在R中使用二分法查找函数的根

Question

I'm trying to find the root of the following function in R f <- x^3 + 2 * x^2 - 7 using the bisection method and the repeat function. This code results in an error:

x <- 1.3
tolerance <- 0.000001

repeat {
  f <- x^3 + 2 * x^2 - 7
  if (abs(f) < tolerance) break
  x <- (x^3 + 2 * x^2 - 7)/2
}

Error in if (abs(f) < tolerance) break : 
  missing value where TRUE/FALSE needed

I've set the initial x to be 1.3, the tolerance to be 0.000001 and I know that the root lies between 1 and 2. I have already tried to substitute the last line of the code for f instead of retyping the function, but the same error appears. Can someone help me?

Answer 1

Based on a very brief reading of the bisection method, I think you're adjusting x incorrectly. You should be bisecting the domain of x (the x value fed into f), not the range of f .

There are many reasons your function does not do what you want, but a primary one is that you are not even using the information you have about reasonable values for x, that is values of x that are near the root of the function. You should never be setting your x value to some value of the function for which you are trying to find a root...there's no reason these two values need to be related. For example, if the root of a function is near 100, the value of the function, f, for f(100) will be some low number. Then perhaps the value of f near 0 is some very high number. So if you start with f(x=100), you'll move x to around 0, then run f(0) and get some very big number so you'll move x to that big number, and so on. You'll be bouncing around according to f's values but not in a way that has anything to do with finding the root.

Answer 2

Let's try this:

x <- 1.3
tolerance <- 0.001

repeat {
  message(x)
  f <- x^3 + 2 * x^2 - 7
  if (abs(f) < tolerance) break
  x <- (x^3 + 2 * x^2 - 7)/2
  }
#1.3
#-0.7115
#-3.1738598729375
#-9.4123720943868
#-331.84120816267
#-18160827.4603406
#-2.99486226359293e+21
#-1.34307592742051e+64
#-1.21135573473788e+192
#Error in if (abs(f) < tolerance) break : 
#  missing value where TRUE/FALSE needed

As you see the x value becomes more and more negative until it's absolute value is too large for double:

x <- -1.21135573473788e+192
x^3 + 2 * x^2 - 7
#[1] NaN

You should look up the algorithm of the bisection method, because what you have implemented here is clearly not the correct algorithm.