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

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

[Andrew Newell]

I think you need to start by clarifying what the functions are. What values do and have? Are they vectors like ? Also, as formulated, f and return vectors with the same length as . Is that what you intended, or are they supposed to be summed to get a single value?

[HMZ]

Thanks for replying, I've input what A1 and B1 are. The intention is that f and fd do return vectors of same length as A1, B1 and Theta_f (I do not wish to sum them). I wish to show how x (my variable) changes as A1, B1 and Theta_f change, but must calculate the root of x via N-R

[Andrew Newell]

In that case, your initial guess should be a vector of length 15 as well.

[Andrew Newell]

Note also that the N-R method doesn't work well if your guess is a long way from the correct answer. Do you have any reason to think that you've got a good initial guess?

[Paul Hoffrichter]

For the NR method, you should figure out how to plot the 15 functions that you are defining before proceeding. Then for each NR step, you can add the new estimate of the zero crossing. Are you required in this assignment to process all 15 cases simultaneously? Even if this is your requirement, I recommend trying just one or two functions to get a baseline for your working method, and then expand to doing all of them in one shot if required.

[Andrew Newell]

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]

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
]

Answer 1

[Andrew Newell]

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.

[HMZ]

Thank you so much for you help! It is extremely appreciated

Answer 2

[Paul Hoffrichter]

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)

[HMZ]

This also worked, so thank you very much as well for all your help!

Answer 3

[Andrew Newell]

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