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发表于 2019-11-28 23:57| 字数 283
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[091]江雪 发表于 2019-11-28 23:47
emm
让我用一个 naive code 来算 π
ang = 0:0.01:2*pi;
r = zeros(size(ang));
N = length(ang);
r(1) = 1;
syms x y
f = (x^2-1)^3/(x^5) - sin(y)^3 * cos(y)^2;
for i = 2 : N
f1 = subs(f, y, ang(i));
rlt = solve(f1);
rlt = eval(rlt);
[~, ind] = min(abs(rlt-r(i-1)));
r(i) = rlt(ind);
end
x = r .* cos(ang);
y = r .* sin(ang);
plot(x,y); |
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