defining a range as a value for while loop

hi everybody I'm trying real hard for finding a way to define a range as a value to put it in a while loop and get a result but I have failed i'm doing it like this:

FX=(5*3.1416*M);

FM=(3*3.1416*M);

F=Wt/(KV*O*M*Y*(10^(-3)));

U=linspace(FM,FX,100);

while F~=U;

do some mathematics;

FX=(5*3.1416*M);

FM=(3*3.1416*M);

U=linspace(FM,FX,100);

F=Wt/(KV*O*M*Y*(10^(-3))); 

end

when i insert the values it just says matlab is busy and no result shows up. i don't which part i am doing wrong. i'll be extremely thankful if you give a help

[Adam]

You haven't really given enough information.

Is F a scalar?

U certainly isn't a scalar so you can't sensibly test:

F ~= U

in a loop. A scalar will never equal a vector and that isn't what you want to test anyway.

Are you trying to test if F is not a member of U or outside of the range it defines or something else?

[ali najafi]

sorry

my bad

thank you for your answer

yes i want to test if F is not the a member of U

Answer 1

[Walter Roberson]

while ~ismember(F, U)

is the test you asked for in your request "yes i want to test if F is not the a member of U". However, remember that ismember() does equality tests, so if F was even just 2^(-52) different than any member of U then ismember would not find it. I suspect what you want is

while F < FM | F > FX

[ali najafi]

thank you for your answer

i'd much thankful if you answer me another question here, preventing me from making another thread

i want to define a set of numbers like this one:

x = [1,1.25,2,2.5,1.5,3,4,5,6,8,10,12,16,20,25,32,40,50];

and want to choose one of them randomly each time what should i do?

i have used this command for this job:

M=randi(length(x))

but it doesn't choose decimal numbers or sometimes gives me numbers that are not defined in my set of numbers

[Adam]

Use that to give you indices into your array instead of the actual values. Then it will always give you one of your valid values as

x(M)

[ali najafi]

thank you it really helped me