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MathGroup Archive 2005

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Solve Limitations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62963] Solve Limitations
  • From: "Marcelo Mayall" <mmayall at bol.com.br>
  • Date: Fri, 9 Dec 2005 05:10:42 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Let's suppose that we are interested in the roots analytic expression of 
the following function:
In[1] :=  f = a x + b x^(3/2) + c;
The function Solve could be used:
In[2] :=  sol = Solve[f==0, x];
Defining the values of the constants a, b, c would return the following numeric values:
In[3] :=  froots = Solve[f==0, x]/. {a->1, b->1, c->1} //N
Out[3] =  {{x-> 2.1479}, {x-> -0.57395 + 0.368989 I}, {x-> -0.57395 - 0.368989 I}}
However, f is not null for those values and therefore, these are not the roots of f:
In[4] :=  f/. froots/. {a-> 1, b-> 1, c-> 1} //Chop
Out[4] =  {6.2958, 0, 0}
At first, it seems that the function Solve doesn't take appropriately in 
consideration the term in square root.
Some idea to obtain the correct analytic solution of f ??? Or, in fact, this a limitation of the function Solve???

Thanks,
Marcelo Mayall



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