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Re: Real Solutions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg95812] Re: Real Solutions
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 28 Jan 2009 06:27:52 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <glmssi$mkr$1@smc.vnet.net>

In article <glmssi$mkr$1 at smc.vnet.net>, Kowalczyk-Schröder wrote:

> I need only the real solutions from Solve or NSolve. I found a 
> complicated way with a lot of "ifs" to get, what I want. Is there an 
> easier way to do this? (I don't want to use Reduce.)

I think that one of the easiest ways to achieve what you are looking for 
is by using *DeleteCases* as in 

    DeleteCases[sols, {s_Rule} /; 
      Head[Chop[N[Last[s]]]] === Complex, Infinity]

Note that Solve may return complex solutions as Complex[] expressions, 
fractional powers of negative bases, Root[] objects, or a mix of two or 
three of those constructs. So we are looking for the lists that contain 
a rule for which its second element numerically evaluates to Complex[].

For NSolve,

    DeleteCases[sols, {s_Rule} /; 
      Head[Chop[Last[s]]] === Complex, Infinity]

would suffice.

In[1]:= sols = Solve[x^2 + 1 == 0]
sols // N
DeleteCases[sols, {s_Rule} /; 
  Head[Chop[N[Last[s]]]] === Complex, Infinity]

Out[1]= {{x -> -I}, {x -> I}}

Out[2]= {{x -> 0.- 1. I}, {x -> 0.+ 1. I}}

Out[3]= {}

In[4]:= sols = Solve[x^5 + x^4 + x^3 + x^2 + x + 1 == 0]
sols // N
DeleteCases[sols, {s_Rule} /; 
  Head[Chop[N[Last[s]]]] === Complex, Infinity]

Out[4]= {{x -> -1}, {x -> -(-1)^(1/3)}, {x -> (-1)^(
   1/3)}, {x -> -(-1)^(2/3)}, {x -> (-1)^(2/3)}}

Out[5]= {{x -> -1.}, {x -> -0.5 - 0.866025 I}, {x -> 
   0.5+ 0.866025 I}, {x -> 
   0.5- 0.866025 I}, {x -> -0.5 + 0.866025 I}}

Out[6]= {{x -> -1}}

In[7]:= sols = Solve[x^5 + 2 x + 1 == 0, x]
sols // N
DeleteCases[sols, {s_Rule} /; 
  Head[Chop[N[Last[s]]]] === Complex, Infinity]

Out[7]= {{x -> Root[1 + 2 #1 + #1^5 &, 1]}, {x -> 
   Root[1 + 2 #1 + #1^5 &, 2]}, {x -> 
   Root[1 + 2 #1 + #1^5 &, 3]}, {x -> 
   Root[1 + 2 #1 + #1^5 &, 4]}, {x -> Root[1 + 2 #1 + #1^5 &, 5]}}

Out[8]= {{x -> -0.486389}, {x -> -0.701874 - 
    0.879697 I}, {x -> -0.701874 + 0.879697 I}, {x -> 
   0.945068- 0.854518 I}, {x -> 0.945068+ 0.854518 I}}

Out[9]= {{x -> Root[1 + 2 #1 + #1^5 &, 1]}}

Regards,
--Jean-Marc


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