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Hi! My problem is that I don't want to use simplify_full(). It messes up the way my formulea look preety bad. Yet I have in them terms of the form:

f(x,y)=(x^y)^(1/y)

and these a left untouched by simplify(). Is there any way for sage to recognize, that f=x without simplify_full()?

.

(Here's what i mean by messed up formula:

var('x,b2,b1')

f1(x,b1,b2)= e^(-x/b2)/b2 - e^(-x/b1)/b1

f2(x,b1,b2)= x

f1=f1( x=log( (b1/b2)^(b1*b2/(b1-b2)) ) )

f2=f2( x=log( (b1/b2)^(b1*b2/(b1-b2)) ) )

show( f1.simplify_full() + f2.simplify_full()== (f1+f2).simplify_full() )

The lhs looks way better, doesn't it?)

.

Big thanks in advance for any comments!

Hi! My problem is that I don't want to use simplify_full(). It messes up the way my formulea look preety bad. Yet I have in them terms of the form:

f(x,y)=(x^y)^(1/y)

and these a left untouched by simplify(). Is there any way for sage to recognize, that f=x without simplify_full()?

.

(Here's what i mean by messed up formula:

var('x,b2,b1')
  
f1(x,b1,b2)= e^(-x/b2)/b2 - e^(-x/b1)/b1e^(-x/b1)/b1
 f2(x,b1,b2)= x
 
f2(x,b1,b2)= x
 
f1=f1( x=log( (b1/b2)^(b1*b2/(b1-b2)) ) ))
  
f2=f2( x=log( (b1/b2)^(b1*b2/(b1-b2)) ) ))
  
show( f1.simplify_full() + f2.simplify_full()== (f1+f2).simplify_full() ))

The lhs looks way better, doesn't it?)

.

Big thanks in advance for any comments!