Newton's iteration

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darlene yen
darlene yen 2017 年 11 月 8 日
コメント済み: Torsten 2017 年 11 月 8 日
I am trying to do Newton's method of iteration. I have f(TH,R) = @(TH,R)(abs((sqrt(2)/(81*sqrt(pi))).*(6.*R-R.^2).*exp(-R./3).*cos(TH))).^2
%df/dTH = -2*sin(TH)*cos(TH)*(2/(pi*81^2)*(6*R-R^2)^2)*exp(-2*R/3)
%df/dR = -4*exp(-2*R/3)*(R-6)*R*(R^2-12*R+18)*sin(2*TH)/(19683*pi)
%d2f/dTH2 = -4*exp(-2*R/3)*(6*R-R^2)^2*sin(TH)*cos(TH)/(6561*pi)
%d2f/dR2 = -8exp(-2*R/3)*(R^4-24*R^3+171*R^2-378*R+162)*sin(2*TH)/(59049*pi)
%d2f/dRdTH = -2*sin(TH)*cos(TH)*(2/(pi*81^2))*exp(-2*R/3)*(2*(6*R-R^2)*(6-2*R)-(2/3)*(6*R-R^2)^2)
%d2f/dTHdR = 4*exp(-2*R/3)(R-6) R(R^2-12*R+18)*sin(2*TH)/(19683*pi)
x_k = [TH_k;R_k]
i am trying to get x_k+1 = x_k − D[G(x_k)]^−1[G(x_k)] using backslash and stop the iteration when the most recent change D[G(x_k)]^−1[G(x_k)] has a two norm that is less than 1e − 4 times the two norm of the current guess xk+1.
  1 件のコメント
Torsten
Torsten 2017 年 11 月 8 日
One equation for two unknowns is not suited to apply Newton's method.
Best wishes
Torsten.

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