algebraic - Maple Help

convert/algebraic

convert procedures into their algebraic form

 Calling Sequence convert(procedure, algebraic)

Parameters

 procedure - any Maple procedure

Description

 • The convert(procedure, algebraic) command converts a given procedure into its algebraic form, if possible. For example, the procedure z -> sin(z) + 1 is converted into sin + 1. Remember that in Maple 1(z) = 1.
 • When the conversion to the algebraic form is not possible, the received procedure itself is returned. For a conversion to be possible, the given procedure is expected to have a calling sequence, that is, be a procedure of some arguments, and return an algebraic expression of its arguments containing only constants or functions of the given arguments.

Examples

 > $z↦\mathrm{\Gamma }\left(z\right)\cdot \mathrm{\Psi }\left(z\right)+1$
 ${z}{↦}{\mathrm{\Gamma }}{}\left({z}\right){\cdot }{\mathrm{\Psi }}{}\left({z}\right){+}{1}$ (1)
 > $\mathrm{convert}\left(,\mathrm{algebraic}\right)$
 ${\mathrm{\Gamma }}{}{\mathrm{\Psi }}{+}{1}$ (2)

Sometimes, when applying an algebraic expression, to simplify any constant "applied to some variables", use simplify/constants.

 > $\left(a,z\right)↦f\left(a,z\right)+\mathrm{\pi }$
 $\left({a}{,}{z}\right){↦}{f}{}\left({a}{,}{z}\right){+}{\mathrm{\pi }}$ (3)
 > $\mathrm{convert}\left(,\mathrm{algebraic}\right)$
 ${f}{+}{\mathrm{\pi }}$ (4)

Applying the algebraic form generates $\mathrm{\pi }\left(a,z\right)$.

 > $\left(a,z\right)$
 ${f}{}\left({a}{,}{z}\right){+}{\mathrm{\pi }}{}\left({a}{,}{z}\right)$ (5)

Simplify $\mathrm{\pi }\left(a,z\right)$.

 > $\mathrm{simplify}\left(\right)$
 ${f}{}\left({a}{,}{z}\right){+}{\mathrm{\pi }}$ (6)

Procedures where the calling sequence is not given are not converted

 > $\left(\right)↦\mathrm{sin}\left(z\right)$
 $\left(\right){↦}{\mathrm{sin}}{}\left({z}\right)$ (7)
 > $\mathrm{convert}\left(,\mathrm{algebraic}\right)$
 $\left(\right){↦}{\mathrm{sin}}{}\left({z}\right)$ (8)

In other cases, the conversion is not possible: objects which are not constant and not functions of the procedure's arguments are present:

 > $z↦f\left(1,z\right)$
 ${z}{↦}{f}{}\left({1}{,}{z}\right)$ (9)

It cannot be expressed in terms of only $f$.

 > $\mathrm{convert}\left(,\mathrm{algebraic}\right)$
 ${z}{↦}{f}{}\left({1}{,}{z}\right)$ (10)
 > $z↦\frac{\mathrm{LambertW}\left(z\right)}{\left(1+\mathrm{LambertW}\left(z\right)\right)\cdot z}$
 ${z}{↦}\frac{{\mathrm{LambertW}}{}\left({z}\right)}{\left({1}{+}{\mathrm{LambertW}}{}\left({z}\right)\right){\cdot }{z}}$ (11)

$z$ in the procedure body prevents the conversion.

 > $\mathrm{convert}\left(,\mathrm{algebraic}\right)$
 ${z}{↦}\frac{{\mathrm{LambertW}}{}\left({z}\right)}{\left({1}{+}{\mathrm{LambertW}}{}\left({z}\right)\right){\cdot }{z}}$ (12)

Without $z$ the conversion is feasible.

 > $z↦\frac{\mathrm{LambertW}\left(z\right)}{1+\mathrm{LambertW}\left(z\right)}$
 ${z}{↦}\frac{{\mathrm{LambertW}}{}\left({z}\right)}{{1}{+}{\mathrm{LambertW}}{}\left({z}\right)}$ (13)
 > $\mathrm{convert}\left(,\mathrm{algebraic}\right)$
 $\frac{{\mathrm{LambertW}}}{{1}{+}{\mathrm{LambertW}}}$ (14)