Unable to convert expression into double array.

My code snippet is as follows:
syms X mu std ;
sym pi;
definite_int = inline('(2*pi*(std^2))^(-0.5)* exp(-((X-mu).^2)/(2*(std^2)))', 'X','mu','std','pi');
inte = vpaintegral(definite_int(x, mu, std, pi),[20 120])
inte = double(vpa(inte))
figure('Name','plot ');
for i = 2:20
height = 20:i:120;
ri = sum(height .* i);
error = ri - inte
plot(error, i);
hold on
end
I am getting the following error:
integration =
[ vpaintegral(exp(-(mu - 60)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 62)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 64)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 66)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 68)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 70)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 72)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 74)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 76)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 78)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120), vpaintegral(exp(-(mu - 80)^2/(2*std^2))/(2*std^2*pi)^(1/2), std, 20, 120)]
Error using symengine
Unable to convert expression into double array.
Error in sym/double (line 692)
Xstr = mupadmex('symobj::double', S.s, 0);
Error in hw1>question4 (line 125)
integration = double(vpa(integration))
How should I solve this error? please help

2 comentarios

Stephen23
Stephen23 el 1 de Sept. de 2019
Note that the inline documentation states in bold letters at the top of the page:
"inline will be removed in a future release. Use Anonymous Functions instead."
Instead of using obsolete inline, you should use efficient function handles.
Thank you for pointing out.

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Respuestas (1)

Walter Roberson
Walter Roberson el 1 de Sept. de 2019

0 votos

Using double() on the vpaintegral() result can only possibly work if you are using vpaintegral() on a formula that has exactly one unresolved symbolic variable. mu and std are unresolved in your integral as well as X, so you cannot possibly get out a symbolic numeric result from the integral.

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el 1 de Sept. de 2019

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el 1 de Sept. de 2019

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