# I need to change this code into one that can solve complex roots

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Elena Cecilia Valero Garza 2022 年 9 月 12 日
コメント済み: Torsten 2022 年 9 月 13 日
I have this Newton-Raphson method code, but I need to change it into a code that can actually solve polynomials and give as an answer complex roots.
Code:
function [tabla, raiz]=newtonraphsonMN(f,xa,errorD,imax)
tabla=[];
error=Inf;
i=0;
xr=NaN;
df=diff(f);
while error>errorD && i<=imax
i=i+1;
fxa=double(subs(f,xa));
fpxa=double(subs(df,xa));
xr=xa-fxa/fpxa;
error=100*abs((xr-xa)/xr);
tabla=[tabla; [xa fxa fpxa xr error]];
xa=xr;
end
%raiz
raiz=xr;

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### 採用された回答

John D'Errico 2022 年 9 月 12 日

Interesting. It does not work? :) Gosh, You could have fooled me. I'll try an example. (I've attached the code you gave, so it will be used.)
syms X
f = X^2 + X + 1;
Does f has complex roots?
solve(f)
ans =
vpa(ans)
ans =
Of course. I'd not have used an example that lacks complex roots, since that is your question.
[tabla, raiz]=newtonraphsonMN(f,1 + i,1e-12,100);
Strange.
tabla
tabla =
1.0e+02 * 0.0100 + 0.0100i 0.0200 + 0.0300i 0.0300 + 0.0200i 0.0008 + 0.0062i 1.6125 + 0.0000i 0.0008 + 0.0062i 0.0070 + 0.0071i 0.0115 + 0.0123i -0.0052 + 0.0063i 0.7267 + 0.0000i -0.0052 + 0.0063i 0.0035 - 0.0002i -0.0003 + 0.0126i -0.0049 + 0.0091i 0.2687 + 0.0000i -0.0049 + 0.0091i -0.0008 + 0.0001i 0.0001 + 0.0182i -0.0050 + 0.0087i 0.0424 + 0.0000i -0.0050 + 0.0087i -0.0000 + 0.0000i 0.0000 + 0.0173i -0.0050 + 0.0087i 0.0010 + 0.0000i -0.0050 + 0.0087i -0.0000 + 0.0000i 0.0000 + 0.0173i -0.0050 + 0.0087i 0.0000 + 0.0000i -0.0050 + 0.0087i -0.0000 + 0.0000i 0.0000 + 0.0173i -0.0050 + 0.0087i 0.0000 + 0.0000i -0.0050 + 0.0087i 0.0000 + 0.0000i 0.0000 + 0.0173i -0.0050 + 0.0087i 0.0000 + 0.0000i
format long
raiz
raiz =
-0.500000000000000 + 0.866025403784439i
To me, it seems to have worked. But then, what do I know? :)
(Hint: Do you see what I did different?)
##### 2 件のコメント表示非表示 1 件の古いコメント
Torsten 2022 年 9 月 13 日
To solve for polynomial roots, use MATLAB's "roots" command.
Newton's method will only give you one solution per call. Which one you get depends on the initial guess.
Example:
To determine all roots of p(x) = x^2+x+1:
p = [1 1 1];
sol = roots(p)
sol =
-0.5000 + 0.8660i -0.5000 - 0.8660i
polyval(p,sol)
ans = 2×1
1.0e-15 * 0.3331 0.3331

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