Newton's Method missing value

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Isaac Soria
Isaac Soria 2022-9-9
编辑： Torsten 2022-9-9
I wrote Newton's method to be
function [cVec,n] = Newtons_method(f,fp,x0,tol,nmax)
cVec = [];
cVec(1)=x0;
i=0;
while i <=nmax
i=i+1;
cVec(i)=x0;
c = x0-(f(x0)/fp(x0));
if abs(c-x0)<tol
c=x0;
break;
else
x0=c;
end
end
n=i-1;
cVec=cVec';
end
and used the following to call it
f=@(x)(x.^2-3);
fp=@(x)(2.*x);
x0=2;
tol=10^-10;
nmax= 100;
[c,n]=Newtons_method(f,fp,x0,tol,nmax)
From this, I get a [5,1] column vector but I'm supposed to get a [6,1] vector. I've messed around with the placement of certain lines of code and indices, but haven't gotten the desired result.
Thanks in advance.

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回答（1 个）

Torsten
Torsten 2022-9-9
编辑：Torsten 2022-9-9
Ran in:
If n is the number of Newton steps, you must start with n = 0 instead of n = 1 in "Newtons_method".
I counted the number of Newton steps + the initial value, thus the length of the array cVec.
f = @(x) x.^2-3;
fp = @(x) 2.*x;
x0 = 2;
tol = 10^-10;
nmax = 100;
format long
[c,n] = Newtons_method(f,fp,x0,tol,nmax)
c = 6×1
2.000000000000000 1.750000000000000 1.732142857142857 1.732050810014728 1.732050807568877 1.732050807568877
n =
6
function [cVec,n] = Newtons_method(f,fp,x0,tol,nmax)
cVec(1) = x0;
n = 1;
while n <= nmax
c = x0 - f(x0)/fp(x0);
n = n+1;
cVec = [cVec;c];
if abs(c-x0) < tol
break;
else
x0 = c;
end
end
end

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