NDHfzero: Newton Downhill Method

Description Usage Arguments Details Value Note Author(s) References See Also Examples

Description

Newton Downhill Method to Find the Root of Nonlinear Equation

Usage

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NDHfzero(f, f1, x0 = 0, num = 1000, eps = 1e-05, eps1 = 1e-05)

Arguments

f

the objective function which we will use to solve for the root

f1

the derivative of the objective function (say f)

x0

the initial value of Newton iteration method or Newton downhill method

num

num the number of sections that the interval which from Brent's method devide into. num=1000 when it is default

eps

the level of precision that |x(k+1)-x(k)| should be satisfied in order to get the idear real root. eps=1e-5 when it is default

eps1

the level of precision that |f(x)| should be satisfied, where x comes from the program. when it is not satisified we will fail to get the root

Details

eps1 of precision that |f(x)| should be satisfied, where x comes from the program. when it is not satisified we will fail to get the root

Value

a root of the objective function

Note

Maintainer:Zheng Sengui<1225620446@qq.com>

Author(s)

Zheng Sengui,Lu Xufen,Hou Qiongchen,Zheng Jianhui

References

Luis Torgo (2003) Data Mining with R:learning by case studies. LIACC-FEP, University of Porto

See Also

BFfzero,NIMfzero,SMfzero

Examples

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f<-function(x){x^3-x-1};f1<-function(x){3*x^2-1};
NDHfzero(f,f1,2)

##---- Should be DIRECTLY executable !! ----
##-- ==>  Define data, use random,
##--	or do  help(data=index)  for the standard data sets.

## The function is currently defined as
function (f, f1, x0 = 0, num = 1000, eps = 1e-05, eps1 = 1e-05) 
{
    a = x0
    b = a - f(a)/f1(a)
    i = 0
    while ((abs(b - a) > eps)) {
        c = 1
        j = 0
        while (abs(f(b)) >= abs(f(a))) {
            b = a - c * f(a)/f1(a)
            j = j + 1
            c = 1/(2^j)
        }
        a = b
        b = a - f(a)/f1(a)
        c = 1
        j = 0
        while (abs(f(b)) >= abs(f(a))) {
            b = a - c * f(a)/f1(a)
            j = j + 1
            c = 1/(2^j)
        }
        i = i + 1
    }
    print(b)
    print(f(b))
    if (abs(f(b)) < eps1) {
        print("finding root is successful")
    }
    else print("finding root is fail")
  }

NLRoot documentation built on May 2, 2019, 7:31 a.m.

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