equation-solving.rst

The Schemannian Reference

Equation Solving

"Schemannian" can currently a solve equation symbolically, if in which the unknown appears only once.

(solve eqn var) → expression?
    eqn : expression?
    var : expression?

In function solve, the argument eqn is a expression with the outest level operator '=. For example, '(= x 3) describes a equation, while '(+ x 3) does not.

Here is an example.

(require "solve.rkt")
(solve '(= (** x z) y) 'z)

gives you

'(= z (* (log y) (** (log x) -1)))