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

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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 Comments

Andrew Newell
on 19 Apr 2021

Edited: Andrew Newell
on 20 Apr 2021

### Accepted Answer

Andrew Newell
on 20 Apr 2021

Edited: Andrew Newell
on 20 Apr 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.

### More Answers (2)

Paul Hoffrichter
on 20 Apr 2021

Edited: Paul Hoffrichter
on 20 Apr 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
on 20 Apr 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

##### 0 Comments

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