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Newton Raphson - saving all values and using last iteration value as initial for next

13 vues (au cours des 30 derniers jours)
I'm attempting to do a Newton - Raphson calculation but am having trouble starting. A1 and B1 are both matrices [1x15], so for the the values of A1, B1 and Theta_F, I need to perform a Newton - Raphson calculation over 5 iterations, saving all iteration values (for plotting) and using the last iteration value as the initial guess for the next N-R step (with the next set of A1, B1 and Theta_F values)
I'm really not sure where to start or how best to approach it (not been using MATLAB very long), any help would be greatly appreciated!
Many thanks.
M_s = 0; % Other variables in equations
alpha_s = 0;
Theta_F = [1:1:15]
A1 = [0,-30,-120,-270,-480,-750,-1080,-1470,-1920,-2430,-3000,-3630,-4320,-5070,-5880]
% B1 has the same value but is calculated as a matrix
B1 = [-196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200]
f = @(x) A1.*cos(x) + B1.*sin(x) + M_s*(x + (Theta_F ./(180*pi)) + (alpha_s /(180*pi))); % Function
fd = @(x) -A1.*sin(x) + B1.*cos(x) - M_s; % Function derivative
x0 = 0.0001;
x = x0;
% Attempt at N-R
for i = 1:5
x(i+1) = x(i) - f(x(i))/fd(x(i))
i = i+1
end
  7 commentaires
Andrew Newell
Andrew Newell le 19 Avr 2021
Modifié(e) : Andrew Newell le 20 Avr 2021
Actually, for and equal to zero, the solution is . Are you going to need to do this for nonzero parameters? If not, you're done.
HMZ
HMZ le 19 Avr 2021
Yes, I've created a spreadsheet to practice the mathematics behind the process and ensure the values are sensible. The values of x (for corresponding value of A1, B1 and Theta_f) after 5 iterations come out to be;
x = [0
-0.000152905
-0.000611621
-0.001376146
-0.002446478
-0.003822611
-0.005504532
-0.007492215
-0.00978562
-0.012384688
-0.015289328
-0.018499418
-0.022014791
-0.025835229
-0.029960451
]

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Réponse acceptée

Andrew Newell
Andrew Newell le 20 Avr 2021
Modifié(e) : Andrew Newell le 20 Avr 2021
Sorry, I just realized that you wanted to save the iterations. Here's how:
x = -0.001*ones(6,length(A1));
for i=1:5
x(i+1,:) = x(i,:) - f(x(i,:))./fd(x(i,:));
end
The top row of this matrix has the initial guess and the next five rows have the five iterations.

Plus de réponses (2)

Paul Hoffrichter
Paul Hoffrichter le 20 Avr 2021
Modifié(e) : Paul Hoffrichter le 20 Avr 2021
The final x values match your spreadsheet results. Your suggested x0 converges too soon to be interesting. Set it to 1.0 to actually see convergence in action.
format long
M_s = 0; % Other variables in equations
alpha_s = 0;
Theta_F = [1:1:15]
iMax = 5;
A1 = [0,-30,-120,-270,-480,-750,-1080,-1470,-1920,-2430,-3000,-3630,-4320,-5070,-5880]
% B1 has the same value but is calculated as a matrix
B1 = [-196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200, -196200]
f = @(x) A1.*cos(x) + B1.*sin(x) + M_s*(x + (Theta_F ./(180*pi)) + (alpha_s /(180*pi))); % Function
fd = @(x) -A1.*sin(x) + B1.*cos(x) - M_s; % Function derivative
x0 = 0.0001;
% x0 = 1; % more interesting starting point
x = zeros(1, length(A1));
x(:) = x0;
xSave = zeros(iMax, length(A1));
% Attempt at N-R
for i = 1:iMax
x = x - f(x)./fd(x);
xSave(i,:) = x;
i = i+1;
end
% Check;
y = f(x)

Andrew Newell
Andrew Newell le 20 Avr 2021
O.k. So here is how you do it:
x = -0.001*ones(size(A1));
for i=1:5
x = x - f(x)./fd(x);
end

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