Question

我试图通过使用二分法来找到函数的根，说明：

if f(a)*f(b) < 0 then a root exists, 
then you repeat with f(a)*f(c)<0 where c = (a+b)/2

但我不确定如何修复代码以使其正常工作。这是我的代码，但它无法正常工作

from scipy import *
from numpy import *


def rootmethod(f, a, b, tol):


    x = a
    fa = sign(eval(f))

    x = b
    fb = sign(eval(f))

    c = a + b
    iterations = 0

    if fa == 0:
        return a
    if fb == 0:
        return b

    calls = 0         
    fx = 1

    while fx != 0:
        iterations = iterations + 1
        c *= 0.5
        x = a + c
        fc = sign(eval(f))
        calls = calls + 1

        if fc*fa >= 0:
            x = a
            fx = sign(eval(f))
        if fc == 0 or abs(sign(fc)) < eps:
            fx = sign(eval(f))
            return x, iterations, calls





print rootmethod("(x-1)**3 - 1", 1, 3, 10*e-15)

新编辑..但仍然无效

   if fa*fb < 0:

        while fx != 0:
            iterations = iterations + 1
            c = (a + b)/2.0
            x =  c
            fc = sign(eval(f))
            calls = calls + 1

            if fc*fa >= 0:
                x = c
                fx = sign(eval(f))
            if fc == 0 or abs(sign(fc)) < tol:
                fx = sign(eval(f))
                return x, iterations, calls

编辑：在方法说明中将c =（a + b）* 2更改为c =（a + b）/ 2.

Answer 1

我认为你的一个问题是：

    x = a + c

自c = (a + b)*.5以来，您无需在此处添加a ...

<强>更新

您似乎无法检查fa * fb < 0是否可以启动，而且我也看不到您缩小界限的位置：您应该重新指定a或b在循环中c，然后重新计算c。

代码自从我最后一次玩python以来已经有一段时间了，所以带上一粒盐^ _ ^

x = a
fa = sign(eval(f))

x = b
fb = sign(eval(f))

iterations = 0

if fa == 0:
    return a
if fb == 0:
    return b

calls = 0         
fx = 1

while fa != fb:
    iterations += 1
    c = (a + b)/2.0
    x = c
    fc = eval(f)
    calls += 1

    if fc == 0 or abs(fc) < tol:
        #fx = fc not needed since we return and don't use fx
        return x, iterations, calls
    fc = sign(fc)
    if fc != fa:
        b = c
        fb = fc
    else
        a = c
        fa = fc
#error because no zero is expected to be found

Answer 2

我相信你的循环应该是这样的（在伪代码中，并省略一些检查）：

before loop:
a is lower bound
b is upper bound
Establish that f(a) * f(b) is < 0

while True:
    c = (a+b)/2
    if f(c) is close enough to 0:
        return c
    if f(a) * f(c) > 0:
        a = c
    else
        b = c

换句话说，如果中点不是答案，那么根据其符号将其作为新端点之一。

Answer 3

坦率地说，你的代码有点混乱。这是一些有效的。阅读循环中的注释。（顺便说一下你给定函数的解是2，而不是3.75）

from scipy import *
from numpy import *


def rootmethod(f, a, b, tol):


  x = a
  fa = sign(eval(f))

  x = b
  fb = sign(eval(f))

  c = a + b
  iterations = 0

  if fa == 0:
    return a
  if fb == 0:
    return b

  calls = 0         
  fx = 1

  while 1:
    x = (a + b)/2
    fx = eval(f)

    if abs(fx) < tol:
      return x

    # Switch to new points.
    # We have to replace either a or b, whichever one will
    # provide us with a negative 
    old = b # backup variable
    b = (a + b)/2.0

    x = a
    fa = eval(f)

    x = b
    fb = eval(f)

    # If we replace a when we should have replaced b, replace a instead
    if fa*fb > 0:
      b = old
      a = (a + b)/2.0




print rootmethod("(x-1)**3 - 1", 1, 3, 0.01)

Answer 4

请注意，代码有一个由四舍五入错误引起的简单缺陷

a=0.015707963267948963 
b=0.015707963267948967
c=(a+b)*.5

c再次成为b（请查看！）。你可以最终进入无限循环如果容差非常小，如1e-16。

def FindRoot( fun, a, b, tol = 1e-16 ):
  a = float(a)
  b = float(b)
  assert(sign(fun(a)) != sign(fun(b)))  
  c = (a+b)/2
  while math.fabs(fun( c )) > tol:
    if a == c or b == c: 
      break
    if sign(fun(c)) == sign(fun(b)):
      b = c
    else:
      a = c
    c = (a+b)/2
  return c

现在，一遍又一遍地调用eval效率不高。这是你可以做的事情

expr = "(x-1.0)**3.0 - 1.0"
fn = eval( "lambda x: " + expr )
print FindRoot( fn, 1, 3 )

或者您可以将eval和lambda定义放在FindRoot中。

有用吗？

    Reson

查找函数的根/零

4 个答案: