dichotomy: Find root(s) of the equation by dichotomy

In FindAllRoots: Find all root(s) of the equation and Find root(s) of the equation by dichotomy

Description

Find root(s) of the equation by dichotomy.Besides,in dichotomy, more than one interval can be given at a time.

Usage

dichotomy(x1, x2, a, b, pert = 10^(-5), n = 1000, s = 0.1)

Arguments

x1	vector of left end point of interval(s)
x2	vector of right end point of interval(s)
a	vector of coefficients of the equation
b	vector of exponention of the equation,One one corresponding with a mentioned above
pert	precision of root(s)
n	the algorithm runs n times at most in one interval and NA will be returned
s	assuming x0 is midpoint of interval [a,b].If f(x0)*f(a)>0 and f(x0)*f(b)>0,b will minus s.

Details

If you want to find root(s) of the equation in [a1,b1],[a2,b2],“',[an,bn],x1 should be c(a1,a2,“',an) and x2 should be c(b1,b2,“',bn). If there is no root in [a1,b1],but there is a root in [min(a1,b1-n*s),max(a1,b1-n*s)] ,the algorithm can still find the root.So the returned root may not in [an,bn] that you give but must be in [min(a1,b1-n*s),max(a1,b1-n*s)].

Value

the root(s) of the equation that the difference between returned root(s) and the real root(s) of the equation is less than 10e-6

Author(s)

Bingpei Wu

References

a passage about finding root(s) of equation ,whose author is Yong Ling

Examples

##---- Should be DIRECTLY executable !! ----
##-- ==>  Define data, use random,
##--	or do  help(data=index)  for the standard data sets.
a=c(2,-1,-13,-1,-5)
b=c(4:0)
x1=c(1:10)
x2=c(2:11)
dichotomy(x1,x2,a,b)

## The function is currently defined as
function (x1, x2, a, b, pert = 10^(-5), n = 1000, s = 0.1) 
{
    x0 = rep(NA, length(x1))
    for (i in 1:length(x1)) {
        if (f(x1[i], a, b) == 0) 
            x0[i] = x1[i]
        if (f(x2[i], a, b) == 0) 
            x0[i] = x2[i]
        if (f(x1[i], a, b) != 0 & f(x2[i], a, b) != 0) {
            x0[i] = (x1[i] + x2[i])/2
            k = 1
            while ((abs(f(x0[i], a, b)) >= pert) & (k < n)) {
                if (f(x0[i], a, b) == 0) 
                  break
                if (f(x1[i], a, b) * f(x0[i], a, b) < 0) 
                  x2[i] = x0[i]
                if (f(x2[i], a, b) * f(x0[i], a, b) < 0) 
                  x1[i] = x0[i]
                if (x1[i] != x0[i] & x2[i] != x0[i]) 
                  x2[i] = x2[i] - s
                x0[i] = (x1[i] + x2[i])/2
                k = k + 1
                if (k == 1000) 
                  x0[i] = NA
            }
        }
    }
    x0
  }

Example output

 [1] -2.357713  2.901739  2.901739  2.901739  2.901739  2.901739  2.901739
 [8]  2.901739  2.901739  2.901739
function (x1, x2, a, b, pert = 10^(-5), n = 1000, s = 0.1) 
{
    x0 = rep(NA, length(x1))
    for (i in 1:length(x1)) {
        if (f(x1[i], a, b) == 0) 
            x0[i] = x1[i]
        if (f(x2[i], a, b) == 0) 
            x0[i] = x2[i]
        if (f(x1[i], a, b) != 0 & f(x2[i], a, b) != 0) {
            x0[i] = (x1[i] + x2[i])/2
            k = 1
            while ((abs(f(x0[i], a, b)) >= pert) & (k < n)) {
                if (f(x0[i], a, b) == 0) 
                  break
                if (f(x1[i], a, b) * f(x0[i], a, b) < 0) 
                  x2[i] = x0[i]
                if (f(x2[i], a, b) * f(x0[i], a, b) < 0) 
                  x1[i] = x0[i]
                if (x1[i] != x0[i] & x2[i] != x0[i]) 
                  x2[i] = x2[i] - s
                x0[i] = (x1[i] + x2[i])/2
                k = k + 1
                if (k == 1000) 
                  x0[i] = NA
            }
        }
    }
    x0
}

FindAllRoots documentation built on May 2, 2019, 6:34 a.m.