2019年2月25日 星期一

julia語言 牛頓法(Newton - Raphson)以圖形概念主要是不停在取切線斜率。

julia語言 牛頓法以圖形概念主要是不停在取切線斜率。

虛擬碼
Algorithm NewtonRoot

    E0 : 初始化最小誤差 EPS,初始點 xo

    E1 :  x = x0

    E2 :  x0 = x - f(x) / f'(x)

    E3 :  if abs(x-x0) < eps,演算法結束,傳回 x

    E4 :  goto E1

End Algorithm


#===========================================================
> func(-1.781503813804761e+000) = +0.000000000000000e+000
> func(+2.313837835588586e+000) = +0.000000000000000e+000
> func(+4.947665978216175e+000) = +3.552713678800501e-015
> func(-1.781503813804761e+000) = +0.000000000000000e+000


Julia has no do-while construct. Here is one of several ways to implement do-while behavior.

julia> i = 0
0

julia> while true
           println(i)
           i += 1
           i % 6 == 0 && break
       end

===========================================================#

using Printf

#[ -2.00 , -1.00 ] , [ 2.00 , 3.00 ] , [ +4.00 , +5.00 ]

function func(x::Float64)
    x2=x*x
    x3=x2*x;
    return (x3 - 5.48*x2 -  1.4883*x + 20.394828)
end


# funcd(x) = func'(x)
function funcd(x::Float64)
    x2=x*x
    return (3*x2-10.96*x-1.4883)
end

#-------------------------------------------------------

function NewtonRoot(x0::Float64,eps::Float64)     #/*   初點    容許誤差*/
   
    x=x0;
    while true
        x0=x;
        x = x0 - func(x0) / funcd(x0)
        (abs(x-x0)<eps) && break
    end   
   
    return x
end   

eps=1.0E-9
max_iterator=100

x0 = -2.0
y0 = NewtonRoot(x0, eps)
s=@sprintf("\n> func(%+.15e) = %+.15e", y0, func(y0))
println(s)

x0 = 2.0
y0 = NewtonRoot(x0, eps)
s=@sprintf("\n> func(%+.15e) = %+.15e", y0, func(y0))
println(s)


x0 = 4.0
y0 = NewtonRoot(x0, eps)
s=@sprintf("\n> func(%+.15e) = %+.15e", y0, func(y0))
println(s)



x0 = -10.0  #// test
y0 = NewtonRoot(x0, eps)
s=@sprintf("\n> func(%+.15e) = %+.15e", y0, func(y0))
println(s)


輸出畫面
> func(-1.781503813804761e+00) = +0.000000000000000e+00

> func(+2.313837835588586e+00) = +0.000000000000000e+00

> func(+4.947665978216175e+00) = +3.552713678800501e-15

> func(-1.781503813804761e+00) = +0.000000000000000e+00

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