Getting conditions for a dense algebraic second degree equation
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Noemi ZM
el 18 de Mzo. de 2022
Comentada: Noemi ZM
el 25 de En. de 2024
I have a dense algebraic second degree equation, with five parameters in each coefficiente.
I'd like to get conditions about positivity of the solutions.
Do you know if this is factible in MATLAB?
Thank you in advance.
4 comentarios
Torsten
el 18 de Mzo. de 2022
If b^2-4*a*c is positive, a sufficient condition is
b/(2*a) < 1/(2*a)*sqrt(b^2- 4ac) < -b/2a
Respuesta aceptada
Ravi
el 25 de En. de 2024
Hi Noemi ZM,
Let us assume that the coefficients of the algebraic equation are governed by the parameters “p1, p2, p3, p4, and p5”. The algebraic equation be , where the coefficients are all some functions of the parameters, say,
a = computeA(p1, p2, p3, p4, p5);
b = computeB(p1, p2, p3, p4, p5);
c = computeC(p1, p2, p3, p4, p5);
Please note that, when the two solutions of a second-degree equation are to be positive, the following conditions should hold,
- The discriminant value should be greater than or equal to zero. That implies, .
- The product of the two solutions should be greater than zero. That implies, c and a should have same sign.
- The sum of the two solutions should be greater than zero. That implies, b and a should have the opposite signs.
We can design a function in MATLAB, that does the following.
- Input the five parameters.
- Compute the coefficients.
- Verify if the conditions to hold are satisfied.
- Return true if all conditions are met, otherwise false.
function result = areSolutionsPositive(p1, p2, p3, p4, p5)
conditionsSatisfied = 0;
% Compute the coefficients
a = computeA(p1, p2, p3, p4, p5);
b = computeB(p1, p2, p3, p4, p5);
c = computeC(p1, p2, p3, p4, p5);
% Compute discriminant(delta)
delta = b * b - 4 * a * c;
% Condition 1: delta is greater than or equal to zero
if delta >= 0
conditionsSatisfied = conditionsSatisfied + 1;
end
% Condition 2: c and a should have the same sign
% In other words, c * a should be greater than zero
if c * a > 0
conditionsSatisfied = conditionsSatisfied + 1;
end
% Condition 3: b and a should have opposite signs
if b * a < 0
conditionsSatisfied = conditionsSatisfied + 1;
end
% Verify if all the three conditions are satisfied
if conditionsSatisfied == 3
result = true;
else
result = false;
end
end
Hope this answer helps you the solve the issue you are facing.
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