TITLE: 3-term arithmetic progressions of terms as frequent as primes
QUESTION [5 upvotes]: Let $\ p_1\ < p_2 < \ldots\ $ be the sequence of all primes $\ (2\ 3\ 5\ \ldots)$.
Let $\ x_1 < x_2 < \ldots\ $ be an arbitrary increasing sequence of positive integers such that $\ x_n\le p_n\ $ for every $\ n=1\ 2\ldots\,$.
QUESTION: Does sequence $\ (x_1 < x_2 < \ldots)\ $ contain a 3-term arithmetic progression (of not necessary three consecutive members)? Does there exist an infinite number of such 3-term arithmetic progressions?
Acknowledgement: The simple version $\ x_n\le p_n\ $ of the assumption of this conjecture was provided by @zeb in a response to my equivalent original assumption which was clumsy and harder to read.

Reference: my MO-problem is related to a famous Klaus Roth's theorem.
Now I see from the Fedor's answer that this was indeed essentially only an MO-problem, and otherwise not essentially original.

REPLY [2 votes]: Meanwhile, this question has been answered in the positive. -- See
Thomas F. Bloom, Olof Sisask: Breaking the logarithmic barrier in Roth's theorem on arithmetic progressions.


text:
theory Bottom_Up_Computation_Heap
  imports "../state_monad/Bottom_Up_Computation" "../heap_monad/DP_CRelVH"
begin

definition (in iterator_defs)
  "iter_heap f \<equiv>
    wfrec
      {(nxt x, x) | x. cnt x}
      (\<lambda> rec x. if cnt x then do {f x; rec (nxt x)} else return ())"

lemma (in iterator) iter_heap_unfold:
  "iter_heap f x = (if cnt x then do {f x; iter_heap f (nxt x)} else return ())"
  unfolding iter_heap_def
  by (simp add: wfrec_fixpoint[OF iterator.wellfounded,OF iterator.intro,OF terminating] adm_wf_def)

locale dp_consistency_iterator_heap =
  dp_consistency_heap P update lookup dp + iterator cnt nxt sizef
  for lookup :: "'a \<Rightarrow> ('c option) Heap" and update and P dp
    and cnt :: "'a \<Rightarrow> bool" and nxt and sizef
begin

context
  includes lifting_syntax
begin

term iter_heap

term crel_vs

lemma crel_vs_iterate_state:
  "crel_vs (=) () (iter_heap f x)" if "((=) ===> crel_vs R) g f"
  using wellfounded
proof induction
  case (less x)
  have unit_expand: "() = (\<lambda> a f. f a) () (\<lambda> _. ())" ..
  from less show ?case
    by (subst iter_heap_unfold)
       (auto intro:
          bind_transfer[unfolded rel_fun_def, rule_format, unfolded unit_expand]
          crel_vs_return_ext[unfolded Transfer.Rel_def] that[unfolded rel_fun_def, rule_format]
       )
qed

lemma crel_vs_bind_ignore:
  "crel_vs R a (do {d; b})" if "crel_vs R a b" "crel_vs S c d"
proof -
  have unit_expand: "a = (\<lambda> a f. f a) () (\<lambda> _. a)" ..
  show ?thesis
    by (subst unit_expand)
       (rule bind_transfer[unfolded rel_fun_def, rule_format, unfolded unit_expand] that)+
qed

lemma crel_vs_iter_and_compute:
  assumes "((=) ===> crel_vs R) g f"
  shows "crel_vs R (g x) (do {iter_heap f y; f x})"
  by (rule
        crel_vs_bind_ignore crel_vs_iterate_state HOL.refl
        assms[unfolded rel_fun_def, rule_format] assms
     )+

lemma consistent_DP_iter_and_compute:
  assumes "consistentDP f"
  shows "consistentDP (\<lambda> x. do {iter_heap f y; f x})"
  apply (rule consistentDP_intro)
  using assms unfolding consistentDP_def Rel_def
  by (rule crel_vs_iter_and_compute)

end (* Lifting Syntax *)

end (* DP Consistency Iterator Heap *)

end (* Theory *)

subset_name: curated


TITLE: Cumulative distribution function question
QUESTION [1 upvotes]: Possible Duplicate:
Normal random variable $X$ and the cdf of $Y=aX+b$

I'm given a standard random variable $X$, and $Y = aX + b$:
How can I find the cumulative distribution function for Y as an integral of $f(x)=(\frac{1}{\sqrt{2}\pi\sigma}e^{-\frac{(x-u)^2}{\sigma^2}}$?
I know $F_y(y)=F_x(\frac{y-b}{a})$, but cant figure out where to go from there.

REPLY [0 votes]: If $a>0$ then
$$
\begin{align}
& f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy}\Pr(Y\le y) = \frac{d}{dy}\Pr(aX+b\le y) = \frac{d}{dy}\Pr\left(X \le \frac{y-b}{a}\right) \\  \\
& = \frac{d}{dy}F_X\left( \frac{y-b}{a} \right) = f_X\left(\frac{y-b}{a}\right) \cdot \frac{d}{dy} \frac{y-b}{a} = \frac 1 a f_X\left(\frac{y-b}{a}\right) \\  \\  \\
& = \frac 1 a \cdot \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-1}{2} \cdot \left(\frac{y-b}{a}\right)^2\right).
\end{align}
$$
Therefore
$$
F_Y(y) = \frac 1 a \int_{-\infty}^y \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-1}{2} \left(\frac{u-a}{a}\right)^2\right)\; du.
$$
The equivalence between $aX+b\le y$ and $X \le \frac{y-b}{a}$ holds only if $a>0$.


TITLE: Conditions - Linear Combination of 3 Vectors inside a Triangle (Strang P10, 1.1.20)
QUESTION [3 upvotes]: Under what restrictions on $c, d, e,$ will the combinations $c\mathbf{u} + d\mathbf{v} + e\mathbf{w}$ fill in [ie bridle/rein in] the dashed triangle? To stay in the triangle, one requirement is $c \geq 0 \; \& \; d \geq 0 \; \& \; e \geq 0.$
Answer: To fill the triangle keep $c, d, e \geq 0$, and $c + d + e = 1.$

Could someone please demystify how and why $c + d + e = 1$ ? Where did it even loom from?

Supplement to Revanth Kashyap's Answer:
How and why:

¿ If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are coplanar and $k_1\mathbf{u} + k_2\mathbf{v} + k_3\mathbf{w} = 0$, then $ k_1 + k_2 + k_3 = 0$ ?

REPLY [2 votes]: Shift the Origin to the heads of any of the three vectors, say w. The original vectors now become v-w, u-w, and $\mathbf 0$(Zero Vector).
Now, in  order that a vector should lie in the triangle thus formed, it should be of the form :
$\lambda$(v-w)+$\mu$(u-w), with $\lambda \ge 0$, $\mu$$ \ge$$ 0$ and $\lambda+\mu\le1$
Simplifying the above expression, we get :
$\mu\mathbf{u}$+$\lambda\mathbf{v}$-$(\lambda$+$\mu)\mathbf w$
Reverting to the original Origin, the expression becomes :
$\mu\mathbf{u}$+$\lambda\mathbf{v}$-$(\lambda$+$\mu)\mathbf w$ + $\mathbf w$
i.e., $\mu\mathbf{u}$+$\lambda\mathbf{v}$+$(1-(\lambda$+$\mu))\mathbf w$
Comparing it to $c\mathbf{u} + d\mathbf{v} + e\mathbf{w}$,  we get :
$\mu$ = c,  and as $\mu \ge 0$, c $\ge 0$
$\lambda$ = d,  and as $\lambda \ge 0$, d $\ge 0$
$1-(\lambda$+$\mu)$ = e, and as $\lambda+\mu\le1$, e $\ge 0$
From the above three facts we get, c + d + e = 1 and c $\ge 0$, d $\ge 0$, and e $\ge 0$


TITLE: Let $g_n(x)= g(x+\frac{1}{n})$. Prove that $(g_n)$ converges uniformly to $g$.
QUESTION [2 upvotes]: Let $g: \mathbb{R}\longrightarrow \mathbb{R}$ be a uniformly continuous function and, for every $n\in\mathbb{N}$, let $g_n(x)=
g(x+\frac{1}{n})$. Prove that $(g_n)$ converges uniformly to $g$.
My try:
Because $g$ is uniformly continous then we know that $\forall \epsilon>0 \quad\exists \delta>0$ tal que $\forall x,y\in \mathbb{R}, |x-y|<\delta$, then $|g(x)-g(y)|<\epsilon$. Now let's take $n\in\mathbb{N}$ and $x+\frac{1}{n}, x \in \mathbb{R}$. Then $|x+\frac{1}{n}-x|<\delta$, then $|g(x+\frac{1}{n})-g(x)|<\epsilon$. Then $|g_n(x)-g(x)|<\epsilon$, $\forall x\in\mathbb{R}$.
I'm not sure if my poof is correct. Any suggestions would be great!

REPLY [3 votes]: Since $g$ is uniformly continuous, given $\varepsilon>0$ there exists $\delta>0$ such that
$$\forall x, y\in \mathbb R; |x-y|<\delta\Rightarrow |g(x)-g(y)|<\varepsilon.$$
By the Archimedian property of $\mathbb R$ we can find $n_0\in\mathbb N$ such that $n_0\cdot \delta>1$. Then,
$$n\geq n_0 \Rightarrow \left|x-\frac{1}{n}-x\right|=\frac{1}{n}\leq \frac{1}{n_0}<\delta.$$
Hence,
$$n\geq n_0\Rightarrow |g_n(x)-g(x)|<\varepsilon$$
for all $x\in \mathbb R$. Since this holds for every $x\in\mathbb R$, it follow that
$$n\geq n_0 \Rightarrow \sup_{x\in\mathbb R} |g_n(x)-g(x)|\leq \varepsilon.$$
This means, $(g_n)_{n\in\mathbb N}$ is uniformly continuous.


text:
\begin{document}
\pagestyle{fancy}
\fancyhead[RE,LO]{\color{mygray}Preprint of ISIT paper \#1061 --- Andreas Schenk, Robert~F.H.~Fischer:\\Soft-Output Sphere Decoder for Multiple-Symbol Differential Detection of Impulse-Radio Ultra-Wideband}
\fancyfoot[LE,RO]{\color{mygray}Andreas~Schenk and Robert~F.~H.~Fischer, \today}

\maketitle

\begin{abstract}

Power efficiency of noncoherent receivers for impulse-radio ultra-wideband (IR-UWB) transmission systems can significantly be improved, on the one hand, by employing multiple-symbol differential detection (MSDD), and, on the other hand, by providing reliability information to the subsequent channel decoder.
In this paper, we combine these two techniques.
Incorporating the computation of the soft information into a single-tree-search sphere decoder (SD), the application of this soft-output MSDD in a typical IR-UWB system imposes only a moderate complexity increase at, however, improved performance over hard-output MSDD, and in particular, over conventional symbol-by-symbol noncoherent differential detection.
\end{abstract}

\section{Introduction}
\label{sec:intro}

Impulse-radio ultra-wideband (IR-UWB) is widely considered as a promising technique for low-power low-cost short-range wireless communication systems.
One of the main reasons for this is its potential to employ noncoherent, hence low-complexity, receivers even in dense multipath propagation scenarios, where channel estimation required for coherent detection would be overly complex due to the high multipath resolution and relatively large delay spread of UWB signals.

The performance penalty between coherent and noncoherent detection in power efficiency, i.e., in the required signal-to-noise ratio to guarantee a certain bit error rate (BER), can be closed by replacing conventional symbol-by-symbol noncoherent detection with a joint detection of a block of symbols, i.e., performing multiple-symbol differential detection (MSDD) \cite{me:IZS2010,UWB:Guo:MSDD}.
In \cite{UWB:Lottici:MSDD} it has been shown that the underlying tree search problem is efficiently solved using the sphere decoder (SD) algorithm.
However, noncoherent (MSDD-based) IR-UWB receiver design has mainly considered uncoded transmission systems, cf., e.g., \cite{UWB:Chao:TR,UWB:Lottici:MSDD,UWB:Guo:MSDD}.

In this paper, we consider coded IR-UWB transmission employing differentially encoded BPSK (also known as differential transmitted reference (DTR)).
To keep transmitter and receiver design simple, we restrict to the conventional serial concatenation of modulation and coding at transmitter, and detection and decoding at receiver side, i.e., restrain to the bit-interleaved coded modulation (BICM) philosophy.
Employing coding, power efficiency can significantly be improved by delivering reliability information to the soft-input channel decoder \cite{book:WozencraftJacobs:Principles}.
Borrowing from techniques recently introduced for SD-based soft output generation in multiple-input/multiple-output (MIMO) systems \cite{SISOSD:StuderBurgBolcskei:SOSD}, we extend the SD-based MSDD algorithm presented in \cite{UWB:Lottici:MSDD} to incorporate also soft output computation.

To this end, in Section~\ref{sec:ir}, we derive log-likelihood ratios (LLR) for MSDD of IR-UWB based on generalized-likelihood ratio testing (GLRT), and formulate their computation as a tree search problem.
Using the soft-output sphere decoder (SOSD) the  LLRs can be found in a single tree search, thus receiver complexity is increased only moderately, especially in comparison to repeated-tree-search approaches, cf. \cite{SISOSD:StuderBurgBolcskei:SOSD}.

In Section~\ref{sec:results}, we investigate the performance of the proposed MSDD-based soft-output IR-UWB receiver and study the tradeoff between performance and complexity obtained by adjusting the channel code, the MSDD block size, and techniques for SD complexity reduction, such as a stopping criterion \cite{me:ICUWB09} and LLR clipping \cite{SISOSD:StuderBurgBolcskei:SOSD}.
We conclude with a summary in Section~\ref{sec:conclusions}.

\section{Impulse-Radio Ultra-Wideband Transmission}
\label{sec:ir}

\subsection{System Model}
\label{sec:ir:system}

The receive signal of differentially encoded BPSK IR-UWB is given as (cf.\ Fig.~\ref{fig:UWB:ACR})

\begin{align}
        r(t) = \sum_{i=0}^{+\infty} b_i p (t-i\T) + n(t)
        \label{eq:UWB:RX}
\end{align}

where $b_i$ are the differentially encoded, interleaved ($\Pi$) and mapped ($\mathcal{M}$) output symbols $a_i\in\mathcal{A} = \{\pm1\}$ of a channel encoder, such that $b_i = b_0 \prod_{k=1}^{i}a_k$, with $b_0 = 1$, and $\T$ is the symbol duration.
The receive pulse shape $p(t)$ is obtained from the convolution of transmit pulse, receive filter, and channel impulse response, i.e., $p(t) = \hCH(t) * \hRX(t) * \pTX(t)$.
The pulse energy is normalized to one and thus, the energy per bit is given by $\Eb = 1$.
$n(t)$ is white Gaussian noise of two-sided power-spectral density $\No/2$ filtered by $\hRX(t)$.
To preclude inter-symbol interference, the symbol duration $\T$ is chosen sufficiently large, such that each pulse has decayed before the next pulse is received.

Note that the usually applied frame structure used for time-hopping and code-division multiple access \cite{UWB:Win:IR,UWB:Win:TH} is not explicitly taken into account, as it can be regarded as additional linear block coding, or removed prior to further receive signal processing \cite{UWB:Lottici:MSDD}.

\begin{figure*}[!t]
\centering

{\footnotesize
        \input{UWB_LbranchACR.pstex_t}}
\caption{Block diagram of the coded IR-UWB transmission system with $L$-branch ACR and soft-output MSDD.}
\vspace*{-3mm}
\label{fig:UWB:ACR}
\end{figure*}

\subsection{Soft Output Generation}

The reliability information for a single information symbol, which is passed as soft input to the channel decoder, is calculated based on the observation of the receive signal in an $(L+1)$-symbols interval $0\leq t< (L+1)\T$
(without loss of generality we consider the interval starting at $t=0$).
The channel is assumed to be constant in this interval, which, in typical indoor UWB communication scenarios, is fulfilled especially for moderate $L$ \cite{UWB:Molisch:Channel}.

Due to the unknown statistics of the receive pulse shape $p(t)$, we revert to the GLRT approach, thus, in contrast to the ML criterion, include an explicit optimization over all finite-energy pulse shapes $\tilde{p}(t)$ of some assumed duration $\Ti$ \cite{UWB:Lottici:MSDD,UWB:Chao:TR,me:IZS2010}.
In terms of LLRs for the $i^{\mathrm{th}}$ symbol, $i=1,...,L$, this reads

\begin{align}
        \LLR_i &= \log\left( \frac{\max_{\tilde{p}(t)} \Pr\{\tilde{a}_i = +1 | r(t),\tilde{p}(t)\} }{\max_{\tilde{p}(t)} \Pr\{\tilde{a}_i = -1 | r(t),\tilde{p}(t)\}}\right)\;.
\label{eq:LLR_first}
\end{align}

Defining the vector of the information symbol hypotheses ${\veat} = [\tilde{a}_1, ...,\tilde{a}_L]$, applying Bayes' rule, and assuming independent i.i.d.\ information symbols, i.e., a-priori information is not considered as feedback from the channel decoder, we have

\begin{align}
        \LLR_i &= \log\left( \frac{\mathop{\sum_{\veat\in\mathcal{A}^L}}_{,\tilde{a}_i = +1} \max_{\tilde{p}(t)} \f(r(t)|\veat,\tilde{p}(t))}
                                  {\mathop{\sum_{\veat\in\mathcal{A}^L}}_{,\tilde{a}_i = -1} \max_{\tilde{p}(t)} \f(r(t)|\veat,\tilde{p}(t))}\right)
\label{eq:LLR_second}
\end{align}

where $\f(r(t)|\veat,\tilde{p}(t))$ is the probability density function of the additive noise in (\ref{eq:UWB:RX}).
As $\f(r(t)|\veat,\tilde{p}(t)) \geq 0$, maximization and summation could be interchanged.

In a first step, we perform the maximization over the unknown pulse shape, applying similar steps as shown in \cite{UWB:Lottici:MSDD}.
This is summarized, very briefly, as follows.

We define the noise-free receive signal hypothesis $s(t|\veat,\tilde{p}(t)) = \sum_{i=0}^{L} b_0 \prod_{k=1}^{i} \tilde{a}_k \tilde{p}(t-i\T)$, including the  receive pulse shape hypothesis $\tilde{p}(t)$.
Since additive Gaussian noise is assumed, which is bandlimited by a receive filter with equivalent noise bandwidth $B_{\mathsf{eq}}$, and relying on the equivalence of distance in signal space and energy of the difference signal \cite{book:Proakis:DigitalCommunications}, we arrive at the joint probability density function

\begin{align}
        \f(r(t)|\veat,\tilde{p}(t))
      &= c_{\mathsf{f}} \cdot\e^{-\frac{1}{2\sigma_n^2}\int_0^{(L+1)\T} \left( r(t) - s(t|\veat,\tilde{p}(t))\right)^2\dx t}
\label{eq:pdf}
\end{align}

where $\sigma_n^2 = \No B_{\mathsf{eq}}$, and $c_{\mathsf{f}}$ is an irrelevant constant, which cancels in (\ref{eq:LLR_second}).
Applying variational calculus and using the fact that $\tilde{a}_i \in\{\pm1\}$, the optimizing pulse shape is found to be $p_{\mathsf{opt}}(t) = \frac{1}{L+1}\,\sum_{i=0}^{L} b_0\prod_{k=1}^{i}\tilde{a}_i\,r(t+i\T)$ (cf. \cite{UWB:Lottici:MSDD} for a similar derivation). Thus,
\begin{align}
      \max_{\tilde{p}(t)}\f(r(t)|\veat,\tilde{p}(t)) &= c_{\mathsf{f}} \cdot \e^{-\frac{-2\,\Gamma(\veat) - \sum_{i=0}^{L} Z_{i,i}}{2\sigma_n^2(L+1)} }
\label{eq:maxpdf}
\end{align}

where we defined

\begin{align}
        \Gamma(\veat) =  \sum_{i=1}^{L}\sum_{l=0}^{i-1}\prod_{k=l+1}^{i} \tilde{a}_k Z_{l,i}
\label{eq:Gamma}
\end{align}

and, for $i=1,...,L$, $l = 0,...,i-1$,

\begin{align}
        Z_{l,i} &= \int_{0}^{\Ti} r(t+i\T)\cdot r(t+l\T) \dx t\;.

\label{eq:definition_Z}
\end{align}

Using the result in (\ref{eq:maxpdf}), as well as applying the max-log approximation, (\ref{eq:LLR_second}) can be approximated by

\begin{align}

\LLR_i ={}&\frac{1}{\sigma_n^2(L+1)} \left[\mathop{\max_{\veat\in\mathcal{A}^L}}_{\tilde{a}_i = +1} \Gamma(\veat) - \mathop{\max_{\veat\in\mathcal{A}^L}}_{\tilde{a}_i = -1}\Gamma(\veat)\right]\;.
\label{eq:LLR_third}
\end{align}

\subsection{Autocorrelation-Based Detection}
\label{sec:ir:sodd}

Before we turn to an efficient implementation of the computation required for (\ref{eq:LLR_third}) based on the SD algorithm, we note that $Z_{l,i}$, defined in (\ref{eq:definition_Z}), corresponds to the output of an $L$-branch autocorrelation receiver (ACR, shown in Fig.~\ref{fig:UWB:ACR}) with delays being multiples of $\T$ and integration interval $\Ti$, set in the order of the expected receive pulse duration.
$Z_{l,i}$ represents the phase transition from $b_l$ to $b_i$ superposed by an ``information $\times$ noise'' and ``noise $\times$ noise'' term.

Clearly, restricting to a single-branch ACR, i.e., $L=1$, corresponds to symbol-by-symbol differential detection (DD).
In this case, it can directly be seen that the hard-quantized ACR output gives the DD estimate, i.e.,  $a_i^{\mathsf{DD}} = \sign\left(Z_{i-1,i}\right)$.
However, using (\ref{eq:LLR_third}) with $L=1$, also the rather intuitive result follows, to use the unquantized ACR output for soft-output DD, in particular $\LLR_i^{\mathsf{DD}} = \frac{1}{\sigma_n^2\cdot2} [2 Z_{i-1,i}]$.

\subsection{Soft-Output Sphere Decoder (SOSD)}
\label{sec:ir:sosd}

For efficient implementation based on the SD algorithm, we reformulate the maximization problems in (\ref{eq:LLR_third}) into equivalent minimization problems.
Since $\tilde{a}_i \in \{\pm1\}$, $\Gamma(\veat) \leq \sum_{i=1}^{L}\sum_{l=0}^{i-1}|Z_{l,i}|$ holds $\forall \veat$. Subtracting this upper bound from both objective functions in (\ref{eq:LLR_third}) yields

\begin{align}
        \LLR_i ={} \frac{1}{\sigma_n^2(L+1)} \left[
\mathop{\min_{\veat\in\mathcal{A}^L}}_{\tilde{a}_i = -1}  \Lambda(\veat) -\mathop{\min_{\veat\in\mathcal{A}^L}}_{\tilde{a}_i = +1}  \Lambda(\veat) \right]\;.
\label{eq:LLR_four}
\end{align}

Note that $\max(x-y) = -\min(y-x)$, and

\begin{align}
        \Lambda(\veat) = \sum_{i=1}^{L}\sum_{l=0}^{i-1} |Z_{l,i}|\left(1-\sign(Z_{l,i}) \prod_{k=l+1}^{i} \tilde{a}_k \right)\;.
\label{eq:Lambda}
\end{align}

Eventually, it can be seen that the addends of the outer sum in (\ref{eq:Lambda}) are always non-negative and depend solely on the first $i$ (preliminary) decisions of information symbols $\tilde{a}_k$, $k=1,...,i$.
This allows to check the decision metric componentwise, and thus each of the two minimization problems in (\ref{eq:LLR_four}) can be solved using the SD operating on an $L$-dimensional binary tree (see Fig.~\ref{fig:UWB:pseudocode} and also \cite{UWB:Lottici:MSDD} for details).

Clearly, one of the two minima in (\ref{eq:LLR_four}) is the GLRT-optimal metric

\begin{align}
        \Lambda^\GLRT = \Lambda(\ve{a}^\GLRT)= \min_{\veat\in\mathcal{A}^L} \Lambda(\veat)
\label{eq:}
\end{align}

whereas the other one is obtained from the corresponding counter hypothesis, i.e., the minimum metric with the restriction $\tilde{a}_i = -a_i^\GLRT$, such that

\begin{align}
        \Lambda_i^{\overline{\GLRT}} =

\min_{\veat\in\mathcal{A}^L,\, \tilde{a}_i = -{a}_i^\GLRT} \Lambda(\veat)\;.
\label{eq:UWB:Lambdacounter}
\end{align}

Consequently, we have

\begin{align}
        \LLR_i =

\frac{1}{\sigma_n^2(L+1)}\, \left[ a_i^\GLRT\,\left( \Lambda_i^{\overline{\GLRT}} - \Lambda^{{\GLRT}}\right) \right]
\;.
\label{eq:UWB:LLR_cases}
\end{align}

\subsubsection{Single-Tree-Search SOSD}

Calculating the LLRs resorts to finding the minimum of an unrestricted tree search, the corresponding sequence, and $L$ ``next-best'' minima.
One could solve these minimization problems subsequently by rerunning the SD for each counterhypothesis with correspondingly restricted search space.
This requires to run the SD $L+1$ times per block of $L$ information symbols, and hence, imposes a high complexity burden.

This can be alleviated by a modified SD algorithm, as introduced in \cite{SISOSD:JaldenOtersten:ParallelSOSD}, and further refined in \cite{SISOSD:StuderBurgBolcskei:SOSD}, for soft-output signal detection in MIMO systems, which ensures that every node in the search tree is visited at most once.
We incorporate these MIMO-SD modifications into the SD for MSDD of IR-UWB as described in \cite{UWB:Lottici:MSDD} (for a detailed description of the MIMO-SOSD, cf.\ \cite{SISOSD:StuderBurgBolcskei:SOSD}).
Thus, the required $\Lambda^{{\GLRT}}$, $\Lambda_i^{\overline{\GLRT}}$, and $\ve{a}^\GLRT$ result from a single tree search process.

First, the SD search radius $R$ is not updated, whenever a new (preliminary) best sequence has been found, but the search radius update is based on the current values $\Lambda^\GLRT$ and $\Lambda_i^{\overline{\GLRT}}$, $i=1,...,L$. It is set such, that only branches in the search tree are considered, which can lead to an update of either $\Lambda^\GLRT$, or $\Lambda_i^{\overline{\GLRT}}$, $i=1,...,L$.
This is achieved by setting

\begin{align}
        R = \max \bigg\{\max_{k = i,...,L} \Lambda_k^{\overline{\GLRT}},~\mathop{\max_{l=1,...,i-1}}_{\mathrm{with}~\tilde{a}_l \neq a_l^\GLRT} \Lambda_l^{\overline{\GLRT}}\bigg\} \;.
\label{eq:UWB:setSDsearchradius}
\end{align}

Further, in the case a sequence $\veat$ with path metric $\Lambda(\veat)$ is investigated, i.e., a leaf in the search tree has been reached, two cases are distinguished:\\

(i) If $\Lambda(\veat) < \Lambda^\GLRT$, a new (preliminary) best sequence has been found. Then all $\Lambda_i^{\overline{\GLRT}}$, $i=1,...,L$, where $\tilde{a}_i = -a_i^\GLRT$, are set to $\Lambda^\GLRT$, followed by the usual SD update of the current best sequence $\ve{a}^\GLRT := \veat$ and metric $\Lambda^\GLRT := \Lambda(\veat)$.\\

(ii) If $\Lambda(\veat) \geq \Lambda^\GLRT$, only the counterhypotheses have to be checked, i.e., all $\Lambda_i^{\overline{\GLRT}}$, $i=1,...,L$, where $\tilde{a}_i = -a_i^\GLRT$ and $\Lambda_i^{\overline{\GLRT}} > \Lambda(\veat)$, are set to $\Lambda(\veat)$.

\subsubsection{SOSD Complexity Reduction}

A reasonable measure for the search complexity of the SD is the number of visited nodes $C_{\mathsf{SD}}$ in the search tree during the tree search process, which is directly related to hardware implementation complexity, cf., e.g., \cite{SISOSD:StuderBurgBolcskei:SOSD}.
In this paper, we adopt this complexity measure.

In \cite{me:ICUWB09} we introduced a packing-radius-based stopping criterion for the SD in hard-output MSDD of IR-UWB, which reduces the average search complexity, yet ensures optimality of the estimated sequence.
We also apply this stopping criterion for the SOSD.
If any preliminary sequence $\veat$ with path metric $\Lambda(\veat)$ fulfills

\begin{align}
        \Lambda(\veat) \leq \Rstop = L \cdot \mathop{\min_{i=1,...,L}}_{l=0,...,i-1} |Z_{l,i} |
\label{eq:UWB:stopcrit}
\end{align}

the search process is terminated early.
In \cite{me:ICUWB09} it has been shown, that this setting

guarantees to find the GLRT-optimal sequence $\ve{a}^\GLRT$ and $\Lambda^\GLRT$.
However, this does not ensure to find the optimal solution to (\ref{eq:UWB:Lambdacounter}), i.e., $\Lambda_i^{\overline{\GLRT}}$, and hence the correct max-log-approximated LLRs.
However, as will be shown later, this stopping criterion enables a reduction in the average search complexity at only minor performance degradation.

As shown in \cite{SISOSD:StuderBurgBolcskei:SOSD}, a crucial part for complexity reduction of the SOSD in MIMO detection, is to limit the maximum LLR values during the SD search process.
For MIMO detection this LLR clipping enables a tradeoff between the power efficiency of optimal soft-output detection and the complexity of hard-output detection.
Since we aim for a similar tradeoff, we apply this LLR clipping with maximum LLR value $\LLRmax$ during the search process, too.
Thus, after each update of the counterhypotheses metrics $\Lambda_i^{\overline{\GLRT}}$, those are limited to

\begin{align}
        \Lambda_i^{\overline{\GLRT}} = \max\bigg\{\Lambda_i^{\overline{\GLRT}}, \Lambda^{{\GLRT}}+\sigma_n^2(L+1)\LLRmax\bigg\},\;\forall i\;.\raisetag{2mm}
\label{eq:UWB:LLRclipping}
\end{align}

From (\ref{eq:UWB:setSDsearchradius}) it can be seen that this LLR clipping limits the search radius to $R \leq \Lambda^{{\GLRT}}+\sigma_n^2(L+1)\LLRmax$ and, together with (\ref{eq:UWB:LLR_cases}), ensures that $|\LLR_i| \leq \LLRmax$ after the SD search.
Clearly, with $\LLRmax = 0$ hard-output SD-based MSDD (HOSD) results.

\subsubsection{SOSD Algorithm}

The resulting algorithm, based on the SD for MSDD of IR-UWB \cite{UWB:Lottici:MSDD}, is given in pseudo-code representation in Fig.~\ref{fig:UWB:pseudocode}.
The modifications for soft output generation, LLR clipping (cf.\ line {19}), and the stopping criterion (cf.\ line {15}), are highlighted in gray shading.
Note that this pseudo-code representation is based on the pseudo-code representation of the SD for MSDD of DPSK given in \cite{MSDD:Lampe:MSDSD}.
We introduced the counter $n_i$, which is used to check if the two branches emanating from each node have been checked, the branch metric

\begin{align}
        \delta = \sum_{l=0}^{i-1} \left( \left| Z_{l,i} \right| \, \left( 1 - \sign\left(Z_{l,i}\right) \, \prod_{k=l+1}^{i} a_{k}   \right) \right)
\label{eq:UWB:branchmetric}
\end{align}

and the path metric $\pathMw{i}$ at a node in depth $i$.
We used the short-hand writing $\Lambda^{\overline{\GLRT}}_i = \Lambda_i$ and $\Lmax = \sigma_n^2(L+1)\LLRmax$, and omitted the tilde for a hypothesis.
The sub-functions \texttt{findbest} and \texttt{findnext} have been introduced (f1 and f2), to account for branch enumeration according to the Schnorr-Euchner search strategy \cite{SD:Agrell:Closest}.

\begin{figure}[!t]

        \input{pseudocode_UWB}
\caption{Pseudo-code representation of the SOSD algorithm for soft-output MSDD of IR-UWB. Gray-shaded: modifications for soft output generation.}
\vspace*{-5mm}
\label{fig:UWB:pseudocode}
\end{figure}

\section{Performance and Complexity}
\label{sec:results}

We evaluate the performance and complexity of the proposed coded IR-UWB transmission system in a typical UWB scenario, where we assume no inter-symbol interference ($\T$ chosen sufficiently large),
and $\pTX(t)$ is a Gaussian monocycle with $2.25\GHz$ center frequency and $10\,\dB$ bandwidth of $3.3\GHz$.
The propagation channel is modeled according to IEEE-CM\,2 \cite{UWB:Molisch:Channel} with each realization normalized to unit energy.
The receive filter is matched to the transmit pulse shape and a good compromise for the integration time is $\Ti = 30\ns$.
We use maximum-free-distance rate-1/2 convolutional codes with $2^\nu$ states, an interleaver size of 1000 information bits, and the Viterbi algorithm as (soft-input) channel decoder \cite{book:Proakis:DigitalCommunications}.

Fig.~\ref{fig:UWB:BER} shows the BER performance of the coded IR-UWB system employing the proposed SOSD-based MSDD.
With increasing MSDD block size $L$ performance improves compared to DD and approaches (hard-decision) coherent detection.
Soft-output MSDD (using SOSD) achieves an additional gain of up to  $1\;\dB$ over hard-output MSDD ($\LLRmax = 0$).

\begin{figure}[!t]
\centering
\psfrag{xlabel}[]{$10\log(\Eb/\No)~[\dB]$ $\rightarrow$}
\psfrag{ylabel}[]{$\BER$ $\rightarrow$}
\psfrag{L=2}[cl][.99]{\hspace*{-3ex}\small$L\hspace*{-.4ex}=2$}
\psfrag{L=5}[cl][.99]{\hspace*{-3ex}\small$L\hspace*{-.4ex}=5$}
\psfrag{L=10}[cl][.99]{\hspace*{-3.5ex}\small$L\hspace*{-.4ex}=\hspace*{-.3ex}10$}
\psfrag{L=15}[cl][.99]{\hspace*{-3.5ex}\small$L\hspace*{-.4ex}=\hspace*{-.3ex}15$}
\psfrag{L=20}[cl][.99]{\hspace*{-3.5ex}\small$L\hspace*{-.4ex}=\hspace*{-.3ex}20$}
\psfrag{L=25}[cl][.99]{\hspace*{-3.5ex}\small$L\hspace*{-.4ex}=\hspace*{-.3ex}25$}

        \includegraphics[width = .9\columnwidth]{BER_v6}
\caption{BER vs.\ $\Eb/\No$ in $\dB$ of coded IR-UWB with soft-output MSDD using the SOSD without (red) and with stopping criterion (\ref{eq:UWB:stopcrit}) (dashed green), and hard-output MSDD (blue) for different $L$, in comparison to soft and hard DD ($L=1$, black, dashed black), coherent detection (left most, black), and uncoded transmission (coh.\ and DD: dash-dotted gray). Rate-$1/2$ conv. code with $\nu = 6$, $\LLRmax = 10$, IEEE-CM\,2.}
\vspace*{-4mm}
\label{fig:UWB:BER}
\end{figure}

Adjusting the LLR clipping level $\LLRmax$ in SOSD, a tradeoff between the power efficiency of GLRT-optimal max-log-approximated soft-output MSDD and hard-output MSDD is achieved, which almost continuously traverses the performance-complexity plane (cf.\ Fig.~\ref{fig:UWB:tradeoff}).
Additionally setting the stopping criterion (cf.\ (\ref{eq:UWB:stopcrit}) and line 15 in Fig.~\ref{fig:UWB:pseudocode}) to enable early termination of the SOSD search process, especially for small to moderate $L$ reduces the average SD complexity at only minor loss in performance.

\begin{figure}[!t]
\centering
\psfrag{L2}[b]{\small$L=2$}
\psfrag{L3}[b]{\small$L=3$}
\psfrag{L4}[b]{\small$L=4$}
\psfrag{L5}[b]{\small$L=5$}
\psfrag{L6}[b]{\small$L=6$}
\psfrag{L8}[bl]{\small$L=8$}
\psfrag{L10}[bl]{\small$L=10$}
\psfrag{L12}[b]{\small$L=12$}
\psfrag{L15}[b]{\small$L=15$}
\psfrag{LLRmax}[bl]{\rotatebox{-20}{\small$\leftarrow\LLRmax\hspace*{-.8ex}\uparrow$}}
\psfrag{xlabel}[]{$\Eb^{\mathsf{req}}/\No$ $[\dB]$ $\rightarrow$}
\psfrag{ylabel}[]{avg. $C_{\mathsf{SD}}$ $\rightarrow$}

        \includegraphics[width = .9\columnwidth]{tradeoff_LLRmax_v6}
\caption{Tradeoff performance vs.\ average SD complexity at $\BER = 10^{-3}$ for LLR clipping levels $\LLRmax = [0.05\;0.1\;0.25\;0.5\;1\;2\;10]$ (right to left) of MSDD of IR-UWB using the SOSD (red), the SOSD with stopping criterion (\ref{eq:UWB:stopcrit}) (green), and hard-output MSDD (blue) with different MSDD block sizes $L$. Rate-$1/2$ conv. code with $\nu = 6$, IEEE-CM\,2.}
\vspace*{-5mm}
\label{fig:UWB:tradeoff}
\end{figure}

However, using the (soft-input) Viterbi algorithm for channel decoding imposes an increase in overall receiver complexity depending mainly on the number of states $2^\nu$ of the convolutional code \cite{book:Proakis:DigitalCommunications}.
Hence, the question for the optimum
 tradeoff between power efficiency and overall receiver complexity, obtained from adjusting the major parameters code constraint length, MSDD block size, and LLR clipping level, arises.

Since the proposed SOSD requires no multiplications  (note that $\tilde{a}_i \in\{\pm1\}$), we assume that soft-input Viterbi decoding and the SOSD for MSDD are comparable with respect to complexity in the number of considered nodes in the trellis, respectively the binary search tree.
Hence, the overall complexity per processed symbol is $C_{\mathsf{o}} = 2^{\nu} + C_{\mathsf{SD}}/L$, where the first summand is the (fixed) complexity of the Viterbi algorithm and the second summand represents the (varying) SD complexity.
As the SD complexity depends on the ACR output, we denote the average and maximum overall complexity as $C_{\mathsf{o}}^{\mathsf{soft}}$ and $C_{\mathsf{o}}^{\mathsf{max}}$, respectively.
The worst-case or maximum SD complexity is $C_{\mathsf{SD}}^{\mathsf{max}} = \sum_{i=1}^{L} 2^i = 2^{L+1}-2$.

We compare the proposed soft-output MSDD system with a reference system employing a $2^{\nu_{\mathsf{ref}}}$-states convolutional code and symbol-by-symbol ACR-based DD (MSDD with $L=1$) with overall complexity $C_{\mathsf{o}}^{\mathsf{ref}} = 2^{\nu_{\mathsf{ref}}} + 1$.
For soft-output MSDD, at each block size the best setting $\nu,\LLRmax$ is chosen (i.e., lowest $\EbNo$ for a desired $\BER$), that still has less overall complexity than the reference system (i.e., $C_{\mathsf{o}}^{\mathsf{soft}} \leq C_{\mathsf{o}}^{\mathsf{ref}}$).
For $\nu_{\mathsf{ref}} = 7$ and $\BER = 10^{-3}$ this results in the trajectories

shown in Fig.~\ref{fig:UWB:ctotal} (chosen setting indicated
in top part, similar results are obtained for other references $C_{\mathsf{o}}^{\mathsf{ref}}$).
The bottom part depicts the average (solid) and maximum (dashed) overall complexity using the SOSD (red), and the reference $C_{\mathsf{o}}^{\mathsf{ref}}$ (dashed black).
The top part shows the corresponding required $\Eb/\No$ in $\dB$. For comparison hard-output MSDD is included (blue).

The proposed soft-output MSDD in combination with convolutional codes with $2^\nu < 2^{\nu_{\mathsf{ref}}}$ states can---up to MSDD block sizes $L\leq16$---beat the reference system, i.e., DD and $\nu_{\mathsf{ref}} =7$.
The lowest $\Eb/\No$, with approximately $4\;\dB$ gain over the reference, is achieved for $L=12$.
However, for $L\geq15$ the complexity of the reference system is only undercut by falling back to hard-decision decoding ($\LLRmax = 0$).
With respect to the maximum overall complexity, soft-output MSDD is only better than the reference system up to $L=8$.

As for each MSDD block size the required $\Eb/\No$ of hard-output MSDD is higher than that of soft-output MSDD, at lower average, but equal maximum complexity, hard-output MSDD does not utilize the (anyway to be reserved) maximum overall complexity as good as soft-output MSDD.

\newcommand{\labelnulmax}[2]
{${#1}\hspace*{-.4ex},\hspace*{-.4ex}{#2}\hspace*{-.0ex}$}
\begin{figure}[!t]
\centering
\psfrag{xlabel}[]{MSDD block size $L$ $\rightarrow$}
\psfrag{y1}[]{$\Eb^{\mathsf{req}}/\No$ $[\dB]$}
\psfrag{y2}[]{$C_{\mathsf{o}}$}
\psfrag{hosdc}[bl]{\small\sffamily $C_{\mathsf{o}}^{\mathsf{hard}}$}
\psfrag{sosdc}[bl]{\small\sffamily $C_{\mathsf{o}}^{\mathsf{soft}}$}
\psfrag{hosd}[bl]{\small$\mathsf{hard}$
\scriptsize($2\hspace*{-.4ex}\leq\hspace*{-.4ex}L\hspace*{-.4ex}\leq\hspace*{-.4ex}14$: \labelnulmax{6}{0})}
\psfrag{sosd}[bl]{\small$\mathsf{soft}$}
\psfrag{hodd}[cl]{\scriptsize\raisebox{.6mm}{{\tiny$<$}\hspace*{-1ex}\rule[1.1pt]{7mm}{.2pt}}\labelnulmax{7}{0}}
\psfrag{sodd}[bl]{\scriptsize\raisebox{0mm}{{\tiny$<$}\hspace*{-1ex}\rule[1.1pt]{11mm}{.2pt}}\labelnulmax{7}{\infty}}
\psfrag{6,10}[]{\raisebox{1mm}{\scriptsize\hspace*{1ex}\labelnulmax{6}{10}}}
\psfrag{l}[]{\rotatebox{-14}{\scriptsize$\rightarrow$}}
\psfrag{r}[]{\rotatebox{-5}{\scriptsize$\leftarrow$}}
\psfrag{6,1}[l]{\scriptsize\raisebox{2mm}{\hspace*{1mm}\labelnulmax{6}{1}}}
\psfrag{6,2}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{6}{2}}}
\psfrag{3,0.1}[l]{\scriptsize\labelnulmax{3}{\hspace*{-.4ex}\frac{1}{10}}}
\psfrag{6,0}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{6}{0}}}
\psfrag{5,2}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{5}{2}}}
\psfrag{5,1}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{5}{1}}}
\psfrag{4,0.5}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{4}{\hspace*{-.4ex}\frac{1}{2}}}}
\psfrag{6,0.25}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{6}{\hspace*{-.4ex}\frac{1}{4}}}}
\psfrag{5,0.25}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{5}{\hspace*{-.4ex}\frac{1}{4}}}}
\psfrag{4,0}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{4}{0}}}
\psfrag{5,0}[l]{\scriptsize\raisebox{2mm}{\labelnulmax{5}{0}}}
\psfrag{(6)}[]{\scriptsize}
\psfrag{(5)}[l]{}
\psfrag{(4)}[l]{}
\psfrag{dd}[bl]{\small\sffamily $C_{\mathsf{o}}^{\mathsf{ref}}$}
\psfrag{cmax}[bl]{\small\sffamily $C_{\mathsf{o}}^{\mathsf{max}}$}
        \includegraphics[width = .99\columnwidth]{ctotal_Cfix129.eps}
\caption{Trajectories of performance and overall receiver complexity of the soft-output and hard-output MSDD IR-UWB receiver in comparison to a DD reference with $\nu_{\mathsf{ref}} = 7$ ($C_{\mathsf{o}}^{\mathsf{ref}} = 2^7+1$) at $\BER = 10^{-3}$. Lables (\labelnulmax{\nu}{\hspace*{.4ex}\LLRmax}) indicate chosen setting for soft-output MSDD (see text). Rate-1/2 conv.\ codes with $2^\nu$ states, IEEE-CM\,2.}
\vspace*{-5mm}
\label{fig:UWB:ctotal}
\end{figure}

\section{Conclusions}
\label{sec:conclusions}

We have presented a noncoherent SD-based soft-output MSDD receiver for coded IR-UWB transmission systems.
Based on the GLRT approach, we have derived the LLRs and formulated their computation as tree search problems, enabling the application of the SD for efficient implementation.
Incorporating recent results from MIMO detection, we are able to compute the LLRs in a single SD tree search.
In combination with further techniques for SD complexity reduction, the proposed soft-output SD for MSDD of IR-UWB thus imposes only a moderate complexity increase compared to hard-output MSDD.
Employing this soft-output demodulator for IR-UWB, a large part of the gap between conventional noncoherent DD and ideal coherent detection can be closed.

config: arxiv


TITLE: Classifying equilibrium points
QUESTION [0 upvotes]: I am a bit unsure about the following problem:
Given:
$\dot{x} = y^2 - 2y + 1$
$\dot{y} = -x^2 + 2x -1$
Find and classify all equilibrium points of the system.
OK, så we know that equilibrium points occur when:
$y^2 - 2y + 1 = 0$
and
$-x^2 + 2x -1 = 0$
It is easy to see that this can only occur at $x = 1, y = 1$.
Now I find the Jacobi matrix for the system:
$J = \begin{bmatrix} 0 & 2y-2 \\ -2x+2 & 0 \end{bmatrix}$
By plotting $x = 1, y = 1$ into the matrix we are left with:
$J = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
I know that I am now supposed to find the eigenvalues, and from this deduct whether we are dealing with node, spiral, center, etc.  But I have never before encoutered a zero matrix in these calcuations before.  Basically, if I find the eigenvalue here, I get $\lambda^{2} = 0$, and I don't see how this can tell me anything about the nature of the equilibrium point.
Any help will be truly appreciated!

REPLY [3 votes]: You're right, you need more than the linearization to see what's going on here.  Note that everywhere off the lines $x=1$ and $y=1$ we have $\dot{x}> 0$ and $\dot{y} < 0$.  So, for example, if you start at $x=1+\epsilon$ and $y = 1-\epsilon$ with $\epsilon>0$ what will happen?  What does this say about stability of the critical point?


TITLE: Basic property of schemes?
QUESTION [0 upvotes]: Let $X$ be a scheme. If $X$ happens to be affine then is it true that the canonical morphism
$$X\to\mathrm{Spec}\;\Gamma(X,\mathcal{O}_X)$$
is an isomorphism?.

REPLY [4 votes]: Yes. Moreover, this property characterizes affine schemes among all locally ringed spaces. See this MO comment for a proof sketch.


TITLE: Does the series $\frac{2}{4-1}+\frac{4}{16-1}+\dots+\frac{2k}{4k^2-1}$ have a sum up to $\infty$?
QUESTION [2 upvotes]: If there's one, then how do I find the sum?
I tried to rewrite the common term into partial fractions to see if it's a telescoping series but got a dead end there. How can I proceed now?

REPLY [2 votes]: If you know Taylor series, one thing which could be interesting is to notice that
$$\sum_{k=1}^\infty \frac {2k}{4k^2-1}x^{2k}=\frac{1}{2} x \tanh ^{-1}(x)+\frac{\tanh ^{-1}(x)}{2 x}-\frac{1}{2}$$ and when $x\to 1$, $\tanh ^{-1}(x)\to \infty$.


TITLE: Number of binary sequences in which the number of $(1, 1)$ and $(0, 0)$ is prespecified
QUESTION [1 upvotes]: Consider a binary sequence $\mathbf{a}_n$ consisting of 1s and 0s.
Let us denote by $f(\mathbf{a}_n)$ the number of $(1, 1)$ and $(0, 0)$ in $\mathbf{a}_n$; I am not sure whether there is a formal name for $f(\mathbf{a}_n)$.
Is there a closed expression for the number of $\mathbf{a}_n$'s such that $f(\mathbf{a}_n) = i$  for
$i = 0, 1, \dotsc, n - 1$?

REPLY [4 votes]: Consider the related binary sequence $\widetilde{a_n}$ of length $n-1$ defined by $\widetilde{a_n}(k) := (a_n(k+1)-a_n(k))\text{ mod }2$ for $k = 0\ldots n-2$; the number of 0's in $\widetilde{a_n}$ is exactly the number of $(0,0)'s$ and $(1,1)'s$ in $a_n$. Now, the number of binary sequences of length $(n-1)$ which have exactly $i$ 0's is just $\binom{n-1}{i}$, and each $\widetilde{a_n}$ arises from two distinct $a_n$'s (one with $a_n(0) = 0$ and one with $a_n(0) = 1$), so it follows that $\left|\{a_n|f(a_n) = i\}\right| = 2\binom{n-1}{i}$.


TITLE: Let $x_n$ be a an unbounded sequence of non-zero real numbers
QUESTION [0 upvotes]: Let $x_n$ be a an unbounded sequence of non-zero real numbers. Then

it must have a convergent subsequence.
it can not have a convergent subsequence.
$\frac{1}{x_n}$ must have a convergent subsequence.
$\frac{1}{x_n}$ can not have a convergent subsequence.

Well, 1. is not true as say $x_n=n\ \forall n$, 2. I am not sure, but I guess there may be an unbounded sequence which may have an convergent subsequence, 3. is true due to Bolzano-Weierstrass as it is bounded sequence, 4. is false.
Am I right? Please correct me if I am wrong anywhere. Thank you.

REPLY [2 votes]: is False. $x_n = n$
is False, consider $x_n = n$ if $n$ is even and $x_n = 0$ if $n$ is odd. It has a subsequence converging to $0$. $(x_n)$ is unbounded.
True. Since $x_n$ is unbounded, you can choose a subsequence $y_n$ such that $|y_n| > n$. Then the sequence $\frac{1}{y_n}$ is bounded. By Bolzano Wierstass, it has a convergent subsequence. Clearly, this is subsequence of the original $\{\frac{1}{x_n}\}$.
False. Look at 3.


TITLE: For x < 5 what is the greatest value of x
QUESTION [21 upvotes]: It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be?

REPLY [19 votes]: The answer is $4$, assuming the domain of $x$ is $\Bbb Z$.


TITLE: How is the suggested method used to estabish the stated result? In proof that a polynomial of degree n has at most n roots
QUESTION [0 upvotes]: This question is based on the discussion in B4-1.2 of Fundamentals of Mathematics, Volume 1 Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle.
See the bold text for the part of the proof I am seeking help with.  My question is: how do I use the method suggested to obtain the needed result?
Technically this proof is applied to $\mathcal{R}$ which is an arbitrary
commutative ring with unit element 1, and without null divisors. And the expression
$$
f\left[x\right]=\sum_{i=0}^{n}a_{i}x^{i}
$$
is equivalent to an entire rational function of one argument
in $\mathcal{R}.$ Here we have defined $0^{0}\equiv1$ to support
our notation. For the purpose of this question we can simply call
$f$ a polynomial of degree $n$ in $\mathbb{R}.$
Our objective is to prove the following theorem: If $f$ is expressible
in the form
$$
f\left[x\right]=\sum_{i=0}^{n}a_{i}x^{i}
$$
with $a_{n}\ne0,$ then $f$ has at most $n$ roots.
Proof: First we introduce the constant expressions
$$
\bar{a}_{k}=\sum_{i=k+1}^{n}a_{i}\alpha^{i-k-1},
$$
and the function
$$
f_{1}\left[x\right]\equiv\sum_{k=0}^{n-1}x^{k}\bar{a}_{k}=\sum_{k=0}^{n-2}x^{k}\bar{a}_{k}+a_{n}x^{n-1}.
$$
The right-most expression follows from $\bar{a}_{n-1}=a_{n},$ and
shows that $f_{1}$ has at most $n-1$ roots by the induction hypothesis.
Next we observe that for the real number constant $\alpha$ we have
$$
(x-\alpha)\sum_{k=0}^{i-1}x^{k}\alpha^{i-k-1}=\sum_{k=0}^{i-1}x^{k+1}\alpha^{i-k-1}-\sum_{k=0}^{i-1}x^{k}\alpha^{i-k}
$$
$$
=\sum_{k=1}^{i}x^{k}\alpha^{i-k}-\sum_{k=0}^{i-1}x^{k}\alpha^{i-k}=x^{i}-\alpha^{i},
$$
and apply this result to obtain, for $n>0$
$$
f\left[x\right]-f\left[\alpha\right]=\left(x-\alpha\right)f_{1}\left[x\right].
$$
The case of $n=0$ is the constant function $f\left[x\right]=a_{0}x^{0},$
which is of the required form. If $\alpha$ is a root of $f,$ we have
$$
f\left[x\right]-f\left[\alpha\right]=\left(x-\alpha\right)f_{1}\left[x\right]=0.
$$
If $x\ne\alpha,$ is a root of $f$ then $f_{1}\left[x\right]=0.$ It follows that $f$
has at most one more root than $f_{1}$ Since $f_{1}$ has at most $n-1$ roots this which completes the proof.
The step I am not following is the establishment of $f\left[x\right]-f\left[\alpha\right]=\left(x-\alpha\right)f_{1}\left[x\right]$
using
$$(x-\alpha)\sum_{k=0}^{i-1}x^{k}\alpha^{i-k-1}=x^{i}-\alpha^{i}.$$
I am able to establish the result using brute force as follows: Expand
$f_{1}$
$$
f_{1}\left[x\right]\equiv\sum_{k=0}^{n-1}\left(\sum_{i=k+1}^{n}a_{i}\alpha^{i-k-1}\right)x^{k}
$$
$$
=\left(a_{1}+a_{2}\alpha+a_{3}\alpha^{2}+a_{4}\alpha^{3}+\ldots+a_{n}\alpha^{n-1}\right)
$$
$$
+\left(a_{2}+a_{3}\alpha+a_{4}\alpha^{2}+a_{5}\alpha^{3}+\ldots+a_{n}\alpha^{n-2}\right)x
$$
$$
+\left(a_{3}+a_{4}\alpha+a_{5}\alpha^{2}+a_{6}\alpha^{3}+\ldots+a_{n}\alpha^{n-3}\right)x^{2}
$$
$$
+\dots+a_{n}x^{n-1}
$$
$$
=a_{1}+a_{2}(\alpha+x)+a_{3}\left(\alpha^{2}+\alpha x+x^{2}\right)
$$
$$
+a_{4}\left(\alpha^{3}+\alpha^{2}x+\alpha x^{2}+x^{3}\right)
$$
$$
+\ldots+a_{i}\left(\alpha^{i-1}+\alpha^{i-2}x+\ldots+\alpha x^{i-2}+x^{i-1}\right)
$$
$$
+\ldots+a_{i}\left(\alpha^{n-1}+\alpha^{n-2}x+\ldots+\alpha x^{n-2}+x^{n-1}\right).
$$
Multiplying by $\left(x-\alpha\right)$ and canceling like terms of opposite
sign produces the desired result
$$
\left(x-\alpha\right)f_{1}\left[x\right]=a_{1}\left(x-\alpha\right)+a_{2}(\alpha^{2}-x^{2})
$$
$$
+a_{3}\left(\alpha^{2}x+\alpha x^{2}+x^{3}-\alpha^{3}-\alpha^{2}x-\alpha x^{2}\right)
$$
$$
+\ldots+a_{i}\left(\alpha^{i-1}x+\alpha^{i-2}x^{2}+\ldots+\alpha^{2}x^{i-2}+\alpha x^{i-1}+x^{i}\right)
$$
$$
-a_{i}\left(\alpha^{i}+\alpha^{i-1}x+\alpha^{i-2}x^{2}+\ldots+\alpha^{2}x^{i-2}+\alpha x^{i-1}\right)
$$
$$
+\ldots+a_{n}\left(\alpha^{n-1}x+\alpha^{n-2}x^{2}+\ldots+\alpha^{2}x^{n-2}+\alpha x^{n-1}+x^{n}\right)
$$
$$
-a_{n}\left(\alpha^{n}+\alpha^{n-1}x+\alpha^{n-2}x^{2}+\ldots+\alpha^{2}x^{n-2}+\alpha x^{n-1}\right)
$$
$$
=f\left[x\right].
$$

REPLY [1 votes]: Well,
\begin{align*}
f[x]-f[\alpha]&=\sum_{i=0}^na_ix^i-\sum_{i=0}^na_i\alpha^i \\
&=\sum_{i=0}^na_i(x^i-\alpha^i) \\
&=\sum_{i=0}^na_i(x-\alpha)\sum_{k=0}^{i-1}x^{k}\alpha^{i-k-1} \\
&=(x-\alpha)\sum_{i=0}^n\sum_{k=0}^{i-1}a_i\alpha^{i-k-1} x^{k}.
\end{align*}
Now we interchange the two sums to get
\begin{align*}
f[x]-f[\alpha]&=(x-\alpha)\sum_{k=0}^{n-1}\sum_{i=k+1}^{n}a_i\alpha^{i-k-1}x^{k} \\
&=(x-\alpha)\sum_{k=0}^{n-1}\bar{a}_kx^k \\
&=(x-\alpha)f_1[x].
\end{align*}


TITLE: Finding multivariable limits for the function $\frac{3x^2y}{x^2+y^2}$
QUESTION [2 upvotes]: Could anyone help me with this
Find $$\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^2+y^2}$$ if this limit exists
I tried using the squeeze theorem, but I could not find a suitable expression for the squeeze theorem as I could not find any number that allow me to squeeze the limit in between and cancel the denominator

REPLY [5 votes]: $x^2\leq x^2+y^2$, then $0\leq\dfrac{x^2}{x^2+y^2}\leq 1$, now we obtain
$\left|\dfrac{3x^2y}{x^2+y^2}\right|\leq 3|y|$, since  $|y|\to 0$ we get  $\dfrac{3x^2y}{x^2+y^2}\to 0$.


text:
\begin{document}

\thispagestyle{plain}

\noindent $\;$

\vspace*{10.00 mm}

\centerline{\large \rm SOME CONSIDERATIONS IN CONNECTION}

\smallskip

\centerline{\large \rm WITH ALTERNATING KUREPA'S FUNCTION}

\medskip

\bigskip
\centerline{\it Branko J. Male\v sevi\' c${\,}^{1}$}
\footnotetext{$\!\!\!\!\!\!\!\!{}^{1}$Research partially supported by the MNTRS, Serbia, Grant No. 144020.}

\begin{center}
${}^{1}$Faculty of Electrical Engineering \\
University of Belgrade, 11000 Belgrade, Serbia \\
E-mail: malesh@eunet.yu
\end{center}

\noindent
{\small {\bf ABSTRACT.}
In this paper we consider the functional equation for alternating factorial sum and some of its particular solutions
{\normalsize (}alternating Kurepa's function $A(z)$ from \cite{Petojevic_02} and function $A_{1}(z)${\normalsize )}.
We determine an extension of domain of functions $A(z)$ and $A_{1}(z)$ in the sense of the principal value at point
\cite{Slavic_73}, \cite{MijajlovicMalesevic_07}. Using the methods from \cite{Slavic_73} and \cite{Malesevic_03}
we give a new representation of alternating Kurepa's function $A(z)$, which is an analog of Slavi\' c's representation
of Kurepa's function $K(z)$ \cite{Slavic_73}, \cite{Marichev_83}. Also, we consider some representations of functions
$A(z)$ and $A_{1}(z)$ via incomplete gamma function and we consider differential transcendency of previous functions~too.}

\medskip
\leftline{KEYWORDS: Gamma function, Kurepa's function, Casimir energy.}

\medskip
\leftline{MSC (2000): 30E20, 11J91, 81V99.}

\bigskip

\bigskip
\noindent
\section{\bf \boldmath \hspace*{-7.0 mm}
1 The functional equation for alternating factorial sum and its particular solutions}

The main object of consideration in this paper is the functional equation for
alternating factorial sum
\begin{equation}
\label{A_FE_1}
A(z) + A(z-1) = \Gamma(z+1),
\end{equation}
with respect to the function $A : \mbox{\newmat{D}} \longrightarrow \mbox{\newmat{C}}$ with domain
$\mbox{\newmat{D}} \subseteq \mbox{\newmat{C}} \backslash \mbox{\newmat{Z}}^{-}$,  where $\Gamma$
is the gamma function, $\mbox{\newmat{C}}$ is the set of complex numbers and $\mbox{\newmat{Z}}^{-}$
is set of negative integer numbers. A solution of functional equation (\ref{A_FE_1})
over the set of natural numbers ($\mbox{\newmat{D}} = \mbox{\newmat{N}}$)
is the function of alternating left factorial $A_n$. R. Guy introduced this function,
in the book \cite{Guy_94} (p.$\:$100), as an alternating sum of factorials
\mbox{$A_n = n! - (n-1)! + \ldots + (-1)^{n-1}1! \, .$}
Let us use the notation
\begin{equation}
\label{A_SUM_1}
A(n) = \displaystyle\sum\limits_{i=1}^{n}{(-1)^{n-i}i!} \; .
\end{equation}

\break

\noindent
Sum (\ref{A_SUM_1}) corresponds to the sequence $A005165$ in \cite{Sloane_06}.
We call the functional equation (\ref{A_FE_1}) {\em the functional equation for
alternating factorial sum}. In consideration which follows we consider two
particular solutions of the functional equation (\ref{A_FE_1}).

\bigskip
\noindent
{\bf \boldmath 1.1$\;$The function $A(z)$} An analytical extension of the function
(\ref{A_SUM_1}) over the set of complex numbers is determined by integral~\cite{Petojevic_02}:
\begin{equation}
\label{A_INT_1}
A(z)
=
\displaystyle\int\limits_{0}^{\infty}{
e^{-t} \displaystyle\frac{t^{z+1}-(-1)^{z}t}{t+1} \: dt},
\end{equation}
which converges for $\mbox{Re} \: z > 0$. For the function $A(z)$ we use the
term {\em alternating Kurepa's function} and it is a solution of the functional
equation (\ref{A_FE_1}). Let us observe that since \mbox{$A(z-1) = \Gamma(z+1) -A(z)$},
it is possible to make analytical continuation of alternating Kurepa's function $A(z)$
for \mbox{$\mbox{Re} \, z \leq 0$}. In that way, the alternating Kurepa's function
$A(z)$ is a meromorphic function with simple poles at $z \!=\! -n$ $(n \!\geq\! 2)$.
At a point $z = \infty$ alternating Kurepa's function has an essential singularity.
Alternating Kurepa's function has the following residues
\begin{equation}
\mathop{\mbox{\rm res}}\limits_{z = -n}{A(z)} =
(-1)^n \displaystyle\sum\limits_{k=0}^{n-2}{\displaystyle\frac{1}{k!}}
\quad (n\!\geq\!2).
\end{equation}
Previous results for alternating Kurepa's function are given according to \cite{Petojevic_02}.

\bigskip
\noindent
{\bf \boldmath 1.2$\;\,$The function $A_{1}(z)\,$}
The functional equation (\ref{A_FE_1}), besides alternating Kurepa's function $A(z)$,
has another solution which is given by the following statement.
\begin{th}
\label{A_lema_4}
Let $\mbox{\newmat{D}} = \mbox{\newmat{C}} \backslash \mbox{\newmat{Z}}$. Then, series
\begin{equation}
\label{Def_A1}
A_{1}(z) = \displaystyle\sum\limits_{n=0}^{\infty}{(-1)^n\Gamma(z+1-n)}
\end{equation}
absolutely converges and it is a solution of the functional equation
{\rm (\ref{A_FE_1})}~over~$\mbox{\newmat{D}}$.
\end{th}

\noindent
{\bf Proof.} Statement of the Theorem is a consequence of the Theorem 1.1
from \cite{Malesevic_03}.~\stop
\begin{re}
Function $A_{1}(z)$, defined by {\rm (\ref{Def_A1})} over $\mbox{\newmat{C}}$,
has poles at integer points $z\!=\!m \in \mbox{\newmat{Z}}$.
\end{re}

\section{\bf \boldmath \hspace*{-7.0 mm}
2 Extending the domain of functions $A(z)$ and $A_{1}(z)$ in the sense of the principal value at point}

Let us observe a possibility of extending the domain of the functions $A(z)$ and $A_{1}(z)$,
in the sense of the principal value at point, over the set of complex numbers. Namely, for
a meromorphic function $f(z)$, on the basis of Cauchy's integral formula, we define
{\em the principal value at point $a$} as follows \cite{Slavic_70}, \cite{Slavic_73}:
\begin{equation}
\label{GAMMA_PV_a}
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{f(z)}
=
\lim\limits_{\rho \rightarrow 0_{+}}{\displaystyle\frac{1}{2 \pi i}
\!\!\!\!\displaystyle\oint\limits_{|z-a|=\rho}{\!\!\!\!\displaystyle\frac{f(z)}{z-a}\,dz}}.
\end{equation}
It is obvious that the principal value at pole $z = a$ exists as a finite complex number
$\mathop{\mbox{res}}\limits_{z=a}{{\Big (}\displaystyle\frac{f(z)}{z-a}}{\Big )}$.
For two meromorphic functions $f_1(z)$ and $f_2(z)$ additivity is true~\cite{Slavic_70}:
\begin{equation}
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{{\Big (}f_{1}(z)+f_{2}(z){\Big )}}
=
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{f_{1}(z)}
+
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{f_{2}(z)}.
\end{equation}
In the paper \cite{Slavic_70} it is proved that multiplicativity
of the principal value does not hold. The following statement is proved in \cite{MijajlovicMalesevic_07}.
\begin{th}
Let $f_{1}(z)$ be a holomorphic function at the point $a$ and let $f_{2}(z)$ be a meromorphic function with
pole of the $m$-th order at the same point $a$.$\;$Then
\begin{equation}
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{ {\Big (} f_{1}(z) \cdot
f_{2}(z) {\Big )} } =
\displaystyle\sum\limits_{k=0}^{m}{\displaystyle\frac{f_{1}^{(k)}(a)}{k!}
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{ {\Big (} (z-a)^k \cdot
f_{2}(z) } {\Big )} }.
\end{equation}
\end{th}
\begin{co}
Let $f_{1}(z)$ be a holomorphic function at the point $a$ and let $f_{2}(z)$ be a meromorphic function with
simple pole at the same point $a$.$\;$Then
\begin{equation}
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{{\Big (}f_{1}(z) \cdot f_{2}(z){\Big )}}
=
f_{1}(a) \cdot \mathop{\mbox{\rm p.v.}}\limits_{z = a}{f_{2}(z)}
+
{f_{1}\!}^{'}\!(a) \cdot \mathop{\mbox{\rm res}}\limits_{z = a}{f_{2}(z)}.
\end{equation}
The previous formula, in the case of the zeta function $f_2(z) = \zeta(z)$,
is also given~in {\rm \cite{Blau_Visser_Wipf_88_b}}.
\end{co}
The following statement is proved in \cite{Malesevic_03}.
\begin{lm}
\label{TH_Gamma_PV_f}
For the function $f(z)$ with simple pole at point $z = a$ the following is true
\begin{equation}
\label{PV_f_1}
\mathop{\mbox{\rm p.v.}}\limits_{z = a}{f(z)}
=
\lim\limits_{\varepsilon \rightarrow 0}{
\displaystyle\frac{f(a-\varepsilon)+f(a+\varepsilon)}{2}}.
\end{equation}
\end{lm}
\begin{no}
For any meromorphic function $f(z)$, that possesses at worst simple poles,
by formula {\rm (\ref{PV_f_1})} {\em the principal value at point} is defined
as {\em principal part}~{\rm \cite{Blau_Visser_Wipf_88_b}}, which comes
from the quantum field theory {\rm \cite{Blau_Visser_Wipf_88_a}}. Especially,
using formula {\rm (\ref{PV_f_1})}, for the zeta function, {\em the Casimir energy} in physics
is given {\rm \cite{Blau_Visser_Wipf_88_a}}, {\rm \cite{Blau_Visser_Wipf_88_b}}. For a precise
de\-finition of the Casimir energy see {\rm \cite{Elizalde_94}}, {\rm \cite{Elizalde_96}}, {\rm \cite{Elizalde_04}}.
\end{no}
\begin{co}
\label{TH_Gamma_PV}
For gamma function $\Gamma(z)$ it is true {\rm \cite{Slavic_70}},
{\rm \cite{Blau_Visser_Wipf_88_b}}, {\rm \cite{Malesevic_03}}$:$
\begin{equation}
\label{PV_1}
\mathop{\mbox{\rm p.v.}}\limits_{z = -n}{\!\!\Gamma(z)}
=
\lim\limits_{\varepsilon \rightarrow 0}{\!
\displaystyle\frac{\Gamma(-n-\varepsilon)+\Gamma(-n+\varepsilon)}{2}}
=
(-1)^{n}\displaystyle\frac{\Gamma^{'}(n+1)}{\Gamma(n+1)^2}
\quad (n\!\in\!\mbox{\newmat{N}}_{0}).
\end{equation}
\end{co}
\begin{re}
\label{Gamma_Kolicnik_Je_Opadajuci_Niz}
For $n \in \mbox{\newmat{N}}_{0}$ it is true~{\rm \cite{Slavic_70}}$:$

\vspace*{-5.0 mm}

\begin{equation}
\label{H_n}
\displaystyle\frac{\Gamma^{'}(n+1)}{\Gamma(n+1)^2}
=
\displaystyle\frac{\mbox{\small $-\gamma$} + \mbox{\footnotesize $\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}$}}{n!},
\end{equation}

\noindent
where $\gamma$ is Euler's constant.
\end{re}

\noindent
Extension of the domain of the functions $A(z)$ and $A_{1}(z)$, in the sense of
the principal value at point, is given by the following two theorems.
\begin{th}
For alternating Kurepa's function $A(z)$ it is true
\begin{equation}
\label{A_PV}
\mathop{\mbox{\rm p.v.}}\limits_{z = -n}{\!\!A(z)} =\!
\displaystyle\sum\limits_{i=0}^{n-1}{(-1)^{n+1-i}\!\!\!\!\mathop{
\mbox{\rm p.v.}}\limits_{z=-(i-1)}{\!\!\!\!\Gamma(z)}} =\! (-1)^{n+1}
{\bigg (}
1-\!\displaystyle\sum\limits_{i=1}^{n-1}{\displaystyle\frac{\Gamma^{'}(i)}{\Gamma(i)^2}}
{\bigg )} \;\; (n \in \mbox{\newmat{N}}).
\end{equation}
\end{th}
{\bf Proof.} If the equality
$
A(z) = (-1)^{n} A(z+n) + {\big (} \Gamma(z+2) - \ldots + (-1)^{n+1}\Gamma(z+n+1) {\big )}
$
we consider at the point $z = -n$ in the sense of the principal value, on the basis
of (\ref{PV_1}), the equality (\ref{A_PV}) follows. Let us remark that
$\mathop{\mbox{\rm p.v.}}\limits_{z = -1}{\!A(z)} = A(-1) = 1$.~\stop

\medskip
\noindent
The following Ramanujan formula is true:
\begin{equation}
\label{Ramanujan}
\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{1 + \frac{1}{2} + \ldots + \frac{1}{n}}{n!} \, x^n}
=
e^x
\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{(-1)^{n-1}}{n! \, n} \, x^n},
\end{equation}
for $x \in \mbox{\newmat{C}}$ (see \cite{Berndt_85}, page 46., corollary 2.).
On the basis of the previous formula follows:
\begin{lm}
\label{L_2}
Let us define $L_{2} = \displaystyle\sum\limits_{n=0}^{\infty}{
(-1)^n\!\!\mathop{\mbox{\rm p.v.}}\limits_{z =
-(n-1)}{\!\!\Gamma(z)}}$, then
\begin{equation}
\label{Const_L_2}
L_{2}
=
1 + e \gamma - e \displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{(-1)^{n-1}}{n! \, n}}
=
1 + e\mbox{\rm Ei}(-1)
\approx
0.403 \, 652 \, 377 \, ,
\end{equation}
where $\mbox{\rm Ei}$ is the function of exponential integral${\,}^{\mbox{\scriptsize $\ast)$}}$\footnote{
$\!\!{}^{\ast)}\,$see formula (\ref{Ei}) in this paper}.
\end{lm}
{\bf Proof.} On the basis of the corollary \ref{TH_Gamma_PV} and the remark \ref{Gamma_Kolicnik_Je_Opadajuci_Niz} we have
\begin{equation}
\label{Const_L_21}
L_{2}
=
1-\displaystyle\sum\limits_{n=0}^{\infty}{\displaystyle\frac{\Gamma^{'}(n\!+\!1)}{\Gamma(n\!+\!1)^2}}
=
1
-
{\bigg (}\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{1 + \frac{1}{2} + \ldots + \frac{1}{n}}{n!} }
-
\gamma e{\bigg )}.
\end{equation}
By substitution $x=1$ in formula (\ref{Ramanujan}) we obtain:
\begin{equation}
\label{Const_L_22}
\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{1 + \frac{1}{2} + \ldots + \frac{1}{n}}{n!}}
=
e
\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{(-1)^{n-1}}{n! \, n}}.
\end{equation}
Next, the following representation of Gompertz
constant $-e \mbox{\rm Ei}(-1)$ (see sequence A073003 in \cite{Sloane_06}) is true

\vspace*{-3.0 mm}

\begin{equation}
\label{Const_L_23}
-
e \mbox{\rm Ei}(-1)
=
e {\bigg (}\displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{(-1)^{n-1}}{n! \, n}}
-
\gamma {\bigg )}.
\end{equation}
Then, on the basis of (\ref{Const_L_21}), (\ref{Const_L_22}) and (\ref{Const_L_23})
follows (\ref{Const_L_2}).~\stop

\noindent
\begin{th}
\label{A_aux}
For the function $A_{1}(z)$ it is true
\begin{equation}
\mathop{\mbox{\rm p.v.}}\limits_{z = n}{A_{1}(z)}
=
(-1)^n L_{2} + \mathop{\mbox{\rm p.v.}}\limits_{z = n}{A(z)}
\quad (n \!\in\! \mbox{\newmat{Z}}).
\end{equation}
\end{th}
{\bf Proof.} For $n \geq 0$ it is true
\begin{equation}
\mbox{\small $\begin{array}{rcl}
\mathop{\mbox{\rm p.v.}}\limits_{z = n}{A_{1}(z)}
&\!\!=\!\!&
\displaystyle\sum\limits_{i=0}^{\infty}{
(-1)^{i}\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = n+1-i}{\!\Gamma(z)}}
=
(-1)^n
\displaystyle\sum\limits_{i=0}^{\infty}{
(-1)^{i}\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = -(i-1)}{\!\Gamma(z)}}                            \\[1.0 ex]
&\!\!+\!\!&
\displaystyle\sum\limits_{i=1}^{n}{(-1)^{n-i}\Gamma(i+1)}
=
(-1)^{n}L_{2} + A(n).
\end{array}$}
\end{equation}

\noindent
For $n < 0$ it is true
\begin{equation}
\!\!\!\!\!\!\!\mbox{\small $\begin{array}{rcl}
\mathop{\mbox{\rm p.v.}}\limits_{z = n}{A_{1}(z)}
&\!\!\!\!\!=\!\!\!\!\!&
\displaystyle\sum\limits_{i=0}^{\infty}{(-1)^i
\!\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = n+1-i}{\!\!\Gamma(z)}}
= (-1)^{(-n)} {\bigg (}
\displaystyle\sum\limits_{i=(-n)}^{\infty}{(-1)^i
\!\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = -i+1}{\!\!\Gamma(z)}} {\bigg )}                        \\[2.0 ex]
&\!\!\!\!\!=\!\!\!\!\!&
(-1)^{(-n)} {\bigg (}\!
\displaystyle\sum\limits_{i=0}^{\infty}{(-1)^i
\!\!\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = -(i-1)}{\!\!\!\!\Gamma(z)}}\!{\bigg )}
-
(-1)^{(-n)} {\bigg (}\!\displaystyle\sum\limits_{i=0}^{(-n)-1}{\!\!\!(-1)^{-i}
\!\!\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = -(i-1)}{\!\!\!\!\Gamma(z)}}\!{\bigg )}
                                                                               \\[2.0 ex]
&\!\!\!\!\!=\!\!\!\!\!&
(-1)^n L_{2}
+\!
{\bigg (}\!
\displaystyle\sum\limits_{i=0}^{(-n)-1}{
(-1)^{(-n)+1-i} \!\!\!\!\!\mathop{\mbox{\rm p.v.}}\limits_{z = -(i-1)}{\!\!\!\!\Gamma(z)}}
\!{\bigg )}
\!=
(-1)^{n} L_{2} + \mathop{\mbox{\rm p.v.}}\limits_{z = n}{A(z)}.\;\stop
\end{array}$}
\end{equation}

\noindent
\section{\bf \boldmath \hspace*{-7.0 mm}
3 Formula of Slavi\' c's type for alternating Kurepa's function}

\medskip
\noindent
The main result in this section is a new formula of Slavi\' c's type for alternating Kurepa's function
$A(z)$, which is an analog of Slavi\' c's representation of Kurepa's function $K(z)$
\cite{Slavic_73}, \cite{Marichev_83}, \cite{Milovanovic_96}. The following statements are true.
\begin{lm}
\label{A_lema_12}
Function
\begin{equation}
\label{Maple_A_Lema_12}
F(z)
=
\displaystyle\sum\limits_{n=1}^{\infty}{
\!{\bigg (}\displaystyle\sum\limits_{k=2}^{\infty}{
\displaystyle\frac{(-1)^{k-1}(n+k-1)}{(n+k)!} \, z^k {\bigg )}}},
\end{equation}
is entire, whereas the following is true
\begin{equation}
\label{Maple_A_Lema_21}
F(z)
=
\displaystyle\sum\limits_{k=2}^{\infty}{ \!{\bigg (}\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\frac{(-1)^{k-1}(n+k-1)}{(n+k)!} \, z^k {\bigg )}}}
=
-e^{-z}-z+1.
\end{equation}
\end{lm}

\noindent
{\bf Proof.} For $z = 0$ the equality (\ref{Maple_A_Lema_21}) is true. Let us
introduce a sequence of functions
\begin{equation}
\label{Def_f_n}
f_{n}(z)
=
\displaystyle\sum\limits_{k=2}^{\infty}{
\displaystyle\frac{(-1)^{k-1}(n+k-1)}{(n+k)!} \, z^k},
\end{equation}
for $z \in \mbox{\newmat{C}}$ $(n \in \mbox{\newmat{N}})$. Previous series converge over $\mbox{\newmat{C}}$ because,
for $z \neq 0$, it is true that
\begin{equation}
\label{Maple_f_n}
f_{n}(z)
=
\displaystyle\sum\limits_{j=0}^{n+1}{
(-1)^{j+n} {\bigg (}\displaystyle\frac{j}{j!}
\!-\!
\displaystyle\frac{1}{j!} {\bigg )} \, z^{j-n}}
\!+\!
(-1)^{n}e^{-z}(z^{-n+1}\!+\!z^{-n}).
\end{equation}

\vspace*{-2.5 mm}

\noindent
Let us mention that the previous equality is easily checked by the following substi\-tution${}^{C1)}$
\mbox{$e^{-z} = \sum_{k=0}^{\infty}{\frac{(-z)^k}{k!}}$} at the right side of
equality of formula (\ref{Maple_f_n}). Let  $\rho>0$ be fixed. Over the set
\mbox{$\mbox{\newmat{D}}\!=\!\{ z\!\in\!\mbox{\newmat{C}} \, | \, 0\!<\!|z|\!<\!\rho\}$}
let us form an auxiliary function \mbox{$g(z) = (z+1)e^{-z} : \mbox{\newmat{D}} \longrightarrow \mbox{\newmat{C}}$}.
If we denote by $R_{n}(.)$ the remainder of $n$-th order of MacLaurin's expansion, then for $z \in \mbox{\newmat{D}}$
the following representation is true${}^{C2)}$
\begin{equation}
\label{MacLaurin_f_n}
f_{n}(z)
=
\displaystyle\frac{R_{n+1}(g(z))}{(-z)^n}.
\end{equation}
Then, for $|z| < \rho$ it is true
\begin{equation}
\label{Ineq_f_n}
|f_{n}(z)|
\leq
\displaystyle
\frac{e^{\rho}(n+1+\rho)}{(n+2)!}\rho^2.
\end{equation}
Indeed, for $z=0$ the previous inequality is true. Over $\mbox{\newmat{E}} = (0,\rho)$ let us form an auxiliary function
\mbox{$h(t) = (t-1)e^{t} + 2 : \mbox{\newmat{E}} \longrightarrow \mbox{\newmat{R}}^{+}$}. For $z \in \mbox{\newmat{D}}$
and $t = |z| \in \mbox{\newmat{E}}$ there exists $c \in (0,t)$ such that${}^{C3)}$
\begin{equation}
|f_{n}(z)|
\leq
\displaystyle\frac{R_{n+1}(h(t))}{t^n}
=
\displaystyle
\frac{h^{(n+2)}(c)}{(n+2)!}\,t^2
\leq
\displaystyle
\frac{e^{\rho}(n+1+\rho)}{(n+2)!}\rho^2.
\end{equation}
For the function
\begin{equation}
F(z)
=
\displaystyle
\sum\limits_{n=1}^{\infty}{f_{n}(z)},
\end{equation}
it is possible, for $|z|\!<\!\rho$, to apply Weierstrass's double series
Theorem~\cite{Knopp_96} (page~83.). Indeed, on the basis of (\ref{Def_f_n}),
the functions $f_{n}(z)$ are regular for $|z|\!<\!\rho$. On the basis of
(\ref{Ineq_f_n}), the series $\sum_{n=1}^{\infty}{f_{n}(z)}$ is uniformly
convergent for $|z|\!\leq\!r\!<\!\rho$, for every $r\!<\!\rho$. Then on
the basis of the Weierstrass's 's double series Theorem, for $|z|\!<\!\rho$,
the following~is~true
\begin{equation}
F(z)
=
\displaystyle\sum\limits_{k=2}^{\infty}{
\!{\bigg (}\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\frac{(-1)^{k-1}(n+k-1)}{(n+k)!} \, z^k {\bigg )}}}
=
-e^{-z}-z+1,
\end{equation}
because${}^{C4)}$
\begin{equation}
\label{Razmena}
\label{A_sum}
\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\frac{(-1)^{k-1}(n+k-1)}{(n+k)!}} =
\displaystyle\frac{(-1)^{k-1}}{k!}.
\end{equation}
Let us note that $\rho>0$ can be arbitrarily large positive number. Hence, the equality (\ref{Maple_A_Lema_21})
is true for all $z \in \mbox{\newmat{C}}$; i.e. the function $F(z)$ is entire. \stop
\begin{lm}
\label{A_lema_3}
For $z \in \mbox{\newmat{C}}$ it is true
\begin{equation}
\label{A_sum_sum} (z\!+\!1)
\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\sum\limits_{k=1}^{\infty}{
\displaystyle\frac{(-1)^{k}}{(k\!+\!n)!} \, z^{k}}}
=
-e^{-z}+(1\!-\!e)\,z+1.
\end{equation}
\end{lm}

\noindent
{\bf Proof.} On the basis of the Lemma \ref{A_lema_12} it is true that
\begin{equation}
\!\!\mbox{\small $
\begin{array}{rcl}
(z\!+\!1)
\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\sum\limits_{k=1}^{\infty}{
\displaystyle\frac{(-z)^{k}}{(k\!+\!n)!}}}
&\!\!=\!\!&
\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\sum\limits_{k=1}^{\infty}{
\displaystyle\frac{(-1)^{k}}{(k\!+\!n)!}\,(z^{k+1}\!+z^{k})}}                  \\[2.5 ex]
&\!\!=\!\!&
\displaystyle\sum\limits_{n=1}^{\infty}{
\!{\bigg (}\!\!-\displaystyle\frac{z}{(n\!+\!1)!}
+\!\displaystyle\sum\limits_{k=2}^{\infty}{
{\Big (}\displaystyle\frac{(-1)^{k-1}}{(k\!+\!n\!-\!1)!}
-\displaystyle\frac{(-1)^{k-1}}{(k\!+\!n)!}{\Big )}z^{k}}{\bigg )}}            \\[2.5 ex]
&\!\!\mathop{=}\limits_{(\ref{Maple_A_Lema_21})}\!\!&
\displaystyle\sum\limits_{k=2}^{\infty}{
{\bigg (}\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\frac{(-1)^{k-1}(n\!+\!k\!-\!1)}{(n\!+\!k)!}}{\bigg )}\,z^{k}}
-\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\frac{z}{(n\!+\!1)!}}                                            \\[5.0 ex]
&\!\!\mathop{=}\limits_{(\ref{Maple_A_Lema_21})}\!\!&
-e^{-z}+(1\!-\!e)\,z+1.\;\stop
\end{array}$}
\end{equation}
\begin{th}
For alternating Kurepa's function $A(z)$ the following representation is true
\begin{equation}
\label{A_sum_Slavic}
A(z)
=
-
{\big (}1 + e \, \mbox{\rm Ei}(-1){\big )}(-1)^z
+
\displaystyle\frac{\pi e}{\sin \pi z}
+
\displaystyle\sum\limits_{n=0}^{\infty}{(-1)^{n}\Gamma(z+1-n)},
\end{equation}
where the values in the previous formula, in integer points $z$,
are determined in the sense of the principal value at point.
\end{th}

\noindent
{\bf Proof.} For $-(n\!+\!1)\!<\!\mbox{\rm Re}\,z\!<\!-n$ and $n=0,1,2,\ldots$ the following
formula is true~\cite{Bateman_65}:
\begin{equation}
\Gamma(z)
=\!\!
\displaystyle\int\limits_{0}^{+\infty}\!{{\bigg (}\!
e^{-t} - \displaystyle\sum\limits_{m=0}^{n}{
\displaystyle\frac{(-t)^{m}}{m!}}\!{\bigg )}t^{z-1}\:dt}.
\end{equation}
Hence, for $0 < \mbox{\rm Re}\:z < 1$ and $n=1,2,\ldots$ the following formula is true
\begin{equation}
\Gamma(z-n)
=\!\!
\displaystyle\int\limits_{0}^{+\infty}\!{{\bigg (}\!
e^{-t} - \displaystyle\sum\limits_{m=0}^{n-1}{
\displaystyle\frac{(-t)^{m}}{m!}}\!{\bigg )}t^{z-n-1}\:dt}.
\end{equation}
Further we observe the following difference
\begin{equation}
\begin{array}{rcl}
A(z) - \displaystyle\sum\limits_{n=0}^{+\infty}{(-1)^{n}\Gamma(z\!+\!1\!-\!n)}
\!\!&\!\!=\!\!&\!\!
\!\displaystyle\int\limits_{0}^{+\infty}{\!
e^{-t}\displaystyle\frac{t^{z+1}\!-\!(-1)^z t}{t\!+\!1}\:dt}
\;-\!
\displaystyle\int\limits_{0}^{+\infty}{\!
e^{-t}t^{z}\:dt}                                                               \\[2.0 ex]
\!\!&\!\!-\!\!&
\!\displaystyle\sum\limits_{n=1}^{\infty}{
(-1)^{n}\!\!\displaystyle\int\limits_{0}^{+\infty}{\!\!{\bigg (}\!
e^{-t}\!-\!\displaystyle\sum\limits_{m=0}^{n}{\!
\displaystyle\frac{(-t)^{m}}{m!}}\!{\bigg )}t^{z\!-\!n}\:dt}}.
\end{array}
\end{equation}
For $0 < \mbox{\rm Re}\:z < 1$ the following derivation is true
\begin{equation}
\label{A_derive}
\mbox{\small $\begin{array}{rl}
& A(z) - \displaystyle\sum\limits_{n=0}^{+\infty}{(-1)^{n}\Gamma(z\!+\!1\!-\!n)}
                                                                               \\[2.0 ex]
&
= -\!\displaystyle\int\limits_{0}^{\infty}{\!e^{-t}
\displaystyle\frac{t^{z}\!-\!(-1)^{z-1}t}{t\!+\!1} \: dt}
-\!
\displaystyle\int\limits_{0}^{+\infty}{\!
\displaystyle\sum\limits_{n=1}^{\infty}{\!
(-1)^{n}\!{\bigg (}\!e^{-t}\!-\!\displaystyle\sum\limits_{m=0}^{n}{\!
\displaystyle\frac{(-t)^{m}}{m!}}\!{\bigg )}t^{z\!-\!n}\:dt}}                  \\[2.0 ex]
& =
\displaystyle\int\limits_{0}^{\infty}{\!{\bigg (}
\!(-1)^{z-1}e^{-t}\displaystyle\frac{t}{t\!+\!1}
-
e^{-t}\displaystyle\frac{t^{z}}{t\!+\!1}
-
\displaystyle\sum\limits_{n=1}^{\infty}{\!
(-1)^n \!\!\! \displaystyle\sum\limits_{m=n+1}^{\infty}{
\displaystyle\frac{(-t)^{m}}{m!}}\,t^{z\!-\!n}\!{\bigg )}\,dt}}                \\[2.0 ex]
& =
\displaystyle\int\limits_{0}^{\infty}{\!{\bigg (}
\!(-1)^{z-1}e^{-t}\displaystyle\frac{t}{t\!+\!1}
-
\displaystyle\frac{t^{z}}{t\!+\!1}
\!{\bigg (}\!e^{-t}+(t\!+\!1)\!
\mathop{\displaystyle\sum}\limits_{n\!\,=\!\,1}^{\infty}{
\mathop{\!\!\displaystyle\sum}\limits_{\,\,m\!\,=\!\,n+1}^{\infty}{\!\!\!
\displaystyle\frac{(-1)^{m+n}}{m!}\,t^{m-n}}}\!{\bigg )}\!\!{\bigg )}\,dt}     \\[2.0 ex]
& =
\displaystyle\int\limits_{0}^{\infty}{\!{\bigg (}
(-1)^{z-1}e^{-t}\displaystyle\frac{t}{t\!+\!1}
-
\displaystyle\frac{t^{z}}{t\!+\!1}
\!{\bigg (}\!e^{-t}+(t\!+\!1)\!
\displaystyle\sum\limits_{n=1}^{\infty}{
\displaystyle\sum\limits_{k=1}^{\infty}{
\displaystyle\frac{(-t)^{k}}{(k\!+\!n)!}}}\!{\bigg )}\!\!{\bigg )}dt}          \\[2.0 ex]
& \mathop{=}\limits_{(\ref{A_sum_sum})}
\displaystyle\int\limits_{0}^{\infty}{\!{\bigg (}
(-1)^{z-1}e^{-t}\displaystyle\frac{t}{t\!+\!1}
+
\displaystyle\frac{t^{z}}{t+1} {\Big (} (e-1)\,t - 1 {\Big )}\!\!{\bigg )} \, dt}.
\end{array}$}
\end{equation}
Integral at the right side of the equality (\ref{A_derive}), which converges in ordinary sense,
will be substituted by the sum of two integrals which converge in the ordinary sense too.
Namely, using the function of exponential integral, i.e. formula 8.211-1 \cite{Rizik_71}:
\begin{equation}
\label{Ei}
\mbox{\rm Ei}(x) =
\displaystyle\int\limits_{-\infty}^{x}{\!\displaystyle\frac{e^t}{t} \, dt}
\;\; (x<0),
\end{equation}
and using the formulas 3.351-5 and 3.241-2 from \cite{Rizik_71}:
\begin{equation}
\displaystyle\int\limits_{0}^{\infty}{
\displaystyle\frac{e^{-t}}{t+1} \, dt}
=
- e \, \mbox{Ei}(-1)
\quad \; \mbox{and} \; \quad
\displaystyle\int\limits_{0}^{\infty}{
\displaystyle\frac{t^{z-1}}{t+1} \, dt}
=
\displaystyle\frac{\pi}{\sin \pi z}
\end{equation}
\noindent
we can conclude that formula (\ref{A_sum_Slavic}) is true for $0 < \mbox{\rm Re}\:z < 1$.
According to Riemann's Theorem we can conclude that formula (\ref{A_sum_Slavic})
is true for each complex $z$. Namely, formula (\ref{A_sum_Slavic}), in integer points $z$,
is true in the sense of Cauchy's principal value at point on the basis of the Lemma \ref{L_2}
and the Theorem \ref{A_aux}.$\;\stop$

\begin{co}
For alternating Kurepa's function $A(z)$ the following representation is true
\begin{equation}
\label{A_sum_Slavic_2}
\quad
A(z)
=
{\Big (} e \displaystyle\sum\limits_{n=1}^{\infty}{\displaystyle\frac{(-1)^{n-1}}{n! \, n}} - 1 - e \gamma {\Big )}(-1)^z
+
\displaystyle\frac{\pi e}{\sin \pi z}
+
\displaystyle\sum\limits_{n=0}^{\infty}{(-1)^{n}\Gamma(z+1-n)},
\end{equation}
where the values in the previous formula, in integer points $z$,
are determined in the sense of the principal value at point.
\end{co}
\begin{co}
Function $A_{1}(z)$ is a meromorphic function with simple poles in
integer points $z=m$ $(m\!\in\!\mbox{\newmat{Z}})$ and with residue values
\begin{equation}
\mathop{\mbox{\rm res}}\limits_{z = m}{A_{1}(z)}
=
(-1)^{m-1}e + \mathop{\mbox{\rm res}}\limits_{z = m}{A(z)} \quad (m\!\in\!\mbox{\newmat{Z}}).
\end{equation}
At the point $z = \infty$ function $A_{1}(z)$ has an essential singularity.
\end{co}

\noindent
\section{\bf \boldmath \hspace*{-7.0 mm}
4 Some representations of functions $A(z)$ and $A_{1}(z)$ via incomplete gamma function}

In this section we give some representations of functions $A(z)$ and $A_{1}(z)$ via
gamma and incomplete gamma functions, where the last ones
are defined by integrals:
\begin{equation}
\gamma(a,z)
=
\displaystyle\int\limits_{0}^{z}{e^{-t} t^{\alpha-1} \, dt}
\quad \; \mbox{and} \; \quad
\Gamma(a,z)
=
\displaystyle\int\limits_{z}^{\infty}{e^{-t} t^{\alpha-1} \, dt}.
\end{equation}
Parameters $\alpha$ and $z$ are complex numbers and $t^\alpha$ takes its principal value.
Let us remark that the value $\gamma(\alpha,z)$ exists for $\mbox{\rm Re} \, \alpha > 0$
and the value $\Gamma(\alpha,z)$ exists for $|\mbox{\rm arg} \, z| < \pi$. Then,  we have:
$\gamma(a,z) + \Gamma(a,z) = \Gamma(a)$. Analytical continuation can be obtained on
the basis of representation of the $\gamma$ function using series.

\medskip
\noindent
On the basis of the well-known formula 13., page 325., from \cite{Prudnikov_81}:
\begin{equation}
\label{Temme_1}
\displaystyle\int\limits_{0}^{\infty}{e^{-t} \displaystyle\frac{t^{z+1}}{t+1} \, dt}
=
e \Gamma(z+2)\Gamma(-z-1,1)
\end{equation}
we directly get some representations of functions $A(z)$ and $A_{1}(z)$ via
an incomplete gamma function.
\begin{th}
For functions $A(z)$ and $A_{1}(z)$ the following representations are true
\begin{equation}
\label{Repr_A}
A(z)
=
-{\big (}1\!+\!e \, \mbox{\rm Ei}(-1){\big )} (-1)^z
+
e \, \Gamma(z+2) \, \Gamma(-\!z\!-\!1,1)
\end{equation}
and

\medskip

\noindent
\begin{equation}
\label{Repr_A1}
A_{1}(z)
=
-
\displaystyle\frac{\pi e}{\sin \pi z}
+
e \, \Gamma(z+2) \, \Gamma(-\!z\!-\!1,1),
\end{equation}

\smallskip

\noindent
where the values in the previous formula, in integer points $z$,
are determined in the sense of the principal value at point.
\end{th}
\begin{re}
Formula {\rm (\ref{Repr_A})} also is given in {\rm \cite{Petojevic_02}}.
\end{re}

\break

\noindent
\section{\bf \boldmath \hspace*{-7.0 mm}
5 Differential transcendency of functions $A(z)$ and $A_{1}(z)$}

In this section we provide one statement about differential transcendency of some
solutions of functional equation (\ref{A_FE_1}). Namely, using the method for
proving~of~the differential transcendency from papers
\cite{MijajlovicMalesevic_06} and \cite{MijajlovicMalesevic_07}
we can conclude that the following statement is true:
\begin{th}
Let $\mbox{$\cal M$}_{\mbox{\newmatsm{D}}}$ be a differential field of the meromorphic functions over
a domain $\mbox{\newmat{D}} \!\subseteq\! \mbox{\newmat{C}} \backslash \mbox{\newmat{Z}}^{-}$.
If $g \!=\! g(z) \!\in\! {\cal M}_{\mbox{\newmatsm{D}}}$ is one solution of the functional \mbox{equation}~\mbox{\rm
(\ref{A_FE_1})}, then $g$ is not a solution of any algebraic-differential equation over the field of rational functions
$\mbox{\newmat{C}}(z)$.
\end{th}
\begin{co}
\label{Posledica_A_K1} Especially the functions $A(z)$ and $A_{1}(z)$ are not solutions of
any algebraic-differential equation over the field of rational functions $\mbox{\newmat{C}}(z)$.
\end{co}

\bigskip

\bigskip


TITLE: $f(x)=\begin{cases} 0, x\in I \cap [0,1] \\ \frac{1}{q} , x=\frac{p}{q}, p,q \in \mathbb{N}, (p,q)=1\end{cases}$ Is $f$ measurable? Continuous?
QUESTION [1 upvotes]: $$ f:[0,1]\to \mathbb{R} \\ \text{defined as:}\\f(x)=\begin{cases} 0, x\in I \cap [0,1]  \\ \frac{1}{q} , x=\frac{p}{q}, p,q \in \mathbb{N}, (p,q)=1\end{cases}$$
I think and am pretty sure that $(,)$ means that $p,q$ aren't divisable. This function is at first glance similar to the characteristic function of $Q \cap[0,1].$ I have the definition of measurability as $f:X\to\mathbb{R}$ is measurable iff $A=\{x\in X: f(x)<c\}\in \mathcal{B}(X)$, $\forall c\in \mathbb{R}$. I have difficulties using this definition with this function with $c\in[0,1]$. I know that for a fixed $c$ in this range I would have $A$ as a set of infinity countable points, being a subset of $\mathbb{Q}\cap[0,1]$. is this $A\in \mathcal{B}([0,1])?$ I an not sure because it is a countable union of closed sets?

REPLY [1 votes]: Consider the set $f^{-1}(\{1/q\})$ with $q$ a positive integer. This is a finite set and its cardinality is $\leq q$.
Let $0<c\leq 1$ and let $n=\lfloor 1/c\rfloor$, then
$$[0,1]\setminus A=\{x\in [0,1]: f(x)\geq c\}=\bigcup_{1\leq q\leq n} f^{-1}(\{1/q\})\quad\mbox{is a finite set}$$
where $A=\{x\in [0,1]: f(x)< c\}$.  Hence $A$ is $[0,1]$ minus a finite set which means that $A\in \mathcal{B}([0,1])$.
P.S. $f$ is not continuous at $\frac{1}{2}$: $f(\frac{1}{2})=\frac{1}{2}$ but
$f(x_n)=0\to 0$ for $x_n=\frac{1}{2}+\frac{\sqrt{2}}{n}\in [0,1]\setminus \mathbb{Q}$ for $n\geq 3$.


TITLE: Can one modify $1 - e^{-\frac{1}{x^2}}$ so that all derivatives at $x=0$ are $1$ and support is $[-1,1]?$
QUESTION [2 upvotes]: It is well-known that $f(x) = 1 - e^{-\frac{1}{x^2}}$ satisfies $f^{(i)}(0) = 1$ for all natural numbers $i.$
However, its support $\{x\in \mathbb{R}:f(x)\neq 0\} = \mathbb{R}.$

Question: Can one modify $1 - e^{-\frac{1}{x^2}}$ so that all derivatives at $x=0$ are $1$  and support is $[-1,1]?$

I tried $1 - e^{1-\frac{1}{x^2}}$ because it has value $0$ at both $x=-1$ and $x=1.$
However, it is not differentiable at those points.

REPLY [2 votes]: Following the advice of Ivan Neretin and his comment on my former answer, a possible example of the function you are searching for is the following one
$$
f(x)=
\begin{cases}
e^x \cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}} & x\in]-1,+1[\\
\\
0 & x\notin]-1,+1[
\end{cases}
$$
You have that
$$
\begin{split}
\frac{\mathrm{d}f}{\mathrm{d}x}&=e^x\cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}+e^x\cdot\frac{\mathrm{d}}{\mathrm{d}x} e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}\\
&=e^x\cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}- e^x\cdot e^{-\dfrac{e^{-\frac{1}{x^2}}}{1-x^2}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\frac{e^{-\frac{1}{x^2}}}{1-x^2}\\
\end{split}
$$
For $x=0$, the second term of the right side of the equation is $0$ since it is a product of functions one of which is the (infinitely) flat function, while the first one is $1$: proceeding by induction you can prove that
$$
f^{(i)}(0)=1\quad\forall i\in\Bbb N,
$$
again by the properties of the flat function.