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# Edit detail for simplify exponents revision 4 of 4

 1 2 3 4 Editor: test1 Time: 2017/02/17 17:14:43 GMT+0 Note:

changed:
-Howto simplify exponents
-
-  How can I make axiom to do 2^a*2^(2*a) -> 2^(3*a) ?
How to simplify exponents

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

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

changed:
-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.
Unfortunately this seemingly simple transformation does not seem to be easy to perform in FriCAS 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.


## How to simplify exponents

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

fricas
2^a*2^(2*a)
 (1)
Type: Expression(Integer)
fricas
simplify %
 (2)
Type: Expression(Integer)

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

fricas
2^(5*a)/2^(4*a)
 (3)
Type: Expression(Integer)
fricas
simplify %
 (4)
Type: Expression(Integer)

Unfortunately this seemingly simple transformation does not seem to be easy to perform in FriCAS 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
 (5)
Type: RewriteRule?(Integer,Integer,Expression(Integer))
fricas
normalize %% 3
 (6)
Type: Expression(Integer)
fricas
exprule %
 (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)
 (8)
Type: RewriteRule?(Integer,Integer,Expression(Integer))
fricas
fracrule %% 3
 (9)
Type: Expression(Integer)

fricas
2^a*4^a
 (10)
Type: Expression(Integer)
fricas
simplify %
 (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)
fricas
Compiling function powerFac with type (Variable(n), Variable(a)) ->
Expression(Integer)
 (12)
Type: RewriteRule?(Integer,Integer,Expression(Integer))

Now we can use the rule and simplify the result

fricas
simplify powerRule (2^a*4^a)
 (13)
Type: Expression(Integer)

Apparently, simplifyExp yields the desired result

fricas
simplifyExp(2^a*2^(2*a))
 (14)
Type: Expression(Integer)

The desired result was

• 2^a*4^a -> 2^(3a)

fricas
simplifyExp(2^a*4^a)
 (15)
Type: Expression(Integer)

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

fricas
normalize(2^a*4^a)
 (16)
Type: Expression(Integer)
fricas
exprule %
 (17)
Type: Expression(Integer)
fricas
powerRule %
 (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)
 (19)
Type: Expression(Integer)
fricas
simplifyExp(2^(3*a)*2^(4*a))
 (20)
Type: Expression(Integer)