help links subscribe changes refresh edit

	FriCAS Problems
series ←	↑	→ simplifying Expressions

simplify exponents

Edit detail for simplify exponents revision 1 of 4

	1 2 3 4
Editor: 127.0.0.1 Time: 2007/11/15 20:22:56 GMT-8
Note: transferred from axiom-developer

changed:
-
Howto simplify exponents

  How can I make axiom to do 2^a*2^(2*a) -> 2^(3*a) ?

\begin{axiom}
2^a*2^(2*a)
simplify %
\end{axiom}

But I cannot convince Axiom to do 2^(5*a)/2^(4*a) -> 2^a

\begin{axiom}
2^(5*a)/2^(4*a)
simplify %
\end{axiom}

Unfortunately this seemingly simple transformation does not seem to be easy to perform in Axiom but I have found two possible ways to do this. The first involves the normalize() operation. Unfortunately normalize takes the process one step too far. This can be undone with a simple rule.

\begin{axiom}
exprule:=rule exp(x*log(n)) == n^x
normalize %% 3
exprule %
\end{axiom}

The second approach uses a single rule to do the whole job.

\begin{axiom}
fracrule:=rule n^m/n^p == n^(m-p)
fracrule %% 3
\end{axiom}

How about 2^a*4^a ?

\begin{axiom}
2^a*4^a
simplify %
\end{axiom}

Here is one approach. First lets define a function that factors a power and a rule that applies this function.

\begin{axiom}
powerFac(n,a) == reduce(*,[(t.factor)^(a*t.exponent) for t in factors(n)])
powerRule := rule n^a == powerFac(n,a)
\end{axiom}

Now we can use the rule and simplify the result
\begin{axiom}
simplify powerRule (2^a*4^a)
\end{axiom}

Apparently, simplifyExp yields the desired result
\begin{axiom}
simplifyExp(2^a*2^(2*a))
\end{axiom}


From BillPage Mon Oct 11 12:48:08 -0500 2004
From: Bill Page
Date: Mon, 11 Oct 2004 12:48:08 -0500
Subject: But ...
Message-ID: <20041011124808-0500@page.axiom-developer.org>

The desired result was

-  2^a*4^a -> 2^(3a)

\begin{axiom}
simplifyExp(2^a*4^a)
\end{axiom}

doesn't do it.

From unknown Mon Oct 11 22:44:30 -0500 2004
From: H.P.
Date: Mon, 11 Oct 2004 22:44:30 -0500
Subject: In the other hand ...
Message-ID: <20041011224430-0500@page.axiom-developer.org>

... it does have some effect:

\begin{axiom}
2**(3*a)*2**(4*a)
simplifyExp(2**(3*a)*2**(4*a))
\end{axiom}

From unknown Thu Apr 7 18:25:36 -0500 2005
From: unknown
Date: Thu, 07 Apr 2005 18:25:36 -0500
Subject: \
Message-ID: <20050407182536-0500@page.axiom-developer.org>

Howto simplify exponents

How can I make axiom to do 2^a2^(2a) -> 2^(3*a) ?

axiom

2^a*2^(2*a)

$\label{eq1}{{2}^{a}}\ {{2}^{2 \ a}}$

(1)

Type: Expression(Integer)

axiom

simplify %

$\label{eq2}{2}^{3 \ a}$

(2)

Type: Expression(Integer)

But I cannot convince Axiom to do 2^(5a)/2^(4a) -> 2^a

axiom

2^(5*a)/2^(4*a)

$\label{eq3}{{2}^{5 \ a}}\over{{2}^{4 \ a}}$

(3)

Type: Expression(Integer)

axiom

simplify %

$\label{eq4}{{2}^{5 \ a}}\over{{2}^{4 \ a}}$

(4)

Type: Expression(Integer)

Unfortunately this seemingly simple transformation does not seem to be easy to perform in Axiom but I have found two possible ways to do this. The first involves the normalize() operation. Unfortunately normalize takes the process one step too far. This can be undone with a simple rule.

axiom

exprule:=rule exp(x*log(n)) == n^x

$\label{eq5}{{e}^{x \ {\log \left({n}\right)}}}\mbox{\rm = =}{{n}^{x}}$

(5)

Type: RewriteRule?(Integer,Integer,Expression(Integer))

axiom

normalize %% 3

$\label{eq6}{e}^{a \ {\log \left({2}\right)}}$

(6)

Type: Expression(Integer)

axiom

exprule %

$\label{eq7}{2}^{a}$

(7)

Type: Expression(Integer)

The second approach uses a single rule to do the whole job.

axiom

fracrule:=rule n^m/n^p == n^(m-p)

$\label{eq8}{{{n}^{m}}\over{{n}^{p}}}\mbox{\rm = =}{{n}^{- p + m}}$

(8)

Type: RewriteRule?(Integer,Integer,Expression(Integer))

axiom

fracrule %% 3

$\label{eq9}{2}^{a}$

(9)

Type: Expression(Integer)

How about 2^a*4^a ?

axiom

2^a*4^a

$\label{eq10}{{2}^{a}}\ {{4}^{a}}$

(10)

Type: Expression(Integer)

axiom

simplify %

$\label{eq11}{{2}^{a}}\ {{4}^{a}}$

(11)

Type: Expression(Integer)

Here is one approach. First lets define a function that factors a power and a rule that applies this function.

axiom

powerFac(n,a) == reduce(*,[(t.factor)^(a*t.exponent) for t in factors(n)])

Type: Void

axiom

powerRule := rule n^a == powerFac(n,a)

$\label{eq12}{{n}^{a}}\mbox{\rm = =}{{{\tt'}powerFac}\left({n , \: a}\right)}$

(12)

Type: RewriteRule?(Integer,Integer,Expression(Integer))

Now we can use the rule and simplify the result

axiom

simplify powerRule (2^a*4^a)

axiom

Compiling function powerFac with type (PositiveInteger,Variable(a))
       -> Expression(Integer)

axiom

Compiling function powerFac with type (PositiveInteger,Polynomial(
      Integer)) -> Expression(Integer)

$\label{eq13}{2}^{3 \ a}$

(13)

Type: Expression(Integer)

Apparently, simplifyExp yields the desired result

axiom

simplifyExp(2^a*2^(2*a))

$\label{eq14}{2}^{3 \ a}$

(14)

Type: Expression(Integer)

But ... --Bill Page, Mon, 11 Oct 2004 12:48:08 -0500 reply

The desired result was

2^a*4^a -> 2^(3a)

axiom

simplifyExp(2^a*4^a)

$\label{eq15}{{2}^{a}}\ {{4}^{a}}$

(15)

Type: Expression(Integer)

doesn't do it.

In the other hand ... --H.P., Mon, 11 Oct 2004 22:44:30 -0500 reply

... it does have some effect:

axiom

2**(3*a)*2**(4*a)

   There are no library operations named ** 
      Use HyperDoc Browse or issue
                                 )what op **
      to learn if there is any operation containing " ** " in its name.

   Cannot find a definition or applicable library operation named ** 
      with argument type(s) 
                               PositiveInteger
                             Polynomial(Integer)

      Perhaps you should use "@" to indicate the required return type, 
      or "$" to specify which version of the function you need.
simplifyExp(2**(3*a)*2**(4*a))

   There are no library operations named ** 
      Use HyperDoc Browse or issue
                                 )what op **
      to learn if there is any operation containing " ** " in its name.

   Cannot find a definition or applicable library operation named ** 
      with argument type(s) 
                               PositiveInteger
                             Polynomial(Integer)

      Perhaps you should use "@" to indicate the required return type, 
      or "$" to specify which version of the function you need.

\ --unknown, Thu, 07 Apr 2005 18:25:36 -0500 reply