Calculation of two less than equal

Hi

C = 139*1 double

I want to calculate :

Answer 1

[John D'Errico]

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) sqrt(C-1),[1,10])
hold on
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.

syms C
solve((1 + sqrt(2*C-1))/2 == sqrt(C-1))
ans =
5

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_min =
          3.60428787739482
          
m_max = sqrt(min(C ) - 1)
m_max =
          3.07214383883712

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.

Answer 2

[Bastian Arturo Romero Mardones]

For the left side it is:

n = sqrt(C-1);

The other side it is the same.

[Pradip Nath]

Hi

To use sqrt I have to compute them in seperately.

But I want to compute together at the same time. Maybe using a loop.

[Bastian Arturo Romero Mardones]

I don't understand, you need to calculate:

??

some code example?

[Pradip Nath]

Hi

I write this code but its seems the for loop only giving equal output not greater than equal value. Also in for loop I can't assign >=

IQ = [1;2;3];
C = sqrt(IQ);
%((1+sqrt(2.*C-1))/2) <= m <= sqrt(C-1)
for m=((1+sqrt(2.*C-1))/2)
   while m >((1+sqrt(2.*C-1))/2)
       
   end
   disp('m')
   disp (m)
end
for m1 = sqrt(C-1)
       while m1 < sqrt(C-1)
           
       end
     disp('m1')
    disp(m1)
end

[John D'Errico]

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.