initial guess in a loop

B
B
2015 5 6
1 个回答
更新时间：2015 5 6
3 次查看（30 天）

0 个投票

the initial guess in the following loop is actually inside the loop, why is that?
function [root,ea,iter]=newtraph(func,dfunc,xr,es,maxit,varargin)
if nargin<3,error('at least 3 input arguments required'),end
if nargin<4|isempty(es),es=0.0001;end
if nargin<5|isempty(maxit),maxit=50;end
iter = 0;
while (1)
xrold = xr;
xr = xr - func(xr)/dfunc(xr);
iter = iter + 1;
if xr ~= 0, ea = abs((xr - xrold)/xr) * 100; end
if ea <= es | iter >= maxit, break, end
end
root = xr;
=======================
thanks a lot

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回答（1 个）

Mischa Kim
Mischa Kim 2015-5-6
编辑：Mischa Kim 2015-5-6

0 个投票

B, the initial guess is provided through the newtraph call. In other words, in order to execute/call this function you need to provide an xr value. That is the initial value that is then updated in the loop.

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B
B 2015-5-6
thanks but why don't we place xrold=xr outside the loop?
Stephen23
Stephen23 2015-5-6
编辑：Stephen23 2015-5-6
@B: because the variable xrold is not the initial value, it is actually the previous value. On every loop it is assigned to be the value of xr from the last loop iteration. Only on the first iteration is it the "initial" value.
Mischa Kim
Mischa Kim 2015-5-6
With xrold = xr you keep track of the last value of xr so you can compute how the solution changes. If the difference between consecutive solutions of xr becomes smaller than es (= 0.0001) then the problem is declared solved and the loop is exited.
B
B 2015-5-6
Thanks a lot Stephen and Mischa. I got now :)

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