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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 10,252 of 10,432   
   antispam@math.uni.wroc.pl to clicliclic@freenet.de   
   Re: FYI, CAS independent integration tes   
   04 Jul 22 11:50:33   
   
   clicliclic@freenet.de  wrote:   
   >   
   > antispam@math.uni.wroc.pl schrieb:   
   > >   
   > > Nasser M. Abbasi  wrote:   
   > > >   
   > > > Any problems found, please let me know.   
   > > >   
   > >   
   > > Looking at results of Timofeev file results show truncated   
   > > after position 667.  I am not sure if this is due to   
   > > incompatiblity of markup with my browser or shows in   
   > > general.   
   > >   
   >   
   > What does this FriCAS failure on Timofeev Chapter 5, Example 81 (p.   
   > 276 in the book, #417 in the test suite) mean:   
   >   
   >   integrate((-2*sin(2*x)+(cos(x)*sin(x)^3)^(1/2))   
   >   /(-(cos(x)^3*sin(x))^(1/2)+tan(x)^(1/2)), x)   
   >   
   >   Error detected within library code: not invertible   
      
   This is rather excessive example of dependent roots:   
      
   (cos(x)*sin(x)^3)^(1/2) = +-sin(x)*(cos(x)*sin(x))^(1/2)   
   (cos(x)^3*sin(x))^(1/2) = +-cos(x)*(cos(x)*sin(x))^(1/2)   
   tan(x)^(1/2) = +-(cos(x)*sin(x))^(1/2)/cos(x)   
      
   So there are 4 essentially different combinations of branches   
   (8 together, but they are equvalent in pairs).   
      
   As I wrote, dependent roots may lead to division by zero,   
   which happens in this example.   
      
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