Newton's method problem
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clear all; close all;
a=4;
b=3;
x=zeros();
x(1)=1;
for i=1:1000;
f(x(i))=(x(i)-a)^2+b;
x(i+1) =x(i)-(f(x(i))/diff(f(x(i))));
end
4 个评论
回答(3 个)
Andrei Bobrov
2011-10-7
for your case EDITED
f = @(x)(x - 4)^2 + 9;
syms x
ex = (x - 4)^2 + 9;
cf = double(coeffs(ex));
distxp = (4*prod(cf([1 end]))-cf(2)^2)/4/cf(3);
fun = matlabFunction(ex - distxp);
df = matlabFunction(diff(ex));
x = 10;
xout = x;
while abs(fun(x)) > 1e-6
x1 = x - fun(x)/df(x);
x = x1;
xout = [xout;x];
end
ADD corrected
f = @(x)(x - 4)^2 + 9;
syms x
ex = (x - 4)^2 + 9;
cf = fliplr(double(coeffs(expand(ex))));
distxp = (4*prod(cf([1 end]))-cf(2)^2)/4/cf(1);
fun = @(x)f(x)-distxp;
dcf = polyder(cf);
df = @(x)polyval(dcf,x);
x = 10;
xout = x;
while abs(fun(x)) > 1e-6
x1 = x - fun(x)/df(x);
x = x1;
xout = [xout;x];
end
output for the first 10 iterations and optimal x
if numel(xout)<=10
out = xout;
else
out = [xout(1:10);xout(end)];
end
7 个评论
Andreas Goser
2011-10-7
This code isn't working because diff(f(x(i))) returns [] and thus a scalar can't be divide by [].
Steve
2011-10-9
You can use this Newton function implementation in general.
function n = newton(f,fp,x0,tol,Nmax)
n=0;
test_val = abs(poly_val(f,x0));
num1 = 1;
while (test_val > tol && num1<Nmax)
n = x0 - poly_val(f,x0)/poly_val(fp,x0);
test_val = abs(poly_val(f,n));
x0=n;
num1=num1+1;
end
end
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