Prodi : Teknik Informatika
Lembaga : UIN Maulana Malik Ibrahim Malang
root_secant <- function(f, x, tol=1e-7, N=100){
iter <- 0
xold <- x
fxold <- f(x)
x <- xold+10*tol
while(abs(x-xold)>tol){
iter <- iter+1
if(iter>N)
stop("No solutions found")
fx <- f(x)
xnew <- x - fx*((x-xold)/(fx-fxold))
xold <- x
fxold <- fx
x <- xnew
}
root<-xnew
return(list(`function`=f, root=root, iter=iter))
}
jawab:
\(x_0=0\)
root_secant(function(x)
{((x^3)-(2*x)+2)},
x=0)
## $`function`
## function(x)
## {((x^3)-(2*x)+2)}
## <bytecode: 0x000000001438aa08>
##
## $root
## [1] -1.769292
##
## $iter
## [1] 26
\(x_0=\frac{1}{2}\)
root_secant(function(x)
{((x^3)-(2*x)+2)},
x=1/2)
## $`function`
## function(x)
## {((x^3)-(2*x)+2)}
## <bytecode: 0x0000000014ab6d28>
##
## $root
## [1] -1.769292
##
## $iter
## [1] 16
jawab:
root_secant(function(x)
{(sin(x)/x)},
x=0.5)
## $`function`
## function(x)
## {(sin(x)/x)}
## <bytecode: 0x0000000014f1cc70>
##
## $root
## [1] 6.283185
##
## $iter
## [1] 7
root_secant(function(x)
{(sin(x)/x)},
x=1)
## $`function`
## function(x)
## {(sin(x)/x)}
## <bytecode: 0x000000001597a9f8>
##
## $root
## [1] 3.141593
##
## $iter
## [1] 8
jawab:
Mendefinisikan fungsi f(x)f(x)
trapezoid <- function(ftn, a, b, n = 100) {
h <- (b-a)/n
x.vec <- seq(a, b, by = h)
f.vec <- sapply(x.vec, ftn) # ftn(x.vec)
Trap <- h*(f.vec[1]/2 + sum(f.vec[2:n]) + f.vec[n+1]/2)
return(Trap)
}
f <- function(x){
sin(x)^2
}
Menghitung integral menggunakan trapezoid dengan permisalan n=6
trapezoid(f,0,pi,n = 6)
## [1] 1.570796
REFERENSI
https://bookdown.org/moh_rosidi2610/Metode_Numerik/rootfinding.html#latihan-1
https://bookdown.org/moh_rosidi2610/Metode_Numerik/diffinteg.html#latihan-3