Calculation of two less than equal
4 views (last 30 days)
Show older comments
Pradip Nath on 28 Nov 2020
C = 139*1 double
I want to calculate :
You want to solve for a scalar m such that the inequalities are always true? Regardless, does a solution ever exist? Note that unless C > 1, sqrt(C-1) is not real.
fplot(@(C) (1 + sqrt(2*C-1))/2,[1,10])
legend('sqrt(C-1)','1 + sqrt(2*C-1))/2')
You should see that m exists only of C is at least 5. But as long as C is greater than 5, an interval of solutions will exist, for EACH calue of C. We can find the point where the two curves cross.
solve((1 + sqrt(2*C-1))/2 == sqrt(C-1))
So NO element of C may be less than 5. If that ever happens, then m will not exist. Whem C == 5, M can have only a unique value, thus 2, so the interval collapses to a point.
If you want to solve for a scalar m such that this is ALWAYS true for all values of C, then it is equally trivial, although a solution may likely not exist, For example, consider a vector C that lies between 10 amd 20.
C = rand(1,10)*10 + 10;
m_min = (1 + sqrt(max(C )*2 -1))/2
m_max = sqrt(min(C ) - 1)
So we must have m >= 3.6, but also we need m <= 3.07. Clearly no solution exists. We can do further computations to show what range C can have such that a solution can exist.
Setsuna Yuuki. on 28 Nov 2020
For the left side it is:
n = sqrt(C-1);
The other side it is the same.
READ MY ANSWER. In there, I show that if ANY element of C is less than 5, NO solution can exist.
Anyway, you cannot assign an interval to a number, and you cannot do an assignment using an inequality. So it is not at all clear what you wnt to do.
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!Start Hunting!