
Re: are the commercial results continuous?
Posted:
Jul 17, 2012 1:21 AM


On 7/16/2012 4:11 PM, Nasser M. Abbasi wrote: > On 7/16/2012 3:50 PM, clicliclic@freenet.de wrote: > >> > > Here they are, using V 8.04 on windows: > >  first one  > Clear[x]; > integrand=(2+3*x2*x^2)/((1+x+x^2)^2*Sqrt[1x+x^2]); > sol=Integrate[integrand,{x,1,0}] > In[3]:= N[sol] > Out[3]= 1.60465+0. I > > In[5]:= sol2=NIntegrate[integrand,{x,1,0},WorkingPrecision>20] > Out[5]= 0.061431946878726698474 >  > >  second example  > integrand=(1x+3*x^2)/((1+x+x^2)^2*Sqrt[1x+x^2]); > sol=Integrate[integrand,{x,1,0}] > > In[8]:= N[sol] > Out[8]= 1.840161.4803*10^16 I > > In[9]:= sol2=NIntegrate[integrand,{x,1,0},WorkingPrecision>20] > Out[9]= 2.6027195829817573677 > 
I upgraded to Maple 16, and tried these on it. Maple seems to handle these type of integrals better. Not only the expressions came out smaller in size, but evaluating the symbolic result numerically using evalf(sol) produced the correct result without having to do numerical integration with increased precision.
 example 1 Maple 16 integrand:=(2+3*x2*x^2)/((1+x+x^2)^2*sqrt(1x+x^2)); sol:=int(integrand,x=1..0);
3/2+(1/16)*sqrt(2)*sqrt(3)*ln(3*sqrt(2)+4)(1/16)*sqrt(2)*sqrt(3)* ln(3*sqrt(2)4)+(1/2)*sqrt(3)(1/16)*sqrt(2)*sqrt(3)*ln(sqrt(2)* sqrt(3)+2)+(1/16)*sqrt(2)*sqrt(3)*ln(sqrt(2)*sqrt(3)2)+(3/8)* sqrt(2)*arctan(sqrt(2))
evalf(sol); 0.0614319480 
 example 2 Maple 16 integrand:=(1x+3*x^2)/((1+x+x^2)^2*sqrt(1x+x^2)); sol:=int(integrand,x=1..0);
1+(1/12)*sqrt(2)*sqrt(3)*ln(3*sqrt(2)+4)(1/12)*sqrt(2)*sqrt(3)* ln(3*sqrt(2)4)(1/12)*sqrt(2)*sqrt(3)*ln(sqrt(2)* sqrt(3)+2)+(1/12)*sqrt(2)*sqrt(3)*ln(sqrt(2)*sqrt(3)2)+sqrt(2)*arctan(sqrt(2))
evalf(sol); 2.602719583 
I am glad to also say that the copy using the mouse from the Maple worksheet seems to be working very well now in V 16. Now I can copy things out like I did above with no problem. Before, this did not work well at all.
Nasser

