# Revision history [back]

### Why simplify doesn't work?

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?)

.

 2 No.2 Revision calc314 4191 ●22 ●48 ●113

### Why simplify doesn't work?

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)

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')

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?)

.