Newton Downhill Method to Find the Root of Nonlinear Equation

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

`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 |

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

a root of the objective function

Maintainer:Zheng Sengui<1225620446@qq.com>

Zheng Sengui,Lu Xufen,Hou Qiongchen,Zheng Jianhui

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

`BFfzero`

,`NIMfzero`

,`SMfzero`

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 | ```
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")
}
``` |

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