similar to: Plotting one dot in a graph

How would I numerically find the x value where the derivative of the function below is zero? x <- seq(1,2, by = .01) y <- 5*cos(2*x)-2*x*sin(2*x) plot(x,abs(y), type = "l", ylab = "|y|")

Hi all, I'm plotting to get the intersection value of three curves. Defining the x-axis as dsm, the following code works; dsm = c(800,600,NA,525,NA,450,400,NA,NA,NA,0) s3 = seq(0.05,1.05,0.1) plot(dsm,s3,col="blue",las=1,ylab="fraction",xlab="distance (km)") fc <- function(x,a,b){a*exp(-b*x)} fm <- nls(s3~fc(dsm,a,b),start=c(a=1,b=0)) co <- coef(fm)

Hi, I am trying to draw a perpendicular line from a point to two points. Mathematically I know how to do it, but to program it, I encounter some problem and hope can get help. Thanks. I have points, A, B and C. I calculate the slope and intercept for line drawn between A and B. I am trying to check whether I can draw a perpendicular line from C to line AB and get the x,y value for the point D

Hi, Say I'd like to fill in color or by hashing the region above 0.5 and below -0.5 in the following situation. > x <- seq(0,2*pi, length=100) > plot(x, sin(x), type="l") > # Now fill the top part and the bottom part The only mention of the words "fill" or "filled" seem to be in filled.contour. So is this possible in R? Thanks in advance. Yves

Hello, I have a problem finding a root of a function, which I define like this: tuki <- function(u, x, a, lambda){ if((lambda%%1>0) & u<0) {u<-(u+0i)} f <- Re(x-(a*(u)^lambda-(1-(u))^lambda)/lambda) f } What I want to do is to find the root, but without specifying the interval within which to search for it. I can do it easily in MATLAB with fsolve() or fzero() functions. I

Hi, plot(xc, yc, type="l", ylim=c(0,50), xlim=c(0,50), lwd=2, xlab="M", ylab="seconds") abline(a=-think, b=cpustime, lty="dashed", col="red") abline( 2.2, 0, lty="dashed", col = "red") This draws a response time curve an asymptote and a horizontal line. How do I draw a line from the intersection of the asymptote and

Hi, I would like to find out the coordinates of the intersection points of 2 density curves. I did a search but i didn't get any significant results. I really hope some of you have some ideas. here it is an example: set.seed(123) x1 <- rnorm(100, 1, 1) x2 <- rnorm(100, 0, 1) d1 <- density(x1) d2 <- density(x2) plot(d1, col = 2) lines(d2, col = 4) Now i would like to know

On Fri, 25 Aug 2023 22:17:05 +0530 ASHLIN VARKEY <ashlinvarkey at gmail.com> wrote: > Sir, Please note that r-help is a mailing list, not a knight! ?? > I want to solve the equation Q(u)=mean, where Q(u) represents the > quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is > the quantile function of Davies (Power-pareto) distribution. Hence I > want to

Dear friends - occurring in Windows R2.6.2 I am modeling physical chemistry in collaboration with a friend who has preferred working in Excel. I used uniroot, and find a solution to a two buffer problem in acid-base chemistry which I believe is physiologically sensible. Using "goal seek" in Excel my friend found another plausible root, quite close to zero, and a plot of the function

I think the function you need is 'help.search'; try: help.search("binomial") and look for something obvious in the 'stats' package. A good deal quicker and easier than posting to an internet forum! Cheers, Mike. cathelf wrote: > > Dear all, > > I am trying a find the value "p" in binomial. > > X ~ Bin(n,p) > > I want to find the

Hi, Let say I have a normal density X~n(0,1) and I have a line y=0.01x+0.07. the following code generate the plots. x=seq(-10,10,length=100) plot(x,p1,type='n',ylab="Density",main="Overlap Measure",xaxt="n",yaxt="n") pi=dnorm(x,0,1) points(x,p1,type='l') abline(0.07,0.01) you can see that the curves intersects at 3 points. My question is

Hi all, I'm trying to figure out a way to create a data graphic that I haven't ever seen an example of before, but hopefully there's an R package out there for it. The idea is to essentially create a heatmap, but to allow each column and/or row to be a different width, rather than having uniform column and row height. This is sort of like a Marimekko chart in appearance, except that

Does anyone know of a method that I can get the intersection where the red and blue curves meet i.e. the value on the x-axis? x <- 1:10 y <- 10:1 plot(x,y) abline(lm(y~x),col="blue") abline(h=2.5,col="red") Muhammad

Hi, I have a cloud of randomly distributed points in 2-dimensional space and want to set up a grid, with a given grid-cell size, that minimizes the distance between my points and the grid nodes. Does anyone know of an R function or toolbox that somehow addresses this problem? This is a problem of optimizing the location of the grid, not a problem of deciding what should be the grid-cell size,

Hi, I would like to draw horizontal lines above a bar graph, in order to display the p-values of a Fisher test. Here is an example<http://thejns.org/action/showPopup?citid=citart1&id=f3-1060501&doi=10.3171%2Fped.2007.106.6.501>of the type of display I would like to have. Is there a way to draw the horizontal lines and write their associated p-values in R? Thanks for you help!

Sir, I want to solve the equation Q(u)=mean, where Q(u) represents the quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is the quantile function of Davies (Power-pareto) distribution. Hence I want to solve , *(c*u^lamda1)/((1-u)^lamda2)=28353.7....(Eq.1)* where lamda1=0.03399381, lamda2=0.1074444 and c=26104.50. When I used the package 'Davies' and solved Eq 1, I got the

Dear All, I'm looking an appropriate way in R to compute/estimate? points of intersection between a line and a curve and will really appreciate for any suggestion or ideas? Thank you, Fir

Hi, Is this method broken in R? I am using it to find roots of the following function: f(x) = 2.5*exp(-0.5*(2*0.045 - x)) + 2.5*exp(-0.045) + 2.5*exp(-1.5*x) - 100 It is giving an answer of -38.4762403 which is not even close (f(x) = 2.903809e+25 for x=-38.4762403). The answer should be around 0.01-0.1. This function should converge.. Even for a simple function like f(x) = exp(-x) * x, it gives

Hi thank you for talking the time to help me with this. I have a sequence of numbers in a file and an equal sequence of various character, say(a b c d) each occurs more than once. I need to plot the numbers so that numbers corresponding to a in the other sequence would have green dots, those corresponding to b a red dot, nothing on c and blue square for d. i.e 2 a show a green dot 4 b show a

Hi all, This is probably a blindingly obvious question: Why does it matter in the uniroot function whether the f() values at the end points that you supply are of the same sign? For example: f <- function(x,y) {y-x^2+1} #this gives a warning uniroot(f,interval=c(-5,5),y=0) Error in uniroot(f, interval=c(-5, 5), y = 0) : f() values at end points not of opposite sign #this doesn't give a