matlab 求解???

Question

matlab有障礙，老師講都聽嘸，請求協助!!!

題目：y=x+2*exp(-x)-3

請用二分法、牛頓法、遞近法、正割法求解!!!

Answer 1

最佳解答

function [y dy]= func(x) y = x + 2*exp(-x) - 3;dy = 1 - 2*exp(-x);

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1. 二分法

function r = bisect(fun,xb)

xeps = 5*eps;

feps = 5*eps;a = xb(1);

b = xb(2);

xref = abs(b - a);

fa = feval(fun,a);

fb = feval(fun,b);

fref = max([abs(fa) abs(fb)]);if sign(fa)==sign(fb) % Verify sign change in the interval

error(sprintf('Root not bracketed by [%f, %f]',a,b));

endk = 0; maxit = 50;

while k < maxit

k = k + 1;

dx = b - a;

xm = a + 0.5*dx;

fm = feval(fun,xm); if (abs(fm)/fref < feps) | (abs(dx)/xref < xeps)

r = xm; return;

end

if sign(fm)==sign(fa)

a = xm; fa = fm;

else

b = xm; fb = fm;

end

end

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2. 牛頓法

function r = newton(fun,x0)xeps = 5*eps;

feps = 5*eps;x = x0; k = 0; maxit = 15;

while k <= maxit

k = k + 1;

[f,dfdx] = feval(fun,x);

dx = f/dfdx;

x = x - dx;

if ( abs(f) < feps ) | ( abs(dx) < xeps ), r = x; return; end

end

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3. 遞近法

function xnew = fixpoint(func, xold)k = 0; kmaxit = 250; tolx = 1;

while k < kmaxit && tolx > 1e-6

xnew = xold - feval(func,xold);

tolx = abs(xnew - xold);

xold = xnew;

k = k + 1;

end

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4. 正割法

function x = secant(func, a, b)xk = b;

xkm1 = a;

fk = feval(func,b);

fkm1 = feval(func,a);k = 0; maxit = 15; tolful = 1;

while k <= maxit && tolful > 1e-6

x = xk - fk*(xk - xkm1)/(fk - fkm1);

tolful = abs(fk - fkm1);

f = feval(func,x);

xkm1 = xk; xk = x;

fkm1 = fk; fk = f;

k = k + 1;

end

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我可以找出兩個根(-0.5831, 2.8887), 你試看看，關於fixpoint

你可以試試

xnew = xold - feval(func,xold)/ (feval(func,xold) - 3);

可以找出另一根

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參考資料： me