Function handle is giving wrong results!
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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)
3 comentarios
Walter Roberson
el 28 de Mayo de 2016
Any code that uses "clear all" is clearly wrong. "clear all" does a lot more than you expect.
Respuestas (2)
the cyclist
el 28 de Mayo de 2016
Editada: the cyclist
el 28 de Mayo de 2016
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.
0 comentarios
the cyclist
el 28 de Mayo de 2016
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.
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