From 5707bbfb4399a092b1e0fe9263f09e0fa2f0e5c9 Mon Sep 17 00:00:00 2001 From: thevenyp Date: Fri, 24 Jul 2009 16:03:34 +0000 Subject: doc/algorithms.tex: Finish enumeration of cases where the sign of the zero part of x^y is determined. git-svn-id: svn://scm.gforge.inria.fr/svn/mpc/trunk@625 211d60ee-9f03-0410-a15a-8952a2c7a4e4 --- doc/algorithms.tex | 597 +++++++++++++++++++++++++++++++++++------------------ 1 file changed, 391 insertions(+), 206 deletions(-) diff --git a/doc/algorithms.tex b/doc/algorithms.tex index 14b29ef..83dd3ea 100644 --- a/doc/algorithms.tex +++ b/doc/algorithms.tex @@ -1177,217 +1177,402 @@ multiplication, division and square root. \paragraph{Sign of zeroes.} When the output value has a zero real or imaginary part, its sign should be -decided. When the inputs also have a zero real or imaginary part, we -consider all possible limits, and if all those limits give the same sign, -we take this as the sign of the zero part. -Otherwise, we round to $+0$, except for rounding toward $-\infty$, where we -round to $-0$. - -Example~: consider $x = 1 + 0 i$ and $y = -0 + 0 i$, we consider that $x$ -is the limit of $1 + \epsilon i$ for $\epsilon > 0$ tending to zero. -Similarly, $y = -\delta + \gamma i$ with $\delta, \gamma > 0$. -Now $\log x \approx \epsilon^2/2 + \epsilon i$, thus -$y \log x \approx (-\epsilon^2 \delta/2 - \epsilon \gamma) -+ i (\epsilon^2 \gamma/2 - \epsilon \delta)$. -Thus if we neglect terms of order $3$ or more, -whatever the relative growth of $\epsilon, \delta, \gamma$, the real and -imaginary parts of $y \log x$ are negative, thus we decide $x^y$ is rounded -to $1 - 0 i$. - -For $x = 1 - 0 i$ and $y = 0 + 0 i$, we write -$x = 1 - \epsilon i$ and $y = \delta + \gamma i$, -which gives $\log x \approx \epsilon^2/2 - \epsilon i$ -and $y \log x \approx \epsilon \gamma - i \epsilon \delta + O(\epsilon^2)$, -thus we also round $x^y$ to $1 - 0 i$. - -% (1 -0)^(-0 -0): -% x = 1 - epsilon i, y = -delta-gamma*i -% y log(x) = -epsilon gamma + i epsilon delta + O(epsilon^2) - - - - -The sign of zero parts are chosen so that they are consistent with the formula -$x^y = \exp(y\log x)$. -Let $\phi \in [-\pi, +\pi]$ the argument of $x = |x| e^{i\phi}$, $y_1$ -(resp. $y_2$) the real (resp. imaginary) part of $y = y_1 + y_2 i$. -Then +decided, which is not always possible if we want it to be consistent with the +formula $x^y = \exp(y\log x)$ (in the following, we exclude $0^y$). + +Let $x_1$, $x_2$, $y_1$, and $y_2$ real numbers so that $x = x_1 + x_2 i$ and +$y = y_1 + y_2 i$. +Let $\phi \in [-\pi, +\pi]$ the argument of $x = |x| e^{i\phi}$, with the +convention that when $x_1 < 0$ the argument of $x$ is $+\pi$ if $x_2 = +0$ and +$-\pi$ if $x_2 = -0$. +Then \[ -x^y=\exp(A(x,y)) (\cos B(x,y)+\sin B(x,y) i) +x^y=\exp\left(A(x,y)\right) \left(\cos B(x,y)+\sin B(x,y) i\right) \] where -\begin {eqnarray*} -A(x,y) & = & y_1\log|x|-y_2\phi,\\ -B(x,y) & = & y_2\log|x|+y_1\phi. -\end {eqnarray*} -As $|x^y| = \exp(A(x,y))$ is positive, the value of $B(x,y)$ determines the -sign of each part of $x^y$. -Special study is needed around $B(x, y)$ values of the form $k \pi/2$ to -determine the sign of the zero part: let $Q_{x_0}$ (resp. $Q_{y_0}$) the -quadrant where $x_0$ (resp. $y_0$) lies, when $B(x_0,y_0)$ lies around zero, -if $B(x,y)$ stays non negative for $x$ and $y$ in an open domain of $Q_{x_0} -\times Q_{y_0}$ containing $(x_0, y_0)$ then $\sin B(x_0, y_0) = \sin(+0) = -+0$, if it stays non positive then $\sin B(x_0, y_0) = \sin (-0) = -0$. -If $B(x,y)$ tends to $+\pi$ (resp. $-\pi$) staying lesser (resp. greater) than -it when $(x,y)$ tends to $(x_0, y_0)$, then $\sin B(x_0,y_0) = +0$. -Conversely, if $B(x,y)$ tend to $+\pi$ (resp. $-\pi$) remaining greater (resp. -lesser) than it, then $\sin B(x_0,y_0) = -0$. -The same applies to the real part when $B(x,y)$ lies around $\pm \pi/2$. - -First, let us solve $B(x,y) = 0$. +\begin {align*} + A(x,y) & = y_1\log|x|-y_2\phi,\\ + B(x,y) & = y_2\log|x|+y_1\phi. +\end {align*} +As $|x^y| = \exp\left(A(x,y)\right)$ is positive, the value of $B(x,y)$ +determines the sign of each part of $x^y$. +Note that $A(\overline{x},y) = A(x,\overline{y})$ and $B(\overline{x}, +y)=-B(x,\overline{y})$, so $\overline{x}^y = \overline{x^{\overline{y}}}$ and +we can restrict the study below to $x$ with nonnegative imaginary value +(i.e. $x_2 \geq 0$ and $\pi \geq \phi \geq 0$). + +To determine the sign of the zero part of $x^y$ when it is pure real or pure +imaginary, special study is needed around points $(x, y)$ where $B(x, y)$ is a +multiple of $\pi/2$. +Let +\begin {equation} + \label {eqn:Bk} + B_k(x, y) = y_2 \log|x| +y_1\phi -k\frac{\pi}{2} +\end {equation} +where $k$ is an integer and let $S_k$ the set of points $(x, y)$ where $B_k(x, +y) = 0$. + +For any integer $k$, we assume that the surface $S_k$ is orientable and not +reduced to a single point, then each neighborhood of a point $(x_0, y_0)$ of +$S_k$ intersects the region where $B(x, y) > k\pi$ and the region where $B(x, +y) < k\pi$. +Thus for an even $k$, we can make $(x, y)$ tend continuously to $(x_0, y_0)$ +so that $\Re(x^y) > 0$ or we can make it tend to $(x_0, y_0)$ so that +$\Re(x^y) < 0$ (the same applies with $k$ odd and $\Im(x^y)$). +In such cases, the sign of the zero part of $x_0^{y_0}$ is not determined. + +However, when $S_k$ intersects an axis, for example when $\Re(x_0) = 0$, we +have to distinguish two cases: $\Re(x_0) = +0$ and $\Re(x_0) = -0$. +Then, if $Q_{x_0}$ (resp. $Q_{y_0}$) denotes the quadrant where $x_0$ +(resp. $y_0$) lies, it is possible that the sign of $B_k(x,y)$ remains +constant for $(x,y)$ in the intersection $I$ of a neighborhood of $(x_0, y_0)$ +with $Q_{x_0}\times Q_{y_0}$, determining the sign of the zero part of $x^y$. +Let $dB_k(x,y)$ the derivative of $B_k(x, y)$, we have +\begin {equation} + \label {eqn:BkDerivative} + dB_k(x, y)\cdot(\delta_1, \delta_2, \epsilon_1, \epsilon_2) = + \frac{x_1y_2-x_2y_1}{x_1^2+x_2^2}\delta_1 + + \frac{x_1y_1+x_2y_2}{x_1^2+x_2^2}\delta_2 + + \phi\epsilon_1 + + \log(|x|) \epsilon_2 +\end {equation} +If $dB_k(x_0, y_0)\cdot(\delta_1, \delta_2, \epsilon_1, \epsilon_2)$ is not +null and if its sign remains constant for all real numbers $\delta_1$, +$\delta_2$, $\epsilon_1$, and $\epsilon_2$ so that $x = x_0 + \delta_1 + +\delta_2i$, $y = y_0 +\epsilon_1 + \epsilon_2i$, and $(x,y)$ is in the given +neighborhood $I$ of $(x_0, y_0)$ defined above, then the sign of $dB_k(x_0, +y_0)\cdot(\delta_1, \delta_2, \epsilon_1, \epsilon_2)$ determines the sign of +the zero part of $x^y$. + +In following discussion, we write $B_k$ as a function of four real arguments, +so that $B_k(x_1, x_2, y_1, y_2) = B_k(x_1+x_2i, y_1+y_2i)$. +Let $\sigma_1 = -1,+1$ (resp. $\sigma_2$, $\rho_1$, $\rho_2$) denote the sign +of $x_1$ (resp. $x_2$, $y_1$, $y_2$). \begin {enumerate} -\item -If -\[ -y_2 \log |x| = y_1 \phi = +0, -\] -then -\[ -x^y = +\exp(A(x,y)) +0i, -\] -the solutions are summarized in the following table: -\[ -\begin {array}{l c r c l l l l} -(x_1 -0i)^{y_1 -0i} & = & x_1^{y_1} +0i -& \text {with} & 0 < x_1 < 1 & & y_1 <0 \\ - -(x_1 -0i)^{-0 -0i} & = & +1 +0i -& \text {with} & |x_1| < 1 \\ - -x^{-0 -0i} & = & +1 +0i -& \text {with} & |x| < 1 & x_2 < 0 \\ - -x^{+0 -0i} & = & +1 +0i -& \text {with} & |x| < 1 & x_2 > 0 \\ - -(x_1 +0i)^{+0 -0i} & = & +1 +0i -& \text {with} & |x_1| < 1\\ - -(x_1 +0i)^{y_1 -0i} & = & x_1^{y_1} +0i -& \text {with} & 0 < x_1 < 1 & & y_1 > 0 \\ - -\hline - -(x_1 -0i)^{y_1 +0i} & = & x_1^{y_1} +0i -& \text {with} & x_1 \geq 1 & & y_1 <0 \\ - -(x_1 -0i)^{-0 +0i} & = & +1 +0i -& \text {with} & |x_1| \geq 1 \\ - -x^{-0 +0i} & = & +1 +0i -& \text {with} & |x| \geq 1 & x_2 < 0 \\ - -x^{+0 +0i} & = & +1 +0i -& \text {with} & |x| \geq 1 & x_2 > 0 \\ - -(x_1 +0i)^{+0 +0i} & = & +1 +0i -& \text {with} & |x_1| \geq 1\\ - -(x_1 +0i)^{y_1 +0i} & = & x_1^{y_1} +0i -& \text {with} & x_1 \geq 1 & & y_1 > 0 \\ - -\hline - -(+1 -0i)^{y_1 +y_2i} & = & +1 +0i -& \text {with} & & & y_1 <0 & y_2 > 0\\ - -(\pm 1 -0i)^{-0 +y_2i} & = & \exp (y_2 \phi) +0i -& \text {with} & & & & y_2 > 0\\ - -x^{-0 +y_2i} & = & \exp (y_2 \phi) +0i -& \text {with} & |x| = 1 & x_2 < 0 & & y_2 > 0 \\ - -x^{+0 +y_2i} & = & \exp (y_2 \phi) +0i -& \text {with} & |x| = 1 & x_2 > 0 & & y_2 > 0\\ - -(\pm1 +0i)^{+0 +y_2i} & = & \exp (y_2 \phi) +0i -& \text {with} & & & & y_2 > 0\\ - -(+1 +0i)^{y_1 +y_2i} & = & +1 +0i -& \text {with} & & & y_1 > 0 & y_2 > 0 -\end {array} -\] - -\item -If -\[ -y_2 \log |x| = y_1 \phi = -0, -\] -then -\[ -x^y = +\exp(A(x,y)) -0i, -\] -the solutions are summarized in the following table: -\[ -\begin {array}{l c r c l l l l} -(+1 +0i)^{y_1 +y_2i} & = & +1 -0i -& \text {with} & & & y_1 < 0 & y_2 < 0\\ - -(\pm 1 +0i)^{-0 +y_2i} & = & \exp(y_2 \phi) -0i -& \text {with} & & & & y_2 <0 \\ - -x^{-0 +y_2i} & = & \exp(y_2 \phi) -0i -& \text {with} & |x| = 1 & x_2 > 0 & & y_2 <0 \\ - -x^{+0 +y_2i} & = & \exp(y_2 \phi) -0i -& \text {with} & |x| = 1 & x_2 < 0 & & y_2 <0 \\ - -(\pm 1 -0i)^{+0 +y_2i} & = & \exp(y_2 \phi) -0i -& \text {with} & & & & y_2 < 0\\ - -(+1 -0i)^{y_1 +y_2i} & = & +1 -0i -& \text {with} & & & y_1 > 0 & y_2 < 0\\ - -\hline - -(x_1 +0i)^{y_1 -0i} & = & x^{y_1} -0i -& \text {with} & x_1 \geq 1 & & y_1 <0 \\ - -(x_1 +0i)^{-0 -0i} & = & +1 -0i -& \text {with} & |x_1| \geq 1 \\ - -x^{-0 -0i} & = & +1 -0i -& \text {with} & |x| \geq 1 & x_2 > 0 \\ - -x^{+0 -0i} & = & +1 -0i -& \text {with} & |x| \geq 1 & x_2 < 0 \\ - -(x_1 -0i)^{+0 -0i} & = & +1 -0i -& \text {with} & |x_1| \geq 1\\ - -(x_1 -0i)^{y_1 -0i} & = & x_1^{y_1} -0i -& \text {with} & x_1 \geq 1 & & y_1 > 0 \\ - -\hline - -(x_1 +0i)^{y_1 +0i} & = & x_1^{y_1} -0i -& \text {with} & 0 < x_1 < 1 & & y_1 <0 \\ - -(x_1 +0i)^{-0 +0i} & = & +1 -0i -& \text {with} & |x_1| < 1 \\ - -x^{-0 +0i} & = & +1 -0i -& \text {with} & |x| < 1 & x_2 > 0 \\ - -x^{+0 +0i} & = & +1 -0i -& \text {with} & |x| < 1 & x_2 < 0 \\ - -(x_1 -0i)^{+0 +0i} & = & +1 -0i -& \text {with} & |x_1| < 1 \\ - -(x_1 -0i)^{y_1 +0i} & = & x_1^{y_1} -0i -& \text {with} & 0 < x_1 < 1 & & y_1 > 0 \\ -\end {array} -\] - -\item -Is it possible that -\[ -\frac{\log |x|}{\phi} = -\frac{y_1}{y_2} \neq 0? -\] -In other words, can $\frac{\log |x|}{\arg (x)}$ be a rational number when $x$ -is a dyadic complex? +\item If $B_k(\sigma_1 0, x_2, y_1, y_2)=0$ for $x_2 > 0$. + Here $\phi = +\frac{\pi}{2}$. + \begin {enumerate} + \item if $y_2=\rho_2 0$, then replacing $x_1$ and $y_2$ by their value in + \ref {eqn:Bk}, we have $y_1= k$ and \ref {eqn:BkDerivative} gives + \[ + dB_k(\sigma_1 0, x_2, k, \rho_2 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = - \frac{k}{x_2} \delta_1 + 0 \delta_2 + \frac{\pi}{2} + \epsilon_1 + \log(x_2) \epsilon_2 + \] + where $\sigma_1 \delta_1 > 0$, and $\rho_2 \epsilon_2 > 0$ so that + $\delta_1 + (x_2 + \delta_2)i$ (resp. $k+\epsilon_1 + \epsilon_2 i$) is in + the same quadrant $Q_{x_0}$ (resp. $Q_{y_0}$) as $x_0=\sigma_1 0 +x_2 i$ + (resp. $y_0=k +\rho_2 0i$). + + We have to eliminate the case $k \neq 0$, because, in last expression, + $\epsilon_1$ would take positive as well as negative values if $k \neq 0$, + so the sign of $\frac{\pi}{2} \epsilon_1$, and therefore the sign of + $dB_k(x, y)\cdot (\delta_1, \delta_2, \epsilon_1, \epsilon_2)$, would not + be constant. + + If $k=0$, then $y_1=\rho_1 0$ and $\rho_1 \epsilon_1 > 0$ . + In this case, + \begin {align*} + dB_0(\pm 0, x_2, +0, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; x_2 \geq 1 \\ + dB_0(\pm 0, x_2, +0, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; 1 \geq x_2 > 0 \\ + dB_0(\pm 0, x_2, -0, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; x_2 \geq 1 \\ + dB_0(\pm 0, x_2, -0, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; 1 \geq x_2 > 0 + \end {align*} + and the sign of $dB_k(\sigma_1 0, x_2, \rho_1 0, \rho_2 0)\cdot(\delta_1, + \delta_2, \epsilon_1, \epsilon_2)$ is not constant in all other + combinations of $k$, $\sigma_1$, $x_2>0$, $\rho_1$, and $\rho_2$. + + \item if $y_2\neq 0$, then from \ref {eqn:Bk} + \[ + x_2 = \exp\left(\frac{k-y_1}{2y_2}\pi\right) + \] + But the number in the right hand side of the last equation is + known to be transcendental unless $k-y_1=0$. + As $x_2$ is dyadic, we have $y_1= k$ and $x_2=+1$. + Using \ref {eqn:BkDerivative}, we write + \[ + dB_k(\sigma_1 0, +1, k, y_2)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = - k \delta_1 + y_2 \delta_2 + \frac{\pi}{2} \epsilon_1 + 0 + \epsilon_2 + \] + where $\sigma_1 \delta_1 > 0$. + + Here, $\delta_2$ can take negative and positive values preventing the sign + of $dB_k(\sigma_1 0, +1, k, y_2)\cdot (\delta_1, \delta_2, \epsilon_1, + \epsilon_2)$ from being constant. + \end {enumerate} + +\item If $B_k(x_1, +0, y_1, y_2)=0$. + \begin {enumerate} + \item if $x_1 >0$, then $\phi = +0$. + From \ref {eqn:Bk}, we have + \[ + y_2 \log x_1 = k \frac{\pi}{2} + \] + \begin {enumerate} + \item if $y_2 = \rho_2 0$, the last equation implies $k = 0$, and from + \ref {eqn:BkDerivative} we have + \[ + dB_0(x_1, +0, y_1, \rho_2 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = 0 \delta_1 + \frac{y_1}{x_1} \delta_2 + 0 \epsilon_1 + + \log(x_1) \epsilon_2 + \] + with $\delta_2 > 0$ and $\rho_2 \epsilon_2 > 0$. + + Then the sign of the last expression in constant only in the following + cases, + \begin {align*} + dB_0(x_1, +0, y_1, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; y_1 \geq 0 \;\text{and}\; x_1 > 1\\ + dB_0(+1, +0, y_1, \pm 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; y_1 > 0\\ + dB_0(x_1, +0, y_1, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; y_1 \geq 0 \;\text{and}\; 1 > x_1 > 0 \\ + dB_0(x_1, +0, y_1, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; y_1 \leq 0 \;\text{and}\; x_1 > 1\\ + dB_0(+1, +0, y_1, \pm 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; y_1 < 0\\ + dB_0(x_1, +0, y_1, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; y_1 \leq 0 \;\text{and}\; 1 > x_1 > 0\\ + \end {align*} + Notice that we cannot conclude when $dB_0(x,y)$ is identically null, + so the cases $x=+1+0i$, $y=y_1 \pm0i$ cannot be determined by this + means. + + \item If $y_2 \neq 0$, then $x_1 = \exp\left(\frac{k}{2y_2}\pi\right)$ + is a dyadic number only if $k=0$ and then $x_1=1$. + We have, using \ref {eqn:BkDerivative}, + \[ + dB_0(+1, +0, y_1, y_2)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = y_2 \delta_1 + y_1 \delta_2 + 0 \epsilon_1 + 0 + \epsilon_2 + \] + with $\delta_2 > 0$. + + But here $y_2 \delta_1$ can take negative as well as positive values + preventing the sign of $dB_0(+1, +0, y_1, y_2)\cdot(\delta_1, + \delta_2, \epsilon_1, \epsilon_2)$ from being constant. + \end {enumerate} + + \item if $x_1 < 0$, then $\phi = \pi$. + Using \ref {eqn:BkDerivative}, we have + \[ + dB_k(x_1, +0, y_1, y_2)\cdot(\delta_1, \delta_2, \epsilon_1, \epsilon_2) + = \frac{y_2}{x_1} \delta_1 + \frac{y_1}{x_1} \delta_2 + \pi \epsilon_1 + + \log(-x_1) \epsilon_2 + \] + with $\delta_2 > 0$. + + If $y_1 \neq 0$, then $\epsilon_1$ can take negative as well as positive + values, preventing $dB_k$ from having a constant sign. + Thus $y_1 = 0$. + As $x_1 \neq 0$, $\delta_1$ can also take negative and positive values, + and $-y_2 \delta_1$ does not have a constant sign unless $y_2 = 0$. + But from \ref {eqn:Bk}, we know that $B_k(x, 0)=0$ implies $k = 0$. + + When $y = 0$, the derivative of $B_k$ is + \[ + dB_0(x_1, +0, \rho_1 0, \rho_2 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = 0 \delta_1 + 0 \delta_2 + \pi \epsilon_1 + \log(-x_1) + \epsilon_2 + \] + with $\delta_2 >0$, $\rho_1 \epsilon_1 >0$, and $\rho_2 \epsilon_2 >0$. + + Then, + \begin {align*} + dB_0(x_1, +0, +0, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; x_1 \leq -1 \\ + dB_0(x_1, +0, +0, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; 1 \leq x_1 < 0 \\ + dB_0(x_1, +0, -0, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; x_1 \leq -1\\ + dB_0(x_1, +0, -0, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; 1 \leq x_1 < 0 + \end {align*} + and the sign of $dB_k(x_1, +0, \rho_1 0, \rho_2 0)\cdot(\delta_1, + \delta_2, \epsilon_1, \epsilon_2)$ is not constant in all other + combinations of $k$, $x_1<0$, $\rho_1$, and $\rho_2$. + \end {enumerate} + +\item If $B_k(x_1, x_2, \rho_1 0, y_2)=0$ for $x_2 \geq 0$. + Here, we have + \[ + y_2\log|x|-k\frac{\pi}{2} = 0 + \] + \begin{enumerate} + \item If $y_2=0$, then from the last equation $k=0$. + Using \ref {eqn:BkDerivative} + \[ + dB_0(x_1,x_2,\rho_10,\rho_20)\cdot(\delta_1, \delta_2, \epsilon_1, \epsilon_2) = + 0\delta_1 + 0\delta_2 + \phi \epsilon_1 +\log|x| \epsilon_2 + \] + where $\rho_1 \epsilon_1 > 0$ and $\rho_2 \epsilon_2 > 0$. + + The case $\phi = 0$, that is $x_1 > 0$ and $x_2 = 0$, has already been + processed above. + Let $\phi > 0$, then $x_2$ is not null or $x_1 < 0$. + Using the expression of the derivative given above, we have + \begin{align*} + dB_0(x_1, x_2, +0, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; |x| \geq 1 \;\text{and}\; x \neq +1+0i\\ + dB_0(x_1, x_2, -0, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; |x| \geq 1 \;\text{and}\; x \neq +1+0i\\ + dB_0(x_1, x_2, +0, -0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &> 0 \;\text{if}\; 1 \geq |x| > 0 \;\text{and}\; x \neq +1+0i\\ + dB_0(x_1, x_2, -0, +0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) &< 0 \;\text{if}\; 1 \geq |x| > 0 \;\text{and}\; x + \neq +1+0i + \end{align*} + \item If $y_2 \neq 0$, from \ref {eqn:Bk}, we have + \[ + |x|^2=\exp\left(\frac{k}{y_2}\pi\right) + \] + The right hand side member of the equation is known to be transcendental + unless $k=0$. + As the left hand side member is dyadic, we have $k=0$, and then $|x|=1$. + Here, \ref {eqn:BkDerivative} gives + \[ + dB_0(x_1,x_2,\rho_10,y_2)\cdot(\delta_1, \delta_2, \epsilon_1, \epsilon_2) + = x_1y_2\delta_1 + x_2y_2\delta_2 + \phi \epsilon_1 + 0\epsilon_2 + \] + with $\rho_1 \epsilon_1 > 0$. + + If $x_2 \neq 0$, then $x_2y_2\delta_2$ can take negative as well as + positive values and the sign of the derivative is not constant. + Thus $x_2 = 0$. + Then, $x_1=\pm 1$ because $|x|=1$, and $x_1y_2\delta_1$ can take negative + and positive values. + + So, the sign of the null part of $B_k(x_1, x_2, \pm 0, y_2)$ (if any) + cannot be determined for all integers $k$ and for all dyadic real numbers + $x_1$, $x_2$, and $y_2 \neq 0$. + \end{enumerate} + +\item If $B_k(x_1, x_2, y_1, \rho_2 0)=0$ for $x_2 \geq 0$. + The case $y_1=0$, $y_2=0$ has already been precessed above. + Let $y_1 \neq 0$, then from \ref {eqn:Bk} we have + \[ + \phi = \frac{k}{2y_1}\pi + \] + which implies that the argument $\phi$ of $x$ can be written as $r \pi$ for + some rational number $r$ and, in the same time, $\cos^2 \phi$ ($= + |x|^2/x_1^2$ if $x_1 \neq 0$, else $=0$) and $\sin^2 \phi$ ($= |x|^2/x_2^2$ + if $x_2 \neq 0$, else $=0$) are rationnal. + + The five only possibilities (with $x_2 \geq 0$) are: + \begin{enumerate} + \item $\phi = 0$, but then $x_2=0$. + This case has been processed above. + \item $\phi = \frac{\pi}{4}$, then $x_1 = x_2 > 0$. + From \ref {eqn:Bk}, we have $y_1 = 2k \neq 0$, and from \ref + {eqn:BkDerivative} + \[ + dB_k(x_1, x_1, 2k, \rho_2 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = -\frac{k}{x_1}\delta_1 + \frac{k}{x_1}\delta_2 + + \frac{\pi}{4}\epsilon_1 + \log(\sqrt{2}x_1)\epsilon_2 + \] + with $\rho_2\epsilon_2>0$. + + As $x_1 \neq 0$ and $k \neq 0$, the term $\frac{k}{x_1}\delta_1$, and + $dB_k(x_1, x_1, 2k, \rho_2 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2)$ as well, has no constant sign. + \item $\phi = \frac{\pi}{2}$, then $x_1 = 0$. + This case has been processed above. + \item $\phi = \frac{3\pi}{4}$, then $x_1 = -x_2$, $x_1 < 0$ and \ref + {eqn:Bk} gives $k \neq 0$ and $y_1 = \frac{2k}{3}$. + As $y_1$ is a dyadic number, the only compatible values for $k$ are + multiple of 3. + Let $n$ be a nonzero integer so that $k = 3n$ and $y_1 = 2n$. + From \ref {eqn:BkDerivative}, we have + \[ + dB_{3n}(-x_2, x_2, 2n, \rho_2 0)\cdot(\delta_1, \delta_2, \epsilon_1, + \epsilon_2) = -\frac{n}{x_2}\delta_1 - \frac{n}{x_2}\delta_2 + + \frac{3\pi}{4}\epsilon_1 + \log(\sqrt{2}x_2)\epsilon_2 + \] + with $\rho_2\epsilon_2 >0$. + + As $n \neq 0$, $\epsilon_1$ can take negative and positive values and + the term $\frac{3\pi}{4}\epsilon_1$ has not a constant sign. + \item $\phi = \pi$, then $x_1<0$ and $x_2 = 0$. + This case has been processed above. + \end{enumerate} \end {enumerate} +To sum up using the inequalities above and deriving those with negative $x_2$ +from them and from the relation $\overline{x}^y = +\overline{x^{\overline{y}}}$, we can give the almost complete list of complex +powers of numbers (for dyadic complex) that have a determined signed zero +part, the only exception being $x=+1 \pm 0i$ raised to a pure real power which +cannot be treated as we have done here. + +\begin{tabular}{rlcrlrl} + $x^{+0 +0i}$ & $=1 +0i$ &and& + $x^{-0 -0i}$ & $= 1 -0i$ & + if $|x| \geq 1$ &and $x \neq +1 \pm 0i$ \\ + $(x_1 +0i)^{y_1 +0i}$ & $= x_1^{y_1} +0i$ &and& + $(x_1 -0i)^{y_1 -0i}$ & $= x_1^{y_1} -0i$ & + if $x_1 > 1$ &and $y_1 > 0$\\ + $(x_1 \pm 0i)^{\pm0 +0i}$ & $= 1 +0i$ &and& + $(x_1 \pm 0i)^{\pm0 -0i}$ & $= 1 -0i$ & + if $x_1 > 1$\\ + $(x_1 \pm 0i)^{-0 +0i}$ & $= 1 +0i$ &and& + $(x_1 \pm 0i)^{+0 -0i}$ & $= 1 -0i$ & + if $|x_1| > 1$\\ + $(x_1 -0i)^{y_1 +0i}$ & $= x_1^{y_1} +0i$ &and& + $(x_1 +0i)^{y_1 -0i}$ & $= x_1^{y_1} -0i$ & + if $x_1 > 1$ &and $y_1 < 0$ \\ + $(+1 +\sigma_20i)^{y_1 \pm0}$ & + \multicolumn{4}{l}{$=1 +\sigma_2\rho_1 0i$} & + if $x_1=+1$ &and $y_1 \neq 0$ \\ + $x^{+0 -0i}$ & $= 1 +0i$ &and& + $x^{-0 +0i}$ & $= 1 -0i$ & + if $1 \geq |x| > 0$ &and $x \neq +1 \pm 0i$ \\ + $(x_1 +0i)^{y_1 -0i}$ & $= x_1^{y_1} +0i$ &and& + $(x_1 -0i)^{y_1 +0i}$ & $= x_1^{y_1} -0i$ & + if $1 > x_1 > 0$ &and $y_1 > 0$ \\ + $(x_1 +0i)^{+0 -0i}$ & $= 1 +0i$ &and& + $(x_1 -0i)^{+0 +0i}$ & $= 1 -0i$ & + if $1 > |x_1| > 0$ \\ + $(x_1 -0i)^{-0 -0i}$ & $= 1 +0i$ &and& + $(x_1 +0i)^{-0 +0i}$ & $= 1 -0i$ & + if $1 > |x_1| > 0$ \\ + $(x_1 \pm0i)^{\pm0 -0i}$ & $= 1 +0i$ &and& + $(x_1 \pm0i)^{\pm0 +0i}$ & $= 1 -0i$ & + if $1 > x_1 > 0$ \\ + $(x_1 -0i)^{y_1 -0i}$ & $= x_1^{y_1} +0i$ &and& + $(x_1 +0i)^{y_1 +0i}$ & $= x_1^{y_1} -0i$ & + if $1 > x_1 > 0$ &and $y_1 < 0$ \\ + $(\pm 0 +x_2i)^{+0 +0i}$ & $= 1 +0i$ &and& + $(\pm 0 +x_2i)^{-0 -0i}$ & $= 1 -0i$ & + if $x_2 \geq 1$ \\ + $(\pm 0 +x_2i)^{+0 -0i}$ & $= 1 +0i$ &and& + $(\pm 0 +x_2i)^{-0 +0i}$ & $= 1 -0i$ & + if $1 \geq x_2 > 0$ \\ + $(\pm 0 +x_2i)^{-0 -0i}$ & $= 1 +0i$ &and& + $(\pm 0 +x_2i)^{+0 +0i}$ & $= 1 -0i$ & + if $ 0 > x_2 \geq -1$ \\ + $(\pm 0 +x_2i)^{-0 +0i}$ & $= 1 +0i$ &and& + $(\pm 0 +x_2i)^{+0 -0i}$ & $= 1 -0i$ & + if $-1 \geq x_2$ \\ + $(-1 +\sigma_2 0i)^{\rho_1 0 \pm0i}$ & + \multicolumn{4}{l}{$= 1 + \sigma_2 \rho_1 0i$} & + if $x_1=-1$ &and $y_1 = \rho_1 0$ +\end{tabular} + +So when $x^y$ is a pure real number, a compatible pattern is: + +\begin{tabular}{ll} + $x^y = x_1^{y_1} + \rho_2 0$ & if $|x| > 1$\\ + $x^y = 1 + \sigma_2 \rho_1 0$ & if $|x| = 1$\\ + $x^y = x_1^{y_1} - \rho_2 0$ & if $|x| < 1$ +\end{tabular} + +where $\sigma_2$ (resp $\rho_1$, $\rho_2$) is the sign of $x_2$ (resp. $y_1$, +$y_2$). + \bibliographystyle{acm} \bibliography{algorithms} -- cgit v1.2.1