Getting different results between Matlab integration and Maple?

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Zeyad Elsayed 2019-5-16

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回答： Sulaymon Eshkabilov 2019-5-16

Hello everyone,

I am having trouble getting the same results for the same equation when using Matlab and Maple. Basically as the value for beta changes, Q should change as well hence its integral, hence T. When putting the same equation in Mable I get values for T that are close to 1 but that decrease as beta increases, however the values I get in Matlab are exactly 1 and do not change by changing beta. Is there something wrong with the level of precision in Matlab? Or a problem with the integration?

length_shaft = 0.4
% Determining 4 values of beta
f = 'cos(x)*cosh(x)-1';                                         
roots = [fzero(f,4.0) fzero(f,7.0) fzero(f,10.0) fzero(f,14.0)];   
beta = roots./length_shaft;
% Determining Constant (T) (a value for each beta)
syms y
Q = ((cos(beta.*y) + cosh(beta.*y)) - (((cos(beta.*length_shaft) - ...
     cosh(beta.*length_shaft))./(sin(beta.*length_shaft) - ...
     sinh(beta.*length_shaft))).*(sin(beta.*y) + sinh(beta.*y)))).^2;
B = int(Q,y,0, length_shaft);
T = (B.*2.5).^(-0.5);
B = double(B)
T = double(T)

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Answer 1

Sulaymon Eshkabilov 2019-5-16

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Hi Zeyad,

Here is an alternative function quad() with its set tolerance to compute integration that changes the solutution a bit.

clearvars
length_shaft = 0.4;
% Determining 4 values of beta
f = 'cos(x)*cosh(x)-1';            
OPTs = optimset('tolX',1e-17);
Roots = [fzero(f,4.0, OPTs) fzero(f,7.0, OPTs) fzero(f,10.0, OPTs) fzero(f,14.0, OPTs)];   
beta = Roots./length_shaft;
% Determining Constant (T) (a value for each beta)
for ii=1:numel(beta)
B(ii) = quad(@(y) ((cos(beta(ii)*y) + cosh(beta(ii)*y)) - (((cos(beta(ii)*length_shaft) - ...
     cosh(beta(ii)*length_shaft))/(sin(beta(ii)*length_shaft) - ...
     sinh(beta(ii)*length_shaft)))*(sin(beta(ii)*y) + sinh(beta(ii)*y)))).^2, 0, length_shaft, 1e-16);
T(ii) = (B(ii)*2.5)^(-0.5);
end
format long
B = double(B)
T = double(T)