text:
Frankenstein

Remember me, remember me, but ah! forget my fate (Dido’s Lament, Henry Purcell)

A Voronoi diagram divides a plane based on a set of original points. Each polygon, or Voronoi cell, contains an original point and all that are closer to that point than any other.

This is a nice example of a Voronoi tesselation. You can find good explanations of Voronoi diagrams and Delaunay triangulations here (in English) or here (in Spanish).

A grayscale image is simply a matrix where darkness of pixel located in coordinates (i, j) is represented by the value of its corresponding element of the matrix: a grayscale image is a dataset. This is a Voronoi diagraman of Frankenstein:

To do it I followed the next steps:

2. Convert it to gray scale
3. Turn it into a pure black and white image
4. Obtain a random sample of black pixels (previous image corresponds to a sample of 6.000 points)
5. Computes the Voronoi tesselation

Steps 1 to 3 were done with imager, a very appealing package to proccess and analice images. Step 5 was done with deldir, also a convenient package which computes Delaunay triangulation and the Dirichlet or Voronoi tessellations.

The next grid shows tesselations for sample size from 500 to 12.000 points and step equal to 500:

I gathered all previous images in this gif created with magick, another amazing package of R I discovered recently:

This is the code:

library(imager)
library(dplyr)
library(deldir)
library(ggplot2)
library(scales)

# Read and convert to grayscale

# This is just to define frame limits
x %>%
as.data.frame() %>%
group_by() %>%
summarize(xmin=min(x), xmax=max(x), ymin=min(y), ymax=max(y)) %>%
as.vector()->rw

# Filter image to convert it to bw
x %>%
threshold("45%") %>%
as.cimg() %>%
as.data.frame() -> df

# Function to compute and plot Voronoi tesselation depending on sample size
doPlot = function(n)
{
#Voronoi tesselation
df %>%
sample_n(n, weight=(1-value)) %>%
select(x,y) %>%
deldir(rw=rw, sort=TRUE) %>%
.$dirsgs -> data # This is just to add some alpha to lines depending on its longitude data %>% mutate(long=sqrt((x1-x2)^2+(y1-y2)^2), alpha=findInterval(long, quantile(long, probs = seq(0, 1, length.out = 20)))/21)-> data # A little bit of ggplot to plot results data %>% ggplot(aes(alpha=(1-alpha))) + geom_segment(aes(x = x1, y = y1, xend = x2, yend = y2), color="black", lwd=1) + scale_x_continuous(expand=c(0,0))+ scale_y_continuous(expand=c(0,0), trans=reverse_trans())+ theme(legend.position = "none", panel.background = element_rect(fill="white"), axis.ticks = element_blank(), panel.grid = element_blank(), axis.title = element_blank(), axis.text = element_blank())->plot return(plot) } # I call the previous function and store resulting plot in jpeg format i=5000 name=paste0("frankie",i,".jpeg") jpeg(name, width = 600, height = 800, units = "px", quality = 100) doPlot(i) dev.off() # Once all images are stored I can create gif library(magick) frames=c() images=list.files(pattern="jpeg") for (i in length(images):1) { x=image_read(images[i]) x=image_scale(x, "300") c(x, frames) -> frames } animation=image_animate(frames, fps = 2) image_write(animation, "Frankenstein.gif")  The Ex Libris Generator Go ahead stomp your feet on the floorboards Clap your hands if that’s really what you came here for (Heaven, The Milk Carton Kids) Inspired by curves created by the harmonograph, I have done a Shiny app to generate random images that you can personalize and use as an Exlibris. You can try the App here. For me, an exlibris (also known as bookplates) can be a nice, original and useful present for book-lovers. This is an example: More examples: I always put the code at the end of my posts. Since I always have doubts about how many people are interested in what I do, today will be different. I will share the code with those who ask it to me in any of the following ways: • Sending me a direct message on Twitter • Droping me an email Cheers! The Somnambulist and Pi How wary we are of something warm and unborn. Something calmly by zero will divide (Unbegotten, The Somnambulist) Some time ago, I assumed the mission to draw a plot for the cover of the new album of The Somnambulist, a music band from Berlin. They wanted a circlization of Pi, which is a graphic where numbers are represented in a circular layout. The idea is connecting each digit of Pi to its successive digit with links to the position of the numerically corresponding external sectors. I used a color palette composed by 10 nuances of the visible spectrum as a tribute for Planck, as Marco (the vocalist) requested me. After a number of attempts: The album is named Unbegotten, a german word which means archaic. As Marco told me, in theology it also means kind of eternal because of being never born and so never dying. I like how π is integrated into the title to substitute the string “tt” in the middle. Pi is also eternal so the association is genuine. The music of The Somnambulist is intense, dark and powerful and is waiting for you here to listen it. My favorite song is the one that gives name to the album. If you want to know more about circlizong numbers, you can visit this post, where you also can see the code I used as starting point to do this plot. Chaotic Galaxies Tell me, which side of the earth does this nose come from? Ha! (ALF) Reading about strange attractors I came across with this book, where I discovered a way to generate two dimensional chaotic maps. The generic equation is pretty simple: $x_{n+1}= a_{1}+a_{2}x_{n}+a_{3}x_{n}^{2}+a_{4}x_{n}y_{n}+a_{5}y_{n}+a_{6}y_{n}^{2}$ $y_{n+1}= a_{7}+a_{8}x_{n}+a_{9}x_{n}^{2}+a_{10}x_{n}y_{n}+a_{11}y_{n}+a_{12}y_{n}^{2}$ I used it to generate these chaotic galaxies: Changing the vector of parameters you can obtain other galaxies. Do you want to try? library(ggplot2) library(dplyr) #Generic function attractor = function(x, y, z) { c(z[1]+z[2]*x+z[3]*x^2+ z[4]*x*y+ z[5]*y+ z[6]*y^2, z[7]+z[8]*x+z[9]*x^2+z[10]*x*y+z[11]*y+z[12]*y^2) } #Function to iterate the generic function over the initial point c(0,0) galaxy= function(iter, z) { df=data.frame(x=0,y=0) for (i in 2:iter) df[i,]=attractor(df[i-1, 1], df[i-1, 2], z) df %>% rbind(data.frame(x=runif(iter/10, min(df$x), max(df$x)), y=runif(iter/10, min(df$y), max(df$y))))-> df return(df) } opt=theme(legend.position="none", panel.background = element_rect(fill="#00000c"), plot.background = element_rect(fill="#00000c"), panel.grid=element_blank(), axis.ticks=element_blank(), axis.title=element_blank(), axis.text=element_blank(), plot.margin=unit(c(-0.1,-0.1,-0.1,-0.1), "cm")) #First galaxy z1=c(1.0, -0.1, -0.2, 1.0, 0.3, 0.6, 0.0, 0.2, -0.6, -0.4, -0.6, 0.6) galaxy1=galaxy(iter=2400, z=z1) %>% ggplot(aes(x,y))+ geom_point(shape= 8, size=jitter(12, factor=4), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=16, size= jitter(4, factor=2), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=46, size= 0, color="#ffff00")+opt #Second galaxy z2=c(-1.1, -1.0, 0.4, -1.2, -0.7, 0.0, -0.7, 0.9, 0.3, 1.1, -0.2, 0.4) galaxy2=galaxy(iter=2400, z=z2) %>% ggplot(aes(x,y))+ geom_point(shape= 8, size=jitter(12, factor=4), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=16, size= jitter(4, factor=2), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=46, size= 0, color="#ffff00")+opt #Third galaxy z3=c(-0.3, 0.7, 0.7, 0.6, 0.0, -1.1, 0.2, -0.6, -0.1, -0.1, 0.4, -0.7) galaxy3=galaxy(iter=2400, z=z3) %>% ggplot(aes(x,y))+ geom_point(shape= 8, size=jitter(12, factor=4), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=16, size= jitter(4, factor=2), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=46, size= 0, color="#ffff00")+opt #Fourth galaxy z4=c(-1.2, -0.6, -0.5, 0.1, -0.7, 0.2, -0.9, 0.9, 0.1, -0.3, -0.9, 0.3) galaxy4=galaxy(iter=2400, z=z4) %>% ggplot(aes(x,y))+ geom_point(shape= 8, size=jitter(12, factor=4), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=16, size= jitter(4, factor=2), color="#ffff99", alpha=jitter(.05, factor=2))+ geom_point(shape=46, size= 0, color="#ffff00")+opt  The Breathtaking 1-Matrix La luna sale a caminar siguiendo tus pupilas (Ojos color sol, Calle 13) This is a 5×5 1-matrix: $\begin{bmatrix} 1 &1 &1 &1 &1 \\ 1 &1 &1 &1 &1 \\ 1 &1 &1 &1 &1 \\ 1 &1 &1 &1 &1 \\ 1 &1 &1 &1 &1 \end{bmatrix}$ And this is a 20×20 1-matrix visualized: Maybe in some other galaxy, aliens represent matrix in this way. par(mar = c(1, 1, 1, 1), bg="violetred4") circlize::chordDiagram(matrix(1, 20, 20), col="white", symmetric = TRUE, transparency = 0.85, annotationTrack = NULL)  Gummy Worms Just keep swimming (Dory in Finding Nemo) Inspired by this post, I decided to create gummy worms like this: Or these: When I was young I used to eat them. Do you want to try? This is the code: library(rgl) library(RColorBrewer) t=seq(1, 6, by=.04) f = function(a, b, c, d, e, f, t) exp(-a*t)*sin(t*b+c)+exp(-d*t)*sin(t*e+f) v1=runif(6,0,1e-02) v2=runif(6, 2, 3) v3=runif(6,-pi/2,pi/2) open3d() spheres3d(x=f(v1[1], v2[1], v3[1], v1[4], v2[4], v3[4], t), y=f(v1[2], v2[2], v3[2], v1[5], v2[5], v3[5], t), z=f(v1[3], v2[3], v3[3], v1[6], v2[6], v3[6], t), radius=.3, color=sample(brewer.pal(8, "Dark2"),1))  Playing With Julia (Set) Viento, me pongo en movimiento y hago crecer las olas del mar que tienes dentro (Tercer Movimiento: Lo de Dentro, Extremoduro) I really enjoy drawing complex numbers: it is a huge source of entertainment for me. In this experiment I play with the Julia Set, another beautiful fractal like this one. This is what I have done: • Choosing the function f(z)=exp(z3)-0.621 • Generating a grid of complex numbers with both real and imaginary parts in [-2, 2] • Iterating f(z) over the grid a number of times so zn+1 = f(zn) • Drawing the resulting grid as I did here • Gathering all plots into a GIF with ImageMagick as I did in my previous post: each frame corresponds to a different number of iterations This is the result: I love how easy is doing difficult things in R. You can play with the code changing f(z) as well as color palettes. Be ready to get surprised: library(ggplot2) library(dplyr) library(RColorBrewer) setwd("YOUR WORKING DIRECTORY HERE") dir.create("output") setwd("output") f = function(z,c) exp(z^3)+c # Grid of complex z0 <- outer(seq(-2, 2, length.out = 1200),1i*seq(-2, 2, length.out = 1200),'+') %>% c() opt <- theme(legend.position="none", panel.background = element_rect(fill="white"), plot.margin=grid::unit(c(1,1,0,0), "mm"), panel.grid=element_blank(), axis.ticks=element_blank(), axis.title=element_blank(), axis.text=element_blank()) for (i in 1:35) { z=z0 # i iterations of f(z) for (k in 1:i) z <- f(z, c=-0.621) df=data.frame(x=Re(z0), y=Im(z0), z=as.vector(exp(-Mod(z)))) %>% na.omit() p=ggplot(df, aes(x=x, y=y, color=z)) + geom_tile() + scale_x_continuous(expand=c(0,0))+ scale_y_continuous(expand=c(0,0))+ scale_colour_gradientn(colours=brewer.pal(8, "Paired")) + opt ggsave(plot=p, file=paste0("plot", stringr::str_pad(i, 4, pad = "0"),".png"), width = 1.2, height = 1.2) } # Place the exact path where ImageMagick is installed system('"C:\\Program Files\\ImageMagick-6.9.3-Q16\\convert.exe" -delay 20 -loop 0 *.png julia.gif') # cleaning up file.remove(list.files(pattern=".png"))  Zooming You don’t have to be beautiful to turn me on (Kiss, Prince) I discovered recently how easy is to create GIFs with R using ImageMagick and I feel like a kid with a new toy. To begin this new era of my life as R programmer I have done this: First of all, read this article: it explains very well how to start doing GIFs from scratch. The one I have done is inspired in this previous post where I take a set of complex numbers to transform and color it using HSV technique. In this case I use this next transformation: $f(z)= -Im(z)+(Re(z)+0.5*Im(z))*1i$ Modifying the range of Real and Imaginary parts of complex numbers I obtain the zooming effect. The code is very simple. Play with it changing the transformation or the animation options. Send me your creations, I would love to see them: library(dplyr) library(ggplot2) dir.create("output") setwd("output") id=1 # label tO name plots for (i in seq(from=320, to=20, length.out = 38)){ z=outer(seq(from = -i, to = i, length.out = 300),1i*seq(from = -i, to = i, length.out = 500),'+') %>% c() z0=z for (k in 1:100) z <- -Im(z)+(Re(z)+0.5*Im(z))*1i df=data.frame(x=Re(z0), y=Im(z0), h=(Arg(z)<0)*1+Arg(z)/(2*pi), s=(1+sin(2*pi*log(1+Mod(z))))/2, v=(1+cos(2*pi*log(1+Mod(z))))/2) %>% mutate(col=hsv(h,s,v)) ggplot(df, aes(x, y)) + geom_tile(fill=df$col)+
scale_x_continuous(expand=c(0,0))+
scale_y_continuous(expand=c(0,0))+
labs(x=NULL, y=NULL)+
theme(legend.position="none",
panel.background = element_rect(fill="white"),
plot.margin=grid::unit(c(1,1,0,0), "mm"),
panel.grid=element_blank(),
axis.ticks=element_blank(),
axis.title=element_blank(),
axis.text=element_blank())
id=id+1
}
system('"C:\\Program Files\\ImageMagick-6.9.3-Q16\\convert.exe" -delay 10 -loop 0 -duplicate 1,-2-1 *.png zooming.gif')
# cleaning up
file.remove(list.files(pattern=".png"))

The Coaster Maker by Shiny

The word you invented is well formed and could be used in the Italian language (The Accademia della Crusca regarding to the word “Petaloso”, recently invented by an eight-year-old boy)

Are you tired of your old coasters? Do you like to make things by your own? Do you have a PC and a printer at home? If you answered yes to all these questions, just follow these simple instructions:

• Install R and RStudio in your PC
• Open RStudio and create a new Shiny Web App multiple file (ui.R/server.R)
• Substitute sample code of each file by the code below
• Press Run App
• Press buttom Get your coaster! until you obtain a image you like
• Print the image
• Cut out the image
• Place on the coaster your favorite drinking

These are some examples:

This is the code of ui.R

#
# This is the user-interface definition of a Shiny web application. You can
# run the application by clicking 'Run App' above.
#
# Find out more about building applications with Shiny here:
#
#    http://shiny.rstudio.com/
#
library(shiny)
shinyUI(fluidPage(
titlePanel("The coaster maker"),
sidebarLayout(
sidebarPanel(
#helpText(),

# adding the new div tag to the sidebar
tags$div(class="header", checked=NA, tags$p("This coasters are generated by hypocycloid curves.The curve is formed by the locus of a point,
attached to a circle, that rolls on the inside of another circle.
In the curve's equation the first part denotes the relative position between the two circles,
the second part denotes the rotation of the rolling circle.")),
tags$div(class="header", checked=NA, HTML( "More info here")), actionButton('rerun','Get your coaster!') ), mainPanel( plotOutput("HarmPlot") ) ) ))  This is the code of server.R # This is the server logic of a Shiny web application. You can run the # application by clicking 'Run App' above. # # Find out more about building applications with Shiny here: # # http://shiny.rstudio.com/ # library(shiny) library(ggplot2) CreateDS = function () { t=seq(-31*pi, 31*pi, 0.002) a=sample(seq(from=1/31, to=29/31, by=2/31), 1) b=runif(1, min = 1, max = 3) data.frame(x=(1-a)*cos(a*t)+a*b*cos((1-a)*t), y=(1-a)*sin(a*t)-a*b*sin((1-a)*t)) } shinyServer(function(input, output) { dat<-reactive({if (input$rerun) dat=CreateDS() else dat=CreateDS()})
output$HarmPlot<-renderPlot({ ggplot(dat())+ geom_point(data=data.frame(x=0,y=0), aes(x,y), color=rgb(rbeta(1, .5, .5), rbeta(1, .5, .5), rbeta(1, .5, .5)) , shape=19, fill="yellow", size=220)+ geom_polygon(aes(x, y), fill=rgb(rbeta(1, 2, 2), rbeta(1, 2, 2), rbeta(1, 2, 2))) + theme(legend.position="none", panel.background = element_rect(fill="white"), panel.grid=element_blank(), axis.ticks=element_blank(), axis.title=element_blank(), axis.text=element_blank()) }, height = 500, width = 500) })  Sunflowers The world is full of wonderful things, like sunflowers (Machanguito, my islander friend) Sunflower seeds are arranged following a mathematical pattern where golden ratio plays a starring role. There are tons of web sites explaining this amazing fact. In general, the arrangement of leaves on a plant stem are ruled by spirals. This fact is called phyllotaxis, and I did this experiment about it some time ago. Voronoi tessellation originated by points arranged according the golden angle spiral give rise to this sunflowers: I know this drawing will like to my friend Machanguito because he loves sunflowers. He also loves dancing, chocolate cookies, music and swimming in the sea. Machanguito loves life, it is just that simple. He is also a big defender of renewable energy and writes down his thoughts on recycled papers. You can follow his adventures here. This is the code: library(deldir) library(ggplot2) library(dplyr) opt = theme(legend.position = "none", panel.background = element_rect(fill="red4"), axis.ticks = element_blank(), panel.grid = element_blank(), axis.title = element_blank(), axis.text = element_blank()) CreateSunFlower <- function(nob=500, dx=0, dy=0) { data.frame(r=sqrt(1:nob), t=(1:nob)*(3-sqrt(5))*pi) %>% mutate(x=r*cos(t)+dx, y=r*sin(t)+dy) } g=seq(from=0, by = 45, length.out = 4) jitter(g, amount=2) %>% expand.grid(jitter(g, amount=2)) %>% apply(1, function(x) CreateSunFlower(nob=round(jitter(220, factor=15)), dx=x[1], dy=x[2])) %>% do.call("rbind", .) %>% deldir() %>% .$dirsgs -> sunflowers
ggplot(sunflowers) +
geom_segment(aes(x = x1, y = y1, xend = x2, yend = y2), color="greenyellow") +
scale_x_continuous(expand=c(0,0))+
scale_y_continuous(expand=c(0,0))+
opt

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## 2020 Vol. 44, No. 7

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Display Method:          |

2020, 44(7): 071001. doi: 10.1088/1674-1137/44/7/071001
Abstract:
We present a model for tail wavelets, a phenomenon known as “echo” in the literature. The tail wavelet may appear in signal reconnaissances in the merger of binary compact objects, including black holes and neutron stars. We show that the dark matter surrounding the compact objects lead to a speculated tail wavelet following the main gravitational wave (GW). We demonstrate that the radiation pressure of the main wave is fully capable of pushing away the surrounding matter to some altitude, and splashing down of the matter excites the tail wavelet after ringing down of the main wave. We illustrate this concept in a simplified model, where numerical estimations are conducted on the specific distribution of dark matter outside the black hole horizon and the threshold values in accordance with observations. We study the full back reaction of the surrounding dark matter to the metric and find that the effect on to the tail wavelets is insignificant. We reveal the fine difference between the tail wavelets of a dressed and a bare black hole. We demonstrate that the tail wavelet can appear as a natural phenomenon in the frame of general relativity, without invoking modified gravities or quantum effects.
2020, 44(7): 071002. doi: 10.1088/1674-1137/44/7/071002
Abstract:
Pion-mass-dependent nucleon-nucleon (NN) potentials are obtained in terms of the one-pion exchange and contact terms from the latest lattice QCD simulations of the two-nucleon system. They assume the forms of the leading order (LO) NN potential from the chiral effective field theory and thus are referred to as the LO chiral potential in this study. We extract the coefficients of contact terms and cut-off momenta in these potentials, for the first time, by fitting the phase shifts of \begin{document}$^1S_0$\end{document} and \begin{document}$^3S_1$\end{document} channels obtained from the HALQCD collaboration with various pion masses from 468.6 to 1170.9 MeV. The low-energy constants in the \begin{document}$^1S_0$\end{document} and \begin{document}$^3S_1$\end{document} channels become weaker and approach each other for larger pion masses. These LO chiral potentials are applied to symmetric nuclear and pure neutron matter within the Brueckner-Hartree-Fock method. Presently, however, we do not yet have the information of the P-wave NN interaction to be provided by the lattice QCD simulations for a complete description of nuclear matter. Our results enhance understanding of the development of nuclear structure and nuclear matter by controlling the contribution of the pionic effect and elucidate the role of chiral symmetry of the strong interaction in complex systems.
2020, 44(7): 073101. doi: 10.1088/1674-1137/44/7/073101
Abstract:
In light of the recently obtained LHC Higgs data, we examine the parameter space of the type II two-Higgs-doublet model, in which the 125 GeV Higgs bosons exhibit wrong sign Yukawa couplings. Combining the relevant theoretical and experimental limits, we find that the LHC Higgs data exclude most of the parameter space of the wrong sign Yukawa coupling. For \begin{document}$m_H=$\end{document} 600 GeV, the allowed samples are mainly distributed across several corners and narrow bands of \begin{document}$m_A<20$\end{document} GeV, 30 \begin{document}$<m_A<120$\end{document} GeV, 240 GeV \begin{document}$<m_A<300$\end{document} GeV, 380 GeV \begin{document}$<m_A<430$\end{document} GeV, and 480 GeV \begin{document}$<m_A<550$\end{document} GeV. For \begin{document}$m_A=$\end{document} 600 GeV, \begin{document}$m_H$\end{document} is required to be lower than 470 GeV. The light pseudo-scalar with a mass of 20 GeV is still permitted in the case of the wrong sign Yukawa coupling of 125 GeV Higgs bosons.
2020, 44(7): 073102. doi: 10.1088/1674-1137/44/7/073102
Abstract:
We study the quasi-two-body decays \begin{document}$B_{(s)} \to \psi [K^*(892), K^*(1410),$\end{document} \begin{document}$K^*(1680)] \to \psi K\pi$\end{document} by employing the perturbative QCD (PQCD) factorization approach, where the charmonia \begin{document}$\psi$\end{document} represents \begin{document}$J/\psi$\end{document} and \begin{document}$\psi(2S)$\end{document}. The corresponding decay channels are studied by constructing the kaon-pion distribution amplitude (DA) \begin{document}$\Phi_{K \pi}^{\rm{P}}$\end{document}, which comprises important final state interactions between the kaon and pion in the resonant region. Relativistic Breit-Wigner formulas are adopted to parameterize the time-like form factor \begin{document}$F_{K\pi}$\end{document} appearing in the kaon-pion DAs. The SU(3) flavor symmetry breaking effect resulting from the mass difference between the kaon and pion is taken into account, which makes significant contributions to the longitudinal polarizations. The observed branching ratios and the polarization fractions of \begin{document}$B_{(s)} \to \psi K^*(892) \to \psi K\pi$\end{document} are accommodated by tuning hadronic parameters for the kaon-pion DAs. The PQCD predictions for \begin{document}$B_{(s)} \to \psi [K^*(1410), K^*(1680)] \to \psi K\pi$\end{document} modes from the same set of parameters can be tested by precise data obtained in the future from LHCb and Belle II experiments.
2020, 44(7): 073103. doi: 10.1088/1674-1137/44/7/073103
Abstract:
We present an improved calculation of the strong coupling constants \begin{document}$g_{D^*D\rho}$\end{document} and \begin{document}$g_{B^*B\rho}$\end{document} in light-cone sum rules, including one-loop QCD corrections of leading power with \begin{document}$\rho$\end{document} meson distribution amplitudes. We further compute subleading-power corrections from two-particle and three-particle higher-twist contributions at leading order up to twist-4 accuracy. The next-to-leading order corrections to the leading power contribution numerically offset the subleading-power corrections to a certain extent, and our numerical results are consistent with those of previous studies on sum rules. A comparison between our results and existing model-dependent estimations is also made.
2020, 44(7): 073104. doi: 10.1088/1674-1137/44/7/073104
Abstract:
We construct holographic Janus solutions, which describe a conformal interface in the theory of M2-branes, in four-dimensional gauged supergravities using a perturbative method. In particular, we study three Einstein-scalar systems and their BPS equations, which are derived by Bobev, Pilch, and Warner (2014). The actions of our interest are all consistent truncations of \begin{document}$D = 11$\end{document} supergravity, chosen to be invariant under \begin{document}$SO(4)\times SO(4)$\end{document}, \begin{document}$SU(3)\times U(1)\times U(1)$\end{document}, and \begin{document}$G_2$\end{document} symmetry subgroups of \begin{document}$SO(8)$\end{document}. The utility of our semi-analytic result is illustrated by the calculation of minimal area surface and the associated holographic entanglement entropy.
2020, 44(7): 073105. doi: 10.1088/1674-1137/44/7/073105
Abstract:
We derive the transverse Ward-Takahashi identities (WTI) of N-dimensional quantum electrodynamics by means of the canonical quantization method and the path integration method, and subsequently attempt to prove that QED3 is solvable based on the transverse and longitudinal WTI, indicating that the full vector and tensor vertices functions can be expressed in terms of the fermion propagators in QED3. Further, we discuss the effect of different γ matrix representations on the full vertex function.
2020, 44(7): 073106. doi: 10.1088/1674-1137/44/7/073106
Abstract:
We analyze different decay observables of semileptonic decays \begin{document}${B}_{c}\to \left({D}_{s,d}^{\left(*\right)}\right){\mu }^{+}{\mu }^{-}$\end{document}, such as the branching ratio, forward-backward asymmetry, polarization fraction, and lepton polarization asymmetry in the non-universal \begin{document}$Z'$\end{document} model. We further study the dependence of the branching fraction to the new model parameters and find that the values of different decay parameters increase in the \begin{document}$Z'$\end{document} model, which indicates a possible approach for the search of new physics as well as for the unknown phenomena of the charm B meson.
2020, 44(7): 074001. doi: 10.1088/1674-1137/44/7/074001
Abstract:
The radionuclide 22Na generates the emission of a characteristic 1.275 MeV \begin{document}$\gamma$\end{document}-ray. This is a potential astronomical observable, whose occurrence is suspected in classical novae. The \begin{document}$^{22}{{\rm{Mg}}}(p,\,\gamma)^{23}{{\rm{Al}}}$\end{document} reaction is relevant to the nucleosynthesis of 22Na in Ne-rich novae. In this study, employing the adiabatic distorted wave approximation and continuum discretized coupled channel methods, the squared neutron asymptotic normalization coefficients (ANCs) for the virtual decay of \begin{document}$^{23}{{\rm{Ne}}}$\end{document} \begin{document}$\to$\end{document} \begin{document}$^{22}{{\rm{Ne}}}$\end{document} + n were extracted, and determined as \begin{document}$(0.483\pm0.060)$\end{document} fm−1 and \begin{document}$(9.7\pm2.3)$\end{document} fm−1 for the ground state and the first excited state from the experimental angular distributions of \begin{document}${}^{22}{{\rm{Ne}}}(d,\,p){}^{23}{{\rm{Ne}}}$\end{document} populating the ground state and the first excited state of \begin{document}$^{23}{{\rm{Ne}}}$\end{document}, respectively. Then, the squared proton ANC of \begin{document}${}^{23}{{\rm{Al}}}_{\rm{g.s.}}$\end{document} was obtained as \begin{document}$C_{d5/2}^{2}({}^{23}{{\rm{Al}}})=(2.65\pm0.33)\times10^{3}$\end{document} fm−1 according to the charge symmetry of the strong interaction. The astrophysical S-factors and reaction rates for the direct capture contribution in \begin{document}${}^{22}{{\rm{Mg}}}(p,\,\gamma){}^{23}{{\rm{Al}}}$\end{document} were also presented. Furthermore, the proton width of the first excited state of \begin{document}$^{23}{{\rm{Al}}}$\end{document} was derived to be \begin{document}$(57\pm14)$\end{document} eV from the neutron ANC of its mirror state in \begin{document}$^{23}{{\rm{Ne}}}$\end{document} and used to compute the contribution from the first resonance of \begin{document}$^{23}{{\rm{Al}}}$\end{document}. This result demonstrates that the direct capture dominates the \begin{document}$^{22}{{\rm{Mg}}}(p,\,\gamma)^{23}{{\rm{Al}}}$\end{document} reaction at most temperatures of astrophysical relevance for \begin{document}$0.33 < T_9<0.64$\end{document}.
2020, 44(7): 074101. doi: 10.1088/1674-1137/44/7/074101
Abstract:
A recent experimental breakthrough identified the last bound neutron-rich nuclei in fluorine and neon isotopes. Based on this finding, we perform a theoretical study of Z = 9, 10, 11, 12 isotopes in the relativistic mean field (RMF) model. The mean field parameters are assumed from the PK1 parameterization, and the pairing correlation is described by the particle number conservation BCS (FBCS) method recently formulated in the RMF model. We show that the FBCS approach plays an essential role in reproducing experimental results of fluorine and neon isotopes. Furthermore, we predict 39Na and 40Mg to be the last bound neutron-rich nuclei in sodium and magnesium isotopes.
2020, 44(7): 074102. doi: 10.1088/1674-1137/44/7/074102
Abstract:
Positive-parity doublet bands were reported in 120I. Based on these, we discuss the corresponding experimental characteristics, including rotational alignment, and re-examine the corresponding configuration assignment. The self-consistent tilted axis cranking relativistic mean-field calculations indicate that the doublet bands are built on the configuration \begin{document}$\pi h _{11/2}\otimes \nu h ^{-1}_{11/2}$\end{document}. By adopting the two quasiparticles coupled with a triaxial rotor model, the excitation energies, energy staggering parameter S(I), \begin{document}$B(M1)/B(E2)$\end{document}, effective angles, and K plots are discussed and compared with available data. The obtained results support the interpretation of chiral doublet bands for the positive-parity doublet bands in 120I, and hence identify this nucleus as the border of the A ≈ 130 island of chiral candidates.
2020, 44(7): 074103. doi: 10.1088/1674-1137/44/7/074103
Abstract:
By incorporating an isospin-dependent form of the momentum-dependent potential in the ultra-relativistic quantum molecular dynamics (UrQMD) model, we systematically investigate effects of the neutron-proton effective mass splitting \begin{document}$m_{n-p}^{*}$\end{document}=\begin{document}$\frac{m_{n}^{*}-m_{p}^{*}}{m}$\end{document} and the density-dependent nuclear symmetry energy \begin{document}$E_{\rm{sym}}(\rho)$\end{document} on the elliptic flow \begin{document}$v_2$\end{document} in \begin{document}$^{197}{{\rm{Au}}}$\end{document} + \begin{document}$^{197}{{\rm{Au}}}$\end{document} collisions at beam energies from 0.09 to 1.5 GeV/nucleon. It is found that at higher beam energies (\begin{document}$\geqslant$\end{document} 0.25 GeV\begin{document}$/$\end{document}nucleon) with the approximately 75 MeV difference in slopes of the two different \begin{document}$E_{\rm{sym}}(\rho)$\end{document}, and the variation of \begin{document}$m_{n-p}^{*}$\end{document} ranging from –0.03 to 0.03 at saturation density with isospin asymmetry \begin{document}$\delta=(\rho_{n}-\rho_{p})/\rho=0.2$\end{document}, the \begin{document}$E_{\rm{sym}}(\rho)$\end{document} has a stronger influence on the difference in \begin{document}$v_{2}$\end{document} between neutrons and protons, i.e., \begin{document}$v_{2}^{n}-v_{2}^{p}$\end{document}, than \begin{document}$m_{n-p}^{*}$\end{document} has. Meanwhile, at lower beam energies (\begin{document}$\leqslant$\end{document} 0.25 GeV\begin{document}$/$\end{document}nucleon), \begin{document}$v_{2}^{n}-v_{2}^{p}$\end{document} is sensitive to both the \begin{document}$E_{\rm{sym}}(\rho)$\end{document} and the \begin{document}$m_{n-p}^{*}$\end{document}. Moreover, the influence of \begin{document}$m_{n-p}^{*}$\end{document} on \begin{document}$v_{2}^{n}-v_{2}^{p}$\end{document} is more evident with the parameters of this study when using the soft, rather than stiff, symmetry energy.
2020, 44(7): 074104. doi: 10.1088/1674-1137/44/7/074104
Abstract:
In this study, we apply a self-consistent mean field approximation of the three-flavor Nambu–Jona-Lasinio (NJL) model and compare it with the two-flavor NJL model. The self-consistent mean field approximation introduces a new parameter, \begin{document}$\alpha$\end{document}, that cannot be fixed in advance by the mean field approach itself. Due to the lack of experimental data, the parameter, \begin{document}$\alpha$\end{document}, is undetermined. Hence, it is regarded as a free parameter and its influence on the chiral phase transition of strong interaction matter is studied based on this self-consistent mean field approximation. \begin{document}$\alpha$\end{document} affects numerous properties of the chiral phase transitions, such as the position of the phase transition point and the order of phase transition. Additionally, increasing \begin{document}$\alpha$\end{document} will decrease the number densities of different quarks and increase the chemical potential at which the number density of the strange quark is non-zero. Finally, we observed that \begin{document}$\alpha$\end{document} affects the equation of state (EOS) of the quark matter, and the sound velocity can be calculated to determine the stiffness of the EOS, which provides a good basis for studying the neutron star mass-radius relationship.
2020, 44(7): 074105. doi: 10.1088/1674-1137/44/7/074105
Abstract:
We investigate the formation distance (R0) from the center of the radioactive parent nucleus at which the emitted cluster is most probably formed. The calculations are performed microscopically starting with the solution to the time-independent Schrödinger wave equation for the cluster-core system, using nuclear potentials based on the Skyrme-SLy4 nucleon-nucleon interactions and folding Coulomb potential, to determine the incident and transmitted wave functions of the system. Our results show that the emitted cluster is mostly formed in the pre-surface region of the nucleus, under the effect of Pauli blocking from the saturated core density. The deeper α-formation distance inside the nucleus allows less preformation probability and indicates a more stable nucleus for a longer half-life. Furthermore, the α-particle tends to be formed at a slightly deeper region inside the nuclei, with larger isospin asymmetry, and in the closed shell nuclei. Regarding the heavy clusters, we observed that the formation distance of the emitted clusters heavier than α-particle increased via increasing the isospin asymmetry of the formed cluster rather than by increasing its mass number. The partial half-life of a certain cluster-decay mode increased with increase of either the mass number or the isospin asymmetry of the emitted cluster.
2020, 44(7): 074106. doi: 10.1088/1674-1137/44/7/074106
Abstract:
The chiral magnetic effect is concisely derived by employing the Wigner function approach in the chiral fermion system. Subsequently, the chiral magnetic effect is derived by solving the Landau levels of chiral fermions in detail. The second quantization and ensemble average leads to the equation of the chiral magnetic effect for righthand and lefthand fermion systems. The chiral magnetic effect arises uniquely from the contribution of the lowest Landau level. We carefully analyze the lowest Landau level and find that all righthand (chirality is +1) fermions move along the direction of the magnetic field, whereas all lefthand (chirality is −1) fermions move in the opposite direction of the magnetic field. Hence, the chiral magnetic effect can be explained clearly using a microscopic approach.
2020, 44(7): 074107. doi: 10.1088/1674-1137/44/7/074107
Abstract:
The relativistic mean field (RMF) model has achieved great success in describing various nuclear phenomena. However, several serious defects are common. For instance, the pseudo-spin symmetry of high-l orbits is distinctly violated in general, leading to spurious shell closures \begin{document}$N/Z = 58$\end{document} and \begin{document}$92$\end{document}. This leads to problems in describing structure properties, including shell structures, nuclear masses, etc. Guided by the pseudo-spin symmetry restoration [Geng et al., Phys. Rev. C, 100: 051301 (2019)], a new RMF Lagrangian DD-LZ1 is developed by considering the density-dependent meson-nucleon coupling strengths. With the newly obtained RMF Lagrangian DD-LZ1, satisfactory descriptions can be obtained for the bulk properties of nuclear matter and finite nuclei. In particular, significant improvements on describing the single-particle spectra are achieved by DD-LZ1. In particular, the spurious shell closures \begin{document}$Z = 58$\end{document} and \begin{document}$92$\end{document}, commonly found in previous RMF calculations, are eliminated by the new effective interaction DD-LZ1, and consistently the pseudo-spin symmetry (PSS) around the Fermi levels is reasonably restored for both low-l and high-l orbits. Moreover, the description of nuclear masses is also notably improved by DD-LZ1, as compared to the other RMF Lagrangians.
2020, 44(7): 074108. doi: 10.1088/1674-1137/44/7/074108
Abstract:
In this study, the multi-quasiparticle triaxial projected shell model (TPSM) is applied to investigate \begin{document}$\gamma$\end{document}-vibrational bands in transitional nuclei of \begin{document}$^{118-128}{\rm{Xe}}$\end{document}. We report that each triaxial intrinsic state has a \begin{document}$\gamma$\end{document}-band built on it. The TPSM approach is evaluated by the comparison of TPSM results with available experimental data, which shows a satisfactory agreement. The energy ratios, B(E2) transition rates, and signature splitting of the \begin{document}$\gamma$\end{document}-vibrational band are calculated.
2020, 44(7): 074109. doi: 10.1088/1674-1137/44/7/074109
Abstract:
We study the structure of nonstrange baryons by analytically calculating the electromagnetic transition helicity amplitudes of the nucleon and Δ resonances. We employ an improved hypercentral constituent quark model and obtain the corresponding eigenenergies and eigenfunctions in closed forms. Then, we calculate the transverse and longitudinal helicity amplitudes for nucleon and Δ resonances. The comparison of evaluated observables and experimental data indicates good agreement between the proposed model and available data.
2020, 44(7): 074110. doi: 10.1088/1674-1137/44/7/074110
Abstract:
The exclusive photoproduction of vector mesons (\begin{document}$J/\psi$\end{document} and \begin{document}$\phi$\end{document}) is investigated by considering the next-to-leading order corrections in the framework of the color glass condensate. We compare the next-to-leading order modified dipole amplitude with the HERA data, finding a good agreement. Our studies show that the \begin{document}$\chi^2/d.o.f$\end{document} from the leading order, running coupling, and collinearly improved next-to-leading order dipole amplitudes are 2.159, 1.097, and 0.932 for the elastic cross-section, and 2.056, 1.449, and 1.357 for the differential cross-section, respectively. The results indicate that the higher-order corrections contribute significantly to the vector meson productions, and the description of the experimental data is dramatically improved once the higher order corrections are included. We extend the next-to-leading order exclusive vector meson production model to LHC energies using the same parameters obtained from HERA. We find that our model provides a rather good description of the \begin{document}$J/\psi$\end{document} and \begin{document}$\phi$\end{document} data in proton-proton collisions at 7 TeV and 13 TeV in LHCb experiments.
2020, 44(7): 075101. doi: 10.1088/1674-1137/44/7/075101
Abstract:
We propose a new method to test the cosmic distance duality relation using the strongly lensed gravitational waves. The simultaneous observation of the image positions, relative time delay between different images, redshift measurements of the lens and the source, together with the mass modelling of the lens galaxy, provide the angular diameter distance to the gravitational wave source. On the other hand, the luminosity distance to the source can be obtained from the observation of the gravitational wave signals. To our knowledge this is the first time a method is proposed to simultaneously measure the angular diameter distance and the luminosity distance from the same source. Hence, the strongly lensed gravitational waves provide a unique method to test the cosmic distance duality relation. With the construction of the third generation gravitational detectors such as the Einstein Telescope, it will be possible to test the cosmic distance duality relation with an accuracy of a few percent.
2020, 44(7): 075102. doi: 10.1088/1674-1137/44/7/075102
Abstract:
The Ryu-Takayanagi (RT) formula plays a large role in the current theory of gauge-gravity duality and emergent geometry phenomena. The recent reinterpretation of this formula in terms of a set of “bit threads” is an interesting effort in understanding holography. In this study, we investigate a quantum generalization of the “bit threads” based on a tensor network, with particular focus on the multi-scale entanglement renormalization ansatz (MERA). We demonstrate that, in the large c limit, isometries of the MERA can be regarded as “sources” (or “sinks”) of the information flow, which extensively modifies the original picture of bit threads by introducing a new variable ρ: density of the isometries. In this modified picture of information flow, the isometries can be viewed as generators of the flow. The strong subadditivity and related properties of the entanglement entropy are also obtained in this new picture. The large c limit implies that classical gravity can emerge from the information flow.
2020, 44(7): 075103. doi: 10.1088/1674-1137/44/7/075103
Abstract:
We construct an alternative uniformly accelerated reference frame based on a 3+1 formalism in adapted coordinates. In this frame, a time-dependent redshift drift exists between co-moving observers, which differs from that in Rindler coordinates. This phenomenon can be tested in a laboratory and improve our understanding of non-inertial frames.


text:
# Derivative of The Sine Function

In my last note I proved that $\quad\displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = 1$

If you have not read that I would recommend reading that first as this limit is used in proving the derivative of the sine function. Here is the link.

$\qquad \qquad\qquad\qquad\qquad\qquad \displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = ?$

To prove the derivative of the sine function we will be using the first principles of derivatives.

$\qquad\qquad\qquad\qquad\qquad\displaystyle \lim_{\Delta\theta\to 0}\frac{\sin(\theta + \Delta\theta) - \sin\theta}{\Delta\theta}$

Use the addition formula for $\sin(\theta + \Delta\theta)$.

$=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta \times \cos \Delta\theta + \sin\Delta\theta \times \cos\theta - \sin\theta}{\Delta\theta}$

rearrange the equation and factor out $\sin\theta$

$=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta \times \cos \Delta\theta - \sin\theta + \sin\Delta\theta \times \cos\theta }{\Delta\theta}$

$=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta( \cos \Delta\theta - 1) + \sin\Delta\theta \times \cos\theta }{\Delta\theta}$

break apart the fraction

$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\frac{\sin\theta( \cos \Delta\theta - 1)}{\Delta\theta} + \frac{\sin\Delta\theta \times \cos\theta }{\Delta\theta}$

$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} + \frac{sin\Delta\theta}{\Delta\theta} \times cos\theta$

We wIll now evaluate the two limits seperately

$\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} \frac{sin\Delta\theta}{\Delta\theta} \times cos\theta$

In my last note I proved that $\displaystyle\lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta}=1$

$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} 1 \times cos\theta$

$=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} cos\theta$

That solves that limit now the other limit.

$\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta}$

divide out the $\sin\theta$

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta}$

multiply by the conjugate

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta} \times \frac{\cos\Delta\theta+1}{\cos\Delta\theta+1}$

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos^{2} \Delta\theta - 1)}{\Delta\theta(\cos\Delta\theta+1)}$

Rewrite the the equation using the Pythagorean identity. $\cos^{2}\theta + \sin^{2}\theta=1$

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{-\sin^{2} \Delta\theta }{\Delta\theta(\cos\Delta\theta+1)}$

break apart the fraction

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} (-sin\Delta\theta) \times \frac{\sin\Delta\theta}{\Delta\theta} \times \frac{1}{\cos\Delta\theta + 1}$

evaluate the limits

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times (-sin0) \times 1 \times \frac{1}{\cos0 + 1}$

$=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times 0 \times 1 \times \frac{1}{2}$

$=\qquad\qquad\qquad\quad\displaystyle 0$

$\displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} + \lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta} \times cos\theta$

$\displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad0 + \cos\theta$

$\displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \cos\theta$

Note by Brody Acquilano
5 years, 11 months ago

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text:
ACTA MATHEMATICA UNIVERSITATIS COMENIANAE

Vol. LXXI, 2(2002)
p. 221

Pointwise weighted vector ergodic theorem in L1(X)
K. El Berdan

Abstract.  In this paper we prove the almost everywhere convergence of weighted multiparameter averages of linear surjective isometries in $L^{1}(X)$ and power bounded in $L^{p}(X),1\le p\leq \infty .$

AMS subject classification:  47A35
Keywords:  Norm resolution, ergodic theorem, Banach space


text:
# Winding Around – ms finished

Last week I produced a more or less final manuscript of the “Winding Around” book – the book based on my MASS course from last year – and sent it off to the publisher.  There is still time if a few people would like to read the manuscript and let me know of any errors or inconsistencies – just email me and I will be happy to send you a copy.

# Artin’s Criterion, Part II

In this post I want to sketch the proof of “Artin’s criterion”, following Ahlfors’ book as referred to in my previous post (and presumably following Artin himself, though I still have not come up with any original reference to him).  The argument generalizes easily to $$(n-1)$$-dimensional cycles in an open subset $$\Omega\subseteq {\mathbb R}^n$$, but for simplicity I will stick to the planar case.

# Artin’s Criterion

Emil Artin, picture from Wikipedia

There’s been a long lull in updates to this page.  I’ve posted elsewhere about the reason for that: in brief, I was found to have cancer, and subsequent treatments have kept me busy for months.  You can read about this on my personal website, but I’m not going to write more about it here.

Meanwhile though I have been slowly writing up a book-length version of my MASS 2013 course, “Winding Around”, whose central theme is “The Winding Number in Topology, Geometry and Analysis”.  As I was “winding around” myself and trying to complete Chapter 5 in a way that was satisfactory, I ran into an interesting “gap” in my own understanding.  This is related to the homology version of Cauchy’s theorem.  This is usually stated in the following way.

Theorem  Let $$f$$ be a function that is holomorphic on an open subset $$\Omega\subseteq\mathbb C$$, and let $$\Gamma$$ be a cycle in $$\Omega$$ that is nullhomologous, this being defined to mean that the winding number of $$\Gamma$$ about each point of $${\mathbb C}\setminus\Omega$$ is zero.   Then

$\int_\Gamma f(z)dz = 0.$

# The Big Question About Limit Operators II

In the first post in this series, I gave some background to the “Big Question” on limit operators which it appears that Lindner and Seidel have solved for the case of free abelian groups.  In the next couple of posts I want to sketch some of the key ideas of their proof and to explore to what extent it can also be generalized to all exact groups (in the same way that I generalized the basic theory of limit operators to all exact groups in my 2005 paper).

There are two components to the L-S argument, it seems to me.

• a localization property for the “lower norm” of a finite propagation operators, and
• a “condensation of singularities” argument.

In this post we’ll look at the first of those. Continue reading

# The Big Question about Limit Operators I

A recent paper on the ArXiv (see bibliography below) is entitled “An Affirmative Answer to the Big Question about Limit Operators”.   I want to do a series of posts about this paper.  In this one I will try to explain the background, at least in the most elementary (Hilbert space) case.  In this introduction I will begin by re-expressing matters in the language of coarse geometry, following my paper (also in the bibliography below).

The basic setting is a discrete group $$\Gamma$$ with a (left-invariant) word metric; in the usual literature about limit operators this group is $$\mathbb Z$$ or $${\mathbb Z}^n$$, but there is no particular need for this restriction.  Let $$A$$ denote the rough algebra of $$\Gamma$$, that is the C*-algebra of operators on $$H=\ell^2(\Gamma)$$ generated by $$\ell^\infty(\Gamma)$$ together with the right translations by elements of $$\Gamma$$; equivalently, the closure of the bounded finite propagation operators on $$H$$.  If $$T\in A$$, then all the translates $$L_\gamma T L_\gamma^*$$ also belong to $$A$$, and indeed it is easy to see that they form a precompact subset of $$A$$ in the strong (or *-strong) operator topology.  The set of *-strong limit points of this subset is called the \emph{operator spectrum} of $$A$$ and denoted $$\sigma_{op}(A)$$. Continue reading


text:
# Vector calculus

Vector calculus, or vector analysis, is concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space ${\displaystyle \mathbb {R} ^{3}.}$ The term "vector calculus" is sometimes used as a synonym for the broader subject of multivariable calculus, which spans vector calculus as well as partial differentiation and multiple integration. Vector calculus plays an important role in differential geometry and in the study of partial differential equations. It is used extensively in physics and engineering, especially in the description of electromagnetic fields, gravitational fields, and fluid flow.

Vector calculus was developed from quaternion analysis by J. Willard Gibbs and Oliver Heaviside near the end of the 19th century, and most of the notation and terminology was established by Gibbs and Edwin Bidwell Wilson in their 1901 book, Vector Analysis. In the conventional form using cross products, vector calculus does not generalize to higher dimensions, while the alternative approach of geometric algebra which uses exterior products does (see § Generalizations below for more).


text:
Mind-blowing Sums: Compute $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}$ and $\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}$

How to prove the following two sums

\begin{align} \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}&=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\&\quad-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52 \end{align}

\begin{align} \sum_{n=1}^\infty\frac{H_n^3}{n^22^n}&=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\&\quad-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52 \end{align} where $$H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$$ is the $$n$$th generalized harmonic number of order $$p$$ and $$\operatorname{Li}_s(x)=\sum_{n=1}^\infty\frac{x^n}{n^s}$$ is the polylogarithmic function.

Edit: These two sums were proposed by Cornel Ioan Valean here but no solution was submitted. I am presenting my solution in the answer section and would like to see different approaches.

Thanks.

I am going to use the following results :

$$V_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)$$

$$V_2=\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac14\ln^42$$

$$V_3=\int_0^{1/2}\frac{\ln^3(1-x)\ln x}{x(1-x)} dx=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{285}{16}\zeta(5)-3\zeta(2)\zeta(3)$$

$$-\frac{21}{4}\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52$$

$$V_4=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^22^n}=4\operatorname{Li}_5\left(\frac12\right)+3\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{64}\zeta(5)+\frac{5}{16}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)$$ $$+\frac78\ln^22\zeta(3)-\frac{5}{12}\ln^32\zeta(2)+\frac{11}{120}\ln^52$$

We are going to establish two relations and solve them by eliminations

The first relation:

From here we have

$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$

Multiply both sides by $$\frac{1}{n2^n}$$ then sum both sides from $$n=1$$ to $$\infty$$ we get

$$\sum_{n=1}^\infty\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=-\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{(x/2)^n}{n}=\int_0^1\frac{\ln^3(1-x)\ln(1-x/2)}{x}\ dx\\\overset{x\mapsto 1-x}{=}\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=-6V_1$$

Note that for the last step, we used this rule $$\int_0^1\frac{\ln^ax\ln(1+x)}{1-x}\ dx=(-1)^aa!\left(\ln2\zeta(a+1)+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}\right)$$

Then

$$R_1=\sum_{n=1}^\infty\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=-6V_1$$

The second relation:

From here we have

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\tag{1}$$

Divide both sides of (1) by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}\ dx=\frac1{n2^n}$$ to get

$$\sum_{n=1}^\infty\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n2^n}=-\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx=-V_2$$

Now multiply both sides of (1) by $$-\frac{\ln x}{x}$$ then integrate from $$x=0$$ to $$1/2$$ and use that fact that $$-\int_0^{1/2}x^{n-1}\ln x\ dx=\frac{\ln2}{n2^n}+\frac1{n^22^n}$$ to get

$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\left(\frac{\ln2}{n2^n}+\frac1{n^22^n}\right)=\int_0^{1/2}\frac{\ln^3(1-x)\ln x}{x(1-x)}\ dx=V_3$$

$$-\ln2V_2+\sum_{n=1}^\infty\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=V_3$$

Then

$$R_2=\sum_{n=1}^\infty\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=V_3+\ln2V_2$$

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}=\frac{R_1-R_2}{6}=-\frac{6V_1+\ln2V_2+V_3}{6}$$

and by plugging the results of $$V_1$$, $$V_2$$ and $$V_3$$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}&=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\&\quad-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52 \end{align}

As for the second sum :

$$\sum_{n^1}^\infty\frac{H_n^3}{n^22^n}=\frac{R_1+R_2}{2}=\frac{-6V_1+\ln2V_2+V_3-4V_4}{2}$$

and by plugging the results of $$V_1$$, $$V_2$$, $$V_3$$ and $$V_4$$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_n^3}{n^22^n}&=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\&\quad-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52 \end{align}

.

Proofs of the results:

$$V_1$$ can be found here and $$V_2$$ can be found using wolfram but here is the steps

\begin{align} V_2&=\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx\overset{x\mapsto 1-x}{=}\int_{1/2}^1\frac{\ln^3x}{x(1-x)}\ dx\\ &=\int_{1/2}^1\frac{\ln^3x}{x}\ dx+\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\ &=-\frac14\ln^42+\sum_{n=1}^\infty \int_{1/2}^1x^{n-1}\ln^3x\ dx, \quad \text{apply integration by parts}\\ &=-\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=-\frac14\ln^42+\ln^42+3\ln^22\operatorname{Li}_2\left(\frac12\right)+6\ln2\operatorname{Li}_3\left(\frac12\right)+6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)\\ &\boxed{=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac14\ln^42} \end{align}

where we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

\begin{align} V_3&=\int_0^{1/2}\frac{\ln^3(1-x)\ln x}{x(1-x)}\ dx\overset{x\mapsto 1-x}{=}\int_{1/2}^{1}\frac{\ln^3x\ln(1-x)}{x(1-x)}\ dx\\ &=-\sum_{n=1}^\infty H_n\int_{1/2}^1 x^{n-1}\ln^3x\ dx\\ &=-\sum_{n=1}^\infty H_n\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=-\ln^32\sum_{n=1}^\infty\frac{H_n}{n2^n}-3\ln^22\sum_{n=1}^\infty\frac{H_n}{n^22^n}-6\left(\color{blue}{\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}+\sum_{n=1}^\infty\frac{H_n}{n^42^n}}\right)+6\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align}

I managed here to prove

$$\color{blue}{\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}+\sum_{n=1}^\infty\frac{H_n}{n^42^n}}=-\frac12\ln^22\sum_{n=1}^\infty\frac{H_n}{n^22^n}-\frac16\ln^32\sum_{n=1}^\infty\frac{H_n}{n2^n}+\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac{47}{32}\zeta(5)+\frac1{15}\ln^52+\frac13\ln^32\operatorname{Li}_2\left(\frac12\right)+\ln^22\operatorname{Li}_3\left(\frac12\right)+2\ln2\operatorname{Li}_4\left(\frac12\right)+2\operatorname{Li}_5\left(\frac12\right)$$

and $$V_3$$ simplifies into

$$\small{V_3=3\sum_{n=1}^\infty\frac{H_n}{n^4}+\frac{141}{16}\zeta(5)-\frac25\ln^52-2\ln^32\operatorname{Li}_2\left(\frac12\right)-6\ln^22\operatorname{Li}_3\left(\frac12\right)-12\operatorname{Li}_4\left(\frac12\right)-12\operatorname{Li}_5\left(\frac12\right)} \\ \small{\boxed{=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{285}{16}\zeta(5)-3\zeta(2)\zeta(3)-\frac{21}{4}\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52\quad}}$$

where we substituted $$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ along with the values $$\operatorname{Li}_2\left(\frac12\right)$$ and $$\operatorname{Li}_3\left(\frac12\right)$$.

To prove the last result of $$V_4$$, we use this generating function

$$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)$$

Divide both sides by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2} x^{n-1}\ dx=\frac1{n2^n}$$ we get

\begin{align} V_4&=\int_0^{1/2}\frac{\operatorname{Li}_4(x)}{x}\ dx-\underbrace{\int_0^{1/2}\frac{\ln(1-x)\operatorname{Li}_3(x)}{x}\ dx}_{IBP}-\frac12\int_0^{1/2}\frac{\operatorname{Li}^2_2(x)}{x}\ dx\\ &=\operatorname{Li}_5\left(\frac12\right)+\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\frac32\int_0^{1/2}\frac{\operatorname{Li}^2_2(x)}{x}\ dx \end{align}

@Song proved in this solution

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}\ dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$

by plugging this result along with the values of $$\operatorname{Li}_2(1/2)$$ and $$\operatorname{Li}_3(1/2)$$ the closed form of $$V_4$$ follows.


text:
# How do you keep yourself motivated?

1. Sep 1, 2015

### EngWiPy

I believe there are many people here doing research on different fields. Research most of the time is very tedious stressful and frustrating. How do you keep yourself motivated to work hard until you get some satisfactory results? and are you willing to sacrifice your mental/physical health to get some results no matter what or at some point you would change your attitude?

2. Sep 2, 2015

### Bystander

Depends on the definition you choose for the word "satisfactory." Negative results demoralize some people and elate others --- learning what won't work is as good as it gets sometimes.

3. Sep 2, 2015

### EngWiPy

I would say satisfactory results are something you can publish in a well recognized journal in your field.

4. Sep 2, 2015

### DiracPool

That's exactly it. I personally believe that in order to create great art, you really have to put yourself on the line and sacrifice everything. There's different schools of thought here, and a good exemplar comes from the music industry.

I believe in the policy of "total immersion." If you want to be a player in a certain field you have to commit yourself 100%. To the exclusion of everything else. Otherwise, you're just going to be an "also-ran"

5. Sep 2, 2015

### EngWiPy

But how to keep yourself up to it all the time? This is the question. Not all people are passionate about what they are doing as some projects may dictate their current research.

6. Sep 2, 2015

### DiracPool

Well, I think you are answering your own question. Are you passionate about it or not? If you are not sure, I would either question what I'm doing or develop a passion for it if I have to. It doesn't really matter if you're constrained in some way. Case in point, I had to take a job at a car rental agency. I initially worked at the San Diego airport, because, at the time, I was hired by these mercenaries who offered me 10% on upgrades and 15% on insurance. This might not sound like much, but when you're booking groups of 75 from Nortel networks for a week each of which get the "supplementary" insurance with the personal effects coverage, it adds up real quick. Especially when I upgrade them from a compact to a midsize

But I digress. I gave the position up at the airport and took a sleepy job at a hotel on Coronado beach for a huge paycut because I didn't want the pressure anymore and because I could write my brain manifesto behind a sleepy counter with few customers and still make $15 an hour. Which I gladly accepted even though I averaged abut$85 an hour at the airport.

But the point is, is that I did write the manifesto there, but I also conjured up a way to make a killing in the rental car business. Each car they were sitting on I calculated was costing them $18 dollars a day. Every car that was sitting idle was a liability, didn't matter if it was a compact, midsize, or fullsize, they all cost the company$18 a day. But you could easily make upwards of $60+ a day (including supplemental insurance) on each rental you worked it right. And there was plenty of customers and resources to do this in San Diego in the 90's. But I gave it up because I was more interested in slacking behind the counter for$15 an hour while I wrote my manifesto. But the point is that you can self-motivate yourself to do anything, even sell rental cars!

7. Sep 2, 2015

### EngWiPy

You are lucky enough to be in a country where you can live without a college degree. Which is to say that if you don't find your passion in education you can do whatever you want to do and be just fine. In other countries even with a college degree you won't be able to cover the basic living expenditures for years. Illiterate people are literally suffering. That means that people are forced to go to college to survive, and some of them continue to PhD. Once you are in PhD and you are not really interested is a big problem. If you manage to finish, you won't be distinguished, and the sense of under achievement is terrible. Everyone wants to be distinguished and have the sense of achievement.

8. Sep 2, 2015

### DiracPool

Well, if that's the case I'd highly suggest you get interested. I don't think you told us what your field is. If you don't like it, try to change it. But that's what my earlier post was designed to create. If you are locked in a situation that you don't like or feel helpless in, there's always a way out. Use your mind to turn the situation around and take charge. Maybe this is not so easy under a dictatorial regime but that's all I got for you..

9. Sep 2, 2015

### EngWiPy

My field is Electrical Engineering. It was more a general question, and I don't want to divert the discussion into any particularities. Do you find research easy or hard? If easy what makes it easy, and if hard how do you keep going in doing it (what motivates you)? Is passion the answer or an answer? Is passion something we discover or create? Do you create incentives to yourself to keep going?

10. Sep 2, 2015

### DiracPool

If you're looking at EE as a means to an end that is financially motivated, as it seems to me what you are saying, then the prognosis is not good. That's the way I look at it.

You create incentives by creating incentives. If your primary reason for doing science isn't to contribute some original work to the field, then you are an imposter in my opinion. You are a janitor. Which isn't bad, per se. somebodys got to clean the petri dishes. But what I see in you is a lack of focus. Just try to figure out what it is you want to do and who you want to be. I, personally, was NEVER in doubt as to what that was, and maybe I was luckier than I thought in that respect.

11. Sep 2, 2015

### EngWiPy

Are you telling me that if you are passionate about science then you will have no problems in doing research and keep yourself motivated? I doubt all PhD holders are passionate about their fields. That is why I think the education level is higher in developing countries than in developed countries. Money is a big derive for education in many cases.

I love science. I have read many scientific books for the public reader, and I have watched many science documentaries. I like mathematics, that is why I have chosen Electrical Engineering/Communications. I have the capacity to understand and do very complicated problems. But I have a problem with doing research as to set down for long hours a day which is isolating me and compromises my health, while in my opinion in life there are more important things like family and peace of mind. Does research necessary mean to give up life and health for science? Is this what it means to be passionate about science?

12. Sep 2, 2015

### micromass

No, that is not what research should be about at all.

13. Sep 2, 2015

### Geofleur

Actually, having hobbies and outside things that you do can help. While I was working on my Ph.D., I took up the study of formal logic as a hobby. I would work a while, study a while and come back refreshed. Doing things that are healthy also helps. Sitting around for hours, your head gets foggy. I had an adviser once out at one of the national labs who would get up and walk around the lab complex at least twice a day. I started doing it too; when I would get back my head was always clearer. Concentrating on one thing to the exclusion of all else leads to burnout. At least it would for me. And I am not known as lacking focus, for sure!

14. Sep 2, 2015

### WWGD

I think a general suggestion is to track down your mental and psychological states and learn to manage them, manage yourself, so you can remain in good overall condition.

15. Sep 2, 2015

### EngWiPy

While I was doing my PhD, and I was working on something and stuck, I couldn't think of anything else. That is why I had such a hard times in my PhD, especially I am not that outgoing type of person. So, I would dwell on the problem day and night. I also have a problem in focusing (unlike you) due to suspecting ADHD. Now things are more relaxing since if you don't get something done, I don't lose anything, and thus I don't have the kind of stress I had during my PhD. But still doing research is a huge part of my life and it is boring most of the time, except when you have a glimpse of optimistic results, which will energize me for a while and then fade away, which is usually quick.

16. Sep 2, 2015

### Geofleur

What you just said reminds me of the making of the stop-motion animation film The Nightmare Before Christmas. For each scene, people had to move these little figures just a little, take a picture, then move them just a little, take a picture, and on and on, etc., ad infinitum. One of those people was saying that you can't just do it for the end product, you have to actually enjoy the process of it. Research is in some ways such an intellectual activity, but in others it's like making stop-motion films! If you are spending the majority of your time unhappy, I don't personally see the point of it. Can you find good things about the process itself? If not, maybe you just need a different kind of project.

17. Sep 2, 2015

### Geofleur

Disclaimer: Err... I didn't mean to suggest that making stop motion films can't be intellectual.

18. Sep 2, 2015

### EngWiPy

So, is it something I alone feel about research and not something prevalent? I thought what I feel is common among many researchers and I am just honest about it!!

19. Sep 2, 2015

### Geofleur

It may be common, but it's still pointless to keep doing something that makes you unhappy over and over. And it's certainly not that way for everyone.

20. Sep 2, 2015

### Geofleur

Let me put it this way. You only live once, and even the greatest discoveries by the greatest scientists will be unknown someday (unless you think humanity will somehow last forever). Landau used to say that you shouldn't set out to do something great because then nothing will happen! Maybe some people can be motivated by "being great" or "doing something great", but actually enjoying what you're doing is a much better recipe. Then, even if you don't do anything great, you were still happy trying.

21. Sep 2, 2015

### EngWiPy

I agree. But, I believe many people are not happy with their jobs. What should they do? Quit and be in a worse situation? I think it is more reasonable to manage the difficulties in any job until you find another viable alternative, and I am here trying to benefit from others' experiences in research and managing its difficulties, which I haven't gotten for the most part of it.

22. Sep 2, 2015

### DiracPool

I think electrical engineering is a great field, whether you like it or not. I have great respect for engineers. In fact, I'm watching a series now trying to learn physics from an engineering perspective rather than from a traditional theoretical physics perspective. When the knowledge is "applied" and practical, you gain a different understanding of it. I wouldn't go as far as saying "deeper," but having multiple perspectives helps me especially, tremendously.

Just as an example, I got much more out of this one lecture on lagrangians than I did in the whole classical physics course given by Susskind:

I took a class at my undergraduate college called "physiological psychology." It was given by the psychology department. This was a small college in the wine country of northern California and, believe it or not, this class was the closest thing to neuroscience that you were going to find there. And I was a biology major.

In any case, the professor was great and actually became a personal friend of mine over the years. But he used to like to "preach" to the class at times. And these were great moments. Real human learning. One of the things I remember was that he actually went to medical school to become a psychiatrist. What he learned almost instantly, though, is that medical "doctors" on the inside are basically divided up into two categories--surgeons and non-surgeons. If you're a surgeon then you are cool and you have street cred. If you're an internist or a family practitioner, you're basically a nobody. So he quit medical school and became a clinical psychologist. But that didn't work out either. He gave this mimeographed handout at the end of each semester called "Looking forward backwards," which was basically an alert to really think about what you want to be doing day to day in your chosen profession. And to really think about it. For him, he thought he would enjoy being Freud with patients sitting on the couch. But what he found out, the hard way, was that he was spending 8 hours a day, day after day, listening to people whine about their problems that he didn't really care about. So after yet another painful and expensive career change, he finally found what he loved, being a psychology professor.

But my point is that I drew a correlation between the "surgeon" and the "engineer." I think the engineer is to the physics world what the surgeon is to the medical world. The engineer is the one that gets things done and applies the science. The theoretical physicist has his head in the clouds. So I have great respect for engineers even though I'm more of the theoretical type myself. Bottom line, though, is that, unless you are really unhappy being an EE, this is not a bad career choice. Btw, speaking of engineers, this series really cemented by admiration for engineers:

23. Sep 2, 2015

### EngWiPy

Actually, I did my Bachelor in Computer Engineering which was more about concepts and algorithms than mathematics and I didn't like it very much because I am not good in memorizing, but then I took two elective courses in communications and I loved them. So, I switched to Electrical Engineering with specialty in communications in my master and PhD. I am not passionate about it, but I don't hate it. The process of research is what I am interested in how to be done properly with minimal effort and stress. I am a fresh PhD graduate, and still have long way to learn about research, but I think the lack of organization and focus is a major part of the problem.

Last edited: Sep 2, 2015
24. Sep 2, 2015

### Choppy

I'll address the second question first. Sacrificing your physical or mental health for research is not going to do you any favours in the long run. While in the short term you may find advantages to say skipping workouts or to complete a set of measurements or sacrificing sleep to get paper corrections in before a deadling, in the long term these behaviours compound your problems making a less effective researcher. That's not to say that successful researchers don't ever do this. Rather, that being a successful researcher is something that comes over years and years not day to day and patterns of poor (or good) choices tend to have cumulative effects.

The next question is how one stays motivated through the tedious labour of day-to-day research.
1. Sometimes it's tough staying motivated. Lots of researchers struggle with this. I'm sure most researchers have days where they would much rather be doing something fun. The same is true for just about any job though. Self-discipline allows people to work through the drudgery because it's a means to achieving a long term goal.
2. Most really successful researchers have more than one project on the go. That allows them to switch gears. If one project doesn't work out there are others. If one gets too boring, the day can be broken up by moving on to something else.
3. In my experience, the most boring stuff is the stuff you don't understand. The more you know about a project, the more exciting it tends to be. so a lot of boredom is eliminated with experience and understanding.


text:
MathSciNet bibliographic data MR1606918 (99b:33036) 33D20 (05A30 11B65 33C70 33D65) Schlosser, Michael Multidimensional matrix inversions and \$A\sb r\$$A\sb r$ and \$D\sb r\$$D\sb r$ basic hypergeometric series. Ramanujan J. 1 (1997), no. 3, 243–274. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text: Cellular automata (CAs) are good examples of systems in which large numbers of autonomous entities exhibit emergent behaviour. Using the occam-pi and JCSP communicating process systems, we show how to construct lazy'' and just-in-time'' models of cellular automata, which permit very efficient parallel simulation of sparse CA populations on shared-memory and distributed systems.


text:
# Where can I obtain the Courier 10 Pitch font for use with LO? [closed]

I really need Courier 10 Pitch to write spec scripts--not the same as your Courier Final Draft--can you help me? Wondering if I can load Courier 10 Pitch (font) onto the Libre program. Or can you? Need this to write spec scripts and the Courier Final Draft (font) is just not the same. best Diana Merrill 415-921-4320

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### Closed for the following reason the question is answered, right answer was accepted by Alex Kemp close date 2016-03-02 12:13:32.036485

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A free version of this font (in the form of Type1) is distributed with most GNU/Linux distributions as part of X11 e.g.,

\$ fc-list "Courier 10 pitch"
/usr/share/fonts/X11/Type1/c0419bt_.pfb: Courier 10 Pitch:style=Regular
/usr/share/fonts/X11/Type1/c0611bt_.pfb: Courier 10 Pitch:style=Bold Italic
/usr/share/fonts/X11/Type1/c0582bt_.pfb: Courier 10 Pitch:style=Italic
/usr/share/fonts/X11/Type1/c0583bt_.pfb: Courier 10 Pitch:style=Bold

I would suggest downloading the Debian xfonts-scalable package here (select any release and then click the "all" link in the table), unzipping it, and installing the required fonts.

more

Fonts are in general NOT provided by LO but by the operating system and only used by programs.

On my win7 box I have Courier New and Courier Regular (from Microsoft, not TrueType but bitmap). I don't know what Courier Final Draft is, maybe an OpenFont which can not be identical to a licenced Font like Courier New.

You might run a Google search for "Courier 10 Pitch" to find a font (hopefully for free). Good luck!

EDIT: Courier New 12 Point looks like 1o Pitch.

more


text:
# My RSA decryption doesnt match answer

Break the RSA code whose key is $$(n, e) = (8369428283, 1234567)$$.

Find the deciphering key and then decipher the message under the assumption that the plaintext consists of 7-letter blocks in the alphabet, converted to an integer between 0 and 26^(7) −1 in the usual way, and the ciphertext consists of 8-letter blocks in the same alphabet.

The given answer was CLAIMYOURPRIZE, but this doesn't match the one I got.

I converted the first 7-Letter block in CLAIMYOURPRIZE and I also got the first 8-letter block of the encrypted message. Then in order for them to match the value of the private key (d) must be 2 and that is not possible

• $$n = 8369428283 = 81799×102317$$ (2 distinct prime factors by Wolfram Alpha)

The $$d$$ can be found in two ways;

1. if Euler's totient function $$\varphi(n)= (p-1)(q-1) = 8369244168$$ is used then
• $$d = 5788687615 = e^{-1} \bmod 8369244168$$
2. If Carmichael Function $$\lambda(n) = \text{LCM}(p-1,q-1) = 4184622084$$ is used then
• $$d=1604065531 = e^{-1} \bmod 4184622084$$

Carmichael Function $$\lambda$$ provides us the smallest $$d$$ (lcm versus phi in RSA) and this can be helpful for reducing the timing of decrypting and signing. Here, $$d$$ calculated with $$\lambda(n)$$ is two-bit less than $$d$$ calculated with $$\varphi(n)$$.

I've encrypted end decrypted both parts by $$\varphi(n)$$ and $$\lambda$$ successfully.

[ C =  2 ][ L = 11 ][ A =  0 ][ I =  8 ][ M = 12 ][ Y = 24 ][ O = 14 ]
plaintext       =     CLAIMYO
plaintext int   =   748676046

ciphertext int  =  1773907495
ciphertext char =    AFTHVVNJ
plaintext char  =     CLAIMYO

[ U = 20 ][ R = 17 ][ P = 15 ][ R = 17 ][ I =  8 ][ Z = 25 ][ E =  4 ]
plaintext       =     URPRIZE
plaintext int   =  6387458406

ciphertext int  =  5283907511
ciphertext char =    ARCSTZFZ
plaintext char  =     URPRIZE

note: the A in the ciphertext means 0 as an integer.

• Hey Kelalaka thank you for the help today, I finally was able to get the correct answer, there was a bug in my code at the final step. I now also get CLAIMYOURPRIZE as the answer. Dec 25 '19 at 2:59


text:
### FabiOquendo's blog

By FabiOquendo, history, 6 years ago,

When you submit a code, codeforces shows the testcases of the problem, but when the input is too long it can't be copy because the input is INCOMPLETE. Someone knows HOW TO SEE OR COPY A COMPLETE INPUT?

• +5

 » 6 years ago, # |   +20 you can't .
•  » » 6 years ago, # ^ |   0 thanks, and sorry for the unlike, it was an accident.
•  » » » 6 years ago, # ^ |   +1 you welcome , it's not a big deal .
 » 6 years ago, # |   +45 I guess you are interested in your last submission 12190150, which fails 19th test case. Here's how you can abuse the system: Find some distinctive feature of the failing test. This particular one is easy: it is the first test case starting with 'd'. Here's my submission to be sure: 12197471 Once you encountered the failing test case, output raw input instead of the answer: 12197631 Codeforces will output only first 511 characters. It means you are to repeat step 2, writing to the output another portion of the input: 12197650, 12197674, 12197689 And here's extracted test case itself: http://pastebin.com/JxkE7pDV
•  » » 6 years ago, # ^ |   0 Yes, that was for my last submission. Thanks for the trick.
•  » » 6 years ago, # ^ |   +4
•  » » » 6 years ago, # ^ |   0 Exactly. Nobody at Codeforces got time to implement direct download links for every large test case (are they bigger than the average blogpost with comments?). Instead they seem to favor a ridiculous excuse and a load caused by such extraction submissions. If they think that getting full test case is useless, why do they bother showing it at all?
•  » » » » 6 years ago, # ^ |   0 I thounght that my code was good, only I needed the testcase for to know when I was failing.
 » 6 years ago, # |   0 You can't and it's the reason why codeforces doesn't allow to do it is intentional
 » 6 years ago, # |   0 This should really be implemented as a new feature.
»
19 months ago, # |
-56

# include<bits/stdc++.h>

using namespace std; int main() { long int num,base,i=0;

int t;
cin>>t;
while(t--)
{
int n,k,flg=0;
cin>>n>>k;
long long int a[n];
for(int i=0;i<n;i++)
cin>>a[i];

vector<long long int>v;
set<long int>st;
for(int j=0;j<n;j++)
{     num=a[j];base=k,i=0;
while(num>0)
{
if(num%base==1)
{ v.push_back(i);
}
if(num%base>1)
{cout<<"NO"<<endl;
flg=1;
break;}
num=num/base;
i++;
}

}
if(flg==0)
{
for(int i=0;i<v.size();i++)
st.insert(v[i]);

if(v.size()==st.size())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}

st.clear();
v.clear();
}
return 0;

}

•  » » 19 months ago, # ^ |   -28 getting wrong answer on test 5, token 15.
•  » » » 19 months ago, # ^ |   +30 What's wrong with you? Why did you bump a 5 year old blog with an unrelated badly formatted code with no context? You don't deserve help, you deserve a ban.
 » 17 months ago, # |   0 Hi for[ https://codeforces.com/problemset/problem/143/A ] what is the wrong here[ https://ideone.com/pk8VRQ ]


text:
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# Corporate Finance Study Set 2

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## Quiz 18 : Multinational Financial Management

If the spot rate of the Israeli shekel is 5.51 shekels per dollar and the 180-day forward rate is 5.97 shekels per dollar, then the forward rate for the Israeli shekel is selling at a ____ to the spot rate.
Free
Multiple Choice

D

In 1985, a given Japanese imported automobile sold for 1,476,000 yen, or $8,200.If the car still sold for the same amount of yen today but the current exchange rate is 144 yen per dollar, what would the car be selling for today in U.S.dollars? Free Multiple Choice Answer: Answer: C Suppose it takes 1.82 U.S.dollars today to purchase one British pound in the foreign exchange market, and currency forecasters predict that the U.S.dollar will depreciate by 12.0% against the pound over the next 30 days.How many dollars will a pound buy in 30 days? Free Multiple Choice Answer: Answer: D Legal and economic differences among countries, although important, do NOT pose significant problems for most multinational corporations when they coordinate and control worldwide operations of subsidiaries. True False Answer: Exchange rate risk is the risk that the cash flows from a foreign project, when converted to the parent company's currency, will be worth less than was originally projected because of exchange rate changes. True False Answer: When the value of the U.S.dollar appreciates against another country's currency, we may purchase more of the foreign currency with a dollar. True False Answer: The United States and most other major industrialized nations currently operate under a system of floating exchange rates. True False Answer: Suppose one U.S.dollar can purchase 144 yen today in the foreign exchange market.If the yen depreciates by 8.0% tomorrow, how many yen could one U.S.dollar buy tomorrow? Multiple Choice Answer: The Eurodollar market is essentially a long-term market; most loans and deposits in this market have maturities longer than one year. True False Answer: Calculating a currency cross rate involves determining the exchange rate for two currencies by using a third currency as a base. True False Answer: A Eurodollar is a U.S.dollar deposited in a bank outside the United States. True False Answer: Exchange rate quotations consist solely of direct quotations. True False Answer: Suppose the exchange rate between U.S.dollars and Swiss francs is SF 1.41 =$1.00, and the exchange rate between the U.S.dollar and the euro is $1.00 = 1.64 euros.What is the cross-rate of Swiss francs to euros? Multiple Choice Answer: If it takes$0.71 U.S.dollars to purchase one Swiss franc, how many Swiss francs can one U.S.dollar buy?
Multiple Choice
If 1.64 Canadian dollars can purchase one U.S.dollar, how many U.S.dollars can you purchase for one Canadian dollar?
Multiple Choice
If the inflation rate in the United States is greater than the inflation rate in Britain, other things held constant, the British pound will
Multiple Choice