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Empty sym: 0-by-1

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I don't understand why matlab gives as an output "Empty sym: 0-by-1" to the following equation which I want to solve. I have used this equation many times with other numbers and it has always worked since. Thank you for the help!
syms x1
eqn = (10.5145 - 0.5 + 6.96.*x1 + (6-(31*(232.87.*(x1*12).^(-0.46)))/1000).*x1/60*1000 + 31.*(232.87.*(x1*12).^(-0.46))/1000.*0/60*1000)*0.0086 - 0.1475 - x1 == 0
vpasolve(eqn, x1)

Accepted Answer

Star Strider
Star Strider on 24 May 2021
Exploring the function is often worthwhile. It apparently has a real root at about -0.45, so share that information with vpasolve:
syms x1
eqn = (10.5145 - 0.5 + 6.96.*x1 + (6-(31*(232.87.*(x1*12)^(-0.46)))/1000)*x1/60*1000 + 31*(232.87*(x1*12)^(-0.46))/1000*0/60*1000)*0.0086 - 0.1475 - x1 == 0;
f(x1) = (10.5145 - 0.5 + 6.96.*x1 + (6-(31*(232.87.*(x1*12)^(-0.46)))/1000)*x1/60*1000 + 31*(232.87*(x1*12)^(-0.46))/1000*0/60*1000)*0.0086 - 0.1475 - x1;
x1_s = vpasolve(real(f), -0.5)
x1_s = 
fplot(real(f), [-1 1])
hold on
fplot(imag(f), [-1 1])
hold off
legend('Real','Imag', 'Location','best')
Star Strider
Star Strider on 26 May 2021
As always, my pleasure!
One problem is that solve identified 50 roots of that function, so there could be a number of them. However plotting it revealed only one .real root. It is a relatively complicated function. I have no idea what the other roots were, since solve did not calculate them. Plotting the real and imaginary parts of the function demonstrated its behaviour and led to the solution.

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More Answers (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov on 24 May 2021
There is a typo in your eqn:
Good luck
  1 Comment
Delia Bosshart
Delia Bosshart on 24 May 2021
does "Empty sym: 0-by-1" mean that it can't solve the equation and that is why you assume that there has to be a typo?

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