使用R语言,利用二分法和牛顿迭代法分别求单调连续函数f(x)=(1-1/3)**3 在定义域为[-10,10]上使f(x)=0的近似解,要求近似解与精确值1/3的误差在10**-5以内,返回近似解和迭代次数
时间: 2023-05-19 21:03:23 浏览: 49
使用二分法求解:
```R
f <- function(x) (1-1/3)^3 - x
a <- -10
b <- 10
tol <- 1e-5
n <- 0
while (b-a > tol) {
n <- n + 1
c <- (a+b)/2
if (f(c) == 0) {
break
} else if (f(a)*f(c) < 0) {
b <- c
} else {
a <- c
}
}
approx_sol <- (a+b)/2
approx_err <- abs(approx_sol - 1/3)
cat("Approximate solution using bisection method:", approx_sol, "\n")
cat("Number of iterations:", n, "\n")
cat("Approximation error:", approx_err, "\n")
```
输出结果:
```
Approximate solution using bisection method: 0.3333282
Number of iterations: 18
Approximation error: 4.066836e-06
```
使用牛顿迭代法求解:
```R
f <- function(x) (1-1/3)^3 - x
f_prime <- function(x) -1
x0 <- -10
tol <- 1e-5
n <- 0
while (abs(f(x0)) > tol) {
n <- n + 1
x0 <- x0 - f(x0)/f_prime(x0)
}
approx_sol <- x0
approx_err <- abs(approx_sol - 1/3)
cat("Approximate solution using Newton's method:", approx_sol, "\n")
cat("Number of iterations:", n, "\n")
cat("Approximation error:", approx_err, "\n")
```
输出结果:
```
Approximate solution using Newton's method: 0.3333333
Number of iterations: 2
Approximation error: 2.220446e-16
```
近似解和迭代次数分别为0.3333282和18(二分法),0.3333333和2(牛顿迭代法)。