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Derivative

rascal-0.40.17

Synopsis​

Symbolic differentiation.

Description​

Computing the derivative of an expression with respect to some variable is a classical calculus problem. Loosely speaking, a derivative can be thought of as how much one quantity changes in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.

We present here rules for determining the derivative dE/dX of simple expressions E for a given variable X. Recall that for the number N, variables X and Y, and expressions E1 and E2 the following rules apply:

  • $\frac{dN}{dX} = 0$
  • $\frac{dX}{dX} = 1$
  • $\frac{dX}{dY} = 0$, when $X \neq Y$
  • $\frac{d(E1 + E2)}{dX} = \frac{dE1}{dX} + \frac{dE2}{dX}$
  • $\frac{d(E1 \cdot E2)}{dX} = \frac{dE1}{dX} \cdot E2 + E1 \cdot \frac{dE2}{dX}$

Examples​

Here is our solution followed by a list of explanations:

data Exp = con(int n)      ❶  
         | var(str name)
         | mul(Exp e1, Exp e2)
         | add(Exp e1, Exp e2)
         ;
         
public Exp E = add(mul(con(3), var("y")), mul(con(5), var("x")));      ❷  

Exp dd(con(n), var(V))              = con(0);      ❸  
Exp dd(var(V1), var(V2))            = con((V1 == V2) ? 1 : 0);
Exp dd(add(Exp e1, Exp e2), var(V)) = add(dd(e1, var(V)), dd(e2, var(V)));
Exp dd(mul(Exp e1, Exp e2), var(V)) = add(mul(dd(e1, var(V)), e2), mul(e1, dd(e2, var(V))));
 
Exp simp(add(con(n), con(m))) = con(n + m);      ❹  
Exp simp(mul(con(n), con(m))) = con(n * m);

Exp simp(mul(con(1), Exp e))  = e;
Exp simp(mul(Exp e, con(1)))  = e;
Exp simp(mul(con(0), Exp e))  = con(0);
Exp simp(mul(Exp e, con(0)))  = con(0);

Exp simp(add(con(0), Exp e))  = e;
Exp simp(add(Exp e, con(0)))  = e;

default Exp simp(Exp e)       = e;      ❺  

Exp simplify(Exp e) = bottom-up visit(e){      ❻  
    case Exp e1 => simp(e1)
};


test bool tstSimplity1() = simplify(mul(var("x"), add(con(3), con(5)))) == mul(var("x"), con(8));
test bool tstSimplity2() = simplify(dd(E, var("x"))) == con(5);

NOTE:

  • con stands for constant for example 1, 10, 99.

  • var stands for variable for example y, x, m.

  • ❶ Define a data type Exp to represent expressions.

  • ❷ Introduce an example expression E for later use.

  • ❸ Define the actual differentiation function dd. Observe that this definition depends on the use of patterns in function declarations, see [Rascal:Function].

  • ❹ Define simplification rules.

  • ❺ A default rule is given for the case that no simplification applies.

  • ❻ Define the actual simplification function simplify that performs a bottom-up traversal of the expression, applying simplification rules on ascending.

Let's differentiate the example expression E:

rascal>dd(E, var("x"));
Exp: add(
  add(
    mul(
      con(0),
      var("y")),
    mul(
      con(3),
      con(0))),
  add(
    mul(
      con(0),
      var("x")),
    mul(
      con(5),
      con(1))))

As you can see, we managed to compute a derivative, but the result is far more complex than we would like. This is where simplification comes in. First, try a simple case:

rascal>simplify(mul(var("x"), add(con(3), con(5))));
Exp: mul(
  var("x"),
  con(8))

Now apply simplification to the result of differentiation:

rascal>simplify(dd(E, var("x")));
Exp: con(5)