Finding a variable in a very complicated equation

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Nikodem Podlaszewski 2021-4-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/812260-finding-a-variable-in-a-very-complicated-equation

评论： Star Strider 2021-5-2

采纳的回答： Star Strider

There are two equations:

y=P*exp(-alfa*x)*cos(alfa*x)/(2*alfa^3*EI)

alfa=(k/(4*EI))^(1/4)

x, y, P, EI are known, I need to find k using these equations, accuracy of 0,0001 is enough

I think that the only way to find k is to use the trial and error (iteration) method, unfortunately I can't manage to write a good script to calculate that

Thank You for all Your help in advance

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Answer 1

Star Strider 2021-4-24

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https://ww2.mathworks.cn/matlabcentral/answers/812260-finding-a-variable-in-a-very-complicated-equation#answer_683915

在 MATLAB Online 中打开

It would be nice if the constants were supplied. Using random numbers for them —

x = rand;
y = rand; 
P = rand;
EI = rand;
alfa = @(k,EI) (k./(4*EI)).^(1/4);
yfcn = @(x,y,P,EI,k) P.*exp(-alfa(k,EI).*x).*cos(alfa(k,EI).*x)./(2*alfa(k,EI).^3.*EI) - y;
[k_est,fv] = fsolve(@(k)yfcn(x,y,P,EI,k), 1)
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
k_est = 2.1218
fv = 4.1983e-06

.

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Star Strider 2021-4-24

在 MATLAB Online 中打开

Sure!

xy = [400 -0.01

300 -0.09

200 -0.11

100 0.14

0 0.87

400 -0.02

300 -0.09

200 -0.10

100 0.12

0 0.80];

E=210000; % [N/mm^2]

I=25*3^3/12; % [mm^4]

EI=E*I; % [Nmm^2]

P=19.62; % [N]

alfa = @(k,EI) (k./(4*EI)).^(1/4);

yfcn = @(x,y,P,EI,k) P.*exp(-alfa(k,EI).*x).*cos(alfa(k,EI).*x)./(2*alfa(k,EI).^3.*EI) - y;

for kk = 1:size(xy,1)

[k_est(kk,:),fv(kk,:)] = fsolve(@(k)yfcn(xy(kk,1),xy(kk,2),P,EI,k), 1);

end

Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. No solution found. fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the value of the function tolerance. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. No solution found. fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the value of the function tolerance. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.

Out = array2table([xy k_est, fv], 'VariableNames',{'x','y','k_est','fv'})

Out = 10×4 table

x y k_est fv ___ _____ _______ ___________ 400 -0.01 0.47147 -5.6502e-06 300 -0.09 0.23986 -3.4322e-08 200 -0.11 0.42319 0.062807 100 0.14 0.53096 1.0563e-06 0 0.87 0.4441 7.7512e-08 400 -0.02 0.36061 -5.5235e-06 300 -0.09 0.23986 -3.4322e-08 200 -0.1 0.42319 0.052807 100 0.12 0.59015 1.3671e-07 0 0.8 0.49665 3.0442e-11

figure

stem3(Out.x, Out.y, Out.k_est, 'filled')

grid on

xlabel('x')

ylabel('y')

zlabel('k_{est}')

It would have been nice to have all that information at the outset.

Nikodem Podlaszewski 2021-5-2

I have an another problem. After inputting the code to MatLab and hitting "Run and Time" the results don't show anywhere. Instead, I see Profiler and a warning that says "The variable "k_est" appears to change size on every loop iteration (within a script). Consider preallocating for speed" The same appears for the variable "fv". I've read a lot about it and it appears to be just warnings but I can't see the results (nothing is being showed in Command Windows nor the output Array2table. Could you help me to get the results from that?

Star Strider 2021-5-2

在 MATLAB Online 中打开

Apparently, there is more to your code than what you posted.

If you are using a for loop, it is relatively straightforward to preallocate ‘k_est’ and ‘fv’ —

x = rand;
y = rand; 
P = rand;
EI = rand;
alfa = @(k,EI) (k./(4*EI)).^(1/4);
yfcn = @(x,y,P,EI,k) P.*exp(-alfa(k,EI).*x).*cos(alfa(k,EI).*x)./(2*alfa(k,EI).^3.*EI) - y;
num_iter = 10;                                                      % Use The Appropriate Value
k_est = zeros(num_iter,1);                                          % Preallocate Vector
fv = zeros(num_iter,1);                                             % Preallocate Vector
for k1 = 1:num_iter
    [k_est(k1),fv(k1)] = fsolve(@(k)yfcn(x,y,P,EI,k), 10);
end
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
Output = table(k_est,fv)
Output = 10×2 table
     k_est          fv    
    ________    __________

    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08

If you are using a while loop, preallocate the vectors to be greater than what you will likely require, then after the looop finishes (since you will have used a counter to create the ‘k1’ values) re-define —

k_est = k_est(1:k1);
fv = fv(1:k1);

This discards the rest of the vector, saving memory.

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Finding a variable in a very complicated equation

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