How to simplify exponents
How can I make FriCAS to do 2^a2^(2a) -> 2^(3*a) ?
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2^a*2^(2*a)
Type: Expression(Integer)
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simplify %
Type: Expression(Integer)
But I cannot convince FriCAS to do 2^(5a)/2^(4a) -> 2^a
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2^(5*a)/2^(4*a)
Type: Expression(Integer)
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simplify %
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.
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exprule:=rule exp(x*log(n)) == n^x
Type: RewriteRule
?(Integer,
Integer,Expression(Integer))
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normalize %% 3
Type: Expression(Integer)
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exprule %
Type: Expression(Integer)
The second approach uses a single rule to do the whole job.
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fracrule:=rule n^m/n^p == n^(m-p)
Type: RewriteRule
?(Integer,
Integer,Expression(Integer))
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fracrule %% 3
Type: Expression(Integer)
How about 2^a*4^a ?
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2^a*4^a
Type: Expression(Integer)
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simplify %
Type: Expression(Integer)
Here is one approach. First lets define a function that factors a power and a rule that applies this function.
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powerFac(n,a) == reduce(*,[(t.factor)^(a*t.exponent) for t in factors(n)])
Type: Void
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powerRule := rule n^a == powerFac(n,a)
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Compiling function powerFac with type (Variable(n), Variable(a)) ->
Expression(Integer)
Type: RewriteRule
?(Integer,
Integer,Expression(Integer))
Now we can use the rule and simplify the result
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simplify powerRule (2^a*4^a)
Type: Expression(Integer)
Apparently, simplifyExp yields the desired result
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simplifyExp(2^a*2^(2*a))
Type: Expression(Integer)
The desired result was
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simplifyExp(2^a*4^a)
Type: Expression(Integer)
doesn't do it. But normalize plus cleanup works:
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normalize(2^a*4^a)
Type: Expression(Integer)
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exprule %
Type: Expression(Integer)
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powerRule %
Type: Expression(Integer)
simplifyExp does have some effect:
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2^(3*a)*2^(4*a)
Type: Expression(Integer)
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simplifyExp(2^(3*a)*2^(4*a))
Type: Expression(Integer)