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 = 1e-7
?i = 1; x1 = x0
?p = numeric(N)
?while (i <= N) {
?df.dx = (f(x + h) - f(x)) / h
?x1 = (x0 - (f(x0) / df.dx))
?p[1] = x1
?i = i + 1
?if (abs(x1 - x0) < tol) break
?x0 = x1
?}
?return(p[1 : (i-1)])
?}
?app <- newton(f, x0 = 3/2)
but i cann't find this approximation
please can you help me?
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This list has a no homework policy. Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sun, Feb 24, 2019 at 9:22 AM malika yassa via R-help < r-help at r-project.org> wrote:> 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 = 1e-7 > i = 1; x1 = x0 > p = numeric(N) > while (i <= N) { > df.dx = (f(x + h) - f(x)) / h > x1 = (x0 - (f(x0) / df.dx)) > p[1] = x1 > i = i + 1 > if (abs(x1 - x0) < tol) break > x0 = x1 > } > return(p[1 : (i-1)]) > } > app <- newton(f, x0 = 3/2) > > but i cann't find this approximation > please can you help me? > > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >[[alternative HTML version deleted]]
On 2/24/2019 5:25 AM, malika yassa via R-help wrote:> 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 = 1e-7 > ?i = 1; x1 = x0 > ?p = numeric(N) > ?while (i <= N) { > ?df.dx = (f(x + h) - f(x)) / h > ?x1 = (x0 - (f(x0) / df.dx)) > ?p[1] = x1 > ?i = i + 1 > ?if (abs(x1 - x0) < tol) break > ?x0 = x1 > ?} > ?return(p[1 : (i-1)]) > ?} > ?app <- newton(f, x0 = 3/2) > > but i cann't find this approximation > please can you help me? > > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.As Bert pointed out there is a no homework policy on this list. But I will point out that when I ran your presented code, I got a lot of warnings.? They were not immediately informative (at least to me) because of a subtle programming error in your code.? Your main problem (but not your only one) is in the calculation of the derivative of the function inside the while loop. The following line is the problem: df.dx = (f(x + h) - f(x)) / h I will leave the solution of the problem to you. Dan -- Daniel Nordlund Port Townsend, WA USA