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	FriCAS Problems
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simplify exponents

Edit detail for simplify exponents revision 3 of 4

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Editor: test1 Time: 2016/05/28 14:50:45 GMT+0
Note:

changed:
-doesn't do it.
doesn't do it.  But normalize plus cleanup works:
\begin{axiom}
normalize(2^a*4^a)
exprule %
powerRule %
\end{axiom}


changed:
-... it does have some effect:
simplifyExp does have some effect:

added:

Howto simplify exponents

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

fricas

2^a*2^(2*a)

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

(1)

Type: Expression(Integer)

fricas

simplify %

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

(2)

Type: Expression(Integer)

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

fricas

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

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

(3)

Type: Expression(Integer)

fricas

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.

fricas

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))

fricas

normalize %% 3

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

(6)

Type: Expression(Integer)

fricas

exprule %

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

(7)

Type: Expression(Integer)

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

fricas

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))

fricas

fracrule %% 3

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

(9)

Type: Expression(Integer)

How about 2^a*4^a ?

fricas

2^a*4^a

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

(10)

Type: Expression(Integer)

fricas

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.

fricas

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

Type: Void

fricas

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

fricas

simplify powerRule (2^a*4^a)

fricas

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

fricas

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

fricas

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)

fricas

simplifyExp(2^a*4^a)

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

(15)

Type: Expression(Integer)

doesn't do it. But normalize plus cleanup works:

fricas

normalize(2^a*4^a)

$\label{eq16}{{e}^{a \ {\log \left({2}\right)}}}^{3}$

(16)

Type: Expression(Integer)

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exprule %

$\label{eq17}{8}^{a}$

(17)

Type: Expression(Integer)

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powerRule %

$\label{eq18}{2}^{3 \ a}$

(18)

Type: Expression(Integer)

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

simplifyExp does have some effect:

fricas

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

$\label{eq19}{{2}^{3 \ a}}\ {{2}^{4 \ a}}$

(19)

Type: Expression(Integer)

fricas

simplifyExp(2^(3*a)*2^(4*a))

$\label{eq20}{2}^{7 \ a}$

(20)

Type: Expression(Integer)