text:
COLOMBIA: Lawsuit to Constitutional Court to extend the exception to perform military service to all ethnic communities
In July, the Constitutional Court in Colombia received a lawsuit to extend the exception to perform compulsory military service to all ethnic communities in Colombia, an exception that currently applies to indigenous communities only. This claim also asks for these communities to be exonerated of the military compensation fee.
Problematic use of gender equality discourse by governments when considering conscription for women
When in some countries conscientious objection has been recognised and in others, conscientious objectors continue to be imprisoned, some governments are considering introducing compulsory military service for women covering it up with gender equality and anti-discrimination discourses. This has been the case for countries like México, Colombia, the United States and more recently in South Korea and Switzerland.
CO Day 2021: Showing solidarity with COs in Turkey and around the world
International Conscientious Objection Day (CO Day), 15th May, was celebrated with actions and events by activists from around the world in solidarity with those who have and those who continue to resist war. This year we had a particular focus on Turkey.
Colombia: For the anti-riot squad dismantling #SpreadingAntimilitarism
The antimilitarist movement in Colombia describes the situation that has been experienced in Colombia since April 28, 2021, the reasons why demonstrations have been held since that date and the militarized response of the government. This is an urgent call for solidarity to challenge he discourses of war and fear that today find an absolutely opposite correlate to the cultural resistance, the bets for change and the possibility of building horizons more in line with the reality of the Colombian people and their proposals to materialize peace from the territories, in diversity and dignified life.
Who arms the National Police of Colombia?
Social movements across Colombia are facing severe repression and violence from the police and military. The demonstrations began on 28th April 2021 as a general strike was called in response to an unpopular tax reform and the mishandling of the Covid-19 pandemic.
Featured Affiliate: The Collective Action of Conscientious Objectors - ACOOC (Colombia)
The Collective Action of Conscientious Objectors (Spanish: la Acción Colectiva de Objetores y Objetoras de Conciencia, or ACOOC) is a social organization which, based around the concepts of nonviolence and conscientious objection, supports alternatives to Colombian militarism, militarization, and the patriarchy, via legal, public campaigning, and educational strategies. Since 2006, ACOOC has worked alongside young people and their communities with to cultivate a peaceful culture.
Webinar: Campaigning for Conscientious Objection – Strategic Litigation
The first webinar of our series, Campaigning for Conscientious Objection to Military Service, co-organised with QUNO and IFOR, is taking place on Wednesday, 3rd Feb. In our first event, we will have campaigners from Geneva (Switzerland), Seoul (S.Korea) and Bogotá (Colombia), explaining the role of litigation in their campaigns supporting conscientious objectors.
The struggle against militarism and the commodification of the territory in Colombia
State projects and state actions centred on the commercialization and extractive industries of the territories have been historically consolidated as the preferred scenario for the militarization of these industries. In the face of this, community struggle over access to land and the guarantee of human rights is legitimate.
Colombia: militarization goes viral
Up to now, Colombia's response to the pandemic - the Common Enemy - has been one of a familiar nationalist and militarist rhetoric, a staunchly-upheld, militarized response that is unfolding in Colombia's towns and cities.
CO-Alert: Colombian CO continues resisting at the Battalion
In February of this year, we sent out a CO-Alert of a Colombian Conscientious Objector, Brayan Gonzalez, who was irregularly recruited by the Army. His CO application wasn't recognised by the military Interdisciplinary Commission. To avoid being charged for desertion, he decided to come back to the battalion. He continues refusing inside the battalion. Read Brayan's story and consider sending the support letter.
Colombia: Vice-presidency proposes promoting women recruitment to stop violence against women by the army and police members
In response to the many known cases of sexual abuse and harassment of women by members of the armed forces and the police in different places in Colombia, at the end of July, the Colombian vice-president Martha Lucía Ramirez held a meeting with the Defence Ministry and the Military Forces. During this meeting, the Vice-president proposed the inclusion and recruitment of women as a strategy that could prevent violence against women by members of the military and police forces.
Subscribe to Colombia


text:
package org.springframework.cloud.localconfig;

import org.springframework.cloud.service.common.MongoServiceInfo;

/**
 * @author Christopher Smith
 */
public class MongoServiceInfoCreator extends LocalConfigServiceInfoCreator<MongoServiceInfo> {

	public MongoServiceInfoCreator() {
		super(MongoServiceInfo.MONGODB_SCHEME);
	}

	@Override
	public MongoServiceInfo createServiceInfo(String id, String uri) {
		return new MongoServiceInfo(id, uri);
	}
}

redpajama_set_name: RedPajamaGithub


text:
Judge Orders a Pennsylvania College to Reinstate Women's Sports
A federal judge ruled on Friday that Slippery Rock University intentionally discriminated against its female student-athletes by eliminating the women's swimming and water polo teams and ordered the school to reinstate the two women's teams immediately. The president of Slippery Rock University in Pennsylvania, Robert Smith, cut three women's and five men's sports teams in January, saving $350,000 in an attempt to make a dent in the $2 million deficit in the university budget.
Field hockey was one of the three women's sports initially cut, but was reinstated after lawyers for the students informed the school of their intentions of challenging the cuts under Title IX of the Education Amendments of 1972. The two other teams, the women's water polo and swimming, were still cut. The school was brought to court over the women's water sports, leading to a reinstatement of the teams on the grounds of sex discrimination.
Said SRU junior Jessica Student, co-captain of the swim team, about the benefits of participating in women's sports, "Swimming gives me a sense of self-worth and improves my academic performance as well. I am overjoyed that I will be swimming when I get back to school," according to the Women's Law Project.
Pittsburgh Post-Gazette 7/26; Associated Press 7/31; WomenÕs Law Project Press Release 7/24

redpajama_set_name: RedPajamaCommonCrawl


Someone once said that anger is like drinking poison and hoping your enemy will die. Impatience is similarly ridiculous. You're in a rush, a salesperson is moving in slow motion, and you're ready to kill. The problem is, the person you're torturing is your own pitiful self: Your nerves are shredding, stress hormones are streaking through your body, and your heart's racing with scenarios of the tragedies that will ensue from being late. And you're not accomplishing a thing. Your frenzy is useless. You're trying to squirm out of the present into the future: Good luck with that. You are irrevocably and inescapably stuck in the moment—your feet are tied to it; you are a spoke in time's wheel. Writhe and holler as much as you want, you're not going anywhere. You are here. Immobile behind the customer with the endless questions, on the street with a tantrum-throwing child, at dinner with your husband who's telling that story yet again while you almost expire of lethal boredom.
There is an antidote, however, and its very name—patience—makes me ache with boredom. It speaks of self-control, restraint, delayed gratification. What is less fun? Impulsivity, intensity, drama—now we're talking! But after years of twisting and burning in the fires of impatience, I've come to appreciate patience as the supreme medicine. Apply patience, and frustration goes away, outrage cools, peace is yours.
1. First thing: Just stop. Catch the mind ranting that you shouldn't be in this situation—because you are. Give up the fight. You've lost the battle, but not the war.
2. Settle into the moment. You might feel your body ease down, yielding to gravity (wise move). Your shoulders and belly relax, your jaw too.
3. Go into your body with your mind's eye and find out how you know you're impatient. Are you tight, tense, breathing shallowly, clenching, jiggling? Where exactly? Focus on those sensations as closely as you can. Touch them with your mind.
4. See if you can open any tightness, breathe into any clenching. With a really ornery knot, give up trying to fix it and see if you can welcome it, make room for it.
Once your brain cools down, your powers of reason return. You thank God for cell phones and call to say you'll be late. You figure out how you can make up the minutes or hours lost to traffic. You understand your husband is telling the story again because he's nervous, and that kind of touches you. And really, it's all going to be okay.
When you give up the fight, you get time. Time stretches. You sink into the moment, and it seems infinite. You have all the time in the world.


to be realistic to remember the limitations of being human.
very nice, I have a copy of that I have saved from late 90 s after I divorced. There was a silhouette of a woman.
Hey KCH, thanks for sharing this, I hadn't see it before.
Alas "I want to know if you can be alone with yourself": that's something at which I still fail.


Tiffany Car Sales is a Used Car Dealer Springs situated at cnr 38 Bushell Crescent & Nigel Road, Selection Park, Springs.
With a great variety of Used Cars in stock, our stock varies from Cars to Bakkies at affordable prices. Tiffany Car Sales was established by Managing Director Fanus Duvenhage, to service the market with quality Used Cars, Bakkies & other vehicles.


text:
\section{Introduction}

Let $S_k(N)$ denote the space of cusp forms of even integral weight $k$ for the group
$\Gamma_0(N)$.
Let $f\in S_k(N)$ be a normalized newform with Fourier expansion
\begin{equation}\label{fourier}
f(z)=\sum_{n=1}^{\infty}a_f(n)e^{2\pi i nz}\; =\; \sum_{n=1}^{\infty}n^{\frac{k-1}{2}}\lambda_f(n)e^{2\pi inz},
\end{equation}
where $z$ is in the complex upper-half plane $\mathcal{H}$.
It is well known that the Fourier coefficients $\lambda_f(n)$ are real numbers.
The signs of the Fourier coefficients $\lambda_f(n)$ have been studied by several authors due to their various number theoretic applications. A standard result of the classical theorem of Landau on Dirichlet series with non-negative coefficients implies that the sequence $\{\lambda_f(n)\}_{n\geq 1}$ changes sign infinitely often.
It is natural to consider the signs of $\lambda_f(n)$ where $n$ varies over a sparse subset $\mathcal{S}$ of the set of natural numbers. The case when $\mathcal{S}$ is the set of prime numbers was studied by M. Ram Murty \cite{Ram} in detail. When $\mathcal{S}$ is the set of squares, cubes or fourth powers of the natural numbers, the sign changes were studied by the authors in \cite{JKV}.
Using \cite[Corollary 3]{MR} about sign change of multiplicative functions established by
K. Matom{\"a}ki and M. Radziwi{\l}{\l} one deduces that for any fixed positive integer $m$, the sequence
$\{\lambda_f(n^m)\}_{n\geq 1}$ has infinitely many sign changes, and the number of sign changes up to $x$ is $\gg x$.
Let $m$ be a fixed positive integer. In this article first we discuss the distribution of the signs of
$\{\lambda_f(p^m)\}$, where $p$ runs over all prime numbers. One has
$\lambda_f(p^m)=\lambda_{{\rm sym}^m f}(p)$, where $\lambda_{{\rm sym}^m f}(p)$ denotes the
$p$-th coefficient of the  $m$-th symmetric power $L$-function $L(s, sym^m f)$ attached to $f$.

Let us denote the set of all primes by $\mathbb{P}$. For a subset $A \subseteq \mathbb{P}$, the density of $A$
in $\mathbb{P}$, denoted by $d(A)$, is defined as
\begin{equation}
d(A)\;=\; \lim_{x\to \infty}\frac{\#\{p\leq x\mid p\in A\}}{\#\{p\leq x \mid p \in \mathbb{P}\}},
\end{equation}
provided the limit exists.
For a fixed integer $m\geq 1$, let
$$
P_m :=\{p \in \mathbb{P} \mid p \nmid N, \lambda_f(p^m)>0\},
$$
and
$$
P^{'}_{m} :=\{p \in \mathbb{P} \mid p \nmid N, \lambda_f(p^m)<0\}.
$$
In our first two theorems which are stated below, we calculate the densities of the two sets $P_m$ and $P^{'}_{m}$. From the two theorems, we see that the densities depend on whether a form is with complex multiplication or without complex multiplication as well as the parity of $m$.
\begin{theorem}\label{thm:noncm}
Let $m\geq 1$ be an integer.
Let $f\in S_k(N)$ be a normalized newform without complex multiplication.
Then the sequence $\{\lambda_f(p^m)\}_{p \in \mathbb{P}}$ changes signs infinitely often.
Moreover, we have
\begin{equation}
    d(P_m)=
    \begin{cases*}
      \frac{1}{2} & \rm{if} $m$ is odd, \\
      \frac{m+2}{2(m+1)} -\frac{1}{2\pi}\tan{(\frac{\pi}{m+1})} & \rm{if} $m$ is even,
    \end{cases*}
  \end{equation}
  and
  \begin{equation}
    d(P^{'}_{m})=
    \begin{cases*}
      \frac{1}{2} & \rm if $m$ is odd,\\
      \frac{m}{2(m+1)} +\frac{1}{2\pi}\tan{(\frac{\pi}{m+1})} & \rm if $m$ is even.
    \end{cases*}
  \end{equation}
\end{theorem}
\begin{theorem}\label{thm:cm}
Let $m\geq 1$ be an integer.
Let $f\in S_k(N)$ be a normalized newform with complex multiplication. Then
the sequence $\{\lambda_f(p^m)\}_{p \in \mathbb{P}}$ changes signs infinitely often.
Moreover, we have
\begin{equation}
    d(P_m)=
    \begin{cases*}
      \frac{1}{4} & \rm if $m$ is odd, \\
      \frac{m+2}{4(m+1)}+\frac{1}{2}    & \rm if $m\equiv0\pmod 4$, \\
      \frac{m+2}{4(m+1)} & \rm if $m\equiv 2\pmod 4$,
    \end{cases*}
  \end{equation}
  and
  \begin{equation}
    d(P^{'}_m)=
    \begin{cases*}
      \frac{1}{4} & \rm if $m$ is odd, \\
      \frac{m}{4(m+1)} & \rm if $m\equiv 0\pmod 4$,\\
      \frac{m}{4(m+1)} +\frac{1}{2}   & \rm if $m\equiv 2\pmod 4$.
    \end{cases*}
  \end{equation}
\end{theorem}
Our next result is on the abscissa of absolute convergences of two Dirichlet series attached to $f$.
Assume that $f$ is a normalized Hecke eigenform of even integral weight $k$ for the full modular group  $SL_2(\mathbb{Z})$ with Fourier expansion given in \eqref{fourier}. For a fixed integer $m\ge 1$, let
\begin{equation}\label{l_1(s)}
L_m(s,f) := \sum_{n \geq 1}\lambda_f(n^m) n^{-s}\;
 \end{equation}
and
\begin{equation}\label{l_2(s)}
L(s, sym^m f) := \sum_{n \geq 1} \lambda_{{\rm sym}^m f}(n) n^{-s}.
\end{equation}
We prove the following theorem.
\begin{theorem}\label{thm:abscissa_absolute}
For any positive integer $m\ge 1$, each of the Dirichlet series $L_m(s, f)$ and $L(s, sym^m f)$ defined in \eqref{l_1(s)} and \eqref{l_2(s)}, has abscissa of absolute convergence exactly equal to $1$.
\end{theorem}
\begin{remark}
The abscissa of absolute convergence of the series $L(s, sym^m f)$ has been found out in \cite{KMP}. But the method of proof of the same result is different in this paper. However, using the method of \cite{KMP}, one will not be able to find the abscissa of absolute convergence of the series $L_m(s, f)$ for all integers $m\ge 1$.
\end{remark}
\begin{remark}
We feel that it may be possible to prove \thmref{thm:abscissa_absolute} for higher level case. But in the proof of \thmref{thm:abscissa_absolute}, we use the result of \cite{TW} which is proved only in level $1$ case.
\end{remark}

This article is organised as follows. In the next section we provide the formulas for certain Fourier coefficients. We recall certain results regarding the distribution of the Fourier coefficients and
write down certain trigonometric formulas for the prime power Fourier coefficients by using the Hecke relation among the coefficients. Also, we recall certain theorems including Sato-Tate conjecture which will be used in establishing our results. In Section 3, we prove Theorem \ref{thm:noncm} by using the distribution results for forms without complex multiplication. In Section 4, we prove Theorem \ref{thm:cm} by using the distribution results for forms with complex multiplication.
In Section 5, we prove Theorem \ref{thm:abscissa_absolute} by using certain results established in \cite{TW}.

\section{Preliminary results}

Let
$$f=\sum_{n=1}^{\infty}n^{\frac{k-1}{2}}\lambda_f(n)e^{2\pi inz}\in S_k(N)$$
be a normalized newform.
The Ramanujan-Petersson conjecture which was proved by Deligne, states that
$$|\lambda_f(p)|\le 2,$$
where $p$ is any prime number not dividing $N$.
Thus for any prime  $p, p \nmid N$, we have
$$
\lambda_f(p)= 2\cos{\theta_p}
$$
for some $\theta_p\in [0,\pi]$. The next result which is well-known, gives a trigonometric formula for
$\lambda_f(p^m)$. For the sake of completeness we provide a proof of the result.

\begin{lemma}\label{lemma:lambdaformula}
Let $p$ be any prime number not dividing $N$ and $m\geq 1$ be any positive integer. We have
\begin{equation}\label{eq:lambdapm}
\lambda_f(p^m)\;=\; \frac{\sin{((m+1)\theta_p)}}{\sin{\theta_p}}\;,
\end{equation}
with the interpretation that the values of $\lambda_f(p^m)$ are $m+1$ and $(-1)^m (m+1)$ when the values of $\theta_p$ are $0$ and $\pi$ respectively.
\end{lemma}

\begin{proof}
To establish the above lemma we use induction on $m$. The lemma is true for $m=1$ since
$\sin{2\theta_p}=2 \sin{\theta_p}\cos{\theta_p}$.
Assume that the lemma is true for all positive integers $\le m$.
Using the Hecke relation among the Fourier coefficients, we have
$$
\lambda_f(p^{m+1})=\lambda_f(p^m)\lambda_f(p)-\lambda_f(p^{m-1}).
$$
Applying the induction hypothesis, we have
$$
\lambda_f(p^{m+1})=\frac{\sin{((m+1)\theta_p)}}{\sin{\theta_p}}
2\cos{\theta_p}-\frac{\sin{m\theta_p}}{\sin{\theta_p}}.
$$
Using the trigonometric formulas
$$\sin{((m+1)\theta_p)}=\sin{m\theta_p}\cos{\theta_p}+\cos{m\theta_p}\sin{\theta_p},$$
and
$$
\cos{2\theta_p}=2\cos^2{\theta_p}-1
$$
in the above expression, we obtain
$$
\lambda_f(p^{m+1})=\frac{\sin{m\theta_p}\cos{2\theta_p}+
\cos{m\theta_p}\sin{2\theta_p}}{\sin{\theta_p}}=\frac{\sin{((m+2)\theta_p)}}{\sin{\theta_p}}.
$$
This proves the lemma by induction.
\end{proof}

\begin{definition}\label{defn:satotatemeasure}{\rm (Sato-Tate measure)}
The Sato-Tate measure $\mu_{ST}$ is the probability measure on $[0,\pi]$ given by
$\frac{2}{\pi}\sin^2\theta d\theta$.
\end{definition}
At this point we state certain crucial results which will be used to establish \thmref{thm:noncm} and \thmref{thm:cm}.
For any interval $I\subseteq [0,\pi]$, let
$$\pi_I(x)=\# \{p \leq x \mid p\in\mathbb{P}, p\nmid N, \theta_p\in I\}.$$
 Taking $\zeta=1$ in case 3 of \cite[Theorem B]{BGHT}, we get the following equidistribution result for newforms without complex multiplication.
\begin{theorem}\label{thm:satotate}{\rm (Barnet-Lamb, Geraghty, Harris, Taylor)}
Let $f\in S_k(N)$ be a normalized newform without complex multiplication. The sequence
$\{\theta_p\}_{p\nmid N}$ is equidistributed in $[0,\pi]$ with respect to the Sato-Tate measure $\mu_{ST}$. In particular, for any sub-interval $I\subseteq[0,\pi]$ we have
$$
\lim_{x\to
\infty}\frac{\pi_I(x)}{\pi(x)}=\mu_{ST}(I)=\frac{2}{\pi}\int_I\sin^2{\theta} d\theta.
$$
\end{theorem}
The above theorem implies that if $A$ is a finite set, then the density of the set
$\{p\in \mathbb{P} ~~~|~~~ \theta_p\in A\}$ is $0$.
For any interval $I$, let $\vert I \vert$ denote the length of the interval $I$. The following theorem is the analog of Theorem \ref{thm:satotate} for the newforms with complex multiplication. For a proof we refer to \cite[Theorem 3.1.1 (a)]{AIW}.
\begin{theorem}\label{thm:deuring}{\rm (Deuring equi-distribution)}
Let $f\in S_k(N)$ be a newform with complex multiplication.
Let $I\subseteq [0,\pi]$ be an interval such that  $\frac{\pi}{2}\notin I$. Then we have
\[
\lim_{x\to \infty}\frac{\pi_I(x)}{\pi(x)•}\;=\; \frac{\vert I\vert}{2\pi},
\]
and the set $\{p \ {\rm prime} \mid \theta_p=\frac{\pi}{2}\}$ has density $\frac{1}{2}$ in the set of primes.
\end{theorem}
In this case, if $A$ is a finite set not containing $\pi/2$, then by \thmref{thm:deuring} the density of the set $\{p\in \mathbb{P} ~~~|~~~ \theta_p\in A\}$ is $0$.
Next, we state the following theorem established by Tang and Wu \cite{TW}, which will be used to prove
Theorem \ref{thm:abscissa_absolute}.
\begin{theorem}\label{thm:TW}
We have
\begin{equation}\label{eq:3}
\begin{split}
\sum_{n\leq x} \vert \lambda_{{\rm sym}^m f}(n) \vert & \sim C_m(f) x (\log
x)^{-\delta_m},\\
\sum_{n\leq x} \vert \lambda_f(n^m) \vert & \sim D_m(f) x (\log
x)^{-\delta_m},
\end{split}
\end{equation}
unconditionally for $x \rightarrow \infty$, where $C_m(f), D_m(f)$ are two positive
constants depending on $f, m$ and
\begin{equation*}
\delta_m = 1-\frac{4 (m+1)}{\pi m(m+2)} \cot \left( \frac{\pi}{2(m+1)}\right).
\end{equation*}
\end{theorem}

\section{Proof of Theorem \ref{thm:noncm}}
By Lemma \ref{lemma:lambdaformula}, we have
$$\lambda_f(p^m)= \frac{\sin{((m+1)\theta_p)}}{\sin{\theta_p}}, \ \theta_p \in [0,\pi],$$
with the obvious interpretation in the limiting case $\theta_p=0, \pi$.
Since the set $\{p\in \mathbb{P} ~~~|~~~ \theta_p=0 ~~~{\rm or}~~~\pi\}$ has density $0$,
we may assume that $\theta_p$ is different from $0$ and $\pi$.
For $\theta_p$ in the interval $(0,\pi)$, we know that $\sin{\theta_p} > 0$. Therefore the sign of
$\lambda_f(p^m)$ is the same as the sign of $\sin{((m+1) \theta_p)}$.
\subsection{Even case}
First assume that $m$ is even. Then we have
\begin{equation}\label{eq:1}
\begin{split}
\sin{((m+1) \theta_p)} >0 & \iff
(m+1)\theta_p\in \bigcup_{j=0}^{\frac{m}{2}}(2j\pi , (2j+1)\pi)\\
& \iff
\theta_p\in \bigcup_{j=0}^{\frac{m}{2}}\left(\frac{2j\pi}{m+1}, \frac{(2j+1)\pi}{m+1}\right).
\end{split}
\end{equation}
Similarly,
\begin{equation}\label{eq:2}
\sin{((m+1) \theta_p)} <0 \iff
\theta_p\in \bigcup_{j=1}^{\frac{m}{2}}\left(\frac{(2j-1)\pi}{m+1}, \frac{2j\pi}{m+1}\right).
\end{equation}
From Theorem \ref{thm:satotate} we know that the sequence $\{\theta_p\}_{p\in \mathbb{P}}$ with
$p\not|N$ is
equi-distributed in $[0,\pi]$ with respect to the Sato-Tate measure $\mu_{ST}$.
Therefore there exist infinitely many primes $p$ such that
$\theta_p\in \bigcup_{j=0}^{\frac{m}{2}}\left(\frac{2j\pi}{m+1}, \frac{(2j+1)\pi}{m+1}\right)$, and there exist infinitely many primes $p$ such that
$\theta_p\in \bigcup_{j=1}^{\frac{m}{2}}\left(\frac{(2j-1)\pi}{m+1}, \frac{2j\pi}{m+1}\right)$.
Thus the sequence $\{\lambda_f(p^m)\}_{p\in \mathbb{P}}$ changes signs infinitely often.
Next we calculate the densities of the sets $P_m$ and $P^{'}_m$.
Let
$$
A=\bigcup_{j=0}^{\frac{m}{2}}\left(\frac{2j\pi}{m+1}, \frac{(2j+1)\pi}{m+1}\right)
 \subseteq [0,\pi].
$$
From the definitions of the density $d$ and Sato-Tate measure $\mu_{ST}$, we have
$$
d(P_m)= \mu_{ST}(A)= \frac{2}{\pi}\int_{A}\sin^2t \ dt=
\frac{2}{\pi}\sum_{j=0}^{\frac{m}{2}}\int_{\left(\frac{2j\pi}{m+1},\frac{(2j+1)\pi}{
m+1}\right)}\sin^2t \ dt.
$$
Using the fact $\int\sin^2t \ dt= \frac{t}{2}-\frac{\sin 2t}{4}$ in the above expression, we obtain
$$
d(P_m)=\frac{2}{\pi}\sum_{j=0}^{\frac{m}{2}}
\left(\frac{\pi}{2(m+1)}-\frac{\sin{\left(\frac{(2j+1)2\pi}{m+1}\right)}}{4}
+\frac{\sin{\left(\frac{4j\pi}{m+1}\right)}}{4}\right).
$$
Denoting $\frac{2\pi}{m+1•}$ by $x$ in the above expression, we get
\begin{equation}\label{j1}
d(P_m)=\frac{m+2}{2(m+1)}-\frac{1}{2\pi}\{\sin{x}-\sin{2x}+\sin{3x}-\cdots+
\sin{((m+1)x)}\}.
\end{equation}
Since $\sin{((m+1)x)}=0$, the trigonometric term $\sin{x}-\sin{2x}+\sin{3x}-\cdots+\sin{((m+1)x)}$
in \eqref{j1} can be rewritten as
$$
\{\sin{x}+\sin{2x}+\sin{3x}+\cdots+\sin{mx}\}-2\{\sin{2x}+\sin{4x}+\cdots+\sin{mx}\}.
$$
Now using the trigonometric  identity
$$
\sin{x}+\sin{2x}+\cdots+\sin{nx}\;=\; \frac{\sin{\frac{(n+1)x}{2}}\sin{\frac{nx}{2}}}
{\sin{\frac{x}{2}}}
$$
in \eqref{j1},
we obtain
\[
d(P_m)=\frac{(m+2)}{2(m+1)}-\frac{1}{2\pi•}\left(\frac{\sin
\frac{(m+1)x}{2•}\sin\frac{mx}{2•}}{\sin \frac{x}{2}•}-2\frac{\sin
((\frac{m}{2•}+1)x)\sin\frac{mx}{2•}}{\sin x}\right)\; .
\]
Since $\sin \left(\frac{m+1•}{2•}\right)x=0$, we deduce that
\[
d(P_m)=\frac{m+2}{2(m+1)•}+\frac{1}{\pi}\frac{\sin
((\frac{m}{2•}+1)x)\sin\frac{mx}{2•}}{\sin x}\;.
\]
Putting $x= \frac{2\pi}{m+1}$ in the above equation and simplifying it, we get
\[
d(P_m)\;=\; \frac{m+2}{2(m+1)•}-\frac{1}{2\pi•}\tan\left(\frac{\pi}{m+1•}\right)\;
.
\]

Since the density of the set
$$
\{p \in \mathbb{P} \mid \lambda_f(p^m)=0\}
=\{p \in \mathbb{P} \mid \theta_p =\frac{i\pi}{m+1}, 1\le i\le m\}
$$
is $0$
in the set of primes, the density $d(P^{'}_m)$ of $P^{'}_m$ is $1-d(P_m)$. Thus
\[
d(P^{'}_m)\;=\;
\frac{m}{2(m+1)•}+\frac{1}{2\pi}\tan\left(\frac{\pi}{m+1•}\right)\; .
\]

\subsection{Odd case}
We assume that $m$ is an odd positive integer. Using the arguments similar to the even case, we have
$$
\lambda_f(p^m)>0 ~~~~\mbox{if~~~and~~~~only~~~~if}~~~~
\theta_p\in \bigcup_{j=1}^{\frac{m+1}{2}}\left(\frac{(2j-2)\pi}{m+1}, \frac{(2j-1)\pi}{m+1}\right),
$$
and
$$\lambda_f(p^m)<0 ~~~~\mbox{if~~~and~~~~only~~~~if}~~~~
\theta_p\in \bigcup_{j=1}^{\frac{m+1}{2}}\left(\frac{(2j-1)\pi}{m+1}, \frac{2j\pi}{m+1}\right).
$$
Let
$$
A= \bigcup_{j=1}^{\frac{m+1}{2}}\left(\frac{(2j-2)\pi}{m+1}, \frac{(2j-1)\pi}{m+1}\right)
$$
and
$$
B=\bigcup_{j=0}^{\frac{m-1}{2}}\left(\frac{(2j+1)\pi}{m+1}, \frac{(2j+2)\pi}{m+1}\right)
=\bigcup_{j=1}^{\frac{m+1}{2}}\left(\frac{(m-2j+2)\pi}{m+1}, \frac{(m-2j+3)\pi}{m+1}\right).
$$
To prove that the sets $P_m$ and $P_m^{'}$ have same density $\frac{1}{2}$,
we prove that the Sato-Tate measures of the two sets $A$ and $B$ are equal. Since
each of the sets $A$ and $B$ are union of $\frac{m+1}{2}$ disjoint intervals given in
the above form, to prove that $\mu_{ST}(A)=\mu_{ST}(B)$, it is sufficient to prove that
$\mu_{ST}(I_j)=\mu_{ST}(I_j^{'})$ for each $j$ with $1\le j\le \frac{m+1}{2}$, where
$$
I_j=\left(\frac{(2j-2)\pi}{m+1}, \frac{(2j-1)\pi}{m+1}\right)~~~~~~~~~~~~~
\mbox{and}~~~~~~~~~~~~~I_j^{'}=\left(\frac{(m-2j+2)\pi}{m+1}, \frac{(m-2j+3)\pi}{m+1}\right).
$$

Using the integral evaluation $\int \sin^2 t \ dt= \frac{t}{2}-\frac{\sin 2t}{4}$, we get
\begin{equation}\label{j2}
\int_{I_j}\sin^2x \ dx=\frac{\pi}{2(m+1)}+\frac{1}{4}
\left(\sin\left(\frac{2(2j-2)\pi}{m+1}\right)-\sin\left(\frac{2(2j-1)\pi}{m+1}\right)\right).
\end{equation}
On the other hand we have
\begin{equation}\label{j3}
\int_{I_j^{'}}\sin^2x \ dx=\frac{\pi}{2(m+1)}+\frac{1}{4}
\left(\sin\left(\frac{2(m-2j+2)\pi}{m+1}\right)-\sin\left(\frac{2(m-2j+3)\pi}{m+1}\right)\right).
\end{equation}
We know that
$$
\sin{\left(\frac{2(m-2j+2)\pi}{m+1}\right)}=\sin{\left(2\pi-\frac{2(2j-1)\pi}{m+1}\right)}
=-\sin{\left(\frac{2(2j-1)\pi}{m+1}\right)}.
$$
Similarly,
$$\sin\left(\frac{2(m-2j+3)\pi}{m+1}\right)=-\sin\left(\frac{2(2j-2)\pi}{m+1}\right).$$
Thus the expressions in the right hand side of \eqref{j2} and \eqref{j3} are same.
Therefore from the above discussion, we deduce that
$$
\int_{I_j}\sin^2x \ dx=\int_{I_j^{'}}\sin^2x \ dx.
$$
This proves the theorem.
\section{Proof of Theorem \ref{thm:cm}}
Let $f$ be a normalized newform with complex multiplication.
\subsection{Odd case}
Let $m\ge1$ be an odd positive integer. If $\theta_p=\frac{\pi}{2}$, then
$$
\lambda_f(p^m)=\frac{\sin{\frac{(m+1)\pi}{2}}}{\sin{\frac{\pi}{2}}}=0.
$$
Therefore from Theorem \ref{thm:deuring}, we have
$$d(\{p \ {\rm prime} \mid \lambda_f(p^m)=0\})= d(\{p \ {\rm prime} \mid \theta_p=\pi/2\}) = \frac{1}{2}.$$
Following the arguments similar to the odd case in the previous section, we have
$$
\lambda_f(p^m) > 0 \iff \theta_p \in
\bigcup_{j=1}^{\frac{m+1}{2}}\left(\frac{(2j-2)\pi}{m+1}, \frac{(2j-1)\pi}{m+1}\right)
\bigg \backslash \{\pi /2\},
$$
and
$$
 \lambda_f(p^m)<0 \iff
 \theta_p \in \bigcup_{j=1}^{\frac{m+1}{2}}\left(\frac{(2j-1)\pi}{m+1}, \frac{2j\pi}{m+1}\right)
\bigg \backslash \{\pi /2\}.
$$
Using Theorem \ref{thm:deuring}, we have
$$
d(P_m)=\frac{1}{2\pi}\sum_{j=1}^{\frac{m+1}{2}}\frac{\pi}{m+1}=\frac{1}{4}.
$$
Similarly, $d(P_m^{'})=\frac{1}{4}$.
\subsection{Even case}
Let $m\ge 1$ be an even integer. Then for primes $p$ such that $\theta_p=\frac{\pi}{2}$, we have
\begin{equation*}
\lambda_f(p^m)= \sin{\frac{(m+1)\pi}{2}}=
    \begin{cases*}
      1 & if $m\equiv 0\pmod 4$, \\
      -1       & if $m\equiv 2\pmod 4$.
    \end{cases*}
  \end{equation*}
  From Theorem \ref{thm:deuring}, we have
  \begin{equation}\label{eq:03}
  d(\{p \ {\rm prime} \mid \theta_p=\pi/2\}) = \frac{1}{2}.
  \end{equation}
  Suppose that $m\equiv 0 \pmod{4}$.
  Following the arguments similar to even case in the previous section, we have
$$
\lambda_f(p^m)>0  \iff
 \theta_p \in \bigcup_{j=0}^{\frac{m}{2}}\left(\frac{2j\pi}{m+1}, \frac{(2j+1)\pi}{m+1}\right)
 \bigcup \{\pi/2\},
 $$
and
$$
\lambda_f(p^m)<0 \iff
 \theta_p \in \bigcup_{j=1}^{\frac{m}{2}}\left(\frac{(2j-1)\pi}{m+1}, \frac{2j\pi}{m+1}\right)
 \bigg\backslash \{\pi/2\}.
$$
Now using Theorem \ref{thm:deuring} and \eqref{eq:03}, we have
$$
d(P_m)=\frac{m+2}{4(m+1)}+\frac{1}{2}
$$
and
$$
 d(P^{'}_m)=\frac{m}{4(m+1)}.
$$
If $m\equiv 2 \pmod{4}$, then similarly we can prove that
$$
d(P_m)=\frac{m+2}{4(m+1)}
$$
and
$$
 d(P^{'}_m)=\frac{m}{4(m+1)} +\frac{1}{2}.
$$
This proves the theorem.

\section{Proof of Theorem \ref{thm:abscissa_absolute}}\label{abscissa}
First we prove the following result. If
$$L(s) := \sum_{n=1}^{\infty} \frac{a(n)} {n^s}$$
is a Dirichlet series such that $\sum_{n \geq 1} | a(n) |$ diverges then
the abscissa of absolute convergence
$\sigma_{a}$ of
$L(s)$ is given by
\begin{equation}\label{eq:4}
\sigma_a= \inf \left\{\alpha \in \mathbb{R} \mid \sum_{n \leq N} \vert a(n)
\vert = O_\alpha(N^\alpha) \right\}.
\end{equation}
Let
$$
\gamma= \inf \left\{\alpha \in \mathbb{R} \mid \sum_{n \leq N} \vert a(n)
\vert = O_\alpha(N^\alpha) \right\},
$$
and
$$
A(N)=\sum_{n=1}^{N}|a(n)|.
$$
By partial summation we have
\begin{equation}\label{well}
\sum_{n=1}^{N}\frac{|a(n)|}{n^\beta}=A(N)\frac{1}{N^\beta}+\beta\int_{1}^N A(u)\frac{1}{u^{\beta+1}} du
\end{equation}
for any $\beta\in \mathbb{R}$.
If $\beta>\gamma$, then
$$
A(N)\frac{1}{N^\beta} \to 0 ~~~~{\rm as}~~~~~ N\to \infty
$$
and
$$
\int_{1}^\infty A(u)\frac{1}{u^{\beta+1}} du< \infty.
$$
Thus by taking $N\to \infty$ in both the sides of \eqref{well}, we deduce that
if $\beta>\gamma$, the series
$$
\sum_{n=1}^{\infty}\frac{|a(n)|}{n^\beta}
$$
is convergent. If $\beta<\gamma$, then $N^\beta=o(A(N))$. Therefore
$$
A(N)\frac{1}{N^\beta} \to \infty ~~~~{\rm as}~~~~~ N\to \infty
$$
and
$$
\int_{1}^\infty A(u)\frac{1}{u^{\beta+1}} du
$$
is infinite. Thus by taking $N\to \infty$ in both the sides of \eqref{well}, we deduce that
if $\beta<\gamma$, the series
$$
\sum_{n=1}^{\infty}\frac{|a(n)|}{n^\beta}
$$
is divergent. This proves that $\sigma_a=\gamma$.

Now we establish the abscissa of absolute convergence of the series $L_m(s, f)$. The establishment of the abscissa of absolute convergence for the series $L(s, sym^m f)$ can be treated exactly along the same lines as $L_m(s, f)$. First, observe that the series $\sum_{n \geq 1} \vert \lambda_f(n^m) \vert$ is divergent by the asymptotic formula given in \eqref{eq:3}.  Therefore the abscissa of absolute convergence $\sigma_{a,m}$ of $L_m(s,f)$ is equal to
$$
 \inf \left\{\alpha \in \mathbb{R} \mid \sum_{n \leq N} \vert \lambda_f(n^m) \vert
= O_\alpha(N^\alpha) \right\}.
$$
For any $\epsilon>0$, we have  $ \lambda_f(n) = O_\epsilon(n^\epsilon)$ by Deligne bound.
Using this bound, we have $\sigma_{a,m} \leq 1$.
If $\sigma_{a,m} < 1$ then there exists  $\delta>0$ such that
\begin{equation}\label{eq:5}
\sum_{n \leq N} \vert \lambda_f(n^m) \vert = O_\delta (N^{1-\delta}).
\end{equation}
On the other hand, using the asymptotic formula \eqref{eq:3} we have
$$
\frac{N}{(\log N)^{\delta_m}} \ll \sum_{n \leq N} \vert \lambda_f(n^m) \vert \
\ \ \ {\rm as} \ N \rightarrow \infty.
$$
This is a contradiction to \eqref{eq:5}. Therefore
$\sigma_{a,m} =1$.
In the case of $L(s,sym^m f)$, we need to use the bound
$\lambda_{{\rm sym}^m f}(n) = O_\epsilon(n^\epsilon)$, which holds for each  $\epsilon >0$, and the
asymptotic formula for the sum
$\sum_{n \leq x} \vert \lambda_{{\rm sym}^m f}(n) \vert$ presented in Theorem
\ref{thm:TW} to get the required result. The bound $\lambda_{{\rm sym}^m f} (n) = O_\epsilon(n^{\epsilon})$ follows from the observation that
$\vert \lambda_{{\rm sym}^m f} (n) \vert \leq d_{m+1}(n)$ \cite[Equation (1.8)]{TW},
where $d_{m+1}(n)$ denotes the number of ways one can write $n$ as a product of $m + 1$ natural numbers.\\

\noindent{\bf Acknowledgements.}
We would like to thank Prof. M. Ram Murty for his valuable suggestions which were helpful in establishing the density results. We also thank Prof. J.-P. Serre for his comments on the paper and correcting one of the definitions. The research of the second author was partially supported by a DST-SERB grant ECR/2016/001359.

redpajama_set_name: RedPajamaArXiv


text:
Journal of Signal and Information Processing
Vol.5 No.3(2014), Article ID:48303,9 pages DOI:10.4236/jsip.2014.53010
An Intonation Speech Synthesis Model for Indonesian Using Pitch Pattern and Phrase Identification
Yohanes Suyanto, Subanar, Agus Harjoko, Sri Hartati
Department of Computer Science and Electronics, Gadjah Mada University, Yogyakarta, Indonesia
Email: yanto@ugm.ac.id, subanar@ugm.ac.id, aharjoko@ugm.ac.id, shartati@ugm.ac.id
Received 25 May 2014; revised 20 June 2014; accepted 16 July 2014
Prosody in speech synthesis systems (text-to-speech) is a determinant of tone, duration, and loudness of speech sound. Intonation is a part of prosody which determines the speech tone. In Indonesian, intonation is determined by the structure of sentences, types of sentences, and also the position of the word in a sentence. In this study, a model of speech synthesis that focuses on its intonation is proposed. The speech intonation is determined by sentence structure, intonation patterns of the example sentences, and general rules of Indonesian pronunciation. The model receives texts and intonation patterns as inputs. Based on the general principle of Indonesian pronunciation, a prosody file was made. Based on input text, sentence structure is determined and then interval among parts of a sentence (phrase) can be determined. These intervals are used to correct the duration of the initial prosody file. Furthermore, the frequencies in prosody file were corrected using intonation patterns. The final result is prosody file that can be pronounced by speech engine application. Experiment results of studies using the original voice of radio news announcer and the speech synthesis show that the peaks of F0 are determined by general rules or intonation patterns which are dominant. Similarity test with the PESQ method shows that the result of the synthesis is 1.18 at MOS-LQO scale.
Keywords:Speech Synthesis, PESQ, Intonation, Indonesian
Modeling emotions in speech synthesis are made based on a number of parameters such as place, the level of the fundamental frequency , voice quality, articulation or accuracy. A different speech synthesis technique aimed for parameters in the different levels Schroder 2001. Formant synthesis, also known as the rule-based synthesis, creates the sound of speech based on the rules alone. No people's voice recording is involved in it. The result is a speech sound like a robot voice.
In the concatenation synthesis, voice recording from speaker strung together to produce synthetic speech. In this process, diphone which cuts the signal in mid phoneme until next mid phoneme is often used. Diphone is recorded with a monotonous tone. At the synthesis, required contour is constructed with signal processing techniques that result in a slight distortion. But the final result is more natural than formant synthesis [1] . In most diphone synthesis systems, only and duration can be controlled, while intensity for each segment is not easy to control.
Emotions of speaker such as neutral, joy, boredom, anger, sadness, fear, and indignation are reflected in the duration and intonation of speech from a person [2] . By copying pitch and duration of the original utterances to a monotonous one, it can be proved that both factors are sufficient to express various emotions. Research results show that emotion can be expressed simply by manipulating pitch and duration. Intonation is a set of syntactic prosody inherent in the utterance sentence [2] . The intensity of sound is not manipulated.
Contour tones containing duration and pitch are calculated by converting an intonation plan into a sequence of prosodic tone and highly dependent on the model of the speaker [3] . Contour patterns derived from the speaker's voice are applied at the time of speech synthesis.
Research on Indonesian speech synthesis which makes the coupling phoneme based speech synthesis has been done [4] . In general, the results can pronounce words in Indonesian quite fluently and can be understood by most listeners. However, speech synthesis has not produced intonation patterns (prosody) as the original speech [4] . Still there are some inaccuracies in the assembly of phonemes [4] .
In Indonesian, utterance position on word stress does not depend on the number of syllables, but the stress falls on the penultimate syllable [5] . Prosodic type is characterized by and intensity. The pressure of nouns group on preverb position will be marked by the duration instead of the frequency. In this group, two peaks of are found.
2. Speech Synthesis Model Using Pitch Pattern and Phrase Identification
Speech synthesis is built to converts the input text into speech. Conversion of text into speech considering a sentence structure, intonation patterns, as well as a normalization. The results are in the form of text that qualifies as input of voice generator.
Prosodic rules which determine the frequency, duration, and intonation was compiled from the results of studying materials about intonation in Indonesian [6] . This rule is encoded to ease the implementation later. Speech synthesis parameters are grouped into four categories, namely pitch, duration, quality and articulation. The pitch parameter determines the value of [7] . The duration parameter determines the duration of rhythm control and rate of speech. Usually the pitch and duration parameters are associated with the phenomenon of linguistic (words or phrases) [3] [8] . The sound quality of speech is a parameter that determines the overall sound quality. The articulation parameter determines clearance of speech. Based on these parameters the model of speech synthesis is composed and shown in Figure 1.
The text comes from standard input or the file is read by the text normalization module in order to obtain text which is free of numbers and signs so that the whole text can be pronounced. The output of a normalization module text-norm will be used by the selector pattern, syntax analyst, and synthesis module. The selector pattern module will produce a pattern based on text-norm and an existing pattern database. This module is responsible for the pitch of each phoneme. Syntax analyst module will generate sentence syntax structure of the text-norm with the phrases and words interval duration information.
The synthesis module combined text-norm from normalization module, phonemes with prosody information in it based on the pattern from pattern selector, and attributes of sentence structure from syntax analysts into phonemes and prosody form. This module is responsible for the articulation and the quality of speech. Finally, this form will be voiced by DSP module.
2.1. Normalization Module
The text is a phrase that came from the standard input or files that may be contains numbers or punctuation
Figure 1. The proposed model of speech synthesis.
marks that can be pronounced. If there are numbers then this module will convert it into text. For example, in the text there is a number "12345" then it will be converted to "dua belas ribu tiga ratus empat puluh lima". Similarly, if there is an abbreviation, like "UGM" then it will be changed to "u-ge-em". Symbols need to be changed too, e.g. " ", " ", or " ".
2.2. Fundamental Frequency Pattern Module
Intonation patterns in the form of pairs of time and fundamental frequency obtained by extraction of radio announcer voice. Based on the previously defined parameters the intonation pattern chose from a set of intonation patterns. The selection is done by paying attention on the pattern duration and length of the text. The duration of the speech text can be calculated to about sum of the duration of all phonemes. Results of pattern selector are for time from 0 to . The pattern have been adapted to the text by interpolation method.
2.3. Syntactic Module
This module perform the text-parsing norm based standard grammatical structures. An important outcome of this module is the determination of the duration of phonemes and duration of pauses between phrases or between words.
Syntax analysis that will be used is Bahasa Indonesia structure rules [9] -[11] . The rules are converted into a context-free grammar structure (CFG). One notation is often used to write the CFG is Backus-Naur Form—BNF. Here are the example segments of Bahasa Indonesia BNF:
Noun phrase ("frasa-nominal") can be construct from noun ("nomina") or noun phrase followed by another phrase. For example, noun phrase and verb phrase ("frasa-verbal"). In another segment verb phrase can be construct from verb ("verba") and another phrase. Completed BNF for Bahasa Indonesia has been made.
2.4. Synthesis Module
Sentence in the text is converted into the format of pho with attention to intonation patterns and phoneme duration and duration of pauses between phrases. Intonation patterns derived from the module selector pattern, while the duration of phonemes and pause duration is derived from syntax analyst module. The pho format is ready to be fed to the DSP module to be voiced.
From pattern selector module resulting array of phonemes, includes spaces, and pho notation for the corresponding phonem. On the other hand, analyst module generates duration of pauses between words from the input sentence. The length of this pause is then used to correct the lag length of the pattern selector module output. It is expected the end result is in accordance with the intonation patterns and also in accordance with the structure of the sentence.
Illustration of the results of output from the intonation pattern module, the sentence structure analyst module, and synthesis module are shown in Figure 2. Output of intonation patterns module focused on intonation while sentence analyst module more focused on the pause duration between words.
Fist, a pattern that contents of series of is got from a recorded voice by Praat application. Then, the input text was normalized by a normalized module written in PHP. Its output is called as text-norm. The analyst syntax running on text-norm gets the sentence structure with duration of phrases and words. The synthesis module composed a series of phonemes and prosody from the norm-text, the pattern, and the sentence structure. Finally, a voice will be generated by MBROLA application from this series of phonemes and prosody.
In this research some tools are used e.g. a Praat and an MBROLA applications. The Praat application is used to get from voice files while MBROLA application used to generate voice from phonemes + prosody form. MBROLA is not a speech synthesis complete software. MBROLA requires input of phonemes and prosody in a form that matches to the MBROLA form called pho. That is why the text to be synthesized must be converted first into MBROLA form (pho). MBROLA made by TCTS Lab of Facult? é Polytechnique de Mons Belgium [12] .
Table 1 shows the result of normalization module. Symbols, numbers, and abbreviation converted correctly. The punctuation marks: comma , period , and hyphenation are still unchanged.
of sentence "Empat LSM lembaga swadaya masyarakat diantaranya ICW indonesia corruption watch PSHK pusat studi hukum dan kebijakan menyampaikan aspirasi kepada panitia ad hock PAH empat DPD di
Figure 2. Relation between selector pattern and syntax analyst (sentence structure) modules. Top: result of intonation pattern, bottom: result of syntax analyst.
komplek parlemen senayan Jakarta hari ini" is shown in Table 2 but there are only partial results. A full result contained about 1200 data. The table obtained by Praat application [7] [13] .
If a sentence "DPD kemudian tetap melakukan uji kepatutan dan kelayakan" fed into the syntax analyst module, a sentence structure diagram obtained as shown in Figure 3.
Voice synthesis module in this study is part of the pho file processing module according to the rules of normalization followed the general tone combined with syntax analysts and intonation patterns. For comparison, the results will be presented waveform synthesis with flat intonation, intonation with the general rule, and intonation pattern.
Table 1. Result of normalization module for complete sentences.
Table 2. Excerpts F0 results for sentence.
Figure 3. Example of sentence structure.
Firstly, the sentence "Wimar tidak dapat mengikuti pemakaman Gus Dur di Jombang" is synthesized with flat intonation. The results of the graph are shown in Figure 4. Secondly, for the same sentence but intonation synthesized according to the general rules, the results of the graph are shown in Figure 5. And, Figure 6 shows the synthesis of the same as the second but improved by incorporating elements of intonation pattern. The selected intonation pattern has as shown in Figure 7.
Similarity checking beetween synthesis result and original voice can be done by quantity approach. In this study, the checking similarity uses the PESQ method [14] [15] . The result of PESQ for several sentences is shown in Table3
Table 1 shows that the integer number of year (2013) and the amount of money (179) can be converted correctly. Likewise fractions (182,57) can be converted to text correctly. Because 57 is a fraction, so it is converted into a "lima tujuh" (five seven) instead of "lima puluh tujuh" (fifty seven). The comma in 182,57 is decimal separator instead of thousand separator. Abbreviations and acronyms can be converted correctly. KJRI, BPS, and AS are converted to "ka je er i", "be pe es", and "a es". The "ybs." and "Sdri." acronyms are converted to "yang bersangkutan" and "Saudari".
Praat application output has 2 columns i.e time and fundamental frequency . In this study, the time has 0.01 seconds interval so there are 100 data in every second of voice. See Table 2, a graph like Figure 7 can be obtained from this table. The graph shows that it has 2 peaks in second and second. During the first 7.5 seconds, there are several small peaks. Those peaks can be dominated when combination of general rules and intonation patterns has made.
A flat intonation was obtained from MBROLA when the frequency was set in one value for a long of speech
Figure 4. Graph F0 with flat intonation.
Figure 5. Graph F0 with intonation according to general rules.
Figure 6. Graph F0 with the general rules and intonation patterns.
synthesis. Figure 4 shows the result of flat intonation graph when the frequency set to 115 Hz. Although several small peaks were shown in a flat intonation, it happened because of MBROLA characteristic.
In gerenal rules of intonation according to [5] the peaks of appeared at the syllables of nouns. For nouns before a verb peak occur in a syllable before the last syllable called the penultimate. In a sentence "Wimar tidak bisa mengikuti pemakaman Gus Dur di Jombang" "Wimar" is a noun before the verb "mengikuti". In that case peak occur in syllable "Wi" from the word "Wimar". That is in the start of sentence. See the first seconds of graph in Figure 4. For nouns after a verb peak occur in the first syllable of the first noun. In the sentence the first syllable of the first noun is "Gus". This syllable is the syllable of 19 syllables in the sentence. We can calculate that this syllable occur in about
Figure 7. Graph of selected pattern.
Table 3. Values of PESQ for several sentences.
This peak can be seen at the fourth second of the graph in Figure 5. After that the values of continues to fall until the end of the sentence.
The pattern has graph as shown in Figure 7. It has two peaks that occur at and second from 17.5 s speech synthesis a long. We can calculate the segments of its value
This selected pattern then combined to speech voice that has as Figure 5 obtained a speech voice that has as Figure 6. The peaks of the syntesis result can be predict from Equation (9) occur at
This results is not match with the graph second peak in Figure 6. There is no peak in this value. If we used Equation (10) and (11) we get
There are peaks that match with peaks in Figure 6. The second peak can take a place the wrong result as Equation (9). Thus Figure 6 is combination of Figure 5 and Figure 7.
From Table 3, we get that the average value of PESQ is 1.096. This value can be converted to MOS-LQO scale by the following formula [14]
where is MOS-LQO in the range 1.02 to 4.56, is PESQ value in the range to 4.5. The value of MOS-LQO is
Speech synthesis model with a combination of general rules and intonation patterns of the fundamental frequency can be implemented. The peaks of determined by general rules or intonation patterns are dominant. Indonesian has intonation general rules, but should not be adhered strictly. Similarity test with the PESQ method shows that the synthesis results are about 1.18 at MOS-LQO scale. A speech synthesis based on the announcer voice intonation patterns is expected to be used as an alternative method of Indonesian speech synthesis refinement.
Schroder, M. (2001) Emotional Speech Synthesis: A Review. Proceedings of Eurospeech 2001, 1, 561-564.
Vroomen, J., Collier, R. and Mozziconacci, S. (1993) Duration and Intonation in Emotional Speech. Proceedings of the Third European Conference on Speech Communication and Technology, Berlin, 22-25 September 1993, 577-580.
Campbell, W.N., Isard, S., Monaghan, A.L.C. and Verhoeven, J. (1990) Duration, Pitch and Diphones in the CSTR TTS System. Proceedings of the International Conference on Spoken Language Processing, Kobe, 1 January 1990, 825-828.
Tritoasmoro, I.I. (2006) Text-to-Speech Bahasa Indonesia Menggunakan Concatenation Synthesizer Berbasis Fonem. Seminar Nasional Sistem dan Informatika, Bali, 17 November 2006, 171-176.
Laksman, M. (1995) Realisasi Tekanan Kata dalam Bahasa Indonesia. PELLBA 8, pages 179{215. Lembaga Bahasa Unika Atmajaya, Jakarta, 1995.
Halim, A. (1975) Intonation in Relation to Syntax in Bahasa Indonesia. Proyek Pengembangan Bahasa dan Sastra Indonesia dan Daerah, Departemen Pendidikan dan Kebudayaan.
van Lieshout, P.P. (2003) PRAAT Short Tutorial. University of Toronto, Graduate Department of Speech-Language Pathology, Faculty of Medicine, Oral Dynamics Lab (ODL), Toronto.
Heuven, V.J.V. and Zanten, E.V. (2007) Prosody in Indonesian Languages. LOT, Utrecht.
Sakri, A. (1994) Bangun Kalimat Bahasa Indonesia. 2nd Edition, Penerbit ITB, Bandung.
McCune, K.M. (1985) The Internal Structure of Indonesian Roots. Number v. 2 in the Internal Structure of Indonesian Roots. Badan Penyelenggara Seri Nusa, Universitas Katolik Indonesia Atma Jaya, Jakarta.
Vamarasi, M.K. (1986) Grammatical Relations in Bahasa Indonesia. Cornell University, Ithaca.
Mbrola, T. (2009) The MBROLA Home Page. http://tcts.fpms.ac.be/synthesis/mbrola/
Boril, H. and Pollák, P. (2004) Direct Time Domain Fundamental Frequency Estimation of Speech in Noisy Conditions. EUSIPCO 2004, 2, 1003-1006.
ITU (2001) ITU-T Recommendation P.862: Perceptual Evaluation of Speech Quality (PESQ): An Objective Method for End-to-End Speech Quality Assessment of Narrow-Band Telephone Networks and Speech Codecs. Technical Report, ITU.
Rix, A.W., Beerends, J.G., Hollier, M.P. and Hekstra, A.P. (2001) Perceptual Evaluation of Speech Quality (PESQ)— A New Method for Speech Quality Assessment of Telephone Networks and Codecs. IEEE International Conference on Acoustics, Speech, and Signal Processing, Salt Lake City, 7-11 May 2001, 749-752.
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Q: How to split an array into two arrays, one for even and one for odd in C I am doing an exercise that consists of a program that generates 100 random numbers ranging from 1 to 1000 and saves them in an array.
I have to separate this array into two arrays, one for even numbers and one for odd numbers and print the lists.
I have managed to separate into two arrays, but when they are empty, they invent them. How can i avoid it? Any clue to improve it?
Thank you so much.
That's my code:
#include<stdio.h>
#include<conio.h>

int main(){
    int array100[100];
    int arraypar[100];
    int arrayimpar[100];

    int i, n=0, m=0;

    for(i=0;i<100;i++){
        array100[i] = rand() % (1000 + 1);
        printf("%i ", array100[i]);
    }

    printf("\n\n");

    for(i=0;i<100;i++){
        if(array100[i]%2==0){
            m++;
            arraypar[m] = array100[i];
        }
        else{
            n++;
            arrayimpar[n] = array100[i];
        }
    }

    printf("\n\n Numeros pares: \n");

    for(i=1;i<=100;i++){
        printf("%i ", arraypar[i]);
    }

    printf("\n\n Numeros impares: \n");

    for(i=1;i<=100;i++){
        printf("%i ", arrayimpar[i]);
    }

    getch();
    return 0;
}

A: Un-initialized variables can contain anything, it is better to start of by initializing.
This
int array100[100];
int arraypar[100];
int arrayimpar[100];

Should be:
int array100[100] = {0};
int arraypar[100] {0};
int arrayimpar[100] = {0};

And since C uses 0 base array indexing, you are probably seeing a run time error; Dereference of out-of-bounds pointer (or similar) if the code shown in your post is the actual code you are using.
This:
for(i=1;i<=100;i++){
      ^  ^^
    printf("%i ", arraypar[i]);
}

Should be this:
for(i=0;i<100;i++){
      ^  ^
    printf("%i ", arraypar[i]);
}

Same for the next for loop.
Also, a suggestion to help you troubleshoot. I would start off by initializing array100 with a known set of sequential values to easily test whether the values are being split properly:
int array100[100] = {1,2,6,4,5,6,7,8,9,10,
                     11,12,16,14,15,16,17,18,19,
                     20,21,22,26,24,25,26,27,28,29,
                     60,61,62,66,64,65,66,67,68,69,
                     40,41,42,46,44,45,46,47,48,49,
                     50,51,52,55,54,55,56,57,58,59,
                     60,61,62,66,64,65,66,67,68,69,
                     70,71,72,77,74,75,77,77,78,79,
                     80,81,82,88,84,85,88,87,88,89,
                     90,91,92,99,94,95,99,97,98,99,100};

Now run your program with the rand() section commented out to see what you get.  If there are problems, they will be easier to see then looking at a collection of randomly generated numbers.

A: You allocate 100 places for the arrayimpar and arraypar (with indexes 0..99). Usable number of elements (those you want to read) in those arrays are: m(for arraypar) and n (for arrayimpar). Why not use those variables when printing those arrays:
printf("\n\n Numeros pares: \n");

for(i = 1; i <= m; i++){
    printf("%i ", arraypar[i]);
}

printf("\n\n Numeros impares: \n");

for(i = 1; i <= n; i++){
    printf("%i ", arrayimpar[i]);
}

Practically, this should work, but theoretically you could get one of the arrays completely full, and the other empty. When that happens, eventually you will access 101-st element (with index 100) which doesn't exist. Try to fill these arrays from index 0 instead of 1, so you could never access 101-st element.

A: Index of array starts from 0 not 1, so your code below:
   for(i=0;i<100;i++){
        if(array100[i]%2==0){
            m++;
            arraypar[m] = array100[i];
        }
        else{
            n++;
            arrayimpar[n] = array100[i];
        }
    }

should change to:
   for(i=0;i<100;i++){
        if(array100[i]%2==0){
            arraypar[m] = array100[i];
            m++;
        }
        else{
            arrayimpar[n] = array100[i];
            n++;
        }
    }

So, when you print, you should print m even numbers and n even numbers instead of print all values in arraypar and arrayimpar.:
    printf("\n\n Numeros pares: \n");

    for(i=0;i < m;i++){ // if m==0, you print nothing
        printf("%i ", arraypar[i]);
    }

    printf("\n\n Numeros impares: \n");

    for(i=0;i< n;i++){ // if n == 0, then you print nothing
        printf("%i ", arrayimpar[i]);
    }

redpajama_set_name: RedPajamaStackExchange


text:
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