matlab 方程式求根

Question

好心的大大幫我看一下

最後一行,我想求方程式的根Dfalpha0 = -68+520*alpha0

但顯示錯誤,不知哪裡搞錯了

clc

clear

syms f Mx0 x1 x2 a b alpha0 falpha0 Dfalpha0 alpha0

function1 = input('please input a function :');

a1 = diff(function1,x1)

b1 = diff(function1,x2)

x1 = input('please input x1:')

x2 = input('please input x2:')

Mx0 = [x1;x2]

x11 = subs(a1,x1)

x22 = subs(b1,x2)

Mv0 = -[x11;x22]

x1 = x1-alpha0*x11

x2 = x2-alpha0*x22

falpha0 = subs(function1,{x1,x2},{sym('x1'),sym('x2')})

Dfalpha0 = diff(falpha0,alpha0)

alpha0 = roots(Dfalpha0)

執行

please input a function :(x1)^2+4*(x2)^2

a1 =

2*x1

b1 =

8*x2

please input x1:1

x1 =

1

please input x2:1

x2 =

1

Mx0 =

1

1

x11 =

2

x22 =

8

Mv0 =

-2

-8

x1 =

1-2*alpha0

x2 =

1-8*alpha0

falpha0 =

(1-2*alpha0)^2+4*(1-8*alpha0)^2

Dfalpha0 =

-68+520*alpha0

??? Error using ==> find

Function 'find' is not defined for values of class 'sym'.

Error in ==> C:\MATLAB6p5\toolbox\matlab\polyfun\roots.m

On line 22 ==> inz = find(c);

Error in ==> C:\MATLAB6p5\toolbox\new data\text1.m

On line 18 ==> alpha0 = roots(Dfalpha0)

已更新項目:

剛剛有用fzero試看看可以還是一樣

沒辦法滴說~

請大大在幫忙一下嚕~感激!!

Answer 1

最佳解答

roots指令不適用於 symbolic math

2007-11-11 18:35:54 補充：

而且 roots 指令只能用於求多項式的根

非線性方程式求根 請用 fzero 指令

2007-11-11 22:37:36 補充：

fzero 也不能用在 symbolic math

你要用 solve

2007-11-11 22:59:12 補充：

clc

clear

syms f Mx0 x1 x2 a b alpha0 falpha0 Dfalpha0 alpha0

function1 = input('please input a function :');

a1 = diff(function1,x1)

b1 = diff(function1,x2)

x1 = input('please input x1:')

x2 = input('please input x2:')

Mx0 = [x1;x2]

x11 = subs(a1,x1)

x22 = subs(b1,x2)

Mv0 = -[x11;x22]

x1 = x1-alpha0*x11

x2 = x2-alpha0*x22

falpha0 = subs(function1,{x1,x2},{sym('x1'),sym('x2')})

Dfalpha0 = diff(falpha0,alpha0)

alpha0 = solve(Dfalpha0)

最後一行改成 solve 就可

我存檔成為 find_root.m

執行 >>find_root

結果如下

please input a function :x1^2+x2^2

a1 =

2*x1

b1 =

2*x2

please input x1:2

x1 =

2

please input x2:3

x2 =

3

Mx0 =

2

3

x11 =

4

x22 =

6

Mv0 =

-4

-6

x1 =

2-4*alpha0

x2 =

3-6*alpha0

falpha0 =

(2-4*alpha0)^2+(3-6*alpha0)^2

Dfalpha0 =

-52+104*alpha0

alpha0 =

1/2

就如同我在意見中所說 roots 只適用於求多項式的根

例如 求 x^3+3*x^2-2*x-10=0 利用 roots 求根

>> p=[1,3,-2,-10];

>> x=roots(p) 就可求出3個根

或是利用 fzero 一次求一個根

>> f=@(x) x^3+3*x^2-2*x-10;

>> x1=fzero(f,0.1) %0.1是隨便猜的

fzero 只能求出實數解 無法求出虛數解

其實 fzero 使用了 bisection, Newton-Raphson, inverse quadratic interpolation 等方法 MathWorks公司已經最適化了

詳細可以參考數值分析課本