How do I create a loop for this calculation in MATLAB?

Question

Nick Lavanture 2022-5-11

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https://ww2.mathworks.cn/matlabcentral/answers/1716775-how-do-i-create-a-loop-for-this-calculation-in-matlab

评论： Nick Lavanture 2022-5-11

%I need to create a loop to calculate d in the following manner, any help is greatly appreciated, thank you!

%constants
c = 1.1e-10;
m = 3.3;
p500 = [ -0.3142    2.3172   -7.0249   11.2867];
p2000 = [ -1.2577    9.2728  -28.1056   45.1501];
p3150 = [ -1.9807   14.6042  -44.2653   71.1096];
d = 0.015; % initial value of d
%the following needs to happen 25 times
%the following needs to happen 1 time
dd_dN = c*((0.706*(p3150(1)+3150)+0.537*(2*d/pi)*p3150(2)+0.448*(d^2./2)*p3150(3)+0.393*(4*d^3/(3*pi))*p3150(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
%the following needs to happen 240 times, picking up d from the previous
%calculation
dd_dN = c*((0.706*(p2000(1)+2000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
dd_dN = c*((0.706*(p2000(1)+2000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
dd_dN = c*((0.706*(p2000(1)+2000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
%(three example iterations ^)
%the following needs to happen 1800 times, picking up d from the previous
%calculation
dd_dN = c*((0.706*(p500(1)+500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
dd_dN = c*((0.706*(p500(1)+500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
dd_dN = c*((0.706*(p500(1)+500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
d = d + dd_dN
%(three example iterations ^)

%the final output should be d, after (1+240+1800)*25 total calculations

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Answer 1

Torsten 2022-5-11

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1716775-how-do-i-create-a-loop-for-this-calculation-in-matlab#answer_961830

在 MATLAB Online 中打开

%constants
c = 1.1e-10;
m = 3.3;
p500 = [ -0.3142    2.3172   -7.0249   11.2867];
p2000 = [ -1.2577    9.2728  -28.1056   45.1501];
p3150 = [ -1.9807   14.6042  -44.2653   71.1096];
d = 0.015; % initial value of d
%the following needs to happen 25 times
for i=1:25
    %the following needs to happen 1 time
    dd_dN = c*((0.706*(p3150(1)+3150)+0.537*(2*d/pi)*p3150(2)+0.448*(d^2./2)*p3150(3)+0.393*(4*d^3/(3*pi))*p3150(4))*(sqrt(pi*d)))^m;
    d = d + dd_dN
    
    for j=1:240
        %the following needs to happen 240 times, picking up d from the previous
        %calculation
        dd_dN = c*((0.706*(p2000(1)+2000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
        d = d + dd_dN
    end
    %(three example iterations ^)
    
    
    for j=1:1800
        %the following needs to happen 1800 times, picking up d from the previous
        %calculation
        dd_dN = c*((0.706*(p500(1)+500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
        d = d + dd_dN
    end
end
%(three example iterations ^)

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Nick Lavanture 2022-5-11

Thank you!

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Answer 2

Jon 2022-5-11

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https://ww2.mathworks.cn/matlabcentral/answers/1716775-how-do-i-create-a-loop-for-this-calculation-in-matlab#answer_961840

在 MATLAB Online 中打开

Slightly different interpretation of how you want to do loops, I ignored the first comment about 25 iterations. Regardless it seems that you have a problem with convergence as your variables become infinite after a small number of loops. You will need to check more on that.

%constants
c = 1.1e-10;
m = 3.3;
p500 = [ -0.3142    2.3172   -7.0249   11.2867];
p2000 = [ -1.2577    9.2728  -28.1056   45.1501];
p3150 = [ -1.9807   14.6042  -44.2653   71.1096];
d = 0.015; % initial value of d
% % % %the following needs to happen 25 times ** ignore this comment??
%the following needs to happen 1 time
dd_dN = c*((0.706*(p3150(1)+3150)+0.537*(2*d/pi)*p3150(2)+0.448*(d^2./2)*p3150(3)+0.393*(4*d^3/(3*pi))*p3150(4))*(sqrt(pi*d)))^m;
d = d + dd_dN;
%the following needs to happen 240 times, picking up d from the previous
%calculation
for k = 1:240
    dd_dN = c*((0.706*(p2000(1)+2000)+0.537*(2*d/pi)*p2000(2)+0.448*(d^2./2)*p2000(3)+0.393*(4*d^3/(3*pi))*p2000(4))*(sqrt(pi*d)))^m;
    d = d + dd_dN;
end
%the following needs to happen 1800 times, picking up d from the previous
%calculation
for k = 1:1800
    dd_dN = c*((0.706*(p500(1)+500)+0.537*(2*d/pi)*p500(2)+0.448*(d^2./2)*p500(3)+0.393*(4*d^3/(3*pi))*p500(4))*(sqrt(pi*d)))^m;
    d = d + dd_dN;
end
%the final output should be d, after (1+240+1800)*25 total calculations

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Nick Lavanture 2022-5-11

The issue you pointed out was a mistake on my part, the constant I added in those long equations were off by a factor of 1000 (0.5 instead of 500, etcetera.) Thank you for your time

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