Why does the anonymous function not evaluate correctly

Question

Hari 2014-6-29

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https://ww2.mathworks.cn/matlabcentral/answers/138678-why-does-the-anonymous-function-not-evaluate-correctly

评论： John D'Errico 2014-7-5

I posted this in stack exchange comp science site 2 days back but couldn't get a response. Will restate with a simplified example and hope I get some directions.

Consider the Rosenbrock function in http://www.mathworks.com/help/optim/ug/writing-objective-functions.html#brhkghv-6

Defining the following in a script (modifying matlab example for how x is inputted) and run it,

clc; clear;
anonrosen = @(x1,x2)(100*(x2 - x1^2)^2 + (1-x1)^2);

and then typing the following in command window,

anonrosen(-1,2)

yields,

ans =
     104

But if I define the following in a script and run it,

clc; clear;syms x1 x2;
p = (100*(x2 - x1^2)^2 + (1-x1)^2);
anonrosen = @(x)p;

and then typing the following in command window,

anonrosen(-1,2)

yields,

ans =
(x1 - 1)^2 + 100*(- x1^2 + x2)^2

Why don't I get 104 as answer in 2nd approach?

The reason I want to make the 2nd approach work is the functional form "(100*(x2 - x1^2)^2 + (1-x1)^2)" in my application is being generated on the fly during application run depending on user inputs using symbolic toolbox so I cannot hard-code it the way it is done in matlab example code.

Thanks in advance Hari

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Answer 1

Star Strider 2014-6-29

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https://ww2.mathworks.cn/matlabcentral/answers/138678-why-does-the-anonymous-function-not-evaluate-correctly#answer_142934

编辑：Star Strider 2014-6-29

在 MATLAB Online 中打开

You have to define functions differently in the Symbolic Math Toolbox:

syms x1 x2
anonrosen = symfun(100*(x2 - x1^2)^2 + (1-x1)^2, [x1 x2])
anonrosen(-1,2)

produces:

anonrosen(x1, x2) =
(x1 - 1)^2 + 100*(- x1^2 + x2)^2
ans =
104

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Hari 2014-6-30

编辑：Star Strider 2014-6-30

在 MATLAB Online 中打开

Thanks for the tip on symfun. I wasn't aware of this.

I have one follow-up question. I had simplified my original requirement and in my actual application the functional form is generated dynamically based on certain user inputs and this functional form is expressed in terms of decision variables x1, x2 as well as parameters a and b. Then for specific combination of parameters a and b, I'm passing my functional form to fmincon.

For example, see the parameterized Rosenbrock function below

clc;clear;
syms x1 x2 a b;
DynamicallyGeneratedParamFunc =(100*(x2 - x1^2)^a + (1-x1)^b);
a= 2; b=3;% providing values for parameters
anonrosen = symfun(DynamicallyGeneratedParamFunc, [x1 x2]);
anonrosen(-1,3);
ans =
100*2^a + 2^b

How do I get anonrosen to evaluate to numeric value by using the values of a and b?

One way, I could do is to use subs function in below,

clc;clear;
syms x1 x2 a b;
DynamicallyGeneratedParamFunc =(100*(x2 - x1^2)^a + (1-x1)^b);
anonrosen = symfun(subs(DynamicallyGeneratedParamFunc,[a b],[2 3]), [x1 x2]);
anonrosen(-1,3);
ans =
408

Is using subs the only way here?

PS: I do need to acknowledge that the same approach of using subs in anonymous function does not work while with symfun it works fine.

Hari 2014-6-30

编辑：Star Strider 2014-6-30

在 MATLAB Online 中打开

I have another follow-up question. Even with using subs within symfun for my application, even though all the parameters and the decision variables x1, x2 get substituted correctly, the answer I get is not simplified to one numeric value but rather a long numerical expression. If I use double function on this long numerical expressions then it evaluates correctly. Why is this happening?

For ex, my dynamically generated function is as follows (In above a and b are the parameters and x1/x2 the decision/independent variables): -

clc; clear;
syms a b x1 x2;
DynamicallyGeneratedParamFunc = (exp(2*a*x1)*exp(2*a*x2)*exp(2*b*x1)*exp(2*b*x2)*(a - b)^6)/(a^2*(a*x1*exp(a*x2)*exp(b*x1) - b*x1*exp(a*x1)*exp(a*x2) - a*x1*exp(b*x1)*exp(b*x2) - a*x2*exp(a*x1)*exp(b*x2) + b*x1*exp(a*x1)*exp(b*x2) + b*x2*exp(a*x1)*exp(a*x2) + a*x2*exp(b*x1)*exp(b*x2) - b*x2*exp(a*x2)*exp(b*x1) + a^2*x1*x2*exp(a*x1)*exp(b*x2) - a^2*x1*x2*exp(a*x2)*exp(b*x1) - a*b*x1*x2*exp(a*x1)*exp(b*x2) + a*b*x1*x2*exp(a*x2)*exp(b*x1))^2);

Now for a = 0.1213 and b= 0.07, I'm creating the symbolic function

MySymbolicFunc = symfun(subs(DynamicallyGeneratedParamFunc,[a b],[0.1213  0.07]), [x1 x2]);

Now in command window when I input the following,

MySymbolicFunc(2.5,5)
ans =
(18226538222455809*exp(5739/2000))/(14713690000000000000000*((1213*exp(21/40))/4000 - (996269*exp(1563/2000))/8000000 - (2829731*exp(2613/4000))/8000000 + (7*exp(3639/4000))/40)^2)

I have to use double function on above to force it to give a numerical value

double(MySymbolicFunc(2.5,5))
ans =
      1.0507

In summary, why do I need to use subs (my first comment) and then why do i need to use double (my second comment)

Thanks

Hari

Star Strider 2014-6-30

编辑：Star Strider 2014-6-30

在 MATLAB Online 中打开

My pleasure!

If you are going to use it with fmincon, I would use matlabFunction to create an anonymous function (or function file) out of it, then call fmincon with (for example):

x = fmincon(@(x) MySymbolicFunction(x,a,b) ... )

It will then pick up (a,b) from the workspace as anonymous functions usually do, and will optimise with respect to x = [x(1) x(2)]. You may have to do a bit of editing with the Editor’s ‘Search and Replace’ ability (changing x1 to x(1) and similarly for x(2)) to create a vector of parameters for x in your function if you use it as an anonymous function, but it should be relatively easy after that. If you use a function file, simply add the lines:

x1 = x(1);
x2 = x(2);

before the function statement itself to do the same thing.

The Symbolic Math Toolbox is generally not good for such recursive operations. It does single calculations and some short loops well, but is highly inefficient for other procedures that MATLAB does well.

Star Strider 2014-7-5

My pleasure!

I wanted to be sure I continued to monitor your question in the event we had not resolved it.

I’ll look for your next Question. Good call on creating it separately, since once an Answer is Accepted (THANKS!) it tends to not be followed-up.

John D'Errico 2014-7-5

Don't close the question. Don't delete what you wrote. Don't do anything of the sort.

Leave the question there, for others to read and learn from it.

Then accept the answer if it solved your problem. Vote it up.

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Why does the anonymous function not evaluate correctly

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