資源簡介
可視化(GUI) 非線性方程組 牛頓求解法 matlab ecjtu
代碼片段和文件信息
function?x=newton_solve(Fvx0e)?
max1=20;
if?(~ischar(F)?||?~ischar(v))?%定義類型
????error(‘Parameter?F?and?v?should?be?char?type!‘);?
end?
if?(~isnumeric(x0)?||?~isnumeric(e))?
????error(‘Parameter?x0?and?e?should?be?numeric?type!‘);?
end?
F=sym(F);v=sym(v);
dF=jacobian(Fv);
for?index=1:numel(v)?
????%把x分量化,把x1變成x(1),把x2變成x(2)
????F=subs(Fv(index)[‘X(‘?num2str(index)?‘)‘]);
????dF=subs(dFv(index)[‘X(‘?num2str(index)?‘)‘]);
end
%dFF
X=x0‘;
P=X;
Y=eval(F‘)
for?k=1:max1
????X=P
????J=eval(dF)
????J\Y‘
????Q=P-(J\Y‘)‘;
????X=Q;
????Z=eval(F‘);
????QP
????err=norm(Q-P);
????P=Q;
????Y=Z;
????k
????if?(err ????????break
????end
end
X=P;
eval(F‘)
x=P
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件?????????726??2012-06-20?18:11??linfeng1.m
?????文件????????1287??2012-06-20?17:28??wangqiu.m
?????文件?????????336??2012-06-20?18:17??新建文本文檔.txt
?????文件??????385024??2012-06-19?15:03??課程設計.doc
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