[sage-support] Re: Simplifying log expressions

2012-01-13 Thread JamesHDavenport 
Unfortunately, full_simplify has its own problems, notably with branch
cuts.
sage: f = (1/2)*log(2*t) + (1/2)*log(-t)
sage: f.full_simplify()
1/2*log(2)
Unfortunately, when t=-1, we have the sum of the logarithms of two
negative numbers, and therefore the imaginary part is 2i pi, not 0
On Jan 12, 10:24 pm, Michael Orlitzky mich...@orlitzky.com wrote:
 On 01/12/12 17:16, Tom Judson wrote:

  I would like to simplify the difference of two log expressions to show
  that I get a constant, but

  simplify((1/2)*log(2*t) - (1/2)*log(t))

  just returns the expression.  Does anyone know of an easy fix for
  this?  Preferably, I would like something that Calculus II students
  could easily use.

 There's no global function for it, but what you want is to call
 full_simplify() on the expression.

   sage: f = (1/2)*log(2*t) - (1/2)*log(t)
   sage: f.full_simplify()
   1/2*log(2)

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Re: [sage-support] Re: Simplifying log expressions

2012-01-13 Thread Michael Orlitzky 

On 01/13/2012 07:38 PM, JamesHDavenport wrote:

Unfortunately, full_simplify has its own problems, notably with branch
cuts.
sage: f = (1/2)*log(2*t) + (1/2)*log(-t)
sage: f.full_simplify()
1/2*log(2)


In my session, I had the difference of two logarithms. In yours above, 
you've got the sum. Is that an actual sage session? I get something 
different on 4.8.alpha6:


  sage: f = (1/2)*log(2*t) + (1/2)*log(-t)
  sage: f.full_simplify()
  1/2*I*pi + 1/2*log(2) + log(t)


In the example below, with t=-1, both logs should have imaginary part pi 
and real parts log(2) and zero respectively?




There's no global function for it, but what you want is to call
full_simplify() on the expression.

   sage: f = (1/2)*log(2*t) - (1/2)*log(t)
   sage: f.full_simplify()
   1/2*log(2)




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