MATLAB: How to calculate the value

equation

I have the following values and want to calculate value of alpha such that alpha >=0;

a=rand(1,49);
b=rand(49,49);
c=rand(1,49);
alpha = (-b+real(sqrt(b^2-4*a*c')))/(2*a);

however I'm getting the value of alpha same which should not be true. Is there any error in my code.

Best Answer

You really need to learn the difference between matrix operations and array operations:

https://www.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html

and then review what you have written (hint: you need array operations, not matrix operations).

To give you a working solution you should also tell us what MATLAB version you are using.

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if b+sqrt(b^2-4*a*c)>b-sqrt(b^2-4*a*c)
d=g1;
end
if b+sqrt(b^2-4*a*c)<b-sqrt(b^2-4*a*c)
d=g2;
end

That code will not assign d under a few circumstances:

b+sqrt(b^2-4*a*c) == b-sqrt(b^2-4*a*c) -- which would occur if sqrt(b^2-4*a*c) == 0
one of the values is nan
sqrt(b^2-4*a*c) is imaginary but b is real; in that case, the > operator would compare only the real parts, and the real parts would be equal

Note:

b+sqrt(b^2-4*a*c)<b-sqrt(b^2-4*a*c) implies b + sqrt(b^2-4*a*c) - b < b - sqrt(b^2-4*a*c) - b implies sqrt(b^2-4*a*c) < - sqrt(b^2-4*a*c) implies 2*sqrt(b^2-4*a*c) < 0 implies sqrt(b^2-4*a*c) < 0 . However, algebraically, sqrt() is defined as "principle square root", which is positive when the value is positive, and is 1i * sqrt(-(b^2-4*a*c)) when b^2-4*a*c < 0 -- so sqrt(b^2-4*a*c) < 0 cannot happen unless you define an ordering of imaginary values with respect to 0. The MATLAB < operator ignores the imaginary component... but then you would be comparing 0 to 0, which would not be < .

Which is to say that your second test can never succeed, and your first test is always true unless b^2-4*a*c == 0 (a ligitimate test) or b^2-4*a*c < 0 (because the imaginary component is ignored)

Best Answer

Related Solutions

MATLAB: How to make output single column in the same line as your calculation

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