Julia語言例題 EX2-8 定點回路法 求非線性方程式 f(x)=exp(x) - 3x^2 = 0

Julia語言例題 EX2-8 定點回路法 求非線性方程式 f(x)=exp(x) - 3x^2 = 0

'''
 定點回路法
/* ex2-8.jl is used for solving nonlinear equation
 * based on Fixed-Point Algorithm g(x)=x with initial
 * approximation P0.
 */

  定點回路法 求非線性方程式 f(x)=math.exp(x) - 3x^2 = 0
  可以改寫成
    y= x
    y= (esp(x)/3) ^ 0.5

    與
    y= x
    y= - (esp(x)/3) ^ 0.5

using Printf

#=======================================================
 * ex2-8.jl is used for solving nonlinear equation
 * based on Fixed-Point Algorithm g(x)=x with initial
 * approximation P0.
=========================================================#
MAX=50
TOL=0.0001

function g1(x0::Float64)
    return ( (exp(x)/3)^0.5)
end   

function g2(x0::Float64)
    return (-(exp(x)/3)^0.5)
end   


i=1
x0=0.0
x=0.0
while(i<=MAX)
    x=g1(x0)
    s=@sprintf("%-2d  %10.7lf",i-1,x0)
    println(s)
    if(abs(x-x0) < TOL)
        s=@sprintf("The Root=%10.7lf  x-x0=%10.7lf",x,abs(x-x0))
        println(s)
        break
    end   
    i+=1
    x0=x
end
s=@sprintf("Fixed-point failed after %d iteration.\n",i)
println(s)


i=1
x0=0.0
x=0.0
while(i<=MAX)
    x=g2(x0)
    s=@sprintf("%-2d  %10.7lf",i-1,x0)
    println(s)
    if(abs(x-x0) < TOL)
        s=@sprintf("The Root=%10.7lf  x-x0=%10.7lf",x,abs(x-x0))
        println(s)
        break
    end   
    i+=1
    x0=x
end
s=@sprintf("Fixed-point failed after %d iteration.\n",i)
println(s)




輸出畫面

0    0.0000000
1    0.5773503
2    0.7705652
3    0.8487220
4    0.8825453
5    0.8975975
6    0.9043784
7    0.9074499
8    0.9088446
9    0.9094786
10   0.9097669
11   0.9098981
The Root= 0.9099578  x-x0= 0.0000597
Fixed-point failed after 12 iteration.

0    0.0000000
1   -0.5773503
2   -0.4325829
3   -0.4650559
4   -0.4575660
5   -0.4592828
6   -0.4588887
The Root=-0.4589791  x-x0= 0.0000904
Fixed-point failed after 7 iteration.