TITLE: $n\in N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $\lim\limits_{n\to\infty} a_n\cdot n = 1$.
QUESTION [2 upvotes]: $n\in N,   a_{n+1} = a_n  (1 - a_n)$ , $0 < a_0 < 1$. Prove that $\lim\limits_{n\to\infty} a_n\cdot n = 1$. I proved that $\lim\limits_{n\to\infty} a_n = 0$ and that $a_n$ is strictly decreasing. I tried Cezaro Stolz but didn't get anywhere. Thanks in advance.

REPLY [3 votes]: Prove that $\frac{1}{a_{n+1}}-\frac{1}{a_n} \rightarrow 1$.


TITLE: Question about compact sets and coverings
QUESTION [1 upvotes]: Let $K \subseteq \mathbb{R}^n$ be sequentially compact. Then for every $\epsilon > 0$ there exists $x_1,...,x_m \in K $ such that $K \subseteq \bigcup_{i=1}^n B(x_i, \epsilon ) $.
Proof
Suppose by contradiction the conclusion is false and fix $\epsilon > 0$. We build a sequence as follows:
Pick $x_1 \in K$ such that $K \nsubseteq B(x_1,\epsilon) $
Pick $x_2 \in K \setminus B(x_1, \epsilon) $ such that $K \nsubseteq B(x_1,\epsilon) \cup B(x_2, \epsilon) $
Continuing in this fashion, suppose $x_1,...,x_m \in K $ have been selected so that for all $i \leq m$, $x_i \in K \setminus \bigcup_{j=1}^{m-1} B(x_j, \epsilon) $ so that
$$ K \nsubseteq \bigcup_{j=1}^m B(x_j, \epsilon) $$
So, picking $x_{m+1} \in \bigcup_{j=1}^m B(x_j, \epsilon)$
we obtain a sequence $(x_n)$ in $K$. By construction, we see $(x_n)$ cannot be cauchy. Using compactness, there exists a subsequence $(x_{n_k})$ of $(x_n)$ that converges to an element $L \in K$. My question: How does this contradict my hypothesis?
Is it because
since $(x_n)$ is not cauchy, then $(x_n)$ is not convergent?
I remember from calculus we can have a sequence that is not convergent but can have a convergent subsequence.

REPLY [0 votes]: The crux is that not only is this sequence not Cauchy, but no subsequence of it can be Cauchy, because $d(x_i,x_j) \geq \varepsilon$ for all $i \neq j$.  But since $K$ is compact some subsequence must be convergent (and therefore Cauchy).

Incidentally, if your assumption is that $\varepsilon > 0$ is such that there are not finitely many $x_1 , \ldots , x_m \in K$ such that $K \subseteq \bigcup_{i=1}^m B(x_i,\varepsilon)$, then in your induction step you just need to pick any $x_{m+1} \in K \setminus \bigcup_{i=1}^m B(x_i,\varepsilon)$ (since $x_1 , \ldots , x_m , x_{m+1}$ will continue to share the desired property by assumption).


TITLE: Does the $5x + 1$ sequence for 7 reach a power of 2 or does it get stuck in a period?
QUESTION [6 upvotes]: This is much like the $3x + 1$ iteration, except that if $x$ is odd, you do $5x + 1$ [and $\frac{x}{2}$ if $x$ is even]. If $x = 7$, then we have 7, 36, 18, 9, 46, 23, 116, 58, 29, 146, 73, 366, 183, 916, 458, 229, ...
I've iterated this twenty thousand times and found no power of 2. It's also possible that there is a period but it's so large that I'm not spotting it. And it's also possible that I have made a mistake somewhere along the way.
Surely someone else has also calculated this, even though $5x + 1$ is not as famous as $3x + 1$. I have tried other starting $x$ and seen that they quickly reach a period. Has anyone determined what happens with $x = 7$?

REPLY [0 votes]: this is a copy of an older answer on math.SE, see the duplicate question here. Your number has the number $35$ as pre-precedessor and its route can be seen on the pictures below.
Here are some pictures for your/our intuition. I graphed the trajectories for initial values $x=5,15,25,...$ for the first $256$ steps of $x_{k+1}=(5x_k+1)/2^A$.
To get the curves to a meaningfully visual interval I show logarithmic scales. The pictures show how most trajectories begin to diverge (not really a safe indication of what characteristic the infinite curves really have) but some show cycling already at early iteration indexes $k$ .
I find $2$ cycles besides the "trivial" one.

$x=5,15,25,35,...,95$ detail of the first few iterations . At the bottom we see the "trivial" cyle (brown curve):

$x=5,15,25,35,...,95$ first $2^8 = 256$ iterations. At later iteration-indexes $k$ a first "non-trivial" cycle occurs (red line):

$x=105,115,125,135,...,195$ first $2^8 = 256$ iterations .

$x=205,215,225,235,...,295$ first $2^8 = 256$ iterations . Here a second "non-trivial" cycle becomes visible:

$x=205,215,225,235,...,295$ first $2^{11} = 2048$ iterations
It seems really that all trajectories which are divergent up to iteration $k=256$ are also divergent up to iteration $k=2048$ . In general: I doubt that there are "later" cycles:


text:
\section{Limit of Function by Convergent Sequences/Real Number Line}
Tags: Real Analysis, Limits of Sequences

\begin{theorem}
Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Open Real Interval|open interval]] $\openint a b$, except possibly at the point $c \in \openint a b$.
Then $\displaystyle \lim_{x \mathop \to c} \map f x = l$ {{iff}}:
:for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \mathop \infty} x_n = c$
it is true that:
:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
\end{theorem}

\begin{proof}
=== Necessary Condition ===
Let $\displaystyle \lim_{x \mathop \to c} \map f x = l$.
Let $\epsilon > 0$.
Then by the definition of the [[Definition:Limit of Real Function|limit of a function]]:
:$\exists \delta > 0: \size {\map f x - l} < \epsilon$
provided $0 < \size {x - c} < \delta$.
Now suppose that $\sequence {x_n}$ is a [[Definition:Real Sequence|sequence]] of points of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$
and:
:$\displaystyle \lim_{n \mathop \to \infty} x_n = c$
Since $\delta > 0$, from the definition of the [[Definition:Limit of Real Function|limit of a function]]:
:$\exists N: \forall n > N: \size {x_n - c} < \delta$
But:
:$\forall n \in \N_{>0}: x_n \ne c$
That means:
:$0 < \size {x_n - c} < \delta$
But that implies:
:$\size {\map f {x_n} - l} < \epsilon$
That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
:$\forall n > N: \size {\map f {x_n} - l} < \epsilon$
Thus:
:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
{{qed|lemma}}
=== Sufficient Condition ===
Suppose that for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = c$, it is true that:
:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
{{AimForCont}} it is ''not'' true that:
:$\displaystyle \lim_{x \mathop \to c} \map f x = l$
Thus:
:$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < \size {x - c} < \delta: \size {\map f {x_n} - l} \ge \epsilon$
In particular, if $\delta = \dfrac 1 n$, we can find an $x_n$ where $0 < \size {x - c} < \dfrac 1 n$ such that:
:$\size {\map f {x_n} - l} \ge \epsilon$
But then $\sequence {x_n}$ is a [[Definition:Real Sequence|sequence]] of points of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = c$
but for which it is ''not'' true that:
:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
The result follows by [[Proof by Contradiction]].
{{qed}}
\end{proof}

config: wiki


TITLE: How to Prove Plancherel's Formula?
QUESTION [7 upvotes]: I have difficulty in proving Plancherel's formula in Fourier transform. Here is what I have thought:
In this question, I denote complex conjugation by an overline and (inverse) Fourier transform is defnined as
\begin{align}
\mathcal{F}[f](\omega)&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt, \\
\mathcal{F}^{-1}[F](t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(t)e^{i\omega t}\,d\omega.
\end{align}
Let $F=\mathcal{F}[f]$ and $G=\mathcal{F}[g]$.
\begin{align}
    \int_{-\infty}^{\infty}f(t)\overline{g(t)}\,dt&=\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\right)\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}\overline{G(-\omega')}e^{i\omega' t}\,d\omega'\right)\,dt\\
    &=\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\right)\left(-\frac{1}{2\pi}\int_{-\infty}^{\infty}\overline{G(\omega')}e^{-i\omega' t}\,d\omega'\right)\,dt%\quad\text{(let $\omega''=-\omega'$ and re-write $\omega''$ to $\omega'$)}
\\
    &=-\left(\frac{1}{2\pi}\right)^2\int_{-\infty}^{\infty}F(\omega)\int_{-\infty}^{\infty}\overline{G(\omega')}
    \int_{-\infty}^{\infty}e^{i(\omega-\omega')t}\,dt\,d\omega'\,d\omega \\
    &=-\left(\frac{1}{2\pi}\right)^2\int_{-\infty}^{\infty}F(\omega)\int_{-\infty}^{\infty}\overline{G(\omega')}2\pi\delta(\omega-\omega')\,d\omega'\,d\omega \\
&=-\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)\overline{G(\omega)}\,d\omega.
\end{align}
So, an undesirable negative sign remains. Where do I make mistakes?

REPLY [5 votes]: $$
\overline{g(t)} = \frac{1}{2\pi} \overline{\int_{-\infty}^\infty G(\omega) e^{i \omega t}\, d \omega } = \frac{1}{2\pi} \int_{-\infty}^\infty \overline{G(\omega)} e^{-i \omega t} \, d \omega
$$


TITLE: Existence of orthogonal vectors
QUESTION [0 upvotes]: In any given vector space, does an orthogonal vector to any given vector actually have to exist? So, if you have a vector a, does there have to exist a vector b so that b is orthogonal to a?
I'm sorry if it's a dumb question, but I couldn't actually find the answer anywhere.

REPLY [0 votes]: Take two linearly independent vectors $a$ and $b$ and form the combination $\lambda a+b$. Then solve $a\cdot(\lambda a+b)=\lambda a^2+a\cdot b=0$, i.e $\lambda=-\dfrac{a\cdot b}{a^2}$.
$a$ and $b-\dfrac{a\cdot b}{a^2}a$ are orthogonal.


text:
\begin{document}

\maketitle

\section{Introduction}

\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\footnotetext[1]{The first author was partially supported by the Deutsche Forschungsgemeinschaft (DFG) through a funded research~project.}
\footnotetext[3]{The computer part of this work was executed on the Sun Fire V20z Servers of the Gau\ss\ Laboratory for Scientific Computing at the G\"ottingen Mathematisches~Institut. Both~authors are grateful to Prof.~Y.~Tschinkel for the permission to use these machines as well as to the system administrators for their~support.}

\begin{ttt}
In~this note, we will present a method to construct examples of
$K3$~surfaces
of geometric Picard
rank~$1$.
Our~approach is a refinement of that of R.~van~Luijk~\cite{vL}. It~is based on an analysis of the Galois~module structure on \'etale~cohomology. This~allows to abandon the original limitation to cases of Picard
rank~$2$
after reduction
modulo~$p$.
Furthermore,~the use of Galois~data enables us to construct examples which require significantly less computation~time.
\end{ttt}

\begin{ttt}
The Picard group of a $K3$~surface $S$ is a highly interesting~invariant.
In~general, it is
isomorphic to $\bbZ^n$ for some $n=1,\ldots,20$.
The first explicit examples of $K3$~surfaces over $\bbQ$ with geometric
Picard~rank $1$ were constructed by R.~van Luijk~\cite{vL}.
His~method is based on reduction modulo~$p$.
It~works as~follows.\smallskip

\begin{iii}
\item
At a
place~$p$
of good reduction, the Picard
group~$\Pic(S_{\overline{\bbQ}})$
of the surface injects into the
Picard
group~$\smash{\Pic(S_{\overline{\bbF}_{\!p}})}$
of its reduction
modulo~$p$.
\item
On~its part,
$\smash{\Pic(S_{\overline{\bbF}_{\!p}})}$
injects into the second \'etale cohomology group
$\smash{H_\et^2(S_{\overline{\bbF}_{\!p}}\!, \bbQ_l(1))}$.\vspace{0.1mm}
\item
Only roots of unity can arise as eigenvalues of the
Frobenius on the image
of~$\Pic(S_{\overline{\bbF}_{\!p}})$
in~$\smash{H_\et^2(S_{\overline{\bbF}_{\!p}}\!, \bbQ_l(1))}$.
The~number of eigenvalues of this form is therefore an upper bound for the Picard rank
of~$S_{\overline{\bbF}_{\!p}}$.
We~can compute the eigenvalues
of~$\Frob$
by counting the points
on~$S$,
defined
over~$\bbF_{\!p}$
and some finite~extensions.
\end{iii}\smallskip

\noindent
Doing~this for one prime, one obtains an upper bound
for~$\rk \Pic(S_{\overline{\bbF}_{\!p}})$
which is always~even. The~Tate conjecture asserts that this bound is actually~sharp.

When~one wants to prove that the Picard rank
over~$\overline{\bbQ}$
is, in fact, equal
to~$1$,
the best which could happen is to find a prime that yields an upper bound
of~$2$.
There~is not much hope to do better when working with a single~prime.\smallskip

\begin{iii}
\item[iv) ]
In~this case, the assumption that the surface would have
Picard~rank~$2$
over~$\overline{\bbQ}$
implies that the discriminants of both Picard groups,
$\Pic(S_{\overline{\bbQ}})$
and~$\Pic(S_{\overline{\bbF}_{\!p}})$,
are in the same square~class. Note~here that reduction
modulo~$p$
respects the intersection~product.
\item[v) ]
When one combines information from two primes, it may happen
that we get the rank
bound~$2$
at both places but different
square classes for the discriminant do~arise. Then,~these data are incompatible
with Picard
rank~$2$
over~$\overline{\bbQ}$.

On~the other hand, there is a non-trivial divisor known explicitly.
Altogether,
rank~$1$
is~proven.
\end{iii}
\end{ttt}

\begin{rem}
This~method has been applied by several authors in order to construct
$K3$~surfaces with
prescribed Picard rank~\cite{vL,Kl,EJ1}.
\end{rem}

\begin{ttt}
{\bf The refinement.}
In this note, we will refine van~Luijk's~method. Our~idea is the~following. We~do not look at the ranks,~only. We~analyze the Galois module structures on the Picard~groups,~too. The~point here is that a Galois~module typically has submodules by far not of every~rank.

As~an example, we will construct
$K3$~surfaces
of geometric
Picard~rank~$1$
such that the reduction
modulo~$3$
has geometric Picard rank~$4$
and the reduction
modulo~$5$
has geometric Picard
rank~$14$.
\end{ttt}

\begin{rem}
This~work continues our investigations on Galois module structures on the Picard~group. In~\cite{EJdisc,EJ2,EJ3}, we constructed
cubic~surfaces~$S$
over~$\bbQ$
with prescribed Galois module structure
on~$\Pic(S)$.
\end{rem}

\section{The Picard group as a Galois~module}
\label{strat}

\begin{ttt}
Let
$K$
be a field and
$S$
an algebraic surface defined
over~$K$.
Denote~by
$S$
the
$\bbQ$-vector
space~$\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ$.
On~$S$,
there is a natural
\mbox{$\Gal(\overline{K}/K)$-operation.}
The~kernel of this representation is a normal subgroup of finite~index.
It~corresponds to a finite Galois extension
$L$
of~$K$.
In~fact, we have a
\mbox{$\Gal(L/K)$-representation}.

The~group
$\Gal(\overline{K}/L)$
acts trivially
on~$\Pic(S_{\overline{K}})$.
I.e.,
$$\Pic(S_{\overline{K}}) = \Pic(S_{\overline{K}})^{\Gal(\overline{K}/L)} \, .$$
Within~this,
$\Pic(S_L)$
is, in general, a subgroup of finite~index. Equality~is true under the hypothesis
that~$S(L) \neq \emptyset$.
\end{ttt}

\begin{ttt}
Now suppose
$K$
is a number field and
$\calpp$
is a prime ideal
of~$K$.
We~will denote the residue class field
by~$k$.
Further,~let
$S$
be a
$K3$~surface
over~$K$
with good reduction
at~$\calpp$.

There~is an injection
of~$\Pic(S_{\overline{K}})$
into~$\Pic(S_{\overline{k}})$.
Taking~the tensor product, this yields an inclusion of vector spaces
$\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ \hookrightarrow \Pic(S_{\overline{k}}) \!\otimes_\bbZ\! \bbQ$.

Both~spaces are equipped with a Galois~operation.
On~$\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ$,
we have
a~$\Gal(L / K)$-action.
On~$\Pic(S_{\overline{k}}) \!\otimes_\bbZ\! \bbQ$,
only
$\smash{\Gal(\overline{k}/k) = \overline{\langle\Frob\rangle}}$~operates.
\end{ttt}

\begin{lem}
The~field extension\/
$L/K$
is unramified
at\/~$\calpp$.\smallskip

\noindent
Proof.
{\em
Let~$D \in \Div(S_L)$
be an arbitrary~divisor. By~good reduction,
$D$
extends to a divisor on a smooth model
$\calS$
over the integer
ring~$\calO_L$.
In~particular, we have the reduction
$D_\calqq$
of~$D$
on the special fiber
$S_\calqq$.
Here,~$\calqq$
is any prime, lying
above~$\calpp$.

$\calO_L/\calqq$~is
a finite extension
of~$k$.
Correspondingly,~there is an unramified
extension~$L^\prime \subset L$
of~$K$.
Good~reduction implies that every divisor
on~$S_\calqq$
lifts
to~$S_{L^\prime}$.
Consequently,~we have a divisor
$D^\prime \in \Div(S_{L^\prime})$
which has the same reduction
as~$D$.

As~intersection products are respected by reduction, we see that the intersection number
of~$D^\prime_L - D$
with any divisor is~zero. The~standard argument from~\cite[Proposition~VIII.3.6.i)]{BPV} implies
$D^\prime_L - D = 0$.
In~other words,
$D$
is defined over an unramified~extension.
}
\eop
\end{lem}

\begin{ttt}
There~is a Frobenius lift
to~$L$
which is unique up to~conjugation. When~we choose a particular
prime~$\calqq$,
lying
above~$\calpp$,
we fix a concrete Frobenius~lift.
Then,~$\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ$
becomes a
$\Gal(\overline{k} / k)$-submodule
of~$\Pic(S_{\overline{k}}) \!\otimes_\bbZ\! \bbQ$.
\end{ttt}

\begin{ttt}
{\bf Computability of the Galois representation.}
The~simplest way to understand the
$\Gal(\overline{k} / k)$-representation
on ~$\Pic(S_{\overline{k}}) \!\otimes_\bbZ\! \bbQ$
is to use \'etale~cohomology. Counting~the numbers of points,
$S$~has
over finite extensions
of~$k$,
we can compute the characteristic
polynomial~$\Phi$
of the Frobenius
on~$H_\et^2(S_{\overline{k}}\!, \bbQ_l(1))$.
This~is actually a polynomial with rational, even integer,~coefficients and independent of the choice
of~$l\neq p$~\cite[Th\'eor\`eme~1.6]{De}.

Denote
by~$V_{\text{Tate}}$
the largest subspace
of~$H_\et^2(S_{\overline{k}}\!, \bbQ_l(1))$
on which all eigenvalues of the Frobenius are roots of~unity.
On~the other hand,
let~$P_{\text{conj}}$
be the subgroup
of~$\Pic(S_{\overline{K}})$
generated by the conjugates of all the divisors we know~explicitly.

Then,~we have the following chain of
$\Gal(\overline{k}/k)$-modules,
$$
H_\et^2(S_{\overline{k}}, \bbQ_l(1))
\supseteq V_{\text{Tate}} \supseteq
\Pic(S_{\overline{k}}) \!\otimes_\bbZ\! \bbQ_l \supseteq
\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ_l \supseteq
P_{\text{conj}}  \!\otimes_\bbZ\! \bbQ_l \, .
$$
In~an optimal situation, the quotient space
$V_{\text{Tate}} / (P_{\text{conj}} \!\otimes_\bbZ\! \bbQ_l)$
has only finitely many
$\Gal(\overline{k}/k)$-submodules.
This~finiteness condition generalizes the codimension one condition, applied in van Luijk's method, step~v).

Our main strategy will then be as~follows. We~inspect the \mbox{$\Gal(\overline{k}/k)$-submodules}
of~$V_{\text{Tate}} / (P_{\text{conj}} \!\otimes_\bbZ\! \bbQ_l)$.
For~all these, except for the null space, we aim to exclude the possibility that it coincides with
$(\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ_l) / (P_{\text{conj}} \!\otimes_\bbZ\! \bbQ_l)$.
\end{ttt}

\begin{rems}
\begin{abc}\item
A~sufficient criterion for
a~$\Gal(\overline{k}/k)$-module
to have only finitely many submodules is that the characteristic polynomial
of~$\Frob$
has only simple~roots. This~fact, although very standard, is central to our~method.
\item
Only~submodules of the
form~$P \!\otimes_\bbZ\! \bbQ_l$
for
$P$
a
\mbox{$\Gal(\overline{k}/k)$-submodule}
of~$\Pic(S_{\overline{k}})$
are possible candidates
for~$\Pic(S_{\overline{K}}) \!\otimes_\bbZ\! \bbQ_l$.
Such~submodules automatically lead to factors
of~$\Phi$
with coefficients
in~$\bbQ$.
\end{abc}
\end{rems}

\begin{defi}
We will call a
$\Gal(\overline{k}/k)$-submodule
of~$H_\et^2(S_{\overline{k}}, \bbQ_l(1))$
{\em admissible\/} if it is a
$\bbQ_l$-sub\-vector~space
and the characteristic polynomial
of~$\Frob$
has rational~coefficients.
\end{defi}

\begin{rem}
In some sense, we apply the van der Waerden criterion to the
representations
of~$\Gal(\overline{\bbQ}/\bbQ)$
on the Picard group and \'etale~cohomology.
\end{rem}

\begin{rem}
In~the practical computations presented below, we will work
with~$\bbQ_l$
instead
of~$\bbQ_l(1)$.
This~is the canonical choice from the point of view of counting points but not for the image of the Picard~group. The~relevant zeroes of the characteristic polynomial of the Frobenius are then those of the form
$q$
times a root of~unity.
\end{rem}

\section{An example}

\subsection{Formulation}

\begin{exas}
\label{examp}
Let
$S \colon w^2 = f_6(x,y,z)$
be a
$K3$~surface
of
degree~$2$
over~$\bbQ$.
Assume~the congruences
$$
f_6 \equiv y^6 + x^4  y^2 - 2 x^2  y^4 + 2 x^5 z + 3 x z^5 + z^6 \pmod 5
$$
and
\begin{eqnarray*}
f_6 &\equiv&
2 x^6 + x^4 y^2 + 2 x^3 y^2 z + x^2 y^2 z^2 + x^2 y z^3 + 2 x^2 z^4 \\
&&{}+ x y^4 z + x y^3 z^2 + x y^2 z^3 + 2 x z^5 + 2 y^6 + y^4 z^2 + y^3 z^3
\pmod 3\, .
\end{eqnarray*}
Then,~$S$
has geometric
Picard~rank~$1$.
\end{exas}

\subsection{Explicit divisors}

\begin{notas}
We will write
$\pr\colon S \to \bP^2$
for the canonical projection.
On~$S$,
there is the ample divisor
$H := \pi^* L$
for
$L$
a line
on~$\bP^2$.
\end{notas}

\begin{ttts}
Let~$C$
be any irreducible~divisor
on~$S$.
Then,~$D := \pi_* C$
is a curve
in~$\bP^2$.
We~denote its degree
by~$d$.
The~projection from
$C$
to~$D$
is generically 2:1 or~1:1. In~the case it is 2:1, we have
$C = \pi^* D \sim dH$.

Thus, to generate a Picard group of
rank~$>\!\! 1$,
divisors are needed which are generically 1:1 over their~projections. This~means,
$\pi^* D$
must be reducible into two components which we call the {\em splittings\/}
of~$D$.

A~divisor
$D$
has a split pull-back if and only if
$f_6$
is a perfect square
on (the normalization
of)~$D$.
A~necessary condition is that the intersection of
$D$
with the ramification locus is a
$0$-cycle
divisible
by~$2$.
\end{ttts}

\subsection{The Artin-Tate conjecture}

\begin{ttts}
The Picard group of a projective variety is equipped with a
$\bbZ$-valued
bilinear form, the intersection~form.
Therefore,~associated
to~$\Pic(S_{\overline{k}})$,
we have its~discriminant, an~integer. The~same applies to every subgroup
of~$\Pic(S_{\overline{k}})$.

For~a
$\bbQ_l$-vector~space
contained
in~$\Pic(S_{\overline{k}}) \!\otimes_\bbZ\! \bbQ_l$,
the discriminant is determined only up to a factor being a square
in~$\bbQ_l$.
However,~every non-square
in~$\bbQ$
is a non-square
in~$\bbQ_l$
for some suitable
prime~$l \gg 0$.
\end{ttts}

\begin{ttts}
Let us recall the Artin-Tate conjecture in the special case of a
$K3$~surface.\medskip

\noindent
{\bf Conjecture}
(Artin-Tate)
{\bf .}
{\em
Let\/
$Y$
be a\/
$K3$~surface
over\/~$\bbF_{\!q}$.
Denote~by\/
$\rho$
the rank
and by\/
$\Delta$
the discriminant of the Picard group
of\/~$Y$,
defined
over\/~$\bbF_{\!q}$.
Then,
\begin{eqnarray}
\label{AT}
\lim_{T \rightarrow q} \frac{\Phi(T)}{(T-q)^{\rho}} =
q^{21 -\rho} \#\!\Br(Y) |\Delta| \, .
\end{eqnarray}
Here,
$\Phi$
is the characteristic polynomial of the Frobenius
on\/~$H_\et^2(Y_{\overline{\bbF}_{\!q}}\!, \bbQ_l)$.
Finally,~$\Br(Y)$
denotes the Brauer group
of\/~$Y$.
}
\end{ttts}

\begin{remass}
\begin{abc}
\item
The Artin-Tate conjecture allows to compute the square class
of the discriminant of the Picard group over a finite field without any knowledge of explicit~generators.
\item
Observe~that
$\#\!\Br(Y)$
is always a perfect square~\cite{LLR}.
\item
The~Artin-Tate conjecture is proven for most
$K3$~surfaces.

Most~notably, the Tate conjecture implies the Artin-Tate conjecture~\cite{Mi2}. We~will use the Artin-Tate conjecture only in situations where the Tate~conjecture is~true. Thus,~our final result will not depend on
unproven~statements.
\end{abc}
\end{remass}

\subsection{The modulo 3 information}

\begin{ttts}
The sextic curve given by
``$f_6=0$''
has three conjugate conics, each tangent in six~points. Indeed,~note that, for
$$f_3 := x^3 + 2 x^2 y + x^2 z + 2 x y^2 + x y z + x z^2 + y^3 + y^2 z + 2 y z^2 + 2 z^3,$$
the term
$f_6 - f_3^2$
factors into three quadratic forms
over~$\bbF_{\!27}$.
Consequently,~we have three divisors
on~$\bP^2_{\bbF_{\!27}}$
the pull-backs of which~split.
\end{ttts}

\begin{ttts}
Counting~the points
on~$S$
over~$\bbF_{\!3^n}$
for~$n=1, \ldots, 11$
yields the numbers
$-2, -8, 28, 100, 388, 2\,458, 964, -692$, $26\,650$,
$-20\,528$,
and
$-464\,444$
as the traces of the iterated Frobenius
on~$H_\et^2(S_{\overline{\bbF}_3}\!,\bbQ_l)$.
Taking~into account the fact that
$p$~is
a root of the characteristic
polynomial~$\Phi$,
these data determine this polynomial~uniquely,
\begin{eqnarray*}
\Phi(t) & = &t^{22} + 2 t^{21} + 6 t^{20} - 27 t^{18} - 162 t^{17} - 729 t^{16}
- 1458 t^{15} - 2187 t^{14} \\
&&{} + 19683 t^{12}+ 118098 t^{11} + 177147 t^{10} - 1594323 t^8 - 9565938 t^7 \\
&&{} \hspace{0.6cm} - 43046721 t^6- 86093442 t^5 -  129140163 t^4 + 2324522934 t^2 \\
&&{} \hspace{5.8cm} + 6973568802 t \!+\! 31381059609 \, .
\end{eqnarray*}
The~functional equation holds with the plus~sign. We~factorize and~get
\begin{eqnarray*}
\Phi(t) & = &(t - 3)^2
(t^2 + 3 t + 9) \\
&&(t^{18} + 5 t^{17} + 21 t^{16} + 90 t^{15} + 297 t^{14} + 891 t^{13}
+ 2673 t^{12}  + 7290 t^{11} \\
&&{} \hspace{1.3cm} + 19683 t^{10} + 59049 t^9 + 177147 t^8 + 590490 t^7 + 1948617 t^6 \\
&&{} \hspace{1.8cm} + 5845851 t^5  + 17537553 t^4 + 47829690 t^3 + 100442349 t^2 \\
&&{} \hspace{6.8cm} + 215233605 t + 387420489) \, .
\end{eqnarray*}
\end{ttts}

\begin{ttts}
From this, we derive an upper bound
of~$4$
for the rank of the Picard~group.

In~the notation of section~\ref{strat},
$V_{\text{Tate}}$
is
a~$\bbQ_l$-vector~space
of dimension~four. On~the other hand,
$P_{\text{conj}}$
is generated
by~$H$.
As~$H$
corresponds to one of the factors
$(t-3)$,
the characteristic polynomial of the Frobenius
on~$V_{\text{Tate}} / (P_{\text{conj}} \!\otimes_\bbZ\! \bbQ_l)$
is~$(t - 3)(t^2 + 3 t + 9)$.
It~has only simple~roots.

Consequently,~for each of the dimensions
$1$,
$2$,
$3$,
and~$4$,
there is precisely one admissible
$\Gal(\overline{\bbF}_{\!3}/\bbF_{\!3})$-submodule
in~$H_\et^2(S_{\overline{\bbF}_{\!3}}\!, \bbQ_l)$
containing the Chern class
of~$H$.

\end{ttts}

\begin{ttts}
Let us compute the corresponding~discriminants.

\begin{iii}
\item
In the one-dimensional case, we have
discriminant~$2$.
\item
It~turns out that splitting the divisors given by the conics which are six times tangent yields a rank~three
submodule~$M$
of~$\Pic(S_{\overline{\bbF}_{\!3}})$.
Its~discriminant~is
$$\disc M = \det
\left(
\begin{array}{rrr}
-2 & 6 & 0 \\
 6 &-2 & 4 \\
 0 & 4 &-2 \\
\end{array}
\right)
= 96 \, .
$$
Hence,~in the three-dimensional case, the discriminant is in the square class
of~$6$.
\item
For~the case of dimension four, we may suppose that
$\Pic(S_{\overline{\bbF}_{\!3}})$
is of rank~four.
As~$\smash{\Gal(\overline{\bbF}_3/\bbF_{\!27})}$
acts trivially on
$\Pic(S_{\overline{\bbF}_{\!3}})$,
the group
$\Pic(S_{\bbF_{\!27}})$
is of rank four,~already. This~means, the Tate conjecture is true
for~$S_{\bbF_{\!27}}$.

We~may compute the square class of the corresponding discriminant according to the Artin-Tate~conjecture. The~result
is~$(-163)$.
\item
Finally,~consider the two-dimensional~case. We~may suppose that
$\Pic(S_{\overline{\bbF}_{\!3}})$
has a
$\smash{\Gal(\overline{\bbF}_{\!3}/\bbF_{\!3}})$-submodule~$N$
which is of rank two and
contains~$H$.
The~corresponding characteristic polynomial of the Frobenius is, necessarily, equal
to~$(t-3)^2$.

On~the other hand, there is the rank~three
submodule~$M$
generated by the splittings of the conics which are six times~tangent. As~the corresponding factors
are~$(t-3)(t^2+3t+9)$,
the modules
$N$
and~$M$
together generate
rank~$4$.
The~Tate conjecture is~true
for~$S_{\bbF_{\!27}}$.
Consequently,~it is true
for~$S_{\bbF_{\!3}}$,~too.

Using~the Artin-Tate conjecture, we can compute the square class of the~discriminant.
It~turns out to
be~$(-489)$.
\end{iii}
\end{ttts}

\begin{remas}
The~Tate conjecture predicts Picard
rank~$2$
for~$S_{\bbF_{\!3}}$.
Let~$C$
be an irreducible divisor linearly independent
of~$H$.
Then,~$C$
is a splitting of a
curve~$D$
of
degree~$d \geq 23$.
Indeed,~$H$
is a genus
$2$
curve.
Hence,~$H^2 = 2$.
For~the discriminant, we find
$-489 \geq 2C^2 - d^2$.
As~$C^2 \geq -2$,
the assertion~follows.

Further,~$D$
is highly singular on the ramification~locus. In~fact, we have
$\smash{C^2 \leq \frac{d^2 - 489}2}$
and~$D^2 = d^2$.
Hence,~going from
$D$
to~$C$
lowers the arithmetic genus by at
least~$\frac{d^2 + 489}4$.
\end{remas}

\subsection{The modulo 5 information}

\begin{ttts}
The sextic curve given~by
``$f_6 = 0$''
has six tritangent~lines. These~are given by
$L_a \colon t \mapsto [1 : t : a]$
where
$a$
is a zero
of~$a^6 + 3 a^5 + 2 a$.
The pull-back of each of these lines splits on the
$K3$~surface
$S_{\overline{\bbF}_{\!5}}$.
\end{ttts}

\begin{ttts}
On~the other hand, counting points yields the following traces of the iterated
Frobenius on~$H_\et^2(S_{\overline{\bbF}_{\!5}}\!, \bbQ_l)$,
$$15, 95, -75, 2\,075, -1\,250, -14\,875, 523\,125, 741\,875, 853\,125, 11\,293\,750 \, .$$
This~leads to the characteristic polynomial
\begin{eqnarray*}
\Phi(t) & = & t^{22} - 15t^{21} + 65t^{20} + 175t^{19} - 3000t^{18} + 11437t^{17} + 10630t^{16} \\
&& {} - 385950t^{15} + 2445250t^{14} - 4530625t^{13} - 38478125t^{12} \\
&& {} + 305656250t^{11} - 566328125t^{10} - 4809765625t^9 + 38207031250t^8 \\
&& {} - 101308593750t^7 - 143457031250t^6 + 2792236328125t^5 \\
&& {} - 14189453125000t^4 + 16400146484375t^3 + 247955322265625t^2 \\
&& {} - 1430511474609375t + 2384185791015625 \\
& = & (t - 5)^2
(t^4 + 5 t^3 + 25 t^2 + 125 t + 625) \\
&& (t^8 - 5 t^7 + 125 t^5 - 625 t^4 + 3\,125 t^3 - 78\,125 t + 390\,625) \\
&& (t^8 - 5 t^7 - 10 t^6 + 75 t^5 - 125 t^4 + 1\,875 t^3 - 6\,250 t^2 - \ldots \\
&& {} \hspace{7.7cm} \ldots - 78\,125 t + 390\,625) \, .
\end{eqnarray*}
Observe here the first two factors correspond to the part of the
Picard group generated by the splittings of the six tritangent~lines.
They~could have been computed directly from the intersection matrix
of these~divisors.
\end{ttts}

\begin{remas}
The knowledge of these two factors allows to compute the characteristic
polynomial only from the numbers of points
over~$\bbF_{\!5},\ldots,\bbF_{\!5^8}$.
Counting~them takes approximately five~minutes when one uses the method described in~\cite[Algorithm~15]{EJ1}.
\end{remas}

\begin{ttts}
Here,~$V_{\text{Tate}}$
is a vector space of
dimension~$14$.
Again,~$P_{\text{conj}}$
is generated
by~$H$.
The~characteristic polynomial of the Frobenius
on~$V_{\text{Tate}} / (P_{\text{conj}} \!\otimes_\bbZ\! \bbQ_l)$~is
\begin{eqnarray*}
(t - 5)(t^4 + 5 t^3 + 25 t^2 + 125 t + 625) \hspace{6.8cm} \\
\hspace{3.5cm} (t^8 - 5 t^7 + 125 t^5 - 625 t^4 + 3\,125 t^3 - 78\,125 t + 390\,625)
\end{eqnarray*}
having only simple~roots. This~shows that, in each of the dimensions
$1$,
$2$,
$5$,
$6$,
$9$,
$10$,
$13$,
and~$14$,
there is precisely one admissible
$\Gal(\overline{\bbF}_{\!5}/\bbF_{\!5})$-submodule
in~$H_\et^2(S_{\overline{\bbF}_{\!5}}\!, \bbQ_l)$
containing the Chern class
of~$H$.
\end{ttts}

\begin{ttts}
For~the cases of low rank, let us compute the square classes of the~discriminant.

\begin{iii}
\item
In the one-dimensional case, we have
discriminant~$2$.
\item
For the two-dimensional case, recall that we know six tritangent lines of the ramification~locus. One~of them,
$L_0$,
is defined
over~$\bbF_{\!5}$.
Splitting~$\pi^* L_0$
yields rank two~alone. For~the discriminant, we find
$$
\det
\left(
\begin{array}{rr}
-2 &  3 \\
 3 & -2 \\
\end{array}
\right)
= -5.
$$
\end{iii}
\end{ttts}

\begin{remas}
Using~the Artin-Tate conjecture, we may compute conditional values for the square classes of the discriminant for the
$6$-
and
$14$-dimen\-sional~modules.
Both~are actually equal
to~$(-1)$.
\end{remas}

\subsection{The situation
over~$\overline{\bbQ}$}

\begin{ttts}
{\em Proof of~\ref{examp}.}
Now we can put everything together and show that the
$K3$~surfaces
described in
Example~\ref{examp}
indeed have geometric Picard~rank~1.

The Picard
group~$\Pic(S_{\overline{\bbQ}}) \!\otimes_\bbZ\! \bbQ_l$
injects as a Galois~submodule into the second \'etale cohomology
groups~$H_\et^2(S_{\overline{\bbF}_{\!5}}\!, \bbQ_l)$
for
$p=3$
and~$5$.
The~modulo~$3$
data show that this module has
\mbox{$\bbQ_l$-dimension}
$1,2,3$
or~$4$.
The~reduction
modulo~$5$
allows the
\mbox{$\bbQ_l$-dimensions}
$1,2,5,6,9,10,13$
and~$14$.
Consequently,~the Picard~rank is either
$1$
or~$2$.

To exclude the possibility of
rank~$2$,
we compare the~discriminants. The~reduction
modulo~$3$
enforces
discriminant~$(-489)$
while the reduction
modulo~$5$
yields
discriminant~$(-5)$.
This~is a~contradiction, e.g.,
for~$l = 17$.
\eop
\end{ttts}

config: arxiv


TITLE: Let $A$ be the event that either a $6$ or $2$ is rolled first, followed by an odd number. Then $P(A)=?$
QUESTION [1 upvotes]: $2$ fair dice are rolled in succession.
With $(1,3)$ denoting values $1$ on the first roll and $3$ on the second...etc. Let $A$ be the event that either a $6$ or $2$ is rolled first, followed by an odd number. Then $P(A)=?$. What shall I do first?

REPLY [1 votes]: The probability of rolling a 2 or 6 on a fair die is $1/3$ and the probability of rolling an odd number (so 1, 3 or 5) is $1/2$. Since these two events are independent, $P(A) = 1/2 \cdot 1/3 = 1/6$.
In general with problems like these, you can sometimes split up the event $A$ into more manageable parts (divide and conquer) that are independent and then simply multiply the probabilities.


TITLE: Linear Transformation with $F=\mathbb Q$ or $ F=\mathbb{R}$
QUESTION [3 upvotes]: Please help with following question:
$V$ and $W$ are vector spaces over a field $F$. A mapping $T : V \to W$satisfies the condition $T(v_1 + v_2) = T(v_1) + T(v_2)$. Is $T$ a linear transformation if $(a): F = Q $
$(b) F = R$ ?

REPLY [2 votes]: Outline: (a) Over the rationals, the answer is yes, $T$ is a linear transformation. We need to prove that for any rational $r$, we have $T(rv)=rT(v)$.
Use the given relation to show that if $p$ is a positive integer, then $T(pv)=pT(v)$. This can be done by induction. Suppose that we know that $T(pv)=pT(v)$. We want to show that $T((p+1)v)=(p+1)T(v)$. This is easy, for
$$T((p+1)v)=T(pv+v)=T(pv)+T(v)=pT(v)+T(v)=(p+1)T(v).$$
Now show that if $q$ is a positive integer, then $T((1/q)v)=(1/q)T(v)$. Use the fact that $T((q)((1/q)v)=qT((1/q)v)$.
We also need to take care of negative rationals.
(b) Things are very different over the reals. For information, please search under Hamel Basis. Look also for information about the Cauchy Functional equation $f(x+y)=f(x)+f(y)$. One can show (using the Axiom of Choice) that this equation has solutions other than $f(x)=kx$.


TITLE: Is matrix transpose a linear transformation?
QUESTION [24 upvotes]: This was the question posed to me. Does there exist a matrix $A$ for which $AM$ = $M^T$ for every $M$. The answer to this is obviously no as I can vary the dimension of $M$. But now this lead me to think , if I take , lets say only $2\times2$ matrix  into consideration. Now for a matrix $M$, $A=M^TM^{-1}$ so $A$ is not fixed and depends on $M$, but the operation follows all conditions of a linear transformation and I had read that any linear transformation can be represented as a matrix. So is the last statement wrong or my argument flawed?

REPLY [0 votes]: A linear transformation is a transformation between two vector spaces that preserves addition and scalar multiplication.  Now if $X$ and $Y$ are two $n$ by $n$ matrices then $X^T + Y^T = (X+Y)^T$ and if $a$ is a scalar then $(aX)^T$ = $a(X^T)$ so transpose is linear on the $n^2$ dimensional vector space of $n$ by $n$ matrices.
On the other hand if $A$ and $M$ are $n$ by $n$ matrices satisfying $AM = M^T$ for all $M$ then if $A$ existed and if $I$ represents the identity matrix then $A = AI = I^T = I$ but $I$ does not transpose an arbitrary matrix so $A$ does not exist unless $n$ is 1.
Note that the two situations described here refer to different vector spaces which is why the transpose can exist in the first space but not the second.  In the first case we are dealing with transformations on the $n^2$ dimensional vector space of $n$ by $n$ matrices and in the second one we are dealing with transformations on a vector space of dimension $n$.


text:
\begin{document}

\title{{\Large\bf  Certain identities, connection  and explicit formulas for the Bernoulli, Euler  numbers\\  and Riemann  zeta -values}}

\author{Semyon   YAKUBOVICH}
\maketitle

\markboth{\rm \centerline{ Semyon YAKUBOVICH}}{}

\markright{\rm \centerline{Bernoulli, Euler  numbers and Riemann  zeta - values}}

\begin{abstract} {Various new identities,  recurrence relations,   integral representations, connection  and explicit formulas
are established for the Bernoulli, Euler  numbers and the values of
Riemann's   zeta function $\zeta (s)$. To do this, we explore
properties of some Sheffer's  sequences of polynomials related to
the Kontorovich-Lebedev transform. }

\end{abstract}

\vspace{4mm}

{\bf Keywords}: {\it Bernoulli polynomials, Bernoulli numbers, Euler polynomials, Euler numbers, generalized Euler polynomials, Sheffer sequences,  Von Staudt-Clausen theorem, Riemann zeta function, modified Bessel functions,   Kontorovich-Lebedev transform}

{\bf AMS subject classification}:  11B68,  11B73, 11B83, 11M06,  12E10,  33C10, 44A15

\vspace{4mm}

\section {Introduction and preliminary results}

In 2009 the author \cite{yakpol} introduced a family  of polynomials of degree $n$, which belongs to Sheffer's sequences (cf. \cite{bro}) and related to the Kontorovich-Lebedev transform.  Precisely, it has the form

$$p_n(x)= (-1)^n e^x {\cal A} ^n e^{-x}, \quad n \in \mathbb{N}_0,\eqno(1.1)$$

where

$${\cal A} \equiv x^2- x{d\over dx} x {d\over dx},\eqno(1.2)$$

is the second order differential operator having as an eigenfunction the modified Bessel function $ K_{i\tau}(x),\  \tau \in \mathbb{R},\ $ ($ i$  is the imaginary unit ), i.e.

$${\cal A} \ K_{i\tau}(x)= \tau^2 K_{i\tau}(x).\eqno(1.3)$$

The modified Bessel function $ K_{i\tau}(x)$ is, in turn,  the kernel of the Kontorovich-Lebedev transform (see in
\cite{sned}, \cite{leb},  \cite{yakind})

$$(Gf) (\tau)=\int_0^\infty K_{i\tau}(x)f(x) {dx\over x}, \ \tau \in \mathbb{R}_+.\eqno(1.4)$$

As it is known,  operator (1.4) extends to a bounded invertible map  $G: L_2\left({\mathbb {R}}_+;  x^{-1}dx  \right)
\to  L_2({\mathbb {R}}_+;\tau\sinh\pi\tau d\tau)$ and this map is isometric, i.e.

$$\int_0^\infty \tau\sinh\pi\tau |(Gf)(\tau)|^2 d\tau= {\pi^2\over 2} \int_0^\infty |f(x)|^2 {dx\over x} .\eqno(1.5)$$

Reciprocally,  the inversion formula holds

$$f(x)={2\over  \pi^2}\int_{0}^\infty \tau\sinh\pi\tau K_{i\tau}(x)(Gf)(\tau) d\tau,\  x >0.\eqno(1.6)$$

The modified Bessel function $K_{i\tau}(x)$ has  the asymptotic behavior \cite{leb}

$$ K_\nu(z) = \left( \frac{\pi}{2z} \right)^{1/2} e^{-z} [1+
O(1/z)], \qquad z \to \infty,\eqno(1.7)$$

and near the origin

$$ K_\nu(z) = O\left ( z^{-|{\rm Re}\nu|}\right), \ z \to 0,\eqno(1.8)$$

$$K_0(z) = -\log z + O(1), \ z \to 0. \eqno(1.9)$$

Moreover it can be defined by the  following integral
representations

$$K_{\nu}(x)=\int_0^\infty e^{-x\cosh u}\cosh \nu u du,
x>0, \eqno(1.10)$$

$$K_{\nu}(x)={1\over 2} \left(x\over 2\right)^\nu\int_0^\infty
e^{-t-{x^2\over 4t}}t^{-\nu-1}dt, \ x>0.\eqno(1.11)$$

Returning to the system of polynomials (1.1), we easily observe that all their coefficients $a_{n,k},\  k= 0,1\dots, n$ are integers.  It can be represented by the integral

$$p_n(x)= {2(-1)^n\over \pi} e^{x}\int_{0}^\infty \tau^{2n} K_{i\tau}(x)\ d\tau,\eqno(1.12) $$ and satisfies  the
differential recurrence relation of the form

$$p_{n+1}(x)= x^2p_n^{\prime\prime}(x) + x(1-2x)p_n^\prime(x)- xp_n(x), \ n= 0,1,2,\dots \ .\eqno(1.13)$$

In particular, we derive

$$p_0(x)=1, \ p_1(x)= -x,\ p_2(x)= 3x^2-x, \quad p_3(x)= -15x^3+ 15x^2-x.$$

The leading coefficient $a_{n,n}$ of these polynomials can be calculated by the formula

$$a_{n,n}= (-1)^n (2n-1)!! =  (-1)^n 1\cdot 3\cdot 5\dots \  \cdot (2n-1), \ n \in \mathbb{N}.\eqno(1.14)$$

Moreover, recently we found the explicit formula of coefficients $a_{n,k} $  (see \cite{yakyor})

$$a_{n,k}= {1\over k!}\sum_{r=0}^{k} {(-1)^r\over 2^r} \binom{k}{r}\sum_{j=0}^{k-r} {(-1)^j\over 2^j} \binom{k-r}{j}
(r-j)^{2n} ,\  k= 1,\dots, n\eqno(1.15)$$

and as a consequence of the definition (1.1) $a_{n,k} \in \mathbb{Z}$, i.e. the right- hand side of (1.15) is an integer.
The generating function $\Phi(x,t)$ for this sequence of polynomials is given by the series

$$\Phi(x,t)=  e^{- 2x\sinh^2( t/2)} = \sum_{n=0}^\infty {p_n(x)\over (2n)!}  \ t^{2n},\ |t| < {\pi\over 4}.\eqno(1.16)$$

Letting $x=0$ in the latter equation, we find

$$p_n(0)= 0, \quad n=1,2,\dots.$$

A differentiation with respect to  $x$ in (1.16) yields the equality

$${\partial \Phi \over \partial x } =  (1-\cosh t)\  e^{- 2x\sinh^2( t/2)} = \sum_{n=0}^\infty {p_n^\prime (x)\over (2n)!} \ t^{2n}.\eqno(1.17)$$

Decomposing the left-hand side of (1.17) as a product of series and equating coefficients in front of $t^{2n}$ we come up with  the following recurrence relation

$$p_n^\prime (x)= - \sum_{k=0}^{n-1} {2n \choose 2k} p_k(x), \ n \in
\mathbb{N}.\eqno(1.18)$$

Putting $x=0$ in (1.17) and using  values $p_0(0)= 1, \  p_k(0)=0, \ k \in \mathbb{N}$ we obtain $p_n^\prime (0)= -1, \ n \in  \mathbb{N}$. Analogously, a differentiation with respect to $t$ in  (2.4) leads us to

$${\partial \Phi \over \partial t } =  - x \sinh t \  e^{- 2x\sinh^2( t/2)} = \sum_{n=1}^\infty {p_n (x)\over (2n -1)!}
\ t^{2n -1}.\eqno(1.19)$$

Similarly we derive the relation

$$p_{n+1} (x)= - x \sum_{k=0}^{n} {2n +1 \choose 2k} p_k(x), \ n \in
\mathbb{N}_0.\eqno(1.20)$$

Moreover, differentiating through in (1.19),  we call  (1.17) and using simple
relations for binomial coefficients,  we obtain  the identity

$$ x \sum_{k=0}^{n} {2n +1 \choose 2k} p_k^\prime(x) =  \sum_{k=1}^{n} {2n +1 \choose 2k -1}
p_k(x).\eqno(1.21)$$

Comparing  (1.17), (1.19), we find that $\Phi(x,t)$ satisfies the following first order partial differential equations

$$ {\partial \Phi \over \partial  t } +  x\sinh t\   \Phi = 0,\eqno(1.22)$$

$${\partial \Phi \over \partial x }  +  2   \sinh^2 \left(t\over 2\right)   \Phi =0,\eqno(1.23) $$

$$x {\partial \Phi \over \partial x }  =  \tanh \left(t\over 2\right)   {\partial \Phi \over \partial t }.\eqno(1.24) $$

Further, returning to representation (1.12) and employing the inversion formula (1.6) of the Kontorovich-Lebedev transform, we obtain the equality

$${\tau^{2n-1}\over \sinh\pi\tau} = {(-1)^n\over \pi}\int_0^\infty e^{-x}K_{i\tau}(x)p_n(x){dx\over
x}, \ n \in \mathbb{N}.\eqno(1.25)$$

In particular, it yields

$$\int_0^\infty e^{-x}K_{0}(x)p_n(x){dx\over x} =0, \ n=2,3,\dots \  .$$

Integrating with respect to $\tau$ in (1.25), we call the value of the integral  (2.4.3.1) in \cite{prud},  Vol. I

$$\int_0^\infty {\tau^{\alpha -1} d\tau \over \sinh\pi\tau} =    \frac{2^\alpha - 1}{\pi^\alpha
2^{\alpha-1}}\  \Gamma(\alpha) \zeta(\alpha),\quad {\rm Re} \  \alpha >  1,\eqno(1.26)$$

where $\Gamma(\alpha)$ is Euler's gamma function and $\zeta(\alpha) $ is Riemann's zeta function (cf. \cite{erd}, Vol. 1)
and relation (2.16.48.1) in \cite{prud}, Vol. II to obtain the following representations of zeta-values at  even and odd integers, respectively,

$$\frac{2^{2n}  - 1}{2^{2(n-1)}}\  (-1)^n  (2n-1)! { \zeta(2n)\over \pi^{2n}}  =  \int_0^\infty e^{-2x} p_n(x){dx\over
x}, \ n \in \mathbb{N},\eqno(1.27)$$

$$(-1)^n  (2n)!  \  \left(2^{2n+1} - 1\right)\   {\zeta(2n+1)\over (2 \pi)^{2n} } =  \int_0^\infty \int_0^\infty \tau K_{i\tau} (x) e^{-x} p_n(x){d\tau dx\over  x}, \ n \in \mathbb{N}.\eqno(1.28)$$

Finally in this section we note that recently in \cite{lour} the family  of polynomials (1.1) was generalized on the sequence

$$p_n(x; \alpha)=   (-1)^n e^x x^{-\alpha} {\cal A} ^n e^{-x} x^\alpha , \quad n \in \mathbb{N}_0,\eqno(1.29)$$

involving an arbitrary parameter $\alpha, \  {\rm Re} \  \alpha >  - 1/2. $

\section{Identities for the Bernoulli and Euler numbers}

In this section we will derive a number of recurrence relations,  finite sum, connection and explicit formulas,  series and integral representations, which are related to the Bernoulli and Euler numbers. The Bernoulli numbers $B_n,\  n= 0,1,2,\dots, $ can be defined via the generating function (see in \cite{erd},  Vol. I )

$${x\over e^x- 1} = \sum_{n=0}^\infty B_n {x^n\over n! }, \quad |x| < 2\pi \eqno(2.1)$$

and Bernoulli polynomials $B_n(x)$ by the equality

$${t e^{xt}   \over e^t - 1} = \sum_{n=0}^\infty B_n(x)  {t^n\over n! }, \quad |x| < 2\pi. \eqno(2.1)$$

In particular, we find the values,   $B_0=1,\ B_1= -1/2, B_2= 1/6, B_4= -1/30$ and $B_n=0$ for all odd  $n \ge 3$. Furthermore,   $(-1)^{n-1} B_{2n} > 0$ for all $ n \ge 1$.    We list some important properties of the Bernoulli numbers and polynomials, which will be employed below.  All details and proofs can be found in \cite{erd}, Vol. I.   The basic identity for Bernoulli numbers is

$$ \sum_{k=0}^{n-1} {n \choose k} B_k = 0, \ n \ge 2.\eqno(2.3)$$

Concerning the Bernoulli polynomials,  it has the explicit formula,

$$B_n(x)=   \sum_{k=0}^{n} {n \choose k} B_k x^{n-k}.\eqno(2.4)$$

Hence,  for instance,

$$B_0(x)= 1, \  B_1(x)= x- {1\over 2},\  B_2(x)=  x^2- x+ {1\over 6},$$$$    B_3(x)=  x^3- {3\over 2} x^2+ {1\over 2} x,\
B_4(x) = x^4- 2x^3+ x^2- {1\over 30}.\eqno(2.5)$$

Evidently,  $B_n= B_n(0)$.   The Bernoulli polynomials and their  derivative satisfy the following important relations

$$B_n^\prime (x)=  n B_{n-1} (x),\eqno(2.6)$$

$$B_n(x+1)- B_n(x)= n x^{n-1},\eqno(2.7)$$

$$B_n(1-x)= (-1)^n B_n(x), \eqno(2.8)$$

$$B_n(mx)= m^{n-1} \sum_{k=0}^{m-1} B_n \left(x+  {k\over m}\right).\eqno(2.9)$$

The remarkable Euler formula relates the Bernoulli numbers $B_{2n}$ and Riemann zeta- values $\zeta(2n)$  (see, for instance, in \cite{dwi} )

$$\zeta(2n)=  (-1)^{n-1} \frac{2^{2n-1} B_{2n}}{(2n)!} \pi^{2n}.\eqno(2.10)$$

The Euler numbers $E_n,\  n= 0,1,2,\dots, $ can be defined, in turn,  by the equality  (see in \cite{erd},  Vol. I )

$${1\over \cosh t } = \sum_{n=0}^\infty E_n  {t^n\over n! }, \quad |t| <  {\pi\over 2}. \eqno(2.11)$$

As we see,  $E_{2n+1} = 0$ and, in particular,  $E_0=1,\ E_2=-1, \  E_4=5,\  E_6= -61$.  The basic relation  for Euler  numbers is

$$ \sum_{k=0}^{n} {2n \choose 2k} E_{2k} = 0, \ n \ge 1.\eqno(2.12)$$

In order to obtain new properties of the Bernoulli and Euler numbers, we will employ Sheffer's sequences, which are associated with the Kontorovich-Lebedev transform (1.4).  Indeed,  calling identity (1.27), we immediately derive the integral representation of Bernoulli's numbers $B_{2n}$ in terms of the sequence of polynomials (1.1), namely

$$B_{2n}= \frac{n}{ 1- 2^{2n}} \int_0^\infty e^{-2x} p_n(x){dx\over  x}.\eqno(2.13)$$

For the numbers $B_{4n}$ we have the formula (see in \cite{yakpol})

$$B_{4n} =  \frac{2n}{ 1- 2^{4n}} \int_0^\infty e^{-2x} p^2_n(x){dx\over  x},\eqno(2.14)$$

which is the result of the equality

$$ \int_0^\infty e^{-2x} p_n(x){dx\over  x} =  \int_0^\infty e^{-2x} p^2_n(x){dx\over  x}.\eqno(2.15)$$

But as it is proved in \cite{yakpol},  a more general relation takes place

$$ \int_0^\infty e^{-2x} p_{n+m}(x){dx\over  x} =  \int_0^\infty e^{-2x} p_n(x) p_m(x) {dx\over  x},\eqno(2.16)$$

which holds for any $n, m \in \mathbb{N}_0$ such that at least one is nonzero.   Hence, appealing to (2.13), we derive the identity

$$B_{2(n+m)}= \frac{n+m}{ 1- 2^{2(n+m)}} \int_0^\infty e^{-2x} p_{n-k}(x) p_{m+k}(x) {dx\over  x}\eqno(2.17)$$

being valid for any $k=0,1,\dots, n$.

Another definition of Euler's numbers can be given in terms of  Sheffer's  sequence of polynomials $q_n(x)$ introduced in \cite{yakpol}

$$q_n(x)= e^x\int_x^\infty e^{-t} p_n(t) dt,\quad   n \in \mathbb{N}_0,\eqno(2.18)$$

which, in turn,   is defined via the generating function $F(x,t)= \Phi(x,t) / \cosh t$  (see (1.16))

$${1\over \cosh t} e^{- 2x\sinh^2( t/2)} = \sum_{n=0}^\infty {q_n(x)\over (2n)!}  \ t^{2n},\ |t| < {\pi\over 4}.\eqno(2.19)$$

Hence with the use of (2.11) and the integral representation of Euler numbers \cite{yakpol}

$$E_{2n}=  \int_0^\infty e^{- x} p_n(x) dx\quad  \in \mathbb{Z}\eqno(2.20)$$

 we find

$$E_{2n} = q_n(0), \quad n=0,1,\dots.$$

Moreover, following \cite{yakpol}, the sequence $q_n(x)$ has a relationship with $p_n(x)$.   Indeed,

$$q_n(x)= \sum_{k=0}^n p_n^{(k)}(x),\ n \in \mathbb{N}_0,\eqno(2.21)$$

$$q_n(x)= \sum_{k=0}^n E_{2(n-k)} {2n  \choose 2k }p_{k}(x),\eqno(2.22)$$

where $p_n^{(k)}(x)$ is the $k$ -th derivative of $p_n(x)$ and relation (2.22) can be obtained employing (2.18) and the binomial type identity for the sequence $p_n(x)$ (cf. \cite{bro})

$$p_n(x+y)=  \sum_{k=0}^n {2n \choose 2k}  p_k(x)p_{n-k}(y).\eqno(2.23)$$

For  instance,

$$q_0(x)= 1, \quad  q_1(x)= - (x+1), $$

$$\ q_2(x)= 3x^2 + 5x +5, \quad   q_3(x)= -15x^3-30x^2-61x -61.$$

Differentiating through in (2.21), we find

$$q_n^\prime (x)= \sum_{k=0}^n p_n^{(k+1)}(x)= \sum_{k=1}^{n} p_n^{(k)}(x)= q_n(x)- p_n(x).$$

Thus, $p_n(x)= q_n(x)- q_n^\prime (x)$ and since $p_n(0)=0, n \in
\mathbb{N}$, we get

$$E_{2n}= q_n^\prime (0),\ n \in \mathbb{N}.$$

Moreover,

$$p_n^\prime (x)= q_n^\prime (x)- q_n^{\prime\prime} (x)$$

and since $p_n^\prime (0)= -1, n = 1,2,\dots \ $ (see (1.18)),  it yields

$$q_n^{\prime\prime} (0)= E_{2n}+1,\  n \ge 1.\eqno(2.24)$$

Returning to (2.19), we easily derive analogs of the first order partial differential equations (1.22), (1.23), (1.24) for the generating  function $F(x,t)$,  namely

$$ {\partial F \over \partial  t } +  \left[ x\sinh t\  + \tanh t  \right] F = 0,\eqno(2.25)$$

$${\partial F \over \partial x }  +  2   \sinh^2 \left(t\over 2\right)   F =0,\eqno(2.26) $$

$$x {\partial F \over \partial x }  =  \tanh \left(t\over 2\right)  \left[  {\partial F \over \partial t } + F \tanh t \right].\eqno(2.27) $$

To find the  inverse of relation (2.22), we apply the product of series of  $\cosh t$ and  (2.19), equating its coefficients in front of $t^{2n}$ with the corresponding terms of  series (1.16). This is indeed allowed within the   interval of the absolute convergence $|t| < \pi/4$.  As a result, we deduce

$$p_n(x)= \sum_{k=0}^n  {2n  \choose 2k }q_{k}(x).\eqno(2.28)$$

But $q_n^\prime (x)=  q_n(x)- p_n(x)$.   So, we have

$$q_n^\prime (x) = -  \sum_{k=0}^{n-1}   {2n  \choose 2k }q_{k}(x).\eqno(2.29)$$

Another source of identities for Bernoulli numbers is a formula related to $p^\prime_n(x)$. To derive it, we call the partial differential equation (1.24) and the Taylor series for the hyperbolic tangent

$$\tanh\left({x\over 2}\right)=   2 \sum_{k=0}^\infty \frac{\left(2^{2(k+1)}  -  1\right) B_{2(k+1)} }{(2(k+1))! }
x^{2k+1}. $$

Hence,  substituting it in (1.24), making the product of series and equating the coefficients in front of $t^{2n}$, we obtain

$$x\  p^\prime_n(x)=   \sum_{k=1}^{n}  {2n  \choose 2k-1  }  \frac{2^{2k}  -  1} {k }\ B_{2k } \  p_{n+1 -k} (x).\eqno(2.30)$$

In particular, dividing (2.30) by $x$ and passing $x$ to zero, we take into account the value $p^\prime_n(0)= -1,\ n \ge 1$ to derive the identity

$$\sum_{k=1}^{n}  {2n  \choose 2k-1  }  \frac{\left(2^{2k}  -  1\right)} {k }\ B_{2k }  = 1.\eqno(2.31)$$

Further,  employing  (1.21), we get from (2.30)

$$ \sum_{k=1}^{n} {2n +1 \choose 2k -1}
p_k(x) =    \sum_{r=1}^{n} {2n +1 \choose 2r } \sum_{k=1}^{r}  {2r  \choose 2k-1  }  \frac{\left(2^{2k}  -  1\right)} {k }\ B_{2k } \  p_{r+1 -k} (x).$$

Thus, multiplying  both sides of the latter equality by $e^{-x}$ and integrating over $\mathbb{R}_+$,  we use (2.20) to find the identity

$$ \sum_{r=1}^{n} \frac {E_{2r}}  {(2r-1)!  (2(n-r+1)) ! } =    2 \sum_{r=1}^{n}  \sum_{k=1}^{r}  \frac{\left(2^{2k}  -  1\right) B_{2k } \  E_{2(r-k +1)}} { (2k)! (2(n-r)+1)! (2(r-k)+1)!}.\eqno(2.32)$$

Returning to (2.17) and  letting $m=0$, we make a summation in  the right-hand side of (2.17) by $k$ from zero to $n$

$$\frac{n}{ 1- 2^{2n}} \sum_{k=0}^n {2n \choose 2k}  \int_0^\infty e^{-2x} p_{n-k}(x) p_{k}(x) {dx\over  x}$$

and employ  the binomial type identity (2.23)  to deduce

$$\sum_{k=0}^n {2n \choose 2k}  \int_0^\infty e^{-2x} p_{n-k}(x) p_{k}(x) {dx\over  x}=  \int_0^\infty e^{-x} p_{n}(x) {dx\over  x}.$$

Therefore,   we obtain the identity

$$ B_{2n}\sum_{k=0}^n {2n \choose 2k} = \frac{n \ }{ 1- 2^{2n}} \int_0^\infty e^{-x} p_n(x){dx\over  x}.\eqno(2.33)$$

An interesting question is to express the finite sum in the left-hand side of (2.33) in terms of the values $p_n^{\prime\prime} (0)$. In fact, differentiating two times in  (2.28) with respect to $x$, we let $x=0$ and use (2.12), (2.24) to  obtain

$$p^{\prime\prime}_n(0)= \sum_{k=2}^n  {2n  \choose 2k }q^{\prime\prime}_{k}(0)=  \sum_{k=0}^n  {2n  \choose 2k } - 2,$$

or,

$$  \sum_{k=0}^n  {2n  \choose 2k } = p^{\prime\prime}_n(0) +2,\ n \ge 1.\eqno(2.34)$$

Moreover,  differentiating two times in (2.30) and letting then $x=0$, we get the following recurrence relation for the values $p^{\prime\prime}_n(0)$

$$ p^{\prime\prime}_n(0)=  {1\over 2-n} \sum_{k=2}^{n-1}  {2n  \choose 2k-1  }  \frac{2^{2(n-k+1)}  -  1} {n-k+1 }\
B_{2(n-k+1) } \  p^{\prime\prime}_{k} (0),\ n\ge 1, \ n \neq 2, \eqno(2.35)$$

and $p^{\prime\prime}_2(0)=6.$

Nevertheless, we are able to calculate explicitly the left-hand side of (2.34) due to trigonometric and exponential series technique developed in \cite{gou} and where one can find a great collection of many such formulas.  Precisely, employing relation (3.7) in  Vol. 6, formula (3.7), it gives

$$  \sum_{k=0}^n  {2n  \choose 2k } =  2^{2n-1},  \  n \in \mathbb{N}\eqno(2.36)$$

and therefore,  $p^{\prime\prime}_n(0)=  2 \left(  2^{2(n-1)} -  1\right)$.   Consequently, identity (2.33) becomes

$$ B_{2n} = \frac{2n \ }{2^{2n}-  2^{4n} } \int_0^\infty e^{-x} p_n(x){dx\over  x}.\eqno(2.37)$$

Substituting the value of $p^{\prime\prime}_n(0)$ in (2.35), we obtain a possibly new identity

$$\sum_{k=2}^{n-1}  {2n  \choose 2k-1  }  \frac{\left(2^{2(n-k+1)}  -  1\right)\left( 2^{2(k-1)} -  1\right) } {n-k+1 }\
B_{2(n-k+1) } $$$$ =  \left(2^{2(n-1)} -  1\right) (2-n),\quad   n \ge 1.\eqno(2.38)$$

Meanwhile,

$$p_n(x)= \sum_{k=1}^n a_{n,k} x^k,\  n \ge 1,$$

where $a_{n,k}$ is defined by (1.15).  Thus, substituting it into (2.13) and (2.36), after calculation of the elementary Euler integral  and the use of (2.36), we find the following explicit formulas for the Bernoulli numbers, respectively,

$$B_{2n}= \frac{n}{ 1- 2^{2n}}  \sum_{k=1}^n   {1\over  2^k\   k}\sum_{r=0}^{k} {(-1)^r\over 2^r} \binom{k}{r}\sum_{j=0}^{k-r} {(-1)^j\over 2^j} \binom{k-r}{j} (r-j)^{2n} , \eqno(2.39)$$

$$B_{2n}= \frac{2n}{ 2^{2n} (1- 2^{2n}) }  \sum_{k=1}^n   {1\over k }\sum_{r=0}^{k} {(-1)^r\over 2^r} \binom{k}{r}\sum_{j=0}^{k-r} {(-1)^j\over 2^j} \binom{k-r}{j} (r-j)^{2n} \eqno(2.40)$$

and the equality of integrals

$$\int_0^\infty e^{-x} p_n(x){dx\over  x} =  2^{2n-1} \int_0^\infty e^{-2x} p_n(x){dx\over  x},\  n \ge 1.\eqno(2.41)$$

Concerning other identities and explicit formulas for Bernoulli's numbers see, for instance, a survey article \cite{gould} and in \cite{short}, \cite{ching}.   Further, substituting the right-hand side of (2.41) into (1.26),  it becomes

$$\zeta(2n)  =  \frac{ (-1)^n  \pi^{2n} }{2 (2^{2n}  - 1)(2n-1)!} \int_0^\infty e^{-x} p_n(x){dx\over x}, \ n \in \mathbb{N}.\eqno(2.42)$$

An explicit formula for the Euler numbers can be deduced similarly to (2.39), (2.40) with the use of the integral  (2.20).
Hence,  due to (1.15), we obtain for all $n \in \mathbb{N}$

 $$E_{2n}=  \sum_{k=1}^n  k! \sum_{r=0}^{k} {(-1)^r\over 2^r\ r!} \sum_{j=0}^{k-r} {(-1)^j(r-j)^{2n} \over 2^j\ j! (k-r-j)! } .$$

Other explicit formulas for Euler numbers see, for instance, in \cite{ming}.  The latter equality can give another characteristic of the Euler numbers.  In fact, we have

$$E_{2n}=  {d^{2n}\over d z^{2n} } \sum_{k=1}^n
   \sum_{r=0}^{k} {(-1)^r\over 2^r} \binom{k}{r}\sum_{j=0}^{k-r} {(-1)^j\over 2^j} \binom{k-r}{j} e^{z(r-j)}\ \Bigg|_{z=0} $$$$
   =    {d^{2n}\over d z^{2n} }  \sum_{k=1}^n  (1-\cosh z)^k \Bigg|_{z=0}.$$

Therefore, we find the formula

$$E_{2n} =    {d^{2n}\over d z^{2n} }  \   \frac{(1-\cosh z)(1 - (1-\cosh z )^{n})}{\cosh z} \Bigg|_{z=0}, \  n \ge 1.$$

Analogously, coefficients  (1.15) of the polynomial sequence $p_n(x)$ take the form

$$a_{n,k} = {(-1)^k 2^k\over k!}  {d^{2n}\over d z^{2n} }  \    \sinh^{2k}\left({z\over 2}\right) \Bigg|_{z=0},\   k=1,2, \dots, n .$$

In the meantime,  equality  (2.41) is quite important to derive  connection formulas  for the Bernoulli and Euler numbers. In fact, the integral in the left-hand side of (2.40) is calculated in \cite{yakpol} and we have

$$\int_0^\infty e^{-x} p_n(x){dx\over  x} =   -   \sum_{k=0}^{n-1} {2n -1 \choose 2k} E_{2k}, \  n \ge 1.$$

Consequently,  combining with  (2.13) and (2.37), we established the connection formula between the Bernoulli and Euler numbers.

{\bf Theorem 1}.  {\it  The following identity holds valid}

$$ B_{2n}= \frac{2n }{  2^{2n} (2^{2n}-  1)} \sum_{k=0}^{n-1} {2n -1 \choose 2k} E_{2k} ,\quad n \in \mathbb{N}.$$

Calling (2.42), we get an immediate

{\bf Corollary 1}.  {\it  For all $n \in \mathbb{N}$ one has }

$$ \zeta(2n)  = \frac{(-1)^{n+1} \pi^{2n}  }{2 (2^{2n}  - 1)}
\sum_{k=0}^{n-1} \frac{ E_{2k}}{(2k)! (2(n-k)-1) !}  ,\quad n \in \mathbb{N}.$$

Calling again (2.22), we differentiate through two times and let $x=0$. Hence with the use of (2.12) and (2.24) we derive a curious recurrence relation for  the Euler numbers.   Indeed, it has

{\bf Theorem 2}.  {\it  The following identity holds }

$$ E_{2n}= 1-   \sum_{k=0}^{n-1}  2^{2(n-k) -1} { 2n \choose 2k} E_{2k} ,\quad n \in \mathbb{N}.$$

As an application, we  announce  at the end of this section an interesting result about the structure of the Bernoulli numbers $B_{2n}$ and the rational values $\zeta(2n)/\pi^{2n}$ (see (2.10), (2.42)), which is a immediate  consequence of the Von Staudt- Clausen theorem \cite{von} about the fractional part of Bernoulli numbers and Fermat's Little theorem.

Precisely, it has

{\bf Theorem 3}.  {\it  The Bernoulli numbers $B_{2n}$ and Riemann zeta-values $\zeta(2n)$ satisfy the following properties, respectively,}

$$ 2 (  2^{2n}-  1) B_{2n}   \in \mathbb{Z},\quad n \in \mathbb{N},\eqno(2.43)$$

$$ 2 \left(2^{2n}  - 1\right)   {\zeta(2n) (2n-1)! \over \pi^{2n} }    \in \mathbb{Z},\quad n \in \mathbb{N}.\eqno(2.44)$$

Meanwhile identity (2.13) leads to

{\bf Corollary 2}.  {\it  For all $n \in \mathbb{N}$  }

$$2n  \int_0^\infty e^{-2x} p_{n}(x) {dx\over  x}  \in \mathbb{Z}.$$

\section {Riemann's  zeta-values}

Our main goal here is to establish certain identities, integral and series representations for the Riemann zeta function of positive argument.  Concerning  zeta-values at integers, as we could see in the previous section,  the Euler formula (2.10) gives a direct relationship  of $\zeta(2n),\ n \in \mathbb{N}$ with the Bernoulli numbers.  However,  similar formula for the values of zeta function at odd integers is unknown and probably does not exist.   Our attempts to find a finite relation between $\zeta(2n+1)$ are still  unsuccessful. Nevertheless, we will derive several integral and series representations,  related to these numbers and general positive numbers greater than one, involving our Sheffer's sequences of polynomials.   Some rapidly convergent series for $\zeta(2n+1)$ see, for instance, in \cite{sri}.

In fact, returning to  (1.28) and substituting  the modified Bessel function  by its representation (1.10), we employ the definition of the improper integral, integration by parts, the absolute and uniform convergence and the Riemann- Lebesgue lemma to make the change of the order of integration and motivate the following equalities

$$(-1)^n  (2n)!  \  \left(2^{2n+1} - 1\right)\   {\zeta(2n+1)\over (2 \pi)^{2n} } =  \lim_{N\to \infty} \int_0^N \tau \int_0^\infty  \int_0^\infty  e^{- 2x\cosh^2( u/2)} p_n(x) \cos(\tau u) {du dx d\tau \over  x} $$

$$=    \lim_{N\to \infty} \int_0^\infty  \int_0^\infty  e^{- 2x\cosh^2( u/2)} \sinh u \ p_n(x)\frac{1-\cos(Nu)}{u} du dx  $$

$$=  \int_0^\infty  \int_0^\infty  e^{- 2x\cosh^2( u/2)} \  p_n(x)\frac{\sinh u}{u} du dx = \int_0^1 \int_0^\infty  K_t(x) e^{-x}p_n(x) dx dt. $$

Consequently, we derived the identity for all $n \in \mathbb{N}$

$$(-1)^n  (2n)!  \  \left(2^{2n+1} - 1\right)\   {\zeta(2n+1)\over (2 \pi)^{2n} }  =
 \int_0^1 \int_0^\infty  K_t(x) e^{-x}p_n(x) dx dt.\eqno(3.1) $$

In the meantime, integrals (1.28), (3.1) have relationships with integrals, involving the Bernoulli polynomials owing
to the following representations proved in \cite{yakpol}

$$B_{2n+1}\left({1- t\over 2}\right)= -  {2n+1\over 2^{2n+1}\pi}\sin\pi t \ \int_0^\infty
K_{t}(x)e^{-x}p_n(x)dx,\    |t| < 1,\eqno(3.2)$$

$$B_{2n+1}\left({1- i\tau\over 2}\right)=  {2n+1\over
2^{2n+1}\ \pi i}\ \sinh\pi\tau  \ \int_0^\infty
K_{i\tau}(x)e^{-x}p_n(x)dx, \tau \in \mathbb{R}.\eqno(3.3)$$

Hence, integrating in (3.2) with respect to $t \in (0,1) $  and taking into account (3.1) we derive the identity

$$(-1)^{n+1}  (2n+1)!  \  \left(2-  2^{-2n}\right)\   {\zeta(2n+1)\over (2\pi)^{2n+1} }  =
\int_0^1 B_{2n+1}\left({1- t\over 2}\right)\frac{dt} { \sin\pi t },\   n\ge 1.\eqno(3.4)$$

Moreover, using (2.6), (2.7), (2.8),   after integration by parts with elementary substitutions in (3.4) and elimination of  the integrated terms, we write it in the form

$$(-1)^{n+1}  (2n)!  \  \left(2-  2^{-2n}\right)\   {\zeta(2n+1)\over 2^{2n+1} \pi^{2n}}  =
\int_0^{1/2} B_{2n}\left(t\right)\log\left(\cot \pi t  \right) dt,\   n\ge 1.\eqno(3.5)$$

One can find a similar identity, for instance, in \cite{ito}. The integral in (3.4) can be reduced  via properties for the Bernoulli polynomials to certain integrals considered recently in \cite{fur}.  Furthermore,  appealing to the addition formula for the Bernoulli polynomials \cite{erd}, Vol.  I

$$B_n(x+y)=    \sum_{k=0}^{n}   { n \choose k} B_{k}(x) y^{n-k}\eqno(3.6)$$

the integral (3.4) can be represented as a linear combination of the moment integrals of $1/\sin t$, which we denote by $I_n$ following \cite{fur}

$$I_n= \int_0^{\pi/2} {t^n\over \sin t} dt,\  n \in \mathbb{N}.$$

Hence,

$$\int_0^1 B_{2n+1}\left({1- t\over 2}\right)\frac{dt} { \sin\pi t }=  \int_0^{1/2}  B_{2n+1}\left({1- t\over 2}\right)\frac{dt} { \sin\pi t }+ \int_0^{1/2}  B_{2n+1}\left({t\over 2}\right)\frac{dt} { \sin\pi t } $$

$$= {1\over \pi}  \sum_{m=0}^{n}   (2\pi)^{2(m-n)-1} { 2n +1\choose 2m} \left[B_{2m}- B_{2m} \left({1\over 2}\right)\right]
I_{2(n-m)+1} -   {2n+1\over (2\pi)^{2n+1}}I_{2n}.$$

But

$$B_{2m} \left({1\over 2}\right) =  - \left(1- 2^{1-2m}\right) B_{2m}.$$

Consequently, taking this value into account, we substitute the previous sum into  (3.4) and after simplifications arrive at the Ramanujan- type identity (cf. \cite{fur}) for zeta -values at odd integers $(n \in \mathbb{N}) $

$$(-1)^{n+1}   (2n+1)! \left(1-  2^{-2n-1}\right)   \zeta(2n+1) +  \left(n+{1\over 2}\right) I_{2n}  $$$$
 =  \sum_{m=0}^{n-1}   { 2n+1  \choose 2m+1 }   B_{2(n-m) } \pi^{2(n-m)-1} \left(2^{2(n-m)}- 1\right) I_{2m+1}.\eqno(3.7)$$

In particular, letting $n=1, 2$ and using the well-known formula $I_1= 2G $, where $G$ is the Catalan constant, we get, respectively,

$${7\over 2} \zeta(3) + I_2=  2\pi G,$$

$$ I_4- {93\over 2} \zeta(5) =  2\pi ( I_3-  \pi^2 G).$$

{\bf Remark 1}.  We note, that the latter identities can be also obtained from corresponding equalities in Example I in \cite{fur}.

In the same manner one can obtain a finite sum representation of the zeta-values at odd integers in terms of the moment integrals of  $1/\sin^2 t$.  In this case our starting point will be Ito's identity \cite{ito}

$$(-1)^{n}  (2n)!  \   {\zeta(2n+1)\over (2\pi)^{2n} }  =
\int_0^{1} B_{2n}\left(t\right)\log\left(\sin \pi t  \right) dt,\   n\ge 1.\eqno(3.8)$$

Indeed, with the use of (2.8)  we have,

$$\int_0^{1} B_{2n}\left(t\right)\log\left(\sin \pi t  \right) dt =  \int_0^{1/2} B_{2n}\left(t\right)\log\left(\sin \pi t  \right) dt$$$$+
\int_0^{1/2} B_{2n}\left(1-t\right)\log\left(\sin \pi t  \right) dt= 2 \int_0^{1/2} B_{2n}\left(t\right)\log\left(\sin \pi t  \right) dt.$$

Hence, appealing to (2.4) in the right-hand side of the latter  equality and integrating twice by parts in the obtained integral, we  find

$$\int_0^{1} B_{2n}\left(t\right)\log\left(\sin \pi t  \right) dt  =   2 \sum_{m=0}^{n}  { 2n \choose 2m}  B_{2m}  \int_0^{1/2}  t^{2(n-m)} \log\left(\sin \pi t  \right) dt$$$$ -  2n \int_0^{1/2}  t^{2n-1} \log\left(\sin \pi t  \right) dt ={\pi^{-2n}\ M_{2n+1} \over 2n+1}    -  \sum_{m=0}^{n}  { 2n \choose 2m}  \frac{ B_{2m} \ \pi^{-2(n-m)-1} \  M_{2(n+1-m)}}{(2(n-m)+1)( n-m+1) } ,$$

where

$$M_n=   \int_0^{\pi/2}  {t^{n} \over \sin^2 t} dt,\quad n \ge 2.$$

Thus combining with (3.8), we derived the identity (compare with (3.7))

$$(-1)^{n+1} (2(n+1))! \    2^{-2n-1} \zeta(2n+1)  +  (n+1) M_{2n+1} $$$$  =
\sum_{m=0}^{n}  { 2(n+1)  \choose 2(m+1) }  B_{2(n-m)} \  \pi^{2(n-m)-1} \  M_{2(m+1)},\   n \ge 1.\eqno(3.9)$$

Appealing to relations (2.5.4.7) in \cite{prud},  Vol. I,  we have the values

$$M_2= \pi \log 2,   \quad\quad    M_4=  {\pi^3\over 2}\log 2-  {9\pi\over 4} \zeta(3).$$

Therefore, letting $n=1$ in (3.9),  we get, for instance,

$$ {21\over 8}  \zeta (3) +  M_3= {3\pi^2 \over 4} \log 2.$$

Further, returning to (3.3),  we   multiply its  both sides by $\tau/\sinh\pi\tau $ and integrate over $\mathbb{R}_+$.  Hence

$$\int_0^\infty \tau  B_{2n+1}\left({1- i\tau \over 2}\right)\frac{d\tau} { \sinh\pi \tau }
=  {2n+1\over  2^{2n+1}\ \pi i}\int_0^\infty  \int_0^\infty \tau K_{i\tau}(x)e^{-x}p_n(x)dxd\tau $$$$
= {2n+1\over  2^{2n+1}\ \pi i}\int_0^1 \int_0^\infty  K_{t}(x)e^{-x}p_n(x)\  x  dx dt
,\   n\in \mathbb{N}_0.\eqno(3.10)$$

On the other hand, calling relation (1.20), we deduce from (3.1) and (3.10)

$$\sum_{k=0}^{n} {2n +1 \choose 2k}   {2^{2k+1} \over 2k+1} \int_0^\infty i \tau  B_{2k+1}\left({1- i\tau \over 2}\right)\frac{d\tau} { \sinh\pi \tau }$$$$ = {1\over \pi }  \sum_{k=0}^{n} {2n +1 \choose 2k}
\int_0^1 \int_0^\infty  K_{t}(x)e^{-x}p_k(x)\  x  dx dt$$$$ =  -  {1\over \pi } \int_0^1 \int_0^\infty  K_{t}(x)e^{-x}p_{n+1}(x) dx dt =  (-1)^{n}   (2(n+1))!  \  \left(2- 2^{- 2(n+1)}\right)\   {\zeta(2n+3)\over \pi^{2n+3} } $$

or
$$ \sum_{k=0}^{n} \frac {2^{2k}} {(2k+1)! (2(n-k)+1)! } \int_0^\infty i \tau  B_{2k+1}\left({1- i\tau \over 2}\right)\frac{d\tau} { \sinh\pi \tau } $$$$=   (-1)^{n}  (n+1) \  \left(2- 2^{- 2(n+1)}\right)\   {\zeta(2n+3)\over\pi^{2n+3} } .\eqno(3.11)$$

Hence, recalling  (3.6),  the integral in (3.11) can be rewritten as follows

$$2^{2k} \int_0^\infty i \tau  B_{2k+1}\left({1- i\tau \over 2}\right)\frac{d\tau} { \sinh\pi \tau } =
 \sum_{m=0}^{k}  (-1)^{k-m+1}   { 2k +1\choose 2m}  \left( 2^{2m-1} - 1\right) B_{2m} $$$$\times  \int_0^\infty  \frac{\tau^{2(k-m+1)}\  d\tau} { \sinh\pi \tau }=   (2k+1)!  \sum_{m=0}^{k}  (-1)^{m+1} (m+1) \  \frac{\left( 2^{2(k-m)} - 2\right) \left( 2^{2(m +2)} - 2\right)B_{2(k-m)}} {(2(k-m))! } $$$$\times  { \zeta(2m+3)\over
 (2\pi)^{2m+3} }.$$

Substituting the right-hand side of the latter equality in (3.11), we find the identity

$$ (-1)^{n}  (n+1) \  \left(2- 2^{- 2(n+1)}\right)\   {\zeta(2n+3)\over\pi^{2n+3} } =
 \sum_{k=0}^{n} \sum_{m=0}^{k}  \frac { (-1)^{m+1} (m+1)} {(2(n-k)+1)! } $$$$ \times   \frac{\left( 2^{2(k-m)} - 2\right) \left( 2- 2^{-2(m +1)}\right)B_{2(k-m)}} {(2(k-m))!  \pi^{2m+3} }  \zeta(2m+3)
 = \sum_{m =0}^{n}   (-1)^{m+1} (m+1) \left( 2- 2^{-2(m +1)}\right) $$$$\times  \frac{\zeta(2m+3)} { \pi^{2m+3}}
\left(  \sum_{k=0}^{n-m}   \frac{\left( 2^{2k} - 2\right) B_{2k}} {(2k)! (2(n-m- k)+1)!   }\right) . \eqno(3.12)$$

It would be a great achievement to have here a finite recurrence relation for zeta- values at odd integers. However, unfortunately, this is not the case.  In fact, (3.12) yields for all $n \ge 1$

$$ \sum_{m =0}^{n-1}   (-1)^{m+1} (m+1) \left( 2- 2^{-2(m +1)}\right) \frac{\zeta(2m+3)} { \pi^{2m+3}}
\left(  \sum_{k=0}^{n-m}   \frac{\left( 2^{2k} - 2\right) B_{2k}} {(2k)! (2(n-m- k)+1)!   }\right)= 0. \eqno(3.13)$$

{\bf Theorem 4}. {\it For all $n \in \mathbb{N}$ the following identity holds for Bernoulli numbers}

$$ \sum_{k=0}^{n}   { 2n+1  \choose 2k } \left( 2^{2k-1} - 1\right) B_{2k}   = 0.\eqno(3.14)$$

\begin{proof}  In fact, recalling (2.3), we see that (3.14) is equivalent to the equality

$$\sum_{k=0}^{n}   { 2n+1  \choose 2k }  2^{2k} B_{2k}   = 2n+1,$$

which yields

$$\sum_{k=0}^{2n+1}   { 2n+1  \choose k }  2^{k-2n-1} B_{k}   = 0.$$

But this is true, because the left-hand side is equal (see (2.4), (2.8)) to $B_{2n+1} (1/2) = 0.$

\end{proof}

Theorem 4 says that all coefficients in front of zeta-values $\zeta(2m+3)$ in (3.14) are equal to zero.  Hence such kind of equalities  can be a source to obtain possibly new identities for Bernoulli numbers.

Finally, we will get an integral representation of zeta-values at  positive numbers, which is a direct consequence of the formulas (1.3), (1.25), (1.26).  Precisely, it has

{\bf Theorem 5}. {\it Let $\alpha > 1,\  [\alpha]$ be its integer part and $\{\alpha\}$ be its fractional part.  Then the following identities take place, when $[\alpha]$ is even or odd, respectively, }

$$\frac{2^\alpha - 1}{(2\pi)^{\alpha-1}}\  \Gamma(\alpha) \zeta(\alpha) = (-1)^{[\alpha]/2}
\int_0^\infty \int_0^\infty \tau^{ \{\alpha\}} K_{i\tau}(x)\ e^{-x} p_{[\alpha]/2} (x){dx\over x},$$

$$\frac{2^\alpha - 1}{(2\pi)^{\alpha-1}}\  \Gamma(\alpha) \zeta(\alpha) = (-1)^{([\alpha]-1) /2}
\int_0^\infty \int_0^\infty \tau^{ \{\alpha\}-1} K_{i\tau}(x)\ e^{-x} p_{([\alpha]+1)/2} (x){dx\over x}.$$

\bigskip
\centerline{{\bf Acknowledgments}}
\bigskip
The present investigation was supported, in part,  by the "Centro de  Matem{\'a}tica" of the University of Porto.


TITLE: Example of an even degree polynomial with no roots in $\mathbb{R}$
QUESTION [1 upvotes]: Full question:

Show that if $n$ is even there exists a polynomial of degree $n$ that has no roots in $\mathbb{R}$.

$n=2k$ and $p(x)=x^n-x^{n-1}+2$

for $x>0$, $p(x)>0$
for $x<0$, $p(x)>0$
for $x=0$, $p(x)=2$

Would this polynomial satisfy the equation, I can't think of any real roots?

REPLY [2 votes]: Your polynomial does satisfy the problem (but is not the easiest choice).
Let $f_n(x) = x^n + x^{n-1} + 2$.  Then $f_n'(x) = x^{n-2}(nx + 1-n)$.  Since, for even $n$, $\lim_{x \rightarrow \pm \infty} f_n(x) = \infty$, these $f_n$ each have a global minimum at one of their local minima.  By the first derivative, candidates are $x = 0$ and $x = \frac{n-1}{n}$.  Since $f_n'(-1) < 0$, $f_n'(1/4) < 0$, and $f_n'(2)  > 0$, only $\frac{n-1}{n}$ can be a minimum.  Then
$$  f_n\left(\frac{n-1}{n}\right) = \frac{2n - 2 - \left(1 - \frac{1}{n}\right)^n}{n-1}  $$
The part $\left(1 - \frac{1}{n}\right)^n$ is a standard way to get $\mathrm{e}^{-1}$ (by taking the limit as $n \rightarrow \infty$) and takes values in $[0,1]$.  (We can get tighter bounds, but there is no need.)  Since $n \geq 2$, the numerator is at least $4-2-1 > 0$ and the denominator is also positive, so the global minimum is positive.  Therefore, $f_n(x)$ is positive for even $n$ and all $x \in \mathbb{R}$.
Contrast with: $x^{2k} = (x^2)^k$ is always nonnegative, so $x^{2k}+1$ is always positive.


text:
\begin{document}

\begin{abstract}
\noindent  In computer aided geometric design a polynomial is usually
represented in Bernstein form. The de Casteljau algorithm is the most
well-known algorithm for evaluating a polynomial in this form. Evaluation
via the de Casteljau algorithm has relative forward error proportional to
the condition number of evaluation. However, for a particular family of
polynomials, a curious phenomenon occurs: the observed error is much smaller
than the expected error bound. We examine this family and prove a much
stronger error bound than the one that applies to the general case. Then
we provide a few examples to demonstrate the difference in rounding.
\\ \\
\noindent \textit{Keywords}: Polynomial evaluation, Floating-point
arithmetic, Bernstein polynomial, Round-off error, Condition number
\end{abstract}

\tableofcontents

\section{Introduction}

In computer aided geometric design, polynomials are usually expressed in
Bernstein form. Polynomials in this form are usually evaluated by the
de Casteljau algorithm. This algorithm has a round-off error bound
which grows only linearly with degree, even though the number of
arithmetic operations grows quadratically. The Bernstein basis is
optimally suited (\cite{Farouki1987, Delgado2015, Mainar2005})
for polynomial evaluation. Nevertheless the de Casteljau
algorithm returns results arbitrarily less accurate than the working
precision \(\mach\) when evaluating \(p(s)\) is ill-conditioned.
The relative accuracy of the computed
evaluation with the de Casteljau algorithm (\texttt{DeCasteljau}) satisfies
(\cite{Mainar1999}) the following a priori bound:
\begin{equation}\label{de-casteljau-error}
  \frac{\left|p(s) - \mathtt{DeCasteljau}(p, s)\right|}{\left|p(s)\right|} \leq
  \cond{p, s} \times \bigO{\mach}.
\end{equation}

\begin{figure}
  \includegraphics[width=0.75\textwidth]{against_a_priori.pdf}
  \centering
  \captionsetup{width=.75\linewidth}
  \caption{Comparing relative forward error to a priori bound for
    \(u(s) = (1 - 4s)^5, v(s) = (1 - 5s)^5\) and \(w(s) = (1 - 6s)^5\)}
  \label{fig:against-a-priori}
\end{figure}

For example, consider \(u(s) = (1 - 4s)^5, v(s) = (1 - 5s)^5\) and
\(w(s) = (1 - 6s)^5\). For points of the form
\begin{equation}
s_u = \frac{1}{4} + \frac{6}{16N}, \quad
s_v = \frac{1}{5} + \frac{8}{25N}, \quad
s_w = \frac{1}{6} + \frac{10}{36N}
\end{equation}
which are near the multiple roots, the condition numbers of evaluation are:
\begin{equation}
\cond{u, s_u} = \cond{v, s_v} = \cond{w, s_w} =
  \left|N\right|^5 + \bigO{N^4}.
\end{equation}
As we can see in Figure~\ref{fig:against-a-priori}, one of these is not
like the others. Evaluation of both \(u(s)\) and \(w(s)\) produces
relative forward error very close to the a prior bound\footnote{There
are actually three different error bounds, but the \(\left|N\right|^5\)
term dominates so much that they are not visually discernible.}. However,
the observed relative error when evaluating \(v(s)\) is significantly
lower than expected.

As we'll explore in Section~\ref{sec:improved-bound}, \(v(s)\) belongs
to a family of polynomials that the de Casteljau method can evaluate with
a significantly higher level of accuracy than expected. Notice that
\begin{equation}
v(s) = (1 - 5s)^5 = \left[(1 - s) - 4s\right]^5 =
  B_{0, 5}(s) - 4 B_{1, 5}(s) +
  16 B_{2, 5}(s) + \cdots;
\end{equation}
in particular, the Bernstein coefficients are powers of \(2\) (up to sign).
The aforementioned family of polynomials contains any of the form
\begin{equation}
b_0 \left[(1 - s) - 2^t s\right]^n =
  b_0 B_{0, n}(s) - \left(b_0 2^t\right) B_{1, n}(s) +
  \left(b_0 2^{2t}\right) B_{2, n}(s) + \cdots.
\end{equation}
Polynomials in this family have coefficients that can be represented
exactly (i.e. with no round-off).

The paper is organized as follows. Section~\ref{sec:notation} establishes
notation for error analysis with floating point operations and reviews the
de Casteljau algorithm. In
Section~\ref{sec:improved-bound}, the lowered error bound is proved
for polynomials in the special family and numerical experiments compare
observed relative error to the newly improved bound.
Finally, in Section~\ref{sec:implications} we comment on the impact
that this lowered bound makes on comparisons between the de Casteljau
algorithm and the VS algorithm.

\section{Basic notation and results}\label{sec:notation}

\subsection{Floating Point and Forward Error Analysis}

We assume all floating point operations obey
\begin{equation}
  a \star b = \fl{a \circ b} = (a \circ b)(1 + \delta_1) =
  (a \circ b) / (1 + \delta_2)
\end{equation}
where \(\star \in \left\{\oplus, \ominus, \otimes, \oslash\right\}\), \(\circ
\in \left\{+, -, \times, \div\right\}\) and \(\left|\delta_1\right|,
\left|\delta_2\right| \leq \mach\). The symbol \(\mach\) is the unit round-off
and \(\star\) is a floating point operation, e.g.
\(a \oplus b = \fl{a + b}\). (For IEEE-754 floating point double precision,
\(\mach = 2^{-53}\).) We denote the computed result of
\(\alpha \in \mathbf{R}\) in floating point arithmetic by
\(\widehat{\alpha}\) or \(\fl{\alpha}\) and use \(\mathbf{F}\) as the set of
all floating point numbers (see \cite{Higham2002} for more details).
Following \cite{Higham2002}, we will use the following classic properties in
error analysis.

\begin{enumerate}
  \item If \(\delta_i \leq \mach\), \(\rho_i = \pm 1\), then
      \(\prod_{i = 1}^n (1 + \delta_i)^{\rho_i} = 1 + \theta_n\),
  \item \(\left|\theta_n\right| \leq \gamma_n \coloneqq
      n \mach / (1 - n \mach)\),
  \item \((1 + \theta_k)(1 + \theta_j) = 1 + \theta_{k + j}\),
  \item \(\gamma_k + \gamma_j + \gamma_k \gamma_j \leq \gamma_{k + j}
    \Longleftrightarrow (1 + \gamma_k)(1 + \gamma_j) \leq 1 + \gamma_{k + j}\),
  \item \((1 + \mach)^j \leq 1 / (1 - j \mach) \Longleftrightarrow
  (1 + \mach)^j - 1 \leq \gamma_j\).
\end{enumerate}

\begin{theorem}\label{thm:floating-point-exact}
In the absence of overflow or underflow, for \(a, b, -2^t \in \mathbf{F}\)
\begin{align}
\left(-2^t a\right) \otimes b &= -2^t \left(a \otimes b\right) \\
\left(-2^t a\right) \oplus \left(-2^t b\right) &= -2^t \left(a \oplus b\right).
\end{align}
\end{theorem}

\subsection{Bernstein Basis and de Casteljau Algorithm}

A polynomial written in the Bernstein basis is of the form
\begin{equation}
p(s) = \sum_{j = 0}^n b_j B_{j, n}(s)
\end{equation}
where \(B_{j, n}(s) = \binom{n}{j} (1 - s)^{n - j} s^j\). When
\(s \in \left[0, 1\right]\), the Bernstein basis functions are
non-negative. We refer to \(\widetilde{b}_j \coloneqq \binom{n}{j} b_j\)
as the \textit{scaled Bernstein coefficients}.
The condition number of evaluation for \(p(s)\) is
\begin{equation}
\cond{p, s} = \frac{\widetilde{p}(s)}{\left|p(s)\right|}
\end{equation}
where
\(\widetilde{p}(s) \coloneqq \sum_{j = 0}^n \left|b_j\right| B_{j, n}(s)\).

\begin{breakablealgorithm}
  \caption{\textit{de Casteljau algorithm for polynomial evaluation.}}
  \label{alg:de-casteljau}

  \begin{algorithmic}
    \Function{\(\mathtt{result} = \mathtt{DeCasteljau}\)}{$b, s$}
      \State \(n = \texttt{length}(b) - 1\)
      \State \(\widehat{r} = 1 \ominus s\)
      \\
      \For{\(j = 0, \ldots, n\)}
        \State \(\widehat{b}_j^{(n)} = b_j\)
      \EndFor
      \\
      \For{\(k = n - 1, \ldots, 0\)}
        \For{\(j = 0, \ldots, k\)}
          \State \(\widehat{b}_j^{(k)} = \left(
              \widehat{r} \otimes \widehat{b}_j^{(k + 1)}\right) \oplus
              \left(s \otimes \widehat{b}_{j + 1}^{(k + 1)}\right)\)
        \EndFor
      \EndFor
      \\
      \State \(\mathtt{result} = \widehat{b}_0^{(0)}\)
    \EndFunction
  \end{algorithmic}
\end{breakablealgorithm}

\begin{theorem}\label{thm:de-casteljau-bound}
The de Casteljau algorithm (Algorithm~\ref{alg:de-casteljau}) satisfies
\begin{equation}
\left|p(s) - \mathtt{DeCasteljau}(p, s)\right| \leq \gamma_{3n}
  \widetilde{p}(s).
\end{equation}
\end{theorem}

\begin{proof}
See Appendix~\ref{sec:appendix-proof-details}.
\end{proof}

\section{Improved Bound}\label{sec:improved-bound}

We seek to analyze our family of polynomials of the form:
\begin{equation}
p(s) = b_0 \left[(1 - s) - 2^t s\right]^n.
\end{equation}
When written in the Bernstein basis, \(p(s)\) has coefficients that satisfy
\begin{equation}\label{exact-ratio}
b_{j + 1} = -2^t b_j.
\end{equation}
In a finite precision binary arithmetic, \eqref{exact-ratio} will hold
exactly (i.e. with no round-off) until overflow or underflow makes it
impossible to represent
\(b_j\) in the given arithmetic (the mantissa will always be the same but
the sign and exponent will change). This useful property remains
true for the intermediate terms computed by the de Casteljau method:
\begin{equation}
b_{j + 1}^{(k)} = (1 - s) b_{j + 1}^{(k + 1)} + s b_{j + 2}^{(k + 1)} =
  -2^t\left[(1 - s) b_j^{(k + 1)} + s b_{j + 1}^{(k + 1)}\right] =
  -2^t b_j^{(k)}.
\end{equation}
Remarkably, this also holds true for the \textbf{computed} values:
\(\widehat{b}_{j + 1}^{(k)} =
-2^t \widehat{b}_j^{(k)}\). Following Theorem~\ref{thm:floating-point-exact}
we have
\begin{align}
\widehat{b}_{j + 1}^{(k)} &= \left[\widehat{r} \otimes
  \widehat{b}_{j + 1}^{(k + 1)}\right] \oplus
  \left[s \otimes \widehat{b}_{j + 2}^{(k + 1)}\right] \\
&= \left[-2^t \left(\widehat{r} \otimes
  \widehat{b}_j^{(k + 1)}\right)\right] \oplus
  \left[-2^t\left(s \otimes \widehat{b}_{j + 1}^{(k + 1)}\right)\right] \\
&= -2^t\left(\left[\widehat{r} \otimes
  \widehat{b}_j^{(k + 1)}\right] \oplus
  \left[s \otimes \widehat{b}_{j + 1}^{(k + 1)}\right]\right) \\
&= -2^t \widehat{b}_j^{(k)}.
\end{align}
Thus, for such \(p(s)\), we only need compute
\(\widehat{b}_0^{(k)}\):
\begin{equation}
\widehat{b}_0^{(k)} = \left[\widehat{r} \otimes
  \widehat{b}_0^{(k + 1)}\right] \oplus \left[s \otimes
  \widehat{b}_1^{(k + 1)}\right] = \left[\widehat{r} \otimes
  \widehat{b}_0^{(k + 1)}\right] \oplus \left[\left(-2^t\right) \left(s \otimes
  \widehat{b}_0^{(k + 1)}\right)\right].
\end{equation}

\begin{theorem}\label{thm:better-error}
For a polynomial of the form
\begin{equation}
p(s) = b_0 \left[(1 - s) - 2^t s\right]^n
\end{equation}
the relative accuracy of the computed
evaluation with the de Casteljau algorithm satisfies
the following a priori bound:
\begin{equation}\label{better-de-casteljau-error}
\frac{\left|p(s) - \mathtt{DeCasteljau}(p, s)\right|}{\left|p(s)\right|}
  \leq \left(1 + \left|\phi\right| \gamma_3\right)^n - 1
\end{equation}
where
\begin{equation}
\phi \coloneqq \frac{(1 - s) + 2^t s}{(1 - s) - 2^t s} =
  \frac{1 + \left(2^t - 1\right) s}{1 - \left(2^t + 1\right) s}.
\end{equation}
\end{theorem}

\begin{proof}
Let \(r = 1 - s\) and \(\widehat{r} = 1 \ominus s\).
For \(k < n\),
\begin{equation}
\widehat{b}_0^{(k)} = \left[\widehat{r} \otimes
  \widehat{b}_0^{(k + 1)}\right] \oplus \left[\left(-2^t\right)
  \left(s \otimes \widehat{b}_0^{(k + 1)}\right)\right] =
  \widehat{b}_0^{(k + 1)} \left[r(1 + \theta_3) - 2^t s(1 + \theta_2)\right].
\end{equation}
Note that this can be written as \(\widehat{b}_0^{(k + 1)}
\left[\left(r - 2^t s\right) + E\right]\) where the round-off term
satisfies \(\left|E\right| \leq \left(r + 2^t s\right) \gamma_3\) for
\(s \in \left[0, 1\right]\). Hence we write
\(\widehat{b}_0^{(k)} = \widehat{b}_0^{(k + 1)} \left[\left(r - 2^t s\right) +
  \left(r + 2^t s\right)\theta_3\right]\).
Since \(\widehat{b}_0^{(n)} = b_0\), we have
\begin{equation}
\widehat{b}_0^{(0)} = b_0 \left[\left(r - 2^t s\right) +
  \left(r + 2^t s\right)\theta_3\right]^n.
\end{equation}
Dividing this by \(b_0^{(0)} = b_0 \left(r - 2^t s\right)^n\) we have
\begin{equation}
\widehat{b}_0^{(0)} = b_0^{(0)} \left(1 + \phi \cdot \theta_3\right)^n.
\end{equation}
Hence our relative error can be bound by
\begin{equation}
\left|\left(1 + \phi \cdot \theta_3\right)^n - 1\right| \leq
\left(1 + \left|\phi\right| \gamma_3\right)^n - 1
\end{equation}
as desired.
\end{proof}

Since
\begin{equation}
\widetilde{p}(s) = \sum_{j = 0}^n \left|b_0 \left(-2^t\right)^j\right|
  B_{j, n}(s) = \left|b_0\right| \left[(1 - s) + 2^t s\right]^n
\end{equation}
we see that \(\cond{p, s} = \widetilde{p}(s) / \left|p(s)\right| =
\left|\phi\right|^n\). So we can compare our improved bound
\(\left(1 + \left|\phi\right| \gamma_3\right)^n - 1\)
to the na\"ive bound \(\gamma_{3n} \left|\phi\right|^n\).

\begin{figure}
  \includegraphics[width=0.75\textwidth]{curbed_condition.pdf}
  \centering
  \captionsetup{width=.75\linewidth}
  \caption{Comparing na\"ive relative error bound to improved bound for
    \(p(s) = (1 - 5s)^{5}\) evaluated at \(s = 1/5 + 8/(25N)\)}
  \label{fig:curbed-condition}
\end{figure}

For example, we'll use \(b_0 = 1, n = 5, t = 2\), i.e. \(p(s) = (1 - 5s)^5\)
for a numerical experiment with. We compare the na\"ive and improved bounds to
the observed relative forward error in Figure~\ref{fig:curbed-condition}.
The figure shows the evaluation of \(p(s)\) at the points
\(\left\{\frac{1}{5} + \frac{8}{25 \cdot 2.1^e}\right\}_{e = 1}^{45}\),
which cause the condition number of evaluation to grow exponentially.
The form \(s = 1/5 + 8/(25N)\) is chosen because it simplifies
\(\phi\):
\begin{equation}
\phi = \frac{1 + 3s}{1 - 5s} = -N - \frac{3}{5}.
\end{equation}
As can be seen in the figure, the observed errors closely match the
improved bound and are significantly smaller than
the na\"ive bound. What's more, the actual error is still useful
(i.e. less than \(\bigO{1}\)) when the condition number becomes
very large.

\section{Implications}\label{sec:implications}

In \cite{Delgado2015}, the de Casteljau algorithm is compared to the VS
algorithm (\cite{Schumaker1986}) along with a few other methods. The VS
algorithm relies on two transformations of \(p(s) =
\sum_{j = 0}^n \widetilde{b}_j (1 - s)^{n - j} s^j\) using
\(\sigma_1 = (1 - s) / s\) and \(\sigma_2 = s / (1 - s)\):
\begin{equation}\label{vs-transform}
p(s) = s^n \left[\widetilde{b}_0 \sigma_1^n + \cdots + \widetilde{b}_n\right]
= (1 - s)^n \left[\widetilde{b}_0 + \cdots + \widetilde{b}_n \sigma_2^n\right].
\end{equation}
See Appendix~\ref{sec:vs-algorithm} for more details on the VS algorithm,
in particular Algorithm~\ref{alg:vs-algorithm} which describes the method and
Theorem~\ref{thm:vs-bound} which provides an error bound.

\begin{figure}
  \includegraphics[width=0.9375\textwidth]{compare_dp15.pdf}
  \centering
  \captionsetup{width=.75\linewidth}
  \caption{Comparing the de Casteljau and VS algorithms.}
  \label{fig:dp15-compare}
\end{figure}

In \cite{Delgado2015}, the authors use\footnote{The coefficients of \(f(s)\)
and \(g(s)\) cannot be represented exactly in IEEE-754 floating point
double precision.}
\begin{equation}
f(s) = \prod_{j = 1}^{20} (s - j/20), \quad
g(s) = \prod_{j = 1}^{20} (s - 2/2^j), \quad
h(s) = (s - 1/2)^{20}
\end{equation}
to compare the relative error of various methods.
Figure~\ref{fig:dp15-compare} reproduces the numerical experiments they
performed by evaluating \(f(s)\) at
\(\left\{\frac{2j - 1}{72}\right\}_{j = 1}^{36}\),
\(g(s)\) at \(\left\{\frac{j}{39}\right\}_{j = 1}^{38}\) and
\(h(s)\) at \(\left\{\frac{4j}{100}\right\}_{j = 1}^{24}\).
In addition, we have included a dotted line marking the
a prior error bound for each point of evaluation.
Viewing the errors for \(f(s)\) and \(g(s)\), there is no qualitative
difference between the de Casteljau algorithm and the VS algorithm, so
the authors use \(h(s)\) as the tiebreaker to conclude
\begin{displayquote}
the algorithm with a good behavior everywhere is the de Casteljau algorithm.
\end{displayquote}
However,
\begin{equation}
h(s) = \left(s - 1/2\right)^{20} = 2^{-20} (2s - 1)^{20}
  = 2^{-20} \left[(1 - s) - s\right]^{20},
\end{equation}
i.e. it is a member of our special family of polynomials with
\(b_0 = 2^{-20}, n = 20, t = 0\). This gives the de Casteljau
algorithm an unfair advantage over the VS algorithm in a
test case that is not representative of general polynomials.

Given the quadratic growth in the number of arithmetic operations
performed in the de Casteljau algorithm, an alternative with
linear growth (i.e. the VS method) should not be discounted if
it produces similar results. The special family of polynomials
explored here explains why the de Casteljau algorithm performs
so much better than the VS method and shows that such polynomials
are not indicative of the accuracy of the two methods
relative to one another.

\section{Acknowledgements}

The author would like to thank W. Kahan the publication
(\cite{Kahan1972}) that motivated the title of this work and
is closely related to the topic at hand.

\bibliography{curious-case}
\bibliographystyle{alpha}

\appendix

\section{Proof Details}\label{sec:appendix-proof-details}

\begin{proof}[Proof of Theorem~\ref{thm:de-casteljau-bound}]
When using the de Casteljau method, we have
\(b_j^{(n)} = \widehat{b}_j^{(n)} = b_j\) and for \(k = n - 1, \ldots, 0\)
and \(j = 0, \ldots, k\):
\begin{align}
b_j^{(k)} &= (1 - s) b_j^{(k + 1)} + s b_{j + 1}^{(k + 1)} \\
\widehat{b}_j^{(k)} &= \left[(1 \ominus s) \otimes
  \widehat{b}_j^{(k + 1)}\right]
  \oplus \left[s \otimes \widehat{b}_{j + 1}^{(k + 1)}\right].
\end{align}
This means that
\begin{equation}
\widehat{b}_j^{(k)} = (1 - s) \widehat{b}_j^{(k + 1)} (1 + \theta_3)
+ s \widehat{b}_{j + 1}^{(k + 1)} (1 + \theta_2)
\end{equation}
so that
\begin{align}
\widehat{b}_0^{(0)} &= (1 - s) \widehat{b}_0^{(1)} (1 + \theta_3)
+ s \widehat{b}_{1}^{(1)} (1 + \theta_2) \\
&= \widehat{b}_0^{(2)} B_{0, 2}(s) (1 + \theta_6) +
\widehat{b}_1^{(2)} B_{1, 2}(s) (1 + \theta_5) +
\widehat{b}_2^{(2)} B_{2, 2}(s) (1 + \theta_4) \\
&= \sum_{j = 0}^n \widehat{b}_j^{(n)} B_{j, n}(s) (1 + \theta_{3n - j}) \\
&= b_0^{(0)} + \sum_{j = 0}^n b_j B_{j, n}(s) \theta_{3n - j}.
\end{align}
Hence we have
\begin{equation}
\left|p(s) - \mathtt{DeCasteljau}(p, s)\right| \leq \gamma_{3n}
  \widetilde{p}(s)
\end{equation}
as desired. Note that this differs from the bound given in
\cite{Mainar1999}, Corollary 3.2 because the authors don't consider the
round-off when computing \(\widehat{r}\).
\end{proof}

\section{VS Algorithm}\label{sec:vs-algorithm}

In \cite{Schumaker1986}, a modified form of Horner's method is
described for evaluating a polynomial in Bernstein form.
Following~\eqref{vs-transform}, the algorithm applies Horner's
method to the scaled Bernstein coefficients with an input
related to \(s\):

\begin{breakablealgorithm}
  \caption{\textit{VS algorithm for polynomial evaluation.}}
  \label{alg:vs-algorithm}

  \begin{algorithmic}
    \Function{\(\mathtt{result} = \mathtt{VS}\)}{$b, s$}
      \State \(n = \texttt{length}(b) - 1\)
      \State \(\widehat{r} = 1 \ominus s\)
      \If{\(s \geq 1/2\)}
        \State \(\widehat{\sigma} = \widehat{r} \oslash s\)
        \State \(m = s\)
        \State \(\left[c_0, \ldots, c_n\right] =
            \left[b_n, \ldots, b_0\right]\)
      \Else
        \State \(\widehat{\sigma} = s \oslash \widehat{r}\)
        \State \(m = \widehat{r}\)
        \State \(\left[c_0, \ldots, c_n\right] =
            \left[b_0, \ldots, b_n\right]\)
      \EndIf
      \\
      \State \(\widehat{p}_n = c_n\)
      \For{\(k = n - 1, \ldots, 0\)}
        \State \(\widehat{p}_k = \left[\widehat{\sigma} \otimes
            \widehat{p}_{k + 1}\right] + \left[
            \binom{n}{k} \otimes c_{k}\right]\)
      \EndFor
      \\
      \State \(\widehat{m}_1 = \widehat{m}\)
      \For{\(k = 2, \ldots, n\)}
        \State \(\widehat{m}_k = \widehat{m}_{k - 1} \otimes \widehat{m}\)
      \EndFor
      \\
      \State \(\mathtt{result} = \widehat{m}_n \otimes \widehat{p}_0\)
    \EndFunction
  \end{algorithmic}
\end{breakablealgorithm}

This has the benefit of using a linear number of floating point operations,
as compared to the de Casteljau method, which uses a quadratic
number of floating point operations.

\begin{theorem}\label{thm:vs-bound}
The value computed by the VS algorithm
(Algorithm~\ref{alg:vs-algorithm}) satisifies\footnote{
The coefficients \(\gamma_{5n}\) and \(\gamma_{6n}\) differ from
\(\gamma_{4n}\) in \cite{Delgado2009} (Theorem 4.2) because the authors
don't account for the multiplication by \(\binom{n}{k}\) in computing the
scaled Bernstein coefficients or the round-off in \(1 \ominus s\) when
computing \((1 - s)^n\).}
\begin{equation}
\left|p(s) - \mathtt{VS}(p, s)\right| \leq
  \begin{cases}
    \gamma_{6n} \widetilde{p}(s) \quad
      \text{when } s \in \left[0, 1/2\right) \\
    \gamma_{5n} \widetilde{p}(s) \quad \text{when } s \in \left[1/2, 1\right].
  \end{cases}
\end{equation}
\end{theorem}

\begin{proof}
Since the algorithm has two branches depending on \(s \geq 1/2\), we
have to make a few distinctions throughout. However, most arguments
apply to both branches of the algorithm. The following analysis
\textbf{assumes}
that there is no round-off introduced by the computation of binomial
coefficients. For \(n \leq 56\), \(\binom{n}{k}\) can be represented exactly
in IEEE-754 floating point double precision for all \(k\) but
\(\binom{57}{25}\) is the
``first'' binomial coefficient that must be rounded.

In either case, computing
\(\widehat{\sigma} = (1 \ominus s) \oslash s\) or
\(\widehat{\sigma} = s \oslash (1 \ominus s)\) requires two floating
point operations, so
\(\widehat{\sigma} = \sigma\left(1 + \theta_2\right)\). This round-off
factor contributes to \(2n\) of the \(5n\) (or \(6n\)) in the coefficient
of \(\widetilde{p}(s)\).

When \(s < 1/2\), we apply Horner's method to the scaled Bernstein
coefficients, however when \(s \geq 1/2\) we reverse the order
before performing Horner's method. As a result, we refer to
\(c_k\) and \(\widetilde{c}_k\) instead of \(b_k\) and
\(\widetilde{b}_k\). In the \(s < 1/2\) case, \(c_k = b_k\) and
in the other \(c_k = b_{n - k}\).
When computing \(p(s)\), we start with \(\widehat{p}_n = c_n\)
and then for \(k = n - 1, \ldots, 0\):
\begin{equation}
\widehat{p}_k = \left[\widehat{\sigma} \otimes \widehat{p}_{k + 1}\right]
  \oplus \left[\binom{n}{k} \otimes c_k\right].
\end{equation}
This means that
\begin{equation}
\widehat{p}_k = \sigma \widehat{p}_{k + 1} (1 + \theta_4)
+ \widetilde{c}_k (1 + \theta_2)
\end{equation}
so that\footnote{We could ignore round-off from the multiplication by
\(\binom{n}{0}\), but we don't.}:
\begin{align}
\widehat{p}_0 &= \sigma \widehat{p}_1 (1 + \theta_4)
+ \widetilde{c}_0 (1 + \theta_2) \\
&= \sigma^2 \widehat{p}_2 (1 + \theta_8) +
  \sigma \widetilde{c}_1 (1 + \theta_6) +
  \widetilde{c}_0 (1 + \theta_2) \\

&\mathrel{\makebox[\widthof{=}]{\vdots}} \nonumber \\
&= \sigma^n \widehat{p}_n (1 + \theta_{4n}) +
  \sum_{j = 0}^{n - 1} \widetilde{c}_j \sigma^j (1 + \theta_{4j + 2}).
\end{align}
Defining \(\widehat{m}_0 = 1, \widehat{m}_1 = m\)
and \(\widehat{m}_{k + 1} = \widehat{m}_k \otimes m\) we'll have
\(\widehat{m}_n = m^n (1 + \theta_{n - 1})\).
Hence, in the final step of the VS algorithm we have
\begin{equation}
\widehat{m}_n \otimes \widehat{p}_0 =
m^n \widehat{p}_0 (1 + \theta_n) = m^n \left[
  \sigma^n \widetilde{c}_n (1 + \theta_{5n}) +
  \sum_{j = 0}^{n - 1} \widetilde{c}_j \sigma^j
  (1 + \theta_{4j + 2 + n})\right].
\end{equation}
When \(s < 1/2\), \(m = \widehat{r} = (1 - s)(1 + \theta_1)\), hence
\(m^n = (1 - s)^n (1 + \theta_n)\), \(\sigma = s/(1 - s)\) and
\begin{align}
\widehat{m}_n \otimes \widehat{p}_0 &= (1 - s)^n \left[
  \sigma^n \widetilde{b}_n (1 + \theta_{6n}) +
  \sum_{j = 0}^{n - 1} \widetilde{b}_j \sigma^j
  (1 + \theta_{4j + 2 + 2n})\right] \\
&= p(s) + (1 - s)^n \left[
  \sigma^n \widetilde{b}_n \theta_{6n} +
  \sum_{j = 0}^{n - 1} \widetilde{b}_j \sigma^j
  \theta_{4j + 2 + 2n}\right] \\
&= p(s) +
  b_n B_{n, n}(s) \theta_{6n} +
  \sum_{j = 0}^{n - 1} b_j B_{n - j, n}(s)
  \theta_{4j + 2 + 2n}
\end{align}
hence
\begin{equation}
\left|p(s) - \mathtt{VS}(p, s)\right| \leq \gamma_{6n}
  \widetilde{p}(s).
\end{equation}

When \(s \geq 1/2\), \(\sigma = (1 - s)/s\). Since \(m = s\), no extra
round-off accumulates in \(m^n\), but the order of the coefficients
is reversed:
\begin{align}
\widehat{m}_n \otimes \widehat{p}_0 &= s^n \left[
  \sigma^n \widetilde{b}_0 (1 + \theta_{5n}) +
  \sum_{j = 0}^{n - 1} \widetilde{b}_{n - j} \sigma^j
  (1 + \theta_{4j + 2 + n})\right] \\
&= p(s) + s^n \left[
  \sigma^n \widetilde{b}_0 \theta_{5n} +
  \sum_{j = 0}^{n - 1} \widetilde{b}_{n - j} \sigma^j
  \theta_{4j + 2 + n}\right] \\
&= p(s) +
  b_0 B_{0, n}(s) \theta_{5n} +
  \sum_{j = 0}^{n - 1} b_{n - j} B_{n - j, n}(s)
  \theta_{4j + 2 + n}
\end{align}
hence
\begin{equation}
\left|p(s) - \mathtt{VS}(p, s)\right| \leq \gamma_{5n}
  \widetilde{p}(s)
\end{equation}
as desired.
\end{proof}

\end{document}


TITLE: What does "maximal number of closed cutting curves that do not disconnect the graph into multiple components" mean?
QUESTION [0 upvotes]: What does "maximal number of closed cutting curves that do not disconnect the graph into multiple components" mean?
Particularly, what does closed cutting curves mean?

REPLY [1 votes]: The phrase refers to the topological genus of a surface... informally, the number of "holes" in the surface.
Consider a sphere (genus $0$).  Draw any closed cutting curve (a loop) on the surface of the sphere.  That "cut" will separate the points on the sphere inside the loop from those on the outside of the loop.  Such a curve "disconnects" the graph representing points on the sphere.  This is true no matter which closed loop you draw on the sphere.  Thus the "maximal" number having the property in the phrase is $0$... which is also the genus of a sphere.  (The same holds for any surface of genus $0$, such as a cube, an octohedron, a complicated sculpture without holes, ...)
Now consider a donut (genus $1$).  You can draw a closed loop encircling the "tube" of the donut and you will not "disconnect" the graph representing the points on the donut.  (Of course, if you drew a small closed loop on the surface, you could separate points on the donut, but the statement refers to the maximal number of such closed loops.) Once you cut the donut in this way, the resulting figure has genus $0$... and you cannot make another such closed loop cut and keep the figure connected.  Thus the maximum number of such loops you can create (keeping the figure connected) is $1$.
Now consider two donuts joined in the obvious way (genus $2$).  Now you can draw two separate closed loops and yet keep the surface (graph) connected.
And so forth.