similar to: code for power and suffix for x,y labels in plot( ).

Hello, I want to solve: x*(3^x)*log(4)-x*log(4/3)-(3^x)+1=0 for x. I used the following code, uniroot(function(x) x*(3^x)*log(4)-x*log(4/3)-(3^x)+1, lower = -2, upper = 2, tol = 0.001 ) While using this I am getting the following error. Can anyone please help me out. Error in uniroot(function(x) x * (3^x) * log(4) - x * log(4/3) - (3^x) + : f() values at end points not of opposite sign.

Hello all, I was going to solve (n-m)! * (n-k)! = 0.5 *n! * (n-m-k)! for m when values of n and k are provided n1<-c(10,13,18,30,60,100,500)         # values of n kx<-seq(1,7,1);                               # values of k slv2<-lapply(n1,function(n){    slv1<-lapply(kx,function(k){              lhs<-function(m)              {                

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

Hi, I have tried to use uniroot to solve a value (value a in my function) that gives f=0, and I repeat this process for 10000 times(stimulations). However error occures from the 4625th stimulation - Error in uniroot(f, c(0, 2), maxiter = 1000, tol = 0.001) : f() values at end points not of opposite sign I have also tried interval of (lower=min(U), upper=max(U)) and it won't work as well.

Hi, I was trying to have a graph whose axes are of the following type: X axis: n and Y axis: var[U ((a,b) in suffix, and (n,d) in the power)]. U ((a,b) in suffix, and (n,d) in the power)- U^(n,d) _ (a,b). Actually I require many plots involving different values of a,b,n,d, so need to keep this complicated notation. The code I used: plot(n, hn$h_pg,

With main R releases only happening yearly in spring, now is good time to consider *and* discuss new features for what we often call "R-devel" and more officially is R Under development (unstable) (.....) -- "Unsuffered Consequences" Here is one such example I hereby expose to public scrutiny: A few minutes ago, I've committed the following to R-devel (the

Hi, I wanted to compute the value of the function ifn at certain values of n. But I am receiving the following error when I was using the following code(given at the end). Error: evaluation nested too deeply: infinite recursion / options(expressions=)? I feel that since the function Grx is recursively related, perhaps making the code too complicated too handle. Can anyone let me know if

I am calling the uniroot function from inside another function using these lines (last two lines of the function) : d <- uniroot(k, c(.001, 250), tol=.05) return(d$root) The problem is that on occasion there's a problem with the values I'm passing to uniroot. In those instances uniroot stops and sends a message that it can't calculate the root because f.upper * f.lower is greater

Hi, I wanted to compute the value of the function ifn at certain values of n. But I am receiving the following error when I was using the following code(given at the end). Error: evaluation nested too deeply: infinite recursion / options(expressions=)? I feel that since the function Grx is recursively related, perhaps making the code too complicated too handle. Can anyone let me know if

Hello, I want to draw a histogram of the mean of sample observations drawn from multivariate t distribution. I am getting the following error corresponding to the code I used. Though I am getting the graph, but I am curious to know the warning message. Warning messages: 1: In if (freq) x$counts else { : the condition has length > 1 and only the first element will be used 2: In if (!freq)

Hi, I want to use a more accurate value of exp(1).  The value given by R is 2.718282. I want a value which has more than 15 decimal places. Can anyone let me know how can I increase the accuracy level. Actually there are some large multipliers of exp(1) in my whole expression, and I want a more accurate result at the last step of my program, and for that I need to use highly accurate value

A recent message on ansari.test() prompted me to play with the examples. This doesn't work for me in R version 2.4.1 R> ansari.test(rnorm(100), rnorm(100, 0, 2), conf.int = TRUE) Error in uniroot(ab, srange, tol = 1e-04, zq = qnorm(alpha/2, lower = FALSE)) : object "ab" not found It looks like there's a small typo in ccia() inside ansari.test.default() in which

I am fairly new to R. I find it surprising that f <- function(x,a) {x-a} uniroot(f, c(0,1), a=.5) works, but integrate(f, 0, 1, a=.5) gives an error: Error in integrate(f, 0, 1, a = 0.5) : argument 4 matches multiple formal arguments What is the best way of avoiding such surprises? Is there a way of telling integrate() that the 'a' argument is for f()? If I wrote my own function

I have a strange problem with uniroot() function. Here is the result : > uniroot(th, c(-20, 20)) $root [1] 4.216521e-05 $f.root [1] 16.66423 $iter [1] 27 $estim.prec [1] 6.103516e-05 Pls forgive for not reproducing whole code, here my question is how "f.root" can be 16.66423? As it is finding root of a function, it must be near Zero. Am I missing something? -- View this message

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,

Hi, I want to find a value of n1. I used the following code but I am getting the error - Error in as.vector(x, mode) : cannot coerce type 'closure' to vector of type 'any' n=10 a_g<-(1/(n*(n-1)))*((pi/3)*(n+1)+(2*sqrt(3)*(n-2))-4*n+6) a_s<-function(n1) { t1=(n1-1)/2; (t1*(gamma(t1)/gamma(n1/2))^2)/2-1-a_g } xm<-solve(a_s) Can anyone help me out. Thanks in

Hi, I was plotting or creating a map for Texas school districts using the shape file of Texas. I could not find any other helpful mail in the mailing list. txshp<-read.shape(system.file("S:\\Districts_10_11.shp", package="maptools")) Error- read.shape no found. But read.shape is there in maptools. If anyone can help me out it will be great. Thanks in advance. Shant

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

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

Dear friends - merry Christmas and thanks a lot for much help during the year! In the example below I fail to understand how the calculated value pH is represented in a simple plot - also included. The calculations are useful in practice and likely to be right in principle but I cannot see how this occurs: why a calculated value of 7.4 known as numeric is not simply plotted as such. It