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Re: How to get sqrt(Year^2)===Year?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg75909] Re: How to get sqrt(Year^2)===Year?
*From*: Hatto von Aquitanien <abbot at AugiaDives.hre>
*Date*: Sun, 13 May 2007 05:45:21 -0400 (EDT)
*References*: <f1s1v4$fip$1@smc.vnet.net> <f1upol$8hr$1@smc.vnet.net>
dh wrote:
>
>
> Hi,
>
> a square root has actually two values but by choosing a branch cut, the
>
> function is made unique.
Kronecker was right.
> Therefore, the Sqrt of a real number is by definition positive.
Hu? Do you mean the square root of a positive number is by definition real?
> Therefore, Sqrt[x^2] is not equal to x if x<0.
Wait for the coffee to kick in before you post.
> 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
Well, it helps in this circumstance. I believe Convert will do the same,
but I'm not sure if it works in all circumstances. I do wish there were
some way for Mathematica to know that something such as a physical constant
is real and positive. Perhaps doing that under all circumstances would
turn into a major performance hit, but it would be nice in circumstances in
which it is inappropriate to consider cases of x < 0, or x not real in
Sqrt[x].
>
>
> 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?
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