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Function handle is giving wrong results!

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Ahmad
Ahmad on 28 May 2016
Commented: Walter Roberson on 28 May 2016
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 Comments

the cyclist
the cyclist on 28 May 2016
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
Ahmad on 28 May 2016
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
Walter Roberson on 28 May 2016
Any code that uses "clear all" is clearly wrong. "clear all" does a lot more than you expect.

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Answers (2)

the cyclist
the cyclist on 28 May 2016
Edited: the cyclist on 28 May 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.
If you don't understand why, read this documentation page about array vs. matrix operations.

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the cyclist
the cyclist on 28 May 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.

  1 Comment

Ahmad
Ahmad on 28 May 2016
found it, partialf_a0(a1,x) is not the same as partialf_a0(x,a1)

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