Description Usage Arguments Details Value Author(s) References Examples

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

1 |

`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. |

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)].

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

Bingpei Wu

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

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 | ```
##---- 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
}
``` |

```
[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
}
```

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