (\ref{phi3}),
\begin{eqnarray}\label{phi4}
\varphi_4(x)=-4\sum_{k=2}^{\infty}\frac{k(k+1)-1+(-1)^k}{2k+1}\varphi_{8,\,k}(x)
\end{eqnarray}
Clearly, $\varphi_4(3)=\mu_2$, and $\varphi_4(\infty)=0$ by (\ref{phi4}). Since
\begin{eqnarray*}
\varphi_{8,\,k}'(x)&=&(2k+1)^{-x}[1-(x-3+1/\log5)\log(2k+1)]\\
&\leq& -(x-3)(2k+1)^{-x}\log5<0 \mbox{ ~for ~} x>3,
\end{eqnarray*}
$\varphi_{8,\,k}$ is strictly decreasing from
$[3,\infty)$ onto $(0,\varphi_{8,\,k}(3)]$, so that the monotonicity of
$\varphi_4$ follows from (\ref{phi4}).

Since $3-1/\log5=2.378665\cdots>0$, and since
$$-\varphi_5(x)=[-\varphi_4(x)][1+(3-1/\log5)/(x-3+1/\log5)]$$
which is a product of two positive and strictly decreasing functions on $[3,\infty)$, the monotonicity of $\varphi_5$ follows.

Clearly, $\varphi_6(3)= \mu_4=\varphi_5(5)-\mu_3=5[5\lambda(6)-\lambda(4)-4\beta(6)]-\mu_3=0.337348\cdots$ and $\varphi_6(\infty)=0$. It follows from (\ref{phi3}) that
\begin{eqnarray}\label{phi6}
\varphi_6(x)=(x+2)\varphi_3(x+2)-x\varphi_3(x)=8\sum_{k=2}^{\infty}\frac{k(k+1)-1+(-1)^k}{(2k+1)^{x+3}}[2k(k+1)x-1].
\end{eqnarray}
Differentiation gives
\begin{eqnarray*}
\frac{d}{dx}\left[\frac{2\left(k^2+k\right)x-1}{(2k+1)^{x+3}}\right]
&=&\frac{2k(k+1)\log(2k+1)}{(2k+1)^{x+3}}\left[\frac{1}{\log(2k+1)}+\frac{1}{2k(k+1)}-x\right]\\
&\leq&\frac{2k(k+1)\log(2k+1)}{(2k+1)^{x+3}}\left(\frac{1}{\log5}+\frac{1}{12}-3\right)<0
\end{eqnarray*}
for $x\in[3,\infty)$, since $(1/\log5)+(1/12)=0.704668\cdots$. Hence the monotonicity of $\varphi_6$ follows from (\ref{phi6}).

(4) Clearly, $b_{n+1}=\varphi_3(2n+2)$, $(n+1)b_{n+1}=\varphi_5(2n+2)/2$ and $(n+2)b_{n+2}-(n+1)b_{n+1}=\varphi_6(2n+2)/2$. Hence part (4) follows from part (3). $\Box$

\begin{lemma}\label{Lem2}
For $x\in(0,1/2]$,
\begin{eqnarray}
R(x)&=&\frac{1}{x}+\sum_{n=1}^{\infty}\left[1+(-1)^n\right]\zeta(n+1)x^n =\log
16+4\sum_{n=1}^{\infty}\lambda(2n+1)(1-2x)^{2n},\label{SerOfR}\\
B(x)&=&\frac{1}{x}+\sum_{n=1}^{\infty}\left[1+(-1)^{n+1}\right]\eta(n+1)x^n
=4\sum_{n=0}^{\infty}\beta(2n+1)(1-2x)^{2n},\label{SerOfB}\\
R\left(\frac14\right)&=&-2\gamma-\psi\left(\frac14\right)-\psi\left(\frac34\right)=6\log2=4.158883\cdots,\label{R(1/4)}\\
R'\left(\frac14\right)&=&\psi'\left(\frac34\right)-\psi'\left(\frac14\right)=-16\beta(2)=-14.655449\cdots.\label{R'(1/4)}
\end{eqnarray}

\end{lemma}

{\it Proof.} It is well known that
\begin{eqnarray}
\psi(1+x)&=&\psi(x)+\frac{1}{x},\label{psi1}\\
\psi(1+x)+\gamma&=&\sum_{n=1}^{\infty}(-1)^{n+1}\zeta(n+1)x^n, ~|x|<1,\label{psi2}\\
\psi^{(n)}(x)&=&(-1)^{n+1}n!\sum_{k=0}^{\infty}\frac{1}{(k+x)^{n+1}}.\label{n-Dpsi}
\end{eqnarray}
(Cf. \citep[6.3.5, 6.3.14 \& 6.4.10]{1}.) By (\ref{psi1}) and (\ref{psi2}),
\begin{eqnarray*}
R(x)&=&\frac{1}{x}-[\gamma+\psi(1-x)]-[\gamma+\psi(1+x)]\\
&=&\frac{1}{x}+\sum_{n=1}^{\infty}(-1)^n\zeta(n+1)(-x)^n+\sum_{n=1}^{\infty}(-1)^n\zeta(n+1)x^n,
\end{eqnarray*}
yielding the first equality in (\ref{SerOfR}). By differentiation,
\begin{eqnarray}\label{n-DR}
R^{(n)}(x)=(-1)^{n+1}\psi^{(n)}(1-x)-\psi^{(n)}(x).
\end{eqnarray}
 Hence by (\ref{n-Dpsi}) and (\ref{n-DR}), $R^{(2n-1)}(1/2)=0$ and
\begin{eqnarray*}
R^{(2n)}\left(\frac12\right)=-2\psi^{(2n)}\left(\frac12\right)=\sum_{k=0}^{\infty}\frac{2(2n)!}{(k+1/2)^{2n+1}}=4^{n+1}(2n)!\lambda(2n+1),
\end{eqnarray*}
so that $R(x)$ has the following power series
expansion
\begin{eqnarray*}
R(x)&=&R\left(\frac12\right)+\sum_{n=1}^{\infty}\frac{R^{(n)}(1/2)}{n!}\left(x-\frac{1}{2}\right)^n
=\log16+\sum_{n=1}^{\infty}\frac{R^{(2n)}(1/2)}{(2n)!}\left(x-\frac{1}{2}\right)^{2n}\\
&=&\log16+4\sum_{n=1}^{\infty}\lambda(2n+1)(1-2x)^{2n}
\end{eqnarray*}
at $x=1/2$. This yields the second equality in (\ref{SerOfR}).

By \citep[4.3.68 \& 23.1.18]{1},
\begin{eqnarray}\label{SerOfSin}
\frac{1}{\sin
z}=\frac{1}{z}+2\sum_{n=1}^{\infty}\frac{\eta(2n)}{\pi^{2n}}z^{2n-1}
=\frac{1}{z}+\sum_{n=1}^{\infty}\left[1+(-1)^{n+1}\right]\frac{\eta(n+1)}{\pi^{n+1}}z^{n},
\end{eqnarray}
for $|z|<\pi$. Hence the first equality in (\ref{SerOfB}) follows from (\ref{B})
and (\ref{SerOfSin}). By \citep[4.3.69 \& 23.1.18]{1}, we have
\begin{eqnarray}\label{SerOfCos}
\frac{1}{\cos
z}=2\sum_{n=0}^{\infty}\left(\frac{2}{\pi}\right)^{2n+1}\beta(2n+1)z^{2n}, ~|z|<\frac{\pi}{2},
\end{eqnarray}
from which it follows that
$$B(x)=\frac{\pi}{\sin(\pi x)}=\frac{\pi}{\cos[\pi(1-2x)/2]}
=4\sum_{n=0}^{\infty}\beta(2n+1)(1-2x)^{2n},$$
yielding the
second equality in (\ref{SerOfB}).

Next, by \citep[6.3.3 \& 6.3.8]{1},
\begin{eqnarray}\label{psi(2x)}
\psi\left(\frac12\right)=-\gamma-\log4, ~\psi(2x)=\frac12\psi(x)+\frac12\psi\left(x+\frac12\right)+\log 2,
\end{eqnarray}
from which we obtain
\begin{eqnarray*}
R\left(\frac14\right)=-2\gamma-\left[\psi\left(\frac14\right)+\psi\left(\frac34\right)\right]
=-2\gamma-2\left[\psi\left(\frac12\right)-\log2\right]=6\log2.
\end{eqnarray*}

Finally, by (\ref{lam-beta}), it is easy to see that
\begin{eqnarray}\label{beta(2)}
\beta(2)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=\sum_{n=0}^{\infty}\frac{1}{[2(2n)+1]^2}-\sum_{n=0}^{\infty}\frac{1}{[2(2n+1)+1]^2}.
\end{eqnarray}
It follows from (\ref{n-Dpsi}), (\ref{beta(2)}) and \citep[Table 23.3]{1} that
\begin{eqnarray*}
R'\left(\frac14\right)=\psi'\left(\frac34\right)-\psi'\left(\frac14\right)
=16\left[\sum_{n=0}^{\infty}\frac{1}{(4n+3)^2}-\sum_{n=0}^{\infty}\frac{1}{(4n+1)^2}\right]
=-16\beta(2). ~\Box
\end{eqnarray*}

\begin{remark}\label{Rem1}
In \citep[Theorem 2.2]{16}, it was proved that
$R(x)=(1/x)+2\sum_{n=1}^{\infty}\zeta(2n+1)x^{2n}$ for $x\in(0,1/2]$,
which is consistent with the first equality in (\ref{SerOfR}). However, its proof given in \cite{16} is quite
complicated.
\end{remark}

\section{\normalsize Proof of Theorem \ref{th1}}

(1) Since $f(x)=R(x)-B(x)+x(1-x)R(x)$, it follows from Lemma \ref{Lem2},
(\ref{lambda(n+1)}) and (\ref{an}) that
\begin{eqnarray*}
f(x)&=&\sum_{n=1}^{\infty}\left\{\left[1+(-1)^n\right]\zeta(n+1)-\left[1+(-1)^{n+1}\right]\eta(n+1)\right\}x^n\\
&{}&+(1-x)\left\{1+\sum_{n=1}^{\infty}\left[1+(-1)^n\right]\zeta(n+1)x^{n+1}\right\}\\
&=&1-x+\sum_{n=1}^{\infty}\left\{\left[1+(-1)^n\right]\zeta(n+1)-\left[1+(-1)^{n+1}\right]\eta(n+1)\right\}x^n\\
&{}&+\sum_{n=1}^{\infty}[1+(-1)^n]\zeta(n+1)x^{n+1}-\sum_{n=1}^{\infty}[1+(-1)^n]\zeta(n+1)x^{n+2}\\
&=&1-[1+2\eta(2)]x+2\zeta(3)x^2+\sum_{n=3}^{\infty}\left\{\left[1+(-1)^n\right]\zeta(n+1)-\left[1+(-1)^{n+1}\right]\eta(n+1)\right\}x^n\\
&{}&+\sum_{n=2}^{\infty}\left[1+(-1)^{n+1}\right]\zeta(n)x^n-\sum_{n=3}^{\infty}\left[1+(-1)^n\right]\zeta(n-1)x^n\\
&=&1-[1+2\eta(2)]x+2\zeta(3)x^2+\sum_{n=3}^{\infty}\left\{\left[1+(-1)^n\right]\zeta(n+1)+\left[1+(-1)^{n+1}\right]\zeta(n)\right.\\
&{}&\left.-\left[1+(-1)^{n+1}\right]\eta(n+1)-\left[1+(-1)^n\right]\zeta(n-1)\right\}x^n\\
&=&1-[1+\zeta(2)]x+2\zeta(3)x^2+\sum_{n=3}^{\infty}a_nx^n=\sum_{n=0}^{\infty}a_nx^n,
\end{eqnarray*}
which yields the first power series expansion in (\ref{SerOff(x)}).

Clearly, $1+x(1-x)=[5-(1-2x)^2]/4$. It follows from (\ref{SerOfR}),
(\ref{SerOfB}), (\ref{lambda(n+1)}), (\ref{zeta(2)}) and (\ref{bn}) that
\begin{eqnarray*}
f(x)&=&\left[\frac{5}{4}-\frac{(1-2x)^2}{4}\right]\left[\log 16
+4\sum_{n=1}^{\infty}\lambda(2n+1)(1-2x)^{2n}\right]-4\sum_{n=0}^{\infty}\beta(2n+1)(1-2x)^{2n}\\
&=&b_0-(1-2x)^2\log2-\sum_{n=1}^{\infty}\lambda(2n+1)(1-2x)^{2n+2}+\sum_{n=1}^{\infty}[5\lambda(2n+1)-4\beta(2n+1)](1-2x)^{2n}\\
&=&b_0+[5\lambda(3)-4\beta(3)-\log2](1-2x)^2+\sum_{n=2}^{\infty}[5\lambda(2n+1)-\lambda(2n-1)-4\beta(2n+1)](1-2x)^{2n}\\
&=&b_0+b_1(1-2x)^2+\sum_{n=2}^{\infty}b_n(1-2x)^{2n}=\sum_{n=0}^{\infty}b_n(1-2x)^{2n},
\end{eqnarray*}
yielding the second series expansion in (\ref{SerOff(x)}).

(2) Clearly, $f(0^+)=a_0=1$ and $f(1/2)=b_0$ by (\ref{SerOff(x)}). By (\ref{SerOff(x)}) and differentiation, we
obtain
\begin{eqnarray}
f'(x)&=&-2\sum_{k=1}^{\infty}(2k)b_k(1-2x)^{2k-1},\label{SerOff'}\\
f''(x)&=&(-2)^2\sum_{k=0}^{\infty}(2k+2)(2k+1)b_{k+1}(1-2x)^{2k},\label{SerOff''}\\
f'''(x)&=&(-2)^3\sum_{k=1}^{\infty}(2k+2)(2k+1)(2k)b_{k+1}(1-2x)^{2k-1}.\label{SerOff'''}
\end{eqnarray}
Generally, by the mathematical induction, it is not difficult to
obtain the following two expressions
\begin{eqnarray}
f^{(2n-1)}(x)&=&-2^{2n-1}\sum_{k=1}^{\infty}\frac{(2n+2k-2)!}{(2k-1)!})b_{k+n-1}(1-2x)^{2k-1},\label{SerOf(2n-1)-Df}\\
f^{(2n)}(x)&=&2^{2n}\sum_{k=0}^{\infty}\frac{(2n+2k)!}{(2k)!}b_{n+k}(1-2x)^{2k},\label{SerOf(2n)-Df}
\end{eqnarray}
for $n\in\IN$. By (\ref{SerOf(2n-1)-Df}) and (\ref{SerOf(2n)-Df}), $f^{(2n-1)}(1/2)=0$ and
$f^{(2n)}(1/2)=4^n(2n)!b_n$ for $n\in\IN$. The limiting value
$f^{(n)}(0^+)=n!a_n$ follows from the first equality in (\ref{SerOff(x)}). Hence it follows from (\ref{IneqOfbn}),
(\ref{SerOf(2n-1)-Df}) and (\ref{SerOf(2n)-Df}) that for all $n\geq 2$, $f^{(2n-1)}$ (
$f^{(2n)}$ ) is strictly decreasing and convex (increasing and
concave) from $(0,1/2]$ onto $[0,(2n-1)!a_{2n-1})$ (
$((2n)!a_{2n},4^n(2n)!b_n]$, respectively).

It follows from (\ref{IneqOfbn}) and (\ref{SerOff''}) that $f''$ is strictly increasing and concave on $(0,1/2]$,
with $f''(0^+)=2a_2>0$ by (\ref{SerOff(x)}). Hence $f'$ is strictly increasing and
convex on $(0,1/2]$ with $f'(1/2)=0$ by (\ref{SerOff'}) and $f'(0^+)=a_1$ by (\ref{SerOff(x)}), and $f$ is strictly
decreasing and convex from $(0,1/2]$ onto $[b_0,1)$.

Clearly, $F(0^+)=b_0+b_1-f(0^+)=b_0+b_1-1=0.0142104\cdots$ and $F(1/2)=b_0-f(1/2)=0$.
 It follows from (\ref{SerOff(x)}) that $F$ has the following power series expansion
\begin{eqnarray}\label{SerOfF(x)}
F(x)=-\sum_{k=2}^{\infty}b_k(1-2x)^{2k}.
\end{eqnarray}
Applying Lemma \ref{Lem1}(4) and (\ref{SerOfF(x)}), one can easily see that
$F$ is strictly completely monotonic on $(0,1/2]$.

Next, for $m, n, k\in\IN$ and $x\in(0,1/2)$,
\begin{eqnarray}\label{F1n}
(-1)^mF_{1,\,n}^{(m)}(x)=(-1)^{m+n+1}f^{(n+m)}(x)=
\begin{cases}
-f^{(2k)}(x), \mbox{~~if ~} m+n=2k,\\
f^{(2k-1)}(x), \mbox{~if ~} m+n=2k-1.\\
\end{cases}
\end{eqnarray}
If $n\geq2$, then $m+n\geq3$ and $k\geq2$. It follows from Lemma \ref{Lem1}(4) and the monotonicity properties of $f^{(2k-1)}$ and
$f^{(2k)}$, $k\geq 2$, that for $x\in(0,1/2)$,
\begin{eqnarray*}
-f^{(2k)}(x)&>&-f^{(2k)}(1/2)=-4^k(2k)!b_{k}>0,\\
f^{(2k-1)}(x)&>&f^{(2k-1)}(1/2)=0.
\end{eqnarray*}
Hence by (\ref{F1n}), $(-1)^mF_{1,\,n}^{(m)}(x)>0$ for each $m\in\IN$, $n\in\IN\setminus\{1\}$ and for all $x\in(0,1/2)$. This yields the strictly complete monotonicity of $F_{1,\,n}$.

Clearly, $F_{1,\thinspace n}(0^+)=(-1)^{n+1}f^{(n)}(0^+)=(-1)^{n+1}n!a_n$. By (\ref{SerOf(2n-1)-Df}) and (\ref{SerOf(2n)-Df}), $F_{1,\thinspace n}(1/2)=0$ if $n$ is odd, and $F_{1,\thinspace n}(1/2)=-2^nn!b_{n/2}$ if $n$ is even.

Finally, it follows from the monotonicity and convexity properties of $f$ and $F$ that
\begin{eqnarray}
\frac{b_0+B(x)+(1-b_0)(1-2x)}{1+x(1-x)}&\geq& R(x)\geq \frac{b_0+B(x)}{1+x(1-x)}
,\label{Ineq0OfR}\\
\frac{b_0+b_1(1-2x)^2+B(x)}{1+x(1-x)}&\geq& R(x)\geq \frac{b_0+(1-b_0-b_1)(1-2x)+b_1(1-2x)^2+B(x)}{1+x(1-x)}
,\label{Ineq00OfR}
\end{eqnarray}
with equality in each instance if and only if $x=1/2$. Hence (\ref{Ineq1OfR}) and its equality case follow. $\Box$

\begin{corollary}\label{Col1}
Let $a_n$,
$b_n$ and $c_n$ be as in (\ref{an})--(\ref{cn}), and set $d_n=\sum_{k=0}^nb_k\lambda(2n-2k+2)-(n+1)b_{n+1}$ for $n\in\IN\cup\{0\}$, $D_0=5d_0-b_0$ and $D_n=5d_n-d_{n-1}-b_n$ for $n\in\IN$. Then we have the following conclusions:

(1) The following identities hold
\begin{eqnarray}
&{}&\sum_{k=0}^{\infty}a_k=\sum_{k=0}^{\infty}b_k=1,\label{Id1}\\
&{}&\sum_{k=0}^{\infty}2^{-k}a_k=b_0=5\log2-\pi.\label{Id2}
\end{eqnarray}

(2) For $n\in\IN$, the sequence $\{c_n\}$ is strictly increasing. In particular, for $n\in\IN\setminus\{1\}$,
\begin{eqnarray}
-0.014210\cdots=c_1&<&-0.000398\cdots=c_2\leq c_n<\lim_{n\to\infty} c_n=0,\label{Ineq1Ofcn}\\
1&<&\sum_{k=0}^nb_k=1-c_n\leq1-c_2.\label{IneqOf(1-cn)}
\end{eqnarray}

(3) $d_0=-0.290171\cdots$, $d_1=1.207861\cdots$, $d_2=1.008920\cdots$, $d_3=1.000824\cdots$, $\lim_{n\to\infty}d_n=1$, and
\begin{eqnarray}\label{IneqOfdn}
d<d_n<\widetilde{d}<d_2 \mbox{ ~for ~} n\geq3,
\end{eqnarray}
where $d=(1-\pi^2/8)(b_0+b_1)+\pi^2/8=0.996679\cdots$ and $\widetilde d=d_3-[(\pi^4/96)-1]b_2-[(\pi^2/8)-1]b_3=1.001117\cdots$.

(4) $D_0=[(5\pi^2/8)-1]b_0-5b_1=-1.7750006\cdots$, $D_1=5d_1-d_0-b_1=5.639413\cdots$, $D_2=5d_2-d_1-b_2=3.850551\cdots$, $D_3=5d_3-d_2-b_3=3.995587\cdots$, and
\begin{eqnarray}\label{Dn}
D_n>5d-\widetilde{d}=3.982277\cdots, ~n\geq4.
\end{eqnarray}
\end{corollary}

\smallskip
{\bf Proof.} (1) Let $f$ be as in Theorem \ref{th1}. Then $f(1-x)=f(x)$ by the symmetry. Hence $f(1^-)=f(0^+)=a_0=1$, so that $\sum_{k=0}^{\infty}a_k=\sum_{k=0}^{\infty}b_k=1$. We obtain (\ref{Id2}) by taking $x=1/2$ in (\ref{SerOff(x)}).

(2) By (\ref{Id1}) and (\ref{cn}), $\lim_{n\to\infty}c_n=0$. Since $c_{n+1}-c_n=-b_{n+1}>0$ for $n\in\IN$ by Lemma \ref{Lem1}(4) and (\ref{(cn+1)+(An+1)}), the monotonicity of $c_n$ and (\ref{Ineq1Ofcn}) follow. (\ref{IneqOf(1-cn)}) holds by (\ref{Ineq1Ofcn}).

(3) By (\ref{lambda(n+1)}), (\ref{zeta(2)}), (\ref{bn}) and \citep[Table 23.3]{1}, we obtain
\begin{eqnarray*}
d_0&=&b_0\lambda(2)-b_1=(\pi^2b_0/8)-b_1=-0.290171\cdots,\\
d_1&=&b_0\lambda(4)+b_1\lambda(2)-2[5\lambda(5)-\lambda(3)-4\beta(5)]=1.207861\cdots,\\
d_2&=&b_0\lambda(6)+b_1\lambda(4)+b_2\lambda(2)-3b_3=1.008920\cdots,\\
d_3&=&b_0\lambda(8)+b_1\lambda(6)+b_2\lambda(4)+b_3\lambda(2)-4b_4=1.000824\cdots.
\end{eqnarray*}

Applying Lemma \ref{Lem1}(1) and (4), one can easily show that $\lim_{n\to\infty}\sum_{k=2}^nb_k[\lambda(2n-2k+2)-1]=0$ by the definition of limit. Hence by Lemma \ref{Lem1}(1) and (4),
\begin{eqnarray*}
\lim_{n\to\infty}d_n&=&\lim_{n\to\infty}\left\{b_0\lambda(2n+2)+b_1\lambda(2n)
+\sum_{k=2}^nb_k-(n+1)b_{n+1}\right.\\
&{}&\left.+\sum_{k=2}^nb_k[\lambda(2n-2k+2)-1]\right\}=\sum_{k=0}^{\infty}b_k=1.
\end{eqnarray*}

Since $\lambda(m)>1$ and $b_m<0$ for $m\geq2$ by Lemma \ref{Lem1}(1) and (4), it follows that for $n\geq3$,
\begin{eqnarray*}
d_n&=&b_0\lambda(2n+2)+b_1\lambda(2n)+\sum_{k=2}^nb_k\lambda(2n-2k+2)-(n+1)b_{n+1}\\
&<&\widetilde{d_n}\equiv b_0\lambda(2n+2)+b_1\lambda(2n)+\sum_{k=2}^nb_k-(n+1)b_{n+1}\\
&=& b_0[\lambda(2n+2)-1]+b_1[\lambda(2n)-1]+1-c_n-(n+1)b_{n+1},
\end{eqnarray*}
which strictly decreasing in $n$ by part (2) and Lemma \ref{Lem1}(1) and (4). Hence for $n\geq3$,
\begin{eqnarray*}
\widetilde{d_n}&\leq&\widetilde{d_3}=\widetilde{d}\equiv b_0[\lambda(8)-1]+b_1[\lambda(6)-1]+1-c_3-4b_4\\
&=&d_3-[\lambda(4)-1]b_2-[\lambda(2)-1]b_3=1.001117\cdots<d_2.
\end{eqnarray*}
Hence the second and third inequalities in (\ref{IneqOfdn}) hold.

On the other hand, by part (2), and by Lemma \ref{Lem1}(1) and (4), we have
\begin{eqnarray*}
d_n&\geq&b_0\lambda(2n+2)+b_1\lambda(2n)+\lambda(2)\sum_{k=2}^nb_k-(n+1)b_{n+1}=\overline{d_n}\\
&\equiv& b_0[\lambda(2n+2)-\lambda(2)]+b_1[\lambda(2n)-\lambda(2)]+\lambda(2)(1-c_n)-(n+1)b_{n+1},
\end{eqnarray*}
which strictly decreasing in $n$ by part (2) and Lemma \ref{Lem1}(1) and (4), and hence
\begin{eqnarray*}
d_n\geq\overline{d_n}>\lim_{n\to\infty}\overline{d_n}=d\equiv b_0\left(1-\frac{\pi^2}{8}\right)+b_1\left(1-\frac{\pi^2}{8}\right)+\frac{\pi^2}{8}
=0.996679\cdots.
\end{eqnarray*}
This yields the first inequality in (\ref{IneqOfdn}).

(4) By computation, one can obtain the values of $D_n$ for $0\leq n\leq 3$. It follows from  (\ref{IneqOfbn}) and (\ref{IneqOfdn}) that
\begin{eqnarray*}
D_n=5d_n-d_{n-1}-b_n>5d-\widetilde{d}=3.982277\cdots
\end{eqnarray*}
for $n\geq 4$, which yields the inequality (\ref{Dn}). $\Box$

\section{\normalsize Proof of Theorem \ref{th2}}

(1) Clearly, $f_0(0^+)=f'(0^+)=a_1$ and $f_0(1/2)=2(b_0-1)$. Let $h_0(x)=xf'(x)-[f(x)-a_0]$. Then
\begin{eqnarray*}
h_0(0^+)=0, ~f_0'(x)=\frac{h_0(x)}{x^2}, ~h_0'(x)\left[\frac{d}{dx}(x^2)\right]^{-1}=\frac12 f''(x).
\end{eqnarray*}
By Theorem \ref{th1}(2), $f''$ is strictly increasing on $(0,1/2]$, and hence so is $f_0'$ by \citep[Theorem 1.25]{2}. Since $f_0'(0^+)=f''(0^+)/2=a_2>0$ by l'H\^opital's Rule, the monotonicity and convexity properties of $f_0$ follow.

Let $h_1(x)=f(x)-R_n(x)$ and $h_2(x)=x^{n+1}$.
Then by ({\ref{SerOff(x)}),
\begin{eqnarray}\label{SerOffn}
h_1(x)=\sum_{k=n+1}^{\infty}a_kx^k=x^{n+1}\sum_{k=0}^{\infty}a_{n+k+1}x^k, ~
f_n(x)=\frac{h_1(x)}{h_2(x)}=\sum_{k=0}^{\infty}a_{n+k+1}x^k,
\end{eqnarray}
$h_1^{(m)}(0^+)=h_2^{(m)}(0)=0$ for
$m\in\IN\cup\{0\}$ with $0\leq m \leq n$, and by differentiation,
$$\frac{h_1^{(n+1)}(x)}{h_2^{(n+1)}(x)}=\frac{f^{(n+1)}(x)}{(n+1)!},$$
so that$f_n$ has the same monotonicity
property as that of $f^{(n+1)}$ by \citep[Theorem 1.25]{2}. Hence the monotonicity properties
of $f_{2n}$ and $f_{2n-1}$ follow from Theorem \ref{th1}(2).

Next, differentiation gives
\begin{eqnarray}\label{fn'}
f_n'(x)&=&\frac{1}{x^{n+2}}\left\{x\left[f'(x)-\sum_{k=0}^nka_kx^{k-1}\right]-
(n+1)\left[f(x)-\sum_{k=0}^na_kx^k\right]\right\}\nonumber\\
&=&\frac{1}{x^{n+2}}\left[x
f'(x)-(n+1)f(x)+\sum_{k=0}^n(n+1-k)a_kx^k\right]=\frac{h_3(x)}{h_4(x)},
\end{eqnarray}
where $h_4(x)=x^{n+2}$ and
\begin{eqnarray*}
h_3(x)&=&xf'(x)-(n+1)f(x)+\sum_{k=0}^n(n+1-k)a_kx^k\\
&=&\sum_{k=n+2}^{\infty}(k-n-1)a_kx^k=x^{n+2}\sum_{k=0}^{\infty}(k+1)a_{n+k+2}x^k
\end{eqnarray*}
by (\ref{SerOff(x)}). It is easy to verify that $h_3^{(n+1)}(x)=xf^{(n+2)}(x)$. Hence $h_3^{(m)}(0^+)=h_4^{(m)}(0)=0$ for
$m\in\IN\cup\{0\}$ with $0\leq m \leq n+1$, and
\begin{eqnarray}\label{(n+2)-Df}
\frac{h_3^{(n+1)}(x)}{h_4^{(n+1)}(x)}=\frac{f^{(n+2)}(x)}{(n+2)!},
\end{eqnarray}
which shows that $f_n'$ has the same monotonicity
property as that of $f^{(n+2)}$ by \citep[Theorem 1.25]{2} and (\ref{fn'}). Consequently, the convexity (
concavity ) of $f_{2n}$ ( $f_{2n+1}$, respectively ) follows from
Theorem \ref{th1}(2).

The limiting value $f_n(0^+)=a_{n+1}$ follows from (\ref{SerOffn}). By the definition of $f_n$,
$f_n(1/2)=A_n$. Hence the double inequality (\ref{Ineq2OfR}) and its equality case follow from the
monotonicity and concavity properties of
$f_{2n-1}$ and (\ref{(cn+1)+(An+1)}). Taking $n=1$ in (\ref{Ineq2OfR}), we obtain (\ref{Ineq3OfR})
and its equality case.

(2) Let $h_5(x)=f(x)-S_n(x)=(1-2x)^{2(n+1)}\sum_{k=0}^{\infty}b_{n+k+1}(1-2x)^{2k}$ and
$h_6(x)=(1-2x)^{2(n+1)}$. Then $h_5^{(m)}(1/2)=h_6^{(m)}(1/2)=0$ for
$m\in\IN\cup\{0\}$ with $0\leq m \leq 2n+1$, and
\begin{eqnarray}\label{SerOfgn}
g_n(x)=\frac{h_5(x)}{h_6(x)}=\sum_{k=0}^{\infty}b_{n+k+1}(1-2x)^{2k}.
\end{eqnarray}
By the definition of $g_n$, $g_n(0^+)=h_5(0^+)=c_n$, and by (\ref{SerOfgn}), $g_n((1/2)^-)=b_{n+1}$. Hence it follows from (\ref{SerOfgn}) and Lemma \ref{Lem1}(4) that $g_n$ is strictly increasing and concave from $(0,1/2)$ onto $(c_n,b_{n+1})$.

Next, applying the method used to prove (\ref{SerOf(2n-1)-Df}) and (\ref{SerOf(2n)-Df}), we can obtain the following derivatives
\begin{eqnarray}\label{(2m+1)-Dgn}
g_n^{(2m+1)}(x)&=&-2^{2m+1}\sum_{k=1}^{\infty}\frac{(2m+2k)!}{(2k-1)!}b_{n+m+k+1}(1-2x)^{2k-1},\label{(2m+1)-Dgn}\\
g_n^{(2m)}(x)&=&2^{2m}\sum_{k=0}^{\infty}\frac{(2m+2k)!}{(2k)!}b_{n+m+k+1}(1-2x)^{2k}.\label{(2m)-Dgn}
\end{eqnarray}
Hence by Lemma \ref{Lem1}(4), $g_n^{(2m)}$ ( $g_n^{(2m+1)}$ ) is strictly increasing (
decreasing, respectively ) on $(0,1/2)$ for $m\in\IN\cup\{0\}$. This also yields the concavity ( convexity ) property of $g_n^{(2m)}$ $\left(
g_n^{(2m+1)}, \mbox{respectively} \right)$.

The proof of the complete monotonicity property for $G_{n,\thinspace m}$ is similar to that for $F_{1,\,n}$ in Theorem \ref{th1}(2), and we omit the details. By (\ref{(2m+1)-Dgn}) and (\ref{(2m)-Dgn}), we obtain the following limiting values
\begin{eqnarray*}
G_{n,\,2m+1}(0^+)&=&-2^{2m+1}\sum_{k=1}^{\infty}\frac{(2m+2k)!}{(2k-1)!}b_{n+m+k+1}, ~G_{n,\,2m+1}((1/2)^-)=0,\\
G_{n,\,2m}((1/2)^-)&=&-2^{2m}\sum_{k=0}^{\infty}\frac{(2m+2k)!}{(2k)!}b_{n+m+k+1}, ~G_{n,\,2m+1}(0^+)=-4^m(2m)!b_{n+m+1}.
\end{eqnarray*}

The double inequality (\ref{Ineq4OfR}) and its equality case follow from the
monotonicity and concavity properties of $g_n$. Taking $n=1$ in (\ref{Ineq4OfR}), we obtain (\ref{Ineq5OfR}) and its equality case. $\Box$

\begin{corollary}\label{Col2}
Let $a_n$, $b_n$, $c_n$ and $A_n$ are as in (\ref{an})--(\ref{cn}). Then we have the following conclusions:

(1) For all $n\in\IN\cup\{0\}$, $c_n<b_{n+1}$.

(2) $A_{2n+1}$ ( $A_{2n}$ ) is strictly increasing (decreasing, respectively) in $n\in\IN$, and for all $n\in\IN$,
\begin{eqnarray}\label{Sign1OfAn}
a_{2n+2}<A_{2n+1}<\lim_{n\to\infty}A_n=0<a_{2n+1}+\frac12 a_{2n+2}< A_{2n}<a_{2n+1}.
\end{eqnarray}

(3) $|a_n|$ is strictly decreasing in $n\in\IN\setminus\{1\}$. In particular, for $n\in\IN\setminus\{1\}$ and $\mu=65/108=0.601851\cdots$,
\begin{eqnarray}
\lim_{n\to\infty}a_n=0<-a_{2n+2}&<&a_{2n+1}<-a_{2n}<\cdots<a_3<a_2<-a_1,\label{Ineq1Ofan}\\
\left(1-\frac{1-\mu}{2^{n-1}-\mu}\right)\zeta(n)&<&\zeta(n+1)<\frac{\zeta(n)+1}{2}<\zeta(n).\label{IneqOfzeta}
\end{eqnarray}
\end{corollary}

{\it Proof.} (1) The inequality in part (1) holds by the result for $g_n$ in Theorem \ref{th2}(2).

(2) It follows from the monotonicity properties of $f_{2n+1}$ and $f_{2n}$ stated in Theorem \ref{th2}(1) that
\begin{eqnarray}\label{Ineq1OfA2n}
A_{2n}<a_{2n+1} \mbox{ ~and ~} a_{2n+2}< A_{2n+1}, ~n\in\IN.
\end{eqnarray}
By (\ref{Id1}), $\lim_{n\to\infty}a_n=0$. From (\ref{(cn+1)+(An+1)}) and (\ref{Ineq1OfA2n}), we obtain
\begin{eqnarray}
a_{2n+2}&<& A_{2n+1}=2(A_{2n}-a_{2n+1})<0,\label{Sign2OfAn}\\
a_{2n+1}&>&A_{2n}=a_{2n+1}+\frac12 A_{2n+1}>a_{2n+1}+\frac12 a_{2n+2},\label{Sign3OfAn}
\end{eqnarray}
and hence $\lim_{n\to\infty}A_n=0$ by Pinching Theorem for limits.

It follows from (\ref{zeta}), (\ref{lam-beta}) and (\ref{SignOfan}) that
\begin{eqnarray}\label{Ineq2Ofan}
a_{2n+1}+\frac12 a_{2n+2}
=\zeta(2n+3)+\zeta(2n+1)-2\eta(2n+2)=\sum_{k=2}^{\infty}\frac{[k+(-1)^k]^2}{k^{2n+3}}>0.
\end{eqnarray}
Hence (\ref{Sign1OfAn}) holds by (\ref{Ineq1OfA2n})--(\ref{Ineq2Ofan}).

Next, by (\ref{(cn+1)+(An+1)}) and (\ref{Sign2OfAn}), we obtain
\begin{eqnarray*}
A_{2n+1}-A_{2n+3}&=&2[(A_{2n}-a_{2n+1})-(A_{2n+2}-a_{2n+3})]\\
&=&2[A_{2n}-2(A_{2n+1}-a_{2n+2})+a_{2n+3}-a_{2n+1}]\\
&=&2[2a_{2n+2}+a_{2n+3}-3(A_{2n}-a_{2n+1})]\\
&<& 2\left(2a_{2n+2}+a_{2n+3}-\frac32 a_{2n+2}\right)=a_{2n+2}+2 a_{2n+3},
\end{eqnarray*}
and hence by (\ref{SignOfan}), (\ref{zeta}) and by (\ref{lam-beta}),
\begin{eqnarray}\label{Ineq4OfAn}
A_{2n+1}-A_{2n+3}&<& 2[\zeta(2n+3)-\zeta(2n+1)]+4[\zeta(2n+3)-\eta(2n+4)]\nonumber\\
&=&2[3\zeta(2n+3)-\zeta(2n+1)-2\eta(2n+4)]\nonumber\\
&=&-2\sum_{k=3}^{\infty}\frac{k^3-3k-2(-1)^k}{k^{2n+4}}
<-8\sum_{k=3}^{\infty}\frac{k+1}{k^{2n+4}}<0
\end{eqnarray}
since $k^3-3k\geq 4k+6\geq4(k+1)+2(-1)^k$ for $k\geq3$. This yields the monotonicity of $A_{2n+1}$.

By (\ref{Ineq1OfA2n}) and (\ref{Sign3OfAn}), we have
$$A_{2n+2}-A_{2n}<a_{2n+3}-a_{2n+1}-\frac12 a_{2n+2},$$
and hence it follows from (\ref{zeta}), (\ref{lam-beta}) and (\ref{SignOfan}) that
\begin{eqnarray*}
A_{2n+2}-A_{2n}&<&2[\zeta(2n+3)-\eta(2n+4)-\zeta(2n+1)+\eta(2n+2)]-\zeta(2n+3)+\zeta(2n+1)\\
&=&\zeta(2n+3)-\zeta(2n+1)+2[\eta(2n+2)-\eta(2n+4)]\\
&=&-\sum_{k=2}^{\infty}\frac{k^2-1}{k^{2n+4}}\left[k+2(-1)^k\right]<0.
\end{eqnarray*}
This yields the the monotonicity of $A_{2n}$.

(3) The limiting value of $a_n$ follows from (\ref{Id1}). Computation gives: $a_4=-0.395066\cdots$, $a_3=0.510048\cdots$, $a_2=2.404113\cdots$ and $a_1=-2.644934\cdots$. Hence $-a_4<a_3<a_2<-a_1$.

By (\ref{SignOfan}), in order to prove the monotonicity of $|a_n|$, we need only to prove (\ref{Ineq1Ofan}). By applying the Monotone l'H\^opital's Rule \citep[Theorem 1.25]{2}, one can  easily verify that the function $P_1(x)\equiv [(x+1)^8-1]/[x(x+1)^7]$ is strictly decreasing from $(0,1/2]$ onto $[6305/2187,8)$, by which we see that
$$P_2(m)\equiv 2m+1-2m\left(\frac{2m}{2m+1}\right)^7=\frac{[1+(2m)^{-1}]^8-1}{(2m)^{-1}[1+(2m)^{-1}]^7}=P_1\left(\frac{1}{2m}\right)$$
is strictly increasing in $m\in\IN$ with $P_2(1)=\omega\equiv 6305/2187=2.882944\cdots$ and $P_2(\infty)=8$.
Hence it follows from (\ref{SignOfan}), (\ref{zeta})
and (\ref{lam-beta}) that for $n\in\IN\setminus\{1\}$,
\begin{eqnarray}\label{Ineq3Ofan}
\frac{a_{2n+1}+a_{2n+2}}{2}&=&\zeta(2n+3)-\eta(2n+2)=\sum_{k=2}^{\infty}\frac{(-1)^kk+1}{k^{2n+3}}\nonumber\\
&=&\frac{1}{2^{2n+3}}\left[3-2\left(\frac{2}{3}\right)^{2n+3}\right]+\frac{1}{4^{2n+3}}\left[5-4\left(\frac{4}{5}\right)^{2n+3}\right]
+\cdots\nonumber\\
&{}&+\frac{1}{(2m)^{2n+3}}\left[(2m+1)-2m\left(\frac{2m}{2m+1}\right)^{2n+3}\right]+\cdots\nonumber\\
&\geq&
\sum_{m=1}^{\infty}\frac{P_2(m)}{(2m)^{2n+3}}>\frac{\omega}{2^{2n+3}}\sum_{m=1}^{\infty}\frac{1}{m^{2n+3}}=\frac{\omega}{2^{2n+3}}\zeta(2n+3),
\end{eqnarray}
and
\begin{eqnarray}\label{Ineq4Ofan}
\frac{a_{2n+1}+a_{2n}}{2}&=&2\zeta(2n+1)-\eta(2n+2)-\zeta(2n-1)=\sum_{k=1}^{\infty}\frac{2k-k^3+(-1)^k}{k^{2n+2}}\nonumber\\
&<&-\sum_{k=2}^{\infty}k^{-2n-2}\left(\frac{k^2}{2}+\frac{k}{2}\right)
=1-\frac{\zeta(2n)+\zeta(2n+1)}{2}<0
\end{eqnarray}
since $k^3-k^2/2-5k/2\geq1\geq(-1)^k$ for $k\geq2$.
Hence it follows from (\ref{SignOfan}), (\ref{Ineq3Ofan}) and (\ref{Ineq4Ofan}) that $0<-a_{2n+2}<a_{2n+1}<-a_{2n}$,
so that (\ref{Ineq1Ofan}) holds.

Similarly to (\ref{Ineq3Ofan}), we can easily obtain
\begin{eqnarray*}
\zeta(n+1)-\eta(n)&=&\sum_{k=2}^{\infty}\frac{(-1)^kk+1}{k^{n+1}}>\frac{1}{2^{n+1}}\sum_{m=1}^{\infty}\frac{P_3(m)}{m^{n+1}}
\\
&>&\frac{P_3(1)}{2^{n+1}}\sum_{m=1}^{\infty}\frac{1}{m^{n+1}}=2^{1-n}\mu\zeta(n+1),
\end{eqnarray*}
for $n\in\IN$, where $P_3(m)=2m+1-2m[2m/(2m+1)]^3$ with $P_3(1)=65/27=4\mu$. Hence by (\ref{lambda(n+1)}),
$$\zeta(n+1)>\frac{\eta(n)}{1-2^{1-n}\mu}=\frac{1-2^{1-n}}{1-2^{1-n}\mu}\zeta(n)=\left(1-\frac{1-\mu}{2^{n-1}-\mu}\right)\zeta(n)$$
for $n\in\IN\setminus\{1\}$. This yields the first inequality in (\ref{IneqOfzeta}). On the other hand, we have
\begin{eqnarray*}
\zeta(n+1)-\zeta(n)=-\sum_{k=2}^{\infty}\frac{k-1}{k^{n+1}}<-\frac12\sum_{k=2}^{\infty}\frac{1}{k^n}
=\frac{1-\zeta(n)}{2}
\end{eqnarray*}
for $n\in\IN\setminus\{1\}$, which yields the second inequality in (\ref{IneqOfzeta}). The third inequality in (\ref{IneqOfzeta}) is clear. $\Box$

\section{\normalsize Proof of Theorem \ref{th3}}

(1) Clearly, $F_1(1/2)=(5\log2)/\pi$. By (\ref{SerOfR}) and (\ref{SerOfB}), it is easy to see that $F_1(0^+)=1$.

Let $f$ be as in Theorem \ref{th1}. Then $F_1(x)=[f(x)/B(x)]+1$. Differentiation gives
\begin{eqnarray}\label{F1'}
\frac{B(x)F_1'(x)}{4(1-2x)}=H(x)\equiv
\frac{f'(x)+f(x)H_1(x)}{4(1-2x)},
\end{eqnarray}
where $H_1(x)=\psi(1-x)-\psi(x)$. Clearly, $H_1(1/2)=0$ and
$H_1(0^+)=\infty$.

By \citep[6.3.7, 4.3.67 \&23.2.16]{1} and (\ref{lambda(n+1)}), we have
\begin{eqnarray}
H_1(x)&=&\psi(1-x)-\psi(x)=\pi \cot(\pi x)=\pi\tan\left(\pi\left(\frac12-x\right)\right) ,\label{psi(1-x)}\\
\tan z&=&2\sum_{n=1}^{\infty}\left(\frac{2}{\pi}\right)^{2n}\lambda(2n)z^{2n-1}, ~|z|<\frac{\pi}{2}.\label{tan}
\end{eqnarray}
It follows from (\ref{psi(1-x)}) and (\ref{tan}) that
\begin{eqnarray}\label{SerOfH1}
H_1(x)=\pi\tan\left(\pi\left(\frac12-x\right)\right)
=-\sum_{k=1}^{\infty}2^{2k+1}\lambda(2k)\left(x-\frac12\right)^{2k-1}=4\sum_{k=1}^{\infty}\lambda(2k)(1-2x)^{2k-1}
\end{eqnarray}
by which we obtain the following limiting values
\begin{eqnarray}\label{n-DH1(1/2)}
H_1^{(2n+1)}(1/2)=-2^{2n+3}(2n+1)!\lambda(2n+2) ~\mbox{and }
~H_1^{(2n)}(1/2)=0, ~n\in\IN\cup\{0\}.
\end{eqnarray}

Let $H_2(x)=f(x)H_1(x)$. Then by differentiation and the Leibniz formula,
\begin{eqnarray}\label{n-DH2}
H_2^{(n)}(x)=\sum_{k=0}^nC_n^kf^{(k)}(x)H_1^{(n-k)}(x),
\end{eqnarray}
where $C_n^k=n!/[(n-k)!k!]$. By Theorem \ref{th1}(1),
$f^{(2n+1)}(1/2)=0$ and $f^{(2n)}(1/2)=4^n(2n)!b_n$ for $n\in\IN\cup\{0\}$. Hence
from (\ref{n-DH1(1/2)}) and (\ref{n-DH2}), we obtain the following values
\begin{eqnarray*}
H_2^{(2n)}\left(\frac12\right)&=&\sum_{k=0}^{2n}C_{2n}^kf^{(k)}\left(\frac12\right)H_1^{(2n-k)}\left(\frac12\right)\\
&=&\sum_{m=1}
^nC_{2n}^{2m-1}f^{(2m-1)}\left(\frac12\right)H_1^{(2n-2m+1)}\left(\frac12\right)+\sum_{m=0}^nC_{2n}^{2m}f^{(2m)}\left(\frac12\right)H_1^{(2n-2m)}\left(\frac12\right)=0,\\
H_2^{(2n+1)}\left(\frac12\right)&=&\sum_{k=0}^{2n+1}C_{2n+1}^kf^{(k)}\left(\frac12\right)H_1^{(2n-k+1)}\left(\frac12\right)\\
&=&\sum_{m=1}^{n+1}C_{2n+1}^{2m-1}f^{(2m-1)}\left(\frac12\right)H_1^{(2n-2m+2)}\left(\frac12\right)+\sum_{m=0}^nC_{2n+1}^{2m}f^{(2m)}\left(\frac12\right)H_1^{(2n-2m+1)}\left(\frac12\right)\\
&=&\sum_{m=0
}^nC_{2n+1}^{2m}f^{(2m)}\left(\frac12\right)H_1^{(2n-2m+1)}\left(\frac12\right)=-(2n+1)!2^{2n+3}\sum_{m=0 }^nb_m\lambda(2n-2m+2)
\end{eqnarray*}
for $n\in\IN\cup\{0\}$. Therefore, $H_2(x)$ has the following power series expansion
\begin{eqnarray}\label{SerOfH2}
H_2(x)=\sum_{n=0}^{\infty}\frac{H_2^{(2n+1)}(1/2)}{(2n+1)!}\left(x-\frac{1}{2}\right)^{2n+1}
=4\sum_{n=0}^{\infty}\left[\sum_{k=0
}^nb_k\lambda(2n-2k+2)\right](1-2x)^{2n+1}.
\end{eqnarray}

Let $d_n$ be as in Corollary \ref{Col1}. Then it follows from (\ref{SerOff(x)}), (\ref{F1'}) and (\ref{SerOfH2}) that
\begin{eqnarray}
H(x)&=&\frac{4\sum_{n=0}^{\infty}\left[\sum_{k=0
}^nb_k\lambda(2n-2k+2)\right](1-2x)^{2n+1}-4\sum_{n=1}^{\infty}nb_n(1-2x)^{2n-1}}{4(1-2x)}\nonumber\\
&=&\sum_{n=0}^{\infty}\left[\sum_{k=0
}^nb_k\lambda(2n-2k+2)-(n+1)b_{n+1}\right](1-2x)^{2n}
=\sum_{n=0}^{\infty}d_n(1-2x)^{2n}.\label{SerOfH}
\end{eqnarray}

Clearly, $H((1/2)^-)=d_0=-0.290171\cdots$, and
$H(0^+)=[a_1+a_0H_1(0^+)]/4=\infty$ by (\ref{F1'}) and Theorem \ref{th1}. It follows from Corollary \ref{Col1}(3) and (\ref{SerOfH}) that $H$ is strictly
decreasing and convex from $(0,1/2)$ onto $(d_0,\infty)$. Hance $H$ has a unique zero $x_1\in(0,1/2)$ such that $H(x)>0$ for $0<x<x_1$, and $H(x)<0$ for $x_1<x<1/2$, so that the
piecewise monotonicity property of $F_1$ follows from (\ref{F1'}).

By (\ref{B}), (\ref{F1'}) and (\ref{psi(1-x)}), we have
\begin{eqnarray*}
4(1-2x)H(x)&=&(1-2x)R(x)+[1+x(1-x)]R'(x)+\frac{\pi^2\cos(\pi x)}{\sin^2(\pi
x)}\\
&{}&+\left\{[1+x(1-x)]R(x)-\frac{\pi}{\sin(\pi x)}\right\}H_1(x),
\end{eqnarray*}
and hence
$$H\left(\frac14\right)=\frac12\left\{3\log2+\frac{19}{16}R'\left(\frac14\right)+\pi^2\sqrt2
+\pi\left[\frac{19}{16}R\left(\frac14\right)-\pi\sqrt2\right]\right\}=0.095698\cdots$$
 by (\ref{R(1/4)}) and (\ref{R'(1/4)}). This shows that $x_1\in(1/4,1/2)$.

Next, it follows from (\ref{psi(1-x)}) that
\begin{eqnarray}\label{B'}
B'(x)=-B(x)H_1(x) \mbox{ ~and ~} H_1'(x)=-\frac{\pi^2}{\sin^2(\pi x)}=-B(x)^2.
\end{eqnarray}
By (\ref{F1'}) and (\ref{SerOfH}),
$$F_1'(x)=\frac{4(1-2x)H(x)}{B(x)}=\frac{4}{B(x)}\sum_{n=0}^{\infty}d_n(1-2x)^{2n+1},$$
and hence by (\ref{B'}) and differentiation,
\begin{eqnarray}\label{F1''}
F_1''(x)=\frac{4}{B(x)}\left[H_1(x)\sum_{n=0}^{\infty}d_n(1-2x)^{2n+1}-2\sum_{n=0}^{\infty}(2n+1)d_n(1-2x)^{2n}\right].
\end{eqnarray}
It follows from (\ref{n-DH1(1/2)}), (\ref{F1''}) and Corollary \ref{Col1}(3) that
\begin{eqnarray}\label{F1''(1/2)}
F_1''(1/2)=-8d_0/\pi >0.
\end{eqnarray}
On the other hand, by (\ref{F1'}) and (\ref{psi(1-x)}),
\begin{eqnarray*}
F_1'(x)=\frac{f'(x)+f(x)H_1(x)}{B(x)}=\frac{f'(x)\sin(\pi x)+\pi f(x)\cos(\pi x)}{\pi},
\end{eqnarray*}
and by differentiation and (\ref{psi(1-x)}), we obtain
\begin{eqnarray}\label{1F1''}
F_1''(x)=\frac{1}{\pi}\left[f''(x)\sin(\pi x)+2\pi f'(x)\cos(\pi x)-\pi^2 f(x)\sin(\pi x)\right],
\end{eqnarray}
and hence we have
\begin{eqnarray}\label{F1''(0)}
F_1''(0^+)=2f'(0^+)=2a_1=-(2+\pi^2/3)<0.
\end{eqnarray}
By (\ref{F1''(1/2)}) and (\ref{F1''(0)}), we see that $F_1$ is neither convex nor concave on $(0,1/2)$.

The double inequality (\ref{Ineq6OfR}) is clear. Since $F_1(0^+)=1$, the coefficient $\alpha=1$ is best
possible.

It follows from (\ref{R(1/4)}) and the piecewise monotonicity property of $F_1$ that
$$\delta=F_1(x_1)>F_1\left(\frac14\right)=\frac{19R(1/4)}{16\pi\sqrt2}=\frac{19\log8}{8\pi\sqrt2 }=1.111592\cdots,$$
and hence the first inequality in (\ref{delta}) holds. On the other hand,
it follows from (\ref{Ineq5OfR}) and (\ref{B}) that
\begin{eqnarray}\label{Ineq11OfR}
R(x)\leq \frac{B(x)}{1+x(1-x)}[1+h_8(x)]
\end{eqnarray}
for all $x\in(0,1/2]$, where $h_8(x)=\frac{1}{\pi}S_2(x)\sin(\pi x)$ and $S_2(x)=b_0+b_1(1-2x)^2+b_2(1-2x)^4$. Let $h_{10}(x)=b_1+2b_2(1-2x)^2$ and $h_{11}(x)=(1-2x)\tan(\pi x)$. Then by differentiation,
\begin{eqnarray}\label{h8'}
h_8'(x)/\cos(\pi x)=h_9(x)\equiv
S_2(x)-4h_{10}(x)h_{11}(x)/\pi.
\end{eqnarray}
Clearly, $h_9(0)=b_0+b_1+b_2>0$ and $h_9((1/2)^-)=b_0-8b_1/\pi^2=-0.235204\cdots$. Since $b_2<0$ and since
$b_1+2b_2=0.662442\cdots$, $h_{10}$ is
strictly increasing from $[0,1/2]$ onto $[b_1+2b_2,b_1]$ and
$$S_2'(x)/[4(1-2x)]=-h_{10}(x)<-(b_1+2b_2)<0, ~0<x\leq 1/2,$$
so that $S_2$ is strictly decreasing from $[0,1/2]$ onto
$[b_0,b_0+b_1+b_2]$. It is easy to show that the function
$h_{11}$ is strictly increasing from $(0,1/2)$
onto $(0,2/\pi)$. Hence by (\ref{h8'}), $h_9$ is strictly decreasing from
$(0,1/2)$ onto $(b_0-8b_1/\pi^2, b_0+b_1+b_2)$, and $h_9$ has a unique zero $x_0\in(0,1/2)$ such that $h_8'(x)>0$ for $x\in(0,x_0)$, and $h_8'(x)<0$ for $x\in(x_0,1/2]$. Computation gives:
\begin{eqnarray*}
h_9(0.276937)=0.00000007895\cdots, ~h_9(0.276938)=-0.00000137425\cdots
\end{eqnarray*}
showing that $x_0\in(0.276937,0.276938)$. Therefore
$$h_8(x)\leq h_8(x_0)<\frac{1}{\pi}S_2(0.276937)\sin(0.276938\pi)=0.11214596\cdots<0.112146,$$
which yields the second inequality in (\ref{delta}) by (\ref{Ineq11OfR}).

(2) Clearly, $F_2(x)=f(x)/\{[1+x(1-x)]B(x)\}$. By differentiation,
\begin{eqnarray}\label{F2'}
\frac{[1+x(1-x)]^2}{1-2x}B(x)F_2'(x)=H_3(x)\equiv
4[1+x(1-x)]H(x)-f(x),
\end{eqnarray}
where $H$ is as in (\ref{F1'}). It follows from (\ref{SerOff(x)}) and (\ref{SerOfH}) that
\begin{eqnarray}
H_3(x)&=&\left[5-(1-2x)^2\right]H(x)-f(x)\label{H31}\\
&=&\sum_{n=0}^{\infty}5d_n(1-2x)^{2n}-\sum_{n=0}^{\infty}d_n(1-2x)^{2(n+1)}-\sum_{n=0}^{\infty}b_n(1-2x)^{2n}\nonumber\\
&=&\sum_{n=0}^{\infty}(5d_n-b_n)(1-2x)^{2n}-\sum_{n=1}^{\infty}d_{n-1}(1-2x)^{2n}
=\sum_{n=0}^{\infty}D_n(1-2x)^{2n},\label{H3}
\end{eqnarray}
where $D_n$ is as in Corollary \ref{Col1}. Clearly, $H_3(0^+)=4H(0^+)-a_0=\infty$ and
$H_3(1/2)=D_0<0$. Hence by (\ref{H3}) and Corollary \ref{Col1}(4), it is clear that $H_3$ is strictly
completely monotonic from $(0,1/2)$ onto $(D_0,\infty)$. This shows that $H_3$ has a unique zero $x_2\in(0,1/2)$
such that $H_3(x)>0$ for $x\in(0,x_2)$, and
$H_3(x)<0$ for $x\in(x_2,1/2]$, and hence the piecewise monotonicity
property of $F_2$ follows from (\ref{F2'}).

Next, it follows from (\ref{B}), (\ref{F1'}), (\ref{F2'}) and (\ref{H3}) that
\begin{eqnarray}\label{2F2'}
F_2'(x)&=&\frac{[1+x(1-x)][f'(x)+f(x)H_1(x)]-(1-2x)f(x)}{[1+x(1-x)]^2B(x)}\nonumber\\
&=&\frac{[1+x(1-x)][f'(x)\sin(\pi x)+\pi f(x)\cos(\pi x)]-(1-2x)f(x)\sin(\pi x)}{\pi[1+x(1-x)]^2}\nonumber\\
&=&\frac{\sum_{n=0}^{\infty}D_n(1-2x)^{2n+1}}{[1+x(1-x)]^2B(x)}.
\end{eqnarray}
By (\ref{B'}) and the third equality in (\ref{2F2'}), and by differentiation,
\begin{eqnarray*}
F_2''(x)&=&-\frac{1}{[1+x(1-x)]^2B(x)}\left\{2\sum_{n=0}^{\infty}(2n+1)D_n(1-2x)^{2n}\right.\\
&{}&\left.+\left[2\frac{1-2x}{1+x(1-x)}
-H_1(x)\right]\sum_{n=0}^{\infty}D_n(1-2x)^{2n+1}\right\}
\end{eqnarray*}
from which we obtain
\begin{eqnarray}\label{F2''(1/2)}
F_2''(1/2)=-32D_0/(25\pi)=0.723202\cdots>0.
\end{eqnarray}

On the other hand, by the second equality in (\ref{2F2'}) and by differentiation,
\begin{eqnarray*}
\pi F_2''(x)&=&\frac{f''(x)\sin(\pi x)+2\pi f'(x)\cos(\pi x)-\pi^2 f(x)\sin(\pi x)}{1+x(1-x)}+\frac{2f(x)\sin(\pi x)}{[1+x(1-x)]^2}\\
&{}&-\frac{2(1-2x)}{[1+x(1-x)]^2}\left\{f'(x)\sin(\pi x)+\pi f(x)\cos(\pi x)
-\frac{(1-2x)f(x)\sin(\pi x)}{1+x(1-x)}\right\},
\end{eqnarray*}
by which we have
\begin{eqnarray}\label{F2''(0)}
F_2''(0^+)=2[f'(0^+)-f(0^+)]=2(a_1-a_0)=-(4+\pi^2/3)<0.
\end{eqnarray}

The assertion on the convexity and concavity property of $F_2$ now follows from (\ref{F2''(1/2)}) and (\ref{F2''(0)}).

(3) Since $F_3(x)=f(x)/[1+x(1-x)]$, the monotonicity property of $F_3$
follows from Theorem \ref{th1}(2).

Clearly, $F_3(1/2)=4f(1/2)/5=\log16-4\pi/5=\rho$, and $F_3(0^+)=f(0^+)=a_0=1$. For $x\in(0,1/2]$, let
$$F_4(x)=[1+x(1-x)]^2f''(x)-2(1-2x)[1+x(1-x)]f'(x)+2[2-3x(1-x)]f(x).$$
Clearly, the function $x\mapsto2-3x(1-x)$ is strictly decreasing from $[0,1/2]$ onto $[5/4,2]$. By Theorem \ref{th1}(2), $f(x)>0$, $f'(x)<0$ and $f''(x)>0$ for all
$x\in(0,1/2]$. Hence $F_4(x)>0$ for $x\in(0,1/2]$. Differentiation gives
\begin{eqnarray*}
[1+x(1-x)]^3F_3''(x)=F_4(x)>0
\end{eqnarray*}
for all $x\in(0,1/2]$, which yields the
convexity of $F_3$.

The double inequality (\ref{Ineq7OfR}) and its equality case are clear. $\Box$

\begin{remark}\label{Rem2}
We present the comparison of the bounds of $R(x)$ given in Theorems \ref{th1}--\ref{th3} as follows.

(1) The double inequality (\ref{Ineq1OfR}) is better than (\ref{Ineq7OfR}). In fact, it is clear that
$$\frac{b_0+(1-2x)P(x)}{1+x(1-x)}\geq\frac{b_0}{1+x(1-x)}\geq \frac{4}{5}b_0=\rho$$
for $x\in(0,1/2]$, with equality if and only if $x=1/2$. On the other hand, we have
\begin{eqnarray*}
\frac{b_0+(1-2x)Q(x)+B(x)}{1+x(1-x)}\leq\frac{b_0+(1-b_0)(1-2x)+B(x)}{1+x(1-x)}, ~x\in(0,1/2],
\end{eqnarray*}
with equality if and only if $x=1/2$.
Put $y=1-2x$. Then $y\in[0,1)$, $1+x(1-x)=(5-y^2)/4$, and
\begin{eqnarray*}
\frac{b_0+B(x)+(1-b_0)(1-2x)}{1+x(1-x)}\leq \rho+(1-\rho)(1-2x)+\frac{B(x)}{1+x(1-x)}\Leftrightarrow\\
4[b_0+(1-b_0)y]\leq \left(5-y^2\right)[\rho+(1-\rho)y]\Leftrightarrow (1-\rho)y^2+\rho y-1\leq0,
\end{eqnarray*}
which is true since the function $y\mapsto (1-\rho)y^2+\rho y-1$ is strictly increasing from $[0,1]$ onto $[-1,0]$.

(2) It is clear that the lower bound of $R(x)$ given in (\ref{Ineq1OfR}) is better than that given in (\ref{Ineq6OfR}).

(3) Let $F_5(x)=[b_0+(1-b_0)(1-2x)]\sin(\pi x)$ for $x\in(0,1/2]$. Then $F_5(0)=0$, $F_5(1/2)=b_0$, and
\begin{eqnarray}
b_0+B(x)+(1-b_0)(1-2x)\leq \delta B(x)\Leftrightarrow F_5(x)\leq \pi(\delta-1),\label{F51}\\
b_0+B(x)+(1-b_0)(1-2x)\geq \delta B(x)\Leftrightarrow F_5(x)\geq \pi(\delta-1).\label{F52}
\end{eqnarray}
Differentiation gives
\begin{eqnarray*}
\frac{F_5'(x)}{\cos(\pi x)}=F_6(x)\equiv \pi[b_0+(1-b_0)(1-2x)]-2(1-b_0)\tan(\pi x),
\end{eqnarray*}
which is clearly strictly decreasing from $(0,1/2)$ onto $(-\infty, \pi)$. Hence $F_6$ has a unique zero $x_5\in(0,1/2)$ such that $F_5$ is strictly increasing on $(0,x_5]$, and decreasing on $[x_5,1/2]$. By (\ref{delta}), $b_0<0.35057<\pi(\delta-1)<0.112146\pi=0.35231704\cdots<0.3523171$. Since
 $F_5(1/4)=(1+b_0)/(2\sqrt{2})=0.468155\cdots>0.112146\pi>\pi(\delta-1)$, there exist two numbers $x_6, x_7\in(0,1/2)$ with $x_6<x_7$ such that
\begin{eqnarray}\label{boundOfF5}
F_5(x)
\begin{cases}
<\pi(\delta-1), \mbox{ ~if ~} x\in(0,x_6)\cup(x_7,1/2],\\
=\pi(\delta-1), \mbox{ ~if ~} x=x_6 \mbox{ ~or ~} x_7,\\
>\pi(\delta-1), \mbox{ ~if ~} x\in(x_6, x_7).
\end{cases}
\end{eqnarray}
Consequently, by (\ref{F51})--(\ref{boundOfF5}), the first upper bound of $R(x)$ given in (\ref{Ineq1OfR}) is better (worse) than that given in (\ref{Ineq6OfR}) for $x\in(0,x_6]\cup[x_7,1/2]$ (for $x\in[x_6,x_7]$, respectively).

Next, let $F_7(x)=\left[b_0+b_1(1-2x)^2\right]\sin(\pi x)$. Then $F_7(0)=0$, $F_7(1/2)=b_0$, and
\begin{eqnarray}
b_0+b_1(1-2x)^2+B(x)\leq \delta B(x)\Leftrightarrow F_7(x)\leq \pi(\delta-1),\label{F71}\\
b_0+b_1(1-2x)^2+B(x)\geq \delta B(x)\Leftrightarrow F_7(x)\geq \pi(\delta-1).\label{F72}
\end{eqnarray}
By differentiation,
\begin{eqnarray*}
\frac{F_7'(x)}{\cos(\pi x)}=F_8(x)\equiv \pi\left[b_0+b_1(1-2x)^2\right]-4b_1h_{11}(x),
\end{eqnarray*}
where $h_{11}$ is as in (\ref{h8'}). Clearly, $F_8$ is strictly decreasing on $(0,1/2)$ with $F_8(0)=\pi(b_0+b_1)>0$ and $F_8((1/2)^-)=\pi b_0-8b_1/\pi=-0.738915\cdots$. Hence $F_8$ has a unique zero $x_8\in(0,1/2)$ such that $F_7$ is strictly increasing on $(0,x_8]$, and decreasing on $[x_8,1/2]$. Since
 $F_7(0.28)=0.352694\cdots>0.112146\pi>\pi(\delta-1)$, there exist two numbers $x_9, x_{10}\in(0,1/2)$ with $x_9<x_{10}$ such that
\begin{eqnarray}\label{boundOfF7}
F_7(x)
\begin{cases}
<\pi(\delta-1), \mbox{ ~if ~} x\in(0,x_9)\cup(x_{10},1/2],\\
=\pi(\delta-1), \mbox{ ~if ~} x=x_9 \mbox{ ~or ~} x_{10},\\
>\pi(\delta-1), \mbox{ ~if ~} x\in(x_9, x_{10}).
\end{cases}
\end{eqnarray}
Hence by (\ref{F71})--(\ref{boundOfF7}), the second upper bound of $R(x)$ given in (\ref{Ineq1OfR}) is better (worse) than that given in (\ref{Ineq6OfR}) for $x\in(0,x_9]\cup[x_{10},1/2]$ (for $x\in[x_9,x_{10}]$, respectively).

(4) Let $R_n$ and $S_n$ be as in Theorem \ref{th2}, and $\delta$ as in Theorem \ref{th3}. Then it follows from (\ref{Ineq1OfR}), (\ref{Ineq2OfR}), (\ref{Ineq4OfR}) and (\ref{Ineq6OfR}) that for all $n\in\IN$ and $x\in(0,1/2]$,
\begin{eqnarray}\label{Ineq10OfR}
\frac{B(x)+D_1(x)}{1+x(1-x)}\leq R(x)\leq \frac{B(x)+D_2(x)}{1+x(1-x)},
\end{eqnarray}
with equality in each instance if and only if $x=1/2$, where
\begin{eqnarray*}
D_1(x)&=&\max\left\{b_0+(1-2x)P(x), R_{2n+2}(x)+A_{2n+2}x^{2n+3}, S_{n+1}(x)+c_{n+1}(1-2x)^{2n+3}\right\},\\
D_2(x)&=&\min\left\{b_0+(1-2x)Q(x), R_{2n+1}(x)+A_{2n+1}x^{2(n+1)}, S_{n+1}(x), (\delta-1)
B(x)\right\}.
\end{eqnarray*}

(5) Applying \citep[Theorem 1.25]{2} and Theorem \ref{th3}(3), one can easily show that the function $F_9(x)\equiv[F_3(x)-1]/x$ is strictly increasing from $(0,1/2]$ onto $(-2-\pi^2/6,-2(1-\rho)]$, while $F_{10}(x)\equiv[F_3(x)-\rho]/(1-2x)$ is strictly decreasing from $(0,1/2)$ onto $(0,1-\rho)$, where $F_3$ is as in Theorem \ref{th3}.
\end{remark}

\begin{remark}\label{Rem3}
Theorems \ref{th1}--\ref{th3} show that the function $B(x)/[1+x(1-x)]=\pi/\{[1+x(1-x)]\sin(\pi x)\}$ is a very good approximation of $R(x)$.
\end{remark}

\begin{conjecture}\label{Conj1}
Let $F_3$ be as in Theorem \ref{th3}. Our computation supports the following conjecture: $F_3$ is strictly completely monotonic on $(0,1/2]$.

If this conjecture is true, then we can obtain sharp lower and upper bounds for $R(x)$, which are expressed in terms of $B(x)/[1+x(1-x)]$ and polynomials.
\end{conjecture}

\bigskip

redpajama_set_name: RedPajamaArXiv


This year my goal was to make a blanket for everyone in my family. I made it in at the last second with my final blanket for my dad (Bob!). I wanted to make this one simple and fuss free just like my dad so I let the kiddos pick the colors (I think they did an exceptional job) and I picked a simple stitch.
For my pattern I chose the seed stitch. I've also seen it referred to as the moss or tweed stitch. The kiddos picked out Bernat Super Value in Denim Ragg and Grey Ragg (a worsted weight yarn) and I chose to go with a K hook for a lighter fabric. Like every design I make I first started with a gauge swatch and this is how I did it.
With color 1, ch an even number. In this case I did 22 but make sure you have 5 or more inches.
Row 1 – sc in 2nd ch from hook, * ch 1, sk next ch, sc in next ch; repeat from * across, turn.
Row 2 – ch 1, sc in 1st sc, sc in next ch-1 sp, *ch 1, sk next sc, sc in next ch-1 sp; repeat from * across to last sc, sc in last sc, turn.
Row 3 – ch 1, sc in 1st sc, *ch 1, sk next sc, sc in next ch-1 sp; repeat from * across to 2 last sts, ch 1, sk next sc, sc in last sc, turn.
Repeat rows 2 and 3 changing color every other row until the swatch is 5 or more inches long.
For my swatch I then did a boarder of sc (adding 3 sc in each corner) because I wanted to have an idea of what the finished blanket was going to look like.
Next, I threw my swatch into the wash and ended up with this lovely little square. One quick note, I like to wash my swatches before I measure my gauge or start a new project. For one, it gives me the most accurate gauge and it also gives me an idea of how the yarn will wear overtime. In this case it came out much softer than before and it didn't fuzz up too much so I knew it would work out great for my project. You can read more about gauge here.
At this point you'll need to grab a ruler and do some math (don't worry it's not too bad).
By measuring my gauge I found out I had 14 sts by 14 rows in a 4″ square. Keep in mind you don't have to use 4″. For my swatch 4″ worked out great but if you are using a different hook and/or yarn it may work out better to do 3″ or 5″ or something completely different. Make it easy and pick whatever works best.
Before you start calculating anything pick your blanket size. I decided to go with a throw which is about 38″ by 50″. If you're not sure here's a handy chart to help you out.
Now, I need to look back at my pattern to choose how many chains to start with. I know I need to start with an even number so at this point I choose to go with 132 chains. I will be adding a simple single crochet boarder so that will add the extra width I need to get to 38″. For your calculations you make the call. There's no rules here so do whatever you want.
The next step is to figure out how many rows I will need. You can make this easy and completely skip this calculation. When you're making your blanket you can simply measure as you go but I like to figure it all out.
Here, I get to make another decision. For my pattern I have 2 rows per stripe so I just know I'll need an even number of rows. I decided on 174 rows because again, I know that single crochet boarder will add the length I need.
One last calculation to make it a but easier. I have 2 rows per stripe so I can take 174 rows and divide that by 2 to tell me how many stripes I need.
This just lets me know that I need 87 stripes. That way I don't have to count all the way to 174 when I'm making my blanket.
Hopefully this all makes sense. If you are making a different size simply plug in your numbers to my calculations. Now to the pattern.
If you'd like to sell finished items from this pattern please do! Finished items belong to you and you may do what you wish with them. If you choose to sell them on online please credit me (KT and the Squid) as the designer.
All KT and the Squid patterns are tested by wonderful volunteers. If you have any questions just ask!
Repeat rows 2 and 3 until length desired is reached.
Change color every other row. Carry yarn, do NOT fasten off when changing color.
Repeat rows 2 and 3 (changing color every other row) until you have 174 total rows or 87 total stripes.
With boarder color sc evenly around the blanket adding 3 sc to each corner, fasten off, weave in all ends.


text:
Filip Gavranović (* 2. Oktober 1991 in Bjelovar) ist ein kroatischer Handballspieler.

Karriere
Der 1,98 Meter große und 100 Kilogramm schwere Kreisläufer feierte sein internationales Debüt mit RK Poreč in der Saison 2011/12 im Europapokal der Pokalsieger. Bereits im darauffolgenden Jahr wechselte er nach Slowenien zu RK Koper, dadurch nahm er am Qualifikationsturnier der EHF Champions League teil. Man verlor beide Duelle und spielte in folge im EHC-Cup in welchem man bis in die Gruppenphase kam, dort schied man unter anderem gegen Frisch Auf Göppingen aus. Seit der Saison 2013/14 läuft er in der Handball Liga Austria für Bregenz Handball auf. In seiner  ersten Saison in Österreich war er auf Anhieb unter den zehn besten Torschützen der Liga. Mit den Vorarlbergern nahm er 2013/14 erneut am EHC-Cup teil, konnte in der zweiten Qualifikationsrunde RK Maribor Branik aber nicht überwinden. Im Januar 2015 gab Gavranović bekannt, dass er ab Saisonbeginn 2015/16 für Pfadi Winterthur auflaufen wird. Nach zwei Jahren in der Nationalliga A wechselte der Kreisläufer zurück zu RK Poreč. Nach nur einem Jahr in seiner Heimat wurde er, mit dem HC Kriens-Luzern, erneut von einem Schweizer Verein verpflichtet.

HLA-Bilanz

Weblinks

Einzelnachweise

Handballspieler (Kroatien)
Handballspieler (RK Koper)
Handballspieler (Bregenz Handball)
Handballspieler (Pfadi Winterthur)
Kreisläufer
Sportler (Bjelovar)
Kroate
Geboren 1991
Mann

redpajama_set_name: RedPajamaWikipedia


world-wide brand recognition. He is known by more people than your small business. He has had more interviews and TV appearences than you. Charles Manson is a household name.
He is a brand. And he's never made a dime from it.
But for you and your business trying to build a brand should not be your primary goal. I know I've just peeved off a bunch of people on Madison Avenue and basically gone against what some of my friends and colleagues believe at a core level.
But really … It's NOT about branding your business. It's about focusing on your customers. It's about building products and services that solve problems and meet needs. It's about making money as a business or making a difference as a nonprofit.
Branding puts the focus on you and your business. To grow, your business must focus on your customers and their needs.
Yes you need to stand for something. But that's not branding – that's character and core values.
I'm reminded of a daydream most of us have from time to time in life. We dream about being rich. But money isn't really the end goal. It is what we think is nessasary to live the life we actually dream about on a deeper level. We think that if we only had millions of dollars we'd have less problems, more time and happiness. The problem with this line of thinking is that it leads us to put emphasis on the tools (money) and not the end result (happiness/freedom).
Brand is the same way. It's not really a brand we are after, it's what we think a brand will bring us… more business, growth, sales or nonprofit donations. While having a brand CAN bring all those things, it's better for us to view branding as a function of marketing not the end goal.
If the end goal is to brand oneself, then Charles Manson would be wealthy and free. So would Hitler.
Don't misunderstand…colors , logos, names, fonts, design, slogans and marketing materials all matter. But they should reinforce your marketing and sales strategy for getting new business. The goal of a brand is to be remembered as a trustworthy source of a particular solution or product. As you deliver a quality product you will being to gain authority, trust and … dare I use the word again … brand.
Just don't forget during this process that brand is a function of the end goal … which is S-A-L-E-S.
I think Dockers did an excellent job with their 2010 Superbowl ad. It gave a clear call to action and continued their brand identity. But for a small business or nonprofit you don't have millions of dollars to throw into a branding campaign. Instead of buying Superbowl ads, integrate branding with your response marketing.
You should help people remember you in the midst of giving them a call to action. Make branding part of your process. As my mentor Eben Pagan likes to say, "it's not about communication, it's about conversion." You can communicate your brand all you want, but at the end of the day, conversions matter.
Here is your action step for today. Rename one of your products or services.
Let me give you an example. I have a client who works with average looking people to teach them to become professional actors and models. He has developed an amazing modeling technique that he teaches to all his students. I recommended that he name his technique the " modeling recall method." I then urged him to add his name to the front of it. This is a simple way of "branding" something that he is already doing.
What are some ways that you can name what you do in a way that helps people remember your business? Spend some time today thinking about something you do in your business and how you can (re)name it. Then leave a comment below with your thoughts, questions or results. I'd love to hear what you come up with.
Now go take action and let me know.
Joel Mark Witt is the Publisher of Folk Media and author of 21 Days To Twitter Leadership and How To Use Social Media To Find A Better Job. He is a producer, author, and speaker who consults with businesses and nonprofits on how to use social media in marketing and communications. Get more from Joel on Twitter or Google Buzz.
I don't know, Hitler did some pretty nifty branding and was really successful there for awhile. I mean, what about "Third Reich," and all the swastikas and uniforms. What was that but a very successful branding campaign.
Hitler? I can't tell if you are serious or sarcastic. Hoping it's not the former and guessing it is the latter.
Both Hitler and Mason are despicable characters. But that's not the point.
Just because you BRAND yourself doesn't mean you have an income stream or solid business. In other words, Brand doesn't equal business.


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text:
BTC price, ETH & Blockchain News
Finder Co-Founder Plans To Launch The 'Crypto Bank Of Australia'
BySamantha Mitchell
Aug 6, 2018 cryptocurrency, Finder
Finder co-founder Fred Schebesta has revealed that he is planning to create a crypto bank in Australia. Schebesta is a popular investor in the country and became especially popular after he was listed among the youngest self-made millionaires. He started Creditcardfinder.com.au, a credit card blog which is now known as Finder and worth around $1 billion and has branches in 10 countries. Schebesta is a Cryptocurrency and blockchain enthusiast and believes that creating a crypto bank is a wise idea.
Creating The Crypto Bank Of Australia
Fred Schebesta has made public his upcoming "multi-dollar" idea of starting the first Cryptocurrency bank of Australia. The 26-year old millionaire says that Finder will not seek to start its own bank but will partner with an existing bank. Confirmed reports say that he has already bought "significant" shares at Goldfields Money, a West Australia bank.
Mr. Schebesta said that although the bank is small, is a licensed authorized deposit-taking institution (ADI). He added that the bank will take the advantage of the growing popularity of Cryptocurrencies in the country. The bank will be ready to provide spending, exchange, escrow, cold storage, and custodianship services to Crypto customers.
Why A Cryptocurrency Bank?
Fred Schebesta knows that the nascent Crypto sector is one of the most misunderstood industry. He observes that when Bitcoin hit $10, people said it was a bubble. The same was said when the virtual currency hit $US 100 and even $1,000 and more. When the coin first hit $100, it then dropped to about $12 which is much lower than the recent drop from about $20,000 to $6,000. Moreover, the drop didn't stop an interest in the Cryptocurrency.
Mr. Schebesta has also said the interest in Cryptocurrencies is even "much higher" than what was observed in 2017. He added that last year, such an interest pushed the Bitcoin prices from less than $1,000 to about $20,000 while the Crypto market reached over $800,000. He has said that the recent reduction in the prices of most of the virtual currencies has led to frustration and not a reduction in interest.
To support his claims, Mr. Schebesta observes that at a recent Crypto conference in New York, about 8,000 people attended. He also notes that new technologies are gaining popularity at a very high rate. He compares Bitcoin to gold and is optimistic that the giant coin is not going to die.
The Bank Will Be a Blessing to the Crypto Market
Crypto traders have experienced a lot of hardship in the hands of the traditional financial institutions. Mr. Schebesta observes that most banks have closed the accounts of Crypto traders making it challenging for them to bank. He says this is not fair and that the new bank will introduce enormous opportunities to Crypto traders in the country.
In September last year, Finder started providing Cryptocurrency comparing services. The company has attracted massive interest with the traffic growing at 10 times that of other areas like credit cards, telcos, and insurance. At the beginning of this year, he partnered with Fran Restuccia to introduce Hivex, an over-the-counter (OTC) exchange. The duo also introduced HiveSpend, a Crypto bill-paying service.
Indians Circumvent The RBI Cryptocurrency Ban
Venezuela and Iran – Why it is good that Rogue States adopt Crypto
By Samantha Mitchell
Cryptocurrency Exchanges 101 Cryptocurrency News
Exchange Based In Africa Garners Investment, And Japanese SBI Is Innovating
Aug 31, 2020 Mateo Brooks
Bitcoin News Cryptocurrency News
A New Bitcoin Fund Launched By Fidelity
Aug 30, 2020 Daniel Shaw
Here Comes Whitelist-only Cryptocurrency Withdrawals
BEST BTC CASINO
Best CFD Trading Platform 2020
Nov 19, 2020 Gustaw Brzezinski
Best Crypto Brokers 2020
Nov 18, 2020 Dov Herman
High Leverage Forex Brokers
The Best Regulated Forex Brokers for 2020
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redpajama_set_name: RedPajamaCommonCrawl


Many of us start off with big intentions to do big things. I'm a person who loves to dream big. But I find that there is a power in thinking small when starting on a new adventure. Whenever you're on the tracks to get anything done, big or small, start from where you presently are. Imagine planning a trip from Trivandrum to Goa and putting your starting point in your GPS as Bengaluru. Things just wouldn't work out.
We aren't even close to being perfect, nor do we live in a perfect world. So reaching half way through 2017 wouldn't necessarily mean we're half way through in achieving our goals. We all wonder what exactly is standing in our way? I would argue that the greatest problem that's been preventing you from accomplishing your biggest goals is that you're not thinking small enough.
Yes, you heard me right! You aren't thinking SMALL enough.
Yes, the traditional method for goal setting works, but it's limited by the size of that person's belief. So with that in mind I've come up with a simple solution. Believe me. It works. Start small. I'm talking SUPER small. The reasoning behind this is quite simple. You need momentum, and nothing builds momentum like getting a few wins under your belt.
Let's say you want to start a startup. You have a billion-dollar idea, but do not know if you're up for it. What do you usually do? Think… Think… Think… And, it usually stops there. But, instead try successfully completing something small. Even things unrelated to your dream. Doing them successfully gives you the momentum and fire to get on with your next task. This builds up and gives you the confidence to chase your real dream.
Don't get me wrong, I'm all about thinking big and having big dreams, but I also understand the need for momentum and confidence. Accomplishing these smaller goals is much better than taking a huge initial leap.The purpose of these smaller goals is not to get you closer to your goal, but to develop the skill of belief, the belief that you can accomplish your goals.
I've found good success with this approach. What do you think?


Baron1414 had the most liked content!
VERY GOOD, Loved the mega tree's light movements with music on Christmas Carol. Keep up the great work!
I have it's about the cross, send me your E-mails and I'll get them to you guys. Not super flashy but got good response last two years.
Best lights for LED Mega Tree?
I have used mini's, M6's and C5's for Mega tree during past 3 years. I have come to my conclusion that I like best the C5's for Mega, M6's for mini trees and mini's for bushes and arches. I think the C5's really 'filled' the tree best when viewed about 50 feet away, the others didn't in my opinion.
Very nice, liked your smooth sequencing of arches the way the lit up to/from each section. Great work.
Our "stuck in the past" 2013 display.
Fantastic Job with your sequencing and layout Jeff. But, I have to ask. How did you make those three snaking poles? I haven't seen those before and they were great !! I also really liked the Sarajevo sequence effects with your fades that made props appear to have tails.
Has any one else had issues with being unable to stop sequences after upgrade to 3.10.12? If so is there a fix? I constantly use space bar for start/stop during sequencing. Thanks everyone.
Newbie buying LED's...in need of some guidance and direction.
Agree HLE is the way I'd go and have in the past with their great service and product.
I am in Florida, with alot of rain at times also. I like M6's for bushes, mini trees and C6's for the rest. But, definately sealed lights with no issues last several years.
That's also what I am planning , 2x4 PVC with LED mini's like arches. But... What type of lights are George's, in his above pictures, RGB's ??
Great job, nice and clean with good use of your props. Love your weber tree, just would love it a little better with a beautiful bethleham star to top it off. Hat's off to you, very enjoyable and soothing to watch.
Excellent Job, Looks and sounds GREAT!!! Now the real madness begins. What am I going to do for next year? Best wishes on that.


text:
""" Class to work with transactions. """

import os
import re
import shutil
import tempfile

from repoguard.core import process

class FileNotFoundException(Exception):
    def __init__(self, filename):
        Exception.__init__(self, "No such file in repository or transaction %r." % filename)

class PropertyNotFoundException(Exception):
    def __init__(self, keyword, filename):
        Exception.__init__(self, "Property %r for file %r not set." % (keyword, filename))

class Transaction(object):

    def __init__(self, repos_path, txn_name):
        """
        Initialize the transaction object.
        """

        if txn_name is None: # HEAD revision
            self.type = "revision"
        else:
            txn_name = str(txn_name)
            try:
                int(txn_name)
                self.type = "revision"
            except ValueError:
                self.type = "transaction"

        self.repos_path = repos_path
        self.txn_name = txn_name
        self._profile = re.compile(".*")
        self.tmpdir = tempfile.mkdtemp()
        self.cache = {}

    def _execute_svn(self, command, arg="", split=False):
        if self.txn_name is None:
            command = 'svnlook %s "%s" %s' % (command, self.repos_path, arg)
        else:
            command = 'svnlook --%s %s %s "%s" %s' % (self.type, self.txn_name, command, self.repos_path, arg)

        if command in self.cache:
            return self.cache[command]

        try:
            output = process.execute(command, raw_out=True)
        except process.ProcessException, error:
            if "Transaction '(null)'" in error.output: # Nothing bad happened we just have an empty repository
                output = ""
            else:
                raise

        if split:
            output = [x.strip() for x in output.split("\n") if x.strip()]

        self.cache[command] = output
        return self.cache[command]

    def cleanup(self):
        """
        Delete the temporary directory.
        Do NOT use in your checks or handlers.
        """
        shutil.rmtree(self.tmpdir)

    def _get_profile(self):
        """
        Returns the current profile.
        Do NOT use in your checks or handlers.
        """
        return self._profile

    def _set_profile(self, profile):
        """
        Sets the current profile.
        Do NOT use in your checks or handlers.
        """
        self._profile = re.compile(profile)

    def _get_user_id(self):
        """ Returns a string with the username of the current transaction. """
        user = self._execute_svn("author")
        return user.strip()

    def get_files(self, check_list=[".*"], ignore_list=[]):
        """
        Returns a map of all modified files. The keys of the map
        are the filenames. The values of the map is the
        associated attribute, which can be one of the default
        svnlook changed attributes:
        http://svnbook.red-bean.com/en/1.4/svn-book.html#svn.ref.svnlook.c.changed

        @param check_list List of regular expressions for files which should be included.
        @param ignore_list List of regular expressions for files which should be ignored.
        """

        output = self._execute_svn("changed", split=True)
        files = {}
        for entry in output:
            attributes = entry[0:3].strip()
            filename = entry[4:].strip()

            if self._profile.search(filename) and self.__check(filename, check_list) and not self.__check(filename, ignore_list):
                files[filename] = attributes
        return files

    def __check(self, datei, items):
        for item in items:
            regex = re.compile(item)
            if regex.search(datei):
                return True
        return False

    def get_file(self, filename):
        """ Returns the path to a temporary copy of a file in the repository. """
        if not self.file_exists(filename):
            raise FileNotFoundException(filename)

        tmpfilename = os.path.join(self.tmpdir, filename)
        if os.path.exists(tmpfilename):
            return tmpfilename

        content = self._execute_svn("cat", "\"" + filename + "\"")

        dirname = os.path.dirname(filename)
        tmpdirname = os.path.join(self.tmpdir, dirname)
        if dirname and not os.path.exists(tmpdirname):
            os.makedirs(tmpdirname)

        file_object = open(tmpfilename, "w")
        try:
            file_object.write(content)
        finally:
            file_object.close()
        return tmpfilename

    def file_exists(self, filename, ignore_case=False):
        """
        Returns whether a file exists in the current transaction or revision of
        the repository, optionally case-insensitive.
        """
        exists = False

        if ignore_case:
            filename = filename.lower()
            files = self._execute_svn("tree", "--full-paths", split=True)
            count = 0
            for fname in files:
                if fname.lower() == filename:
                    count += 1
                    if count >= 2:
                        exists = True
                        break

        else:
            try:
                self._execute_svn("proplist", "\"" + filename + "\"", split=True)
                exists = True
            except process.ProcessException:
                pass

        return exists

    def _get_commit_msg(self):
        """
        Returns the commit message.
        """

        output = self._execute_svn("info", split=True)
        temp = output[3:]
        msg = "\n".join(temp)
        return msg.strip()

    def _get_revision(self):
        """
        Returns the id of the revision or transaction.
        """

        revision = "HEAD"
        if not self.txn_name is None:
            try:
                revision = int(self.txn_name.split("-")[0]) + 1
            except ValueError:
                revision = self.txn_name
        return str(revision)

    def get_property(self, keyword, filename):
        """
        Returns a specified property of a file.
        """

        if not self.has_property(keyword, filename):
            raise PropertyNotFoundException(keyword, filename)

        return self._execute_svn("propget", " ".join([keyword, "\"" + filename + "\""]))

    def has_property(self, keyword, filename):
        """
        Checks if a given file has the given property.
        """

        return keyword in self.list_properties(filename)

    def list_properties(self, filename):
        """
        Returns a list of names of the properties for a file.
        """

        if not self.file_exists(filename):
            raise FileNotFoundException(filename)

        return self._execute_svn("proplist", "\"" + filename + "\"", split=True)

    profile = property(_get_profile, _set_profile)
    user_id = property(_get_user_id)
    revision = property(_get_revision)
    commit_msg = property(_get_commit_msg)

redpajama_set_name: RedPajamaGithub


Dust off your weirdest and most wonderful costumes because Wonderfruit is returning for a triumphant third edition. Heading back to its traditional heartland at The Fields at Siam Country Club, the sustainability focussed festival is back, celebrating another year of music, culture and art.
With the stunning natural beauty of Thailand as its backdrop, Wonderfruit takes place from 15 – 18 of December and attrcts attendees from all across the globe with its four stages of world-class international and local music, visionary arts, culinary farm to feasts, health and wellness experiences, engaging talks and workshops, natural adventures and more.
This year's diverse, international lineup includes live drum and bass outfit Rudimental and Mercury Prize-winning hip hop three-piece, Young Fathers. Joining them will be Wolf + Lamb, JUNUN, featuring Shye Ben Tzur & the Rajasthan Express, Kate Simko and the LEO, plus DJs Matador, Mad Professor and Nastia. More artists will be announced closer to the date.
Billing itself as a sustainability focussed festival, the emphasis is beyond just music and the organizers have made a bold ambition to be completely sustainable within five years. The site has implemented water systems, developed a farm and uses a lot of recycled and natural materials in their build.
The three-day festival includes a range of accommodation options including a host of nearby hotels, boutique camping for those that are looking for a bit of luxury and convenience along with the general camp area where attendees can pitch their tent for free.
Tickets are on sale now including 4-day, weekend and Sunday-only passes for adults. Coming with friends or youngsters? There are discounts for groups and teens and under-12s go free.


text:
By: Thomas Schulz 26/02/2020 13/12/2020
Contemporary, PhotobookAddiction, Black and White, Color, Documentary photography, Portrait Photography, Street Photography, Visual diary, Vivian Keulards
Vivian Keulards: To Hans (Schilt Publishing, 2020)
Vivian Keulards is a Dutch conceptual portrait photographer. She graduated in 2009 from the Fotoacademie Amsterdam.
Vivian's portrait series are often created over a longer period of time. Children, adolescents and vulnerable groups are recurring subjects in her photography. Her work is shown on different platforms like photo festivals, galleries, newspapers and magazines. Her portraits have been published, exhibited and rewarded internationally and are part of public and private collections.
Artist: http://www.viviankeulards.nl/
Exhibition: Dutch Full House at Schilt Publishing & Gallery, Amsterdam presents works from Geert Broertjes, Lotte Bronsgeest, Vivian Keulards and Carla Kogelman (23 February – 8 March 2020).
Addiction is still a big taboo. With this photo book and project, Vivian Keulards breaks her silence around the addiction and death of her brother Hans. Years after his death, she went in search for answers. Why did he get addicted and she didn't, how did he feel and could she have done anything? Her brother Hans was only 38 years old when he died in a hotel room in Berlin. His sudden cardiac arrest was caused by drugs. His death is a fact, the way he died a taboo.
The book reflects her feelings, thoughts and memories during her quest.
Publisher's Info
To Hans by Vivian Keulards is published by Schilt Publishing, available in stores and online for €40. You find it here.
The book comes with an essay by writer Ralf Mohren, known from his novel Tonic.
Tolnes Fjellestad; Greve: Starman: Sophus Tromholt Photographs 1882 – 1883 (Forlaget Press, 2019)
Catherine Balet: Moods in a Room (Dewi Lewis Publishing, 2020)
2 replys to
Brad Purinton says:
This sounds very meaningful. A personal journey to understand and come to terms with loss.
Thomas Schulz says:
Hello Brad, kind regards from Germany and thanks a lot for your feedback. Thomas


The Theosophical Teachings have been identified as linked to the Northern Buddhist scriptures, and at the same time not being Buddhist Teachings (David Reigle's recent article – Theosophy and Buddhism – printed in Studies in the Wisdom Tradition – Eastern School Press, 2015). Many important differences are observed between the two traditions. One hypothesis has been proposed to reconcile them : that the Buddhism which is presented in the Theosophical teachings is a pre-Buddha Shakyamuni Buddhism tradition, i.e. which existed thousand of years before the Buddha of our era. The substantiation of this claim remains to be published. The history of Tibetan Buddhism, a tradition which was imported from India in the 7th century of our era with the arrival of Guru Rinpoche (Padma Sambhava) in Tibet, reveals many different lineages of teachers and teachings. The main ones are known as the Nyingmapas, the Sakyapas, the Kargyutpas, and the Gelugpas. A fifth one, which was ostracized and almost extincted in Tibet at their time was the Jonangpas, which David considered recently as maybe the closest possible corpus of teachings, specially vis-a-vis the Theosophical fundamental propositions (see The Doctrinal Position of the Wisdom Tradition : Great Madhyamaka). Part of the differences to be reconciled are linked to the cosmogenesis of human beings and cycles (modes of birth and races).
Among these five traditions, there is one which has not really being deeply researched among theosophists : the so-called "red-hat", the Nyingma tradition. The main reason being probably the strong statement of HPB, considering them as no more than sorcerers (dugpas). Most of the scholars of the 19th century had also this opinion that their teachings was totally degraded by a mixing with the local tradition pre-Guru Rinpoche (Bön). But this tradition has a corpus of doctrine which deserves some of our attention. A good starting point would be the History & Fundamentals of the Nyingma School of Tibetan Buddhism, written by Dudjom Rinpoche. Many root texts are now surfacing in Western language translation, like the Guhyagarbha Tantra (2011).
Inside the Nyingma Tradition there are 9 teachings corpus which are considered as different steps on the ladder of spirituality, to culminate with what is called the Ati-Yoga or Dzogchen Teachings.
These teachings support some uncommon ideas which have some parallel with Theosophy. Let's explore it a little bit.
1) At the time when the life span was incalculable, in the divine dimension called Gaden Tsegpa (Joyous Pagoda), all beings had bodies of light composed of the substance of the elements. They were born miraculously, did not wear clothes, and shone by their own light. To transmit the teaching to them, Buddha Vajradhara manifested as a white, eight-year-old child. His name was Khyeu Nangwa Mikhyappa (Supreme Child Inconceivable Vision).
2) …Thus came the epoch when the life span was ten million years. In the dimension called Sahâ the beings were now born from five-colored eggs composed of the substance of the elements. They were perfectly endowed with the senses and all the limbs, and were vigorous as sixteen-year-old youths. Tall as arrows, they were dressed in leaves and surrounded by a luminous aura. They all had miraculous powers and few passions, did not encounter material obstacles, and their food was of the substances of the four elements. Buddha Khyeu Wöd Mitruppa (Child Imperturbable Light) manifestyed as one of them.
3)… and the life span shortened further to one hundred thousand years. At that time, beings born from heat and humidity fell victim to the first illness, caused by the imbalance of the elements, and started to eat plants. In a place called Trödsher Düpa Wödkyil Pungpa (Mass of Light that Gathers Humidity) there was born among them Buddha Jigpa Kyoppai Yid (Mind that Protects from Fear).
4) The life span became even shorter, and the passions were becoming ever stronger. Beings' bodies lost their light and thus the sun and the moon appeared. When the average life span had become eighty thousand years, because of desire and attachment the sexual organs sprouted forth,… and finally beings coupled and procreated… In that epoch, in the place called Chagjung Ngaldu Nangwa (Apparition in the Womb of Conception), Buddha Zhönnu Rolpa Nampar Tsewa (Young Manifestation of Compassion) was born from the uterus, like everybody else, in the form of a ten-year-old child.
5) When the average life span became sixty thousand years, in the dimension of the Thirty-three Gods, the Buddha Sixth Vajradhara was born as a divine Bodhisattva. In the enchanting garden of the Young Doctor (Tsho byed gzhon nu) he transmitted to the seven heroic Buddhas of our epoch teachings on the six, three, and eighteen paramitas that embraces methods both with and without effort, including the tantras of Dzogpa Chenpo. After having remained among the devas for seventy-five years, he entered parinirvâna where he was absorbed in samadhi for seven thousand years.
6) After seven thousand years he reawakened, and moved by compassion towards beings, he was reborn as the son of a yaksa and a fierce dâkini in the dimension of the Cemetery of the Secret Manifestation in the terrifying place of the yaksas northeast of Mount Meru… His name was Zhönnu Pawo Tobden (Young Powerful Hero). He taught to the seven Bodhisattvas the Tantra of the Spontaneous State of Pure Presence (Rig pa rang shar) and other tantras4. He remained among them for a thousand years, and he entered parinirvana. He remained in samadhi for one hundred thousand years.
7) When the average life span had become ten thousand years, he reawakened from his samadhi and was reborn in the dimension of the râksasas on earth. He was called Trangson Tröpai Gyalpo (Wise Wrathful King) and he transmitted the "ten tantras to subjugate gross negativities" and other teachings5. At the end of his life he became reabsorbed in samadhi. He remained so for fifty thousand years.
8) The average life span became five thousand years, and in the place on this earth called Vulture Peak he was reborn in a royal family and was named Ser Wöd Tampa (Supreme Golden Light)… He transmitted the Vinaya and Prajnaparamita teachings to countless shravakas.
9) When the average life span became one thousand years, in the land called Yui Minmachen (With Turquoise Eyebrows) in northern Mongolia, Tsewe Rolpai Lodrö (Intelligence Manifestation of Compassion) was born. To his disciples, he transmitted the "seven special tantras", including The All-creating King (Kun byed rgyal po) and Total Space (Nam mkha' che)6. He remained there one hundred and twenty years.
10) When the average life span became five hundred years, from the world of the Thirty-three Gods, Buddha Kâsyapa the Elder chose to take birth in the human world to alleviate the suffering of old age. So in the place called Vulture Peak, he gave many teachings, including the anuyoga scriptures to seven disciples who had taken on the form of arhats. He remained there seventy-five years and then went to practice asceticism for seven years. At the end of his life, he left no mortal remains; dissolving into a body of light.
11) When the average life span became three hundred years Buddha Ngöndzog Gyalpo (Perfected King), the son of a brahmin expert in the Vedas, was born at Vajrasana (Bodhgaya). To the Bodhisattvas Manjusri, Avalokitesvara, and Vajrapani he transmitted all the teachings regarding the real conditions and other tantras. After having taught for twenty-five years, he entered parinirvana manifesting the ordinary signs of death in order to display to his disciples of lower capacity the truth of the suffering of birth, old age, sickness, and death.
12) When the average life span became one hundred years, Buddha Sakyamuni was born as the son of King Suddhodana. At Varanasi and other places he taught the four noble truths and the diverse gradual paths and accomplished twelve great deeds.
So, we have here references to various modes of human birth, cycles based on human life duration, and reference to the primordial Buddha (Adi-Buddha or Samantabhadra). He is described as : "Without beginning, middle or end, and indivisible, It is neither two, nor three, taintless and without thought." (Supreme Continuum of the Greater Vehicle – Ch. 2, v.58ab).
4The Ma rig mun sel specifies that he taught the father tantra as the outer teaching, the mother tantras as the inner teaching, the Rig pa rang shar as the secret teaching, and the 'Khor 'das rang grol as the supreme teaching.
5Among the ten tantras to subjugate gross negativities (rags pa 'dul ba'i rgyud bcu) the Ma rig mun sel mentions the tantras of Hayagriva, of Vajrakilaya, of Yamantaka, of Ekajati, etc.
6Although it does not mention the "seven special tantras" (phra bdun rgyud), the Ma rig mun sel quotes, among others, the following titles : Kun byed rgyal po, Khu dbyug rol pa, Sems lung chen mo, Kun tu bzang po che ba rang la gnas pa, Nam mkha' che, Chub pas rol pa, Ye shes mchog, 'Khor lo gcod pa, Thig le bra rgyud, 'Bar ba chen po, Nor bu phra bkod.
In the Voice the usage of Alaya is suggestive of Dzogchen. Even the meaning of the 'Two Paths' may have a Dzogchen depth.
"The two paths lead to the two different results of samsara and nirvana. There is the path of consciousness, which leads to the fruition of samsaric appearances and suffering, and the path of wisdom, which leads to the fruition of enlightenment that is the result of freedom from samsara.