similar to: solving a linear equation

Sorry, I didn't mean a linear equation, of course this equation is not linear, I meant an equation in one unknown. Hi all, I've a linear equation of the form: 0.95=2 ( [3+ln(x/3)]^-13? + 4 [3+2ln(x/3)]^-13? + [3+4ln(x/3)]^-13 ) and I want to solve it for x, can I do this using R? Thanks in advance. Maram. ? ? ? ??? [[alternative HTML version deleted]] -------------- next

I first answered to the email below in r-help, but as I did not see any response, and it looks like a bug/unwanted behavior, I am also posting here. I have observed this in RGui, whereas it seems not to happen in RStudio. Similar to OP, I sometimes have a problem with functions using the progress bar. Frequently, the console is cleared after x iterations when the progress bar is called in a

Hi all, I've a question concerning the R 3.3.1 version. I have a long code that I used to run on versions earlier to the 3.3.1 version, and when I copied the code to the R console, I can still see the code while the loop is executing , along with the output printed after each iteration of the loop. Now, on the 3.3.1 version, after I copy the code to the console, it disappears and I only see

Hi all, I've a question concerning the R 3.3.1 version. I have a long code that I used to run on versions earlier to the 3.3.1 version, and when I copied the code to the R console, I can still see the code while the loop is executing , along with the output printed after each iteration of the loop. Now, on the 3.3.1 version, after I copy the code to the console, it disappears and I only see

Hi all, I want the y-axis label to be ( in symbols) g(sigma given alpha) where given is the conditional sign. I've tried ylab=expression(g(sigma|alpha))) but it gave me g(|(sigma,alpha)) where the sigma and alpha are in greek but the conditional sign is misplaced (before the bracket) Any help would be appreciated. Maram. [[alternative HTML version deleted]]

Dear All, Attached are the codes of a histogram & a kernel density estimate and the output they produced. In fact the variable q is a vector of 1000 simulated values; that is I generated 1000 samples from the pareto distribution, from each sample I calculated the value of q ( a certain fn in the sample observations), and thus I was left with 1000 values of q and I don't know the

Dear all, How can I use the histogram density estimate (hist) to find the value of the cdf at a certain point? Thanks Maram [[alternative HTML version deleted]]

Dear all, I have the cdf of the following power fuction distribution: F(y)=(y/350)^a               ,0<y<350, where " a " is some parameter with range a>0. I want to use it as the argument of the discretize function of the actuar package. So I think I need to define this function to R so that if I entered a=1, I get the following F(y)=(y/350) and if I entered a=4.5, I get the

Hi all, I'm using a certain  procedure to calculate the value of some variable(Bayes risk),B. So I got the values B1, B2, ........, B1000, each under certain input values and using a long procedure. Now, I want to put the values I got in a nummerical vector and find their minimum value. I think c( ) should work.For example if I have only 10 values I could have used

Hi all, I've a trivial question. If (q) is a continous variable,actually a vector of 1000 values. how to calculate the probability that q is greater than a specific value, i.e. P(q>45)?? Thanks Maram [[alternative HTML version deleted]]

I have no idea to be honest. I have never used the package, I simply did a search for it. Hopefully a more experienced user can help --- On Sun, 7/12/09, maram salem <marammagdysalem at yahoo.com> wrote: > From: maram salem <marammagdysalem at yahoo.com> > Subject: Re: [R] (no subject) > To: "John Kane" <jrkrideau at yahoo.ca> > Received: Sunday, July 12,

Hi Group, I've a vector of 1000 numeric values for which I want to draw a histogram. I've read this vector into R with no variable name.I mean only the 1000 values, which makes V1 the name of the variable by default?? Then I tried > hist(V1, breaks = "Sturges", +      freq = NULL, probability = !freq, +      include.lowest = TRUE, right = TRUE, +      density = NULL, angle =

Dear All, I have a matrix m of the form  m             [,1]             [,2]   [1,]  9072.302 0.00000004462366   [2,]  9086.811 0.00000005628169   [3,]  9101.320 0.00000007126347   [4,]  9115.830 0.00000008986976   [5,]  9130.339 0.00000011260000   [6,]  9144.848 0.00000014018405   [7,]  9159.357 0.00000017344229   [8,]  9173.866 0.00000021363563   [9,]  9188.375 0.00000026389996  [10,]  9202.884

I would like to ask once more if this is reproducible also for others? If yes, should I submit it as a bug-report? Best, Jon On 11/28/2016 11:26 AM, Jon Skoien wrote: > I first answered to the email below in r-help, but as I did not see > any response, and it looks like a bug/unwanted behavior, I am also > posting here. I have observed this in RGui, whereas it seems not to >

Dear group, Thank u so much 4 ur help. I've tried the link, http://finzi.psych.upenn.edu/R/library/quantreg/html/akj.html for adaptive kernel density estimation. But since I'm an R beginer and the topic of adaptive estimation is new for me, i still can't figure out some of the arguments of akj(x, z =, p =, h = -1, alpha = 0.5, kappa = 0.9, iker1 = 0) I've a vector of 1000 values

Dear All, I'm trying to plot a histogram (with the relative frequencies as the Y axis), But the scale of the y axis is given by 0e+00, 1e-04, 2e-04, 3e-04,......... Now, I have 2 questions 1- Does (1e-04=0.01831563)? 2- If this true,how can i change the given scale to (0.01,0.03,0.05,0.07,0.09)? [[alternative HTML version deleted]]

Dear All, I have a variable q which is a vector of 1000 simulated positive values; that is I generated 1000 samples from the pareto distribution, from each sample I calculated the value of q ( a certain fn in the sample observations), and thus I was left with 1000 values of q and I don't know the distribution of q. Hence, I used the given code for kernel density estimation to estimate the

Dear all, I want to use the histogtam as a density estimator, with the binwidths calculated using scott's formula which is binwidth = 3.49*ST.dev.*n^(-1/3) for the following data  (30 data points) 12-9-3-6-1-23-21-7-18-16-15-4-19-22-20-2-3-18-8-10-1-7-5-4-11-12-3-9-19-7 so first,I' ve tried this manually, and substituted in the above formula and I got st.dev.=7.02745 and thus the

--- On Tue, 6/30/09, maram salem <marammagdysalem at yahoo.com> wrote: > From: maram salem <marammagdysalem at yahoo.com> > Subject: [R] (no subject) > To: r-help at r-project.org > Received: Tuesday, June 30, 2009, 6:34 AM > Hi Group, > I've a vector of 1000 numeric values for which I want to > draw a histogram. I've read this vector into R with no >

On 06/10/2020 11:00 a.m., Sorkin, John wrote: > Colleagues, > I am trying to learn to use uniroot to solve a simple linear equation. I define the function, prove the function and a call to the function works. When I try to use uniroot to solve the equation I get an error message, > Error in yfu n(x,10,20) : object 'x' not found. > > I hope someone can tell we how I can fix