Inexplicable Symbolic Function problem

When I type :

syms xi 
syms yi 
syms lami 

f(xi,yi,lami) = 1/2.*sum(abs((yi-xi).^2)) + lami.*sum(abs(diff(diff(xi))));

I obtain in the workspace the function " f 1x1 symfun "

but when I type in the command line " f ENTER" :

f(xi, yi, lami) =

abs(xi - yi)^2/2

What is this function I never typed and why does it not return an error?

Instead running the default example it works:

syms x y z
f(x,y,z) = 2*y*z*sin(x) + 3*x*sin(z)*cos(y);
>> f
f(x, y, z) =
2*y*z*sin(x) + 3*x*cos(y)*sin(z)

Where's my error ?

Thanks!

Answer 1

[Walter Roberson]

xi is not a function. diff(xi) is going to search the expression xi looking for symbolic variables, find xi, and take the derivative with respect to that variable getting 1. diff(1) is 0. So the second term becomes 0.

xi and yi are scalar so the sum() only has the single term so the sum() wrapper can be removed.

The overall calculation becomes what you see.

When you create a symbolic function with that kind of assignment, then internal mechanism is to fully evaluate the right hand side, and then to use symfun() to create a symbolic function with the parameters of the left side.

Defining a symbolic function is not like defining an anonymous function with @. The names on the left do not just act as placeholders for passing values positional. Doing the diff(xi) on the right side immediately evalutes diff(xi) that takes the derivative of scalar xi. It does not just record the sequence of steps and apply the diff to whatever gets passed in later

[Emiliano Rosso]

Thanks!

You say "Doing the diff(xi) on the right side immediately evalutes diff(xi) that takes the derivative of scalar xi. It does not just record the sequence of steps and apply the diff to whatever gets passed in later".

I only partly understand the reasons but how can I just record the sequence of steps and apply the diff to whatever gets passed in later?

[Walter Roberson]

use @() anonymous function definition.