HELLOplease I want to approximate the solution of the equation
f(x)=x*(x-2)+log(x)=0
for that i did this program
f <- function(x){x*(x-2)+log(x)}
x <- c(1 : 2)
f(x)
h <- 1e-7
df.dx <- function(x){(f(x + h) - f(x)) / h}
df.dx(3/2);df.dx(2)
newton <- function(f, tol = 1e-7, x0 = 3/2, N = 100){
h
Please, can you help me I have a equation to solve by newton method but I can
not do it
for example
f<-function(x) {
2+X2-X3=0}
this equation have un solution in [1,2]
is there a function in R for solve it or i have to programme it
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helloplease I want to make a sustration of two vectors of a matrix
i have this program
aa<-matrix(outer(0:3,0:4,function(x,y) x+y*2),nrow=4,ncol=5)
for(i in 1:4)
+ {for(j in 2:5)
+ {bb[i,j-1]=aa[i,j]-aa[i,j-1]
+ }
+ }
at the end i obtain the bb=matrix( nrow=4,ncol=4)
but i cann't obtain this mat
hello
for examplei have this matrix
w2<-c(0.1,0.2,0.4,0.2,0.4,0.1)aa<-matrix(w1,nrow=3,ncol=3)aa
[,1] [,2] [,3]
[1,] 0.4 0.4 0.4
[2,] 0.1 0.1 0.1
[3,] 0.2 0.2 0.2
if i use this code
matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],2,3)
i will obtaine this matrix[,1] [,2] [,3]
[
helloi didn't obtaine the matrix after extrat non diagonalmy
programx<-rnorm(6,0,1)
aa<-matrix(x,nrow=6,ncol=6)
matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],5,6)
nrow=5ncol=6thank you
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R-help@r-p
helloplease i have this matrixx<-rnorm(6,0,1)
aa<-matrix(x,nrow=6,ncol=6)
i have to extrat non diagonal value, i use this code
matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],5,6)
but i didn't get the resultthank you
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helloplease can anyone help me, I find it difficult to calculate this integrali
have this programmx<-rnorm(10,0,1)y<-rexp(10,2)z<-exp(10,3)s<-vector()for ( j
in 1:10)s[j]<-x[j+1]+x[j]s1[i]<-s[j]/2f<-function(y,u){exp(y-u)}sapply(x,
function(i){
z[j] integrate(f,lower=s1[j],upper=,s1[j+1] x=i)$
helloplease can anyone help me, I find it difficult to calculate this integrali
have this programmx<-rnorm(10,0,1)y<-rexp(10,2)z<-exp(10,3)s<-vector()for ( j
in 1:10)s[j]<-x[j+1]+x[j]s1[i]<-s[j]/2f<-function(y,u){exp(y-u)}sapply(x,
function(i){
z[j] integrate(f,lower=s1[j],upper=,s1[j+1] x=i)$
hello
please you help me i have this functionx<-rnorm(10,0,1)f<-fuction(u,x)
{exp((x-u)}I want to calculate the integral of this function for each value of
x{for(i in 1:lenght(x)integrate(f,lower=1,upper=4)
thinks
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hello
please you help mei have this functionx<-rnorm(10,0,1)f<-fuction(u,x)
{exp((x-u)}I want to calculate the integral of this function for each value of
x{for(i in 1:lenght(x)
integrate(f,lower=1,upper=4)
}but I can not find the vector of resulatwhere is the errorthinks
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hellowplease,do you help mei have this matrixm<-matrix(( 1:12, nrow = 3 )
I want to delete the diagonal values of this matrix
can anyone do thisthinks
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R-help@r-project.org mailing list -- To UNSUBSCRIBE
hellothis is my programmeyou can help me, i cann't found a solution for H and
this function i calculate for all value for x1thank you
x<-rexp(N,2)
z<-rnorm(0,1,n)
g=k)-k*(Z-m<=-k)}
k1<-function(u,x1){-1/(2*pi)exp((x1-u)/h1}
for (i in 1:n)
{k1(u,x1)=integrate(-1/(2*pi)exp((x1-u)/h1,lower=
hello
> p(X ≤ V) diffetente zero
> look this
> Let X and V be two independent random variables with unknow
> distribution functions (d.f.’s) F and G respectively. Under truncation from
>the right we observe (X, Z) only if X ≤ Z
> I simulate X and Z and I use Lynden bell estimation
> I need now to
had.co.nz/Reproducibility.html
On Thursday, May 10, 2018, 11:07:30 a.m. EDT, malika yassa via R-help
wrote:
Hello
Do You help me, i have the problem in the package DTDA for find the
probability of truncation (alpha)
thank you
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Hello
Do You help me, i have the problem in the package DTDA for find the
probability of truncation (alpha)
thank you
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