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Re: Solve results
*To*: mathgroup at smc.vnet.net
*Subject*: [mg5090] Re: [mg5061] Solve results
*From*: Sherman Reed <Sherman.Reed at worldnet.att.net>
*Date*: Wed, 30 Oct 1996 22:03:38 -0500
*Sender*: owner-wri-mathgroup at wolfram.com
>At 02:48 AM 10/26/96 +0000, you wrote:
>>I recently came upon this and was wondering if it's a Solve bug, some
>>notation I'm not familiar with, or just an unfortunate happenstance.
>>
>>When I give Mathematica the equation x^3-8==0, this is what it returns:
>>
>>In[1]:=
>>Out[1]=
>>
>>But if I first factor the polynomial instead:
>>
>>In[2]:=
>>Out[2]=
>>
>>if I use reduce instead of solve
>>
>>In[3]:=
>>Out[3]=
>>
>>
>>My question is: why the different behaviors, and how can I tell in advance
>>which to use?
>>
>>+----------------------------+
>>| | I understand mine's a risky plan,
>>| Greg Anderson | and your system can't miss
>>| dwarf at wam.umd.edu | but is security, after all a cause
>>| timbwolf at eng.umd.edu | or symptom of happiness.
>>| |
>>+----------------------------+ Dream Theater -- A Matter of Time
>>
>>
>Somehow the Mma input and output did not make it into your message.
>
>When I input Solve[x^3==8,x], I get the following answer
>
>{{x->2}, {x->-2(-1)^1/3},{x->2(-1)^2/3} which is correct.
>
>Most people do not recognize the solution process when
>
>imaginary solutions are present. The 8 on the rhs of the
>
>equation is really 8*Exp[I*2*Pi]. The cube root of this
>
>expression is 2*Exp[I*0], 2*Exp[I*(2/3)*Pi], and 2*Exp[I*(4/3)*Pi].
>
>Note that the 3rd power of each is 8*Exp[I*2*Pi]. Applying Euler's
>
>Formula (Exp[I*theta]=Cos[theta]+I*Sin[theta]) will give
>
>you the more conventional solution, which I suspect is the
>
>second example. The third example, is I believe just the
>
>rectangular version of the Solve solution.
>
>Very few Algebra courses cover the use of Euler's formula,
>
>which is a bridge between polar and rectangular,
>
>which is commonly used in Electrical Engineering for AC circuit
>
>theory problems and I suspect many other places.
>
>I have not seen Solve make a mistake, so I would place my
>
>trust on the initial results. Incidentally, one can always
>
>check the results from all three cases, with ease.
>
>Hope this helps.
>
>Sherman C. Reed
>
>
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