R中的二分法

Question

我正在尝试使用Bisection方法求解方程。但是，当我尝试运行它时，我得到以下错误

"Error in if ((fn(kVec, tVec, b) * fn(kVec, tVec, a) > 0) | (b > a)) { : 
  argument is of length zero"

二分法例程

bisect<-function(kVec,tVec,fn,b,a,tol=1e-15){
  i<-0
  r<-(b+a)/2
  res<-c(i,r,fn(kVec,tVec,r))
  if ((fn(kVec,tVec,b)*fn(kVec,tVec,a)>0)|(b>a)) {
    return('Improper start values') }
  else
    while (abs(fn(kVec,tVec,r))>tol) {
      i<-i+1
      if (fn(kVec,tVec,b)*fn(kVec,tVec,r)>0) {
        b<-r
        r<-(b+a)/2 
      }
      else {
        a<-r
        r<-(b+a)/2 
      }
      res<-rbind(res,c(i,r,fn(kVec, tVec,r)))  
  }
return(res)
}

定义等式的函数

FCfunc<-function(kVec,tVec,b){
for(i in 1:k){
    ((kVec[i]*(tVec[i]*exp(-tVec[i])-tVec[i-1]*exp(-b*tVec[i-1])))
    }
}

使用一些初始估计值调用二分程序

bisect(kVec,tVec,FCfunc,0.00001,10.00001,tol=10e-16) 

Answer 1

二分法

    rm(list=ls())
    x0<-0
    x1<-1
    e<-0.001 #Error of tolerance
    fun<-function(x)(x^3-9*(x^2)+18*x-6) # An example function
    y0<-fun(x0);y0
    y1<-fun(x1);y1
    x2<-x1
    i<-1
    if(sign(y0)==sign(y1)){

        print("Starting vaules are not suitable")
    }else
    {
        while(fun(x2)!=0){
            x2<-(x0+x1)/2

            y2<-fun(x2)

            if(sign(y1)==sign(y2)){x1<-x2}else{x0<-x2}
                y0<-fun(x0)
                y1<-fun(x1)
                cat(i,x2,"\n") #Print the value obtained in each iteration next line 
                i<-i+1
            }
        }

上面的while循环停止为x2的值，函数变得完全等于零，现在如果你想在最短迭代时想要答案的错误你可以替换条件while（abs（x） ））GT; 0.001） 并运行整个程序。其中0.001表示允许的误差范围。

Answer 2

#bisection method
a<--give trial value-;a
b<--give trial value;b
f<-function(x){-give function-}
f(a)
f(b)
c<-(a+b)/2;c
while(f(c)!=0&&b-a>.00002)
{
    if(f(c)==0)
    {
        c
    }
    if(f(c)<0)
    {
        a<-c
    }
    else
    {
        b=c
    }
    {
        c<-(a+b)/2;c
    }
}
c

Answer 3

也许您会发现R中的二等分方法代码很有用

f.acc <- function(x){
1+1/x-log(x)
}
f.acc(0.5)
f.acc(6)
# since f.acc is continuous, it must have a root between 0.5 and 6.
x.left <- 0.5
x.right <- 6
iter <- 1
tol <- 1e-6
max.iter <- 100
while ((abs(x.right-x.left) > tol) && (iter < max.iter)) {
x.mid <- (x.left+x.right)/2
if (f.acc(x.mid)*f.acc(x.right)<0) {
x.left <- x.mid
} else {
x.right <- x.mid
}
iter <- iter + 1
cat("At iteration", iter, "value of x.mid is:", x.mid, "\n")
}

cat("At iteration", iter, "value of x.mid is:", x.mid, "and the function value is",
f.acc(x.mid),"\n")

x.n.minus.1 <- 0.5
x.n <- 6
iter <- 1
while ((abs(x.n-x.n.minus.1) > tol) && (iter < max.iter)) {
x.n.plus.1 <- x.n - f.acc(x.n)*(x.n-x.n.minus.1)/(f.acc(x.n)-f.acc(x.n.minus.1))
x.n.minus.1 <- x.n
x.n <- x.n.plus.1
iter <- iter + 1
cat("At iteration", iter, "value of x.n is:", x.n, "\n")
}

cat("At iteration", iter, "value of x is:", x.n, "and the function value is",
f.acc(x.n),"\n")