Substituting a number for NaN in anonymous function

I'm trying unsuccessfully to substitute a number for NaN in anonymous function. Here it's an example of the problem. Bear with it's silliness please:

clc;
clear all;
f=@(x) x*1/x;
g=@(x) (~isnan(f(x))).*f(x) + (isnan(f(x))).*1;

I'd expect that g(0)=1, but it's still NaN. What is wrong in the way that I defined g?

[John]

Your replies made me understand why it didn't work. However, in attempting to overcome this problem, I still don't grasp why the following happens:

clc;
clear all;
f1=@(x) x.*(1./x);

g(0,f1) %ok, got the desired result
y=linspace(0,1,10);
g(y(1),f1) %good
g(y,f1) %didn't work: first element is NaN

    function out=g(x,f1)
    fval=f1(x);

    if isnan(fval)==1;
    out = 1;
    else
        out=fval;
    end
    end

From this, I can see that I can construct the function "element-wise". But how to do it for the whole vector?

[Guillaume]

It's getting a bit messy. You should have accepted the answer that helped you the most and started a new question. Don't pile questions on top of questions, because now there's no appropriate place to answer your new question.

Anyway, the reason why it does not work in the second case is that isnan(fval) == 1 (which is just the same as isnan(fval)) is a logical vector equal to

1 0 0 0 0 0 0 0 0 0 0

If you pass a vector to if it will only evaluate to true if and only if all elements are not zeros. Therefore your expression is false.

To fix this:

if any(isnan(fval))
   out = 1;
else 
   out = fval;
end

Or if you just want to replace the Nans by 1:

fval(isnan(fval)) = 1; %no need for if

Answer 1

[Alfonso Nieto-Castanon]

There may be "cleaner" ways to do this but one possibility would be:

g = @(x) [f(x) 1]*sparse(1+isnan(f(x)),1,1,2,1);

EDIT: and as others have pointed out the problem with your original g function is that, when f(x) is NaN, you get 0*NaN + 1*1 which still evaluates to NaN...

Answer 2

[John Petersen]

You have .*f(x) in g(x), which is still giving you a NaN

Answer 3

[Star Strider]

If you want to define L’Hospital’s rule, you have to define it specifically. In the IEEE standard that MATLAB implements, 0/0 is NaN.

See if this works in your application:

n = @(x) 2.*x;                                          % Numerator Function
d = @(x) x;                                             % Denominator Function

Lh = @(n,d,x) (n(x+1E-12)-n(x)) ./ (d((x+1E-12)-d(x))); % L’Hospital’s RUle

Lh0 = Lh(n,d,0)                                         % Evaluating AT 0
Lh2 = Lh(n,d,2)                                         % Evalutaing At 2