Metode Root Finding

Berikut adalah user-defined function dari metode-metode Root Finding

Metode Fixed-Point

Fungsi fixedpoint memiliki beberapa argumen yaitu ftn yang berupa function di R, x0 merupakan nilai awal, tol merupakan nilai tolerasi dan max.iter adalah iterasi maksimum. tol dan max.iter masing-masing memiliki nilai default 1e-9 dan 100.

fixedpoint <- function(ftn, x0, tol = 1e-9, max.iter = 100) {
  
  xold <- x0
  xnew <- ftn(xold)
  iter <- 1
  cat("At iteration 1 value of x is:", xnew, "\n")
  

while ((abs(xnew-xold) > tol) && (iter < max.iter)) {
# xold digunakan untuk menyimpan akar-akar persamaan dari iterasi sebelumnya
    xold <- xnew
# xnew digunakan untuk menghitung akar-akar persamaan pada iterasi yang sedang berjalan
    xnew <- ftn(xold) #ingat ftn bentuknya fungsi
# iter digunakan untuk menyimpan banyaknya iterasi
    iter <- iter + 1
  cat("At iteration", iter, "value of x is:", xnew, "\n")
  }
# output bergantung pada kesuksesan algoritma
  if (abs(xnew-xold) > tol) {
      cat("Algorithm failed to converge\n")
      return(NULL)
      } else {
           cat("Algorithm converged\n")
           return(xnew)
           }
 }

Metode Newton-Raphson

Fungsi newtonraphson memiliki beberapa argumen yaitu ftn yang berupa function di R, x0 merupakan nilai awal, tol merupakan nilai tolerasi dan max.iter adalah iterasi maksimum. tol dan max.iter masing-masing memiliki nilai default 1e-9 dan 100.

newtonraphson <- function(ftn, x0, tol = 1e-5, max.iter = 100) {
  
    x <- x0
    fx <- ftn(x)
    iter <- 0
   while ((abs(fx[1]) > tol) && (iter < max.iter)) {
         x <- x - fx[1]/fx[2]
         fx <- ftn(x)
         iter <- iter + 1
         cat("At iteration", iter, "value of x is:", x, "\n")
         }
   # output bergantung pada kesuksesan algoritma
   if (abs(fx[1]) > tol) {
       cat("Algorithm failed to converge\n")
       return(NULL)
       } else {
            cat("Algorithm converged\n")
            return(x)
            }
     }

Metode Secant

Fungsi secant memiliki beberapa argumen yaitu ftn yang berupa function di R, x0 dan x1 merupakan nilai awal, tol merupakan nilai tolerasi dan max.iter adalah iterasi maksimum. tol dan max.iter masing-masing memiliki nilai default 1e-9 dan 100.

secant <- function(ftn, x0, x1, tol = 1e-9, max.iter = 100) 
  {
  xa <- x0
  fxa <- ftn(xa)
  xb <- x1
  fxb <- ftn(xb)
  iter <- 0
  while ((abs(xb-xa) > tol) && (iter < max.iter)) 
    {
    xt <- fxb*((xb-xa)/(fxb-fxa))
    xc <- xb-xt
    # fxc <- ftn(xc)
    iter <- iter + 1
    cat("At iteration", iter, "value of x is:", xc, "\n")
    xa <- xb
    xb <- xc
    fxa <- ftn(xa)
    fxb <- ftn(xb)
  

  if (abs(xb-xa) > tol) {
    cat("Algorithm failed to converge\n")
    return(NULL)
  } else {
    cat("Algorithm converged\n")
    return(xb)
  }
}
}

Metode Bisection

Fungsi bisection memiliki beberapa argumen yaitu fx yang berupa function di R, xl dan xr merupakan nilai awal, tol merupakan nilai tolerasi dan max.iter adalah iterasi maksimum. tol dan max.iter masing-masing memiliki nilai default 1e-9 dan 100.

bisection <- function(f, xl, xr, tol = 1e-7,max.iter = 100) {
  # Jika tanda dari hasil evaluasi fungsi pada titik xl dan xr berbeda maka program dihentikan
if ((f(xl) > 0) && (f(xr) < 0)) {
    stop('signs of f(xl) and f(xr) differ')
  }
  
  for (iter in 1:max.iter) {
    xm <- (xl + xr) / 2 # menghitung nilai tengah

cat("At iteration", iter, "value of x is:", xm, "\n")       
#Jika fungsinya sama dengan 0 di titik tengah atau titik tengah di bawah toleransi yang diinginkan, hentikan dan keluarkan nilai tengah sebagai akar
if ((f(xm) == 0) || abs(xr - xl) < tol) {
      return(xm)
    }
# Jika diperlukan iterasi lain,
# periksa tanda-tanda fungsi pada titik xm dan xl dan tetapkan kembali
# xl atau xr sebagai titik tengah yang akan digunakan pada iterasi berikutnya.    
    ifelse(sign(f(xm)) == sign(f(xl)), 
           xl <- xm,xr <- xm)
  }
}

Latihan Soal

Soal 1

Misalkan diberikan suatu persamaan \(x^3-2x-5=0\). a. Gambarkan fungsi tersebut! b. Carilah akar-akar persamaan dengan menggunakan metode fixed-point, newton-raphson, secant dan bisection!

Jawaban 1

#mendefinikan persamaan dalam bentuk fungsi
fx <- function(x) {
  x^3 - 2 * x - 5
}

#mendefinisikan persamaan dalam bentuk fungsi untuk fixed-iteration
fx_fi <- function(x){
  5/(x^2-2)
}

#mendefinikan persamaan dalam bentuk fungsi untuk newton-raphson
fx_nr <- function(x) {
rumus_fungsi <- function(x) x^3-2*x-5
fungsi <- rumus_fungsi(x)
# mencari turunan dengan menggunakan fungsi Deriv
# dari package Deriv
rumus_turunan <- Deriv::Deriv(fx)
turunan <- rumus_turunan(x)
return(c("fungsi"=fungsi,"turunan"=turunan))
}

Gambarkan fungsi tersebut!

curve(fx,xlim=c(-3,3), col='steelblue',lwd=2)
abline(h=0)
abline(v=0)

b. Carilah akar-akar persamaan dengan menggunakan metode fixed-point, newton-raphson, secant dan bisection!

#fixed-point
fixedpoint(ftn = fx_fi,x0 =0)

## At iteration 1 value of x is: -2.5 
## At iteration 2 value of x is: 1.176471 
## At iteration 3 value of x is: -8.117978 
## At iteration 4 value of x is: 0.07824535 
## At iteration 5 value of x is: -2.507676 
## At iteration 6 value of x is: 1.165925 
## At iteration 7 value of x is: -7.804949 
## At iteration 8 value of x is: 0.08486483 
## At iteration 9 value of x is: -2.509035 
## At iteration 10 value of x is: 1.164075 
## At iteration 11 value of x is: -7.752778 
## At iteration 12 value of x is: 0.08605028 
## At iteration 13 value of x is: -2.50929 
## At iteration 14 value of x is: 1.163728 
## At iteration 15 value of x is: -7.743083 
## At iteration 16 value of x is: 0.08627333 
## At iteration 17 value of x is: -2.509339 
## At iteration 18 value of x is: 1.163662 
## At iteration 19 value of x is: -7.741248 
## At iteration 20 value of x is: 0.08631566 
## At iteration 21 value of x is: -2.509348 
## At iteration 22 value of x is: 1.16365 
## At iteration 23 value of x is: -7.740899 
## At iteration 24 value of x is: 0.08632371 
## At iteration 25 value of x is: -2.50935 
## At iteration 26 value of x is: 1.163647 
## At iteration 27 value of x is: -7.740833 
## At iteration 28 value of x is: 0.08632524 
## At iteration 29 value of x is: -2.50935 
## At iteration 30 value of x is: 1.163647 
## At iteration 31 value of x is: -7.74082 
## At iteration 32 value of x is: 0.08632553 
## At iteration 33 value of x is: -2.50935 
## At iteration 34 value of x is: 1.163647 
## At iteration 35 value of x is: -7.740818 
## At iteration 36 value of x is: 0.08632558 
## At iteration 37 value of x is: -2.50935 
## At iteration 38 value of x is: 1.163647 
## At iteration 39 value of x is: -7.740817 
## At iteration 40 value of x is: 0.08632559 
## At iteration 41 value of x is: -2.50935 
## At iteration 42 value of x is: 1.163647 
## At iteration 43 value of x is: -7.740817 
## At iteration 44 value of x is: 0.0863256 
## At iteration 45 value of x is: -2.50935 
## At iteration 46 value of x is: 1.163647 
## At iteration 47 value of x is: -7.740817 
## At iteration 48 value of x is: 0.0863256 
## At iteration 49 value of x is: -2.50935 
## At iteration 50 value of x is: 1.163647 
## At iteration 51 value of x is: -7.740817 
## At iteration 52 value of x is: 0.0863256 
## At iteration 53 value of x is: -2.50935 
## At iteration 54 value of x is: 1.163647 
## At iteration 55 value of x is: -7.740817 
## At iteration 56 value of x is: 0.0863256 
## At iteration 57 value of x is: -2.50935 
## At iteration 58 value of x is: 1.163647 
## At iteration 59 value of x is: -7.740817 
## At iteration 60 value of x is: 0.0863256 
## At iteration 61 value of x is: -2.50935 
## At iteration 62 value of x is: 1.163647 
## At iteration 63 value of x is: -7.740817 
## At iteration 64 value of x is: 0.0863256 
## At iteration 65 value of x is: -2.50935 
## At iteration 66 value of x is: 1.163647 
## At iteration 67 value of x is: -7.740817 
## At iteration 68 value of x is: 0.0863256 
## At iteration 69 value of x is: -2.50935 
## At iteration 70 value of x is: 1.163647 
## At iteration 71 value of x is: -7.740817 
## At iteration 72 value of x is: 0.0863256 
## At iteration 73 value of x is: -2.50935 
## At iteration 74 value of x is: 1.163647 
## At iteration 75 value of x is: -7.740817 
## At iteration 76 value of x is: 0.0863256 
## At iteration 77 value of x is: -2.50935 
## At iteration 78 value of x is: 1.163647 
## At iteration 79 value of x is: -7.740817 
## At iteration 80 value of x is: 0.0863256 
## At iteration 81 value of x is: -2.50935 
## At iteration 82 value of x is: 1.163647 
## At iteration 83 value of x is: -7.740817 
## At iteration 84 value of x is: 0.0863256 
## At iteration 85 value of x is: -2.50935 
## At iteration 86 value of x is: 1.163647 
## At iteration 87 value of x is: -7.740817 
## At iteration 88 value of x is: 0.0863256 
## At iteration 89 value of x is: -2.50935 
## At iteration 90 value of x is: 1.163647 
## At iteration 91 value of x is: -7.740817 
## At iteration 92 value of x is: 0.0863256 
## At iteration 93 value of x is: -2.50935 
## At iteration 94 value of x is: 1.163647 
## At iteration 95 value of x is: -7.740817 
## At iteration 96 value of x is: 0.0863256 
## At iteration 97 value of x is: -2.50935 
## At iteration 98 value of x is: 1.163647 
## At iteration 99 value of x is: -7.740817 
## At iteration 100 value of x is: 0.0863256 
## Algorithm failed to converge

## NULL

# newton raphson
newtonraphson(ftn=fx_nr,x0=0)

## At iteration 1 value of x is: -2.5 
## At iteration 2 value of x is: -1.567164 
## At iteration 3 value of x is: -0.5025924 
## At iteration 4 value of x is: -3.820706 
## At iteration 5 value of x is: -2.549393 
## At iteration 6 value of x is: -1.608111 
## At iteration 7 value of x is: -0.5761004 
## At iteration 8 value of x is: -4.59771 
## At iteration 9 value of x is: -3.083543 
## At iteration 10 value of x is: -2.022194 
## At iteration 11 value of x is: -1.123764 
## At iteration 12 value of x is: 1.208652 
## At iteration 13 value of x is: 3.58079 
## At iteration 14 value of x is: 2.655233 
## At iteration 15 value of x is: 2.216106 
## At iteration 16 value of x is: 2.102125 
## At iteration 17 value of x is: 2.094584 
## At iteration 18 value of x is: 2.094551 
## Algorithm converged

##   fungsi 
## 2.094551

# secant
secant(fx,x0=1,x1=3)

## At iteration 1 value of x is: 1.545455 
## Algorithm failed to converge

## NULL

# bisection
bisection(fx,xl=-3,xr=3)

## At iteration 1 value of x is: 0 
## At iteration 2 value of x is: 1.5 
## At iteration 3 value of x is: 2.25 
## At iteration 4 value of x is: 1.875 
## At iteration 5 value of x is: 2.0625 
## At iteration 6 value of x is: 2.15625 
## At iteration 7 value of x is: 2.109375 
## At iteration 8 value of x is: 2.085938 
## At iteration 9 value of x is: 2.097656 
## At iteration 10 value of x is: 2.091797 
## At iteration 11 value of x is: 2.094727 
## At iteration 12 value of x is: 2.093262 
## At iteration 13 value of x is: 2.093994 
## At iteration 14 value of x is: 2.09436 
## At iteration 15 value of x is: 2.094543 
## At iteration 16 value of x is: 2.094635 
## At iteration 17 value of x is: 2.094589 
## At iteration 18 value of x is: 2.094566 
## At iteration 19 value of x is: 2.094555 
## At iteration 20 value of x is: 2.094549 
## At iteration 21 value of x is: 2.094552 
## At iteration 22 value of x is: 2.094551 
## At iteration 23 value of x is: 2.094551 
## At iteration 24 value of x is: 2.094552 
## At iteration 25 value of x is: 2.094552 
## At iteration 26 value of x is: 2.094551 
## At iteration 27 value of x is: 2.094551

## [1] 2.094551

Soal 2

Misalkan diberikan suatu persamaan \(x^3-x-1=0\). a. Gambarkan fungsi tersebut! b. Carilah akar-akar persamaan dengan menggunakan metode fixed-point, newton-raphson, secant dan bisection!

Jawaban 2

#mendefinikan persamaan dalam bentuk fungsi
fx <- function(x) {
  x^3 - x - 1
}

#mendefinisikan persamaan dalam bentuk fungsi untuk fixed-iteration
fx_fi <- function(x){
  1/(x^2-1)
}

#mendefinikan persamaan dalam bentuk fungsi untuk newton-raphson
fx_nr <- function(x) {
rumus_fungsi <- function(x) x^3-x-1
fungsi <- rumus_fungsi(x)
# mencari turunan dengan menggunakan fungsi Deriv
# dari package Deriv
rumus_turunan <- Deriv::Deriv(fx)
turunan <- rumus_turunan(x)
return(c("fungsi"=fungsi,"turunan"=turunan))
}

Gambarkan fungsi tersebut!

curve(fx,xlim=c(-3,3), col='steelblue',lwd=2)
abline(h=0)
abline(v=0)

b. Carilah akar-akar persamaan dengan menggunakan metode fixed-point, newton-raphson, secant dan bisection!

#fixed-point
fixedpoint(ftn = fx_fi,x0 =0)

## At iteration 1 value of x is: -1 
## At iteration 2 value of x is: Inf 
## At iteration 3 value of x is: 0 
## At iteration 4 value of x is: -1 
## At iteration 5 value of x is: Inf 
## At iteration 6 value of x is: 0 
## At iteration 7 value of x is: -1 
## At iteration 8 value of x is: Inf 
## At iteration 9 value of x is: 0 
## At iteration 10 value of x is: -1 
## At iteration 11 value of x is: Inf 
## At iteration 12 value of x is: 0 
## At iteration 13 value of x is: -1 
## At iteration 14 value of x is: Inf 
## At iteration 15 value of x is: 0 
## At iteration 16 value of x is: -1 
## At iteration 17 value of x is: Inf 
## At iteration 18 value of x is: 0 
## At iteration 19 value of x is: -1 
## At iteration 20 value of x is: Inf 
## At iteration 21 value of x is: 0 
## At iteration 22 value of x is: -1 
## At iteration 23 value of x is: Inf 
## At iteration 24 value of x is: 0 
## At iteration 25 value of x is: -1 
## At iteration 26 value of x is: Inf 
## At iteration 27 value of x is: 0 
## At iteration 28 value of x is: -1 
## At iteration 29 value of x is: Inf 
## At iteration 30 value of x is: 0 
## At iteration 31 value of x is: -1 
## At iteration 32 value of x is: Inf 
## At iteration 33 value of x is: 0 
## At iteration 34 value of x is: -1 
## At iteration 35 value of x is: Inf 
## At iteration 36 value of x is: 0 
## At iteration 37 value of x is: -1 
## At iteration 38 value of x is: Inf 
## At iteration 39 value of x is: 0 
## At iteration 40 value of x is: -1 
## At iteration 41 value of x is: Inf 
## At iteration 42 value of x is: 0 
## At iteration 43 value of x is: -1 
## At iteration 44 value of x is: Inf 
## At iteration 45 value of x is: 0 
## At iteration 46 value of x is: -1 
## At iteration 47 value of x is: Inf 
## At iteration 48 value of x is: 0 
## At iteration 49 value of x is: -1 
## At iteration 50 value of x is: Inf 
## At iteration 51 value of x is: 0 
## At iteration 52 value of x is: -1 
## At iteration 53 value of x is: Inf 
## At iteration 54 value of x is: 0 
## At iteration 55 value of x is: -1 
## At iteration 56 value of x is: Inf 
## At iteration 57 value of x is: 0 
## At iteration 58 value of x is: -1 
## At iteration 59 value of x is: Inf 
## At iteration 60 value of x is: 0 
## At iteration 61 value of x is: -1 
## At iteration 62 value of x is: Inf 
## At iteration 63 value of x is: 0 
## At iteration 64 value of x is: -1 
## At iteration 65 value of x is: Inf 
## At iteration 66 value of x is: 0 
## At iteration 67 value of x is: -1 
## At iteration 68 value of x is: Inf 
## At iteration 69 value of x is: 0 
## At iteration 70 value of x is: -1 
## At iteration 71 value of x is: Inf 
## At iteration 72 value of x is: 0 
## At iteration 73 value of x is: -1 
## At iteration 74 value of x is: Inf 
## At iteration 75 value of x is: 0 
## At iteration 76 value of x is: -1 
## At iteration 77 value of x is: Inf 
## At iteration 78 value of x is: 0 
## At iteration 79 value of x is: -1 
## At iteration 80 value of x is: Inf 
## At iteration 81 value of x is: 0 
## At iteration 82 value of x is: -1 
## At iteration 83 value of x is: Inf 
## At iteration 84 value of x is: 0 
## At iteration 85 value of x is: -1 
## At iteration 86 value of x is: Inf 
## At iteration 87 value of x is: 0 
## At iteration 88 value of x is: -1 
## At iteration 89 value of x is: Inf 
## At iteration 90 value of x is: 0 
## At iteration 91 value of x is: -1 
## At iteration 92 value of x is: Inf 
## At iteration 93 value of x is: 0 
## At iteration 94 value of x is: -1 
## At iteration 95 value of x is: Inf 
## At iteration 96 value of x is: 0 
## At iteration 97 value of x is: -1 
## At iteration 98 value of x is: Inf 
## At iteration 99 value of x is: 0 
## At iteration 100 value of x is: -1 
## Algorithm failed to converge

## NULL

newtonraphson(ftn=fx_nr,x0=0)

## At iteration 1 value of x is: -1 
## At iteration 2 value of x is: -0.5 
## At iteration 3 value of x is: -3 
## At iteration 4 value of x is: -2.038462 
## At iteration 5 value of x is: -1.390282 
## At iteration 6 value of x is: -0.9116119 
## At iteration 7 value of x is: -0.3450285 
## At iteration 8 value of x is: -1.427751 
## At iteration 9 value of x is: -0.9424179 
## At iteration 10 value of x is: -0.4049494 
## At iteration 11 value of x is: -1.706905 
## At iteration 12 value of x is: -1.155756 
## At iteration 13 value of x is: -0.6941918 
## At iteration 14 value of x is: 0.7424943 
## At iteration 15 value of x is: 2.781296 
## At iteration 16 value of x is: 1.982725 
## At iteration 17 value of x is: 1.536927 
## At iteration 18 value of x is: 1.357262 
## At iteration 19 value of x is: 1.325663 
## At iteration 20 value of x is: 1.324719 
## Algorithm converged

##   fungsi 
## 1.324719

secant(fx,x0=1,x1=3)

## At iteration 1 value of x is: 1.083333 
## Algorithm failed to converge

## NULL

bisection(fx,xl=-3,xr=3)

## At iteration 1 value of x is: 0 
## At iteration 2 value of x is: 1.5 
## At iteration 3 value of x is: 0.75 
## At iteration 4 value of x is: 1.125 
## At iteration 5 value of x is: 1.3125 
## At iteration 6 value of x is: 1.40625 
## At iteration 7 value of x is: 1.359375 
## At iteration 8 value of x is: 1.335938 
## At iteration 9 value of x is: 1.324219 
## At iteration 10 value of x is: 1.330078 
## At iteration 11 value of x is: 1.327148 
## At iteration 12 value of x is: 1.325684 
## At iteration 13 value of x is: 1.324951 
## At iteration 14 value of x is: 1.324585 
## At iteration 15 value of x is: 1.324768 
## At iteration 16 value of x is: 1.324677 
## At iteration 17 value of x is: 1.324722 
## At iteration 18 value of x is: 1.324699 
## At iteration 19 value of x is: 1.324711 
## At iteration 20 value of x is: 1.324717 
## At iteration 21 value of x is: 1.324719 
## At iteration 22 value of x is: 1.324718 
## At iteration 23 value of x is: 1.324717 
## At iteration 24 value of x is: 1.324718 
## At iteration 25 value of x is: 1.324718 
## At iteration 26 value of x is: 1.324718 
## At iteration 27 value of x is: 1.324718

## [1] 1.324718

Tugas Mandiri

Lakukan Modifikasi pada fungsi newtonraphson sehingga output dari fungsi tersebut berupa list dengan tiga object:

Object pertama bernama root berisi akar persamaan.
Object kedua bernama x_values berisi semua nilai x yang dicobakan
object ketiga bernama error berisi nilai dari hasil operasi abs(fx[1]) Gunakan persamaan pada contoh soal 1 untuk menujukkan hasil outputnya!

Selidiki penggunaan fungsi browser() di R. Gunakan fungsi newtonraphson untuk ilustrasi penggunaan. Laporkan hasil penyelidikanmu beserta ilustrasi koding!
Buatlah gambar fungsi pada contoh soal 2 dengan menggunakan package ggplot dengan menggunakan fungsi geom_curve!