Question

使用以下规范实现bisection方法：

Input: Function f, values low and high, error range epsilon.
`enter code here`Precondition: low<high, f(low) and f(high) diffs on their signs; that is,
either: f(low)>0 and f(high)<0
or:     f(low)<0 and f(high)>0.
Output: x where |f(x)| < epsilon.

Test your implementation with the following input values:
f(x) = 2x^3-3x^2-17x-50
low = -10
high = 10
epsilon = 1*10^(-6)  

Run your program, print out the solution (approximation) one each iteration

我的代码，不确定这是否正确：

object IntervalHalving {

def main(args: Array[String]) {
  val low = -10
  val high = 10
  val epsilon = 1*10^(-6)

//top part seems correct//

//Not sure if i defined the function correctly//

  val f(x) = (x: Double) => x*x*x + x*x - 3*x-3

  val answer= halveTheInterval(f(x), low, high, epsilon)

   // print the answer
   println(answer)
}

Answer 1

我对scala了解不多，但val epsilon = 1*10^(-6)可能不是您的想法，而是xor(10,-6) = -16。请改用科学记数法：val epsilon = 1.0e-6。

Answer 2

请注意

scala> val epsilon = 1e-6
epsilon: Double = 1.0E-6

上面定义的epsilon评估为Int，其值为-16：）

和

scala> val f = (x: Double) => x*x*x + x*x - 3*x-3
f: Double => Double = <function1>

而不是val f(x) = ...另请考虑例如Math.pow(x,3)。

回想为object IntervalHalving添加一个结束花括号。

有人可以帮助改善我的Scala二分答案吗？怎么了？

2 个答案: