Use matlab to realize: chord cut method, dichotomy, CG method, find zero point and solve equation

String cutting

Example

seek $x{e}^x-1=0$ The root of the , Initial value $x_0=0.5,x_1=0.6.$

Realization

Its mathematical iteration formula is ：
$x_{k+1}=x_k-\frac{f(x_k)}{f(x_k)-f(x_{k-1})}\cdot (x_k-x_{k-1})$

Use Matlab Realization ：

f = inline('x * e^x - 1');
x = 0;
xx = 0.5;
xxx = 0.6;
i=0;
while abs(xx-xxx) >= 1e-5
	x = xx;
	xx = xxx;
	xxx = xx - f(xx) / (f(xx) - f(x)) * (xx - x);
	i = i + 1;
end

After iteration 4 Time , obtain $x_5=0.56714$.
If you need to see more decimal places , have access to ：

format long g

Dichotomy

Example

Find the equation by dichotomy $2e^{-x}-sinx=0$ In the interval $[0,1]$ The inner root , requirement $|x_k-x^{\ast }|<\frac{1}{2}\times 10^{-5}$ .

Realization

First, find the endpoint value , Derivation , Make sure the function on the left is $[0,1]$ Have solution , And it is a single root interval .
take $c=0,d=1$ , take $[c,d]$ Bisection , Set the bisection point as $x_0=\frac{1}{2}(c+d)$ , Calculation $f(x_0)$ , If $f(x_0)$ And $f(c)$ Same number ,……
$d_n-c_n=\frac{1}{2^n}(d-c)$ , as long as n Large enough , If there is a root, the length of the interval will be small enough , When the length meets the accuracy requirements , take $x_n=\frac{1}{2}(c_n+d_n)$ As an approximation of the root of the equation .

Use Matlab Realization ：

f = inline('2 * e^(-x) - sin(x)');
c = 0;
d = 1;
x = 1/2 * (c + d);
i = 0;
while d - c >= 1e-5/2
    x = 1/2 * (c + d);
    i ++;
    if f(x) > 0
        c = x;
    elseif f(x) < 0
        d = x;
    else
        break;
    end;
end;

after 18 Sub iteration ,$c=x_{17}$ when ,$d-c<\frac{1}{2}\times 10^{-5}$ , The solution of the original equation is ：$x^{\ast }=\frac{1}{2}(c_{18}+d_{18})=0.92103$ .

CG Law

Example

take $x_0=(0,0{)}^T,useCGLawseekExplain：$
$$\left[\begin{array}{cc}6& 3\\ 3& 2\end{array}\right]\cdot \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ -1\end{array}\right]$$

Realization

The equations are reduced to ：$A\cdot x=b$ .

take $x_0=(0,0{)}^T,r_0=b-A{x}_0=(0,-1{)}^T,p_0=r_0$

${\alpha }_0=\frac{(r_0,r_0)}{(p_0,A{p}_0)}=\frac{1}{2}$

$x_1=x_0+{\alpha }_0\cdot p_0=(0,-\frac{1}{2}{)}^T$

$r_1=r_0-{\alpha }_0\cdot A\cdot p_0=(\frac{3}{2},0{)}^T$

${\beta }_0=\frac{(r_1,r_1)}{(r_0,r_0)}=\frac{9}{4}$

$p_1=r_1+\beta \cdot p_0=(\frac{3}{2},-\frac{9}{4}{)}^T$

${\alpha }_1=\frac{(r_1,r_1)}{(p_1,A{p}_1)}=\frac{2}{3}$

$x_2=x_1+{\alpha }_1\cdot p_1=(1,-2{)}^T$

$r_2=r_1-{\alpha }_1\cdot A\cdot p_1=(0,0{)}^T$

stay $r_k=0or(p_k,A{p}_k)=0$ when , Calculation termination , Yes $x_k=x^{\ast }$ .

Use Matlab Realization ：

x = [0;0];
A = [6,3;3,2];
b = [0;-1];
r = b - A * x;
p = r;
i = 0;
while !all(r == [0;0]) && !all(dot(p, A * p) == [0;0])
    alpha = dot(r, r) / dot(p, A * p);
    xx = x + alpha * p;
    rr = r - alpha * A * p;
    beta = dot(rr, rr) / dot(r, r);
    pp = rr + beta * p;
    i ++;
    
    r = rr;
    x = xx;
    p = pp;
end;