Re: sqrt[x^2]
- To: mathgroup at smc.vnet.net
- Subject: [mg9849] Re: sqrt[x^2]
- From: Allan Hayes <hay at haystack.demon.co.uk>
- Date: Fri, 28 Nov 1997 05:36:05 -0500
- Sender: owner-wri-mathgroup at wolfram.com
ZenArcher wrote: > > Well I know this is really elementary, but I just don't understand this > result: > sqrt[x^2]=sqrt[x^2] > sqrt[4^2]=4 > Why doesn't Mathematica tell me that the Square Root of x^2 is x, but it > will tell me that the square root of 4^2 is 4? Zen: Mathematica gives Sqrt[x^2] Sqrt[x^2] because it has no information about x to help it compute further (x is not always the answer - thought this is so for positive real x) For example Sqrt[(-4)^2] 4 Sqrt[(-I)^2] I Why is this? Well, (-4)^2 = 16 and 4^2 = 16 so there is no unique square root of 16. We either give all possibilies or make a choice. When it can calculate further Mathematica makes a choice, it gives the "principal square root". The definition of this is included in the defition of z^p below.: The principal value of z^p (for non-zero z) is Exp[p Log[z]] where Log[z] is the "principal logarithm" of z that is log[Abs[z]] + I Arg[z] where log is the ordinary log and Arg[z] is the "principal argument" of z that is the angle, a, in the range (-pi, +pi] (not including -pi) such that Exp[I a] = z/Abs[z] Put another way, for non-zero z, Log[z] the unique complex number w such that Exp[w] = z and -pi < Im[w] <= pi and z^a is Exp[a w] -- Allan Hayes Mathematica Training and Consulting Leicester, UK hay at haystack.demon.co.uk http://www.haystack.demon.co.uk voice: +44 (0)116 271 4198 fax: +44 (0)116 271 4198
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