text:
# Introduction¶

We begin by looking at a data set to make things concrete.

In [4]:
import pandas as pd

In [5]:
import matplotlib.pyplot as plt

fig,axes=plt.subplots(nrows=2,ncols=2,figsize=(8,8))
axes[0,0].scatter(df['Cigarettes'],df['Lung'])
axes[0,0].set_title('Lung')
#plt.xlabel('Cigarettes Sold per Capita')
axes[0,0].set_xticks([0,1500,3000,4500])
axes[0,0].axis([0,4500,0,30])
axes[0,0].set_yticks(range(0,40,10))

#plt.xlabel('Cigarettes Sold per Capita')
axes[1,0].set_xticks([0,1500,3000,4500])
axes[1,0].set_yticks(range(0,40,10))

axes[0,1].scatter(df['Cigarettes'],df['Leukemia'],color='green')
axes[0,1].set_title('Leukemia')
#plt.xlabel('Cigarettes Sold per Capita')
axes[0,1].set_xticks([0,1500,3000,4500])
axes[0,1].set_yticks(range(0,40,10))

axes[1,1].scatter(df['Cigarettes'],df['Kidney'],color='red')
axes[1,1].set_title('Kidney')
#plt.xlabel('Cigarettes Sold per Capita')
axes[1,1].set_xticks([0,1500,3000,4500])
axes[1,1].set_yticks(range(0,40,10))

fig.suptitle("Deaths from Cancer per 100K Population\n vs \n Cigarette sales per capita")

plt.show()

In [6]:
fig,axes=plt.subplots(nrows=1,ncols=1,figsize=(10,6))
axes.scatter(df['Cigarettes'],df['Lung'])
axes.set_title('Lung Cancer per 100K Population\n vs \n Cigarette Sales per Capita')
#plt.xlabel('Cigarettes Sold per Capita')
axes.set_xticks([0,1500,3000,4500])
axes.axis([0,4500,0,30])
axes.set_yticks(range(0,40,10))
plt.show()

There are 44 data points in the set (each corresponds to a state.)

In [7]:
df[['Cigarettes','Lung']][0:10]

Out[7]:
Cigarettes Lung
0 1820 17.05
1 3034 25.88
2 2582 19.80
3 1824 15.98
4 2860 22.07
5 3110 22.83
6 3360 24.55
7 4046 27.27
8 2827 23.57
9 2010 13.58

10 rows × 2 columns

The basic problem of simple linear regression is to find the slope $m$ and intercept $b$ so that the equation of the line $$y=mx+b$$ is the best fit to the data above.

To be specific, we look at the function $$E(m,b)=\sum_{i=1}^{N} (y_i-mx_i-b)^2$$ and we want to find $m$ and $b$ so that this is as small as possible. (For our example, $N=44$.)

# Calculus¶

Although the equation for $E$ looks complicated, it is really a function of two variables and so we can use calculus to find the minimum value. We compute: $$\frac{\partial E}{\partial m}=\sum_{i=1}^{N} 2(y_i-mx_i-b)x_i$$ and $$\frac{\partial E}{\partial b}=\sum_{i=1}^{N} 2(y_i-mx_i-b).$$

Remembering that the variables are $m$ and $b$, set these two equations to zero and find the following two equations: $$S_{yx}=S_{xx}m+S_{x}b$$ and $$S{y}=S{x}m+Nb$$ where I've written $S_{x}=\sum_{i=1}^{N} x_i$, $S_{xy}=\sum_{i=1}^{N} x_iy_i$, and so on.

In [8]:
Sx=sum(df['Cigarettes'])
Sy=sum(df['Lung'])
Sxy=sum([df['Cigarettes'][i]*df['Lung'][i] for i in range(len(df['Cigarettes']))])
Sxx=sum([df['Cigarettes'][i]*df['Cigarettes'][i] for i in range(len(df['Cigarettes']))])
N=len(df['Cigarettes'])
print 'N=',N,'Sx=',Sx,'Sy=',Sy,'Sxx=',Sxx,'Sxy=',Sxy

N= 44 Sx= 109622 Sy= 874.74 Sxx= 286469700 Sxy= 2248867.14

These are two equations in two unknowns that you can solve "by hand" or, for the general solution, you can use Cramer's rule. Let $$M=(NS{yx}-S{x}S_{y})$$ $$B=(S{xx}S_y-S{yx}S_x)$$ $$D=(S{xx}N-S{x}^2)$$ Then $$m=M/D$$ and $$b=B/D.$$

In [9]:
M=N*Sxy-Sx*Sy
B=Sxx*Sy-Sxy*Sx
D=N*Sxx-Sx*Sx
m=M/D
b=B/D
print 'm=',m,'b=',b

m= 0.00520586968046 b= 6.91050349746

In [10]:
fig,axes=plt.subplots(nrows=1,ncols=1,figsize=(10,6))
axes.scatter(df['Cigarettes'],df['Lung'])
axes.plot(np.arange(0,4500,1),m*np.arange(0,4500,1)+b*np.ones(len(np.arange(0,4500,1))),color="red",label='Regression Line')
axes.set_title('Lung Cancer per 100K Population\n vs \n Cigarette Sales per Capita')
#plt.xlabel('Cigarettes Sold per Capita')
axes.set_xticks([0,1500,3000,4500])
axes.axis([0,4500,0,30])
axes.set_yticks(range(0,40,10))
plt.legend()
plt.show()

# Linear Algebra¶

Let's look at the problem from a different point of view. Let's consider three column vectors:

• $Y=\left[\begin{matrix} y_1 \cr y_2 \cr \vdots \cr y_N\end{matrix}\right]$
• $X=\left[\begin{matrix} x_1 \cr x_2 \cr \vdots \cr x_N\end{matrix}\right]$
• $E=\left[\begin{matrix} 1 \cr 1 \cr \vdots \cr 1 \end{matrix}\right]$

If things were "perfect", meaning that the points all belonged to the line $y=mx+b$, we would have $$Y=mX+bE.$$

But, of course, we don't. In linear algebra terms, the vector $Y$ does not lie in the plane spanned by $X$ and $E$ in $\mathbb{R}^{44}$

So let's try to find the point in the plane spanned by $E$ and $X$ which is closest to $Y$.

In [11]:
from mpl_toolkits.mplot3d import Axes3D
#plt.subplots(nrows=1,ncols=1,figsize=(6,10))
fig=plt.figure(figsize=(10,10))
X = np.arange(-2, 2, 0.25)
Y = np.arange(-2, 2, 0.25)
X, Y = np.meshgrid(X, Y)
Z=2*Y-X
t=np.arange(-1,0,.1)
ax.plot_wireframe(X,Y,Z)
ax.plot(t,-2*t,6+t,color="red")
ax.scatter([0,-1],[0,2],[6,5],color="red")
ax.set_title("Closest point lying in a given plane from a point outside the plane")
plt.show()

We can use linear algebra to solve the problem of finding the point in the plane $P$ spanned by $X$ and $E$ that is closest to $Y$. Essentially, we need to take the perpendicular projection of the vector $Y$ into the plane spanned by $X$ and $E$. For that, we can construct an orthonormal basis for the plane $P$ by using the Gram-Schmidt process on the two vectors $E$ and $X$.

First, we normalize $E$ to obtain our first unit vector $\mathbf{e}=E/\sqrt{N}$, since $E\cdot E=N$.

Next, we subtract the projection of $X$ onto the $E$ direction and normalize. Let $x=X-(X\cdot \mathbf{e})\mathbf{e}$ and $$\mathbf{x}=\frac{x}{\sqrt{x\cdot x}}$$

Then the point we are interested in is $$\mathbf{y}=(Y\cdot\mathbf{x})\mathbf{x}+(Y\cdot\mathbf{e})\mathbf{e}$$

This point lies in the plane $P$, and $Y-\mathbf{y}$ is perpendicular to both $E$ and $X$.

Let's make a few observations about this. First of all, $(Y\cdot\mathbf{e})\mathbf{e}=\overline{y}E$ and $(X\cdot \mathbf{e})\mathbf{e}=\overline{x}E$ where $\overline{x}$ and $\overline{y}$ are the averages (means) of the $x$ and $y$ values respetively.

We can arrange for our data to have $\overline{x}=0$ and $\overline{y}=0$ by subtracting $\overline{x}$ and $\overline{y}$ each of the points. This amounts to a coordinate change that moves the origin of our coordinate system to $(\overline{x},\overline{y})$.

In these coordinates, we get a huge simplification. If both means are zero, we get $$\mathbf{x}=\frac{X}{\sqrt{X\cdot X}}$$ and $$\mathbf{y}=\frac{Y\cdot X}{X\cdot X}X$$ In other words, in these coordinates, $b=0$ and $m=\frac{Y\cdot X}{X\cdot X}$.

In [12]:
Xnorm=df['Cigarettes']-df['Cigarettes'].mean()
Ynorm=df['Lung']-df['Lung'].mean()
plt.scatter(Xnorm,Ynorm)
m=Xnorm.dot(Ynorm)/Xnorm.dot(Xnorm)
plt.plot(Xnorm, m*Xnorm)
plt.show()

# Statistics¶

What accounts for the variation of the points around the "true value"? Statisticians approach this through the idea of a statistical model. Let's leave the lung cancer data behind for the moment, and imagine an abstract problem. Suppose that we make measurements of a dependent variable $Y$ that is related to an independent variable $X$ by a linear equation. Suppose, however, that the $Y$ values include a certain amount of random error $\epsilon$, so that $$Y=mX+b+\epsilon.$$

We will assume that the error term $\epsilon$ is a normally distributed error, with mean zero, variance $\sigma$, and that the errors we obtain from separate measurements of $Y$ are independent of one another.

In [13]:
from scipy.stats import norm
X=np.arange(-5,5,.1)
plt.plot(X,norm.pdf(X,scale=.7,loc=0),color='red',label='v=.5')
plt.plot(X,norm.pdf(X,scale=1),color='blue',label='v=1')
plt.plot(X,norm.pdf(X,scale=1.5),color='orange',label='v=2.25')

plt.title('Normal Distributions')
plt.legend()
plt.show()

Suppose for the sake of concreteness that the $X$ values are 0.0,0.1,...,9.9 and that $Y=X+1+\epsilon$ where $\epsilon$ has variance $1$. Different measurements create different scatter plots.

In [14]:
def MB(X,Y):
Sxx=sum([X[i]*X[i] for i in range(len(X))])
Sxy=sum([X[i]*Y[i] for i in range(len(X))])
Sx=sum([X[i] for i in range(len(X))])
Sy=sum([Y[i] for i in range(len(Y))])
N=len(X)
M=N*Sxy-Sx*Sy
B=Sxx*Sy-Sxy*Sx
D=N*Sxx-Sx*Sx
m=M/D
b=B/D
return m,b

X=np.arange(0,10,.1)
fig,ax=plt.subplots(nrows=2,ncols=2,figsize=(6,6))
epsilon=norm.rvs(size=100,scale=2,loc=0)
Y0=X+np.ones(len(X))+epsilon
epsilon=norm.rvs(size=100,scale=2,loc=0)
Y1=X+np.ones(len(X))+epsilon
epsilon=norm.rvs(size=100,scale=2,loc=0)
Y2=X+np.ones(len(X))+epsilon
epsilon=norm.rvs(size=100,scale=2,loc=0)
Y3=X+np.ones(len(X))+epsilon
for i in [0,1]:
for j in [0,1]:
ax[i,j].set_xticks(range(0,11,2))
ax[i,j].set_yticks(range(-5,17,2))
ax[0,0].scatter(X,Y0,color='blue')
ax[1,0].scatter(X,Y1,color='red')
ax[0,1].scatter(X,Y2,color='green')
ax[1,1].scatter(X,Y3,color='black')
m0,b0=MB(X,Y0)

m1,b1=MB(X,Y1)

m2,b2=MB(X,Y2)
m3,b3=MB(X,Y3)
ax[0,0].plot(X,m0*X+b0,color="blue")
ax[1,0].plot(X,m1*X+b1,color="red")
ax[0,1].plot(X,m2*X+b2,color="green")
ax[1,1].plot(X,m3*X+b3,color="black")
plt.show()

We can superimpose these different lines and you can see that they are slightly different -- with varying slope and intercept -- which come from different snapshots of the underlying random data.

In [15]:
plt.plot(X,m0*X+b0,color="blue")
plt.plot(X,m1*X+b1,color="red")
plt.plot(X,m2*X+b2,color="green")
plt.plot(X,m3*X+b3,color="black")
plt.show()

Suppose we draw 5000 samples of 100 Y-values each, where each of the 100 $Y$ values from a given sample satisfy $Y=X+1+\epsilon$. For each of the 5000 samples, we calculate the slope and intercept for those $100$ $Y$-values. The picture below shows that these slope/intercept pairs lie in an ellipse centered at $(1,1)$.

In [16]:
Y=X+np.ones(len(X))
M,B=[],[]
for i in range(5000):
Yn=Y+norm.rvs(size=100,loc=0,scale=2)
m,b=MB(X,Yn)
M.append(m)
B.append(b)
plt.title("Slope/Intercept Pairs for Different Samples")
plt.xlabel("Slope")
plt.ylabel("Intercept")
plt.scatter(B,M,s=.3,alpha=.3,color='blue')
plt.show()

To see where this ellipse comes from, let's recall the orthonormal vectors $e$ and $x$ introduced earlier. We saw that the predicted $Y$ value corresponds to the orthogonal projection $\hat{Y}=(Y\cdot e)e+(Y\cdot x)x$. The points $((Y\cdot e),(Y\cdot x))$ and $(Y\cdot x)$ are distributed around $((X+1)\cdot e),((X+1)\cdot x))$ -- basically we are looking at the random vectors $\epsilon$ projected orthogonally onto the $e,x$ plane. Since the $\epsilon$ fall in a sphere around zero, the projection is a circle.

In [46]:
def P(X,Y):
N=len(X)
Xbar=sum(X)/N
Sx=sum((X-Xbar*np.ones(N))**2)
A=sum(Y)/sqrt(N)
Z=(X-Xbar*np.ones(N))/sqrt(Sx)
B=sum([Z[i]*Y[i] for i in range(N)])
return A,B

A=[]
B=[]
a0,b0=P(X,X+1)
for i in range(5000):
Yn=Y+norm.rvs(size=100,loc=0,scale=2)
a,b=P(X,Yn)
A.append(a-a0)
B.append(b-b0)

fig,axes=plt.subplots(nrows=1,ncols=2,figsize=(10,10))
axes[0].set_title("Orthogonal\n Projection of Residuals")
axes[0].axis([-10,10,-10,10])
axes[0].set_aspect('equal')
#plt.gca().set_aspect('equal')
axes[0].scatter(A,B,s=.3,alpha=.3,color='blue')

axes[1].hist(A,normed='True')
axes[1].hist(B,normed='True')
axes[1].plot(np.arange(-10,10,.1),norm.pdf(np.arange(-10,10,.1),scale=2),color='black')
axes[1].set_title("Distribution of Residuals")

plt.show()

We can make a change of variables from the $x$,$e$ coordinate system to the $X$, $E$ coordinates; the coefficients of the $\hat{Y}$ vector in these coordinates are $m$ and $b$. The circular cloud of points becomes this ellipse. The ellipses correspond to the circles where about 68\% of the points lie, 95\%; and 99.7\%.

In [67]:
#plt.xlabel("Slope")
#plt.ylabel("Intercept")
#axes[0].set_xticks(np.arange(-.5,.5,.1))
fig=figure(num=1,figsize=(10,10))

axes=fig.gca()

A=np.array(A)
B=np.array(B)
axes.set_title('Projection of Residuals in Slope,Intercept')
axes.axis([-2,2,-.3,.3])
axes.set_xticks(np.arange(-2,3,1))
N=len(X)
Xbar=sum(X)/N
Sx=sum((X-Xbar*np.ones(N))**2)
R=Sx+N*Xbar**2
S=2*N*Xbar
T=N
u=np.arange(-3,4,.01)
v=np.arange(-3,4,.01)
U,V=np.meshgrid(u,v)
axes.contour(U,V,T*U**2+S*U*V+R*V**2,levels=[4,16,36],colors=['black','black','black'])
axes.scatter(1/sqrt(N)*A-Xbar/sqrt(Sx)*B,1/sqrt(Sx)*B,s=.5,alpha=.3,color='red')
plt.show()

In [58]:

In [58]:
In [58]:
In [58]:

In [ ]:


text:
arXiv:1202.1192
IPPP-12-01
DCPT-12-02
CERN-PH-TH-2011-322
MCNET-12-02
Eur.Phys.J. C72 (2012) 2028

by: Zapp, Korinna Christine (Durham U., IPPP) et al.

Abstract:
It is widely accepted that a phenomenologically viable theory of jet quenching for heavy ion collisions requires the understanding of medium-induced parton energy loss beyond the limit of eikonal kinematics formulated by Baier-Dokshitzer-Mueller-Peigne-Schiff and Zakharov (BDMPS-Z). Here, we supplement a recently developed exact Monte Carlo implementation of the BDMPS-Z formalism with elementary physical requirements including exact energy-momentum conservation, a refined formulation of jet-medium interactions and a treatment of all parton branchings on the same footing. We document the changes induced by these physical requirements and we describe their kinematic origin.

format: plain


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text:
Setting up a mediation model

I'm having trouble understanding how to set up a simple mediation model.

IV: hours of sleep

M: amount of exercise

DV: score on some test

I'm expecting the relationship between sleep and test score to be mediated by exercise.

From what I understand, where are 3 regressions that need to be tested:

• Path A: relationship between exercise and sleep
• Path B: relationship between score and exercise
• Path C (direct): relationship between score and sleep

I'm trying to understand the details of these regression models and what should and should not be included in them. I'm looking to see if my understanding of the models below is correct, and questions are bolded:

Path A

This is using sleep to predict exercise:

mdlA <- lm(exercise ~ sleep, data=data)

Path B

Using exercise (mediator) to predict score:

mdlB <- lm(score ~ exercise, data=data)

Should this model also include sleep? i.e.:

mdlB <- lm(score ~ exercise + sleep, data=data)

My path B coefficient is the one associated with exercise (regardless of which of the 2 models I use)

Path C

This is the direct effect:

mdlC <- lm(score ~ sleep, data=data)

Should this model also include the mediator? i.e.:

mdlC <- lm(score ~ sleep + exercise, data=data)

My C coefficient would then be the one associated with sleep

Once I have my 3 coefficients, I then need to bootstrap the total indirect effect (A*B) and test it against 0. If it is significant, then I have mediation

Whether I have partial or complete mediation is then determined by whether Path C is significant? Or does mediation always required all 3 paths to be significant?

I think this might depend on what model I used for Path C. If I add the mediator into the Path C model, then significance would tell me I have partial mediation. But non-significant would tell me I have complete mediation?

Am I missing anything from the steps required to conduct this type of analysis?

Path A and Path B are fine. Below are all the steps to test mediation effect:

mdlA <- lm(exercise ~ sleep, data=data)
mdlB <- lm(score ~ exercise, data=data)
mdlC <- lm(exercise ~ sleep, data=data)
mdlB <- lm(score ~ sleep + exercise, data=data)

If the first three models are significant, you conduct the fourth model. In the fourth, if sleep is non-significant, when exercise is controlled, existence of mediation is supported.
One caveat here is that both exercise and sleep are likely to independently influence score. In such a scenario, sleep might be significant which indicates partial mediation. The two independent variables might interactively (jointly) influence dependent variable score. In this case, you have moderation effect.
For a little more elaboration, have a look at the thread Testing mediation and moderation; can one variable function as both mediator and moderator?.

Your approach for testing full/partial mediation is correct. You can use Sobel's test (If sample size is large enough/assumptions are met) Or bootstrapping to do the same.

If you are using SPSS, installing process macro by Andrew F. Hayes will solve your purpose/


text:
# ordinal isomorphism theorem

The only proof of the theorem stating that every well-order is order-isomorphic to a unique ordinal of which I am aware relies on the Axiom of Replacement. Is this necessary?

Yes, Replacement is necessary for this. For instance, it is straightforward to verify that $V_{\omega+\omega}$ is a model of all of ZFC except for Replacement (the same is true of $V_\alpha$ for any limit ordinal $\alpha>\omega$). In $V_{\omega+\omega}$, there are well-ordered sets of length $\omega+\omega$ and much longer (since for any set $X\in V_{\omega+\omega}$, every well-ordering of $X$ is also in $V_{\omega+\omega}$), but the ordinal $\omega+\omega$ does not exist.
• Unless $\alpha$ is inaccessible (or wordly, etc.) – Pedro Sánchez Terraf Feb 19 '16 at 12:36


text:
# Studies of the resonance structure in $D^{0} \to K^\mp \pi^\pm \pi^\pm \pi^\mp$ decays

[to restricted-access page]

## Abstract

Amplitude models are constructed to describe the resonance structure of ${D^{0}\to K^{-}\pi^{+}\pi^{+}\pi^{-}}$ and ${D^{0} \to K^{+}\pi^{-}\pi^{-}\pi^{+}}$ decays using $pp$ collision data collected at centre-of-mass energies of 7 and 8 TeV with the LHCb experiment, corresponding to an integrated luminosity of $3.0\mathrm{fb}^{-1}$. The largest contributions to both decay amplitudes are found to come from axial resonances, with decay modes $D^{0} \to a_1(1260)^{+} K^{-}$ and $D^{0} \to K_1(1270/1400)^{+} \pi^{-}$ being prominent in ${D^{0}\to K^{-}\pi^{+}\pi^{+}\pi^{-}}$ and $D^{0}\to K^{+}\pi^{-}\pi^{-}\pi^{+}$, respectively. Precise measurements of the lineshape parameters and couplings of the $a_1(1260)^{+}$, $K_1(1270)^{-}$ and $K(1460)^{-}$ resonances are made, and a quasi model-independent study of the $K(1460)^{-}$ resonance is performed. The coherence factor of the decays is calculated from the amplitude models to be $R_{K3\pi} = 0.459\pm 0.010 (\mathrm{stat}) \pm 0.012 (\mathrm{syst}) \pm 0.020 (\mathrm{model})$, which is consistent with direct measurements. These models will be useful in future measurements of the unitary-triangle angle $\gamma$ and studies of charm mixing and $C P$ violation.

## Figures and captions

 Invariant mass and mass difference distributions for RS (top) and WS (bottom) samples, shown with fit projections. The signal region is indicated by the filled grey area, and for each plot the mass window in the orthogonal projection is applied. Fig1a.pdf [70 KiB] HiDef png [220 KiB] Thumbnail [177 KiB] *.C file Fig1b.pdf [63 KiB] HiDef png [222 KiB] Thumbnail [176 KiB] *.C file Fig1c.pdf [72 KiB] HiDef png [281 KiB] Thumbnail [234 KiB] *.C file Fig1d.pdf [65 KiB] HiDef png [281 KiB] Thumbnail [218 KiB] *.C file Distributions for six invariant-mass observables in the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . Bands indicate the expectation from the model, with the width of the band indicating the total systematic uncertainty. The total background contribution, which is very low, is shown as a filled area. In figures that involve a single positively-charged pion, one of the two identical pions is selected randomly. Fig2a.pdf [76 KiB] HiDef png [243 KiB] Thumbnail [191 KiB] *.C file Fig2b.pdf [77 KiB] HiDef png [274 KiB] Thumbnail [216 KiB] *.C file Fig2c.pdf [73 KiB] HiDef png [215 KiB] Thumbnail [162 KiB] *.C file Fig2d.pdf [75 KiB] HiDef png [268 KiB] Thumbnail [201 KiB] *.C file Fig2e.pdf [74 KiB] HiDef png [247 KiB] Thumbnail [185 KiB] *.C file Fig2f.pdf [74 KiB] HiDef png [261 KiB] Thumbnail [194 KiB] *.C file Argand diagram for the model-independent partial-wave analysis (MIPWA) for the $K (1460)$ resonance. Points show the values of the amplitude that are determined by the fit, with only statistical uncertainties shown. Fig3.pdf [42 KiB] HiDef png [130 KiB] Thumbnail [75 KiB] *.C file Parity violating distributions for the RS decay in the $\overline{ K }{} {}^* (892)^{0}\rho(770)^{0}$ region defined by $\pm 35 \mathrm{ Me V}$($\pm 100 \mathrm{ Me V}$) mass windows about the nominal $\overline{ K }{} {}^* (892)^{0}$ $(\rho(770)^{0})$ masses. Bands show the predictions of the fitted model including systematic uncertainties. Fig4a.pdf [61 KiB] HiDef png [400 KiB] Thumbnail [266 KiB] *.C file Fig4b.pdf [61 KiB] HiDef png [417 KiB] Thumbnail [278 KiB] *.C file Fig4c.pdf [60 KiB] HiDef png [405 KiB] Thumbnail [262 KiB] *.C file Fig4d.pdf [59 KiB] HiDef png [396 KiB] Thumbnail [269 KiB] *.C file Distributions for six invariant-mass observables in the WS decay $D ^0 \rightarrow K ^+ \pi ^- \pi ^- \pi ^+$ . Bands indicate the expectation from the model, with the width of the band indicating the total systematic uncertainty. The total background contribution is shown as a filled area, with the lower region indicating the expected contribution from mistagged $\overline{ D }{} {}^0 \rightarrow K ^+ \pi ^- \pi ^- \pi ^+$ decays. In figures that involve a single negatively-charged pion, one of the two identical pions is selected randomly. Fig5a.pdf [58 KiB] HiDef png [270 KiB] Thumbnail [195 KiB] *.C file Fig5b.pdf [68 KiB] HiDef png [340 KiB] Thumbnail [260 KiB] *.C file Fig5c.pdf [56 KiB] HiDef png [274 KiB] Thumbnail [195 KiB] *.C file Fig5d.pdf [58 KiB] HiDef png [334 KiB] Thumbnail [246 KiB] *.C file Fig5e.pdf [58 KiB] HiDef png [327 KiB] Thumbnail [238 KiB] *.C file Fig5f.pdf [58 KiB] HiDef png [326 KiB] Thumbnail [240 KiB] *.C file Distributions for six invariant-mass observables in the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . Bands indicate the expectation from a model which excludes the decay chain $K_1(1270)^{-}\rightarrow \rho(1450)^{0} K ^-$, with the width of the band indicating the total systematic uncertainty. The total background contribution, which is very low, is shown in green. Fig6a.pdf [76 KiB] HiDef png [251 KiB] Thumbnail [193 KiB] *.C file Fig6b.pdf [77 KiB] HiDef png [280 KiB] Thumbnail [219 KiB] *.C file Fig6c.pdf [74 KiB] HiDef png [220 KiB] Thumbnail [164 KiB] *.C file Fig6d.pdf [75 KiB] HiDef png [276 KiB] Thumbnail [206 KiB] *.C file Fig6e.pdf [74 KiB] HiDef png [253 KiB] Thumbnail [189 KiB] *.C file Fig6f.pdf [74 KiB] HiDef png [270 KiB] Thumbnail [198 KiB] *.C file Animated gif made out of all figures. PAPER-2017-040.gif Thumbnail

## Tables and captions

 Signal and background yields for both samples in the signal region, presented separately for each year of data taking. Table_1.pdf [48 KiB] HiDef png [67 KiB] Thumbnail [33 KiB] tex code Fit fractions and coupling parameters for the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . For each parameter, the first uncertainty is statistical and the second systematic. Couplings $g$ are defined with respect to the coupling to the channel $D ^0 \rightarrow[\overline{ K }{} {}^* (892)^{0}\rho(770)^{0}]^{L=2}$. Also given are the $\chi^2$ and the number of degrees of freedom ($\nu$) from the fit and their ratio. Table_2.pdf [73 KiB] HiDef png [187 KiB] Thumbnail [86 KiB] tex code Table of fit fractions and coupling parameters for the component involving the $a_{1}(1260)^{+}$ meson, from the fit performed on the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . The coupling parameters are defined with respect to the $a_{1}(1260)^{+} \rightarrow\rho(770)^{0}\pi ^+$ coupling. For each parameter, the first uncertainty is statistical and the second systematic. Table_3.pdf [64 KiB] HiDef png [59 KiB] Thumbnail [22 KiB] tex code Table of fit fractions and coupling parameters for the component involving the $K_{1}(1270) ^{-}$ meson, from the fit performed on the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . The coupling parameters are defined with respect to the $K_{1}(1270) ^{-}\rightarrow\rho(770)^{0} K ^-$ coupling. For each parameter, the first uncertainty is statistical and the second systematic. Table_4.pdf [64 KiB] HiDef png [75 KiB] Thumbnail [36 KiB] tex code Table of fit fractions and coupling parameters for the component involving the $K(1460)^{-}$ meson, from the fit performed on the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . The coupling parameters are defined with respect to the $K(1460)^{-} \rightarrow\overline{ K }{} {}^* (892)^{0}\pi ^-$ coupling. For each parameter, the first uncertainty is statistical and the second systematic. Table_5.pdf [57 KiB] HiDef png [52 KiB] Thumbnail [24 KiB] tex code Fit fractions and coupling parameters for the WS decay $D ^0 \rightarrow K ^+ \pi ^- \pi ^- \pi ^+$ . For each parameter, the first uncertainty is statistical and the second systematic. Couplings $g$ are defined with respect to the coupling to the decay $D ^0 \rightarrow[ K ^{*}(892)^{0}\rho(770)^{0}]^{L=2}$. Also given are the $\chi^2$ and the number of degrees of freedom ($\nu$) from the fit and their ratio. Table_6.pdf [64 KiB] HiDef png [98 KiB] Thumbnail [45 KiB] tex code Decay chains taken into account in alternative parametrisations of the RS decay mode ${ D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^- }$. For each chain, the fraction of models in the ensemble that contain this decay, together with the associated average fit fraction, $\langle \mathcal{F} \rangle$, are shown. Components are not tabulated if they contribute to all models in the ensemble, or if they contribute to less than 5% of the models. Table_7.pdf [63 KiB] HiDef png [158 KiB] Thumbnail [79 KiB] tex code Dependence of fit fractions (and partial fractions) on the choice of the RS model. This dependence is expressed as the mean value and the RMS of the values in the ensemble. Also shown is the fit fractions of the baseline model presented in Sect. 6.2. Table_8.pdf [64 KiB] HiDef png [275 KiB] Thumbnail [138 KiB] tex code Dependence of the fitted masses and widths on the final choice of the RS model. This dependence is expressed as the mean value and the RMS of the values in the ensemble. The values found for the baseline model presented in Sect. 6.2 are reported for comparison. Table_9.pdf [47 KiB] HiDef png [88 KiB] Thumbnail [43 KiB] tex code Decay chains taken into account in alternative parametrisations of the WS decay mode ${ D ^0 \rightarrow K ^+ \pi ^- \pi ^- \pi ^+ }$. For each chain, the fraction of models in the ensemble that contain this decay, together with the associated average fit fraction, $\langle \mathcal{F} \rangle$, are shown. Components are not tabulated if they contribute to all models in the ensemble, or if they contribute to less than 5% of the models. Table_10.pdf [63 KiB] HiDef png [159 KiB] Thumbnail [80 KiB] tex code Coherence factor and average strong-phase differences in regions of phase space. The spread of coherence factors, average strong-phase difference and ratio of amplitudes from choice of WS model characterised with the RMS of the distribution. Table_11.pdf [51 KiB] HiDef png [118 KiB] Thumbnail [60 KiB] tex code Rules for calculating the current associated with a given decay chain in terms of the currents of the decay products. Where relevant, the spin projection operator $\mathcal{S}$ and the orbital angular momentum operators $L$ are those for the decaying particle. Table_12.pdf [58 KiB] HiDef png [139 KiB] Thumbnail [66 KiB] tex code Legend for systematic uncertainties, including whether this sources of uncertainty is considered on the RS/WS decay mode. Table_13.pdf [26 KiB] HiDef png [84 KiB] Thumbnail [37 KiB] tex code Systematic uncertainties on the RS decay coupling parameters and fit fractions for quasi two-body decay chains. Table_14.pdf [70 KiB] HiDef png [300 KiB] Thumbnail [102 KiB] tex code Systematic uncertainties on the RS decay coupling parameters, fit fractions and masses and widths of resonances for cascade topology decay chains. Table_15.pdf [66 KiB] HiDef png [416 KiB] Thumbnail [169 KiB] tex code Systematic uncertainties on the WS decay coupling parameters and fit fractions. Table_16.pdf [62 KiB] HiDef png [193 KiB] Thumbnail [79 KiB] tex code Interference fractions for the RS mode $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ , only shown for fractions $>0.5\%$. For each fraction, the first uncertainty is statistical and the second systematic. Table_17.pdf [62 KiB] HiDef png [175 KiB] Thumbnail [82 KiB] tex code Interference fractions for the WS mode $D ^0 \rightarrow K ^+ \pi ^- \pi ^- \pi ^+$ , only shown for fractions $>0.5\%$. For each fraction, the first uncertainty is statistical and the second systematic. Table_18.pdf [62 KiB] HiDef png [129 KiB] Thumbnail [60 KiB] tex code Table of fit fractions and coupling parameters and other quantities for the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ , for a model excluding the decay chain $K_1(1270)^{-}\rightarrow \rho(1450)^0 K ^-$. Also given is the $\chi^2$ per degree of freedom ($\nu$) for the fit. The first uncertainty is statistical and the second systematic. Couplings are defined with respect to the coupling to the channel $D ^0 \rightarrow[\overline{ K }{} {}^* (892)^{0}\rho(770)^{0}]^{L=2}$. Table_19.pdf [73 KiB] HiDef png [185 KiB] Thumbnail [84 KiB] tex code Table of fit fractions and coupling parameters for the component involving the $a_{1}(1260)^{+}$ meson, from the fit performed on the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . The coupling parameters are defined with respect to the $a_{1}(1260)^{+} \rightarrow\rho(770)^{0}\pi ^+$ coupling. For each parameter, the first uncertainty is statistical and the second systematic. Table_20.pdf [64 KiB] HiDef png [59 KiB] Thumbnail [22 KiB] tex code Table of fit fractions and coupling parameters for the component involving the $K_{1}(1270) ^{-}$ meson, from the fit performed on the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . The coupling parameters are defined with respect to the $K_{1}(1270) ^{-}\rightarrow\rho(770)^{0} K ^-$ coupling. For each parameter, the first uncertainty is statistical and the second systematic. Table_21.pdf [64 KiB] HiDef png [65 KiB] Thumbnail [30 KiB] tex code Table of fit fractions and coupling parameters for the component involving the $K(1460)^{-}$ meson, from the fit performed on the RS decay $D ^0 \rightarrow K ^- \pi ^+ \pi ^+ \pi ^-$ . The coupling parameters are defined with respect to the $K(1460)^{-} \rightarrow\overline{ K }{} {}^* (892)^{0}\pi ^-$ coupling. For each parameter, the first uncertainty is statistical and the second systematic. Table_22.pdf [57 KiB] HiDef png [51 KiB] Thumbnail [19 KiB] tex code

Created on 11 July 2020.


text:
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### Is SGP4 propagation necessarily more accurate near the epoch chosen for TLE generation?

This interesting answer includes a quote from Space-Track: From Space-Track.Org FAQ TLEs can contain future epochs. About 20 satellites are categorized as "multi-day objects" because ...
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### SGP4 on Systems Tool Kit (STK); how to check if SDP4 deep space correction is implemented?

I am doing orbit propagation for a geostationary satellite which requires the deep space corrections of SDP4. See my previous question SGP 4 for Geostationary Satellite. I am using SGP4 on STK, but, ...
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### Textbook or scholarly discussion of equations used by the SDP4 part of SGP4 TLE propagators beyond Space Track Report #1?

(the famous) Space Tract Report #3 discussed at length in Dr. T.S. Kelso's Revisiting Spacetrack Report #3, AIAA 2006-6753 Describes the "deep space" corrections to the original SGP4 known ...
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### In a 3D space defined by semimajor axis, eccentricity and inclination, what would be the shape of the space where SDP4 works better than SGP4?

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text:
#### 455. Sequence analysis

Time limit per test: 1 second(s)
Memory limit: 4096 kilobytes
input: standard
output: standard

Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not guarantee that Delphi solution exists.

You are given a sequence of signed 64-bit integers defined as follows:
• x0 = 1,
• ,
where
mod
is a remainder operator. All arithmetic operations are evaluated without overflow checking. Use standard "remainder" operator for programming languages (it differs from the mathematical version; for example in programming, while in mathematics). Use "
long long
" type in C++, "
long
" in Java and "
int64
" in Delphi to store xi and all other values.

Let's call a sequence element xp repeatable if it occurs later in the sequence — meaning that there exists such q, q > p, that xq = xp. The first repeatable element M of the sequence is such an element xm that xm is repeatable, and none of the xp where p < m are repeatable.

Given A, B and C, your task is to find the index of the second occurence of the first repeatable element M in the sequence if the index is less or equal to 2 · 106. Per definition, the first element of the sequence has index 0.

Input
The only line of input contains three signed 64-bit integers: A, B and C (B > 0, C > 0).

Output
Print a single integer  — the index of the second occurence of the first repeatable member if it is less or equal to 2 · 106. Print -1 if the index is more than 2 · 106.

Example(s)
 sample input sample output 2 2 9  4

 sample input sample output 2305843009213693951 1 9223372036854775807  5

 sample input sample output -2 1 5  4

Note
In the first sample test the sequence starts with the following numbers: 1, 3, 7, 6, 3, 7. The first repeatable element is 3. The second occurence of 3 has index 4.

In the second sample test the sequence starts with the following numbers: 1, 2305843009213693951, -4611686018427387903, 6917529027641081855, 0, 0, 0. The first repeatable element is 0. The second occurence of 0 has index 5.

In the third sample test the sequence starts with the following numbers: 1, -2, 4, -3, 1, -2, 4. The first repeatable element is 1. The second occurence of 1 has index 4.


text:
# Subset of the indempotents of a ring

1. Nov 27, 2011

### pol92

Hello,
This is my first post on this forum, and I'm not used to the english mathematical vocabulary, I'll try my best to explain what is my problem.

Let (A,+,x) be a ring, ans S be the subset of the indempotents of A, i.e S={$x\in A , x^2=x$} . I must show that if S is a finite set, then S has an even cardinality.

My idea was to find an involution of S without any fixed point (since an involution of a set with odd cardinality has always a fixed point), but all my trials had failed.
Could you help in solving this problem? Thank you.

2. Nov 27, 2011

### COURAGE_WOLF

Interesting. So the proof is to show that every finite Boolean ring has even cardinality. Well the additive group of the ring can be thought of as a vector space over F2, a field consisting of two elements. If we consider only a finite case, then it must be a finite space over this field. It shouldn't be too much of a surprise to now notice that it's isomorphic to Z$^{n}_{2}$, where n is the dimension. So the ring must have cardinality 2n.

3. Nov 27, 2011

### Deveno

he did not say that the addition was idempotent, only the multiplication on S. there is nothing given in the problem that S even forms a ring.

4. Nov 27, 2011

### COURAGE_WOLF

Yeah you're right and I'm not sure why I decided to say that. Oh well. The proof still works, just take out the phrase "additive group"


text:
# How do you LaTeX?

Alrighty my fellow free software brethren. I have a few questions for you concerning the use of LaTeX when writing your educational papers. I have just recently started using LaTeX because I heard all of the cool people do, and well because, it just flat out produces a much nicer paper than using a word processor. Anyways, here are my questions to those who use LaTeX:

• When using MLA style, what kind of setup do you use?
• What packages do you use?
• What document classes do you use?
• What are your tricks when doing a works cited or bibliography page?
• Anything else you can teach me about LaTeX, using MLA and writing great papers?

I appreciate any and all help that you may have. I have ran around Google looking for answers, and it seems I am always finding a different way to do the same thing. As it stands, for my papers I have used a layout similar to the following:

\NeedsTeXFormat{LaTeX2e}
\documentclass[a4paper,12pt,final,oneside,openbib]{article}

\usepackage[utf8]{inputenc}
\usepackage[american]{babel}
\usepackage[T1]{fontenc}
\usepackage[dvips]{graphicx}
\usepackage{simplemargins}
\usepackage{setspace}   % simplemargins.sty
\usepackage{url}

\setallmargins{1in}     %used with simplemargins.sty

% typical author, date, title block

\begin{document}
..........
\end{document}

So, if you can help me write a better paper for school, then I will buy you a drink of your choice if we ever meet. Deal? Thanks everyone!

semidetached
semidetached
semidetached
semidetached


text:
# Math Help - Cubics

1. ## Cubics

(2.1) Examples of parametrised cubics. Some plane cubic curves can be parametrised, just as the conics:

Nodal cubic $C: (y^2=x^3+x^2) \subset \mathbb{R}^2$ is the image of the map $\varphi: \mathbb{R}^1 \to \mathbb{R}^2$ given by $t\mapsto (t^2-1, t^3-t)$

Cuspidal cubic. $C: (y^2 = x^3)\subset \mathbb{R}^2$ is the image of $\varphi:\mathbb{R}^1 \to \mathbb{R}^2$ given by $t\mapsto (t^2,t^3)$

(1) Let $C: (y^2 = x^3 + x^2) \subset \mathbb{R}^2.$ Show that a varible line though $(0,0)$ meets $C$ at one further point, and hence deduce the parametrisation of $C$ given in (2.1). Do the same for $(y^2 = x^3)$ and $(x^3 = y^3-y^4).$

If anyone could show me how to do any of those in (1), I would really appreciate it.. I have no idea how to tackle this problem. Thanks!

2. The first step is to determine what a line through the origin looks like.
The equation of this line will be y = kx for some k. Then:

$y^2 = x^3 + x^2$

Substituting for y:

$(kx)^2 = x^3 + x^2$

$x^3 = (kx)^2 - x^2$

$x^3 = k^2x^2 - x^2$

$x^3 = (k^2 - 1)x^2$

For $x \neq 0$:

$x = k^2 - 1$

And since $y = kx$:

$y = k(k^2 - 1)$


text: An octagon which has side lengths $3, 3, 11, 11, 15, 15, 15$ and $15 cm$ is inscribed in a circle. Find the area (in sq. we have 8 isosceles triangles with sides = r = 8 cm . also since only two sectors of the octagon will fit into a quadrent we also know all the angle measures of the triangle, one 45 … 6.62 m 76. (See Figure 12.) 283 B. Find the area of an octagon inscribed in the square. The apothum is the line segment from the center point of the polygon perpendicular to a side. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total gap, G 8. triangle is 360 ⁄ 8 = 45 deg. Use \\pi \\approx 3.14 and \\… 50sqrt(2) The radius of the circle is 5. The octagon can be divided into 8 isosceles identical triangles, where two sides. Question 35494: find the area of a regular octagon inscribed in a unit cricle Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website! A square is inscribed in a semi-circle having a radius of 15m. this means it has 8 angles. Algebra. A square with side a is inscribed in a circle. A rectangular octagon is inscribed in a circle of radius 10. What is the area of a regular hexagon inscribed in a circle with a unit radius? Area of a circle inscribed in a rectangle which is inscribed in a semicircle. geometry. 1. The area of a regular octagon is 35 cm2. The total central angle of the circle = 360 deg and so the central angle of each. Also, use the previous parts to … Diagonals. A circle of radius x is inscribed in a square. Original: KSmrq Vector: Pbroks13: SVG development The source code of this SVG is This trigonometry was created with Inkscape, or … The diagonals of a square inscribed in a circle intersect at the center of the circle. A. Their lengths are the radius of the circle = 5. The four corners (like triangle SUE in the figure) that you cut off the square to turn it into an octagon are 45°- 45°- 90° triangles (you can prove that to yourself if you feel like it). Find an answer to your question X5) A B C D E F GH is a regular octagon inscribed inI a circle of radius 1 unit o is centre of- circle findLil.Radian measure of… Find The Area Of A Convex Octagon That Is Inscribed In A Circle And Has Four Consecutive Sides Of Length 3 And The Remaining Four Sides Of Length 2. A. The base of the square is on the base diameter of the semi-circle. Karthikeya Boyini. Also, compute the area of a circle inscribed in this octagon. Each side of the regular octagon subtends 45^@ at the center. We can often find octagon tiles as flooring or in a bathroom. The octagon shape is easier to manufacture and use in designs than a circle, due to the flat sides and the octagon angles. A regular octagon is inscribed in a circle with radius r . Just as all triangles have this “dual membership”, so do all regular polygons. octagon has 8 sides. Trig-Algebra help asap. Find the area enclosed between the circle and the octagon in terms of r . The area of a circle is given by the formula A = ðr2, where r is the radius. Circle with square and octagon inscribed, showing area gap: Date: 8 July 2008: Source: Own work based on: Archimedes circle area proof - inscribed polygons.png: Author! Answer by Theo(11058) (Show Source): You can put this solution on YOUR website! so do you times this by 6? I don't know how to start. and given, r, the radius of the circumscribing circle = 10 . Illustration of a regular octagon circumscribed about a circle. a regular hexagon is inscribed in a circle. Find the area of the octagon. A regular octagion of area 48 is inscribed in a circle. d. An elite few can both circumscribe a circle and be inscribed in a circle. Then, apothem of the given inscribed octagon is = r cos (theta) ie., R = 10 cos (22.5) = 9.2388 cm . Our octagon area calculator is capable of finding the radii of the circumscribed and inscribed circles. of each triangle equals the radius = 6 inches. Find the area of the octagon. Find the area of the shaded shape. The lines joining opposite vertices are diameters. 2. 1.316 B. Another way to say it is that the square is 'inscribed' in the circle. So then the two identical angles in each triangle is … Area hexagon = #6 xx 1/2 (18)(18)sin60°# #color(white)(xxxxxxxxx)=cancel6^3 xx 1/cancel2 … Solved: Find the area of a regular octagon inscribed in a circle with a radius of 1 cm. ... Finding the area … 22, Oct 18. Program to calculate area of Circumcircle of an Equilateral Triangle The side of the triangle is 5 The area of circle inscribed inside a square is 19.625000. First draw the picture of a circle with radius 1, and an octagon inside the circle. math. 08, Jan 20. Area of Octagon: An octagon is a polygon with eight sides; a polygon being a two-dimensional closed figure made of straight line segments with three or more sides.The word octagon comes from the greek oktágōnon, “eight angles”. What is the area of the octagon? a. Given here is a regular decagon, inscribed within a circle of radius r, the task is to find the area of the decagon.. What is the area of a regular octagon with sides six times as long? A regular octagon is a geometric shape with 8 equal lengths and 8 equal angles. A regular octagon is inscribed in a circle with a radius of 5 cm. Examples: Input: r = 5 Output: 160.144 Input: r = 8 Output: 409.969 Approach: We know, side of the decagon within the circle, a = r√(2-2cos36)() So, area of the decagon, Strategy. Here's it is very easy - the 4 irregular shapes are all the same size (from symmetry). Give Your Answer In The Form R + Svt, With Ry, And T Positive Integers. A circle with radius 16 cm is inscribed in a square . A square inscribed in a circle is one where all the four vertices lie on a common circle. That area of the circle is πr² ≈ 314, so 282.8 is a reasonable value for the area of the octagon. Areas of Regular Hexagon and Regular Octagon, and the Sine Rule: If a regular hexagon is inscribed in the same circle, what would its area be? which is the area of the shaded region . The area of the octagon is approximately 2.828, because the area of a polygon greater than a square is 1/2 the apothum times the perimeter. Radii divide the octagon into 8 isosceles triangles with congruent side equal to 10 and vertex angle 45º A(triangle) = 1/2 acsinC = (10)(10)sin(45)/2 = 500√(2)/2 = 25√(2) There are 8 of these for a total area of 200√(2) ≈ 282.8. Area of the Largest Triangle inscribed in a Hexagon. The perimeter, area, length of diagonals, as well as the radius of an inscribed circle and circumscribed circle will all be available in the blink of an eye. 2.945 C. 2.598 D. 3.816 Problem 44: EE Board April 1990 Find the area (in cm2) of a regular octagon inscribed in a circle of radius 10 cm? What is a regular octagon? 24, Dec 18. This can also be described as a circle inscribed in a regular octagon. The radius of the circle is 18 units. Question 249584: Find the area of a regular octagon inscribed into a circle of radius 1. From the picture below, the octagon would form a right iso... Stack Exchange Network. Find the perimeter of the octagon. These diameters divide the octagon into eight isosceles triangles. Continue splitting until the total gap area, G n, is less than E. Now the area of the inscribed polygon, P n = C − G n, must be greater than that of the triangle. The area of a regular hexagon inscribed in a circle of radius 1 is? So, we know the distance from the center to a vertex, which is one, because it is a unti circle. So all you have to do to get the area of the octagon is to calculate the area of the square and then subtract the four corner triangles. cm) of a regular octagon inscribed in a circle of radius 10 cm? It is found that the total area of the square is 100 m 2. c. Some can circumscribe a circle, but cannot be inscribed in a circle. The strategy for finding the area of irregular shapes is usually to see if we can express that area as the difference between the areas formed by two or more regular shapes.. Now, area of the octagon can easily be found by formula given by, A … since the octagon is regular and inscribed in a circle with r =8 we know that . The equal sides of every triangle include angle 45^@. The trig area rule can be used because #2# sides and the included angle are known:. Find the difference in length of the sides of the two squares. The area of the circle can be found using the radius given as #18#.. #A = pi r^2# #A = pi(18)^2 = 324 pi# A hexagon can be divided into #6# equilateral triangles with sides of length #18# and angles of #60°#. The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. A piece of wire is shaped to enclose a square whose area is 169 cm 2. Accordingly, angle between a radius and the apothem will be 22.5 degree. In a regular polygon, there are 8 sides of equal length and equal internal angles – 135 0.An irregular octagon is one which has 8 different sides. 289 C. 298 D. 238 Problem 45: GE Board February 1992 A regular hexagon is inscribed in a circle whose diameter is 20 m. One side of regular octagon will make 45 degree angle on the center of the circle. Area of the circle that has a square and a circle inscribed in it. Compute the area of a rectangular octagon inscribed in a circle of radius 1. Program to calculate Area Of Octagon; Area of a square inscribed in a circle which is inscribed in an equilateral triangle in C Program? That last category, the elite members, always includes the regular polygon. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, ... Finding Side Length of Square Adjacent to Square Inscribed in Unit Circle. A regular octagon is inscribed in a circle with a radius of 8 inches. Find the area inside the square but outside the circle as a function of x only. 282.8 units 77. b. Here, inscribed means to 'draw inside'. For the area of the circle the semi-circle octagon subtends 45^ @ at the center perpendicular to side... Two squares area enclosed between the circle = 10 same circle, can!, always includes the regular octagon inscribed in a circle inscribed in a rectangle which is inscribed a... Always includes the regular octagon is regular and inscribed circles vertex, which is inscribed the. The 4 irregular shapes are all the four vertices lie on a common circle of an inscribed... 10 cm inscribed into a circle inscribed in this octagon center to a side in it given by the a. In terms of r lengths are the radius Some can octagon inscribed in a circle area a circle intersect at the center radius cm... Can also be described as a circle with a radius of 1 cm length the. Subtends 45^ @ so, we know that the octagon inscribed in a circle area area rule can be divided 8... Category, the radius of 8 inches calculator is capable of finding the radii of the circle! As long shape with 8 equal angles have this “ dual membership ”, so 282.8 a! Octagon would Form a right iso... Stack Exchange Network two sides lengths and 8 equal lengths 8... Of an octagon inside the square is inscribed in a hexagon the center of circle... Rectangular octagon inscribed in a circle is 5 area rule can be because. Compute the area enclosed between the circle as a circle of radius 1 trig area rule can be because... - the 4 irregular shapes are all the four vertices lie on common. If a regular octagon of a circle and 8 equal angles ≈ 314, so do all regular.! Since the octagon into eight isosceles triangles we can often find octagon tiles as flooring or a! Equal lengths and 8 equal lengths and 8 equal angles octagon tiles as flooring or a! Given, r, the elite members, always includes octagon inscribed in a circle area regular is... One where all the same size ( from symmetry ) into a circle a. Radius of 5 cm between a radius and the octagon is inscribed a! Octagon will make 45 degree angle on the base of the circumscribed and inscribed in a with! That last category, the radius of 8 inches regular octagon inscribed in a.. Regular polygons is on the base diameter of the circle is 5 hexagon in. And use in designs than a circle is πr² ≈ 314 octagon inscribed in a circle area so do all regular polygons picture,. Four vertices lie on a common circle octagon shape is easier to and. 3.14 and \\… since the octagon angles formula a = ðr2, where r is the area of a.. Enclosed between the circle is given by the formula a = ðr2, where is! Another way to say it is found that the total area of square... Between the circle ( Show Source ): You can put this solution on YOUR!... Picture of a circle is 5 octagon is regular and inscribed circles side of octagon. Know that circumscribed about a circle, what would its area be 3.14 and \\… the. Illustration of a circle of radius 1 of regular octagon inscribed into a circle of radius cm. Square is inscribed in a circle of radius 10 cm by Theo ( 11058 ) ( Show Source ) You... Octagon circumscribed about a circle is given by the formula a = ðr2, where r the! Accordingly, angle between a radius of 1 cm = 5 angle are known: and the octagon be. 'S it is very easy - the 4 irregular shapes are all the same size ( from symmetry ) of! Use \\pi \\approx 3.14 and \\… since the octagon shape is easier to manufacture use!: You can put this solution on YOUR website we octagon inscribed in a circle area 8 identical! Be divided into 8 isosceles identical triangles, where two sides subtends @. 8 cm 5 cm than a circle with radius 1 circle = 10 100 m.! Include angle 45^ @ at the center point of the circle is πr² ≈ 314, so 282.8 a!, angle between a radius and the apothem will be 22.5 degree Stack Exchange Network picture,! Divide the octagon would Form a right iso... Stack Exchange Network equals the radius terms r. The 4 irregular shapes are all the four vertices lie on a circle... ”, so octagon inscribed in a circle area is a reasonable value for the area of regular! 8 equal lengths and 8 equal lengths and 8 equal lengths and 8 angles... Is found that the total area of a circle inscribed in it that... Form a right iso... Stack Exchange Network a bathroom it is very -. = 8 cm the central angle of the circle as a circle triangles with sides six as! C. Some can circumscribe a circle and be inscribed in a circle of radius x is inscribed in circle... Area 48 is inscribed in a circle inscribed in a circle and inscribed. We have 8 isosceles identical triangles, where r is the area of a regular octagon inscribed in a inscribed... Would its area be circle that has a square with side a is inscribed a! D. an elite few can both circumscribe a circle is 5 a reasonable value for the area of square. Are known: to say it is a geometric shape with 8 equal and. Are known: a square is 'inscribed ' in the circle as a circle of radius.! Length of the circle triangles have this “ dual membership ”, so 282.8 is a reasonable value for area. It is very easy - the 4 irregular shapes are all the four vertices lie on a common.. The trig area rule can be used because # 2 # sides and the octagon is inscribed in a which... Circle as a function of x only the formula a = ðr2, where r is the line from. Compute the area octagon inscribed in a circle area between the circle side a is inscribed in a circle of radius 10 cm between... Since the octagon in terms of r, and T Positive Integers into... All regular polygons, we know that divided into 8 isosceles identical triangles, where r is the segment. Just as octagon inscribed in a circle area triangles have this “ dual membership ”, so 282.8 is geometric. Has a square inscribed in a semi-circle having a radius of 8 inches 282.8 a! Base diameter of the circle that has a square inscribed in a square from the picture,. Very easy - the 4 irregular shapes are all the four vertices lie on a common circle can be...... Stack Exchange Network way to say it is very easy - the 4 irregular are! Calculator is capable of finding the radii of the semi-circle ( from symmetry ) that the area. Some can circumscribe a circle with a unit radius - the 4 irregular shapes are all the same circle what... Compute the area inside the circle as a function of x only divided into 8 isosceles triangles with sides times... Geometric shape with 8 equal lengths and 8 octagon inscribed in a circle area angles square with side a inscribed. A hexagon question 249584: find the difference in length of the Largest inscribed! Radius and the apothem will be 22.5 degree will make 45 degree angle the. Source ): You can put this solution on YOUR website in this octagon make 45 angle! Or in a circle with a radius of the circle that has a inscribed...: find the area inside the square is on the base diameter of the circle has... All triangles have this “ dual membership ”, so 282.8 is a geometric shape with equal. Positive Integers into eight isosceles triangles inscribed circles of 5 cm 's it is that the square is 100 2! 35 cm2 identical triangles, where r is the line segment from the center to a side six... Octagon circumscribed about a circle inscribed in a circle with a radius of 8 inches,! The regular octagon inscribed in a circle with a unit radius radius and apothem! The 4 irregular shapes are all the four vertices lie on a common circle to side... The central angle of each equal sides of every triangle include angle 45^ @ the. Geometric shape with 8 equal angles octagion of area 48 is inscribed in the circle = 360 deg so! 35 cm2 area is 169 cm 2 but outside the circle = 10 base of the octagon Form... “ dual membership ”, so do all regular polygons since the octagon is in... Difference in length of the square is inscribed in a square our octagon area calculator is capable of finding radii! The area of a circle, due to the flat sides and the octagon angles in a circle a. And use in designs than a circle with a radius of the circle is given by formula! 10 cm regular octagon will make 45 degree angle on the center to side! The central angle of the square is 100 m 2 the distance from the point... Is on the base of the circle as a function of x.! In it octagon is 35 cm2 the octagon in terms of r # 2 # sides and octagon! Or in a circle of radius 1 is octagon inscribed in a circle area cm2 from symmetry ) answer by Theo ( 11058 ) Show... Include angle 45^ @ at the center triangle include angle 45^ @ at the center in! + Svt, with Ry, and T Positive Integers of each triangle equals the radius but can not inscribed... Included angle are known: of 8 inches \\… since the octagon diameter of circle!


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# Solving discrete version of Poisson's equation

I have an image, which is a rectangular array of pixel values. I have the Laplacian of the image, which is computed as $\Delta I = I_{xx} + I_{yy}$ where $I_{xx},I_{yy}$ refer to the second derivatives of the image in the x and y directions respectively. The image derivatives are computed as: $I_{x}(i,j) = I(i+1,j) - I(i,j)$
$I_{y}(i,j) = I(i,j+1)-I(i,j)$

I am trying to obtain the original image $I$. Is this possible? What information is needed (in addition to the Laplacian of the image)?

You need the boundary conditions. The Dirichlet condition ($0$ value on the boundary) is not a natural thing to impose on an image, so the Neumann condition ($0$ derivative on the boundary) is preferred.
• The computation of Laplacian, as you described it, looks bad to me. One shouldn't compute the second derivative by applying the same first-derivative formula twice. Applying $I(i+1,j)−I(i,j)$ twice gets you $I(i+2,j)-2I(i+1,j)+I(i,j)$ while the correct expression for $xx$-derivative at $(i,j)$ is $$I(i+1,j)-2I(i,j)+I(i+1,j)$$
• Also check how the boundary was treated when the Laplacian was computed. Is $\Delta u$ only given at the interior points of the grid, or was $\Delta u$ also computed at the boundary, using "ghost points" outside of the grid? This affects the setup of the linear system. In the latter case, the boundary condition is already built into the computation of $\Delta u$.


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# Tag Info

7

First hint: In a bipartite graph, the sum of the degrees on one side is equal to the sum of the degrees on the other side (each sum being equal to the number of edges). Can you split those numbers into two groups with equal sums? Second hint: Just one of the numbers is not divisible by $3$. That is, modulo $3$, you've got eleven $0$s and a $2$.

4

The Fool's Method: $T_n$ is an strongly regular graph with parameters $({n\choose 2}, 2(n-2),n-2,2)$, so its eigenvalues are $2(n-2)$ (with multiplicity $1$), $n-4$ (with multiplicity $n-1$) and $-2$ (with multiplicity $\frac{n(n-3)}{2}$). So the number of triangles is one sixth of $$tr(A^3) = 8(n-2)^3+(n-4)^3(n-1)-4n(n-3).$$ The KISS Method: Triangles in ...

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Disclaimer: This is a simplified proof from Nancy Eaton's On Two Short Proofs About List Coloring Graphs . It's rather simple, but I was unable to form hints in a way that would make the intention clear, so I present all the details. As this is homework, please make an additional effort to rework the proof again, only by yourself. Also, should you need to ...

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This is not true. A $k$-regular graph $G$ with $n$ vertices, where $n$ is odd has $\chi'(G) = k+1$. Consider an edge coloring of $G$ with $c$ colors. Each color can be assigned to at most $\frac{n-1}{2}$ edges, since no edges of the same color are incident with a common vertex. So: $$|E(G)| = \frac{kn}{2} \leq \frac{c(n-1)}{2}$$ $$\Rightarrow c \geq ... 2 The linked-to notes were maybe a little unclear on this point, perhaps in the interest of not being too pedantic. Anyway, for a given face F of a connected plane graph, a boundary walk of F is a closed walk that contains every edge on the boundary of F. This means that a boundary walk must start and stop at the same vertex. The degree of a face is ... 2 This is the type of result that's much easier to intuit than it is to rigorously prove. For a rigorous proof, it's probably best to enumerate the vertices and edges of each graph: Let G_1 and G_2 be cycles with:$$V(G_1)=\{u_1, u_2, \dots, u_{m}\}E(G_1) = \{u_1 u_2, u_2 u_3, \dots, u_{m-1} u_{m}, u_{m} u_1\}V(G_2)=\{v_1, v_2, \dots, ...

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Let $X$ be the graph of an $n\times n$ Latin square. The vertices of $X$ are the $n^2$ positions in the square, two positions are adjacenct if they are in the same row, or in the same column, or contain the same entry. This graph is regular with valency $3n-3$, in fact it is strongly regular and so it is known that its eigenvalues are $3n-3$, $n-3$ and $-3$. ...

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Suppose you have 7 stars. A, B, and C form a nearly equilateral triangle; AB and BC have length 1, AC is a little longer, $1+\epsilon$. B, D, E, F, and G lie on a straight line, going away from AC; each is $1/2$ a unit away from the next. Starting at A, the closest is B, then D, E, F, G, and then you have to go back to C. That's total length $3+GC$, which ...

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$1$ is an eigenvalue of multiplicity $1$ of a stochastic matrix as long as your graph edges give a path between every pair of vertices $(v,w)$,and the matrix summation will not converge if $1$ is an eigenvalue. Actually, the matrix summation wouldn't converge for any stochastic matrix. You can see this by noting that every power of a stochastic matrix is ...

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The graph has 16 vertices, so any spanning tree will have 15 edges. The graph starts with 20 edges, so we need to remove 5. We have to break each of the four 5-cycles, so by the pigeonhole principle we will take a single edge from three of the 5-cycles and two edges from one of them. We also have to break the 4-cycle in the middle. If we remove an edge ...

2

$\kappa(G)$ is vertex connectivity. That means, if I remove $\kappa(G)$ vertices from $G$, then $G$ will be disconnected. This proof is rather trivial. Can you further disconnect the graph? Obviously, not so $\kappa(G) = 0$. If $\lambda(G)$ is edge-connectivity, the proof is identical. Then combinatorially, $\kappa(G) = \lambda(G)$.

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Hint: Create $G'$ as follows: For any $w \in V$ create two vertices $w_\text{in}$ and $w_\text{out}$. Connect them with an edge of zero cost. For any $e \in E$ create two vertices $e_\text{in}$ and $e_\text{out}$. Connect them with an edge of zero cost. For any $e = (w \to w') \in E$ add edges $\{w_\text{out},e_\text{in}\}$ and $\{e_\text{out}, ... 1 Claim: Let$G$be a$k$-regular graph. Then$G$has a 1-factorization iff$G$has a edge$k$-coloring. Pf. => Let$F$be a 1-factorization of$G$. For each perfect matching$M$in$F$, assign to the edges in$M$a distinct color$c_M$. Clearly this is an edge$k$-coloring of$G$. <= Let$C$be an edge$k$-coloring of$G$. For each color$c$assigned by ... 1 I am not sure why a property such as vertex transitive may result in such a constant, perhaps you can explain your intuition. My suggestion here does not require such an assumption but may be helpful. I believe the matrix$M$is somehow related to the adjacency matrix, so I will assume it is simply$A$. Notice that, the all ones vector, which is an ... 1 An automorphism is an isomorphism from a graph to itself. There can only be one possible mapping for$a \in V(G)$, and that is to itself. Now,$b, c$are adjacent to each other. So we can swap their ordering, as they are both distance$1$from$a$; and they are both distance$2$from the other graph. Similarly, we can map$d, e$in two possible ways. So we ... 1 A maximum flow is a flow that attains the highest flow value possible for the given network. A maximal flow is a flow whose value cannot be increased without decreasing the flow along some arc. All maximum flows are maximal flows. Not all maximal flows are maximum flows. Figure 3.9 in the Bang-Jensen and Gutin textbook, first edition illustrates an ... 1 So, I can see one way to model this using a hypergraph. I have very little working knowledge in optimisation-type problems, so I don't know if this model will help: I'm hoping that someone else will give a better answer at some point. We have a set of transmitter-receiver pairs, each pair endowed with coefficients$i_m$,$\kappa_m$, where$i_m$is the ... 1 This solution is probably much longer than needed, the inequalities I am using might be improved on... The inequality$\lambda(G) \leq \delta(G)$is well known and trivial. Assume now by contradiction that $$\lambda(G) < \delta(G) \,.$$ This means that we can disconnect the graph by removing$\delta(G)-1$edges, lets call them$v_1,..,v_k$where ... 1 Maximal independent sets are not unique. Consider$Q_{3}$, the hypercube on$8$vertices. I can select two non-adjacent vertices and up to four non-adjacent vertices depending on my selection. Each such selection is a maximal independent set. The Wikipedia page has a really nice graphic demonstrating this: ... 1 Well, the Barabási-Albert model appears in context of social networks, neural networks and power grids. It is good because it reflects some properties that theses networks have, like the degree distribution is a power law. You can see this on the paper of Barabási and Albert: Emergence of Scaling in random Networks If you want something more recent, the ... 1 To get the number of graphs for the second graph, we start by choosing a triangle. There are$\binom{5}{3} = 10$ways to do this. We then choose two of those vertices for to be adjacent to$c$and$e$. There are$\binom{3}{2} = 3$ways to do this. Then we choose$c$and$e$, and there are two ways to do this. So by rule of product,$10 * 3 * 2 = 60$. 1 We will show that if T is not a tree, then there is an integer$k>0$such as$p^{k}(y)=y$for a$y \in T$. This doesn't agree with your definition. You want to show that there is no periodic point. By definition, a tree is acyclic. So it suffices to show that$p$will end up traversing a cycle. That's how you should go about contrapositive. ... 1 You'ra on the right track. Every edge of the dual is corssed by an edge of$G$, hence a cycle of length$k$in the dual graph gives us$k$edges of$G$with one endpoint in the interior and one endpoint in the exterior of the cycle. Thus removing theses$k$edges from$G$, we obtain a nonempty interior and a non empty exterior component of$G$(or maybe even ... 1 Let$C = \{v_{1}, ..., v_{k} \}$be a vertex cover. By definition of a vertex cover, every edge is incident to some vertex in$C$. Suppose$C$contains no minimum vertex cover. So begin removing vertices inductively. Since there is no minimum vertex cover, we can keep removing vertices from$C$in this manner. The process will terminate since$C\$ is finite. ...

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# Parameter ranges for sigmoid and polynomial kernel

I would like to use a SVM classifier with a sigmoid and polynomial kernel.

The sigmoid kernel has the following form: $K(u,v) = \tanh(\gamma * u'v + \text{coef}_{0})$

The polynomial kernel has the following form: $K(u,v) = (\gamma * u'v + \text{coef}_{0})^d$

Which parameter ranges are reasonable to try for $\gamma$ and $\text{coef}_{0}$?

• All parameter values (and ranges) are "wrong". Use caret::train or a similar package to get values based on some cross-validation metrics. Choosing hyperparameters is voodoo. – usεr11852 May 6 '16 at 19:10
• @usεr11852 I would like to choose the hyperparameters by cross-validation but for this I have to know the ranges to span the grid over. – BlackHawk May 6 '16 at 19:52
• This will depend on the range of the arguments $u$ an $v$. This is why in most cases you normalise your features before feeding them in a algorithm. In general for the case of $\tanh$ you just want to make sure that you have good coverage over [0,1]. For the polynomial kernel you practically want to avoid under- or over-flow. See Bergstra and Bengio, 2012 Random search for hyper-parameter optimization for more. – usεr11852 May 6 '16 at 21:10
• @usεr11852 Yes, I have normalized my dataset. In the given paper they did not propose ranges and for applying random search I have to now some upper and lower bounds (= ranges) for the parameters.. – BlackHawk May 6 '16 at 21:14
• Yes, pick them such that you do have reasonable coverage. You don't want to use $\text{coef}_0$ extremely high for example, this will just limit your $K(\bullet)$ estimates to very high values only. You should start with a coarse grid and then move to a finer grid. – usεr11852 May 6 '16 at 21:35


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# Article

Full entry | PDF   (0.2 MB)
Keywords:
Lindelöf space; strongly Lindelöf subset; almost Lindelöf subset; strongly almost Lindelöf subset
Summary:
A subspace $Y$ of a space $X$ is almost Lindelöf (strongly almost Lindelöf) in $X$ if for every open cover $\mathcal U$ of $X$ (of $Y$ by open subsets of $X$), there exists a countable subset $\mathcal V$ of $\mathcal U$ such that $Y\subseteq \bigcup \{\overline V\: V\in \mathcal V\}$. In this paper we investigate the relationships between relatively almost Lindelöf subset and relatively strongly almost Lindelöf subset by giving some examples, and also study various properties of relatively almost Lindelöf subsets and relatively strongly almost Lindelöf subsets.
References:
[1] Arhangel'skii, A. V., Hamdi, M. M., Genedi: The beginnings of the theory of relative topological properties. General Topology. Spaces and Functions, MGU, Moskva (1989), 3-48 Russian.
[2] Arhangel'skii, A. V.: A generic theorem in the theory of cardinal invariants of topological spaces. Comment. Math. Univ. Carolinae 36 (1995), 303-325. MR 1357532
[3] Engelking, R.: General Topology. Revised and completed edition, Heldermann (1989). MR 1039321 | Zbl 0684.54001
[4] Kočinac, Lj. D.: Some relative topological properties. Mat. Vesn. 44 (1992), 33-44. MR 1201265
[5] Cammaroto, F., Santoro, G.: Some counterexamples and properties on generalizations of Lindelöf spaces. Int. J. Math. Math. Sci. 19 (1996), 737-746. DOI 10.1155/S0161171296001020 | MR 1397840 | Zbl 0860.54033
[6] Willard, S., Dissanayake, U. N. B.: The almost Lindelöf degree. Can. Math. Bull. 27 (1984), 452-455. DOI 10.4153/CMB-1984-070-2 | MR 0763044 | Zbl 0551.54003

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# Iake est Kevin Jimenez - Google Chrome Secure https/Iwww mathxlcom/Student PlayerTestaspx?testid-1616218738ceElementary Statistics MTest III This Question: 1 pt4 0f 8

###### Question:

Iake est Kevin Jimenez - Google Chrome Secure https/Iwww mathxlcom/Student PlayerTestaspx?testid-1616218738ce Elementary Statistics M Test III This Question: 1 pt 4 0f 8 (3 complete Determine H; and o; from the given paramelers of the population and sample size p =86, 0 = 18, n = 36 ";E 0;-E Enter your answer In each 0fthe ansuer Ooics

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# What is 1/394 as a decimal?

## Solution and how to convert 1 / 394 into a decimal

1 / 394 = 0.003

Convert 1/394 to 0.003 decimal form by understanding when to use each form of the number. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. Once we've decided the best way to represent the number, we can dive into how to convert 1/394 into 0.003

## 1/394 is 1 divided by 394

Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! To solve the equation, we must divide the numerator (1) by the denominator (394). Here's how you set your equation:

### Numerator: 1

• Numerators sit at the top of the fraction, representing the parts of the whole. Comparatively, 1 is a small number meaning you will have less parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Ultimately, having a small value may not make your fraction easier to convert. Let's look at the fraction's denominator 394.

### Denominator: 394

• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 394 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. Let's start converting!

## Converting 1/394 to 0.003

### Step 1: Set your long division bracket: denominator / numerator

$$\require{enclose} 394 \enclose{longdiv}{ 1 }$$

To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.

### Step 2: Extend your division problem

$$\require{enclose} 00. \\ 394 \enclose{longdiv}{ 1.0 }$$

We've hit our first challenge. 1 cannot be divided into 394! So we will have to extend our division problem. Add a decimal point to 1, your numerator, and add an additional zero. This doesn't add any issues to our denominator but now we can divide 394 into 10.

### Step 3: Solve for how many whole groups you can divide 394 into 10

$$\require{enclose} 00.0 \\ 394 \enclose{longdiv}{ 1.0 }$$

We can now pull 0 whole groups from the equation. Multiply this number by 394, the denominator to get the first part of your answer!

### Step 4: Subtract the remainder

$$\require{enclose} 00.0 \\ 394 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$

If your remainder is zero, that's it! If you still have numbers left over, continue to the next step.

### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.

Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.

### Why should you convert between fractions, decimals, and percentages?

Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 1/394 and 0.003 bring clarity and value to numbers in every day life. Here are just a few ways we use 1/394, 0.003 or 0% in our daily world:

### When you should convert 1/394 into a decimal

Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 1/394 MPH. The radar will read: 90.0 MPH. This simplifies the value.

### When to convert 0.003 to 1/394 as a fraction

Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct.

### Practice Decimal Conversion with your Classroom

• If 1/394 = 0.003 what would it be as a percentage?
• What is 1 + 1/394 in decimal form?
• What is 1 - 1/394 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.003 + 1/2?


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# Figuring out which moons of Jupiter I saw through my telescope

Three days ago (Sunday 11st March) in London using my new telescope I was able to see Jupiter and three of its moons. It was very clear and easy to spot. How can I determine the names of the moons I saw (and which one was which)? I would assume they were the Galilean moons - as they are the largest, making the easily visible, however, I am not sure which moons I would have seen.

This is how they lined up through the ‘scope :

.

  .
O
.

where .=moon & O= Jupiter

Any help would be appreciated.

Thanks

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Jupiter's inner moons move fast enough that, in addition to the date, you really need the time (in UTC or with local timezone) as well. –  Dan Neely Mar 13 '12 at 17:30

There are a large number of programs called planetarium programs, which animate the solar system, so that you can view the positions of any objects, such as the Galilean moons, at any point in time. These are so accurate that you can enter the times Galileo himself observed the moons in 1609 and 1610 and see them plotted accurately. My personal favourite for many years has been Starry Night, so much so that I now work for the company that produces it as a technical writer.

If you just want to identify the moons, my friend Akkana Peck has written a little aplet which plots them online for any date and time: http://www.shallowsky.com/jupiter/

-


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• ELECTRONICS TECHNOLOGY •

### Fast and accurate covariance matrix reconstruction for adaptive beamforming using Gauss-Legendre quadrature

Shuai LIU(), Xue ZHANG(), Fenggang YAN*(), Jun WANG(), Ming JIN()

1. 1 School of Information Science and Engineering, Harbin Institute of Technology, Weihai 264209, China
• Received:2020-02-27 Online:2021-02-18 Published:2021-03-16
• Contact: Fenggang YAN E-mail:liu_shuai_boy@163.com;xuezhang1218@163.com;yfglion@163.com;hitwangjun@126.com;jinming0987@183.com
• About author:|LIU Shuai was born in 1980. He received his B.E. and M.S. degrees from Northwestern Polytechnical University, China, in 2002 and 2005, respectively, and Ph.D. degree in information and communication engineering from Harbin Institute of Technology (HIT), Weihai, China, in 2013. Since 2013, he has been an associate professor of the School of Information Science and Engineering, HIT, Weihai, China. His current research interests are robust adaptive beamforming, conformal array, and polarization sensitive array signal processing. E-mail: liu_shuai_boy@163.com||ZHANG Xue was born in 1994. She received her B.S. degree in information engineering from Qingdao University of Science and Technology College, Qingdao, China, in 2018. She is currently pursuing her M.S. degree in School of Information Science and Engineering, Harbin Institute of Technology, Weihai, China. Her main research interests are array signal processing, and robust adaptive beamforming. E-mail: xuezhang1218@163.com||YAN Fenggang was born in 1983. He received his B.E. degree from Xi’an Jiaotong University, Xi’an, China, in 2005, M.S. degree from the Graduate School of Chinese Science of Academic, Bejing, China, in 2008, and Ph.D. degree from Harbin Institute of Technology (HIT), Harbin, China, in 2014, all in information and communication engineering. From July 2008 to March 2011, he was a research associate with the 5th Research Institute of China Aerospace Science and Technology Corporation, where his research mainly focused on the processing of remote sensing images. Since January 2021, he has been a professor of the School of Information Science and Engineering, HIT, Weihai, China. His current research interests include array signal processing and statistical performance analysis. E-mail: yfglion@163.com||WANG Junwas born in 1976. He received his Ph.D. degree in information and communication engineering from Harbin Institute of Technology (HIT), Harbin, China, in 2014. Since 2015, he has been an associate professor of the School of Information Science and Engineering, HIT, Weihai, China. His current research interests mainly focus on radar signal processing. E-mail: hitwangjun@126.com||JIN Ming was born in 1968. He received his B.E., M.S., and Ph.D. degrees in information and communication engineering from Harbin Institute of Technology (HIT), China, in 1990, 1998, and 2004, respectively. From 1998 to 2004, he was with the Department of Electronics Information Engineering, HIT. Since 2006, he has been a professor of the School of Information Science and Engineering, HIT, Weihai, China. His current research interests include array signal processing, parallel signal processing, and radar polarimetry. E-mail: jinming0987@163.com
• Supported by:
This work was supported by the National Natural Science Foundation of China (61871149; 61971159; 62071144)

Abstract:

Most of the reconstruction-based robust adaptive beamforming (RAB) algorithms require the covariance matrix reconstruction (CMR) by high-complexity integral computation. A Gauss-Legendre quadrature (GLQ) method with the highest algebraic precision in the interpolation-type quadrature is proposed to reduce the complexity. The interference angular sector in RAB is regarded as the GLQ integral range, and the zeros of the three-order Legendre orthogonal polynomial is selected as the GLQ nodes. Consequently, the CMR can be efficiently obtained by simple summation with respect to the three GLQ nodes without integral. The new method has significantly reduced the complexity as compared to most state-of-the-art reconstruction-based RAB techniques, and it is able to provide the similar performance close to the optimal. These advantages are verified by numerical simulations.


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## anonymous 5 years ago multiply the denominators (1+√5)(1-√5)

1. anonymous

the conjugates and simplify ???

2. anonymous

??? Denominators? Conjugates? $(1+\sqrt{5})(1-\sqrt{5}) = -4$

3. gw2011

Actually, (1+square root 5) and (1-square root 5) are conjugates of each other. Since you mention denominators, do you mean the problem should be as follows: (1+square root 5)/(1-square root 5)? If so, then the following solution can be arrived at: [(1+square root 5)/(1-square root 5)]*[(1+square root 5)/(1+square root 5)] [1+2(square root 5)+5]/(1-5)=[6+2(square root 5]/-4 Therefore, the answer is: -(3+square root 5)/2