Can someone tell me why the E (x) (equation below) work perfect up to E (416)? If I try to calculate the E(x) for any x > 416 the answer that I get is NaN.

E=  @(x)1.2840.*(x<=55)+...
    exp(((0.5+((5.9)*((x-55)./(112.3-55)))).^2)./2).*(x>55 & x<=112.3)...
     +24.5325.*(x>112.3);

Thanks for the help.

Answer 1

[Matt J]

Because for x>416, the expression

exp(((0.5+((5.9)*((x-55)./(112.3-55)))).^2)./2)

evaluates to Inf due to overflow. Then you multiply this Inf by 0 resulting in NaN. To avoid this, you'll need to use if statements to implement each piece of the function instead of implementing as a sum of the different pieces.

[Oscar]

Thanks Matt Can you send me a link or example of How to use the if statements to implement each piece of the function?

[Matt J]

As an alternative to if-statements, you could do this,

function E=Ecalc(x)

   E=nan(size(X));

     idx=(x<=55);

   E(idx)=1.2840;

     idx=(x>55 & x<=112.3);

   E(idx)=exp(((0.5+((5.9)*((x(idx)-55)./(112.3-55)))).^2)./2);

    idx=(x>112.3);

   E(idx)=24.5325;