Function handle is giving wrong results!

This code is clearly giving wrong results:

clc;clear all
a0 = 2;
a1 = 2;
y = [1 1.4 2.3];
x = [1 2 3];
f = @(x,a0,a1) (a0.*x)/(a1+x);
partialf_a0 = @(x,a1) (x./(x+a1));
partialf_a1 = @(x,a0,a1) (-(a0.*x)/((a1+x).^2));
partialf_a0(a1,x)
partialf_a1(x,a0,a1)
f(x,a0,a1)

Substituting a0 and a1 with their values into the equations, gives correct results for partialf_a0:

clc;clear all
a0 = 2;
a1 = 2;
y = [1 1.4 2.3];
x = [1 2 3];
f = @(x,a0,a1) (a0.*x)/(a1+x);
partialf_a0 = @(x,a1) (x./(2+x));
partialf_a1 = @(x,a0,a1) (-(a0.*x)/((a1+x).^2));
partialf_a0(x)
partialf_a1(x,a0,a1)

[the cyclist]

Can you be more specific about what is "clearly wrong"? Do you mean the value of f(x,a0,a1) is not what you expect? What value did you get, and what did you expect to get?

[Ahmad]

What I mean is substituting x and a1 and a2 values by hand, gives different results than matlab, for example, x = 1, a1 =2, x/(x+a1)=1/3, please run the code and see what u get!

[Walter Roberson]

Any code that uses "clear all" is clearly wrong. "clear all" does a lot more than you expect.

Answer 1

[the cyclist]

Just a guess, but where you did

f = @(x,a0,a1) (a0.*x)/(a1+x);

maybe you really want

f = @(x,a0,a1) (a0.*x)./(a1+x);

[Notice the "./" instead of "/".]

Similarly for your other functions.

If you don't understand why, read this documentation page about array vs. matrix operations.

Answer 2

[the cyclist]

Please, as I asked earlier, tell us what you get and what you expect to get. I ran this simplified version of your code

f = @(x,a0,a1) (a0.*x)./(a1+x);
f(1,1,2)

The displayed answer is 0.3333. This is the answer I would expect to get.

I am trying to understand what you think would be different. One possibility is that you are getting confused about variable values that are defined before you define the function, which may be irrelevant because the function arguments will be used instead.

[Ahmad]

found it, partialf_a0(a1,x) is not the same as partialf_a0(x,a1)