MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: How to get sqrt(Year^2)===Year?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75796] Re: How to get sqrt(Year^2)===Year?
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 10 May 2007 05:27:24 -0400 (EDT)
  • References: <f1s1v4$fip$1@smc.vnet.net>


Hi,

a square root has actually two values but by choosing a branch cut, the 

function is made unique. Therefore, the Sqrt of a real number is by 

definition positive.  Therefore, Sqrt[x^2] is not equal to x if x<0. 

However, if you tell Mathematica that Year is positive you will get what you 

want. E.g.

Simplify[x==Sqrt[x^2],x>0]

hope this helps, Daniel



Hatto von Aquitanien wrote:

> If I have some expression which takes the square root of a square, it

> evaluates leaving the whole square root expression unchanged.  

> 

> Year == (Year^2)^(1/2)

> 

> just returns what I entered when evaluated.

> 

> Year == Year

> 

> evaluates to True.

> 

> How do I persuade Mathematica to evaluate the first expression completely?




  • Prev by Date: Re: How to get sqrt(Year^2)===Year?
  • Next by Date: Re: Simplify by Recurrence Relations
  • Previous by thread: Re: How to get sqrt(Year^2)===Year?
  • Next by thread: Re: How to get sqrt(Year^2)===Year?