How we can calculate using MATLAB for small values?

R = 1000;
cf = 0.001;
f(cf) = 4.04*cf.^0.5 + (cf.^0.5)*3.94*log(R*cf.^0.5) - 1
??? Attempted to access (0.001); index must be a positive integer or logical.

Answer 1

[Image Analyst]

You don't need the (cf) if you just want the value of f for that value. Just use f by itself, like this:

R = 1000;
cf = 0.001;
f = 4.04*cf.^0.5 + (cf.^0.5)*3.94*log(R*cf.^0.5) - 1

You can turn it into a function if you want to pass it various values of cf as input arguments to a function, but it's not necessary if you just want the value of f and you can just specify what cf is before you calculate f, like I did above.

Answer 2

[Akiva Gordon]

You must first create an anonymous function to be able to evaluate the function like that:

This means "f" is a function with an input of "x". You would commonly see this on paper as f(x) = x+2;

f = @(x)(x+2);

a = 2;

This line evaluates "f" at the point "a" and stores it in "y".

y = f(a)

[Akiva Gordon]

Please visit the following link for more information on anonymous functions:

http://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

[Jan]

@Akiva: You can edit your answer instead of appending a comment.

Answer 3

[Star Strider]

MATLAB considers f as a matrix as you have written it. You need to define f as a function. It is easiest to write it as an Anonymous Function:

R = 1000; 
cf = 0.001; 
f = @(cf) 4.04*cf.^0.5 + (cf.^0.5)*3.94*log(R*cf.^0.5) - 1;
fcf = f(cf)

results in:

fcf =

    -441.9125e-003

You can also define it as a function of both cf and R if you want to:

f = @(cf,R) 4.04*cf.^0.5 + (cf.^0.5)*3.94*log(R*cf.^0.5) - 1;