One of the most popular websites for online activists launched a new service on Wednesday that allows lawmakers to create official pages where constituents can petition them on everything from potholes to major national policies.

A handful of prominent legislators have already signed up for the new service from Change.org, including Sen. Elizabeth Warren, D-Mass., House Budget Committee Chairman Paul Ryan, R-Wisc., and Minority Whip Steny Hoyer, D-Md., said Jake Brewer, the site’s external affairs director and developer of Change.org for Decision Makers.

Any of Change.org’s 50 million users will be able to create or sign an appeal directed at a lawmaker who creates an account on the site, but lawmakers will be able to see how many of the signers are actual constituents based on information they provided when they registered with the site, Brewer said.

People browsing the site will be able to see nearly everything the politicians can see, he said, including the petition text, the number of people who signed it and comments people left after signing.

The site’s designers have twin goals. They hope collecting constituent demands into a single petition rather than hundreds or thousands of emails will give politicians a positive incentive to respond and engage with them. They also hope that making the entire process public will give politicians a “negative incentive” to engage with petitioners because they don’t want to look unresponsive in the public eye.

“This will help members of Congress to prioritize,” Brewer said. “It will also give citizens an opportunity to join in the process because it’s transparent and online. It’s not just happening in the black hole of an inbox somewhere.”

Change.org is classified as a B Corporation by the state of California, meaning it’s a for-profit company dedicated to social good. The company’s income comes from sponsored petitions bought mainly by nonprofits and advocacy organizations that receive prominent status on the site similar to promoted posts on social media.

Change.org for Decision Makers bears some similarities to the White House petition website We the People. Petitioners on that site are promised an official White House response if they can gain more than 100,000 signatures in one month.

Unlike We the People, however, Change.org petitions won’t stop gathering signatures once an official responds, Brewer said. That means petitioners can write back if they believe a politician hasn’t responded adequately or if they don’t like what she said.

For many politicians, Brewer said, he expects responding to Change.org petitions will be less formal than We the People responses. Politicians will be able to defend their responses when petitioners don’t like them or ask for input about how to address an issue. They can also tell petitioners when they think a petition should be directed at another public figure, for instance a petition on U.S. intervention in Syria that’s directed at a senator who doesn’t sit on the foreign relations committee.

“This isn’t just about having an antagonistic relationship,” Brewer said. “It’s really an opportunity to start a conversation.”

Change.org isn’t taking a position on whether politicians should or shouldn’t set a threshold for when they’ll respond to a petition, Brewer said. If politicians do set a threshold, however, organizers are asking -- but not demanding -- that they not set it at a higher ratio of petition signers per constituents than the We the People threshold, he said.

In other words, the 100,000-signature We the People threshold divided by the U.S. population of roughly 300 million works out to about one signature per every 3,000 constituents. That same ratio would equal about 230 signatures for the average congressional district.

Other members of Congress have expressed interest in using the site but registering many of them was delayed by the partial government shutdown that ended last week, Brewer said. The decision maker profiles will also be available to non-federal officials including governors, state legislators, city council members and mayors.

“For years people have been trying to use technology to improve the way citizens can interact and engage with Congress or with mayors or with governors,” Brewer said. “The answer has always been trying to get more citizen voices inside of government, it’s ‘how do we allow for citizens to be heard in the halls of Congress.’ With Change.org for Decision Makers, what we’re doing is saying ‘no, instead of bringing citizens into government, we’re bringing government leaders out where the people already are’ . . . We’re inviting elected members into our quote-unquote offices on the Internet where we can talk about the issues we want to talk about.”


---
address:
- ', , '
- ', '
author:
- Oana Brandibur
- 'Eva Kaslik\*'
bibliography:
- 'bibliografie.bib'
- 'fractional\_applications.bib'
title: 'Stability of two-component incommensurate fractional-order systems and applications to the investigation of a FitzHugh-Nagumo neuronal model [^1]'
---

Introduction
============

In many real world applications [@Cottone; @Engheia; @Henry_Wearne; @Heymans_Bauwens; @Mainardi_1996], fractional-order dynamical systems have proven to provide more accurate and realistic results than their classical integer-order counterparts, due to the fact that fractional-order derivatives, as non-local operators, are able to reflect memory and hereditary properties. However, it is important to emphasize that important qualitative differences may appear when generalizing properties of integer-order dynamical systems to the fractional-order case, and such generalizations have to be done with great care.

Stability analysis is one of the most important research topics of the qualitative theory of fractional-order systems. Two recent surveys [@Li-survey; @Rivero2013stability] provide comprehensive overviews of stability properties of fractional-order systems. In the particular case of linear autonomous commensurate fractional order systems, the most important starting point is Matignon’s stability theorem [@Matignon], which has been generalized in [@Sabatier2012stability]. Linearization theorems (or analogues of the classical Hartman-Grobman theorem) for fractional-order systems have been recently proved in [@Li_Ma_2013; @Wang2016stability]. Up to this date, incommensurate order systems have not received as much attention as their commensurate order counterparts. Linear incommensurate fractional order systems with rational orders have been analyzed in [@Petras2008stability]. Oscillations in two-dimensional incommensurate fractional order systems have been investigated in [@Datsko2012complex; @Radwan2008fractional]. BIBO stability of systems with irrational transfer functions has been recently investigated in [@trachtler2016bibo].

The first aim of this paper is to explore necessary and sufficient conditions for the asymptotic stability of two-dimensional linear autonomous incommensurate fractional-order systems with Caputo derivatives of different orders. An extended summary of these results has been given in [@Brandibur_CMMSE2017]. The theoretical results included in the following sections generalize the previous findings reported in [@Brandibur_2016], concerning two-dimensional systems composed of a fractional-order differential equation and a classical first-order differential equation. These results are later applied to investigate stability properties of a fractional-order FitzHugh-Nagumo neuronal model. It is worth emphasizing that the fractional-order formulation of neuronal dynamics is strongly justified by experimental results concerning biological neurons [@Anastasio; @Lundstrom]. Moreover, [@du2013measuring] suggests that a possible physical meaning of the order of a fractional derivative is that of an index of memory, which further justifies its use in mathematical models arising from neuroscience.

Preliminaries
=============

Consider the $n$-dimensional fractional-order system with Caputo derivatives $$\label{sys.gen}
^c\!D^\mathbf{q}\mathbf{x}(t)=f(t,\mathbf{x})$$ where $\mathbf{q}=(q_1,q_2,...,q_n)\in(0,1)^n$ and $f:[0,\infty)\times\mathbb{R}^n\rightarrow \mathbb{R}^n$ is continuous on the whole domain of definition and Lipschitz-continuous with respect to the second variable, such that $$f(t,0)=0\quad \textrm{for any }t\geq 0.$$ Let $\varphi(t,x_0)$ denote the unique solution of (\[sys.gen\]) which satisfies the initial condition $x(0)=x_0\in\mathbb{R}^n$. The existence and uniqueness of the initial value problem associated to system (\[sys.gen\]) is guaranteed by the properties of the function $f$ stated above [@Diethelm_book].

Generally, the asymptotic stability of the trivial solution of system (\[sys.gen\]) is not of exponential type [@Cermak2015; @Gorenflo_Mainardi], because of the presence of the memory effect. A special type of non-exponential asymptotic stability concept has been defined for fractional-order differential equations [@Li_Chen_Podlubny], called Mittag-Leffler stability. In this paper, we are concerned with $\mathcal{O}(t^{-\alpha})$-asymptotic stability, which reflects the algebraic decay of the solutions.

\[def.stability\] The trivial solution of (\[sys.gen\]) is called *stable* if for any $\varepsilon>0$ there exists $\delta=\delta(\varepsilon)>0$ such that for every $x_0\in\mathbb{R}^n$ satisfying $\|x_0\|<\delta$ we have $\|\varphi(t,x_0)\|\leq\varepsilon$ for any $t\geq 0$.

The trivial solution of (\[sys.gen\]) is called *asymptotically stable* if it is stable and there exists $\rho>0$ such that $\lim\limits_{t\rightarrow\infty}\varphi(t,x_0)=0$ whenever $\|x_0\|<\rho$.

Let $\alpha>0$. The trivial solution of (\[sys.gen\]) is called *$\mathcal{O}(t^{-\alpha})$-asymptotically stable* if it is stable and there exists $\rho>0$ such that for any $\|x_0\|<\rho$ one has: $$\|\varphi(t,x_0)\|=\mathcal{O}(t^{-\alpha})\quad\textrm{as }t\rightarrow\infty.$$

Stability results for linear systems with two Caputo derivatives of different orders
====================================================================================

The following two-dimensional linear autonomous incommensurate fractional-order system is considered: $$\label{linearsys}
\left\{
\begin{array}{l}
  ^c\!D^{q_1}x(t)=a_{11}x(t)+a_{12}y(t) \\
  ^c\!D^{q_2}x(t)=a_{21}x(t)+a_{22}y(t)
\end{array}
\right.$$ where $A=(a_{ij})$ is a real 2-dimensional matrix and $q_1,q_2\in(0,1)$ are the fractional orders of the Caputo derivatives. We may assume $a_{12}a_{21}\neq 0$, because otherwise, one equation would be decoupled.

Applying the Laplace transform to system (\[linearsys\]) we obtain the following system: $$\begin{bmatrix}
s^{q_1}X(s)-s^{q_1-1}x(0)\\s^{q_2}Y(s)-s^{q_2-1}y(0)
\end{bmatrix}=A\cdot \begin{bmatrix}
X(s)\\Y(s)
\end{bmatrix},$$ where $X(s)={\mathcal{L}}(x)(s)$ and $Y(s)={\mathcal{L}}(y)(s)$ represent the Laplace transforms of the functions $x$ and $y$, respectively, and $s^{q_1}$, $s^{q_2}$ represent the principal values (first branches) of the corresponding complex power functions [@Doetsch]. Therefore: $$\left(\text{diag}(s^{q_1},s^{q_2})-A\right)\cdot\begin{bmatrix}
X(s)\\Y(s)
\end{bmatrix}=\begin{bmatrix}
s^{q_1-1} x(0)\\ s^{q_2-1}y(0)
\end{bmatrix}.$$ In the following, we denote $$\Delta_A(s)=\det\left(\text{diag}(s^{q_1},s^{q_2})-A\right)=s^{q_1+q_2}-a_{11}s^{q_2}-a_{22}s^{q_1}+\det(A)$$ and therefore, we obtain $$\label{transfer.functions}
X(s)=\frac{s^{q_1}(s^{q_2}-a_{22})x(0)+a_{12}s^{q_2}y(0)}{s\Delta_A(s)}\quad\text{and}\quad Y(s)=\frac{s^{q_2}(s^{q_1}-a_{11})y(0)+a_{21}s^{q_1}x(0)}{s\Delta_A(s)}$$

The following result provides necessary and sufficient conditions for the global asymptotic stability of system (\[linearsys\]) as well as sufficient conditions for the instability of system (\[linearsys\]). The proof is based on the Final Value Theorem and asymptotic expansion properties of the Laplace transform [@Doetsch; @Bonnet_2000; @Brandibur_2016].

\[thm.lin.stab\]$ $

1.  Denoting $q=\min\{q_1,q_2\}$, system (\[linearsys\]) is $\mathcal{O}(t^{-q})$-globally asymptotically stable if and only if all the roots of $\Delta_A(s)$ are in the open left half-plane ($\Re(s)<0$).

2.  If $\det(A)\neq 0$ and $\Delta_A(s)$ has a root in the open right half-place ($\Re(s)>0$), system (\[linearsys\]) is unstable.

Without loss of generality, we consider $0<q_1\leq q_2<1$.

*Part 1 - Necessity.* Assuming that (\[linearsys\]) is $\mathcal{O}(t^{-q})$-globally asymptotically stable and letting $(x(t),y(t))$ denote the solution of system (\[linearsys\]) which satisfies the initial condition $(x(0),y(0))=(x_0,y_0)\in\mathbb{R}^2$ such that $x_0,y_0\ne 0$, there exist $M>0$ and $T>0$ such that $$|x(t)|\leq\|(x(t),y(t))\|\leq Mt^{-q},\ \forall \ t\geq T.$$ Hence, the Laplace transform $$X(s)=\frac{x_0s^{q_1}(s^{q_2}-a_{22})+a_{12}y_0s^{q_2}}{s\Delta_A(s)}$$ is absolutely continuous and holomorphic in the open right half-plane, therefore, it does not have any poles in the open right half-plane. The function from the numerator is holomorphic on $\mathbb{C}\setminus\{s\in\mathbb{R},s\leq 0 \}$ and $\Delta_A(s)\ne 0$, for any $s\in\mathbb{C}$, $\Re(s)>0$.

Assuming that $\Delta_A(0)=0$, we obtain that $\det(A)=0$ and $\Delta_A(s)=s^{q_1+q_2}-a_{11}s^{q_2}-a_{22}s^{q_1}.$ Hence $$\begin{aligned}
    \lim\limits_{s\rightarrow 0}sX(s)&=\lim\limits_{s\rightarrow 0}s\cdot \frac{x_0s^{q_1}(s^{q_2}-a_{22})+a_{12}y_0s^{q_2}}{s\Delta_A(s)}=\\
    &=\lim\limits_{s\rightarrow 0}\frac{x_0s^{q_1+q_2}-x_0a_{22}s^{q_1}+a_{12}y_0s^{q_2}}{s^{q_1+q_2}-a_{11}s^{q_2}-a_{22}s^{q_1}}=\\
    &=\lim\limits_{s\rightarrow 0}\frac{x_0s^{q_2}-x_0a_{22}+a_{12}y_0s^{q_2-q_1}}{s^{q_2}-a_{11}s^{q_2-q_1}-a_{22}}=\\
    &=\left\{
    \begin{array}{ll}
    x_0, & \textrm{if }a_{22}\neq 0 \\
    -\frac{a_{12}}{a_{11}}y_0, & \textrm{if }a_{22}=0
    \end{array}
    \right.\quad\neq 0,
    \end{aligned}$$ This contradicts the Final Value Theorem for the Laplace transform $X(s)$ because $x(t)\rightarrow 0$ as $t\rightarrow\infty$. In conclusion, $\Delta_A(0)\neq 0$.

Let us now consider the solution $(x(t),y(t))$ of system (\[linearsys\]) which satisfies the initial condition $(x(0),y(0))=\left(0,\frac{1}{a_{12}}\right)$. For $x(t)$ we obtain the Laplace transform $X(s)=\left[s^{1-q_2}\Delta_A(s)\right]^{-1}$. Assuming that $\Delta_A(s)$ has a root on the imaginary axis (but not at the origin), it follows that $X(s)$ has a pole on the imaginary axis, which implies that $x(t)$ has persistent oscillations, contradicting the convergence of $x(t)$ to the limit $0$, as $t\rightarrow\infty$.

Therefore, we obtain $\Delta_A(s)\neq 0$, for any $s\in\mathbb{C}$, $\Re(s)\geq 0$.

*Part 1 - Sufficiency.* Let $(x(t),y(t))$ denote the solution of system (\[linearsys\]) which satisfies the initial condition $(x(0),y(0))=(x_0,y_0)\in\mathbb{R}^2$. Assuming that all the roots of $\Delta_A(s)$ are in the open left half-plane, it follows that all the poles of the Laplace transforms functions $X(s)$ and $Y(s)$ given by (\[transfer.functions\]) are either in the open left half-plane or at the origin, and $X(s)$ and $Y(s)$ have at most a single pole at the origin. A simple application of the Final Value Theorem of the Laplace transform [@Chen_Lundberg] yields $$\begin{aligned}
\lim_{t\rightarrow\infty}x(t)&=\lim_{s\rightarrow 0}sX(s)=\lim_{s\rightarrow 0}\frac{s^{q_1}(s^{q_2}-a_{22})x(0)+a_{12}s^{q_2}y(0)}{\Delta_A(s)}=0;\\
\lim_{t\rightarrow\infty}y(t)&=\lim_{s\rightarrow 0}sY(s)=\lim_{s\rightarrow 0}\frac{s^{q_2}(s^{q_1}-a_{11})y(0)+a_{21}s^{q_1}x(0)}{\Delta_A(s)}=0.\end{aligned}$$ The Laplace transform $X(s)$ is holomorphic in the right half-plane, except at the origin and has the asymptotic expansion $$X(s)\sim \sum_{n=0}^\infty c_n s^{\lambda_n},\quad \textrm{as } s\rightarrow 0,$$ where $\lambda_0=q_1-1<\lambda_1<...<\lambda_n<...$. Based o Theorem 37.1 from [@Doetsch], this leads to the following asymptotic expansion: $$x(t)\sim \sum_{n=0}^\infty \frac{c_n}{\Gamma(-\lambda_n)}\frac{1}{t^{\lambda_n+1}} ,\quad \textrm{as } t\rightarrow \infty,$$ where $\Gamma$ represents the Gamma function with the convention $$\frac{1}{\Gamma(-\lambda_n)}=0\quad \textrm{if }\lambda_n\in\mathbb{Z}_+.$$ As $\lambda_0+1=q_1$, it follows that $x(t)$ converges to $0$ as $t^{-q_1}$. Similarly, we also obtain that $y(t)$ converges to $0$ as $t^{-q_1}$. Combining the convergence results for the two components $x(t)$ and $y(t)$, it follows, based on Definition \[def.stability\] that system (\[linearsys\]) is $\mathcal{O}(t^{-q})$-globally asymptotically stable, where $q=\min\{q_1,q_2\}$.

*Part 2.* Assume that $\det(A)\neq 0$, which is equivalent to $\Delta_A(0)\neq 0$. Consider the solution of $(x(t),y(t))$ of system (\[linearsys\]) which satisfies the initial condition $(x(0),y(0))=(0,y_0)$, with an arbitrary $y_0\in\mathbb{R}^\star$. The Laplace transform of $x(t)$ is $X(s)=a_{12}y_0\left[s^{1-q_2}\Delta_A(s)\right]^{-1}$. Based on Proposition 3.1 from [@Bonnet_2002], it follows that $\Delta_A(s)$ has a finite number of roots in $\mathbb{C}\setminus\mathbb{R}_{-}$, and in particular, in the open right half-plane. Obviously, the Laplace transform $X(s)$ is analytic in $\mathbb{C}\setminus\mathbb{R}_{-}$, except at the poles given by the roots of $\Delta_A(s)$.

If $\Delta_A(s)$ has at least one root in the open right half-plane, let us denote by $\rho>0$ the real part of a dominant pole of $X(s)$, i.e. $\rho=\max\{\Re(s):\Delta_A(s)=0\}$, and by $\nu\geq 1$ the largest order of a dominant pole. Following Theorem 35.1 from [@Doetsch], we obtain that $|x(t)|$ is asymptotically equal to $k~t^{\nu-1} e^{\rho t}$ (with $k>0$) as $t\rightarrow \infty$. Hence, $x(t)$ is unbounded and therefore, system (\[linearsys\]) is unstable.

\[lem.a.star\] Let $c>0$, $0<q_1<q_2\leq 1$, and consider the smooth parametric curve in the $(b,a)$-plane defined by $$\Gamma_{c,q1,q2}~:\quad
\begin{cases}
b=b(\omega)=\rho_1\omega^{q_2}-c\rho_2\omega^{-q_1}\\
a=a(\omega)=c\rho_1\omega^{-q_2}-\rho_2\omega^{q_1}
\end{cases},\quad \omega>0,$$ where: $$\rho_1=\frac{\sin\frac{q_1\pi}{2}}{\sin\frac{(q_2-q_1)\pi}{2}}> 0\quad,\quad \rho_2=\frac{\sin\frac{q_2\pi}{2}}{\sin\frac{(q_2-q_1)\pi}{2}}>1~.$$ The curve $\Gamma_{c,q1,q2}$ is the graph of a smooth, decreasing and convex bijective function $a^\star=a^\star_{c,q_1,q_2}:\mathbb{R}\rightarrow\mathbb{R}$ which satisfies the inequalities $$\label{ineg.a.star}
\begin{cases}
a^\star(b)\leq (-b)^\frac{q_2}{q_1}c^{\left(1-\frac{q_2}{q_1}\right)} &\textrm{, if }b<0\\
a^\star(b)\leq -b^{\frac{q_1}{q_2}} &\textrm{, if }b\geq 0
\end{cases}$$

It is easy to see that for any $\omega>0$ one has: $$\begin{aligned}
b'(\omega)&=\rho_1q_2\omega^{q_2-1}+c\rho_2q_1\omega^{-q_1-1}>0\\
a'(\omega)&=-c\rho_1q_2\omega^{-q_2-1}-\rho_2q_1\omega^{q_1-1}<0\end{aligned}$$ Moreover, $a(0_+)=b(\infty)=\infty$ and $a(\infty)=b(0_+)=-\infty$, and hence, $\Gamma_{c,q_1,q_2}$ is the graph of a smooth decreasing bijective function $a^\star=a\circ b^{-1}:\mathbb{R}\rightarrow \mathbb{R}$ such that $a^\star(\pm\infty)=\mp\infty$.

Furthermore, we compute: $$a''(\omega)b'(\omega)-a'(\omega)b''(\omega)=\omega^{-q_1-q_2-3} \left(\rho_1\rho_2q_1 q_2(q_2-q_1)(c^2+\omega^{2(q_1+q_2)})+2 c \omega^{q_1+q_2} (q_2^3 \rho_1^2-q_1^3 \rho_2^2)\right).$$ This expression is strictly positive, as $\frac{q_1}{q_2}\leq \frac{\rho_1}{\rho_2}\leq 1$ (which follows from the fact that the function $x\mapsto\frac{\sin x}{x}$ is decreasing on $(0,\pi)$). Therefore, as $b'(\omega)>0$, it follows by the chain rule that $\frac{d^2a^\star}{db^2}>0$, for any $b\in\mathbb{R}$, and hence, $a^\star$ is a convex function.

Let us denote $\rho=\frac{\rho_1}{\rho_2}\in(0,1)$. We note that the root of the equation $a(\omega)=0$ is $\omega_a=(c\rho)^{\frac{1}{q_1+q_2}}$, while the root of $b(\omega)=0$ is $\omega_b=(c\rho^{-1})^{\frac{1}{q_1+q_2}}$. We have $\omega_a<\omega_b$ and hence, $a^\star(0)<0$.

We will first prove the inequality $a^\star(b)\leq -b^{\frac{q_1}{q_2}}$ for any $b\geq 0$, which is equivalent to $(-a(\omega))^{q_2}\geq b(\omega)^{q_1}$, for any $\omega\geq \omega_b$. Denoting $u=c\omega^{-q_1-q_2}\in(0,\rho]$, this further simplifies to proving the inequality $$\label{ineg.dem}
(\rho_1-u\rho_2)^{q_1}\leq (\rho_2-u\rho_1)^{q_2},\quad\forall~u\in(0,\rho].$$ To prove inequality (\[ineg.dem\]), we first apply Bernoulli’s inequality and obtain $$(\rho_1-u\rho_2)^{\frac{q_1}{q_2}}\leq 1+\frac{q_1}{q_2}(\rho_1-u\rho_2-1),\quad \forall~~u\in(0,\rho].$$ We now have to show that $$1+\frac{q_1}{q_2}(\rho_1-u\rho_2-1)\leq \rho_2-u\rho_1,\quad \forall~~u\in(0,\rho]$$ or equivalently $$u(\rho_1-r\rho_2)\leq \rho_2-1+r(1-\rho_1),\quad \forall~~u\in(0,\rho].$$ where $r=\frac{q_1}{q_2}\leq\rho$. This inequality holds if and only it is true for $u=\rho$, or equivalently: $$\sin\left(\frac{(q_1+q_2)\pi}{2}\right)\geq (1-r)\sin\left(\frac{q_2\pi}{2}\right),~\quad\forall~0<q_1<q_2\leq 1.$$ The above inequality follows from Jensen’s inequality applied to the concave sine function: $$(1-r)\sin\left(\frac{q_2\pi}{2}\right)+r\sin(q_2\pi)\leq \sin\left((1-r)\frac{q_2\pi}{2}+r q_2\pi\right)=\sin\left(\frac{(q_1+q_2)\pi}{2}\right).$$

The other inequality, $a^\star(b)\leq (-b)^\frac{q_2}{q_1}c^{\left(1-\frac{q_2}{q_1}\right)}$ for any $b<0$, is equivalent to $a(\omega)^{q_1}\leq (-b(\omega))^{q_2}c^{q_1-q_2}$, for any $\omega<\omega_a$. Denoting $u=c^{-1}\omega^{q_1+q_2}\in(0,\rho)$, this further simplifies to inequality (\[ineg.dem\]) which has been proved above.

With the aim of exploring the distribution of the roots of the characteristic function $\Delta_A(s)$ given above, the following result is given, which is a generalization of Proposition 2 from [@Brandibur_2016].

\[prop112\] Consider the complex-valued function $$\Delta (s)=s^{q_1+q_2}+as^{q_2}+bs^{q_1}+c,$$ where $0<q_1<q_2<1$, $s^{q_1}$ and $s^{q_2}$ represent the principal values (first branches) of the corresponding complex power functions and $a,b,c\in\mathbb{R}$.

1.  If $c<0$, then $\Delta(s)$ has at least one positive real root.

2.  $\Delta(0)=0$ if and only if $c=0$.

3.  Assume that $c>0$.

    1.  If $a\geq 0$ and $b\geq 0$ then all the roots of $\Delta(s)$ satisfy $\Re(s)<0$.

    2.  $\Delta(s)$ has a pair of pure imaginary roots if and only if $$a=a^\star_{c,q_1,q_2}(b):=a^\star(b,c,q_1,q_2),$$ where $a^\star_{c,q_1,q_2}$ is the function given by Lemma \[lem.a.star\].

    3.  If $s(a,b,c,q_1,q_2)$ is one of the roots of $\Delta(s)$ such that $$\Re(s(a^\star,b,c,q_1,q_2))=0,$$ where $a^\star=a^\star(b,c,q_1,q_2)$ defined at (b), the following transversality condition is satisfied: $$\frac{\partial \Re(s)}{\partial a}\Big|_{a=a^*}<0.$$

    4.  All roots of $\Delta(s)$ are in the left half-plane if and only if $a>a^\star(b,c,q_1,q_2)$.

    5.  $\Delta(s)$ has a pair of roots in the right half-plane if and only if $a<a^\star(b,c,q_1,q_2)$.

1\. Since $\Delta(0)=c<0$ and $\Delta(\infty)=\infty$, due to the fact that $\Delta(s)$ is continuous on $(0,\infty)$, it results that it has at least one strictly positive real root.

2\. As $\Delta(0)=c$, it is easy to see that $\Delta(0)=0\Leftrightarrow c=0$.

3.(a) Let $a\geq 0$, $b\geq 0$ and $c>0$. If $\Delta(s)$ had a root $s$ with $\Re(s)\geq 0$, then $$|\arg(s)|\leq\frac{\pi}{2}\Rightarrow
    |\arg(s^{q_i})|=q_i\cdot |\arg(s)|\leq\frac{q_i\pi}{2}<\frac{\pi}{2},~~ i\in\{1,2\}$$ So $\Re(s^{q_1})>0$ and $\Re(s^{q_2})>0$. Moreover, we have $$s^{q_1+q_2}+as^{q_2}+bs^{q_1}+c=0\Leftrightarrow s^{q_1}=\frac{-as^{q_2}-c}{s^{q_2}+b}$$ and we obtain $$\begin{aligned}
 \Re(s^{q_1})&=\Re\Big(\frac{-as^{q_2}-c}{s^{q_2}+b}\Big)=\Re\left[\frac{(-as^{q_2}-c)(\bar{s}^{q_2}+b)}{|s^{q_2}+b|^2}\right]\\
    &=\frac{\Re\big[(-as^{q_2}-c)(\bar{s}^{q_2}+b)\big]}{|s^{q_2}+b|^2}\\
    &=\frac{\Re(-as^{q_2}-c)\Re(\bar{s}^{q_2}+b)-\Im(-as^{q_2}-c)\Im(\bar{s}^{q_2}+b)}{|s^{q_2}+b|^2}\\
    &=\frac{(-a\Re(s^{q_2})-c)(\Re(s^{q_2})+b)+a\Im(s^{q_2})(-\Im(s^{q_2}))}{|s^{q_2}+b|^2}\\
    &=\frac{-a|s|^{2q_2}-(ab+c)\Re(s^{q_2})-bc }{|s^{q_2}+b|^2}.
    \end{aligned}$$ Because $a\geq 0, b>0, c\geq 0$, we have that $-a|s|^{2q_2}-(ab+c)\Re(s^{q_2})-bc\leq 0$ and we obtain that $\Re(s^{q_1})\leq 0$, which contradicts $\Re(s^{q_1})>0$. In conclusion, the equation $\Delta(s)=0$ does not have any roots with $\Re(s)\geq 0$.

3.(b) $\Delta(s)$ has a pair of pure imaginary roots if and only if there exist $\omega>0$ such that $\Delta(i\omega)=0$. As $i^q=\cos \frac{q\pi}{2}+i\sin\frac{q\pi}{2}$, taking the real and the imaginary parts of the equation $\Delta(i\omega)=0$, we obtain: $$\label{sistemrealimaginar}
    \begin{cases}
    &\omega^{q_1+q_2}\cos\frac{(q_1+q_2)\pi}{2}+a\omega^{q_2}\cos\frac{q_2\pi}{2}+b\omega^{q_1}\cos \frac{q_1\pi}{2}+c=0\\  &
    \omega^{q_1+q_2}\sin\frac{(q_1+q_2)\pi}{2}+a\omega^{q_2}\sin\frac{q_2\pi}{2}+b\omega^{q_1}\sin \frac{q_1\pi}{2}=0
    \end{cases}$$ Solving this system for $a$ and $b$, using the notations from Lemma \[lem.a.star\], it follows that $\Delta(s)$ has a pair of pure imaginary roots if and only if $(b,a)\in\Gamma_{c,q_1,q_2}$, or equivalently $a=a^\star_{c,q_1,q_2}(b):=a^\star(b,c,q_1,q_2)$.

3.(c) Let $s(a,b,c,q_1,q_2)$ be the root of $\Delta(s)$ such that $s(a^\star,b,c,q_1,q_2)=i\omega$, with $\omega>0$, where $a^\star=a^\star(b,c,q_1,q_2)$. Differentiating with respect to $a$ in the equation $$s^{q_1+q_2}+as^{q_2}+bs^{q_1}+c=0$$ we obtain $$(q_1+q_2)s^{q_1+q_2-1}\frac{\partial s}{\partial a}+s^{q_2}+aq_2s^{q_2-1}\frac{\partial s}{\partial a}+bq_1s^{q_1-1}\frac{\partial s}{\partial a}=0$$ which is equivalent to $$\frac{\partial s}{\partial a}=\frac{-s^{q_2}}{(q_1+q_2)s^{q_1+q_2-1}+aq_2s^{q_2-1}+bq_1s^{q_1-1}}.$$ Therefore $$\frac{\partial \Re(s)}{\partial a}=\Re \left(\frac{\partial s}{\partial a}\right)=\Re\left( \frac{-s^{q_2}}{(q_1+q_2)s^{q_1+q_2-1}+aq_2s^{q_2-1}+bq_1s^{q_1-1}} \right).$$ which leads to $$\frac{\partial \Re(s)}{\partial a}\Big|_{a=a^*}=\Re\left( \frac{-\left(i\omega\right)^{q_2}}{(q_1+q_2)\left(i\omega\right)^{q_1+q_2-1}+a^*q_2\left(i\omega\right)^{q_2-1}+bq_1\left(i\omega\right)^{q_1-1}} \right).$$ Denoting $P(\omega)=(q_1+q_2)\left(i\omega\right)^{q_1+q_2-1}+a^*q_2\left(i\omega\right)^{q_2-1}+bq_1\left(i\omega\right)^{q_1-1}$, we have $$\label{real}
    \frac{\partial \Re(s)}{\partial a}\Big|_{a=a^*}=\Re \left( \frac{-\left(i\omega\right)^{q_2}}{P(\omega)} \right)=\omega^{q_2}\Re \left( \frac{-i^{q_2}\overline{P(\omega)}}{|P(\omega)|^2} \right)=-\frac{\omega^{q_2}}{|P(\omega)|^2}\cdot \Re\left(i^{q_2}\overline{P(\omega)}\right)$$ In what follows, we compute $\Re\left(i^{q_2}\overline{P(\omega)}\right)$ and we obtain: $$\begin{aligned}
    \Re\left(i^{q_2}\overline{P(\omega)}\right)&=\Re \left(\overline{i}^{q_2} P(\omega)\right)\\
    &=\Re \Big[ (q_1+q_2)\overline{i}^{q_2} i^{q_1+q_2-1}\omega^{q_1+q_2-1}+a^*q_2 \overline{i}^{q_2} i^{q_2-1}\omega^{q_2-1}+bq_1\overline{i}^{q_2} i^{q_1-1}\omega^{q_1-1} \Big]\\
    &=\omega^{q_1-1}\Re \Big[ (q_1+q_2)i^{q_1-1}\omega^{q_2}-a^\star q_2 i\omega^{q_2-q_1}+b q_1\overline{i}^{q_2}i^{q_1-1} \Big]\\
    &=\omega^{q_1-1} \Big[ (q_1+q_2)\omega^{q_2}\Re(i^{q_1-1})+b q_1\Re(\overline{i}^{q_2}i^{q_1-1}) \Big]\\
    &=\omega^{q_1-1}\Big[ (q_1+q_2)\omega^{q_2}\sin \frac{q_1\pi}{2}-b q_1\sin \frac{(q_2-q_1)\pi}{2} \Big]\\
    &=\omega^{q_1-1}\sin \frac{(q_2-q_1)\pi}{2}\Big[ (q_1+q_2)\omega^{q_2}\rho_1-q_1(\rho_1\omega^{q_2}-c\rho_2\omega^{-q_1})\Big]\\
    &=\omega^{q_1-1}\sin \frac{(q_2-q_1)\pi}{2}\Big[ q_2\rho_1\omega^{q_2}+c~q_1\rho_2\omega^{-q_1})\Big]>0
    \end{aligned}$$ From (\[real\]) it results that $$\frac{\partial \Re(s)}{\partial a}\Big|_{a=a^*}<0.$$

3.(d,e) From the transversality condition obtained at 3.(c), we observe that $\Re(s)$ is decreasing in a neighborhood of $a^*$, so when $a$ decreases below the critical value $a^*=a^*(b,c,q_1,q_2)$, the pair of conjugated roots $(s,\overline{s})$ crosses the imaginary axis from the left half-plane to the right half-plane. Therefore, taking into consideration 3.(a), we obtain the conclusions.

Furthermore, sufficient stability conditions which do not depend on the fractional orders $q_1$ and $q_2$ will be obtained using the following:

\[prop113\] Let $c>0$ and $0<q_1<q_2<1$. For the complex-valued function $\Delta(s)$ defined in Proposition \[prop112\] we have:

1.  If $a+1>0$, $a+b>0$ and $b+c>0$, then all roots of $\Delta(s)$ are in the open left-half plane, regardless of $q_1$ and $q_2$.

2.  If $a+b+c+1\leq 0$ then the equation $\Delta(s)$ has at least one positive real root, regardless of $q_1$ and $q_2$.

1\. Assume that $a+1>0$, $a+b>0$ and $b+c>0$.

If $b\geq 0$, then $a>-1$ and $a>-b$ and inequality \[ineg.a.star\] from Lemma \[lem.a.star\] implies that $$a> -\min\{b,1\}\geq -b^{\frac{q_1}{q_2}}\geq a^\star(b,c,q_1,q_2).$$

If $b<0$, we have $a>-b$ and $c>-b$ and inequality \[ineg.a.star\] from Lemma \[lem.a.star\] provides: $$a> -b=(-b)^{\frac{q_2}{q_1}}(-b)^{1-\frac{q_2}{q_1}}\geq (-b)^{\frac{q_2}{q_1}}c^{1-\frac{q_2}{q_1}}\geq a^\star(b,c,q_1,q_2).$$

Hence, in both cases, based on Proposition \[prop112\].(d) it follows that all the roots of $\Delta(s)$ are in the open left half-plane.

2\. As $\Delta(1)=1+a+b+c\leq 0$ and $\Delta(\infty)=\infty$, the function $\Delta(s)$ has at least one positive real root belonging to the interval $[1,\infty)$.

Based on Theorem \[thm.lin.stab\] and Propositions \[prop112\] and \[prop113\], the following conditions for the stability of system (\[linearsys\]) are obtained, with respect to its coefficients and the fractional orders $q_1,q_2$:

\[cor.stab.lin\] Consider the linear system (\[linearsys\]) with $q_1,q_2\in(0,1)$, $q_1<q_2$, the fractional orders of the Caputo derivatives. Denoting $a=-a_{11}$, $b=-a_{22}$, $c=\det(A)$, the following hold:

1.  If $c<0$, system (\[linearsys\]) is unstable, regardless of the fractional orders $q_1$ and $q_2$.

2.  Assume that $c>0$ and $q_1,q_2$ are arbitrarily fixed. Consider $a^\star(b,c,q_1,q_2)$ given by Lemma \[lem.a.star\].

    -   System (\[linearsys\]) is $\mathcal{O}(t^{-q_1})$-asymptotically stable if and only if $a>a^\star(b,c,q_1,q_2)$.

    -   If $a<a^\star(b,c,q_1,q_2)$ system (\[linearsys\]) is unstable.

3.  If $c>0$, the following sufficient conditions for the asymptotic stability and instability of system (\[linearsys\]), independent of the fractional orders $q_1$, $q_2$ are obtained:

    -   If $a+1>0$, $a+b>0$ and $b+c>0$ system (\[linearsys\]) is asymptotically stable, regardless of the fractional orders $q_1$ and $q_2$.

    -   If $a+b+c+1\leq 0$ then system (\[linearsys\]) is unstable, regardless of the fractional orders $q_1$ and $q_2$.

Let us consider $c=4$, $q_1=0.4$ and $q_2=0.8$. The curve $\Gamma_{c,q1,q2}$ given by Lemma \[lem.a.star\], which represents the graph of the function $a^\star(b,c,q_1,q_2)$ is shown in Fig. \[fig.stab\]. System (\[linearsys\]) is $\mathcal{O}(t^{-0.4})$-asymptotically stable if and only if the parameters $(a,b)$ take values above the plotted curve $\Gamma_{c,q1,q2}$, based on Corollary \[cor.stab.lin\]. On the other hand, if $(a,b)$ take values below the curve $\Gamma_{c,q1,q2}$, system (\[linearsys\]) is unstable. For any values of $(q_1,q_2)$, if $(a,b)$ belong to the red shaded region shown in \[fig.stab\], system (\[linearsys\]) is asymptotically stable. For any values of $(q_1,q_2)$, if $(a,b)$ belong to the blue shaded region shown in \[fig.stab\], system (\[linearsys\]) is unstable.

![Curve $\Gamma_{c,q1,q2}$ given by Lemma \[lem.a.star\], for fixed values of $c=4$, $q_1=0.4$ and $q_2=0.8$. The red/blue shaded regions represent the combination of parameters $(a,b)$ for which system (\[linearsys\]) is asymptotically stable /unstable, regardless of the fractional orders $q_1$ and $q_2$.[]{data-label="fig.stab"}](fig_stab){width="0.5\linewidth"}

Investigation of a fractional-order FitzHugh-Nagumo model
=========================================================

The FitzHugh-Nagumo neuronal model [@FitzHugh] is a simplification of the well-known Hodgkin-Huxley model, which describes a biological neuron’s spiking behavior. In this paper, we consider an extension of the classical FitzHugh-Nagumo model, by replacing the integer-order derivatives by fractional-order Caputo derivatives:

$$\label{FN1}
\left\{
\begin{array}{l}
^c\!D^{q_1}v(t)=v-\dfrac{v^3}{3}-w+I\\
^c\!D^{q_2}w(t)=r(v+c-dw)
\end{array}
\right.$$

where $v$ represents the membrane potential, $w$ is a recovery variable, $I$ is an external excitation current and $0< q_1\leq q_2\leq 1$. A similar model has been investigated by means of numerical simulations in [@Armanyos2016fractional].

The second equation of system (\[FN1\]) can be rewritten as follows: $$^c\!D^{q_2}w(t)=rd\Big(\frac{1}{d}v+\frac{c}{d}-w\Big)=\phi(\alpha v+\beta -w)$$ where $\phi=rd\in (0,1)$, $\alpha=\dfrac{1}{d}>1$ and $\beta=\dfrac{c}{d}$. Therefore, system (\[FN1\]) is equivalent to the following two-dimensional conductance-based model: $$\label{FN2}
\begin{cases}
 ^c\!D^{q_1}v(t)=I-I(v,w)\\
 ^c\!D^{q_2}w(t)=\phi(w_{\infty}(v)-w)
\end{cases}$$ where $I(v,w)=w-v+\dfrac{v^3}{3}$ and $w_{\infty}(v)=\alpha v+\beta $ is a linear function.

The equilibrium states of the fractional-order neuronal model (\[FN2\]) are the solutions of the following algebraic system $$\begin{cases}
 I=I_\infty(v)\\
 w=w_{\infty}(v)
\end{cases}$$ where $$\begin{aligned}
I_{\infty}(v)=I(v,w_{\infty}(v))=w_{\infty}(v)-v+\frac{v^3}{3}=(\alpha-1)v+\dfrac{v^3}{3}+\beta.\end{aligned}$$ We observe that $I_\infty\in C^1$, $\lim\limits_{v\to -\infty}I_\infty(v)=-\infty$ and $\lim\limits_{v\to \infty}I_\infty(v)=\infty$. Moreover, $I'_{\infty}(v)=v^2+\alpha-1.$

As it is assumed that $\alpha>1$, it follows that $I'_{\infty}(v)=v^2+\alpha-1>0$, so the function $I_\infty$ is increasing and, as it is also continuous, it results that $I_\infty$ is bijective. Therefore, there exists a unique solution for the equation $I_\infty(v)=I$, which we denote by $v^*=v^*(I,\alpha,\beta)$.

For the investigation of the stability of equilibrium states, we consider the Jacobian matrix associated to system (\[FN2\]) at an arbitrary equilibrium state $(v^*,w^*)=(v^*,w_{\infty}(v^*))$: $$J=\begin{bmatrix}
1-(v^*)^2 & -1\\
\phi ~\alpha & -\phi
\end{bmatrix}$$ The characteristic equation at the equilibrium state $(v^*,w^*)$ is $$\label{car.eq}
s^{q_1+q_2}+a(v^*)s^{q_2}+b(v^*)s^{q_1}+c(v^*)=0$$ where $$\begin{aligned}
& a(v^*)=-1+(v^*)^2\\
& b(v^*)=\phi >0 \\
& c(v^*)=\det(J)=\phi\cdot I'_{\infty}(v^*)>0.\end{aligned}$$

Any equilibrium state $(v^*,w^*)$ with $|v^*|>\sqrt{1-\phi}$ is asymptotically stable, regardless of the fractional orders $q_1$ and $q_2$.

As the function $I_\infty$ is increasing and $$\begin{aligned}
& a(v^*)+1=(v^*)^2>0;\\
& a(v^*)+b(v^*)=-1+(v^*)^2+\phi>0;\\
& b(v^*)+c(v^*)=\phi+\phi\cdot I'_{\infty}(v^*)>0;
   \end{aligned}$$ it follows from Corollary \[cor.stab.lin\] 3.a, that the equilibrium state $(v^*,w^*)$ is asymptotically stable, regardless of the fractional orders $q_1$ and $q_2$.

The stability of any equilibrium state $(v^*,w^*)$ with $|v^*|\leq \sqrt{1-\phi}$ depends on the fractional orders $q_1$ and $q_2$ (see Figs. \[fig.branch\] and \[fig.regiuni\]).

![Membrane potential ($v^*$) of the equilibrium states $(v^*,w^*)$ of system (\[FN1\]) (with parameter values: $r = 0.08$, $c = 0.7$, $d = 0.8$) with respect to the external excitation current $I$ and their stability: red represents asymptotic stability, regardless of the fractional orders $q_1$ and $q_2$; green represents equilibrium states whose stability depends on the fractional orders $q_1$ and $q_2$.[]{data-label="fig.branch"}](fig_branch){width="0.6\linewidth"}

![Stability regions (shaded) in the $(q_1,q_2)$-plane for equilibrium states $(v^*,w^*)$ of system (\[FN1\]) (with parameter values: $r = 0.08$, $c = 0.7$, $d = 0.8$), with different values of the membrane potential $v^*$ satisfying the inequality $|v^*|\leq \sqrt{1-\phi}\approx 0.98$. In each case, the part of the blue curve strictly above the first bisector represents the Hopf bifurcation curve in the $(q_1,q_2)$-plane.[]{data-label="fig.regiuni"}](fig_regiuni_q12){width="0.95\linewidth"}

Let us know consider an arbitrarily fixed equilibrium state $(v^*,w^*)$ of system (\[FN1\]), such that $|v^*|\leq \sqrt{1-\phi}$. According to Proposition \[prop112\], at the critical values of the fractional orders $(q_1^\star,q_2^\star)$ defined implicitly by the equality $$a(v^\star)=a^\star(b(v^\star),c(v^\star),q_1,q_2),$$ a Hopf bifurcation is expected to occur (see Fig. \[fig.regiuni\]).

Indeed, considering the following values for the system parameters: $r = 0.08$, $c = 0.7$, $d = 0.8$ and $I=1.24567$, the equilibrium state is $(v^*,w^*)=(0.8,1.875)$. In Fig. \[fig.spikes\], the evolution of the state variables is shown, considering an initial condition in a small neighborhood of the equilibrium point. For a fixed value $q_2=0.8$, the critical value of the fractional order $q_1$ for which a Hopf bifurcation occurs is $q_1^*=0.599$. Indeed, for $q_1=0.58$, asymptotically stable behavior is observed. For $q_1=0.63$, numerical simulations show quasi-periodic behavior, corresponding to the existence of a stable limit cycle. As $q_1$ is increased, the frequency of the oscillations increases. Numerical simulations suggest that fractional-order versions of the FitzHugh-Nagumo system provide a more realistic modeling of individual spikes than the corresponding integer-order counterpart (as seen in the last image from Fig. \[fig.spikes\]).

![Evolution of the state variables of system (\[FN1\]) (with parameter values: $r = 0.08$, $c = 0.7$, $d = 0.8$ and $I=1.24567$) for different values of the fractional orders.[]{data-label="fig.spikes"}](fig_spikes){width="0.6\linewidth"}

Conclusions
===========

Necessary and sufficient conditions have been obtained for the asymptotic stability of a two-dimensional incommensurate order linear autonomous system with Caputo derivatives of different fractional orders. These results can be regarded as a generalization of the classical Routh-Hurwitz stability conditions. As an application, the stability properties of a fractional-order FitzHugh-Nagumo system have been explored. Numerical simulations are provided to exemplify the theoretical findings, additionally revealing the occurrence of Hopf bifurcations when critical values of the fractional orders are encountered.

[^1]: This work was supported by a grant of the Romanian National Authority for Scientific Research and Innovation, CNCS-UEFISCDI, project no. PN-II-RU-TE-2014-4-0270.


Q:

AnswerHandler and PlayerCountHandler mixed up

I'm building an Alexa buttons skill using this template from Amazon for a trivia game. In the game, Alexa asks questions and users respond with an answer from the slot type "answers", which has the following values from the template:
fox,
wolf,
cat,
etc.

I want users to be able to answer trivia questions with a number, instead of a word.
I copied the interaction model and connected a Lambda function to my Alexa skill (as per the readme). When I test it without changing any of the code, it works fine. The dialog goes like this:

Alexa: Welcome to Beginning Addition. This game supports up to 4 players. How many players are there?
User: two
Alexa: Ok. Players, press your buttons now, so I'll know which buttons you will be using.

Debugging info from the device log shows that Alexa thinks the User is responding with PlayerCount intent, which is correct:
"request": {
    "type": "IntentRequest",
    "intent": {
        "name": "PlayerCount",
        "confirmationStatus": "NONE",
        "slots": {
            "players": {
                "name": "players",
                "value": "2",
                "confirmationStatus": "NONE"
            }
        }
    }
}

When I change the slot type of AnswerQuestionIntent and AnswerOnlyIntent to "AMAZON.NUMBER" (instead of the slot type "answers"), the dialog goes like this:

Alexa: Welcome to Beginning Addition. This game supports up to 4 players. How many players are there?
User: two
Alexa: Sorry, I didn't get that. Please say again!

(note: if the user responds with anything else, Alexa will reprompt with "Welcome to Better with Buttons Trivia. This game supports up to 4 players. How many players are there?")
Looking at debugging info from the Device Log, I can see that Alexa thinks that the intent is AnswerOnlyIntent, instead of PlayerCount:
 "request": {
    "type": "IntentRequest",
    "intent": {
        "name": "AnswerOnlyIntent",
        "confirmationStatus": "NONE",
        "slots": {
            "answers": {
                "name": "answers",
                "value": "2",
                "confirmationStatus": "NONE"
            }
        }
    }
}

Why does Alexa think that the user wants AnswerOnlyIntent?
According to the CloudWatch logs, it isn't the AnswerHandler which is handling the response, but the Global DefaultHandler, which is set up to reprompt the user. AnswerHandler is supposed to only run in the following conditions:
return requestEnvelope.request.type === 'IntentRequest' &&
        (requestEnvelope.request.intent.name === 'AnswerQuestionIntent' ||
          requestEnvelope.request.intent.name === 'AnswerOnlyIntent') &&
          validPlayerCount &&
          (sessionAttributes.STATE === settings.STATE.BUTTON_GAME_STATE ||
          sessionAttributes.STATE === settings.STATE.BUTTONLESS_GAME_STATE);

PlayerCountHandler returns canHandle if the following is true:
return requestEnvelope.request.type === 'IntentRequest' &&
        (requestEnvelope.request.intent.name === 'PlayerCount' ||
        requestEnvelope.request.intent.name === 'PlayerCountOnly') &&
        attributesManager.getSessionAttributes().STATE === settings.STATE.START_GAME_STATE;

A:

This is happening because both your AnswerOnlyIntent, AnswerQuestionIntent and PlayerCount all has  number slots. They get mapped when users say a number, which is as expected. A good idea is to have just one intent which takes number input from the user. Ex: a NumberInputIntent. And Alexa will map it whenever users say a number. This way you can always expect this intent to be mapped when there is a number involved.
Use sessionAttributes to keep a track of what you are excepting from the user like
STATE = "playerCount" or STATE = "answer".
You can use two different handlers with these canHandle() conditions.
return requestEnvelope.request.type === 'IntentRequest'
       && requestEnvelope.request.intent.name === 'NumberInputIntent'
       && validPlayerCount
       && sessionAttributes.STATE === 'answer'
       && <add your extra conditions>

and for player count
return requestEnvelope.request.type === 'IntentRequest'
       && requestEnvelope.request.intent.name === 'NumberInputIntent'
       && validPlayerCount
       && sessionAttributes.STATE === 'playerCount'
       && <add your extra conditions>

The idea is simple, use  only one intent for all inputs of that particular kind. For number inputs, just use one intent which maps a number.


Q:

Realm Cocoa can't update property of realm objects

I'm using Realm Cocoa and met one problem here, I can't update property of realm objects,
RLMRealm *realm = [RLMRealm defaultRealm];

SomeObject *obj = [SomeObject objectForPrimaryKey: objectID];
if(obj){
  NSError *error;
  [realm transactionWithBlock:^{
    obj.someProperty = @YES;
  } error: &error];
}

SomeObject is subclass of RLMObject, someProperty is a property of type RLMBool
NSNumber<RLMBool> *someProperty

I checked the error which is nil, which means the transaction was succeeded.
But someObject is never updated as expected.
I also checked the property in Realm browser, which is not updated either.(it's always false).
Please help.

A:

Sorry about this question, I just found out that I accidentally Modified someProperty to @NO in another thread which may happen after the code block in the question, the code block is actually working as expected. Maybe I shouldn't suspect Realm at the beginning.


Geranylgeranyl-pyrophosphate (GGPP) synthase is down-regulated during differentiation of osteoblastic cell line MC3T3-E1.
Isoprenylation of geranylgeranyl-pyrophosphate (GGPP) is critical for activation of small GTPases. We examined the roles of GGPP synthase (GGPPS) during the differentiation induced by the cell-to-cell contact in osteoblastic cell line MC3T3-E1 cells. We found that (1) both mRNA and protein expression of GGPPS was reduced with decrement of its activity during the differentiation, (2) GGOH, which is converted to GGPP in the cells, inhibited differentiation. These results suggest that the decrement of GGPP is critical for the cell-to-cell contact-induced differentiation, in which the down-regulation of GGPPS might be involved.


Clinic of Advanced Biological Medicine

The Clinic of Advanced Biological Medicine in Frankfurt am Main aims to boost the immune system, to maintain and improve the quality of life in case of disease, minimize side effects of toxic treatment, to increase chances for recovery as well as to increase life span through the use of effective treatments.

University Hospital Halle (Saale)

9.6/10 from
78 Votes

Department of Urology

Diagnostic tests for urinary bladder cancer #254719

The Department of Urology at the University Hospital of Halle (Saale) provides treatment of the entire spectrum of urological diseases as well as non-malignancies of the urogenital tract in accordance with the latest achievements in medicine and technology.


In 1979 when the Chinese attacked Vietnam they barely penetrated inside Vietnamese territory before retreating back into China. The ability of the Vietnamese to hold back the Chinese invasion was all the more remarkable as unlike the Chinese, Vietnam's front line troops were elsewhere in Cambodia fighting the Khmer Rouge who's government they had mercifully overthrown.

The Chinese failed in their objectives and the Khmer Rouge never regained control (despite also receiving support from Britain and the US)

I cycled the border last year. Its very much changed now lots of Chinese holiday makers. The locals took more photos of us than we did of them. Also visited the Hanoi war museum. The Vietnam war with the US is covered very honestly but its interesting to note that North Vietnams interventions in Cambodia and China’s attack receive just 4 very poor black and white photos. Their so bad its impossible to tell what’s even in the photos. No names or locations. The descriptions don’t even name China despite it being a victory (or more accurately a Chinese failure). Simply says Northern and Western Border 1979. It’s a sensitive period for them something I think they genuinely want to see firmly in the past.


Not content to end up in a punky post-apocalyptic world because of an energy shortage, a la the great Australian film The Road Warrior, the Australian version of AAA is requesting the government implement some sort of average fuel limits similar to our CAFE standards. The NRMA thinks the move could save the country 1.4 billion gallons of fuel a year, as well as drop costs for the consumer. Lacking these standards, the Australians have produced vehicles such as the V8-powered FPV Super Pursuit Ute and HSV Tourer. As an alternative they could just stop driving with Koalas in their grilles.

Said the NRMA president, "For too long, Australia's automotive industry has been allowed to play by its own rules, setting paltry voluntary targets and consistently building and importing cars that consume more fuel than their overseas counterparts." That's ridiculous; voluntary standards always work... [Drive.com.au]


October26, 2017

Last week, Klobuchar sent Vice President Mike Pence and Secretary Chris Kobach a letter regarding data security and reports that a staffer employed by the Commission was charged with possession and distribution of child pornography; Klobuchar received an auto-reply from the Commission’s official email address directed to the staffer she expressed concerns about, raising serious questions about the vetting process for the staff the Commission has hired, what sensitive data the staffer had access to, and whether the Commission will adequately address the issue

In her sixth letter to the Commission, Klobuchar asks the individual members of the Commission to provide a timeline for responding to previous letters from members of Congress, detail the vetting process for hiring staffers, and to explain how sensitive voter data will be protected

WASHINGTON, DC — U.S. Senator Amy Klobuchar, Ranking Member of the Senate Rules Committee, today said that the Presidential Commission on Election Integrity’s lack of transparency, questionable vetting procedures for staffing, and serious vulnerability in voter system data undermine confidence in our election infrastructure. Last week, Klobuchar sent Vice President Mike Pence and Secretary Chris Kobach a letter regarding data security and reports that Ronald Williams III, a staffer employed by the Commission, was charged with possession and distribution of child pornography. Klobuchar received an auto-reply from the Commission’s official email address directed to the staffer she expressed concerns about, raising serious questions about the vetting process for the staff the Commission has hired, what sensitive data Williams had access to, and whether the Commission will adequately address the issue. In her sixth letter to the Commission, Klobuchar asks the individual members of the Commission to provide a timeline for responding to previous letters from members of Congress, detail the vetting process for hiring staffers, and to explain how sensitive voter data will be protected.

“The Commission's lack of transparency in dealing with members of Congress, its own Commission members and the press exacerbates the overwhelming concern raised by both Republicans and Democrats regarding the Commission's unprecedented request for sensitive voter data. American voters have no way of knowing who has access to their voter information, what it is being used for, or how it is being stored and protected. This is unacceptable,” Klobuchar wrote. “The Commission's stated goal is to address factors that ‘undermine the American people's confidence in the integrity of the voting processes.’ The Commission's lack of transparency, questionable vetting procedures for staffing, and the serious vulnerability in its voter data system present clear examples of practices that undermine confidence in our election infrastructure. This is evidenced by the fact that voters across the country have elected to un-register to vote because of the Commission's actions.”

Klobuchar continued, “Recent revelations from Commission members suggest that our communications may not be reaching them, therefore I am now directing all communication to individual members of the Commission in addition to the Commission’s official correspondence account. It is my sincere hope that by raising these concerns directly with individual members of the Commission, the concerns will be swiftly addressed.”

Last week, Klobuchar and a group of seventeen Senate Democrats demanded responses to unanswered letters and raised new concerns with the Commission. Following the Commission’s original request for sensitive voter data from state officials, Klobuchar and a group of twenty-five senators wrote a letter in July to request the Commission rescind its demand for voter registration information. Klobuchar has also requested details about the Commission’s compliance with the Federal Advisory Commission Act (FACA) and demanded details on the Commission’s procedures and protocols to ensure covered records are properly preserved and made accessible.

The full text of Klobuchar’s letter is below:

Dear Members of the Presidential Commission on Election Integrity,

Last week, I sent Vice President Pence and Secretary Kobach a letter regarding data security and reports that a staffer employed by the Commission was charged with possession and distribution of child pornography. I am alarmed by the fact that the day the letter was sent, we received an auto-reply from the Commission's official email address that directed correspondence to the attention of Ronald Williams II, the very staffer I wrote expressing concerns about.121 The week of October 9, Williams was arrested and charged with eleven counts of possession and distribution of child pornography. A week later the Commission still had not updated its accounts to ensure that correspondence was not directed to Williams. This raises serious questions about the vetting process for the staff the Commission has hired, what sensitive data Williams had access to and whether the Commission will adequately address the issue.

Further troubling is that fact that the Commission's own members report that they are “completely uninformed” or “in the dark” about its activities, which suggests a systematic failure to be transparent. Multiple members of the Commission have indicated they do not have access to complete staff lists, and reporters ' requests for these lists have been ignored. In a letter to Commission Executive Director Kossack, Commissioner Dunlap reported that he has received no information about the Commission's research or activities since September 12th. If the Commission is not transparent and accountable to its own members, it raises serious concerns about its compliance with federal transparency laws which govern the Commission’s actions to ensure it is accountable to the American people.

The Commission is either unwilling or unable to respond to repeated requests for information. This is the sixth letter that l have sent to the Commission in my capacity as Ranking Member of the Senate Rules Committee. My colleagues in the Senate have also sent letters to the Commission. None have been answered.

The Commission's lack of transparency in dealing with members of Congress, its own Commission members, and the press exacerbates the overwhelming concern raised by both Republicans and Democrats regarding the Commission's unprecedented request for sensitive voter data. American voters have no way of knowing who has access to their voter information, what it is being used for, or how it is being stored and protected. This is unacceptable.

The Commission's stated goal is to address factors that “undermine the American people's confidence in the integrity of the voting processes.” The Commission's lack of transparency, questionable vetting procedures for staffing, and the serious vulnerability in its voter data system present clear examples of practices that undermine confidence in our election infrastructure. This is evidenced by the fact that voters across the country have elected to un-register to vote because of the Commission's actions.

Recent revelations from Commission members suggest that our communications may not be reaching them, therefore I am now directing all communication to individual members of the Commission in addition to the Commission’s official correspondence account. It is my sincere hope that by raising these concerns directly with individual members of the Commission, the concerns will be swiftly addressed. I respectfully request that each individual Commissioner:

Provide an estimated timeline for its response to previous letters from members of Congress.


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Show and tell time

There's a new game in town that may soon rival Monopoly in popularity.

The game, actually more an exercise, doesn't have an official name, yet. I like to think of it as, "You show me yours, but I won't show you mine!" Sounds fun, right. It is, for some people. For others, not so much.

The deal is this: Your salary, bonuses and other financial data (like stocks) are available to anybody with a computer. All federal workers are pawns in the game (meaning their data is up for grabs) unless they work for the CIA, the Defense Department, or other security-related agencies.

[Congressional Staff Salaries - Correction]

In an earlier column about making federal salaries public, I took an unfair shot at congressional staffers. I said that their salaries, unlike those of executive branch employees, were not public as part of the new freedom of information action. But that was wrong. Congressional staff salaries have been available, online, for several years now. Bottom line: Their pay has been in the public domain for a long time and nothing has changed.

As yesterday's column pointed out, the FOI (freedom of information) request by a New Jersey newspaper now makes public the salaries of millions of federal workers, unless the agencies are exempt for reasons of national security — or they are the personal employees of members of the House and Senate.

Some feds are stoic. They say they are public employees and can understand (even if they aren't wild about the idea) that their compensation is fair game. Others are furious. Some see it as a bigger issue. Or the foot in the door. As in, what next, how much do you weigh?

Carol Bonosaro, president of the Senior Executives Association, said, "As aggravating as publishing salaries and awards may be, the STOCK Act requires that the financial disclosure forms which executives file be published on agency web sites — and those reports include a great deal more than salaries. This is causing a lot of people a great deal of anguish — because the reports also include spousal info, for one thing. Would you want to be the subject of phishing expeditions? And possible identity theft?"

As you can imagine, lots of comments and emails on this issue. We will have more to come. For now, TGIF day, here are some interesting thoughts from a fed who thinks the pool of people who must play salary-show-and-tell should be expanded dramatically. She writes:

Regarding your question about how we (federal employees) feel about our salaries being exposed — the problem I have is this right here, which is listed on the website where this database resides:

"Employees involved in security work, the FBI, CIA, Defense Department, nuclear materials, IRS, and jobs essential to national security are excluded. The list contains most executive branch employees but does not cover the White House, Congress, the Postal Service, and independent agencies and commissions.

I get the idea behind hiding CIA employees or jobs essential to national security, but the rest of these groups? I can't understand why IRS salaries are secret, nor why all DoD jobs are withheld. And why would the White House and Congress want to hide their employees' salaries from their constituents? And what's the rationale for citizens not knowing their postal carrier's salary? These groups do not make sense. If you are going to expose the salaries of government workers, expose them all — put them all out there, whether they are hired, or elected or contracted. Yes, even contractors' salaries should be public. The taxpayers' dollars are being spent on their salaries in the form of federal contracts, so citizens should have the right to know where their money is being spent.

"And while we're at it, I'd like to know the salaries of other folks where I spend my money. I'd like to know the salaries of each employee of every charity I send a check to, including the church I belong to. I'm spending money, shouldn't I know how much the clerk who opens my check makes? And, oh yeah, I spend money on investments through the TSP and other means. I should have the right to know what every one in those private firms that I invest in are making. Not just the CEOs, but everyone in that firm, seeing as I'm spending my money investing in their firm. Oh, and by the way, I also shop at stores, gas stations, malls and spend money there too. Shouldn't I have the right to see the salaries posted for every employee at every store I shop in? And what about my insurance companies? They're taking my money (oh how they're taking my money!), I need to see all of their salaries as well. I also subscribe to numerous magazines (those dang school fundraisers!) and newspapers, I think I'm owed all of the employees' salaries at those establishments.


Osprey Wheeled Transporter Review & First Impression

The Osprey Wheeled Transporter Duffel might be the best unconventional duffel bag you’ll ever use. It is the ultimate versatile, durable and practical bag to haul your gear to your outdoor adventure activities.

If you’re tired of lugging 50 kilos of gear on your back across gravel roads, cities or airports – I’m with you. That’s why I was so excited when I found out Osprey was elevating their duffel game and bringing in a heavy hitter like no other: The Osprey Wheeled Transporter Duffel. I was more than excited to test it out for action adventure travel – whether you’re heading off for outdoor trekking, adventure climbing, mountain adventure, trail and ultra running or a trip that encompasses it all, Osprey’s new duffel will ease getting your essential outdoor gear from A to B.

Before we continue on with the phenomenal specs and overall user friendliness of the Transporter, for a limited time Paddy Pallin is offering you a chance to win 1 of 6 $500 OSPREY SHOPPING SPREES! If you’re a keen gear lover like I am, your eyes have probably already lit up.

I know I know, hear me out – I’m aware of the Osprey Sojourn and Osprey Ozone. These bags have been on the market for some time and serve their purpose well. They’re fully functional travel machines built for years of abuse, but there’s just something about the Sojourn and Ozone that missed the mark for my personal needs. They’re more complicated than what I need my duffel for, and relatively heavy. I don’t need the shoulder straps. I don’t want extra padding or multiple compartments. I simply want a safe place to put my belongings while I travel. Plus, the added versatility to do what I want with the bag, compartmentalising it with stuff sacks, dry bags, etc.

The Wheeled Transporter Duffels on the other hand, do the job like no other. Osprey has successfully stripped out everything you don’t need in a rolling bag and left only simplicity, efficiency and durability behind.

It’s the gear-hauling machine so many adventurers have been asking for.

The one caveat of any duffel is its sheer size. It truly is a blessing and a curse. When opting for a duffel for long-haul travelling, I want it for carrying an enormous amount of gear from Point A to Point B. The downside to that is, if you have the space, you’ll probably use it. And the more you pack, the heavier that duffel becomes.

Cue mankind’s greatest invention: The wheel.

With the transporter’s oversized monster wheels, I hardly noticed uneven surfaces while dragging it over curbs, down stairwells, across gravel roads and muddy fields. The duffels wheels are ‘field serviceable’, which means if you have to, you can remove them with a hex key. (But you really shouldn’t have to, as Osprey will repair or replace any product that gets damaged within reason.)

Stairs don’t care: Drag the Wheeled Duffel up and over anything. The beefy wheels and High Road Chassis protect the duffel when dragging it through cities or the outback.

The overlapping rain flaps cover the zipper. Combined with the Transporter Duffel’s TPU coated nylon and ultra durable face fabrics, you’d be hard-pressed to find bad enough weather to dampen the gear inside. There’s one large pocket on the handle end of the bag for storing small items. This pocket is accessible while the bag is vertical or horizontal, making it a nice place to store items that you might need easy access to, like battery banks or travel guides.

There are numerous handles which make tossing the Transporter Duffel easy. There’s also a card slot for easy bag identification, as well as a good sized zippered pocket found just below.

While I had no need for the external lashing points found across the outside of the bag, they’d theoretically come in handy when strapping the pack to a rooftop cargo rack, just like Osprey’s more traditional duffels.

The duffel has internal tie down straps to keep gear in place while in transit, and has a reinforced backing and bottom, which makes the duffel easy to pack without the sides collapsing in on themselves. When used on climbing expeditions we found we always had ample room for tents, ropes, climbing helmets and camping gear. Having one place to store all your equipment makes it easier to hit the road without delay.

Just like your regular Osprey hiking packs, the Wheeled Transporter series offers more than enough variation to keep most people happy. I opted for the larger 120 Litre version for the purpose of this review, but Osprey also makes 90 Litre and 40 Litre wheeled versions. No matter which size you choose, they all share the same simple design, fabrics, and colours.

The Osprey Wheeled Transporter duffel proved to be the perfect go-with-you-anywhere, take-what-you-want, and get there comfortably bag. With heaps of space for your gear (and your partner’s gear), you’ll be ready to set up your base camp for a hike up Mt Feathertop, or your expedition to Mt. Kilimanjaro.

If there’s one downside to the Wheeled Transporter it’s the weight. Noticeably heavier than their predecessors, the Osprey Wheeled Transporter Duffel 120 Litre, tipped my scale at 3.85kg compared to theolder, more traditional Osprey Transporter 130 Litre Duffel, which came in at 1.8kg (The Wheeled Transporter 90 Litre and 40 Litre Duffels weighed in at 3.5kg and 2.7kg, respectfully).

The PTU coated nylon sheds water like a dream and the overlapping rainflaps keep water from reaching the inside of your bag. The zipper also boasts a TSA approved design, allowing you to secure valuables while in transit.

While I find this a small price to pay for its comfort and ease of use – this could chew into your overall checked luggage weight limit and should be noted in advance. The 40 Litre Wheeled Duffel is carry on size, while the 90 Litre will have to be checked if you plan on flying. The 120 Litre duffel and the largest of the three, while massive, comes approximately 7 cm shy of being considered oversize luggage, so you should be able to check it while travelling on most airlines without additional fees (as long as it doesn’t exceed the weight limits).

Whether you’re looking for the best wheeled luggage around, something that’s built like a tank and able to take a tumble, the Osprey Wheeled Transporter literally has your back and will save your back and shoulders from having to carry all the weight all the time.

Fill them up and take them anywhere. The Osprey Wheeled Transporter Duffels will get you off the beaten track – getting you and your essential outdoor gear where you need to go.

Bring it everywhere.

Are you a gear hauler who could use a new Osprey duffel to elevate your gear collection? Paddy Pallin has got your back! Shop the Osprey Transporter range in-store or online. Plus, if you haven’t already, enter our giveaway for a chance to win 1 of 6 $500 Osprey Shopping Sprees at Paddy Pallin HERE (available for a limited time only!).

Welcome to the Blog

The name Paddy Pallin certainly means many things to many people. For some it brings to mind the genial personality of one of Australia’s pioneer bushwalkers. Others will recognise the name from the label of a sleeping bag or rucksack that has given a generation of faithful service in the great outdoors. Or perhaps it brings to mind the colourful history of Australia’s greatest cross-country ski race. For most people however, the name is associated with a group of specialist retail stores catering for travellers and outdoor enthusiasts.

The Experience Blog is a place to collaboratively share knowledge, adventures and inspire each other to enjoy the outdoors. We stand firmly by the motto that 'Experience is Everything' & strive to embody that philosophy within our community whenever possible.


Q:

How do I decompress an NSString I have that was compressed with zlib

I have an NSString (e.g. eNoLycgsVgChvILSEoWS1IoSPQBFZgb4) that I get from a web server that is just a normal string compressed using zlib. How do I decompress this in Objective-C? I've found some libraries but they only accept full files, not a simple string.

A:

OK based on the above it seems you didn't really know what the format is - but seems that it's ZLib deflate, Base64 encoded. Cribbing shamelessly off other StackOverflow answers, here's a chunk of code that'll decompress your data:
#import <Foundation/Foundation.h>
#import <zlib.h>

int main( int argc, const char * argv[] )
{
    @autoreleasepool
    {
        NSLog( @"Starting..." );

        NSString      * base64String     = @"eJwLycgsVgChvILSEoWS1IoSPQBFZgb4";
        NSData        * compressedData   = [ [ NSData alloc ] initWithBase64EncodedString: base64String options: 0 ];
        NSUInteger      full_length      = [ compressedData length ];
        NSUInteger      half_length      = [ compressedData length ] / 2;
        NSMutableData * decompressedData = [ NSMutableData dataWithLength: full_length + half_length ];
        BOOL            done             = NO;
        int             status;
        z_stream        strm;

        strm.next_in   = ( Bytef * )[ compressedData bytes  ];
        strm.avail_in  = ( unsigned int ) full_length;
        strm.total_out = 0;
        strm.zalloc    = Z_NULL;
        strm.zfree     = Z_NULL;

        if ( inflateInit2( &strm, ( 15 + 32 ) ) != Z_OK )
        {
            NSLog( @"Could not initialise ZLib" );
            return EXIT_FAILURE;
        }

        while ( done == NO )
        {
            // Make sure we have enough room and reset the lengths.

            if ( strm.total_out >= [ decompressedData length ] )
            {
                [ decompressedData increaseLengthBy: half_length ];
            }

            strm.next_out  = [ decompressedData mutableBytes ] + strm.total_out;
            strm.avail_out = ( unsigned int )( [ decompressedData length ] - strm.total_out );

            // Inflate another chunk.

            status = inflate( &strm, Z_SYNC_FLUSH );

            if ( status == Z_STREAM_END )
            {
                done = YES;
            }
            else if ( status != Z_OK )
            {
                NSLog( @"Decompression failed with status %i", status );
                break;
            }
        }

        if ( inflateEnd( &strm ) != Z_OK )
        {
            NSLog( @"Could not complete decompression" );
            return EXIT_FAILURE;
        }

        // Set real length.

        if ( done )
        {
            [ decompressedData setLength: strm.total_out ];

            NSString * string = [ [ NSString alloc ] initWithData: decompressedData
                                                         encoding: NSUTF8StringEncoding ];

            NSLog( @"String: %@", string );
        }
        else
        {
            return EXIT_FAILURE;
        }
    }

    NSLog( @"...Success" );

    return EXIT_SUCCESS;
}

When run, this prints something like:
2016-06-16 12:40:14.454 Objective Z[82569:7532295] Starting...
2016-06-16 12:40:14.455 Objective Z[82569:7532295] String: This is input text.
2016-06-16 12:40:14.455 Objective Z[82569:7532295] ...Success
Program ended with exit code: 0

You'll need to ensure you link against ZLib. From your project's Build Phases section in XCode, under "Link Binary With Libraries", the "+" button gives a searchable popup; look for "libz". This might not be present on iOS; it is present in the OS X SDK; that's beyond scope of this answer and Google will doubtless present an answer swiftly if you have trouble.


Location, Location, Location! This house is absolutely charming! Remodeled kitchen (2016) with clean white cabinets, stainless appliances, and quartz counters plus new flooring. Dine at the large breakfast bar or in the combined living and dining room with wood fireplace and tons of natural light. Enjoy wood floors throughout the main level, including in the two sizable bedrooms. The lower level has been finished recently with a spacious recreational room and a very private master suite with egress window, walk-in closet, and full bathroom with dual vanity and tiled shower. The home has two large side yards, one with a stone patio. 2014 updates include: Tear down roof, whole house window replacement, basement waterproofing. Bath remodel (tile, plumbing fixture, lighting), whole house re-piping w/ PEX, install of AO Smith 50 gal. water heater, new smart furnace/nest thermostat and smoke/carbon detector, air conditioner, door installed in living room leading to large brick paved patio, and gutter replacement, basement water drainage system installed and a Protech radon mitigation system.

Taxes

Ann Arbor

Ann Arbor, the hub of Washtenaw County, is the sixth largest city in Michigan. A lively downtown and charming tree-lined neighborhoods make Ann Arbor an ideal place to live, work, and raise a family.

Known for academic excellence, Ann Arbor Public Schools District is consistently ranked one of the best in the country. Ann Arbor is also home to the University of Michigan, Cleary University, Concordia University and Washtenaw Community College.

You will find a wonderful mix of city advantages and scenic beauty in Ann Arbor. Nicknamed “Tree Town”, Ann Arbor has more than 2,000 acres of tree lined city parks. Ann Arbor provides endless recreation opportunities including trails, golf courses, canoeing, tennis courts, athletic fields and more. One of the most walkable communities in the country, Ann Arbor prides itself on being environmentally friendly with a progressive recycling program and conscious effort to preserve the city’s green space.

While surrounded by all of this natural beauty, you'll also find that Ann Arbor is less than an hour from the big city advantages Detroit, Michigan has to offer. Detroit is home to the Detroit Symphony Orchestra, several museums, extraordinary shopping and professional sports team including the Detroit Lions, Pistons, Red Wings and Tigers.

Within a four-hour drive of Ann Arbor are Chicago and Toronto, or take a 25-minute trip to Detroit Metropolitan Airport and the world is at your feet. Amtrak makes several stops daily in Ann Arbor, connecting passengers to over 450 locations across the country.

There is a very broad range of major employers countywide, many of them located in and near Ann Arbor. The University of Michigan, the University of Michigan Health System, and the St. Joseph Mercy Health System are among the area's leading employers. These opportunities allow many of the county's residents to work quite close to home; other residents commute less than an hour to greater Detroit or Lansing.

Whether it's life in a college town that you prefer, or the appeal of a small town conveniently located to a larger urban area, Ann Arbor and Washtenaw County has it all. Come see why our home is well loved by so many!

The data relating to real estate for sale on this web site comes in part from the Internet Data eXchange ("IDX") program of Ann Arbor Area Board of Realtors®. Real estate listings held by brokers other than Reinhart Realtors are marked with the IDX Logo. All information is subject to change and should be independently verified.


Z-palatopharyngoplasty combined with 70-degree endoscopy-assisted coblator partial medial glossectomy on severe obstructive sleep apnea.
Background: Patients with severe obstructive sleep apnea (OSA) require safe and effective surgical treatment. Objectives: To evaluate the effectiveness of combined Z-palatopharyngoplasty (ZPPP) and partial glossectomy via 70-degree endoscopy-assisted coblation (Eco-TBR) on severe OSA. Materials and methods: Twenty-two consecutive patients with severe OSA were enrolled between September 2014 and July 2016. The lingual artery was evaluated by contrasted computed tomography (CT). Drug-induced sleep endoscopy (DISE) was performed to determine the necessity of multilevel surgery. Combined ZPPP and Eco-TBR were performed. Polysomnographic parameters were collected pre-operatively and 6 months post-operatively. Results: No adverse events were observed postoperatively. The short-term ( < 12 months) rate of total surgical effectiveness was 63.6% (14/22), with Friedman classifications as follows: I (1/1), Friedman II (6/10) and Friedman III (7/11). There were no differences between Friedman classification groups. The differences in AHI, apnea index (AI), mean blood oxygen, and percentage of cumulative time with oxygen saturation < 90% (CT 90%) were different compared to preoperative levels. Tonsil size was also significantly larger in the multi-level surgery group (2.14 ± 1.03) compared with the control group (1.13 ± 0.64). Conclusions and significance: Combined ZPPP and Eco-TB is safe and effective, with good surgical effectiveness for the treatment of severe OSA patients.


MONTPELIER, Vt. -- A growing number of police departments across the country are taking action over concerns that carbon monoxide fumes from Ford Explorer patrol vehicles are seeping inside the SUVs, potentially sickening officers.

Vermont State Police and several other departments in the Northeast are inspecting their fleets or have installed carbon monoxide monitors in the vehicles. At least two departments in Texas have gone further, pulling their entire fleets Ford Explorers off the road.

The actions come as police in Auburn, Massachusetts, confirmed an officer who passed out behind the wheel of his cruiser and crashed had tested positive for exposure to carbon monoxide.

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The department said on its Facebook page that three officers have had to go to the hospital because of exposure to carbon monoxide. Auburn police also said that the town's fire and public works departments had also taken two Explorers out of service.

These are photos of the two involved vehicles from the earlier crash involving our officer who suffered a carbon... Posted by Auburn MA Police Department on Wednesday, August 2, 2017

"We would urge other departments to have their cruisers tested and/or purchase detectors to ensure everyone's safety," the Auburn police department said in a Facebook post.

A spokeswoman for Ford Motor Co. (F) said Thursday it sent engineers to Auburn to inspect the vehicles and will go "to any department that asks for assistance."

Reports of the problems aroused public ire on social media.

"The community should be outraged that our officers that put it all on the line to keep us safe have substandard equipment," said one person who commented on the Auburn police department's Facebook page.

Earlier this month, dashcam video from March appears to show Austin police Sgt. Zachary LaHood sickened by what his department says was carbon monoxide seeping into his Ford Explorer police cruiser.

A spokesman for the department said Thursday that it has received more than 70 reports from officers of fumes entering the vehicles. Eighteen officers who have been tested were found to have elevated levels of carbon monoxide, according to a department memo sent to Austin's mayor and city council.

The Austin police department has pulled 397 Explorers off the road because of the concerns, while other local agencies have taken some 40 of the vehicles out of service. Carbon monoxide detectors have since been added to the police department's Explorer fleet, and the city is working with Ford and the National Highway Traffic Safety Administration to rectify the problem, the spokesman said.

In February, CBS News identified more than 450 complaints involving 2011 to 2017 model year Ford Explorers, not just police units. Federal regulators acknowledge that number has grown and are investigating.

Ford has known about the problem since at least 2012. A company representative acknowledged in a deposition that it appears to be a "design issue" that may allow exhaust, which contains carbon monoxide, to seep in, likely through unsealed seams in the rear of the SUV.


Brain Injury Lawyer in Edinburg

According to some estimates, an American suffers a brain injury every 15 seconds. Brain damage and head injuries can be devastating and life-altering not only for the victim but also for the individual’s loved ones. The brain is the most complex organ in the body and mere seconds of oxygen deprivation can mean the difference between normal function and profound disability.

A vast range of situations can lead to brain injuries, including car accidents, smoke inhalation, near-drowning incidents, falls and construction accidents, carbon monoxide exposure, drug interactions, medical mistakes during labor and delivery or even a surgery gone wrong. Brain injuries can have a diverse range of effects on an individual, from paralysis, limited motor control and an inability to speak, to numbness and loss of bodily functions, loss of memory and identity, and even coma or death.

The experienced team of hypoxia and traumatic brain injury lawyers with Brasure Law Firm, PLLC are here to help guide you through the process of seeking compensation for your injuries and losses.

Founded nearly a decade ago by Attorney Christopher R. Brasure, Brasure Law Firm, PLLC handles personal injury law cases throughout Texas. Our law firm maintains offices in Grapevine, Texas and in Edinburg, Texas.

To discuss your personal injury case in a free and confidential initial consultation session, contact us by phone at 956-686-3555.

Common Questions and Concerns Regarding a Brain Injury Lawsuit

Personal injury lawsuits can provide you with a chance to recoup the funds you need to cover medical expenses, care costs, lost wages and other losses stemming from an incident involving hypoxia, brain injury or head trauma.

The legal team at Brasure Law Firm, PLLC has compiled information on a few of the most common questions that we receive from clients who are seeking to proceed with this type of case.

How much will I receive in compensation if I file a brain injury lawsuit?

Personal injury lawsuits for brain injury and hypoxia injuries can result in some significant settlements and jury awards due to the profound, life-altering nature of many brain injuries. It is not uncommon for a head injury victim to suffer permanent disability, which subsequently impacts the individual’s ability to perform everyday tasks, live independently or hold down a job. Many head trauma victims require ongoing medical care and other services, such as physical therapy and home care services. This care is often required for a lifetime.

For this reason, your brain injury lawyer will evaluate your case to determine what losses you have suffered to date, including economic losses (such as hospital bills or lost wages) and non-economic losses (such as pain and suffering.)

Additionally, your personal injury lawyer will consider what costs you will incur in the future, including medical costs, future lost wages, and home care expenses, amongst others. It can take some skill and experience to accurately predict these future expenses, which underscores the importance of trusting your case to an experienced attorney.

These figures will be combined to arrive at a figure for your lawsuit.

What is a Coma? A Vegetative State? How are Coma and Vegetative State different from brain death and how can a brain injury lawyer help?

A coma can result from a serious brain injury. In a coma, the patient appears asleep, but they cannot be awakened. The individual is unconscious and unaware of their surroundings. They are unable to communicate or respond in a purposeful manner. Coma may be permanent or the patient may gradually improve, awakening gradually. Often, the patient will pass through a vegetative state on the way to recovery.

A vegetative state may arise in cases of profound brain injury. The patient may exhibit body movements, sleeping and waking cycles, and they may even open their eyes. But there is no capacity for communication or awareness of their surroundings. A person in a vegetative state can typically breathe independently, but they are reliant upon a feeding tube for nutrition. This type of patient is incapable of meaningful, purposeful and voluntary responses, including responses to pain.

Sometimes, a comatose patient may pass through a vegetative state as they recover. In other cases, the vegetative state is permanent and there is no hope for recovery.

Brain death occurs when the upper portions of the brain (cerebrum) and the lower portion of the brain (the brain stem) ceases functioning, and the individual is unable to regain consciousness.

But the body may remain functioning for a period of time while sustained on life support. But once life support is removed, the body dies because the area of the brain that controls vital functions such as breathing and heart rate is irreparably damaged.

Coma, vegetative state and brain death all have very different care requirements and each individual’s prognosis is unique. So it’s best to discuss your case with an experienced brain injury attorney.

Is there a limit to how much I can receive via a brain injury lawsuit?

According to Texas law, there are only statutory limits on damages on medical malpractice cases; all other cases are without this limit.

The maximum is typically $250,000 for non-economic damages like pain and suffering and up to $500,000 per case (in situations where there are multiple people involved.)

For wrongful death cases that arise from a case involving medical malpractice, the maximum is set at $1.9 million. This figure routinely updated to account for inflation.

How long do I have to file a personal injury lawsuit?

Texas law requires that personal injury lawsuits must be filed within two years of the date of injury, with the exception of cases against the state or government employees. In these cases, the timeframe is six months from the date of injury.

If the statute of limitations expires, the courts will typically refuse to hear the case.

What causes a traumatic brain injury?

Brain injuries have many causes. Brain injuries can be caused by blunt force trauma direct blows to the head, gunshot wounds, violent shaking of the head, or force from a whiplash type injury, electrical current, or any other loss of oxygenated blood to the brain.

What are the symptoms of a traumatic brain injury?

Symptoms of a TBI include loss of consciousness, memory loss, confusion, mood changes, and anxiety. Sleepiness, confusion, or feeling like you are in a fog. Changes in behavior, attention, or memory, intense headaches, feeling of fullness, or pressure in the head. Loss of consciousness. Dizziness. Ringing in your ears. Difficulty remembering the event that caused the injury. Vomiting. Slurred speech. Changes in speech.Difficulty responding to questions. Difficulty understanding conversations.

How long does a traumatic brain injury last?

A TBI can last forever. Often a TBI involves the loss of brain tissue that cannot be replaced. In many instances, the TBI patient does not realize they suffered a TBI.

Why are head injuries considered life threatening?

Head injuries can be life threatening because your brain controls all parts of your body. If your brain is impaired, your brain may not tell your heart to beat, which may cause death. If your brain is impaired, your brain may not tell your lungs to take in oxygen. Your body cannot function without oxygen.

Which is worse concussion or contusion?

A concussion can be worse than a contusion. A contusion is a localized injury like a bruise on your arm. Contusions range from relatively minor to life-threatening. Because the blood must clot to stop the bleeding a bruise results from, contusions also increase cardiovascular risks. A concussion is a wider-reaching injury due to different types of brain trauma. According to Dr. Richard Smayda, D.O. A concussion is an immediate but reversible traumatic paralysis of the nervous function of the brain. A concussion may not be as obvious to someone as a contusion; a concussion may have wide-ranging effects on a person’s body. A contusion can be worse if the contusion causes neurological or psychological deficits.

What happens to the brain during a concussion?

During a concussion, the brain the head is typically jolted in one direction. The force of the movement causes the brain to contact some portion of the skull. The brain can bruise or swell where it made contact with the skull. Bruising or swelling can actually harm your brain cells.

Trust Your Brain Injury Claim to Brasure Law Firm, PLLC

Brain injuries have the power to alter more than your life; they can alter who you are as a person, including your personality, your memory, your physical capabilities and your ability to understand and interact with the world around you. Head injuries can quite literally impact every element of your life.

At Brasure Law Firm, PLLC, our legal team is experienced in all aspects of personal injury law, including hypoxia and brain injury lawsuits. We also handle wrongful death cases, which often arise in situations where a personal injury case turns fatal.

Established more than a decade ago, Attorney Christopher R. Brasure maintains a diverse legal background that means he’s well-positioned to help you achieve your objectives for your case.

At Brasure Law Firm, PLLC, we are ready to discuss your case in a confidential, no-cost initial consultation session. Just call 956-686-3555.

*Court costs, litigation expenses, and medical bills are paid from your share of the recovery. If there is no recovery, you will not be responsible for any court costs or litigation expenses, except for unpaid medical bills. Copyright 2018. Brasure Law Firm, PLLC. All rights reserved. | Terms of Service | Privacy Policy | Sitemap | Attorney Marketing by : Social Firestarter, LLC


Measuring the impact of genetic knowledge on intentions and attitudes of the community towards expanded preconception carrier screening.
Preconception carrier screening (PCS) provides the potential to empower couples to make reproductive choices before having an affected child. An important question is what factors influence the decision to use or not use PCS. We analysed the relationship between knowledge, attitudes and intentions to participate in PCS using logistic regression in 832 participants in Western Australia. Two-thirds of participants said they would take the test, with 92% of these supporting screening for diseases reducing the lifespan of children and infants. Those who had good genetic knowledge were seven times more likely to intend to use PCS (p≤0.001), while those with high genetic knowledge were four times more likely to (p=0.002) and raised concerns such as insurance and confidentiality.Decreasing genetic knowledge correlated positively with religiosity and apprehension (p≤0.001), which correlated negatively with intention to use PCS (p≤0.001). Increasing genetic knowledge correlated positively with factors representing positive attitudes (p≤0.001), which correlated positively with intention to use PCS (p≤0.001). Many participants with good genetic knowledge nevertheless answered questions that tested understanding incorrectly.80% of participants stated they would prefer to access the test through their general practitioners and 30% would pay up to $A200. Knowledge is instrumental in influencing participation. Having good genetic knowledge may not be enough to understand core concepts of PCS and may impact informed decision-making. This study recommends that continuous education of health professionals and thus the community, in PCS is crucial to reduce misconceptions.


name: Deploy docs

on:
  push:
    branches: [ master ]

jobs:
  build:

    runs-on: macos-latest

    steps:
    - uses: actions/checkout@v2
    - name: Switch Xcode version
      run: sudo xcode-select -s /Applications/Xcode_11.4.1.app/Contents/Developer
    - name: Generate microsite
      run: |
        brew install nef
        brew install sourcekitten
        gem install bundler -v 2.0.2
        gem install cocoapods -v 1.9.1
        ./scripts/gen-docs.rb
    - name: Deploy microsite
      uses: peaceiris/actions-gh-pages@v3
      with:
        deploy_key: ${{ secrets.DEPLOY_NEF_TOKEN }}
        publish_branch: gh-pages
        publish_dir: ./pub-dir
        enable_jekyll: true


tag:blogger.com,1999:blog-5052525513598295339.post2029714897917930467..comments2015-02-10T18:55:37.467-08:00Comments on Misce-Looney-ous: "Feather Finger" (1966) Commentary by Daffy DuckJon Cookehttp://www.blogger.com/profile/11420721141591983997noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5052525513598295339.post-22115384801451576072009-03-04T07:24:00.000-08:002009-03-04T07:24:00.000-08:00Finally managed to take a listen to this, and I've...Finally managed to take a listen to this, and I've gotta say, that Daffy commentary was spot on! If that was without using a little audio editing to alter your original voice Matt, then bravo for the supoib work! Hope to see something similar done again in the future - personally, I'm rather keen to hear Sylvester's views on his final outings before the "Mysteries" series cropped up some years later ;)Christopherhttp://qwertypictures.deviantart.com/noreply@blogger.comtag:blogger.com,1999:blog-5052525513598295339.post-44166132804505039902008-10-05T20:48:00.000-07:002008-10-05T20:48:00.000-07:00The voiceover is brilliant. Spot-on -- whoever di...The voiceover is brilliant. Spot-on -- whoever did this has got a great ear for Daffy's voice as it changed over the years, and totally nailed what Daffy *should* sound like as an older actor reminiscing over his work.Jimhttp://www.blogger.com/profile/07840228636027306303noreply@blogger.comtag:blogger.com,1999:blog-5052525513598295339.post-17844838938568190472008-10-04T21:13:00.000-07:002008-10-04T21:13:00.000-07:00Great commentary! This is sorta the thing I had i...Great commentary! This is sorta the thing I had in mind when I told Jerry about my idea on how to present the Speedy vs. Daffy films.Nic kramernoreply@blogger.com


Reduced interference from memory testing: A postretrieval monitoring account.
Initial learning can interfere with subsequent learning (proactive interference [PI]), but recent work indicates initial testing can reduce PI. Here, we tested 2 alternative hypotheses of this effect: Does testing reduce PI by constraining retrieval to the target list, or by facilitating a postretrieval monitoring process? Participants first studied 4 lists of unrelated words. The study-only group performed a distractor task following each list, whereas the tested group recalled each list. After these initial lists, both groups studied and were tested on a final list. Replicating prior work, the tested group recalled more of the final list items and had fewer prior-list intrusions than the study-only group (i.e., initial testing reduced subsequent PI). To test the 2-alternative hypotheses, Experiment 1 used a modified recall test for the final list, whereby participants were asked to recall the final list of words and also report any items from prior lists that inadvertently came to mind. Contrary to the constrained retrieval hypothesis, initial testing did not reduce the number of prior list items that came to mind, but consistent with the postretrieval monitoring hypothesis, testing increased the likelihood that the intrusions would be correctly attributed to prior lists. Experiments 2 and 3 further tested the postretrieval monitoring hypothesis by testing the final list twice. According to the hypothesis, testing all of the lists should render prior testing nondiagnostic of list membership, thereby impairing retrieval monitoring in the test group and minimizing its ability to reduce PI. This prediction was confirmed. (PsycINFO Database Record


Q:

jquery Wookmark Plugin Infinite Load/Scoll

Im using the Wookmark jQuery grid layout for my website.
https://github.com/GBKS/Wookmark-jQuery
I need the auto load/endless scroll working.
In the example it simply just loads the first 10 images again, how do you make it continue loading the rest of a list as on Wookmark.com and Pinterest.com?
The code part that ive been trying to change:
/**
 * When scrolled all the way to the bottom, add more tiles.
 */
function onScroll(event) {
  // Check if we're within 100 pixels of the bottom edge of the broser window.
  var closeToBottom = ($(window).scrollTop() + $(window).height() > $(document).height() - 100);
  if(closeToBottom) {
    // Get the first then items from the grid, clone them, and add them to the bottom of the grid.
    var items = $('#tiles li');
    var firstTen = items.slice(0, 10);
    $('#tiles').append(firstTen.clone());

    // Clear our previous layout handler.
    if(handler) handler.wookmarkClear();

    // Create a new layout handler.
    handler = $('#tiles li');
    handler.wookmark(options);
  }
};

Thanks in advance. Ive looked around everywhere and tried a few different things, but no luck. Cheers.

A:

I just added an example to the GitHub repository on exactly how to do this. The example loads data from the Wookmark API and displays the images on the page. When you scroll down and reach the bottom, it loads more.
Here's the direct link to the example: https://github.com/GBKS/Wookmark-jQuery/tree/master/example-api
Let me know how this works out for you.